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12100	https://stackoverflow.com/questions/9076336/how-do-you-implement-the-divisor-function-in-code	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog How do you implement the divisor function in code? Ask Question Asked Modified 10 years, 7 months ago Viewed 8k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. Overall Problem: Project Euler 12 - What is the value of the first triangle number to have over five hundred divisors? Focus of problem: The divisor function Language: Python Description: The function I used is brute and the time it take for the program to find a number with more divisors than x increases almost exponentially with each 10 or 20 numbers highers. I need to get to 500 or more divisors. I've identified that the divisor function is what is hogging down the program. The research I did lead me to divisor functions and specifically the divisor function which is supposed to be a function that will count all the divisors of any integer. Every page I've looked at seems to be directed toward mathematics majors and I only have high-school maths. Although I did come across some page that mentioned allot about primes and the Sieve of Atkins but I could not make the connection between primes and finding all the divisors of any integer nor find anything on the net about it. Main Question: Could someone explain how to code the divisor function or even provide a sample? Maths concepts make more sense to me when I look at them with code. So much appreciated. brute force divisor function: ``` def countdiv(a): count = 0 for i in range(1,(a/2)+1): if a % i == 0: count += 1 return count + 1 # +1 to account for number itself as a divisor ``` python Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Jan 22, 2015 at 17:05 Zero Piraeus 59.6k2828 gold badges155155 silver badges164164 bronze badges asked Jan 31, 2012 at 8:12 RussWRussW 43744 silver badges1111 bronze badges 2 1 stackoverflow.com/questions/110344/… duplicate? – joni Commented Jan 31, 2012 at 8:15 You could divide a by the divisor you find, try that one again, and continue. That would reduce the number of possibilities greatly. If I'm not missing something. – rplnt Commented Jan 31, 2012 at 8:16 Add a comment \| 4 Answers 4 Reset to default This answer is useful 5 Save this answer. Show activity on this post. If you need a bruteforce function to calculate Number of Divisors (also known as tau(n)) Here's what it looks like ``` def tau(n): sqroot,t = int(n0.5),0 for factor in range(1,sqroot+1): if n % factor == 0: t += 2 # both factor and N/factor if sqrootsqroot == n: t = t - 1 # if sqroot is a factor then we counted it twice, so subtract 1 return t ``` The second method involves a decomposing n into its prime factors (and its exponents). tau(n) = (e1+1)(e2+1)....(em+1) where n = p1^e1 p2^e2 .... pm^em and p1,p2..pm are primes More info here The third method and much more simpler to understand is simply using a Sieve to calculate tau. ``` def sieve(N): t = (N+1) for factor in range(1,N+1): for multiple in range(factor,N+1,factor): t[multiple]+=1 return t[1:] ``` Here's it in action at ideone Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Jan 31, 2012 at 8:33 st0lest0le 33.6k88 gold badges9292 silver badges8989 bronze badges Add a comment \| This answer is useful 4 Save this answer. Show activity on this post. I agree with the two other answers submitted here in that you will only need to search up to the square root of the number. I have one thing to add to this however. The solutions offered will get you the correct answer in a reasonable amount of time. But when the problems start getting tougher, you will need an even more powerful function. Take a look at Euler's Totient function. Though it only indirectly applies here, it is incredibly useful in later problems. Another related concept is that of Prime Factorization. A quick way to improve your algorithm is to find the prime factorization of the number. In the Wikipedia article, they use 36 as an example, whose prime factorization is 2^2 3^2. Therefore, knowing this, you can use combinatorics to find the number of factors of 36. With this, you will not actually be computing each factor, plus you'd only have to check divisors 2 and 3 before you're complete. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Feb 1, 2012 at 12:09 Daniel Fischer 184k1919 gold badges318318 silver badges436436 bronze badges answered Jan 31, 2012 at 20:41 Rey AbolofiaRey Abolofia 1,02499 silver badges1414 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. When searching for divisors of n you never have to search beyond the square root of the number n. Whenever you find a divisor that's less than sqrt(n) there is exactly one matching divisor which is greater than the root, so you can increment your count by 2 (if you find divisor d of n then n/d will be the counterpart). Watch out for square numbers, though. :) The root will be a divisor that doesn't count twice, of course. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Jan 31, 2012 at 8:20 Johannes CharraJohannes Charra 30k66 gold badges4444 silver badges5353 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. If you're going to solve the Project Euler problems you need some functions that deal with prime numbers and integer factorization. Here is my modest library, which provides primes(n), is_prime(n) and factors(n); the focus is on simplicity, clarity and brevity at the expense of speed, though these functions should be sufficient for Project Euler: ``` def primes(n): """ list of primes not exceeding n in ascending order; assumes n is an integer greater than 1; uses Sieve of Eratosthenes """ m = (n-1) // 2 b = [True] m i, p, ps = 0, 3, while pp < n: if b[i]: ps.append(p) j = 2ii + 6i + 3 while j < m: b[j] = False j = j + 2i + 3 i += 1; p += 2 while i < m: if b[i]: ps.append(p) i += 1; p += 2 return ps def is_prime(n): """ False if n is provably composite, else True if n is probably prime; assumes n is an integer greater than 1; uses Miller-Rabin test on prime bases < 100 """ ps = [2,3,5,7,11,13,17,19,23,29,31,37,41, 43,47,53,59,61,67,71,73,79,83,89,97] def is_spsp(n, a): d, s = n-1, 0 while d%2 == 0: d /= 2; s += 1 if pow(a,d,n) == 1: return True for r in xrange(s): if pow(a, dpow(2,r), n) == n-1: return True return False if n in ps: return True for p in ps: if not is_spsp(n,p): return False return True def factors(n): """ list of prime factors of n in ascending order; assumes n is an integer, may be positive, zero or negative; uses Pollard's rho algorithm with Floyd's cycle finder """ def gcd(a,b): while b: a, b = b, a%b return abs(a) def facts(n,c,fs): f = lambda(x): (xx+c) % n if is_prime(n): return fs+[n] t, h, d = 2, 2, 1 while d == 1: t = f(t); h = f(f(h)) d = gcd(t-h, n) if d == n: return facts(n, c+1, fs) if is_prime(d): return facts(n//d, c+1, fs+[d]) return facts(n, c+1, fs) if -1 <= n <= 1: return [n] if n < -1: return [-1] + factors(-n) fs = [] while n%2 == 0: n = n//2; fs = fs+ if n == 1: return fs return sorted(facts(n,1,fs)) ``` Once you know how to factor a number, it is easy to count the number of divisors. Consider 76576500 = 2^2 3^2 5^3 7^1 11^1 13^1 17^1. Ignore the bases and look at the exponents, which are 2, 2, 3, 1, 1, 1, and 1. Add 1 to each exponent, giving 3, 3, 4, 2, 2, 2, and 2. Now multiply that list to get the number of divisors of the original number 76576500: 3 3 4 2 2 2 2 = 576. Here's the function: ``` def numdiv(n): fs = factors(n) f = fs.pop(0); d = 1; x = 2 while fs: if f == fs: x += 1 else: d = x; x = 2 f = fs.pop(0) return d x ``` You can see these functions at work at and learn more about how they work at my blog. I'll leave it to you to work out the solution to Problem 12. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Feb 1, 2012 at 17:40 answered Feb 1, 2012 at 17:17 user448810user448810 17.9k44 gold badges3737 silver badges6161 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions python See similar questions with these tags. 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12101	https://stackoverflow.com/questions/586284/finding-prime-numbers-with-the-sieve-of-eratosthenes-originally-is-there-a-bet	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Finding prime numbers with the Sieve of Eratosthenes (Originally: Is there a better way to prepare this array?) Ask Question Asked Modified 6 years, 3 months ago Viewed 44k times This question shows research effort; it is useful and clear 22 Save this question. Show activity on this post. Note: Version 2, below, uses the Sieve of Eratosthenes. There are several answers that helped with what I originally asked. I have chosen the Sieve of Eratosthenes method, implemented it, and changed the question title and tags appropriately. Thanks to everyone who helped! Introduction I wrote this fancy little method that generates an array of int containing the prime numbers less than the specified upper bound. It works very well, but I have a concern. The Method ``` private static int [] generatePrimes(int max) { int [] temp = new int [max]; temp = 2; int index = 1; int prime = 1; boolean isPrime = false; while((prime += 2) <= max) { isPrime = true; for(int i = 0; i < index; i++) { if(prime % temp [i] == 0) { isPrime = false; break; } } if(isPrime) { temp [index++] = prime; } } int [] primes = new int [index]; while(--index >= 0) { primes [index] = temp [index]; } return primes; } ``` My Concern My concern is that I am creating an array that is far too large for the final number of elements the method will return. The trouble is that I don't know of a good way to correctly guess the number of prime numbers less than a specified number. Focus This is how the program uses the arrays. This is what I want to improve upon. I create a temporary array that is large enough to hold every number less than the limit. I generate the prime numbers, while keeping count of how many I have generated. I make a new array that is the right dimension to hold just the prime numbers. I copy each prime number from the huge array to the array of the correct dimension. I return the array of the correct dimension that holds just the prime numbers I generated. Questions Can I copy the whole chunk (at once) of temp[] that has nonzero elements to primes[] without having to iterate through both arrays and copy the elements one by one? Are there any data structures that behave like an array of primitives that can grow as elements are added, rather than requiring a dimension upon instantiation? What is the performance penalty compared to using an array of primitives? Version 2 (thanks to Jon Skeet): ``` private static int [] generatePrimes(int max) { int [] temp = new int [max]; temp = 2; int index = 1; int prime = 1; boolean isPrime = false; while((prime += 2) <= max) { isPrime = true; for(int i = 0; i < index; i++) { if(prime % temp [i] == 0) { isPrime = false; break; } } if(isPrime) { temp [index++] = prime; } } return Arrays.copyOfRange(temp, 0, index); } ``` Version 3 (thanks to Paul Tomblin) which uses the Sieve of Erastosthenes: ``` private static int [] generatePrimes(int max) { boolean[] isComposite = new boolean[max + 1]; for (int i = 2; i i <= max; i++) { if (!isComposite [i]) { for (int j = i; i j <= max; j++) { isComposite [ij] = true; } } } int numPrimes = 0; for (int i = 2; i <= max; i++) { if (!isComposite [i]) numPrimes++; } int [] primes = new int [numPrimes]; int index = 0; for (int i = 2; i <= max; i++) { if (!isComposite [i]) primes [index++] = i; } return primes; } ``` java arrays primes sieve-of-eratosthenes Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited May 23, 2017 at 11:55 CommunityBot 111 silver badge asked Feb 25, 2009 at 14:54 eleven81eleven81 6,4491111 gold badges3939 silver badges4848 bronze badges 7 why do you need a fixed size array? – Matt Davison Commented Feb 25, 2009 at 14:56 Matt Davison: I don't want to return an array that has a bunch of zero elements at the end, it feels so sloppy. – eleven81 Commented Feb 25, 2009 at 14:57 One micro-optimization I'd make: replace "for (int j = i; ij..." with "for (int j = i; j <= max; j+=i) { isComposite[j] = true;}" – Paul Tomblin Commented Feb 25, 2009 at 23:27 1 Paul Tomblin: The replacement code you provided misses several prime numbers. – eleven81 Commented Feb 26, 2009 at 16:21 Using a BitSet is 8x more efficient than a large boolean[]. – Peter Lawrey Commented Feb 6, 2010 at 11:15 \| Show 2 more comments 13 Answers 13 Reset to default This answer is useful 13 Save this answer. Show activity on this post. Your method of finding primes, by comparing every single element of the array with every possible factor is hideously inefficient. You can improve it immensely by doing a Sieve of Eratosthenes over the entire array at once. Besides doing far fewer comparisons, it also uses addition rather than division. Division is way slower. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications edited Feb 25, 2009 at 15:04 answered Feb 25, 2009 at 14:57 Paul TomblinPaul Tomblin 183k5959 gold badges323323 silver badges411411 bronze badges 7 Yes, that was just a placeholder until I could look it up on Wikipedia. – Paul Tomblin Commented Feb 25, 2009 at 14:59 Paul Tomblin: My method just returns an array of int representing the prime numbers less than the specified upper bound, I don't think I need to sieve anything. – eleven81 Commented Feb 25, 2009 at 14:59 The problem is the use of arrays - you can't actually remove an element from a Java array, just 0/null it out. The Sieve is orthogonal. – Hank Gay Commented Feb 25, 2009 at 15:00 1 But you're checking every value of the array against every possible factor. By doing a Sieve, you check the whole array for primes in one shot, reducing the complexity from O(n^2) to O((nlogn)(loglogn)) – Paul Tomblin Commented Feb 25, 2009 at 15:02 @Hank, no, the problem is that he's trying to optimize the half a millisecond it will take to convert his array, instead of the several seconds it will take to generate primes his way. – Paul Tomblin Commented Feb 25, 2009 at 15:16 \| Show 2 more comments This answer is useful 10 Save this answer. Show activity on this post. ArrayList<> Sieve of Eratosthenes `` // Return primes less than limit static ArrayList<Integer> generatePrimes(int limit) { final int numPrimes = countPrimesUpperBound(limit); ArrayList<Integer> primes = new ArrayList<Integer>(numPrimes); boolean [] isComposite = new boolean [limit]; // all false final int sqrtLimit = (int)Math.sqrt(limit); // floor for (int i = 2; i <= sqrtLimit; i++) { if (!isComposite [i]) { primes.add(i); for (int j = ii; j < limit; j += i) //j+=i` can overflow isComposite [j] = true; } } for (int i = sqrtLimit + 1; i < limit; i++) if (!isComposite [i]) primes.add(i); return primes; } ``` Formula for upper bound of number of primes less than or equal to max (see wolfram.com): ``` static int countPrimesUpperBound(int max) { return max > 1 ? (int)(1.25506 max / Math.log((double)max)) : 0; } ``` Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Feb 26, 2009 at 2:22 jfsjfs 416k209209 gold badges1k1k silver badges1.7k1.7k bronze badges 1 prime counting function was one very interesting thing in your answer – nambastha Commented Oct 18, 2020 at 7:45 Add a comment \| This answer is useful 8 Save this answer. Show activity on this post. Create an ArrayList<Integer> and then convert to an int[] at the end. There are various 3rd party IntList (etc) classes around, but unless you're really worried about the hit of boxing a few integers, I wouldn't worry about it. You could use Arrays.copyOf to create the new array though. You might also want to resize by doubling in size each time you need to, and then trim at the end. That would basically be mimicking the ArrayList behaviour. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Feb 25, 2009 at 14:58 Jon SkeetJon Skeet 1.5m892892 gold badges9.3k9.3k silver badges9.3k9.3k bronze badges 1 I like java.util.Arrays.copyOfRange(int [] original, int from, int to); – eleven81 Commented Feb 25, 2009 at 15:04 Add a comment \| This answer is useful 7 Save this answer. Show activity on this post. Algo using Sieve of Eratosthenes ``` public static List findPrimes(int limit) { List<Integer> list = new ArrayList<>(); boolean [] isComposite = new boolean [limit + 1]; // limit + 1 because we won't use '0'th index of the array isComposite = true; // Mark all composite numbers for (int i = 2; i <= limit; i++) { if (!isComposite[i]) { // 'i' is a prime number list.add(i); int multiple = 2; while (i multiple <= limit) { isComposite [i multiple] = true; multiple++; } } } return list; } ``` Image depicting the above algo (Grey color cells represent prime number. Since we consider all numbers as prime numbers intially, the whole is grid is grey initially.) Image Source: WikiMedia Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Dec 22, 2013 at 14:04 answered Dec 22, 2013 at 13:57 AmitAmit 34.9k9292 gold badges233233 silver badges304304 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. The easiest solution would be to return some member of the Collections Framework instead of an array. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Feb 25, 2009 at 14:58 Hank GayHank Gay 72.3k3636 gold badges163163 silver badges224224 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Are you using Java 1.5? Why not return List<Integer> and use ArrayList<Integer>? If you do need to return an int[], you can do it by converting List to int[] at the end of processing. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications edited Feb 25, 2009 at 15:08 answered Feb 25, 2009 at 15:00 BogdanBogdan 1,16033 gold badges1111 silver badges1717 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. As Paul Tomblin points out, there are better algorithms. But keeping with what you have, and assuming an object per result is too big: You are only ever appending to the array. So, use a relatively small int[] array. When it's full use append it to a List and create a replacement. At the end copy it into a correctly sized array. Alternatively, guess the size of the int[] array. If it is too small, replace by an int[] with a size a fraction larger than the current array size. The performance overhead of this will remain proportional to the size. (This was discussed briefly in a recent stackoverflow podcast.) Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Feb 25, 2009 at 15:06 Tom Hawtin - tacklineTom Hawtin - tackline 147k3030 gold badges221221 silver badges314314 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Now that you've got a basic sieve in place, note that the inner loop need only continue until temp[i]temp[i] > prime. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Feb 25, 2009 at 15:13 dmckee --- ex-moderator kittendmckee --- ex-moderator kitten 102k2525 gold badges146146 silver badges236236 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. I have a really efficient implementation: we don't keep the even numbers, therefore halving the memory usage. we use BitSet, requiring only one bit per number. we estimate the upper bound for number of primes on the interval, thus we can set the initialCapacity for the Array appropriately. we don't perform any kind of division in the loops. Here's the code: ``` public ArrayList sieve(int n) { int upperBound = (int) (1.25506 n / Math.log(n)); ArrayList result = new ArrayList(upperBound); if (n >= 2) result.add(2); int size = (n - 1) / 2; BitSet bs = new BitSet(size); int i = 0; while (i < size) { int p = 3 + 2 i; result.add(p); for (int j = i + p; j < size; j += p) bs.set(j); i = bs.nextClearBit(i + 1); } return result; } ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Mar 1, 2015 at 16:13 answered Mar 1, 2015 at 12:16 Rok KraljRok Kralj 48.9k1010 gold badges7474 silver badges8080 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Restructure your code. Throw out the temporary array, and instead write function that just prime-tests an integer. It will be reasonably fast, since you're only using native types. Then you can, for instance, loop and build a list of integers that are prime, before finally converting that to an array to return. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Feb 25, 2009 at 15:00 unwindunwind 401k6464 gold badges490490 silver badges619619 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Not sure if this will suite your situation but you can take a look at my approach. I used mine using Sieve of Eratosthenes. ``` public static List sieves(int n) { Map numbers = new LinkedHashMap<>(); List<Integer> primes = new ArrayList<>(); //First generate a list of integers from 2 to 30 for(int i=2; i<n;i++){ numbers.put(i,true); } for(int i : numbers.keySet()){ / The first number in the list is 2; cross out every 2nd number in the list after 2 by counting up from 2 in increments of 2 (these will be all the multiples of 2 in the list): The next number in the list after 2 is 3; cross out every 3rd number in the list after 3 by counting up from 3 in increments of 3 (these will be all the multiples of 3 in the list): The next number not yet crossed out in the list after 5 is 7; the next step would be to cross out every 7th number in the list after 7, but they are all already crossed out at this point, as these numbers (14, 21, 28) are also multiples of smaller primes because 7 × 7 is greater than 30. The numbers not crossed out at this point in the list are all the prime numbers below 30: / if(numbers.get(i)){ for(int j = i+i; j<n; j+=i) { numbers.put(j,false); } } } for(int i : numbers.keySet()){ for(int j = i+i; j<n && numbers.get(i); j+=i) { numbers.put(j,false); } } for(int i : numbers.keySet()){ if(numbers.get(i)) { primes.add(i); } } return primes; } ``` Added comment for each steps that has been illustrated in wikipedia Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Aug 15, 2017 at 14:40 KyelJmDKyelJmD 4,7221010 gold badges5858 silver badges7878 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. I have done using HashMap and found it very simple ``` import java.util.HashMap; import java.util.Map; /Using Algorithms such as sieve of Eratosthanas / public class PrimeNumber { public static void main(String[] args) { int prime = 15; HashMap<Integer, Integer> hashMap = new HashMap<Integer, Integer>(); hashMap.put(0, 0); hashMap.put(1, 0); for (int i = 2; i <= prime; i++) { hashMap.put(i, 1);// Assuming all numbers are prime } printPrimeNumberEratoshanas(hashMap, prime); } private static void printPrimeNumberEratoshanas(HashMap<Integer, Integer> hashMap, int prime) { System.out.println("Printing prime numbers upto" + prime + "....."); for (Map.Entry<Integer, Integer> entry : hashMap.entrySet()) { if (entry.getValue().equals(1)) { System.out.println(entry.getKey()); for (int j = entry.getKey(); j < prime; j++) { for (int k = j; k j <= prime; k++) { hashMap.put(j k, 0); } } } } } } ``` Think this is effective Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Feb 24, 2018 at 10:14 reactdontactreactdontact 5311 gold badge11 silver badge1111 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. ``` public static void primes(int n) { boolean[] lista = new boolean[n+1]; for (int i=2;i<lista.length;i++) { if (lista[i]==false) { System.out.print(i + " "); } for (int j=i+i;j<lista.length;j+=i) { lista[j]=true; } } } ``` Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Apr 23, 2019 at 9:14 answered Apr 22, 2019 at 21:01 user7188242user7188242 1 1 Welcome to StackOverflow! You should add an addition explanation to your answer, check out How do I write a good answer? – Dennis Vash Commented Apr 22, 2019 at 21:24 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions java arrays primes sieve-of-eratosthenes See similar questions with these tags. The Overflow Blog Renewing Chat on Stack Overflow AI isn’t stealing your job, it’s helping you find it Featured on Meta Will you help build our new visual identity? 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12102	https://pmc.ncbi.nlm.nih.gov/articles/PMC11315634/	The Clinical Spectrum of HbSC Sickle Cell Disease-Not a Benign Condition - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Published in final edited form as: Br J Haematol. 2024 Jun 19;205(2):653–663. doi: 10.1111/bjh.19523 Search in PMC Search in PubMed View in NLM Catalog Add to search The Clinical Spectrum of HbSC Sickle Cell Disease-Not a Benign Condition M Nelson M Nelson 1 Department of Medicine, Division of Hematology/Center for Sickle Cell Disease, University of Tennessee Health Science Center, Memphis, TN, USA Find articles by M Nelson 1, L Noisette L Noisette 2 Division of Pediatric Hematology and Oncology, Children’s Mercy Hospital, Kansas City, Missouri, USA Find articles by L Noisette 2, N Pugh N Pugh 3 Division of Biostatistics and Epidemiology, RTI International, Research Triangle Park, North Carolina, USA Find articles by N Pugh 3, V Gordeuk V Gordeuk 4 Division of Hematology and Oncology, Department of Medicine, University of Illinois at Chicago, Chicago, IL Find articles by V Gordeuk 4, LL Hsu LL Hsu 5 Division of Pediatric Hematology-Oncology, University of Illinois at Chicago, Chicago, Illinois Find articles by LL Hsu 5, T Wun T Wun 6 Department of Medicine, Division of Hematology and Oncology, UC Davis, Sacramento, California, USA Find articles by T Wun 6, N Shah N Shah 7 Pediatric Hematology/Oncology, Duke University, Durham, North Carolina, USA Find articles by N Shah 7, J Glassberg J Glassberg 8 Department of Emergency Medicine, Icahn School of Medicine at Mount Sinai, New York, New York, USA Find articles by J Glassberg 8, A Kutlar A Kutlar 9 Sickle Cell Center, Augusta University, Augusta, Georgia, USA Find articles by A Kutlar 9, JS Hankins JS Hankins 10 Departments of Global Pediatric Medicine and Hematology, St. Jude Children's Research Hospital, Memphis, Tennessee, USA Find articles by JS Hankins 10, AA King AA King 11 Division of Pediatric Hematology and Oncology, Washington University School of Medicine, St. Louis, Missouri, USA Find articles by AA King 11, D Brambilla D Brambilla 3 Division of Biostatistics and Epidemiology, RTI International, Research Triangle Park, North Carolina, USA Find articles by D Brambilla 3, J Kanter J Kanter 12 Department of Medicine, University of Alabama, Birmingham, Alabama, USA Find articles by J Kanter 12 Author information Article notes Copyright and License information 1 Department of Medicine, Division of Hematology/Center for Sickle Cell Disease, University of Tennessee Health Science Center, Memphis, TN, USA 2 Division of Pediatric Hematology and Oncology, Children’s Mercy Hospital, Kansas City, Missouri, USA 3 Division of Biostatistics and Epidemiology, RTI International, Research Triangle Park, North Carolina, USA 4 Division of Hematology and Oncology, Department of Medicine, University of Illinois at Chicago, Chicago, IL 5 Division of Pediatric Hematology-Oncology, University of Illinois at Chicago, Chicago, Illinois 6 Department of Medicine, Division of Hematology and Oncology, UC Davis, Sacramento, California, USA 7 Pediatric Hematology/Oncology, Duke University, Durham, North Carolina, USA 8 Department of Emergency Medicine, Icahn School of Medicine at Mount Sinai, New York, New York, USA 9 Sickle Cell Center, Augusta University, Augusta, Georgia, USA 10 Departments of Global Pediatric Medicine and Hematology, St. Jude Children's Research Hospital, Memphis, Tennessee, USA 11 Division of Pediatric Hematology and Oncology, Washington University School of Medicine, St. Louis, Missouri, USA 12 Department of Medicine, University of Alabama, Birmingham, Alabama, USA ✉ Corresponding Author: Julie Kanter, MD (on behalf of my co-authors), Director, Adult Sickle Cell Program, Professor of Medicine and Pediatrics, Division of Hematology and Oncology, University of Alabama at Birmingham, 1720 2nd Avenue South, NP 2540, Birmingham, AL 35294-3300, jkanter@uabmc.edu Author Contributions MN, NL, NP, JK drafted the manuscript and directed the statistical analysis. NP performed the statistical analyses. MN, NL, VRG, LLH, JG, AK, TW, JH, AAK, DB and JK contributed to recruiting and clinically studying the registry. MN, LN, NP, DB, JK contributed to the concept and design of the study. All authors critically reviewed and approved the manuscript. Issue date 2024 Aug. PMC Copyright notice PMCID: PMC11315634 NIHMSID: NIHMS1992172 PMID: 38898714 The publisher's version of this article is available at Br J Haematol Abstract Sickle cell disease (SCD) includes a group of heterogenous disorders that result in significant morbidities. HbSS is the most common type of SCD and HbSC is the second most common type of SCD. The prevalence of HbSC disease in the United States and United Kingdom is approximately 1 in 7174 births and 1 in 6174 births respectively. Despite its frequency, however, HbSC disease has been insufficiently studied and was historically categorized as a more “mild” form of SCD. We conducted this study of HbSC disease as part of the NHLBI funded Sickle Cell Disease Implementation Consortium (SCDIC). The SCDIC registry included 2282 individuals with SCD, ages 15-45 years of whom 502 (22%) had HbSC disease. Compared to people with sickle cell anemia (SCA), the study found that people with HbSC disease had a higher frequency of splenomegaly (n (%)= 169 (33.7) vs 392 (22.1)) and retinopathy (n (%)= 116 (23.1) vs 189 (10.6)). A Many people with HbSC also had avascular necrosis (n(%)= 112 (22.3)), pulmonary embolism (n(%)= 43 (8.6)) and acute chest syndrome (n(%)= 228 (45.4)) demonstrating significant disease severity. HbSC disease is more clinically severe than was previously recognized and deserves additional evaluation and targeted treatments. Graphical Abstract The most frequent complications of adults with HbSC disease: results from the SCDIC Registry INTRODUCTION Sickle cell disease (SCD) includes a group of heterogenous disorders that share the presence of the gene for hemoglobin S (HbS). The homozygous condition, HbSS, is the most common accounting for approximately 65% of cases in the United States (1). Those who are double heterozygotes inherit HbS with another abnormal hemoglobin, most commonly hemoglobin C resulting in HbSC disease, the second most common type of SCD. Despite its frequency, however, HbSC disease has been rarely studied. In people with HbSC disease, their red blood cells contain approximately 50% HbS and 50% hemoglobin C. Individually, these single gene abnormalities do not have significant clinical significance (as people who co-inherit a HbS trait or HbC trait with normal HbA). However, in people with HbSC disease, there is more HbS present than in people with HbAS (sickle cell trait) and HbC is substantially less soluble. Previous studies have demonstrated that the red cells in people with Hb SC disease have high intracellular hemoglobin concentration and that they are more sensitive to dehydration(2). The prevalence of HbSC disease in the United States and United Kingdom is approximately 1 in 7174 births and 1 in 6174 births respectively (Streetly et al 2008 and Therell, et al 2015). People with SCD present with a wide range of clinical manifestations even within the distinct genotypes. Historically, the disease has been characterized as mild, moderate, or severe according to the genotype; however, data shows that there are significant variations in clinical manifestations within genotypes as well as between genotypes. As all affected individuals are at lifelong risk of vaso-occlusion and hemolysis both of which cause endothelial damage, there may be significant worsening of disease as individuals age. Moreover, disease modifying therapies such as hydroxyurea are used less in HbSC disease compared to SCA. This is due to the lack of participants with HbSC disease in prospective drug clinical trials, which results in a paucity of efficacy and safety data in this disease subtype. (Ware 2010). Several previous retrospective and case cohort studies have examined people with HbSC disease(3-5). These studies have demonstrated that the childhood course of people with HbSC is usually milder than those with HbSS disease, substantiated by the relative lower risk of sepsis or stroke. In addition, the children with HbSC disease have more normal growth and development with less renal concentrating defects than is seen in those with HbSS disease.(6-10) During episodes of acute chest syndrome, HbSC disease is associated with higher levels of nitric oxide compared to patients with HbSS. (Piccin et al 2015) Nitric oxide reduces cell adhesion and inflammation in the endothelium which ameliorates nitric oxide pathology in sickle cell mice models. (Kim-Shapiro DB, et al 2018) These findings have resulted in this sub-type of SCD often being labeled as “less severe.” Many of these earlier studies were performed before universal newborn screening was instituted across the United States and prior to the widespread use of the protein conjugate anti-pneumococcal vaccines were recommended for all children (with and without SCD). Standardization of penicillin prophylaxis for all children with sickle cell disease resulted in monumental improvements in morbidity and mortality due to pneumococcal sepsis. (Gaston et al, 1986). These early child interventions have dramatically improved the survival of all children with SCD such that the majority (>98%) are living into adulthood in high-resource countries.(11) Thus, the objective of this study was to better characterize the clinical complications and laboratory manifestations in people with the HbSC genotype in a large prospective contemporary cohort. METHODS Study Population. This analysis was conducted as part of the Sickle Cell Disease Implementation Consortium (SCDIC), a cooperative research program aimed at using implementation science research to accelerate the translation of evidence-based therapies into clinical care among individuals with SCD ages 15 to 45 years.(12) The SCDIC was funded by the National Heart, Lung, Blood Institute and the principal investigators from each clinical site participated on the Steering Committee, which oversaw the operations of the SCDIC. Adolescents and adults aged 15 to 45 years, enrolled in the SCDIC registry as a convenience sample from 8 comprehensive sickle cell centers, were included in the study population. As previously described,(13) baseline data were gathered between 2016 and 2019 preceding two annual follow-up surveys. The focus of this analysis was on participants with Hb SC disease. This was a cross-sectional study population that ascertained the participants’ baseline co-morbidities, laboratory assessments and surveys as well as their acute care utilization. Final data was obtained at a median follow-up of 4.7 years and included vital status. This study was approved by the institutional review board of all participating institutions and written informed consent was obtained from all participants or their legal guardian if the participant was a minor. Measures. Demographic characteristics including gender, age, SCD genotype (Hb SC, other), race, ethnicity, primary language, marital status (married or living as married, other), number of children and adults living in the household, household income, education, and occupation were collected. Other data included laboratory results, transfusion history, SCD complications, number of acute pain/infusion center visits, emergency department visits and hospitalizations in the year prior to registry enrollment and pain history. For qualitative outcomes such as chronic pain, information was abstracted from the medical history/problem list which were generally based on the medical provider’s independent assessment. Pregnancy characteristics were also collected and examined in this analysis. Outcomes of Interest. The primary outcome of interest was the frequency of clinical manifestations and complications among Hb SC participants compared to those individuals with sickle cell anemia (SCA) types (HbSS and HbSß 0). Secondary outcomes included select laboratory values and their association with SCD complications and pain frequency within the HbSC cohort. Statistical Analyses Baseline characteristics of the study population, including pregnancy characteristics, were summarized by frequency and percentage or by mean and standard deviation. Initially, three sets of odds ratios for the presence/absence of history of each SCD complication vs genotype (HbSC vs SCA) were calculated, with odds ratios <1 indicating lower prevalence in HbSC patients relative to SCA. The first set of odds ratios and confidence limits were generated using a case control approach with 2 controls (genotypes Hb SS or other) age matched to each case (HbSC). There were sufficient controls to match each case 2:1 for most of the outcomes. Where there were insufficient controls, cases were randomly dropped. We then averaged the odds ratios and confidence limits from 100 separate analyses, each with a separate set of randomly selected controls and, if needed, randomly excluded cases. A second set of odds ratios and confidence limits were generated similarly except that age was added to each model as a continuous predictor. Before fitting these models, the data were checked for any age-genotype interactions. A third set of odds ratios and confidence limits were generated by a single run for each outcome using all available subjects with the genotypes of interest. For final reporting purposes, we decided to use the age-adjusted, case-control models, and adjust them for multiple imputation using PROC MIANALYZE. To examine select laboratory results and their association with complications and pain frequency, within the Hb SC cohort, we used either linear or logistic regression, as appropriate. All analyses were conducted in SAS Version 9.4 (SAS Institute Inc., Cary, NC, USA). RESULTS Demographics A total of 2282 SCDIC registry participants with HbSC or SCA (HbSS and HbSß 0) were included in the analysis. Baseline characteristics of the Hb SC cohort (n (%)= 502 (22)) and others (n (%)= 1780 (78)) are presented in Table 1. The Hb SC cohort had a mean age of 28.2 (8.0) years and was 58% female. There were 403 (82.1%) with an education level of high school graduate or higher, but only 174 (37.7%) were engaged in the workforce. These findings were very similar to the other participants who had a mean age of 28.1 (7.8) years and was 55.9% female. Table 1: Baseline Demographics of the HbSC and SCA (HbSS/HbSB0 thalassemia) cohorts \| Characteristic \| Hb SC (N=502) \| SCA (N=1780) \| Total (N=2282) \| ---: ---: \| \| Demographics \| \| \| \| \| Age group (years) \| \| \| \| \| <18 \| 46 (9.2%) \| 154 (8.7%) \| 200 (8.8%) \| \| 18-24 \| 139 (27.7%) \| 490 (27.5%) \| 629 (27.6%) \| \| 25-34 \| 199 (39.6%) \| 729 (41.0%) \| 928 (40.7%) \| \| 35-45 \| 118 (23.5%) \| 407 (22.9%) \| 525 (23.0%) \| \| Age (years) \| \| \| \| \| N \| 502 \| 1780 \| 2282 \| \| Mean (std) \| 28.2 (8.0) \| 28.1 (7.8) \| 28.1 (7.9) \| \| Median \| 28.0 \| 28.0 \| 28.0 \| \| Min, Max \| 15, 45 \| 15, 45 \| 15, 45 \| \| Gender \| \| \| \| \| Male \| 211 (42.0%) \| 785 (44.1%) \| 996 (43.6%) \| \| Female \| 291 (58.0%) \| 995 (55.9%) \| 1,286 (56.4%) \| \| Race \| \| \| \| \| Black or African American \| 487 (98.2%) \| 1,694 (96.9%) \| 2,181 (97.1%) \| \| Other \| 9 (1.8%) \| 55 (3.1%) \| 64 (2.9%) \| \| Ethnicity \| \| \| \| \| Hispanic or Latino \| 21 (4.2%) \| 91 (5.1%) \| 112 (4.9%) \| \| Not Hispanic or Latino \| 479 (95.8%) \| 1,678 (94.9%) \| 2,157 (95.1%) \| \| Language you feel most comfortable speaking \| \| \| \| \| English \| 498 (99.6%) \| 1,737 (99.3%) \| 2,235 (99.3%) \| \| Other \| 2 (0.4%) \| 13 (0.7%) \| 15 (0.7%) \| \| Marital Status \| \| \| \| \| Married or Living as married \| 78 (17.8%) \| 239 (15.5%) \| 317 (16.0%) \| \| Other \| 359 (82.2%) \| 1,301 (84.5%) \| 1,660 (84.0%) \| \| Household income \| \| \| \| \| $25,000 or less \| 207 (47.2%) \| 865 (56.1%) \| 1,072 (54.1%) \| \| $25,001+ \| 232 (52.8%) \| 678 (43.9%) \| 910 (45.9%) \| \| Education \| \| \| \| \| Less than high school graduate \| 88 (17.9%) \| 298 (17.3%) \| 386 (17.4%) \| \| High school graduate or higher \| 403 (82.1%) \| 1,429 (82.7%) \| 1,832 (82.6%) \| \| Employment \| \| \| \| Open in a new tab Clinical Complications Although the HbSC participants had a lower prevalence of most SCD-related complications, they had a higher frequency of splenomegaly (n (%)= 169 (33.7) vs 392 (22.1) for SCA) and retinopathy (n (%)= 116 (23.1) vs 189 (10.6) for SCA) as shown in Figure 1, Tables 2 and 4. Age related analysis was based on the age of the person at the time of registry enrollment. Splenomegaly was found to increase with age; therefore, it was excluded from the odds ratio analysis. However, the age at the time of actual diagnosis (of all complications) is not known. Additionally, the Hb SC participants had a higher prevalence of anxiety (n (%)= 74 (14.7) vs 222 (12.5) for SCA) and depression (n (%)= 109 (21.7) vs 316 (17.8) for SCA). Notably, this cohort also had many people with a history of avascular necrosis (n(%)= 112 (22.3) vs. 489 (27.5) for SCA), pulmonary embolism (n(%)= 43 (8.6) vs 152 (8.5) for SCA), and acute chest syndrome (n(%)= 228 (45.4) vs. 1,054 (59.2)) demonstrating significant disease severity even if the frequency was slightly lower than for those with SCA (Table 2). Figure 1: Distribution of SCD complications in people with HbSC disease compared to those with SCA (HbSS and HbSB 0). Open in a new tab Graph includes conditions that are present in 10% or more of total study Population Avascular necrosis includes hip, shoulder, and knee Pulmonary arterial hypertension was reported by medical history and did not include method of diagnosis Splenomegaly includes splenic sequestration, splenic infarcts, hypersplenism, and splenectomy Table 2. Baseline Clinical Complications of the HbSC and SCA (HbSS/HbSB0 thalassemia) cohorts \| Characteristic \| Hb SC (N=502) \| SCA (N=1780) \| Total (N=2282) \| ---: ---: \| \| Avascular necrosis1 \| \| \| \| \| Yes \| 112 (22.3%) \| 489 (27.5%) \| 601 (26.3%) \| \| No \| 246 (49.0%) \| 760 (42.7%) \| 1,006 (44.1%) \| \| Missing/Not in record \| 144 (28.7%) \| 531 (29.8%) \| 675 (29.6%) \| \| Osteomyelitis \| \| \| \| \| Yes \| 13 (2.6%) \| 93 (5.2%) \| 106 (4.6%) \| \| No \| 256 (51.0%) \| 813 (45.7%) \| 1,069 (46.8%) \| \| Missing/Not in record \| 233 (46.4%) \| 874 (49.1%) \| 1,107 (48.5%) \| \| Chronic Kidney Disease \| \| \| \| \| Yes \| 16 (3.2%) \| 177 (9.9%) \| 193 (8.5%) \| \| No \| 291 (58.0%) \| 876 (49.2%) \| 1,167 (51.1%) \| \| Missing/Not in record \| 195 (38.8%) \| 727 (40.8%) \| 922 (40.4%) \| \| End stage renal disease \| \| \| \| \| Yes \| 3 (0.6%) \| 19 (1.1%) \| 22 (1.0%) \| \| No \| 264 (52.6%) \| 880 (49.4%) \| 1,144 (50.1%) \| \| Missing/Not in record \| 235 (46.8%) \| 881 (49.5%) \| 1,116 (48.9%) \| \| Priapism \| \| \| \| \| Yes \| 36 (7.2%) \| 203 (11.4%) \| 239 (10.5%) \| \| No \| 321 (63.9%) \| 1,070 (60.1%) \| 1,391 (61.0%) \| \| Missing/Not in record \| 145 (28.9%) \| 507 (28.5%) \| 652 (28.6%) \| \| Stroke: Ischemic \| \| \| \| \| Yes \| 9 (1.8%) \| 103 (5.8%) \| 112 (4.9%) \| \| No \| 8 (1.6%) \| 90 (5.1%) \| 98 (4.3%) \| \| Missing/Not in record \| 485 (96.6%) \| 1,587 (89.2%) \| 2,072 (90.8%) \| \| Stroke: Hemorrhagic \| \| \| \| \| Yes \| 1 (0.2%) \| 21 (1.2%) \| 22 (1.0%) \| \| No \| 13 (2.6%) \| 140 (7.9%) \| 153 (6.7%) \| \| Missing/Not in record \| 488 (97.2%) \| 1,619 (91.0%) \| 2,107 (92.3%) \| \| Pulmonary arterial hypertension2 \| \| \| \| \| Yes \| 32 (6.4%) \| 245 (13.8%) \| 277 (12.1%) \| \| No \| 277 (55.2%) \| 847 (47.6%) \| 1,124 (49.3%) \| \| Missing/Not in record \| 193 (38.4%) \| 688 (38.7%) \| 881 (38.6%) \| \| Left ventricular dysfunction \| \| \| \| \| Yes \| 7 (1.4%) \| 68 (3.8%) \| 75 (3.3%) \| \| No \| 265 (52.8%) \| 864 (48.5%) \| 1,129 (49.5%) \| \| Missing/Not in record \| 230 (45.8%) \| 848 (47.6%) \| 1,078 (47.2%) \| \| Acute chest syndrome \| \| \| \| \| Yes \| 228 (45.4%) \| 1,054 (59.2%) \| 1,282 (56.2%) \| \| No \| 181 (36.1%) \| 403 (22.6%) \| 584 (25.6%) \| \| Missing/Not in record \| 93 (18.5%) \| 323 (18.1%) \| 416 (18.2%) \| \| Asthma \| \| \| \| \| Yes \| 108 (21.5%) \| 467 (26.2%) \| 575 (25.2%) \| \| No \| 231 (46.0%) \| 725 (40.7%) \| 956 (41.9%) \| \| Missing/Not in record \| 163 (32.5%) \| 588 (33.0%) \| 751 (32.9%) \| \| Gallstones/cholelithiasis, cholecystitis \| \| \| \| \| Yes \| 141 (28.1%) \| 896 (50.3%) \| 1,037 (45.4%) \| \| No \| 216 (43.0%) \| 445 (25.0%) \| 661 (29.0%) \| \| Missing/Not in record \| 145 (28.9%) \| 439 (24.7%) \| 584 (25.6%) \| \| Splenomegaly3 \| \| \| \| \| Yes \| 169 (33.7%) \| 394 (22.1%) \| 563 (24.7%) \| \| No \| 186 (37.1%) \| 707 (39.7%) \| 893 (39.1%) \| \| Missing/Not in record \| 147 (29.3%) \| 679 (38.1%) \| 826 (36.2%) \| \| Pulmonary embolism \| \| \| \| \| Yes \| 43 (8.6%) \| 152 (8.5%) \| 195 (8.5%) \| \| No \| 6 (1.2%) \| 78 (4.4%) \| 84 (3.7%) \| \| Missing/Not in record \| 453 (90.2%) \| 1,550 (87.1%) \| 2,003 (87.8%) \| \| Deep vein thrombosis (DVT) \| \| \| \| \| Yes \| 23 (4.6%) \| 102 (5.7%) \| 125 (5.5%) \| \| No \| 12 (2.4%) \| 76 (4.3%) \| 88 (3.9%) \| \| Missing/Not in record \| 467 (93.0%) \| 1,602 (90.0%) \| 2,069 (90.7%) \| \| Skin ulcers (Integumentary) \| \| \| \| \| Yes \| 11 (2.2%) \| 81 (4.6%) \| 92 (4.0%) \| \| No \| 288 (57.4%) \| 920 (51.7%) \| 1,208 (52.9%) \| \| Missing/Not in record \| 203 (40.4%) \| 779 (43.8%) \| 982 (43.0%) \| \| Retinopathy \| \| \| \| \| Yes \| 116 (23.1%) \| 189 (10.6%) \| 305 (13.4%) \| \| No \| 249 (49.6%) \| 957 (53.8%) \| 1,206 (52.8%) \| \| Missing/Not in record \| 137 (27.3%) \| 634 (35.6%) \| 771 (33.8%) \| \| Chronic refractory pain \| \| \| \| \| Yes \| 96 (19.1%) \| 369 (20.7%) \| 465 (20.4%) \| \| No \| 191 (38.0%) \| 630 (35.4%) \| 821 (36.0%) \| \| Missing/Not in record \| 215 (42.8%) \| 781 (43.9%) \| 996 (43.6%) \| \| Anxiety \| \| \| \| \| Yes \| 74 (14.7%) \| 222 (12.5%) \| 296 (13.0%) \| \| No \| 255 (50.8%) \| 853 (47.9%) \| 1,108 (48.6%) \| \| Missing/Not in record \| 173 (34.5%) \| 705 (39.6%) \| 878 (38.5%) \| \| Depression \| \| \| \| \| Yes \| 109 (21.7%) \| 316 (17.8%) \| 425 (18.6%) \| \| No \| 242 (48.2%) \| 822 (46.2%) \| 1,064 (46.6%) \| \| Missing/Not in record \| 151 (30.1%) \| 642 (36.1%) \| 793 (34.8%) \| Open in a new tab 1 Avascular necrosis includes hip, shoulder, and knee 2 Pulmonary Arterial Hypertension was reported by medical history and did not include method of diagnosis 3 Splenomegaly includes splenic sequestration, splenic infarcts, hypersplenism, and splenectomy Table 4. Odds ratios for presence/absence of each complication vs genotype (HbSC vs SCA) \| Complication \| Odds ratio, CI \| :--- \| \| avascular necrosis \| 0.8634 (0.756, 0.986) \| \| osteomyelitis \| 0.7009 (0.484, 1.02) \| \| chronic kidney disease \| 0.5424 (0.401, 0.735) \| \| end stage renal disease \| 0.7564 (0.352, 1.623) \| \| priapism (males only) \| 0.7536 (0.597, 0.951) \| \| ischemic stroke \| 0.5531 (0.370, 0.826) \| \| hemorrhagic stroke \| 0.402 (0.136, 1.189) \| \| intracranial bleeding \| 0.3592 (0.122, 1.060) \| \| pulmonary arterial hypertension \| 0.6417 (0.514, 0.802) \| \| left ventricular dysfunction \| 0.6073 (0.376, 0.982) \| \| acute chest syndrome \| 0.7586 (0.648, 0.888) \| \| asthma \| 0.8858 (0.774, 1.014) \| \| gallstones/cholelithiasis, cholecystitis \| 0.6063 (0.523, 0.703) \| \| splenomegaly \| age-genotype interaction, p<0.05 \| \| pulmonary embolism \| 0.9862 (0.782, 1.243) \| \| venous thromboembolism \| 0.8906 (0.652, 1.217) \| \| pneumococcal sepsis \| 0.6677 (0.441, 1.011) \| \| skin ulcers \| 0.6854 (0.460, 1.021) \| \| retinopathy \| 1.5846 (1.33, 1.888) \| \| chronic refractory pain \| 0.9626 (0.83, 1.118) \| \| anxiety \| 1.1192 (0.93, 1.347) \| \| chronic pain \| 1.1354 (0.979, 1.316) \| Open in a new tab Odds ratios <1 indicate lower prevalence in HbSC patients. Bolded findings note significance. The odds ratios and confidence limits were generated using a case control approach with 2 controls (genotypes HbSS or HbSS/HbBeta 0) age matched to each case (HbSC). There were sufficient controls to match each case 2:1 for most of the outcomes. Where there were insufficient controls, cases were randomly dropped. The total number of cases in the data set is shown in column C and the total number used in column D. For each complication, we ran 100 separate logistic regression analyses, each with a separate set of randomly selected controls and, if needed, randomly excluded cases. Laboratory Findings: The mean (standard deviation) hemoglobin level was 11.5 (1.5) g/dL among those with Hb SC versus the mean hemoglobin in the SCA cohort of 8.9 (1.5) g/dL. People with HbSC disease also had lower markers of hemolysis with mean LDH of 314.5 (294.5) compared to 497.8 (352.5) in SCA genotypes and a total bilirubin of 1.6 (1.6) mg/dl compared to 3.3 (2.9)mg/dl (Table 3). Table 3. Pregnancy Characteristics (for all Pregnancies among 687 unique women with HbSC or SCA) \| \| \| :--- \| \| Characteristic \| Hb SC genotype (N=486) \| SCA genotypes (N=1179) \| Total (N=1665) \| \| Pregnancy outcome \| \| \| \| \| Live birth \| 304 (63.5%) \| 687 (58.9%) \| 991 (60.2%) \| \| Still birth \| 7 (1.5%) \| 31 (2.7%) \| 38 (2.3%) \| \| Miscarriage \| 91 (19.0%) \| 287 (24.6%) \| 378 (23.0%) \| \| Abortion \| 59 (12.3%) \| 132 (11.3%) \| 191 (11.6%) \| \| Currently pregnant \| 18 (3.8%) \| 30 (2.6%) \| 48 (2.9%) \| \| Taking hydroxyurea at time of conception \| \| \| \| \| Yes \| 34 (7.5%) \| 237 (21.6%) \| 271 (17.5%) \| \| No \| 419 (92.5%) \| 859 (78.4%) \| 1,278 (82.5%) \| \| Fertility drugs or any procedure to help get pregnant \| \| \| \| \| Yes \| 8 (1.7%) \| 37 (3.2%) \| 45 (2.7%) \| \| No \| 469 (98.3%) \| 1,126 (96.8%) \| 1,595 (97.3%) \| \| Premature birth \| \| \| \| \| Yes \| 92 (30.6%) \| 286 (42.9%) \| 378 (39.0%) \| \| No \| 209 (69.4%) \| 381 (57.1%) \| 590 (61.0%) \| \| Baby less than 5.5 pounds at birth \| \| \| \| \| Yes \| 68 (22.4%) \| 233 (34.8%) \| 301 (30.9%) \| \| No \| 235 (77.6%) \| 437 (65.2%) \| 672 (69.1%) \| \| LBW/birth defect/genetic condition/other serious medical problem related to birth \| \| \| \| \| Yes \| 66 (21.9%) \| 196 (29.4%) \| 262 (27.0%) \| \| No \| 236 (78.1%) \| 471 (70.6%) \| 707 (73.0%) \| \| -Pain Crisis \| \| \| \| \| Yes \| 168 (34.6%) \| 431 (36.6%) \| 599 (36.0%) \| \| No \| 318 (65.4%) \| 748 (63.4%) \| 1,066 (64.0%) \| \| -Acute chest syndrome \| \| \| \| \| Yes \| 20 (4.1%) \| 52 (4.4%) \| 72 (4.3%) \| \| No \| 466 (95.9%) \| 1,127 (95.6%) \| 1,593 (95.7%) \| \| -Preeclampsia \| \| \| \| \| Yes \| 37 (7.6%) \| 107 (9.1%) \| 144 (8.6%) \| \| No \| 449 (92.4%) \| 1,072 (90.9%) \| 1,521 (91.4%) \| \| -Maternal diabetes \| \| \| \| \| Yes \| 6 (1.2%) \| 5 (0.4%) \| 11 (0.7%) \| \| No \| 480 (98.8%) \| 1,174 (99.6%) \| 1,654 (99.3%) \| \| -Blood clots \| \| \| \| \| Yes \| 9 (1.9%) \| 43 (3.6%) \| 52 (3.1%) \| \| No \| 477 (98.1%) \| 1,136 (96.4%) \| 1,613 (96.9%) \| Open in a new tab Among HbSC participants, there was no association between hemoglobin and retinopathy or avascular necrosis. When examining hemoglobin as a continuous variable, the p-value for association with retinopathy was p=0.3615 (β=0.2) and with avascular necrosis was p=0.5650 (β=0.1). The parameter estimate, or β, is the estimated difference in mean hemoglobin between those with and without the specific complication. Using a categorical hemoglobin cutoff of 10 g/dL, the association with retinopathy was OR [95% CI] = 1.37, [0.695, 2.859], p=0.3821. The association with avascular necrosis was OR [95% CI] = 1.54, [0.755, 3.408], p=0.2567. Using a hemoglobin cutoff of 11 g/dL yielded a similar, non-significant association. Additionally, there was no association within the HbSC cohort between hemolysis biomarkers and the number of clinical complications, using our calculated marker for hemolysis and the clinical complications considered in this study population (p=0.3178, β=0.04). Nor was there any significant association between hemolysis and pain frequency over the previous 6 months (p=0.6917, β=−0.1 (pain=rarely), β=−0.1 (pain=sometimes), β=−0.3 (pain=often), β=−0.1 (pain=always)). Acute Care Utilization and Patient Reported Outcomes Acute care data are shown in table 4. Over the year prior to enrollment, Hb SC participants also experienced more acute pain/infusion center visits for acute pain (mean (sd)=6.9 (17.1) vs 5.7 (11.1)) and more emergency department visits for acute pain (mean (sd)=3.4 (7.5) vs 2.9 (5.7)) than people with SCA. The mean number of hospitalizations in the prior year was similar 2.4 (6.3) for the HbSC cohort and 2.6 (3.9) for the SCA cohort. Among patient reported outcomes regarding pain (table 5), Hb SC participants had a lower prevalence of taking daily pain medication for SCD (n (%)= 225 (44.8) vs 837 (47) for SCA), and were more likely to report no instances of very severe pain over the previous 6 months (n (%)= 154 (31.0) vs 511 (29.3) for SCA), and more likely to report never having pain so bad that it was hard to finish what they were doing over the previous 6 months (n (%)= 184 (37.4) vs 567 (32.7) for others). Table 5. Patient Reported Outcomes \| Patient Reported Outcome \| Genotype \| Total (N=2439) \| :---: \| Hb SC (N=502) \| SCA Genotypes (N=1937) \| \| Take daily pain medicine for SCD \| \| \| \| \| Yes \| 225 (44.8%) \| 904 (46.7%) \| 1,129 (46.3%) \| \| No \| 272 (54.2%) \| 1,001 (51.7%) \| 1,273 (52.2%) \| \| Number pain attacks, past 12 months \| \| \| \| \| None \| 70 (14.1%) \| 230 (12.1%) \| 300 (12.5%) \| \| 1 \| 45 (9.0%) \| 206 (10.8%) \| 251 (10.4%) \| \| 2 \| 66 (13.3%) \| 256 (13.4%) \| 322 (13.4%) \| \| 3 \| 74 (14.9%) \| 298 (15.6%) \| 372 (15.5%) \| \| 4 or more \| 243 (48.8%) \| 917 (48.1%) \| 1,160 (48.2%) \| \| When was last pain attack \| \| \| \| \| Never had a pain attack \| 13 (2.6%) \| 30 (1.6%) \| 43 (1.8%) \| \| More than 5 years ago \| 26 (5.2%) \| 73 (3.8%) \| 99 (4.1%) \| \| 1-5 years ago \| 51 (10.2%) \| 212 (11.1%) \| 263 (10.9%) \| \| 7-11 months ago \| 35 (7.0%) \| 110 (5.8%) \| 145 (6.0%) \| \| 1-6 months ago \| 125 (25.1%) \| 540 (28.3%) \| 665 (27.6%) \| \| 1-3 weeks ago \| 101 (20.3%) \| 406 (21.3%) \| 507 (21.1%) \| \| Less than a week ago \| 83 (16.7%) \| 302 (15.8%) \| 385 (16.0%) \| \| I have one right now \| 64 (12.9%) \| 236 (12.4%) \| 300 (12.5%) \| \| How often did you have very severe pain, past 6 months \| \| \| \| \| Never \| 154 (31.0%) \| 556 (29.3%) \| 710 (29.6%) \| \| Rarely/Sometimes \| 178 (35.8%) \| 700 (36.9%) \| 878 (36.7%) \| \| Often/Always \| 165 (33.2%) \| 642 (33.8%) \| 807 (33.7%) \| \| How often did you have pain so bad that it was hard to finish what you were doing, past 6 months \| \| \| \| \| Never \| 184 (37.4%) \| 614 (32.6%) \| 798 (33.6%) \| \| Rarely/Sometimes \| 182 (37.0%) \| 681 (36.1%) \| 863 (36.3%) \| \| Often/Always \| 126 (25.6%) \| 591 (31.3%) \| 717 (30.2%) \| Open in a new tab Pregnancy Outcomes Pregnancy outcomes were collected using a patient-reported survey assessment tool, results show in table 3). The HbSC cohort included 486 (29%) pregnancies, while other genotypes accounted for 1197 (71%) pregnancies. Among the Hb SC cohort, only 34 (7.5%) pregnancies were conceived while the mother was taking hydroxyurea, versus 237 (21.6%) of those with SCA as anticipated given less hydroxyurea use in the HbSC genotype. There were a comparable number of people with HbSC disease who had miscarriage (n (%) = 91(19%) vs. 302 (23.5%) compared to those with other genotypes. Similarly, women with HbSC disease frequently had babies prematurely (92, 30.6%), had babies who weighed less than 5.5 pounds at birth (68, 22.4%) or had those reported as low birth weight/genetic condition/other serious medical problem (66, 21.9%). LIMITATIONS The authors acknowledge several limitations of this study. First, patients were obtained via convenience samples and were established clinic patients at SCD centers. Those patients who were not in care or lost to care were not included in the study. It is possible that individuals with mild disease who do not engage in care regularly were excluded from this study. Therefore, there are several potential selection biases. Given the use of surveys in the study, we cannot exclude recall bias by participants. Additionally, it is not possible to ascertain from this survey study if concurrent illnesses such as anxiety, depression or chronic pain were specifically caused by SCD or were comorbid conditions with SCD. Finally, the diagnosis of chronic pain was obtained from the medical record (as like other co-morbidities in the SCDIC registry) and was not rigorously defined. DISCUSSION Data from the SCDIC registry suggest that HbSC disease is more clinically severe in people over age 15 than was previously recognized. In the majority of the SCD literature, HbSC disease is regarded as milder or “less severe” on several reports(14). This perception might be a carryover from when SCD was almost exclusively a disease of children. Data from the SCDIC registry demonstrates that adults affected by HbSC disease continue to experience significant life-limiting complications as well as disease-related acute pain resulting in frequent acute care utilization. Although organ-specific complications such as stroke and renal disease are higher in those with sickle cell anemia (HbSS and HbSβ 0-thalassemia) than in people with HbSC disease, these complications do occur in this population. Further, there are also other significant complications including bone necrosis and retinopathy that can lead to detriments in quality of life for this patient population. People with HbSC disease are often excluded from clinical trials of novel pharmaceuticals due to clinical and laboratory differences from those with SCA. Many clinical trials in SCD starting with the multi-center trial of hydroxyurea and including the transfusion trials for stroke prevention were limited to those individuals with HbSS disease or to those with all forms of sickle cell anemia(15, 16). Other more recent clinical trials include the HOPE study of voxelotor and the ongoing HIBISCUS study of etavopivat are open to people with SCD of all genotypes but limit the upper limit of hemoglobin of 10.5 g/dL allowed at baseline, which excludes the majority of people with HbSC disease as noted by the data included in this publication and other prior papers (17-19). While this population was included in the recent SUSTAIN trial of crizanlizumab, there were an insufficient number of participants with HbSC disease to specifically assess the efficacy of the study drug in this patient population.(20, 21) If the study been powered to ensure sufficient enrollment of this sub-population, a more specific finding could have been concluded. While the world is seeing a more robust pharmaceutical interest in SCD than ever before, most of the research and novel therapeutics remain focused on people living with SCA including the recently approved gene therapies ( However, 25% of the population of people with SCD are affected by HbSC disease and have not been included in these breakthrough trials. As noted here, there are similarities in disease complications and manifestations but there are also differences including the higher frequency and severity of retinopathy as well as several reports of people with HbSC disease who have succumbed to multi-organ failure syndrome thought to be triggered by severe dehydration (22). Additionally, as people age, there are reports of more adults suffering from splenic sequestration due to long-term hypersplenism with HbSC disease (23, 24). HbSC disease is not less severe (than SCA)-rather it is different and needs attention both in the lab and in the clinic to improve outcomes for affected individuals, especially in settings where disease-modifying therapies are primarily used by people with SCA. The lack of treatment data for this sub-population with SCD is especially concerning given recent studies show loss of care engagement during pediatrics and subsequent failure to transition to an adult provider. Specifically, there are reports detailing children with SCD that become lost to follow-up in childhood during periods of less severe disease and re-emerge into the system as adults. These individuals are thought of as failure of the transition system, but the data has shown that (part of) the problem is they were not seen after early childhood (25). These individuals may not see a SCD specialist until they have frequent acute care or new onset organ complications that potentially could have been identified earlier with continuous longitudinal screening. Conclusion The findings this study demonstrate that most people older than 16 years of age with HbSC disease have at least one organ-related complication and many with multi-organ disease and/or significant pain events. It is important that we ensure these individuals with HbSC maintain care engagement with SCD specialists as children to ensure they successfully transition to adult care before they develop these higher risk complications. It is important to educate medical providers and people with HbSC disease on the critical importance of regular follow up with an experienced SCD provider. Similarly, it is very important that these individuals receive appropriate, critical assessment when presenting for acute care complications. Lastly, individuals with HbSC should be included in more clinical trials for disease modification to improve their access and potential benefit. Supplementary Material Supinfo NIHMS1992172-supplement-Supinfo.docx (22.4KB, docx) FUNDING SOURCES The Sickle Cell Disease Implementation Consortium (SCDIC) has been supported by US Federal Government cooperative agreements HL133948, HL133964, HL133990, HL133996, HL133994, HL133997, HL134004, HL134007, and HL134042 from the National Heart Lung and Blood Institute and the National Institute on Minority Health and Health Disparities (Bethesda, MD). Footnotes Declarations Conflict of interest VRG has served as a consultant for CSL Behring, Forma, Vertex, Takeda, Pfizer/GBT. LLH is a consultant for Dupont/Nemours Children’s Hospital, Novartis, Emmaus, Forma Therapeutics, Hilton Publishing Company (HPC). He is also on the NHLBI sickle cell advisory committee. NS has received research funding from Pfizer/GBT, Akirabio and is a consultant for Pfizer/GBT, Forma, Agios, Bluebird bio, Vertex, Emmaus Pharmaceuticals. NS also serves as speaker for Pfizer/GBT, Emmaus Pharmaceuticals, Alexion. AK has received research funding from Novartis, Novo-Nordisk, Akira Bio, is a consultant for Novartis, serves on the DSMB for Bluebird Bio, and is on an external advisory committee for Pfizer and Vertex. TW is a consultant for Pfizer inc. JH serves on the NHLBI advisory committee and on the DSMB for the NHLBI. AAK has received research funding from GBT, ASH, NIH, HRSA, CDC JK is a consultant for Novartis, Bluebird Bio, Fulcrum Tx, Beam Tx, GLG, Guidepoint Global, and Glaxo-Smith Kline and has received research funding from NIH, HRSA, and CDC. CONFLICT OF INTEREST STATEMENT NS receives research funding from Global Blood Therapeutics, is a consultant for Novartis, Emmaus Medical and Forma Therapeutics, Agios and a speaker for Global Blood Therapeutics and Novartis. VRG has served as a consultant for CSL Behring. JK has served as a consultant for Bluebird Bio, Beam Therapeutics, and Fulcrum Therapeutics Ethics Approval Statement This study was approved by the institutional review boards of the participating institutions (Approval Numbers: University of California San Francisco 20–31161; Research Triangle Institute 14157; Washington University, St. Louis 201706016; Mount Sinai PPHS 16–01047; University of Illinois, Chicago 17–061401; Duke University Pro00073506; Augusta University 1078191). The authors certify that all procedures in the study were performed in accordance with the ethical standards as laid down in the 1964 declaration of Helsinki and its latter amendments or comparable ethical standards. 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The data that support the findings of this study are available from Department of Clinical Research, SSES, RTI International, Research Triangle Park, North Carolina, USA, upon reasonable request. ACTIONS View on publisher site PDF (538.3 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Graphical Abstract INTRODUCTION METHODS RESULTS LIMITATIONS DISCUSSION Conclusion Supplementary Material FUNDING SOURCES Footnotes Data Availability Statement REFERENCES Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
12103	https://tradukka.com/unit/power/horsepower/foot-pound-force-second	Convert horsepower (hp) to foot pound-force/second (lbfft/s) \| Tradukka - [x] Translate Dictionary Questions Units Currency My account Preferences Preferences Convert horsepower (hp) to foot pound-force/second (lbfft/s) UnitsPowerhorsepower hp foot pound-force/second lbfft/s horsepower watt foot pound-force/second exawatt Convert Clear Swap Copy Share ⇣ hp lbfft/s ⇣ 1 550.0000365393 10 5,500.000365393 20 11,000.000730786 50 27,500.001826965 100 55,000.00365393 ⇣ hp lbfft/s ⇣ 1,000 550,000.0365393 10,000 5,500,000.365393 25,000 13,750,000.913482 50,000 27,500,001.826965 100,000 55,000,003.65393 Commonly converted units Common conversions kilowatt volt amperemegawatt gigawattwatt Btu (IT)/hourvolt ampere kilowattwatt newton meter/secondkilowatt Btu (IT)/hourkilowatt Btu (th)/hour Recently converted units Recently converted foot pound-force/second horsepowerwatt kilojoule/secondMBH Btu (IT)/hourterawatt megawattkilowatt Btu (IT)/hourwatt Btu (IT)/hourjoule/minute watt Search words Search Upgrade your experience watt exawatt petawatt terawatt gigawatt megawatt kilowatt hectowatt dekawatt deciwatt centiwatt milliwatt microwatt nanowatt picowatt femtowatt attowatt horsepower horsepower (metric) horsepower (boiler) horsepower (electric) horsepower (water) Btu (IT)/hour Btu (IT)/minute Btu (IT)/second Btu (th)/hour Btu (th)/minute Btu (th)/second MBtu (IT)/hour MBH ton (refrigeration) kilocalorie (IT)/hour kilocalorie (IT)/minute kilocalorie (IT)/second kilocalorie (th)/hour kilocalorie (th)/minute kilocalorie (th)/second calorie (IT)/hour calorie (IT)/minute calorie (IT)/second calorie (th)/hour calorie (th)/minute calorie (th)/second foot pound-force/hour foot pound-force/minute foot pound-force/second pound-foot/hour pound-foot/minute pound-foot/second erg/second kilovolt ampere volt ampere newton meter/second joule/second exajoule/second petajoule/second terajoule/second gigajoule/second megajoule/second kilojoule/second hectojoule/second dekajoule/second decijoule/second centijoule/second millijoule/second microjoule/second nanojoule/second picojoule/second femtojoule/second attojoule/second joule/hour joule/minute kilojoule/hour kilojoule/minute Type to find a unit Close Terms of Service Terms / Privacy Policy Privacy / Contact us Contact Tradukka © 2025 — Made with Preferences Language preferences Close Share this content Send in WhatsApp Send Share on Facebook Share Post on X Post Close
12104	https://www.youtube.com/watch?v=mrst8ZZu9vU	Closure property \| Binary operations Tambuwal Maths Class 309000 subscribers 6 likes Description 123 views Posted: 24 Aug 2025 🚀 Unlock the first and most crucial rule of algebra! 🚀 Ever added two whole numbers and gotten a fraction? Of course not! That's the magic of the Closure Property at work. In this video, we're breaking down this fundamental concept of binary operations that keeps numbers neatly inside their sets. What you'll learn in this video: ✅The simple definition of the Closure Property (it's easier than you think!). ✅How to quickly test if a set is closed under an operation like +, -, ×, or ÷. ✅Real examples with integers, natural numbers, and even polynomials. ✅Common mistakes and misconceptions to avoid. Understanding closure is the first step to mastering higher math like group theory and abstract algebra. Whether you're prepping for an exam or just curious about how math works, this video will give you that "Aha!" moment. Don't forget to: 🔔Subscribe and turn on notifications so you never miss a lesson! 👍Like this video if it helped you. 💬Comment below with an example of a set that is NOT closed under subtraction! Math #Algebra #ClosureProperty #BinaryOperations 2 comments Transcript: Hello everyone. In this tutorial, I am going to talk about the properties of binary operations. Don't forget to like, comment, and subscribe. Let's get started. The first property we have is called the closure property. So, we have the closure property. The second one we have the commitative property. And we have the associative property. We have the inverse and identity elements. So these are the five properties we have under binary operations. To make the video comprehensive and intuitive, I will be taking one at a time. So in this tutorial, I'm going to focus on the closure property. Let's go through that. Uh what is closure? Uh before we talk about closure property, let me remind you the definition of binary operation which we say that uh it is a rule that combine two elements from non-MPT set and generate another element which is also a member of the same set. Uh for example, if you have an element A and B, a trump from a given set which is non- empty S. Um there exist another element C uh which which comes as a result of combining these two elements which is also from the same set S. But how do we get this uh as a result of combining A and B together with an operation? Right? So this is a clear definition of a binary operation. So for every two elements we pick from non-mpt set there exist another one C from the same set such that A combined with B with a particular operation gives an element C which is a member of the non-mpt set S. So when this property holds with an operation this an operation then we say that the set is closed under that particular operation. This operation which is an asterisk has to be defined. Therefore in this tutorial I'm only going to pay attention to the four basic operations. We have the plus, we have the subtraction, we have the multiplication and lastly the division. All right, we would like to verify whether some set of numbers are closed under addition, subtraction, multiplication and division separately. So let me take the set of real numbers and test is the set of real numbers closed under addition. We have to verify every number you know that exists is set to be a real number. Even if it is an e, right? If you combine it with another real number, let's say pi, the result must also be a real number. If you take two, which is a real number, add it to 10, this is going to give us 12. Both the two elements from non-MPT set, which is the real number, combine with an addition sign, which resulted 12. 12 is also a member of the set of real numbers. So when this happens, then we say that the set of real numbers is closed under addition. You can go and verify for all numbers you know once you add them the result must be a real number. So we can see that real number uh is closed under addition. But what about subtraction? go and verify every number you know once you subtract it from another number provided the two numbers are real numbers the result must also be a real number. So because this uh oh I supposed to write for all here so that you should know that there is no exception for every two elements you picked at random from a non empty set this will be true. What about multiplication? Multiplication is also true. Take any two real numbers at random. 2 10 this is 20. Any two real numbers pro whether they are negative, positive, whatever, provided they are real numbers once you multiply them the output must also be a real number. So therefore real numbers is closed under addition, subtraction and multiplication. But set of real numbers is not closed under division. Why? Remember two is a real number by 0 is also a real number but this is undefined. So since this is undefined we say that the set of real numbers is not closed under division despite the fact that all other numbers will satisfy. Yes. 2 / uh 1 this is equal to 2. 1 / 2 this is 0.5. Any number you take you divided by another number provided the denominator is not zero. Even if the numerator is zero right you're going to get a real number except when the denominator is zero. So because one number failed to satisfy which is the denominator of zero then you conclude that the whole set is not closed under such operation which is division. And this is why I said the question number one of y 20 25 mathematics uh for that mathematics examination is wrong because we realize a number which is 10 that has no inverse. For a number to have an inverse, it means that the entire set will have to satisfy that operation. It has to be closed. Yes, if it is closed, then we have to get another uh element that will satisfy it. Let me take another number. We have natural numbers. Natural numbers are the counting numbers 1 2 3 4 5. Every two numbers you take, you add, you have to get another natural number. 1 + 2 is three. 2 + 3 is 5. This plus this is seven. 1 + 4 is six. At random, keep on adding two two numbers from natural numbers. You have to get um what do you call it? Another natural number. But what about subtraction? If I take this number three and I subtract this number four, I'm going to get a negative one. But negative one is not a natural number. Therefore, uh the set of natural numbers is not closed under subtraction. What about multiplication? If you take two numbers, you multiply, you have to get another natural number. So, multiplication is close. Uh what about division? Division will fail. Why? Because remember one is a natural number, two is a natural number, but this is 0.5 which is a real number, not a natural number. Therefore, division and um subtraction will not satisfy natural numbers under closure property. Therefore, natural numbers are only closed under addition and multiplication. Let's take another number. Uh integers set of integers. Let's verify. You know there are numbers on the number line. Um if you have zero here, one, you have two you have three. If you come to the left -1, -2, -3, pick any other number at random you add, you must get an integer. So do that yourself. So addition will satisfy, subtraction will satisfy. Pick two numbers, subtract, you must get an integer. Uh multiplication will also satisfy. But division will fail. Why? Because of two reason. when the denominator is zero. You know, any number you divide by zero will give you an undefined and um for instance these two numbers are all inteious one and two but the result is 0.5 which is not an integer. So fractions are not inteious. Therefore division will not satisfy. Likewise yeah only division. So let us take another number like rational rational numbers. Yeah, rational numbers are close under addition. Let's take rational. What are rational? This is a rational number. A ratio of two integers, right? So if I take one rational number and I add it to another rational number, two is irrational number. Why? Because it can be rationalized. So if you add these together, you're going to get what? 5 / two, which is also a rational number. So addition is true. Subtraction is also true. For example, if I have two which is a rational number minus uh 2 over 5, this is also a rational number. This is going to give us um is it 8 over 5? Yeah, 8 over 5, which is also a rational number. So subtraction will also satisfy. But what about multiplication? Multiplication will also satisfy. Take any rational number multiply by another rational number like zero. Zero is also irrational number because it can be rationalized. This is going to give us zero. So multiplication of rational numbers is also true. But division is not. Why? Because we say that a rational number is a combination of the ratio of two inteious and we know that one is an integer, zero is an integer but this is not a rational number. So if you take two which is the same thing as 2 / 1, you divide by 0 which is also 0 / 1. 2 and 0 are all rational numbers. This is going to give us 2 / 0 which is undefined. So division will not satisfy. So set of rational numbers is closed under uh addition, subtraction and multiplication. So I would like to take two other sets for example a distinct set that has a finite number of elements. For example, uh let's take an an S that has elements 1 2 3 4. We would like to verify whether this set is closed under addition. I know it is not closed under addition because if you take three, you add it to four, you're going to get seven. And seven is not in this set. So this set is not close under addition. What about subtraction? If I take 1 minus 2, uh that will give us negative 1 which is not in the set. So addition and subtraction will not satisfy. What about multiplication? 2 3 is six and six is not in this set. So it will not satisfy. I'm just picking numbers at random. Then division. If I take one, I divide by two that will give us 0.5. You see that this set is not closed under addition, subtraction, multiplication and also division. So the last set I would like to take is Z modulus modulus 5. Z modulus 5 will give us elements 0 to 4. 0 1 2 3 and four. And to verify whether this set is closed under addition, let us make a table. T table of addition. So we have our addition sign here. We have 0 1 2 3 and four. 0 1 2 3 and four. 0 + 0 is 0. 1 + 0 + 1 is 1. 0 + 2 is 2. 0 + 3 is 3. And 0 + 4 is 4. And all of these numbers are inside the set. Let's move to the second one. 1 + 0 is 1. 1 + 1 is 2. 1 + 2 is 3. 1 + 3 is 4. 1 + 4 is 5. Five is not in this set. But you should know that in modulus five is the same thing as one or zero sorry is the same thing as zero. Why? Since we have 1 + 4 is 5. So you take five you divide by 5 which is our modulus. This is equal to one remainder zero. So this is the remainder. You know modulus is talking about the remainder. So if you take two you add it to zero that will give us two. 2 + 1 is going to give us three. 2 + 2 is going to give us four. 2 + 3 is 5. And you know 5 has a remainder of one when divided by five. Uh has a remainder of zero. Sorry. And 2 + 4 is 6. And if you take six, you divide by five. This is going to give us one with remainder of one. So this is one. And uh 3 + 0 is 3. 3 + 1 is 4. 3 + 2 is 5 which is zero. We have one. We have two. This is how it is going to be. 4 + 0 is 4. 4 + 1 is 5 which is 0. We have 1. We have two. We have three. Here look at it. 4 + 4 is 8. 8 / 5 uh is one with a remainder of three. So it is remainder arithmetics. So all these values you are seeing here they are present here. Therefore this set is closed is closed under addition. What about subtraction? Let me clear everything. Under subtraction 0 - 0 is what? 0. 0 - 1 is -1. But is minus1 here? No. But under modus you can keep on adding modus onto the negative number until it becomes positive. So add the mod five. This is going to give us four. So you add four here. So -1 and four are the same. And you can see four is present in our set. 0 - 2 is -2 and -2 + 5 is 3. 0 - 3 is - 3 and -3 + 5 is 2. 0 - 4 is -4 and -4 + 5 is 1. 1 - 0 is 1. 1 - 1 is 0. 1 - 2 is -1. Then -1 + 5 is 4. 1 - 3 is -2 and - 2 and + 5 is 3. 1 - 4 is - 3 and -3 + 5 is 2. 2 - 0 is 2. 2 - 1 is 1. 2 - 2 is 0. 2 - 3 is - 1. And -1 + 5 is 4. 2 - 4 is - 2 and -2 + 5 is 3. Come down here. 3 - 0 is 0. Is 3. Sorry. 3 - 1 is 2. 3 - 2 is 1. 3 - 3 is 0. Then 3 - 4 is -1 + 5 is 4. And lastly 4 - 0 is 4. 4 - 1 is 3. 4 - 2 is 2. 4 - 3 is um 1 and 4 - 4 is 0. So look at this combination of numbers. All of them are present in this particular set. Therefore minus also will satisfy. So the set is closed under addition and subtraction. Now I would like to test for multiplication and division. 0 0 is 0. 0 1 is 0. 0 3 is 0. 0. 1 0 is 0. 1 1 is 1. 1 2 is 2. 1 3 is 3. 1 4 is 4. Then 2. 2 0 is 0. 2 1 is 2. 2 2 is 4. 2 3 is 6. And 6 / 5 is equal to 1 with a remainder of 1. 2 4 is 8 / 5. This is one remainder three. This is three. 3 0 is 0. 3 1 is 3. 3 2 is 6. And you know 6 is equivalent to 1. 3 3 is 9. And 9 in mod 5 is equal to uh four, right? Because 9 / 5 is one remainder of four. 3 4 is 12. Then 12 / 5 is 2 with a remainder of 2. Right? Then 4 0 is 0. 4 1 is 4. 4 2 is 8. But 8 is equivalent to 3. 4 3 is 12. 12 is equivalent to 2. 4 4 is 16. How many fives can you find in 16? I think we have three because 3 5 is 15 with a remainder of one. So this is one. So pay close attention to these numbers. They are all present in this set. Therefore multiplication is also inclusive. Then the last one which is division I will not even test it because from the start 0 / 0 doesn't make sense. Right? So therefore we say that division is not closed. So the set is not closed under division but it is closed under addition, subtraction and multiplication. So this is all about closure property. In our next class we are going to talk about the uh commutative property. Thank you for watching. Do share to your learning colleagues. Bye-bye.
12105	https://brainly.com/question/8679127	[FREE] What is the focus of a parabola with the equation y^2 = 4x? - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +28,8k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +12,1k Ace exams faster, with practice that adapts to you Practice Worksheets +7,5k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified What is the focus of a parabola with the equation y 2=4 x? 2 See answers Explain with Learning Companion NEW Asked by js26151jstiltner • 02/10/2018 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 52665046 people 52M 4.7 22 Upload your school material for a more relevant answer Focus = (1,0) Explanation The equation of parabola with vertex at origin is given by: y 2=4 a x .... where, 4a is the width of the parabola and a is positive. Focus = (a, 0) Given the equation: y 2=4 x on comparing with we have; 4a = 4 Divide both sides by 4 we have; a = 1 Focus = (1, 0) Therefore, the focus of a parabola with equation is, (1, 0) Answered by OrethaWilkison •3.1K answers•52.7M people helped Thanks 22 4.7 (13 votes) Expert-Verified⬈(opens in a new tab) This answer helped 52665046 people 52M 4.7 21 Upload your school material for a more relevant answer The focus of the parabola with the equation y 2=4 x is found to be at the point (1,0). This is determined by comparing the equation to the standard form of a parabola. Hence, the coordinates of the focus are (1,0). Explanation To find the focus of the parabola given by the equation y 2=4 x, we can use the standard form of a parabola. The general equation of a parabola that opens to the right is given by: y 2=4 a x In this equation, a represents the distance from the vertex to the focus. Identify the equation format: We compare the given equation y 2=4 x to the standard equation y 2=4 a x. Determine the value of a: From the equation y 2=4 x, we have: 4 a=4 Dividing both sides by 4 gives us: a=1 Find the focus: The focus of the parabola is located at the point (a,0). Since we found a=1, the coordinates of the focus are: (1,0) Therefore, the focus of the parabola described by the equation y 2=4 x is at the point (1,0). Examples & Evidence For example, if we have another parabola like y 2=8 x, we would follow the same steps to find the focus. In this case, 4 a=8 leads to a=2, so the focus would be at the point (2,0). This process is derived from algebraic principles associated with the geometry of parabolas, confirming that the focus is at the point determined by (a,0) when the equation is in the form y 2=4 a x. Thanks 21 4.7 (12 votes) Advertisement Community Answer 5.0 13 Vertex:(0,0)(0,0)Focus:(1,0)(1,0)Axis of Symmetry:y=0 y=0 Directrix:x=−1 Answered by Brainly User Thanks 13 5.0 (8 votes) 2 Advertisement ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Which is bigger: 9 7 or 6 5? The table represents a linear function. The rate of change between the points (−5,10) and (−4,5) is -5. What is the rate of change between the points (−3,0) and $(-2,-5)? \| x \| y \| :---: \| \| -5 \| 10 \| \| -4 \| 5 \| \| -3 \| 0 \| \| -2 \| -5 \| A. −5 B. −5 1 C. 5 1 D. 5 Is the function represented by the table non-linear? \| x \| y \| :---: \| \| 6 \| 4 \| \| 7 \| 2 \| \| 8 \| 0 \| \| 9 \| -2 \| A. Yes, because it has a constant rate of change. B. Yes, because it does not have a constant rate of change. C. No, because it has a constant rate of change. D. No, because it does not have a constant rate of change. Max is painting a fence. The table shows the number of square feet he painted after various numbers of minutes. What is the rate of change? \| Amount of Fence Max Painted \| \| Minutes \| Square Feet Painted \| \| 1 \| 3.5 \| \| 2 \| 7 \| \| 3 \| 10.5 \| \| 4 \| 14 \| \| 5 \| 17.5 \| -3.5 -1 1 3.5 Max is painting a fence. The table shows the number of square feet he painted after various numbers of minutes. What is the rate of change? \| Amount of Fence Max Painted \| \| Minutes \| Square Feet Painted \| \| 1 \| 3.5 \| \| 2 \| 7 \| \| 3 \| 10.5 \| \| 4 \| 14 \| \| 5 \| 17.5 \| Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
12106	https://simple.wikipedia.org/wiki/LC_circuit	LC circuit - Simple English Wikipedia, the free encyclopedia Jump to content [x] Main menu Main menu move to sidebar hide Getting around Main page Simple start Simple talk New changes Show any page Help Contact us About Wikipedia Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Give to Wikipedia Create account Log in [x] Personal tools Give to Wikipedia Create account Log in [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 Other websites LC circuit [x] 33 languages العربية Беларуская Беларуская (тарашкевіца) Български Català Čeština Dansk Deutsch Ελληνικά English Español فارسی Français 한국어 हिन्दी Hrvatski Italiano עברית Latviešu മലയാളം Nederlands 日本語 Norsk bokmål Polski Português Română Slovenčina Slovenščina Српски / srpski Suomi Svenska தமிழ் 中文 Change links Page Talk [x] English Read Change Change source View history [x] Tools Tools move to sidebar hide Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sandbox Edit interlanguage links Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikidata item From Simple English Wikipedia, the free encyclopedia An LC circuit is an electronic circuit made up of an inductor and a capacitor. LC circuit's resonant frequency is equal to: ω=1 L C{\displaystyle \omega ={\sqrt {1 \over LC}}} The angular frequency ω has units of radians per second. LC circuits are used for creating signals at a particular frequency, or picking out a signal at a particular frequency from a more complex signal. An ideal LC circuit does not have resistance. At LC circuit energy saves in the capacitor's electric field. U=q 2 2 C{\displaystyle {U}={q^{2} \over 2C}} U is energy and q is electric charge. At LC circuit energy also save in the inductor's magnetic field. U=L i 2 2{\displaystyle {U}={Li^{2} \over 2}} U is energy and i is electric current that flows in inductor. Let's analyze an LC circuit's vibration. Vibrating LC circuit's total energy is U. U=q 2 2 C+L i 2 2{\displaystyle U={q^{2} \over 2C}+{Li^{2} \over 2}} Because circuit's resistance is 0, there is no energy that transmits to heat energy, and U is maintained regularity. d U d t=0{\displaystyle {dU \over dt}=0} So LC circuit's vibration is shown like that L d 2 q d t 2+q C=0{\displaystyle {Ld^{2}q \over dt^{2}}+{q \over C}=0} First consider the Electrical impedance of the series LC circuit. The total impedance is given by the sum of the inductive and capacitive impedances Z=Z L+Z C{\displaystyle Z=Z_{L}+Z_{C}} By writing the inductive impedance as Z L=j ω L{\displaystyle Z_{L}=j\omega L} and capacitive impedance as Z C=1 j ω C{\displaystyle Z_{C}={\frac {1}{j{\omega C}}}} Z=j ω L+1 j ω C{\displaystyle Z=j\omega L+{\frac {1}{j{\omega C}}}} . Resultingly the series connected circuit, when connected to a circuit in series, will act as a band-pass filter having zero impedance at the resonant frequency of the LC circuits. The same analysis may be applied to the parallel LC circuit. The total impedance is then given by Z=Z L Z C Z L+Z C{\displaystyle Z={\frac {Z_{L}Z_{C}}{Z_{L}+Z_{C}}}} and after substitution of Z L{\displaystyle Z_{L}} and Z C{\displaystyle Z_{C}} we have Z=L C(ω 2 L C−1)j ω C{\displaystyle Z={\frac {\frac {L}{C}}{\frac {(\omega ^{2}LC-1)j}{\omega C}}}} which simplifies to Z=−L ω j ω 2 L C−1{\displaystyle Z={\frac {-L\omega j}{\omega ^{2}LC-1}}} . Resultingly the parallel connected circuit will act as band-stop filter having infinite impedance at the resonant frequency of the LC circuit. Other websites [change \| change source] Understanding Destructive LC Voltage Spikes This short article about technology can be made longer. You can help Wikipedia by adding to it. Retrieved from " Category: Electronic circuits Hidden category: Technology stubs This page was last changed on 7 February 2025, at 19:10. Text is available under the Creative Commons Attribution-ShareAlike License and the GFDL; additional terms may apply. See Terms of Use for details. Privacy policy About Wikipedia Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents LC circuit 33 languagesAdd topic
12107	https://brilliant.org/wiki/exponential-functions-solving-equations/	Solving Exponential Equations Sign up with Facebook or Sign up manually Already have an account? Log in here. Hemang Agarwal, Paul Ryan Longhas, Ashish Menon, and Andrew Ellinor Abdur Rehman Zahid Aditya Virani Sachin Vishwakarma Mahindra Jain Nguyen Tran Calvin Lin Jimin Khim contributed To solve exponential equations, we need to consider the rule of exponents. These rules help us a lot in solving these type of equations. Contents Same Base Different Base Problem Solving Same Base In solving exponential equations, the following theorem is often useful: If a is a non-zero constant and ax=ay, then x=y. □ We have axayaxax−yx−yx=ay,a=1=1=1=0=y. □ Here is how to solve exponential equations: Manage the equation using the rule of exponents and some handy theorems in algebra. Use the theorem above that we just proved. Solve 5x−11=125. Making the bases on both sides equal to 5 gives 5x−115−(x−1)−(x−1)x=125=53=3=−2. □ Solve 4x−3=0.125. Converting the bases of both sides to 2 gives 4x−34x−322x−622x−62x−6x=0.125=1000125=81=2−3=−3=23. □ Find the value of x when 4x=16. We have 4x=16⇒4x=42. Then the theorem "if a is a non-zero constant and ax=ay, then x=y" gives x=2. □ If 8x=2, what is x? We have [\begin{align} 8^x & = 2\ \big(2^3\big)^{x} & = 2\ 2^{3x} & = 2^{1}. \end{align}] Equating the powers, we get [\begin{align} 3x & = 1\ x & = \dfrac{1}{3}.\ _\square \end{align}] If 6x−1=0, what is x? We have 6x−16x6xx=0=1=60=0. □ If (8)(9x)=9x, what is x? We have (8)(9x)(8)(9x)−9x(7)9x9x=9x=0=0=0. Then the theorem "if a=0 and ax=0, then x=ϕ" gives x=ϕ. □ Different Base If the bases are different, there are still techniques for solving these exponential equations. If the bases are powers of a common base, we need only convert one or both bases to the common base and proceed using the "Same Base" case. Solve 43x=8x−1. We see that while 4 and 8 are different bases, they are both powers of a common base, namely 2. We'll proceed by rewriting 4 and 8 in terms of their common base: 43x(22)3x26x6xx=8x−1=(23)x−1=23x−3=3x−3=−1. □ Solve for x: 162x+x84x−x=22x We have [\begin{align} \dfrac{8^{4x - \sqrt{x}}}{{16}^{2x + \sqrt{x}}} & = 2^{2\sqrt{x}}\ \ \dfrac{{\big(2^3\big)}^{4x - \sqrt{x}}}{{{\big(2^4\big)}}^{2x + \sqrt{x}}} & = 2^{2\sqrt{x}}\ \ \dfrac{2^{12x - 3\sqrt{x}}}{2^{8x + 4\sqrt{x}}} & = 2^{2\sqrt{x}}\ \ 2^{12x - 3\sqrt{x} - 8x - 4\sqrt{x}} & = 2^{2\sqrt{x}}\ \ 2^{4x - 7\sqrt{x}} & = 2^{2\sqrt{x}}. \end{align}] Equating the powers gives [\begin{align} 4x - 7\sqrt{x} & = 2\sqrt{x}\ 4x & = 9\sqrt{x}\ \sqrt{x} & = \dfrac{9}{4}\ x & = \dfrac{81}{16}.\ _\square \end{align}] Solve 243273x−2=813x−6. Reducing the bases of 27, 81, and 243 on both sides to 3 yields 243273x−23539x−639x−6−539x−119x−113xx=813x−6=312x−24=312x−24=312x−24=12x−24=13=313. □ Unfortunately, it won't always be possible to convert to a common base as we did in the examples above. For instance, in solving 5x=3x+2, we note that 5 and 3 are not powers of a nice common base. In this case, we'll need to make use of logarithms. Solve 5x=3x+2. We have 5xlog10(5x)xlog105xlog105xlog105−xlog103x(log105−log103)=3x+2=log10(3x+2)=(x+2)log103=xlog103+2log103=2log103=2log103. Therefore, x=log105−log1032log103≈4.3013. □ Problem Solving Given 1728=2a.3b, find positive integers a and b. We have [\begin{align} 1728 & = {12}^{3}\ & = {(4 × 3)}^{3}\ & = 4^3 × 3^3\ & = {(2^2)}^3 × 3^3\ & = 2^6 × 3^3\ \Rightarrow a & = 6, \ b = 3.\ _\square \end{align}] Solve 3x2=3x. Since both sides of the equation have the same base, their exponents must also be the same: 3x2x2x2−xx(x−1)⇒x=3x=x=0=0=0,1. □ If 2x⋅3y⋅5z=45, what is value of x+y+z? Observe that 45 can be factorized as follows: 45=32⋅5=20⋅32⋅51. Then x=0,y=2,z=1, which gives x+y+z=0+2+1=3. □ If xx⋅yy=108, what is value of x+y? 108 can be factorized as follows: 108=22⋅33. This implies that either x=2,y=3 or x=3,y=2. Therefore the answer is x+y=2+3=5. □ If 2325⋅30⋅31⋅32=2x⋅3y, what is value of x+y? 2325 can be rewritten as 2325=25−3=22.(1) 30⋅31⋅32 can be rewritten as 30⋅31⋅32=30+1+2=33.(2) From (1) and (2), x and y are 2325⋅30⋅31⋅32⇒x=22⋅33=2x⋅3y=2, y=3. Hence, x+y=2+3=5. □ An exponential equation is one in which a variable occurs in the exponent. If both sides of the equation have the same base, then the exponents on both sides are also the same: ax=ay⟹x=y. Here is a list of some rules concerning exponential functions: (1) am×an=am+n (2) am÷an=am−n (3) (am)n=amn (4) (ab)n=anbn (5) a0=b0 (6) 1m=1n, where a=0 and b=0. Always be cautious of (5) and (6); never forget to check if plugging a zero in an exponent works, or if there are any bases that are equal to 1. Cite as: Solving Exponential Equations. Brilliant.org. Retrieved 10:50, September 28, 2025, from
12108	https://brilliant.org/wiki/taylor-series/	Taylor Series Andrew Ellinor, Rahil Sehgal, Abha Vishwakarma, and Steven Zheng Sravanth C. Worranat Pakornrat Haris Hussain Jimin Khim Calvin Lin Eli Ross Zandra Vinegar contributed This article uses summation notation. A Taylor series is a polynomial of infinite degree that can be used to represent many different functions, particularly functions that aren't polynomials. Taylor series has applications ranging from classical and modern physics to the computations that your hand-held calculator makes when evaluating trigonometric expressions. Taylor series is both useful... ∫0xtsintdt=x−3⋅3!x3+5⋅5!x5−7⋅7!x7+⋯=n=0∑∞(−1)n(2n+1)⋅(2n+1)!x2n+1 Here, a Taylor series is being used to evaluate an integral that cannot be computed using known methods. ...and beautiful: n=0∑∞(−1)n2n+14=4−34+54−74+⋯=π Here, elegant use of a Taylor series gives us the exact value of π. Contents Introduction Interval and Radius of Convergence Taylor Polynomial Derivation Using Taylor Series in Approximations See Also Introduction Let f(x) be a real-valued function that is infinitely differentiable at x=x0. The Taylor series expansion for the function f(x) centered around the point x=x0 is given by n=0∑∞f(n)(x0)n!(x−x0)n. Note that f(n)(x0) represents the nth derivative of f(x) at x=x0. n=0∑∞f(n)(x0)n!(x−x0)n It is not immediately obvious how this definition constructs a polynomial of infinite degree equivalent to the original function, f(x). Perhaps we can gain an understanding by writing out the first several terms of the Taylor series for f(x)=cosx centered at x=0. Note that there is nothing special about using x=0 other than its ease in computation, but any other choice of center is allowed and will vary based on need. We will now use the definition above to construct a graceful polynomial equivalency to cosx. Because the formula for the Taylor series given in the definition above contains f(n)(x0), we should build a list containing the values of f(x) and its first four derivatives at x=0: f(0)f′(0)f′′(0)f′′′(0)f(4)(0)=cos0=−sin0=−cos0=sin0=cos0=1=0=−1=0=1. We begin assembling the Taylor series by writing f(x)= [the first number in our list] ⋅0!(x−x0)0 like so: f(x)=1⋅0!(x−0)0=1. So far, our constructed function f(x)=1 looks nothing like f(x)=cosx. They merely have f(0)=1 in common, but we shall add more terms. We add the next term from our list above, this time multiplied by 1!(x−x0)1: f(x)=1⋅0!(x−0)0+0⋅1!(x−0)1=1. Notice the exponent on (x−0) and the argument inside the factorial are both 1 this time, rather than 0 as they were in the previous term. This is because the summation dictates that we increment n from 0 to 1. This process will continue by adding the next term from our list above, but again incrementing the power on (x−0) and the value inside the factorial: f(x)=1⋅0!(x−0)0+0⋅1!(x−0)1+(−1)⋅2!(x−0)2=1−2!x2. Let's stop and look at what we have so far. After three terms, our Taylor series has given us f(x)=1−2x2. Interestingly enough, if we continue taking numbers from our list while appending incremented powers of (x−0) and incremented factorials, then our Taylor series slowly but surely conforms to the cosine curve: f(x)=1⋅0!(x−0)0+0⋅1!(x−0)1+(−1)⋅2!(x−0)2+0⋅3!(x−0)3+1⋅4!(x−0)4=1−2!x2+4!x4. At this point, we can guess at the emerging pattern. The powers on x are even, the factorials in the denominator are even, and the terms alternate signs. Note that more derivatives of the original function may be needed to discover a pattern, but only four derivatives were needed for this example. We encode this pattern into a summation, which finally yields our Taylor series for cosx: cosx=n=0∑∞(−1)n(2n)!x2n. In the animation below, each frame represents an additional term appended to the previous frame's Taylor series. As we add more terms, the Taylor series tends to fit better to the cosine function it's attempting to approximate: Important note: Because this series expansion was centered at x=0, this is also known as a Maclaurin series. A Maclaurin series is simply a Taylor series centered at x=0. So how does this work exactly? What is the intuition for this formula? Let's solidify our understanding of the Taylor series with a slightly more abstract demonstration. For the purposes of this next example, let T(x) represent the Taylor series expansion of f(x). T(x)=n=0∑∞f(n)(x0)n!(x−x0)n=f(x0)+f′(x0)(x−x0)+f′′(x0)2(x−x0)2+f′′′(x0)6(x−x0)3+⋯ It is important to note that the value of this summation at x=x0 is simply f(x0), because all terms after the first will contain a 0 in their product. This means the value of the power series agrees with the value of the function at x0 (or that T(x0)=f(x0)). Surely this is what we'd want from a series that purports to agree with the function! After all, if our claim is that the Taylor series T(x) equals the function f(x), then it should agree in value at x=x0. Granted, there are an uncountable number of other functions that share the same value at x0, so this equivalence is nothing special so far. Let's investigate by taking the derivative of the terms in the power series we have listed: T′(x)=0+f′(x0)+f′′(x0)(x−x0)+f′′′(x0)2(x−x0)2+f(4)(x0)3!(x−x0)3+⋯. If we evaluate the differentiated summation at x=x0, then all terms after f′(x0) vanish (again due to containing 0 in their product), leaving us with only f′(x0). So, in addition to T(x0)=f(x0), we also have that T′(x0)=f′(x0), meaning the Taylor series and the function it represents agree in the value of their derivatives at x0. One can repeatedly differentiate T(x) and f(x) at x=x0 and find that this pattern continues. Indeed, the next derivative T′′(x) takes on the value f′′(x0), the derivative after that T′′′(x) takes on the value f′′′(x0), and so on, all at x=x0. This is a promising result! If we can ensure that the nth derivative of T(x) agrees with the nth derivative of f(x) at x=x0 for all values of n, then we can expect the behavior of the Taylor series and f(x) to be identical. Now, there are rare, pathological examples to this conclusion, but to ensure those don't crop up, we condition this theorem on the function being infinitely differentiable. x+3!x3+5!x5 x−3x3+5x5 x−3!x3+5!x5 x+3x3+5x5 Compute the first three non-zero terms of the Taylor series for f(x)=sinx centered at x=0. The correct answer is: x−3!x3+5!x5 There are already dozens of known Taylor series. Some of them are easy to derive on your own (and you should!) while others are far too complicated for the scope of this wiki: cosxex=n=0∑∞(−1)n(2n)!x2n=n=0∑∞n!xnsinxln(1+x)=n=0∑∞(−1)n(2n+1)!x2n+1=n=1∑∞(−1)n+1nxntan−1x1−x1=n=0∑∞(−1)n(2n+1)x2n+1=n=0∑∞xn. Interval and Radius of Convergence Main Article: Interval and Radius of Convergence The interval of convergence for a Taylor series n=0∑∞an(x−x0)n is the set of values of x for which the series converges. Examine the geometric power series 1−x1=1+x+x2+x3+x4+⋯=n=0∑∞xn. Recall that a geometric progression of infinite terms is Sn=a+a⋅r+a⋅r2+a⋅r3+⋯, which is equal to Sn=1−ra. Taylor Polynomial Derivation Suppose we want to interpolate an infinite number of points on the Cartesian plane using a continuous and differentiable function f. How can this be done? Given n points on the Cartesian plane, the set of points can be interpolated using a polynomial of at least degree n−1. Given an infinite number of points to interpolate, we need an infinite polynomial f(x)=a0+a1(x−x0)+a2(x−x0)2+⋯, where ∣x−x0∣ is within the radius of convergence. Observation: f(x0)f′(x0)f′′(x0)f′′′(x0)f(4)(x0)f(n)(x0)=a0=a1=2a2=6a3=24a4=n!an. Solving for each constant term expands the original function into the infinite polynomial f(x)=n=0∑∞n!1f(n)(x0)(x−x0)n. Using Taylor Series in Approximations Main Article: Taylor Series Approximation Imagine that you have been taken prisoner and placed in a dark cell. Your captors say that you can earn your freedom, but only if you can produce an approximate value of 38.1. Worse than that, your approximation has to be correct to five decimal places! Even without a calculator in your cell, you can use the first few terms of the Taylor series for 3x about the point x=8 as a tool for making a quick and decent approximation. We certainly won't be able to compute an infinite number of terms in a Taylor series expansion for a function. However, as more terms are calculated in the Taylor series expansion of a function, the approximation of that function is improved. Using the first three terms of the Taylor series expansion of f(x)=3x centered at x=8, approximate 38.1. We have f(x)=3x≈2+12(x−8)−288(x−8)2. The first three terms shown will be sufficient to provide a good approximation for 3x. Evaluating this sum at x=8.1 gives an approximation for 38.1: f(8.1)=38.138.1≈2+12(8.1−8)−288(8.1−8)2=2.00829861111…=2.00829885025…. With just three terms, the formula above was able to approximate 38.1 to six decimal places of accuracy. □ See Also Maclaurin Series Taylor Series Manipulation Cite as: Taylor Series. Brilliant.org. Retrieved 03:08, September 28, 2025, from
12109	https://www.ccmagnetics.com/blog/magnetic-field-strength-h-and-magnetic-induction-b-magnetization-m-and-magnetic-polarization-j.html	Published Time: 2023-03-31 Magnetic field strength H and magnetic induction B, magnetization M and magnetic polarization J \| CCmagnetics Skip to content Every new customer will receive a piece of magnetic field viewing film as a gift. FAQ Every new customer will receive a piece of magnetic field viewing film as a gift. Menu Search for: sales@ccmagnetics.com Inquiry Categories Magnetic Couplings Axial Magnetic Couplings Magnetic Disc Couplings Fixed Gap Magnetic Couplings Magnetic Linear Drive Coupling Encoder Magnets Magnetic Encoder Ring IC-MU Magnetic Target Linear Magnetic Scales Auto ABS Magnetic Ring Magnetic Transmission Cross Type Magnetic Gears Parallel Type Magnetic Gears Magnetic Rack Pinion Gears Magnetic Bevel Gears Magnetic Disc Couplings Motor Magnets Arc Magnets Motor Magnetic Ring High Performance Motor Magnetic Ring Magnetic Coupling Magnet Magnetic Wheel for Robots Robot Magnetic Wheel Magnetic Wheel Rubber Tires Home Glossary Blog About Contact How to order ? Inquiry CCmagnetics Blog Magnetic field strength H and magnetic induction B, magnetization M and magnetic polarization J Posted on March 31, 2023by ccmagnetics Magnetic field strength H The magnetic field strength H is actually a physical quantity with no practical meaning. When people defined it earlier, they assumed that there was a magnetic charge, but later found that this thing does not exist, it is just the other side of the current. Scientists made a series of revolutionary discoveries in the 1820s that started the modern theory of magnetism. Danish physicist Hans Oersted discovered in July 1820 that the current in a current-carrying wire would exert a force on a magnetic needle, causing the magnetic needle to deflect. (Oersted experiment - magnetic effect of electric current). In September, just a week after the news reached the French Academy of Sciences, Ampere successfully performed experiments showing that two parallel current-carrying wires attract each other if the currents they carry are flowing in the same direction; will be mutually exclusive. In 1825 Ampere published Ampere's law, which is the law about the relationship between the current and the direction of the magnetic field lines of the magnetic field excited by the current. Through the measurement of mechanics, it can be concluded that the distance from the outside of the long straight wire to the wire is equal, the strength of the "magnetic field" felt by the magnetic needle is the same, and the strength of the "magnetic field" at points with different distances is inversely proportional to the distance. In this way, we define the physical quantity of magnetic field strength H through mechanical measurement and current intensity. Its unit is A/meter m. In the Gauss unit system, the unit of H is Oe Oersted, 1A/m=4π×10-3Oe. There are many explanations about the magnetic field strength H, and we can simply understand H as an external magnetic field (analogous to the electric field strength, for example, using a current I to apply a magnetic field H to an object). Magnetic induction B The magnetic field strength is only a magnetic field given by an external current, and for ferromagnetic substances in the magnetic field, in addition to being affected by the external magnetic field H, the particles inside the material will also generate an induced magnetic field under the action of the external magnetic field. The magnetic induction intensity B means that a particle "feels" the total magnetic field, which is the sum of the external magnetic field H and the induced magnetic field M at this time. In vacuum, the magnetic induction intensity is proportional to the external magnetic field, B=μ0H, where μ0 is the vacuum permeability. The magnetic induction intensity B=μ0(H+M) inside the ferromagnetic material, that is, the total magnetic field is equal to the sum of μ0 multiplied by the "magnetic field H generated by the current" plus the "magnetic field M generated after the medium is magnetized by H". The unit of B is Tesla T, and the unit in the Gauss unit system is Gauss Gs, 1T=10KGs. In fact, magnetic induction is the real "magnetic field strength" of the magnet, but since H has been called the magnetic field strength in history, we can only give B another name called magnetic induction. B and H both refer to "magnetic field strength", but due to the different ways of definition and derivation, their units are different (the unit of B under the Gauss system is Gauss Gs, and the unit of H is Oersted Oe, 1Oe =1×10-4Wb·m-2=1×10-4T=1Gs). The magnetic field strength H is the magnetic field of the empty space, it does not consider the matter in the space, it pays attention to the relationship between the magnetic field and the source of the magnetic field - the current, and the magnetic induction B is considered on the basis of the magnetic field H of the empty space The strength of the final magnetic field after adding the actual matter, it focuses on the actual strength of the magnetic field of the matter. Magnetization M Just now we have mentioned the magnetization M, which is an induced magnetic field generated by the particles inside the material under the action of the external magnetic field. Modern physics proves that every electron in an atom is doing orbital motion and spin motion around the nucleus, both of which generate magnetic effects. If the molecule is regarded as a whole, the sum of the magnetic effects produced by the electrons in the molecule can be represented by an equivalent circular current. This equivalent circular current is called molecular current, and its corresponding magnetic moment is called The molecular magnetic moment, represented by pm, is the vector sum of the magnetic moments and spin magnetic moments of each electron orbit in the molecule. When there is no external magnetic field, the vector sum of all molecular magnetic moments in any volume element inside the magnetic medium is zero, and the substance does not show magnetism to the outside; while when the magnetic medium is in the external magnetic field, each molecule is subjected to a torque, which is The molecular magnetic moment is forced to turn to the direction of the external magnetic field, so under the action of the external magnetic field, the vector sum of all molecular magnetic moments in any volume element is not zero. In this way, the magnetic medium shows a certain degree of magnetism to the outside, or the magnetic medium is magnetized. In order to describe the magnetization state (magnetization degree and magnetization direction) of the magnetic medium, we introduce the magnetization vector M, which represents the vector sum of all molecular magnetic moments in a unit volume, and the unit is A/m (the unit of M under the Gauss system is Gauss Gs ). In order to study the relationship between the induced magnetic field M and the applied field H, we define the magnetic susceptibility χ=M/H. A large magnetic susceptibility means that the same external magnetic field can generate more internal induced magnetic fields; a small magnetic susceptibility means that even if the external magnetic field is very large, the material inside is "too lazy to care about it" and has only a weak response. Magnetic susceptibility can be positive or negative. Positive magnetic susceptibility χ>0 means that the generated internal magnetic field M is in the same direction as the external magnetic field H. Negative magnetic susceptibility χ<0 means that the additional magnetic field M generated by H inside the material is in the opposite direction to the external field H. Magnetic polarization J Above we introduced the magnetic induction intensity B=μ0(H+M) =μ0H+μ0M, we call μ0M the magnetic polarization of the substance, that is, J=μ0M, and its unit is also T (Tesla). The magnetic polarization J is interpreted as the magnetic dipole moment per unit volume of the magnetic medium in a physical sense, also known as the intrinsic magnetic induction. The symbol is Bi or J. It is not difficult to see from J=μ0M that the difference between the magnetic polarization J and the thinning intensity M is only M multiplied by a constant μ0. In soft magnetic materials, the value of the magnetic field strength is usually not more than 1000A/m, μ0 is 4×10-7H/m, and J=B-μ0H, so the difference between the magnetic induction B and the magnetic polarization J is very small; But in hard magnetic materials, this difference is very significant, so two relationship curves, B=f(H) and J=f(H), are usually given. Leave a Reply Cancel reply Your email address will not be published.Required fields are marked Comment Name Email [x] Save my name, email, and website in this browser for the next time I comment. Δ Alternative: WPA About CCmagnetics is a distinguished manufacturer of magnetic encoder rings and contactless magnetic transmission drives. Our products are supplied to numerous renowned automobile, electronics, laboratory, and studio brands. 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12110	https://en.wikipedia.org/wiki/Radian_Group	Jump to content Radian Group Add links From Wikipedia, the free encyclopedia Mortgage insurance company \| \| \| \| \| \| \| --- --- --- \| \| \| This article has multiple issues. Please help improve it or discuss these issues on the talk page. (Learn how and when to remove these messages) \| \| \| --- \| \| \| A major contributor to this article appears to have a close connection with its subject. It may require cleanup to comply with Wikipedia's content policies, particularly neutral point of view. Please discuss further on the talk page. (August 2019) (Learn how and when to remove this message) \| \| \| \| --- \| \| \| This article relies excessively on references to primary sources. Please improve this article by adding secondary or tertiary sources. Find sources: "Radian Group" – news · newspapers · books · scholar · JSTOR (December 2019) (Learn how and when to remove this message) \| (Learn how and when to remove this message) \| Radian Group Inc. \| \| \| \| Company type \| Public \| \| Traded as \| NYSE: RDN Russell 2000 component S&P 600 component \| \| Industry \| Financial technology & Insurance & Finance \| \| Founded \| 1977 (as CMAC) \| \| Headquarters \| Philadelphia, Pennsylvania \| \| Key people \| Richard Thornberry (CEO) \| \| Website \| \| Radian Group Inc. (NYSE: RDN) is a mortgage insurance company with a suite of mortgage, risk, real estate, and title services. The company is headquartered at Centre Square in Philadelphia. Radian Companies [edit] Source: Radian is a group of separately capitalized companies that share a unified strategic focus. Radian's core business, Radian Guaranty Inc., provides private mortgage insurance to protect lenders from default-related losses, facilitate the sale of low-down-payment mortgages in the secondary market and enable homebuyers to purchase homes with down-payments less than 20%. In 2019 it had a net income of $672.3 million. Vanguard Group Inc, FMR LLC, Blackrock Inc are some of the highest stock holders of Radian Group Inc. References [edit] ^ "Radian Home". www.radian.com. Retrieved 2020-10-23. ^ "Radian Announces Fourth Quarter and Full Year 2019 Financial Results". businesswire.com. 5 February 2020. ^ Baldwin, Edison. "Radian Group Inc. [RDN] is -49.72% lower this YTD. Is it still time to buy? \| The DBT News". Retrieved 2020-03-19. External links [edit] Radian corporate website \| v t e Philadelphia-area corporations (including the Delaware Valley) \| \| --- \| \| List of companies based in the Philadelphia area \| \| \| Philadelphia-based Fortune 500 corporations (rank in the 2017 list) \| Comcast (31) Aramark (192) Crown Holdings (333) \| \| Delaware Valley-based Fortune 500 corporations (rank in the 2017 list) \| AmerisourceBergen (11) DuPont (113) Lincoln National Corporation (207) Universal Health Services (276) Campbell Soup (339) UGI (457) Burlington Stores (463) \| \| Other notable Philadelphia-based businesses \| Amoroso's Chemtura Day & Zimmermann FMC Corporation Independence Blue Cross Pep Boys Philadelphia Media Network PREIT Radian Group Urban Outfitters \| \| Notable Philadelphia-based professional partnerships \| Ballard Spahr Blank Rome Cozen O'Connor Dechert Drinker Biddle & Reath Duane Morris Morgan, Lewis & Bockius Pepper Hamilton Saul Ewing White and Williams \| \| Other notable Delaware Valley-based businesses \| Actua Corporation Airgas AlliedBarton Ametek Aqua America Asplundh Bentley Systems Brandywine Realty Trust Boscov's Carpenter Technology Cephalon Chemours Christiana Care Health System Crozer Keystone Health System David's Bridal DuckDuckGo EPAM Systems EnerSys Liberty Property Trust Penn Entertainment Penn Mutual Rita's Italian Ice SEI Investments SLM Susquehanna International Group Vanguard Toll Brothers Triumph Group Unisys ViroPharma Vishay Intertechnology VWR Wawa Wilmington Trust W. L. Gore & Associates WSFS Bank \| \| Notable Delaware Valley-based US headquarters of foreign businesses \| Aberdeen Asset Management AgustaWestland AstraZeneca Chubb Delaware Investments GSK Keystone Foods SAP Siemens Healthineers Shire Pharmaceuticals Subaru Teva Pharmaceuticals TD Bank \| \| Notable Delaware Valley-based division headquarters of US corporations \| Acme (Cerberus Capital Management) Centocor (Johnson & Johnson) Colonial Penn (Conseco) Delmarva Power (Exelon) GSI Commerce (eBay) Hercules (Ashland) MAB Paints (Sherwin-Williams) McNeil Laboratories (Johnson & Johnson) Neoware (Hewlett-Packard) PECO (Exelon) QVC (Liberty Media) Rohm & Haas (Dow Chemical) SunGard (FIS) Tasty Baking (Flowers Foods) \| Retrieved from " Categories: Financial services companies established in 1977 Companies listed on the New York Stock Exchange Financial services companies of the United States Companies based in Philadelphia United States financial services company stubs Hidden categories: Articles with short description Short description matches Wikidata Wikipedia articles with possible conflicts of interest from August 2019 Articles lacking reliable references from December 2019 All articles lacking reliable references Articles with multiple maintenance issues All stub articles
12111	https://allen.in/dn/qna/1460702	Prove that the function f(x)=(log)_a x is increasing on (0,\ oo) if a >1 and decreasing on (0,\ oo) . if 0<a<1 . Courses NEET Class 11th Class 12th Class 12th Plus JEE Class 11th Class 12th Class 12th Plus Class 6-10 Class 6th Class 7th Class 8th Class 9th Class 10th View All Options Online Courses Distance Learning Hindi Medium Courses International Olympiad Test Series NEET Class 11th Class 12th Class 12th Plus JEE (Main+Advanced) Class 11th Class 12th Class 12th Plus JEE Main Class 11th Class 12th Class 12th Plus Classroom Results NEET 2025 2024 2023 2022 JEE 2025 2024 2023 2022 Class 6-10 Scholarships NEW TALLENTEX AOSAT ALLEN E-Store More ALLEN for Schools About ALLEN Blogs News Careers Request a call back Book home demo Login HomeClass 12MATHSProve that the function f(x)=(log)a x is... Prove that the function f(x)=(log)a x is increasing on (0,∞) if a>1 and decreasing on (0,∞) . if `0 To view this video, please enable JavaScript and consider upgrading to a web browser thatsupports HTML5 video Video Player is loading. Play Video Play Skip Backward Skip Forward Mute Current Time 0:00 / Duration-:- Loaded: 0% Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2x 1.5x 1x, selected 0.5x 0.25x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Opacity Text Background Color Opacity Caption Area Background Color Opacity Font Size Text Edge Style Font Family Reset Done Close Modal Dialog End of dialog window. Text Solution AI Generated Solution To prove that the function f(x)=log a⁡x is increasing on (0,∞) if a>1 and decreasing on (0,∞) if 0<a<1, we will follow these steps: ### Step 1: Differentiate the function We start by differentiating the function f(x). The derivative of f(x)=log a⁡x can be expressed using the change of base formula: f′(x)=d d x(log⁡x log⁡a)=1 log⁡a⋅1 x ... Show More \| Share Topper's Solved these Questions HIGHER ORDER DERIVATIVES RD SHARMA\|Exercise Solved Examples And Exercises\|176 Videos View Playlist INDEFINITE INTEGRALS RD SHARMA\|Exercise Solved Examples And Exercises\|1401 Videos View Playlist Similar Questions Explore conceptually related problems Prove that the function f(x)=(log)_(e)x is increasing on (0,oo) Watch solution Prove that the function f(x)=log_(e)x is increasing on (0,oo) Watch solution Function f(x)=(log)_a x is increasing on R , if (a) 0 1 (c) a 0 Watch solution Prove that the function f given by f(x) = log sin x" is increasing on ")0,pi/2) and decreasing on (pi/2,pi) . Watch solution Prove that the logarithmic function is strictly increasing on (0,oo) Watch solution Prove that function f(x)=log_(e)x is strictly increasing in the interval (0,oo) Watch solution Prove that function f(x)=log_(e)x is strictly increasing in the interval (0,oo) Watch solution Prove that the function f given by f(x)=log cos x is strictly increasing on (-pi/2,0) and strictly decreasing on (0,pi/2) Watch solution Show that the function f(x ) = \|x\| is (a) striclty increasing on [0, oo] (b) strictly decreasing on [-oo, 0] Watch solution Without using the derivative,show that the function f(x)=\|x\| is strictly increasing in (0,oo) strictly decreasing in (-oo,0) Watch solution RD SHARMA-INCREASING AND DECREASING FUNCTION -Solved Examples And Exercises Show that the function f(x)=x^2 is neither strictly increasing nor ...01:49 \| Play Prove that the function f(x)=(log)e x is increasing on (0,\ oo) .00:56 \| Play Prove that the function f(x)=(log)a x is increasing on (0,\ oo) if a >...01:32 \| Playing Now Prove that f(x)=a x+b , where a ,\ b are constants and a >0 is an incr...00:57 \| Play Prove that f(x)=a x+b , where a ,\ b are constants and a<0 is a decrea...00:59 \| Play Show that f(x)=1/x is a decreasing function on (0,\ oo) .01:14 \| Play [Show that f(x)=1/(1+x^2) decreases in the interval 0,\ oo) and increa...01:29 \| Play Show that f(x)=1/(1+x^2) is neither increasing nor decreasing on R ...01:28 \| Play Without using the derivative, show that the function f(x)=\|x\| is (a) s...01:50 \| Play Without using the derivative show that the function f(x)=7x-3 is stric...01:17 \| Play Find the intervals in which f(x)=-x^2-2x+15 is increasing or decreasin...01:19 \| Play Find the intervals in which the function f(x)=2x^3-9x^2+12 x+15 is inc...01:41 \| Play Find the intervals in which the function f(x)=2x^3+9x^2+12 x+20 is inc...04:11 \| Play Find the intervals in which f(x)=(x+1)^3(x-1)^3 is increasing or decre...06:29 \| Play Find the intervals in which f(x)=(x-1)^3(x-2)^2 is increasing or decre...05:47 \| Play Find the intervals in which the function f(x)=x^4-(x^3)/3 is increasin...01:02 \| Play Find the intervals in which the function f(x)=log(1+x)-(2x)/(2+x) is i...07:27 \| Play Find the intervals in which f(x)=(4x^2+1)/x is increasing or decreasin...03:14 \| Play Determine the intervals in which the function f(x)=x^4-8x^3+22 x^2-24 ...05:52 \| Play Find the intervals for which f(x)=x^4-2x^2 is increasing or decreasing...01:10 \| Play HomeProfile
12112	https://stackoverflow.com/questions/1373035/how-do-i-scale-one-rectangle-to-the-maximum-size-possible-within-another-rectang	algorithm - How do I scale one rectangle to the maximum size possible within another rectangle? - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How do I scale one rectangle to the maximum size possible within another rectangle? Ask Question Asked 16 years ago Modified3 years ago Viewed 36k times This question shows research effort; it is useful and clear 78 Save this question. Show activity on this post. I have a source rectangle and a destination rectangle. I need to find the maximum scale to which the source can be scaled while fitting within the destination rectangle and maintaining its original aspect ratio. Google found one way to do it but I'm not sure if it works in all cases. Here is my home-brewed solution: Calculate Height/Width for each rectangle. This gives the slopes of the diagonals msrc and mdest. If msrc < mdst, scale source width to fit the destination width (and scale height by the same ratio) Otherwise, scale source height to fit the destination height (and scale width by the same ratio) Looking for other possible solutions to this problem. I'm not even sure if my algorithm works in all cases! algorithm scaling Share Share a link to this question Copy linkCC BY-SA 2.5 Improve this question Follow Follow this question to receive notifications asked Sep 3, 2009 at 12:11 community wiki Agnel Kurian Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 160 Save this answer. Show activity on this post. scale = min(dst.width/src.width, dst.height/src.height) This is your approach but written more cleanly. To use this, the scaled rectangle would have the shape: width = src.width scale height = src.height scale Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 8, 2022 at 20:58 community wiki 5 revs, 3 users 80%tom10 4 Comments Add a comment fregante freganteOver a year ago Sweet! Once you have the scale, use these to get the final dimensions: width = src.width scale and height = src.height scale 2013-09-12T02:04:22.56Z+00:00 10 Reply Copy link Glogo GlogoOver a year ago Change min with max if you want to cover whole destination area. 2014-06-17T10:45:37.373Z+00:00 9 Reply Copy link Peppe L-G Peppe L-GOver a year ago Same solution, but with names more clear to me: scale = min(maxWidth/actualWidth, maxHeight/actualHeight), newWidth = actualWidthscale, newHeight = actualHeightscale. 2014-08-06T08:38:22.09Z+00:00 19 Reply Copy link Yoz YozOct 8, 2024 at 5:44 be aware of the edge cases like 161 (100 / 161) = 99.99999999999999, possible fix if(Math.abs(width - dst.width) <= Number.EPSILON dst.width) width = dst.width 2024-10-08T05:44:14.58Z+00:00 0 Reply Copy link Add a comment This answer is useful 13 Save this answer. Show activity on this post. Another option might be to scale to maximum width and then check if the scaled height is greater then the maximum allowed height and if so scale by height (or vice versa): scale = (dst.width / src.width); if (src.height scale > dst.height) scale = dst.height / src.height; I think this solution is both shorter, faster and easier to understand. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications edited Sep 3, 2009 at 17:24 answered Sep 3, 2009 at 12:18 GussGuss 32.7k 20 20 gold badges 117 117 silver badges 143 143 bronze badges Comments Add a comment This answer is useful 2 Save this answer. Show activity on this post. Work out the smaller of destWidth / srcWidth and destHeight / srcHeight Scale by that edit it's of course the same as your method, with the pieces of the formula moved around. My opinion is that this is clearer semantically, but it's only that - an opinion. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Sep 3, 2009 at 12:14 AakashMAakashM 63.5k 17 17 gold badges 154 154 silver badges 190 190 bronze badges Comments Add a comment This answer is useful 1 Save this answer. Show activity on this post. If all dimensions are non-zero, I would use the following code (that essentially matches your code). scaleFactor = (outerWidth / outerHeight > innerWidth / innerHeight) ? outerHeight / innerHeight : outerWidth / innerWidth This can also be modified to allow any dimension to be zero if required. Share Share a link to this answer Copy linkCC BY-SA 2.5 Improve this answer Follow Follow this answer to receive notifications answered Sep 3, 2009 at 12:41 community wiki Daniel Brückner 4 Comments Add a comment MusiGenesis MusiGenesisOver a year ago I think a rectangle with a zero dimension is called a "line". :P 2009-09-03T12:47:46.093Z+00:00 3 Reply Copy link Guss GussOver a year ago This solution is mathematically identical to mine: multiply your inequality by (innerHeight outerHeight / innerWidth) and you get my inequality. The advantage of my code is that if the inequality fails, then the solution need not be recomputed. 2009-09-03T12:50:15.087Z+00:00 1 Reply Copy link Daniel Brückner Daniel BrücknerOver a year ago The ternariy operator will evaluate the condition only once, too. And, of course, all solutions should be mathematical equivalent ... or wrong ... 2009-09-03T13:35:06.36Z+00:00 0 Reply Copy link Guss GussOver a year ago The ternary operator will evaluate the condition only once, but by formatting the condition differently part of the calculation of the condition can be one of the answers. Notice that you have always 3 multiplication operations in the computation. My solution has 3 operations in the worst case, otherwise it has only 2 because the correct calculation was already done before the test. Hence, 1/6th better performance (assuming random distribution of values). 2009-09-03T22:20:01.177Z+00:00 0 Reply Copy link Add a comment This answer is useful 1 Save this answer. Show activity on this post. The other answers suffer from a risk of generating a division by zero exception when either the sourceWidth or sourceHeight becomes zero. To safeguard against this, we should rewrite the comparison into a mathematically equivalent multiple expression. Also, additional edge condition to catch the infinite scale scenario. Apart from having the scale, I really wanted the dimensions of the target rectangle, so, here I will provide the scale calculation and the target rectangle calculation. Because of the infinity edge condition, I think the target rectangle will be more robust / useful: if (sourceWidth == 0 && sourceHeight == 0) { // scale = Infinity; outputWidth = 0; outputHeight = 0; outputX = destWidth / 2; outputY = destHeight / 2; } else if (destWidth sourceHeight > destHeight sourceWidth) { scale = destHeight / sourceHeight; outputWidth = sourceWidth destHeight / sourceHeight; outputHeight = destHeight; outputX = (destWidth - outputWidth) / 2; outputY = 0; } else { scale = destWidth / sourceWidth; outputWidth = destWidth; outputHeight = sourceHeight destWidth / sourceWidth; outputX = 0; outputY = (destHeight - outputHeight) / 2; } Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Jun 15, 2018 at 4:16 community wiki Stephen Quan 1 Comment Add a comment AUSTX_RJL AUSTX_RJLJul 27 at 22:01 Trying to scale keeping aspect ratio (IsoMorphic) when the source / dest are very different width and height : 656, 1232 => 1254, 668 this answer worked for me but the very simple MIN in the most upvoted answer did not. 2025-07-27T22:01:55.557Z+00:00 0 Reply Copy link Your Answer Thanks for contributing an answer to Stack Overflow! 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12113	https://zhuanlan.zhihu.com/p/2155280310	直答深入浅出数学系列 \| 数列（2）：数列求和与奇偶数列首发于深入浅出解析高中数学切换模式深入浅出数学系列 \| 数列（2）：数列求和与奇偶数列收录于 · 深入浅出解析高中数学 119 人赞同了该文章前述数列求和 <一>.常见幂和公式很多综合题中都会出现幂和公式的身影，如2020年一卷文数的压轴填空（在后面的文章中会对这道题进行解析）。这里给出三个常见的公式： \sum_{k=1}^{n}{k}=\frac{n(n+1)}{2}\ \sum_{k=1}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}\ \sum_{k=1}^{n}{k^3}=(\frac{n(n+1)}{2})^2\我们挑选第二个公式进行证明：观察下面的等式： (n+1)^3-n^3=3n^2+3n+1， (n)^3-(n-1)^3=3(n-1)^2+3(n-1)+1， …… 3^3-2^3=3\cdot 2^2+3\cdot 2+1， 2^3-1^3=3\cdot 1^2+3\cdot 1+1，将以上n个等式相加，左边就可以消去中间项，右边提取公因式合并，于是可以得到： (n+1)^3-1^3=3(1^2+2^2+...+n^2)+3(1+2+...+n)+n\ 将自然数之和公式代入上式，化简就可以得到： 1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\ 这是一种很巧妙的证明方法，也是幂和公式证明最普遍的方法：递推法。下面证明一下普遍的情况：不妨设： S_k=0^k+1^k+2^k+3^k+\cdots+n^k\ 令 a_n=n^{k+1} ，取： \Delta a_n=(n+1)^{k+1}-n^{k+1} ，将 (n+1)^{k+1} 二项式展开，得: Δan=(k+1∑i=0C ik+1ni)−nk+1=k∑i=0C ik+1ni\begin{align} \Delta a_n&=\left(\sum_{i=0}^{{k+1}}C_{k+1}^{\ i}n^{i}\right)-n^{k+1}=\sum_{i=0}^{k}C_{k+1}^{\ i}n^i \end{align}\ 再对其求和： \begin{align} \sum_{j=0}^n\Delta a_j &=\sum_{j=0}^n\sum_{i=0}^{k}C_{k+1}^{\ i}j^{\ i} \end{align}\ 这里与我们上次的累和法很像，思路一样，都是前后相抵最终留下两项： \begin{align} \sum_{j=0}^n\Delta a_j &=a_{n+1}-a_{n}+a_{n}-a_{n-1}+……+a_1-a_0=a_{n+1}-a_0=(n+1)^{k+1} \end{align}\ 对于右侧，进行简单的代数变形，可得: \begin{align} \sum_{j=0}^n\sum_{i=0}^{k}C_{k+1}^{\ i}j^{\ i}&= \sum_{i=0}^k\left(C_{k+1}^{\ i}{\sum_{j=0}^{n}j^{\ i}}\right)= \sum_{i=0}^kC_{k+1}^{\ i}{S_i} \end{align}\ 所以有： \begin{align} (n+1)^{k+1}=\sum_{i=0}^{{k}}C_{k+1}^{\ i}{S_i} \end{align}\ 即: \begin{align} (n+1)^{k+1}=(k+1)S_k+\sum_{i=0}^{{k-1}}C_{k+1}^{\ i}{S_i} \end{align}\故： {\begin{align} S_k=\frac1{k+1}\left[(n+1)^{k+1}-\sum_{i=1}^{k}C_{k+1}^{i+1}S_{k-i}\right] \end{align}}\ 如此，便得到了 S_k 的递推公式，用的思路和上面的一样。你要求 S_k ，就得取 (n+1)^{k+1}-n^{k+1} 然后进行展开求和，裂项抵消。 <二>.错位相减法错位相减法很局限，一般是解决差比数列用的。差比数列是每个学生和老师都会反复练习的题型，早已不是难题。之前网上很火的万能公式，我认为完全没有必要背诵，只需要会推导就行。我们先从最简单的差比数列看起：例1：已知数列 {a_n} 的通项公式为 a_n=n\cdot2^n ，求数列 {a_n} 的前 n 项的和 S_n 解：我们考虑错位相减法： S_n=1\times2^1+2\times2^2+\dots+n\times2^n 2S_n=1\times2^2+\dots+(n-1)\times2^n+n\times2^{n+1} 作差得 -S_n=(2^1+2^2+\dots+2^n)-n\cdot2^{n+1} 故 S_n=(n-1)\cdot2^{n+1}+2 这是一阶的差比数列，我们还可以考虑二阶的形式：例2 :已知数列 {a_n} 的通项公式为 a_n=n^2\cdot2^n ，求数列 {a_n} 的前 n 项的和 S_n 解：这里仍然是利用错位相减法，一次可能不行，我们使用两次。 S_n=1\times2^1+4\times2^2+\dots+n^2\times2^n 2S_n=1^2\times2^2+\dots+(n-1)^2\times2^n+n^2\times2^{n+1} 作差得 -S_n=1\times2^1+3\times2^2+\dots+(2n-1)\times2^n-n^2\times2^{n+1} 到这里我们发现又变成了差比数列的形式，我们再次使用错位相减： -2S_n=1\times2^2+\dots+(2n+1)\times2^{n+1}-n^2\times2^{n+2} 与上式子作差，得： S_n=1\times2+2(2^2+2^3+……+2^n)-(n^2+2n-1)×2^{n+1}+n^2\times2^{n+2} 化简得： S_n=(n^2-2n+3)2^{n+1}-6 如果等比数列前面是 n 次多项式，要想得出答案，我们就需要进行 n 次错位相减。这种方法比较麻烦，但却是最基本的。我们针对差比数列还有裂项法，分组法和阿贝尔变换法，这些方法在后面会叙述。 <三>.倒序相加法这种方法也属于比较基础的方法，没什么特别好说的，直接上例题：例题： a_n=\frac{2n-98}{2n-99} ，求 S_{98} . 解： a_n=1+\frac{1}{2n-99} .这种属于调和类型，没办法直接求和（ a_n=\frac{1}{an+b} ,我们说 a_n 是调和类型），我们发现99这个数字比较特殊。注意到 2n-99=99-2n'\Rightarrow n+n'=99 ,也就是说，只要两项下标和为99，就可以进行抵消。我们考虑进行倒序相加，抵消，进而解决问题。于是 2S_{98}=(a_1+a_{98})+(a_2+a_{97})+……+(a_{97}+a_{2})+(a_{98}+a_{1})=2\times 98 故 S_{98}=98 . <四>.裂项相消法这种方法是求和题的重点方法，也是高考的常客。裂项的意义何在呢？假设我们有一个数列 b_n ,该数列不能直接求和，我们可以尝试把它拆成 b_n=a_{n+1}-a_n ，又是前后相抵的原理，和之前幂和公式的意义相同，这样对其求和我们最终得到 S_n=a_{n+1}-a_1 . 我们先给出几个常见的裂项公式：普通分式型： \frac{1}{n(n+k)}=\frac{1}{k}(\frac{1}{n}-\frac{1}{n+k}) \frac{2}{n(n+1)(n+2)}=\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}. \frac{(a-1)n+1}{(n-1)n\cdot a^n}=\frac{1}{(n-1)\cdot a^{n-1}}-\frac{1}{n\cdot a^n}.(a=1,2,3,...) \frac{1}{(2n-1)(2n+1)}=\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1}) {{\ \frac{1}{n}-\frac{1}{n+1}<\frac{1}{n^2}<\frac{1}{(n-c)(n+c)}=\frac{1}{2c}(\frac{1}{n-c}-\frac{1}{n+c})\leq\frac{1}{n-1}-\frac{1}{n}.(c=\frac{1}{2},1)}} (-1)^{n-1}\frac{4n}{(2n-1)(2n+1)}=(-1)^{n-1}(\frac{1}{2n-1}+\frac{1}{2n+1}) (奇偶数列中的常客) 根式型： \frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\ <\frac{1}{2\sqrt n}<\sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}}<\sqrt n-\sqrt{n-1} {{ \frac{1}{n\sqrt n}<\frac{2}{\sqrt{n-c}\sqrt{n+c}(\sqrt{n-c}+\sqrt{n+c})}=\frac{1}{c}(\frac{1}{\sqrt{n-c}}-\frac{1}{\sqrt{n+c}})(c=\frac{1}{2},1)}} 三角型： \frac{1}{cosncos(n+1)}=\frac{sin[(n+1)-n]}{sin1cosncos(n+1)}=\frac{1}{sin1}[tan(n+1)-tann) 若 a_n 为等差数列，则 \frac{1}{a_na_{n+1}} 可裂项。 \frac{1}{a_na_{n+1}}=\frac{1}{d}[\frac{1}{a_1+(n-1)d}-\frac{1}{a_1+nd}] . 在平时的练习中，应用这些公式的题目很多，大家可以对照自行整理，这里略过常规的裂项试题。很多时候想凑出裂项的形式是比较困难的，通常需要拆项添配或者取倒数才能看出端倪。例1（取倒数）： a_1=\frac{7}{6},a_n^2-a_n+1=a_{n+1}，S_n 是 \left{ \frac{1}{a_n} \right} 的前 n 项和，求证 S_{2024}<6 . 解：与之前说过的相同，我们对问题不是很清楚，初步猜测是一个有界的放缩，那么我们首先从题目条件入手： a_{n+1}=a_n(a_n-1)+1 ，由于 a_1>1 ,带入得到 a_2>1 以此类推很容易得到 a_n>1 。此时我们确定了 a_n 的下界。题目中说 S_n是 \left{ \frac{1}{a_n} \right} 的前 n 项和，我们试着凑出 \frac{1}{a_n} . a_{n+1}-1=a_n(a_n-1) ，我们对两边取倒数： \frac{1}{a_{n+1}-1}=\frac{1}{a_n(a_n-1)}=\frac{1}{a_n-1}-\frac{1}{a_n} 得到题目所给的同时也进行了裂项。这样 \frac{1}{a_n}=\frac{1}{a_{n+1}-1}-\frac{1}{a_{n}-1} 于是 S_n=\frac{1}{a_1-1}-\frac{1}{a_{2025}-1}=6-\frac{1}{a_{2025}-1} 由于 a_{2025}>1 故 S_n<6 . 裂项相消法常伴随着待定系数法一起出现。我们之前说过，求差比数列也可以用裂项法，尝试将 a_n=n\cdot2^n 写成两项的差的形式： n\cdot2^n=f(n+1)\cdot2^{n+1}-f(n)\cdot2^n ，其中 f(n)=An+B 即 n\cdot2^n=(2A(n+1)+2B-An-B)\cdot2^n=(An+2A+B)\cdot2^n 即 n= An+2A+B ，于是 A=1 ， B=-2 ， f(n)=n-2 因此我们有： n\cdot2^n=(n-1)\cdot2^{n+1}-(n-2)\cdot2^n 仍然用这种最基本的题目来举例：例2：已知数列 {a_n} 的通项公式为 a_n=n\cdot2^n ，求数列 {a_n} 的前 n 项的和 S_n 我们上面使用裂项相消得到：a_n=n\cdot2^n=(n-1)\cdot2^{n+1}-(n-2)\cdot2^n 故 S_n=\sum_{k=1}^{n}{(k\cdot2^k)}=\sum_{k=1}^{n}{[(k-1)\cdot2^{k+1}-(k-2)\cdot2^k]}\ 于是： S_n=(n-1)\cdot2^{n+1}+2 之前我们说了差比数列的万能公式，这里使用裂项相消法来推导。对于数列 a_n=(an+b)\cdot q^n ，应用待定系数法可以得到如下列项公式： (an+b)\cdot q^n=f(n+1)\cdot q^{n+1}-f(n)\cdot q^{n}\ 其中 f(n)=\frac{a}{q-1}\cdot n+\frac{q(b-a)-b}{(q-1)^2} ，由此可以得到： \sum_{k=1}^{n}{a_k}=f(n+1)\cdot q^{n+1}-\frac{bq^2-(a+b)q}{(q-1)^2}\ 代入化简得到公式： \begin{alignat}{2} a_n=(an+b)q^{n}\quad &S_n=(A(n+1)+B)q^{n+1}-B \end{alignat}\ 其中 A=\frac{a}{q-1} , B=\frac{b-A}{q-1}\ 这种方法对于 a_n=f(n)q^{n} 的任意式子求和都适用（ f(n) 为多项式），大家可以自行尝试。 <五>.分组求和与并项求和法分组求和一般分为两类，一类是基本的分组，另一类是分奇偶的分组。我们这里讨论一般的分组求和，在下面详细讨论分奇偶的分组求和。并列求和一般也和分奇偶的分组求和有关。这里再次给出这道例题：已知数列 {a_n} 的通项公式为 a_n=n\cdot2^n ，求数列 {a_n} 的前 n 项的和 S_n 我们这次使用分组求和的方法来解决这道问题：我们可以把每一项前的系数拆开，得到： S_n=2^1+(2^2+2^2)+\cdots+(2^n+2^n+\cdots+2^n) =(2^1+2^2+\cdots+2^n)+(2^2+2^3+\cdots+2^n)+\cdots+2^n =\sum_{k=1}^{n}{2^k}+\sum_{k=2}^{n}{2^k}+\cdots+\sum_{k=n}^{n}{2^k}=\sum_{l=1}^{n}\sum_{k=l}^{n}{2^k} 即： S_n=\sum_{l=1}^{n}\sum_{k=l}^{n}{2^k} ，又 \sum_{k=l}^{n}{2^k}=2^{n+1}-2^l 故 S_n=\sum_{l=1}^{n}(2^{n+1}-2^l)=n2^{n+1}-2^{n+1}+2=(n-1)2^{n+1}+2\ 之后我会把奇偶相关的分组求和与并项求和放在分奇偶的题型里讲。 <六>.含有绝对值的数列这种也属于比较基础的题型，核心思想是分段。例题：等差数列 \left{ a_n \right} 中， a_n=11-n ,求 \|a_1\|+\|a_2\|+……+\|a_n\| . 我们将这道题分成两段来讨论： n\leq11 时，绝对值没有起到作用，我们直接套等差数列公式。于是 \|a_1\|+\|a_2\|+……+\|a_n\|=S_n=-\frac{1}{2}n^2+\frac{21}{2}n n>11 时，需要考虑变号的问题。之后符号全为反号。于是 \|a_1\|+\|a_2\|+……+\|a_n\|=S_{11}-(S_n-S_{11})=2S_{11}-S_n=-\frac{1}{2}n^2-\frac{21}{2}n+110 奇偶数列 <一>.隔项等差数列这种题型的特征是 a_{n+2}-a_n=d ，他有一些衍生版本，如 a_{n+1}+a_n=s ，如 a_{n+1}+a_n=dn+s 事实上这三个讲的是一个东西，都是 a_{n+2}-a_n=d 对于此类题目，我们一般的做法是对下标进行换元，进而分成奇数列与偶数列：当 n 为奇数时，令 n=2k-1 ，反解得出 k=\frac{n+1}{2} ， a_{2k+1}-a_{2k-1}=d . 于是 a_n=a_{2k-1}=a_1+(k-1)d=a_1+(\frac{n+1}{2}-1)d=a_1+\frac{n-1}{2}d 当 n 为偶数时，令 n=2k ，反解得出 k=\frac{n}{2} ， a_{2k+2}-a_{2k}=d . 于是 a_n=a_{2k}=a_2+(k-1)d=a_2+(\frac{n}{2}-1)d=a_2+\frac{n-2}{2}d 综上所述： a_n=\left{ \begin{array}{rcl}a_1+\frac{n-1}{2}d , &n=2k-1&k\in N^+ \a_2+\frac{n-2}{2}d, & n=2k&k\in N^+ \end{array} \right. 在进行换元后，我们要将最后的结果还原为关于 n 的表达式。这类题目还有一种待定系数的方法可以解决。观察上面得到的结果，我们可以发现带有 n 项的系数是不变的。我们可以假定 a_n=xn+y+(-1)^nz ，其中 x=\frac{d}{2} ，代入 n=1 和 n=2 即可确定 y 和 z 的值。下面来一道例题：例题： a_1=a ， a_n+a_{n+1}=3n-54 ，求 \left{ a_n \right} 的通项公式。解：这属于我们说的隔项等差数列的类型。我们可以通过项数相减化简成我们熟悉的类型。 a_n+a_{n+1}=3n-54 下标减一，有 a_{n-1}+a_{n}=3(n-1)-54 两式相减，有： a_{n+1}-a_{n-1}=3 代入 n=1 ,有 a_1=a,a_2=-51-a 我们用上面的方法：当 n 为奇数时，令 n=2k-1 ，反解得出 k=\frac{n+1}{2} ， a_{2k+1}-a_{2k-1}=3 . 于是 a_n=a_{2k-1}=a_1+(k-1)d=a+3(\frac{n+1}{2}-1)=a+3\frac{n-1}{2}=\frac{3n+2a-3}{2} 当 n 为偶数时，令 n=2k ，反解得出 k=\frac{n}{2} ， a_{2k+2}-a_{2k}=3 . 于是 a_n=a_{2k}=a_2+(k-1)d=-51-a+3(\frac{n}{2}-1)=-51-a+3\frac{n-2}{2}=\frac{3n-2a-108}{2} 综上所述： a_n=\left{ \begin{array}{rcl}\frac{3n+2a-3}{2} , &n=2k-1&k\in N^+ \\frac{3n-2a-108}{2}, & n=2k&k\in N^+ \end{array} \right. 我们也可以利用待定系数法解决这个问题。令a_n=xn+yn+(-1)^nz，则 a_{n+1}=xn+y+x+(-1)^{n+1}z 相加得 a_n+a_{n+1}=2xn+x+2y=3n-54 对比系数，得： x=\frac{3}{2},y=-\frac{111}{4} 由于 a_1=x+y-z=-\frac{105}{4}-z=a\Rightarrow z=-\frac{105}{4}-a 于是 a_n=\frac{3}{2}n-\frac{111}{4}-(-1)^{n}(\frac{105}{4}+a) 这两种方法得出的结果完全相同。我个人更喜欢第一种方法，方法更加清晰，表示出来更加好看。 <二>.隔项等比数列这类题型和隔项等差数列很像。这种题型的特征是 a_{n+2}=qa_n ，他有一些衍生版本，如 a_{n+1}a_n=p ，如 a_{n+1}a_n=p^{dn+s} 事实上这三个讲的是一个东西，都是 a_{n+2}=qa_n 对于此类题目，我们一般的做法是对下标进行换元，进而分成奇数列与偶数列：当 n 为奇数时，令 n=2k-1 ，反解得出 k=\frac{n+1}{2} ， \frac{a_{2k+1}}{a_{2k-1}}=q . 于是 a_n=a_{2k-1}=a_1q^{k-1}=a_1q^{\frac{n-1}{2}} 当 n 为偶数时，令 n=2k ，反解得出 k=\frac{n}{2} ， \frac{a_{2k+2}}{a_{2k}}=q . 于是 a_n=a_{2k}=a_2q^{k-1}=a_2q^{\frac{n-2}{2}} 综上所述： a_n=\left{ \begin{array}{rcl}a_1q^{\frac{n-1}{2}} , &n=2k-1&k\in N^+ \a_2q^{\frac{n-2}{2}}, & n=2k&k\in N^+ \end{array} \right. 在进行换元后，我们要将最后的结果还原为关于 n 的表达式。这类题目同样还有一种待定系数的方法可以解决。观察上面得到的结果，我们可以发现带有 n 项的系数是不变的。我们可以假定 a_n=q^{xn+y+(-1)^nz} ，通过对比系数，代入 n=1 和 n=2 即可确定 x ，y 和 z 的值。例题： a_1=1,a_na_{n+1}=2^n ,求 \left{ a_n \right} 的通项公式。解：这属于我们说的隔项等比数列的类型。我们可以通过项数相减化简成我们熟悉的类型。 a_na_{n+1}=2^n ，于是 a_{n+1}a_{n+2}=2^{n+1} 两式相除，得到： a_{n+2}=2a_n 当 n 为奇数时，令 n=2k-1 ，反解得出 k=\frac{n+1}{2} ， \frac{a_{2k+1}}{a_{2k-1}}=2 . 于是 a_n=a_{2k-1}=2^{k-1}=2^{\frac{n-1}{2}} 当 n 为偶数时，令 n=2k ，反解得出 k=\frac{n}{2} ， \frac{a_{2k+2}}{a_{2k}}=2 . 于是 a_n=a_{2k}=a_2q^{k-1}=a_2q^{\frac{n-2}{2}} 综上所述： a_n=\left{ \begin{array}{rcl}2^{\frac{n-1}{2}} , &n=2k-1&k\in N^+ \2^{\frac{n-2}{2}}, & n=2k&k\in N^+ \end{array} \right. 同样的我们也可以使用待定系数法来解决这个问题。设 a_n=2^{xn+y+(-1)^nz} ，则 a_{n+1}=2^{xn+x+y+(-1)^{n+1}z} 于是 a_na_{n+1}=2^{2xn+x+2y}=2^n 对比系数，代入特值，得出 x=\frac{1}{2},y=-\frac{1}{4},z=\frac{1}{4} 解得 a_n=2^{\frac{2n-1+(-1)^n}{4}} . <三>.与隔项等差等比相关的求和这种题目我们可以根据上面的隔项等差等比方法，先把通项公式分奇偶地分别求出来，再分别分组进行求和。例1： a_1=1,a_2=4,a_n=a_{n-2}+2 ，求 S_n . 解：用上面的方法，我们可以算出: n 为奇数时，a_n=1+2(\frac{n+1}{2}-1)=n n 为偶数时， a_n=4+2(\frac{n}{2}-1)=n+2 故 n 为偶时， S_n=(1+3+5+……+n-1)+(4+6+……+n+2)=\frac{n^2+3n}{2} 当n 为奇时，我们也可以这样写，也可以利用偶数得到的结论： n 为奇数，那么 n-1 就为偶数，故可以对 S_{n-1} 套用上面得到的公式： S_n=S_{n-1}+a_n=\frac{(n-1)^2+3(n-1)}{2}+n=\frac{n^2+3n-2}{2} 于是 S_n=\left{ \begin{array}{rcl}\frac{n^2+3n-2}{2} , &n=2k-1&k\in N^+ \\frac{n^2+3n}{2}, & n=2k&k\in N^+ \end{array} \right. 对于一些题目，我们也不需要求出通项公式，直接求和即可。例2： a_n+a_{n+1}=4n-3,a_1=2 ，求 S_n . 解：我们可以进行归类求和，不需要求出通项公式。 n 为奇数时，S_n=a_1+(a_2+a_3)+(a_4+a_5)+……+(a_{n-1}+a_n)=2+\frac{(n-1)(4n-2)}{4}=\frac{2n^2-3n+5}{2} n 为偶数时， S_n=(a_1+a_2)+(a_3+a_4)+……+(a_{n-1}+a_n)=\frac{n(4n-6)}{4}=\frac{2n^2-3n}{2} 综上： S_n=\left{ \begin{array}{rcl}\frac{2n^2-3n+5}{2} , &n=2k-1&k\in N^+ \\frac{2n^2-3n}{2}, & n=2k&k\in N^+ \end{array} \right. <四>.分段数列例1： a_1=1,a_{n+1}=\left{ \begin{array}{rcl}a_n+1 , &n=2k-1&k\in N^+ \a_n+2, & n=2k&k\in N^+ \end{array} \right. 求 S_{20} . 解：此类问题一般通过两次迭代可以解决。为了方便讨论，我们写成 n=2k+1 的形式。 a_{2k+1}=a_{2k}+2=a_{2k-1}+3 a_{2k+2}=a_{2k+1}+1=a_{2k}+3 这样隔项的差都为3，满足隔项等差数列。由于 a_2=a_1+1,a_4=a_3+1…… 于是 S_{20}=2(a_2+……+a_{20})-10=300 所有类似的题目都可以用相似的方法来解决。这种题目还可以用其他方法包装。例2： a_{2n}-a_{2n-1}=3^n-1,a_{2n+1}+a_{2n}=3^n+5 ，求 S_{40} . 解：这题和上面的思路一样，也是通过两次迭代来解决。 a_{2n+1}=-a_{2n}+3^n+5=-a_{2n-1}+6 于是 a_{2n-1}+a_{2n+1}=6 同时 a_{2n}=a_{2n-1}+3^n-1 于是 S_{40}=2(a_1+a_3+……+a_{39})+3^1+3^2+……+3^{20}-20=2\times6\times10+\frac{3^{21}-3}{2}-20=\frac{3^{21}+197}{2} <五>.摆动数列中的分组与并项例1（分奇偶后分组求和）： a_n=(-1)^n(3^n+2n) ，求 S_n 解： S_n=(-3+9-27+……+(-3)^n)+(-2+4-6+8+……+(-1)^n2n) 我们将其分成了一个等比数列和等差数列分别求和。当 n 为偶数时， S_n=\frac{-3[1-(-3)^n]}{1-(-3)}+2\frac{n}{2}=n-\frac{3}{4}+\frac{3^{n+1}}{4} 当 n 为奇数时，我们利用偶数得出的结论来简化计算。 S_n=S_{n-1}+a_n=n-1-\frac{3}{4}+\frac{3^n}{4}-3^n-2n=-n-\frac{7}{4}-\frac{3^{n+1}}{4} 综上： S_n=\left{ \begin{array}{rcl}-n-\frac{7}{4}-\frac{3^{n+1}}{4} , &n=2k-1&k\in N^+ \n-\frac{3}{4}+\frac{3^{n+1}}{4},&n=2k&k\in N^+ \end{array} \right. 例2（分奇偶后并项求和）： a_n=(-1)^n(n+1)^2 ，求 S_{2n} 解：本题可以用分组求和，把二次项和一次项和常数项分开求和，但是过于繁琐。由于奇偶交替，我们考虑并项求和，将相邻两项放在一起，可以消去一些东西（至少二次项会被消除）。 a_{2n-1}+a_{2n}=-(2n)^2+(2n+1)^2=4n+1 这下消除了二次项和 (-1)^n ，我们可以放心求和了。于是 S_{2n}=(b_1+b_2)+(b_3+b_4)+……+(b_{2n-1}+b_{2n})=\sum_{i=1}^{n}{(4i+1)}=2n^2+3n 我们再来一道并项求和的题，这次可不止并两项了。例3： a_n=ncos\frac{nπ}{3} ,求 S_{2016} . 解：根据 cosx 的周期性，我们可以把项分为6组： a_{6k+1}=(6k+1)\times\frac{1}{2} a_{6k+2}=(6k+2)\times(-\frac{1}{2}) a_{6k+3}=(6k+3)\times(-1) a_{6k+4}=(6k+4)\times(-\frac{1}{2}) a_{6k+5}=(6k+5)\times\frac{1}{2} a_{6k+6}=(6k+6)\times1 于是 a_{6k+1}+a_{6k+2}+a_{6k+3}+a_{6k+4}+a_{6k+5}+a_{6k+6}=3 得出来一个常数！这正是我们喜欢的类型，这次并项十分成功。于是 S_{2016}=336\times3=1008 . 例4（裂项与并项）： a_n=(-1)^{n-1}\frac{4n}{(2n-1)(2n+1)} ,求 S_n . 解： b_n=(-1)^{n-1}(\frac{1}{2n-1}+\frac{1}{2n+1}) n 为奇数时，S_n=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+……+(\frac{1}{2n-1}+\frac{1}{2n+1})=\frac{2n+2}{2n+1} n 为偶数时， S_n=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+……-(\frac{1}{2n-1}-\frac{1}{2n+1})=\frac{2n}{2n+1} 综上： S_n=\left{ \begin{array}{rcl}\frac{2n+2}{2n+1} , &n=2k-1&k\in N^+ \\frac{2n}{2n+1},&n=2k&k\in N^+ \end{array} \right. 之后我们介绍一种模拟题比较火的题型：隔四项出规律的等差数列。 a_{n+1}+(-1)^na_n=An+B ，则 S_4,S_8-S_4,S_{12}-S_8…… 为 6A+2B 为首项， 8A 为公差的等差数列。容易得出： \left{ \begin{array}{rcl}a_2-a_1=A+B① & \ a_3+a_2=2A+B②& \ a_4-a_3=3A+B③& \ \end{array} \right. ②-①+②+③ ，得 a_1+a_2+a_3+a_4=6A+2B 同理 a_5+a_6+a_7+a_8=14A+2B ，以此类推。这种类型，每四项作为一个整体处理比较容易。例5： a_{n+1}+(-1)^{n+1}a_n=2n ，求 S_{60} . 解：这题和上面的不太一样，把 (-1)^{n} 换成了 (-1)^{n+1} .不过递推规律还是在的。 d=8A ,只是首项变了。取 n=1 和 n=3 ,计算出： S_4=8 。 d=8\times2=16 于是 S_{60}=S_4+(S_8-S_4)+(S_{12}-S_8)+……+(S_{60}-S_{56})=15S_4+\frac{15\times14}{2}\times d=1800 . 最后我们来介绍一种填空压轴题型：递推式为 S_n=(-1)^na_n-q^n 型。例6： S_n=(-1)^na_n-\frac{1}{2^n} ，求 \sum_{k=1}^{100}{S_k} . 解：这回的题目不太一样，他是让我们求和的和。当然，根据 S_n=(-1)^na_n-\frac{1}{2^n} ，我们可以把其轻松地转化为 a_n 相关的求和。无论如何， a_n 的通项公式是必须要求的。 n=1 时， S_1=-a_1-\frac{1}{2} ，得 a_1=-\frac{1}{4} . n\geq2 时，我们可以用阶差法求出 a_n :a_n=S_{n}-S_{n-1}=(-1)^na_n-\frac{1}{2^n}-(-1)^{n-1}a_{n-1}+\frac{1}{2^{n-1}}=(-1)^na_n+(-1)^{n-1}a_{n-1}+\frac{1}{2^n}\此时观察到如果 n 为偶数，那么左右两边的 a_n 可以消去。我们让 n 为偶数，则可得 a_{n-1}=-\frac{1}{2^{n}} . 这里是很多同学会犯错的点。 n 若为偶数，那么 n-1就为奇数。也就是说，我们让 n 为偶数这一步操作实际上是得到了 n 为奇数的通项公式： a_{n}=-\frac{1}{2^{n+1}} . 接着对于一开始的式子，我们让 n 为奇数，得到 a_{n-1}=-2a_n+\frac{1}{2^n} .由于我们得到过 n 为奇数的通项公式： a_{n}=-\frac{1}{2^{n+1}} .于是我们把其带入，得到 a_{n-1}=\frac{1}{2^{n-1}} .n 若为奇数，那么 n-1就为偶数。所以我们让 n 为奇数这一步操作实际上是得到了 n 为偶数的通项公式： a_{n}=\frac{1}{2^{n}} . 是不是很神奇？我们让 n 为奇数的时候，得到的是 n 为偶数的通项公式；我们让 n 为偶数的时候，得到的是 n 为奇数的通项公式；下面我们来正式求和： a_1=-\frac{1}{2^2},a_2=\frac{1}{2^2},a_3=-\frac{1}{2^4},a_4=\frac{1}{2^4}…… 因此: \sum_{k=1}^{100}{S_k}=(-a_1+a_2)+(-a_3+a_4)+……+(-a_{99}+a_{100})-\sum_{k=1}^{100}{\frac{1}{2^k}}=\frac{1}{2^1}+\frac{1}{2^3}+……+\frac{1}{2^{99}}-(\frac{1}{2^1}+\frac{1}{2^2}+……+\frac{1}{2^{100}})=\frac{1}{3}(\frac{1}{2^{100}}-1)\ 后记本文介绍了数列求和的多种方法，并把奇偶数列和数列求和联系在一起，信息量比较大，想要掌握还需要多加练习。下一章将介绍数列递推，会引入不动点蛛网模型等工具，敬请期待。编辑于 2025-02-04 21:55・安徽数列高考数学数列求和写下你的评论... 7 条评论默认最新豆豆28开符号好像反了 02-06 · 四川 axawm 这个地方写错啦 02-04 · 陕西 axawm 豆豆28开如果是对的登号就不成立了吧，你可以再算一下，最后一个多项式也就是答案是对的，但倒数二个不等于答案 02-06 · 陕西豆豆28开对的吧 02-06 · 四川倦梦还 2024-12-28 · 浙江 2024-11-12 · 天津 2024-10-25 · 安徽写下你的评论... 云帆立志成为ML领域的大神 97 94 2,763 推荐阅读 # 数列必须满分系列(3.3)等比等差数列求和--其他方式总结 # 高中数列：你知道二阶等差数列吗，其实还挺有意思的！ # 数列必须满分系列(3.2) 数列求和---裂项相消德先生an... 发表于李老师学数... # 深入浅出数学系列 \| 数列（1）：等差等比数列基础想来知乎工作？请发送邮件到 jobs@zhihu.com 未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App
12114	https://www.wyzant.com/resources/answers/572398/an-example-of-the-math-problem-the-sum-of-the-digits-of-a-certain-two-digit	WYZANT TUTORING Car C. The sum of the digits of a certain two-digit number is 7. Reversing its digits increases the number by 9. What is the number? Can I get an example of this math problem in a real world scenario? 1 Expert Answer Patrick B. answered • 01/14/19 Math and computer tutor/teacher T is the tens digit O is the ones digit T + O = 7. The original number is 10T + O The reverse digits is 10O + T Reversing the digits increases the number by 9 10O + T = 10T + O + 9 10O + T - 10T - O = 9 9O - 9T = 9 9(O - T) = 9 O - T = 1 So O + T = 7 O - T = 1 Adding them together 2O = 8 O = 4 The ones digit is 4, so the tens digit must be 3, since T = 7-O = 7-3 = 4 So the original number is 34. Note that 43 - 34 = 9 You can also figure it out by process of elimination using the following table, as the digits must add up to 7. Original number reverse digits difference 16 61 45 25 52 27 34 43 9 <--- there it is Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. RELATED TOPICS RELATED QUESTIONS what are all the common multiples of 12 and 15 Answers · 10 need to know how to do this problem Answers · 8 what are methods used to measure ingredients and their units of measure Answers · 8 how do you multiply money Answers · 6 spimlify 4x-(2-3x)-5 Answers · 18 RECOMMENDED TUTORS Teri K. David B. Dennis D. find an online tutor Download our free app A link to the app was sent to your phone. Get to know us Learn with us Work with us Download our free app Let’s keep in touch Need more help? Learn more about how it works Tutors by Subject Tutors by Location IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Education.com 35,000 worksheets, games, and lesson plans TPT Marketplace for millions of educator-created resources Vocabulary.com Adaptive learning for English vocabulary ABCya Fun educational games for kids SpanishDictionary.com Spanish-English dictionary, translator, and learning Inglés.com Diccionario inglés-español, traductor y sitio de aprendizaje Emmersion Fast and accurate language certification
12115	https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus__An_Investigation_of_Functions_(Lippman_and_Rasmussen)/07%3A_Trigonometric_Equations_and_Identities/7.03%3A_Double_Angle_Identities/7.3E%3A_Double_Angle_Identities_(Exercises)	7.3E: Double Angle Identities (Exercises) - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 7.3: Double Angle Identities 7: Trigonometric Equations and Identities { } { "7.3E:Double_Angle_Identities(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "7.01:_Solving_Trigonometric_Equations_with_Identities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.02:_Addition_and_Subtraction_Identities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.03:_Double_Angle_Identities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "7.04:_Modeling_Changing_Amplitude_and_Midline" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sat, 02 Jan 2021 21:48:20 GMT 7.3E: Double Angle Identities (Exercises) 13936 13936 admin { } Anonymous Anonymous 2 false false [ "article:topic", "license:ccbysa", "showtoc:no", "authorname:lippmanrasmussen", "licenseversion:40", "source@ ] [ "article:topic", "license:ccbysa", "showtoc:no", "authorname:lippmanrasmussen", "licenseversion:40", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. 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Precalculus - An Investigation of Functions (Lippman and Rasmussen) 5. 7: Trigonometric Equations and Identities 6. 7.3: Double Angle Identities 7. 7.3E: Double Angle Identities (Exercises) Expand/collapse global location 7.3E: Double Angle Identities (Exercises) Last updated Jan 2, 2021 Save as PDF 7.3: Double Angle Identities 7.4: Modeling Changing Amplitude and Midline Page ID 13936 David Lippman & Melonie Rasmussen The OpenTextBookStore ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents No headers Section 7.3 Exercises If sin⁡(x)=1 8 and x is in quadrant I, then find exact values for (without solving for x): a. sin⁡(2⁢x) b. cos⁡(2⁢x) c. tan⁡(2⁢x) If cos⁡(x)=2 3 and x is in quadrant I, then find exact values for (without solving for x): a. sin⁡(2⁢x) b. cos⁡(2⁢x) c. tan⁡(2⁢x) Simplify each expression. cos 2⁡(28⁢∘)−sin 2⁡(28⁢∘) 2⁢cos 2⁡(37⁢∘)−1 1−2⁢sin 2⁡(17⁢∘) cos 2⁡(37⁢∘)−sin 2⁡(37⁢∘) cos 2⁡(9⁢x)−sin 2⁡(9⁢x) cos 2⁡(6⁢x)−sin 2⁡(6⁢x) 4⁢sin⁡(8⁢x)⁢c o s⁡(8⁢x) 6⁢sin⁡(5⁢x)⁢c o s⁡(5⁢x) Solve for all solutions on the interval [0,2⁢π). 6⁢sin⁡(2⁢t)+9⁢sin⁡(t)=0 2⁢sin⁡(2⁢t)+3⁢cos⁡(t)=0 9⁢cos⁡(2⁢θ)=9⁢cos 2⁡(θ)−4 8⁢cos⁡(2⁢α)=8⁢cos 2⁡(α)−1 sin⁡(2⁢t)=cos⁡(t) cos⁡(2⁢t)=sin⁡(t) cos⁡(6⁢x)−cos⁡(3⁢x)=0 sin⁡(4⁢x)−sin⁡(2⁢x)=0 Use a double angle, half angle, or power reduction formula to rewrite without exponents. cos 2⁡(5⁢x) cos 2⁡(6⁢x) sin 4⁡(8⁢x) sin 4⁡(3⁢x) cos 2⁡x⁢sin 4⁡x cos 4⁡x⁢sin 2⁡x If csc⁡(x)=7 and 90⁢∘<x<180⁢∘, then find exact values for (without solving for x): a. sin⁡(x 2) b. cos⁡(x 2) c. tan⁡(x 2) If sec⁡(x)=4 and 270⁢∘<x<360⁢∘, then find exact values for (without solving for x): a. sin⁡(x 2) b. cos⁡(x 2) c. tan⁡(x 2) Prove the identity. (sin⁡t−cos⁡t)2=1−sin⁡(2⁢t) (sin 2⁡x−1)2=cos⁡(2⁢x)+sin 4⁡x sin⁡(2⁢x)=2⁢tan⁡(x)1+tan 2⁡(x) tan⁡(2⁢x)=2⁢sin⁡(x)⁢cos⁡(x)2⁢cos 2⁡(x)−1 cot⁡(x)−tan⁡(x)=2⁢cot⁡(2⁢x) sin⁡(2⁢θ)1+cos⁡(2⁢θ)=tan⁡(θ) cos⁡(2⁢α)=1−tan 2⁡(α)1+tan 2⁡(α) 1+cos⁡(2⁢t)sin⁡(2⁢t)−cos⁡(t)=2⁢cos⁡(t)2⁢sin⁡(t)−1 sin⁡(3⁢x)=3⁢sin⁡(x)⁢cos 2⁡(x)−sin 3⁡(x) cos⁡(3⁢x)=cos 3⁡(x)−3⁢sin 2⁡(x)⁢cos⁡(x) Answer 1. a. 3⁢7 32 b. 31 32 c. 3⁢7 31 cos⁡(56∘) cos⁡(34∘) cos⁡(18⁢x) 2⁢sin⁡(16⁢x) 0, π, 2.4189,3.8643 0.7297, 2.4119, 3.8713, 5.5535 π 6, π 2, 5⁢π 6, 3⁢π 2 a. 2⁢π 9, 4⁢π 9, 8⁢π 9, 10⁢π 9, 14⁢π 9, 16⁢π 9, 0, 2⁢π 3, 4⁢π 3 1+cos⁡(10⁢x)2 3 8−1 2⁢cos⁡(16⁢x)+1 8⁢cos⁡(32⁢x) 1 16−1 16⁢cos⁡(2⁢x)+1 16⁢cos⁡(4⁢x)−1 16⁢cos⁡(2⁢x)⁢cos⁡(4⁢x) a. 1 2+2+7 7 b. 1 2−2+7 7 c. 1 7−4⁢3 This page titled 7.3E: Double Angle Identities (Exercises) is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by David Lippman & Melonie Rasmussen (The OpenTextBookStore) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 7.3: Double Angle Identities 7.4: Modeling Changing Amplitude and Midline Was this article helpful? Yes No Recommended articles 7.1: Solving Trigonometric Equations with IdentitiesIn the last chapter, we solved basic trigonometric equations. In this section, we explore the techniques needed to solve more complicated trig equatio... 7.2: Addition and Subtraction IdentitiesIn this section, we begin expanding our repertoire of trigonometric identities. 7.4: Modeling Changing Amplitude and Midline 1: Functions 2: Linear FunctionsAs we approach day to day life we often need to quantify the things around us, giving structure and numeric value to various situations. This ability ... Article typeSection or PageAuthorDavid Lippman & Melonie RasmussenLicenseCC BY-SALicense Version4.0Show Page TOCno Tags source@ © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? 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12116	https://math.stackexchange.com/questions/4744/formal-power-series-coefficient-multiplication	Skip to main content Formal power series coefficient multiplication Ask Question Asked Modified 6 years, 2 months ago Viewed 4k times This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. Given that I have two formal power series: A(x)=∑k≥0akxk B(x)=∑k≥0bkxk The Cauchy Product gives a series C(x)=∑k≥0ckxk ck=∑n=0kanbk−n Which comes from taking the product of the two series C(x)=A(x)B(x). What then, in terms of A(x) and B(x), is this series? Y(x)=∑k≥0akbkxk generating-functions power-series Share CC BY-SA 2.5 Follow this question to receive notifications edited Sep 16, 2010 at 2:27 Hooked asked Sep 16, 2010 at 2:11 HookedHooked 6,78533 gold badges4141 silver badges6464 bronze badges 1 I think you have a typo in C(x), your coefficients should be c_k. – BBischof Commented Sep 16, 2010 at 2:14 Add a comment \| 3 Answers 3 Reset to default This answer is useful 17 Save this answer. Show activity on this post. It is known as the Hadamard product f⋆g. There is no closed form for general hadamard products, but some classes of functions are known to be closed under such products, e.g. rational power series. However, algebraic series are not generally closed under Hadamard products, the standard example being f=g=(1−4x)−1/2=∑(2nn) xn. However, f rational, g algebraic ⇒f⋆g algebraic. Also D-finite power series are closed under Hadamard products, i.e. power series satisfying a linear differential equation with polynomial coefficients or, equivalently, series whose coefficients satisfy a linear recursive equation with polynomial coefficients. Hadamard products can also be defined in terms of diagonals of multivariate series (and vice versa), e.g. see this paper, where you'll find some interesting connections with finite automata which, e.g. help one to easily observe that algebraic series over Fp are closed under Hadamard product (which fails in characteristic 0) Share CC BY-SA 4.0 Follow this answer to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Sep 16, 2010 at 2:33 Bill DubuqueBill Dubuque 283k4242 gold badges337337 silver badges1k1k bronze badges Add a comment \| This answer is useful 14 Save this answer. Show activity on this post. Y is known as the Hadamard product of A and B, and there is no simple way to find it from A and B in general. (If you don't believe me, set A=B=ex.) The closest thing I know to a formula is (with all the analytic caveats that makes everything converge) Y(r)=12π∫2π0A(eiθ)B(re−iθ)dθ which follows by Parseval's theorem. If A and B are sufficiently simple (for example if they are rational functions) then it is possible to evaluate this integral, but in general there's not much you can do; Y can be much more complicated than A or B. (Another idea is to write the above as a contour integral, and if the integrand ends up being meromorphic you can try to use the residue theorem.) Computing Hadamard products is a special case of computing the diagonal of a two-variable generating function, a problem which I describe with a few examples here using the residue theorem. (Of course for very special A it is possible to say more, e.g. when ak is a polynomial in k.) Share CC BY-SA 2.5 Follow this answer to receive notifications edited Sep 16, 2010 at 2:51 answered Sep 16, 2010 at 2:33 Qiaochu YuanQiaochu Yuan 472k5555 gold badges1.1k1.1k silver badges1.5k1.5k bronze badges 1 Ah; actually if one of A or B is equal to e^x then there is a way to compute Y using a slightly modified Laplace transform. But I still don't think it gives a particularly nice answer if the other of A or B is also equal to e^x. – Qiaochu Yuan Commented Sep 16, 2010 at 3:57 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. For more on Hadamard Products or Hadamard Multiplication Theorem look for papers by Louis R. Bragg in American Mathematical Monthly, Jan. 1999, pp 36-42 or SIAM J. Math. Anal., Vol. 17, 1986, pp 220-230 for starters. Share CC BY-SA 2.5 Follow this answer to receive notifications answered Sep 18, 2010 at 17:08 PaulPaul Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions generating-functions power-series See similar questions with these tags. 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12117	https://math.stackexchange.com/questions/2259829/does-there-always-exist-a-nonzero-idempotent-in-a-ring-r	abstract algebra - does there always exist a nonzero idempotent in a ring $R$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more does there always exist a nonzero idempotent in a ring R R Ask Question Asked 8 years, 5 months ago Modified8 years, 5 months ago Viewed 1k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I was studying about integral domain and wondered that why it should be a commutative ring with unity; that is, I was wondering if there is a noncommutative ring with no zero divisors (1) with/ or (2) without unity, or a (3) commutative ring with no zero divisors with unity. and I guess this page provided a partial answer; it proved that if a ring R R has no zero divisors, a nonzero idempotent(a a is an idempotent if a 2=a a 2=a) of R R must be the unity of R R. so if I can prove that a ring R R with no zero divisors has a nonzero idempotent, then every ring with no zero divisors must have a unity. every ring has an idempotent since it contains 0 0. but (4) is there a nonzero one if the ring has no zero divisors? I guess the commutativity is less connected with the condition having no zero divisors since the division ring can be noncommutative and has no zero divisors. so the case (1) is solved and (2), (3) are relying on (4). hence, what I want to know is just (4) as in the title. abstract-algebra ring-theory idempotents Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked May 1, 2017 at 2:43 user159234user159234 1,027 6 6 silver badges 13 13 bronze badges 1 4 is false if you allow non-unital rings (Fondly called rngs). The set of even integers is one such object and it has no non-zero idempotents.Arkady –Arkady 2017-05-01 02:57:53 +00:00 Commented May 1, 2017 at 2:57 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. No, there does not always exist a nontrivial idempotent in a ring, for example the ring of integers has 0 0 and 1 1 as its only idempotents. If the ring has no zero divisors, there is no nontrivial idempotent since a 2=a a 2=a rearranges as a(a−1)=0 a(a−1)=0, and since there are no zero divisors we can apply cancellation to get a=0 a=0 or a=1 a=1. The place where nontrivial idempotents show up is in products of rings: examples of this are (0,1)(0,1) and (1,0)(1,0) in any product of unitial rings R×S R×S. I know that for unital commutative rings, the presence of a nontrivial idempotent means the ring factorises into a product, and this should be true in more general contexts as well. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 1, 2017 at 2:52 JoppyJoppy 14.1k 22 22 silver badges 45 45 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions abstract-algebra ring-theory idempotents See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 8Idempotents in a ring without unity (rng) and no zero divisors. 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12118	https://artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality?srsltid=AfmBOoqWjAHOEMhf782sjtCI2Q4Q0i-1ef6hV1y12jkgajuGSgIyZ2Ei	Art of Problem Solving Power Mean Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Power Mean Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Power Mean Inequality The Power Mean Inequality is a generalized form of the multi-variable Arithmetic Mean-Geometric Mean Inequality. Inequality For positive real numbers and positive real weights with sum , the power mean with exponent , where , is defined by ( is the weighted geometric mean.) The Power Mean Inequality states that for all real numbers and , if . In particular, for nonzero and , and equal weights (i.e. ), if , then Considering the limiting behavior, we also have , and . The Power Mean Inequality follows from Jensen's Inequality. Proof We prove by cases: for for with Case 1: Note that As is concave, by Jensen's Inequality, the last inequality is true, proving . By replacing by , the last inequality implies as the inequality signs are flipped after multiplication by . Case 2: For , As the function is concave for all , by Jensen's Inequality, For , becomes convex as , so the inequality sign when applying Jensen's Inequality is flipped. Thus, the inequality sign in is flipped, but as , is a decreasing function, the inequality sign is flipped again after applying , resulting in as desired. Retrieved from " Categories: Algebra Inequalities Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
12119	https://www.youtube.com/watch?v=JaZ4DzZQ35g	TL;DR 🔊 Statistics: Episode 12, Linear Regression and Correlation Brandon Foltz 296000 subscribers 18 likes Description 782 views Posted: 26 Sep 2023 🔔 Welcome to Chapter 12: Linear Regression and Correlation! Dive into the intriguing world of data patterns and predictive analysis through linear regression. 📈 0:00 Introduction 0:23 Learning Objectives 0:52 Key Points 1:35 Real-World Application 1:57 Conclusion 🔹 What You'll Learn: 1. Discover the basics of linear regression and how it can be used to predict one variable based on another. 2. Learn to interpret scatter plots to understand the strength and direction of the relationship between two variables. 3. Understand how to calculate and interpret the slope and y-intercept in a regression line. 4. Get acquainted with the least squares regression line and how it's used for making predictions. 5. Recognize the limitations and assumptions that come with linear regression models. 🔹 Why it's Important: - Understanding linear regression and correlation allows you to model and interpret relationships between variables. - The skills you acquire can be directly applied in numerous fields like economics, healthcare, finance, and social sciences to make informed decisions. 🔹 Who Should Watch: - Students looking to understand the foundational concepts in statistical modeling. - Professionals in fields such as real estate, finance, and healthcare, where predictive modeling is frequently used. - Data analysts and researchers interested in exploring relationships between variables. 🔹 Real-World Application: For instance, in the real estate market, linear regression can be used to predict housing prices based on variables like square footage, location, and number of bedrooms. Such models can give buyers and sellers valuable insights into the fair market value of a property. 🔹 Key Points: - Scatter plots are the go-to graphical representation for examining the relationship between two variables. - The least squares regression line is critical for reducing the sum of squared errors and finding the best-fitting line. - The y-intercept and slope have distinct roles in a regression line: the y-intercept predicts the dependent variable when the independent variable is zero, and the slope gives the rate of change. - Residuals help us understand how well our model fits the data. - The correlation coefficient quantifies the strength and direction of the relationship between variables. Linear regression is more than just a statistical tool; it's a lens through which you can explore and understand the world around you. By mastering the concepts in this chapter, you'll gain valuable analytical skills that can be applied in a wide array of contexts. Let's get started on your journey into the fascinating world of linear regression and correlation. 📊📘 Authors: Barbara Illowsky, Susan Dean Publisher/website: OpenStax Book title: Introductory Statistics Publication date: Sep 19, 2013 Location: Houston, Texas Book URL: Section URL: statistics #datascience #education #audio #dataanalytics Transcript: Introduction Chapter 12: Linear Regression and Correlation Introduction Chapter 12 explores the concepts of linear regression and correlation. It discusses the use of linear equations to determine the relationship between two variables and emphasizes the importance of scatter plots in visualizing this relationship. The chapter also introduces the least squares regression line as a tool for making predictions based on data. Learning Objectives Learning Objectives By the end of this chapter, you should be able to, 1 Understand and apply linear regression to predict one variable based on another. 2 Interpret scatter plots and assess the strength of the linear relationship between variables. 3 Calculate and interpret the slope and y-intercept of a regression line. 4 Use the least squares regression line to make predictions and estimate values. 5 Recognize the limitations and assumptions of linear regression analysis. Key Points Key Points Scatter plots are used to visualize the relationship between two variables with a linear pattern indicating a potential linear relationship. Linear regression involves calculating a line of best fit that represents the relationship between variables. The least squares regression line minimizes the sum of squared errors and provides the best fit to the data in the independent variable. The y-intercept of the regression line represents the predicted value of the dependent variable when the independent variable is zero. Residuals are the vertical distances between the actual data points and the predicted points on the regression line. The correlation coefficient measures the strength and direction of the linear relationship between variables. Real-World Application Real-World Application One real-world application of linear regression is in predicting housing prices. By analyzing historical data on factors such as square footage, number of bedrooms, and location, a regression model can be created to estimate the price of a house. This information is valuable for both buyers and sellers as it provides insights into the fair market value of a property. Conclusion Conclusion Chapter 12 introduces the concept of linear regression and correlation, providing a foundation for understanding the relationship between variables. By learning how to calculate and interpret regression lines, you can make predictions and estimate values based on the data. The chapter emphasizes the importance of scatter plots and the least squares regression line in visualizing and analyzing the linear relationship between variables. Understanding these concepts and their real-world applications can enhance your ability to make informed decisions and predictions in various fields. Linear regression is a powerful tool that allows us to uncover relationships and make predictions based on data, and it is widely used in fields such as economics, finance, and social sciences. By mastering the concepts and techniques covered in this chapter, you will be equipped with valuable skills that can be applied to a wide range of real-world problems. So, dive into the world of linear regression and correlation and unlock the potential of data analysis and prediction.
12120	https://en.wikipedia.org/wiki/Animal_locomotion	Jump to content Search Contents 1 Etymology 2 Aquatic 2.1 Swimming 2.2 Benthic 3 Aquatic surface 4 Aerial 4.1 Active flight 4.2 Gliding 4.3 Soaring 4.4 Ballooning 5 Terrestrial 5.1 Jumping 5.2 Peristalsis and looping 5.3 Sliding 5.4 Climbing 5.5 Walking and running 5.6 Powered cartwheeling 6 Subterranean 7 Arboreal locomotion 8 Energetics 9 Passive locomotion 9.1 Hydrozoans 9.2 Mollusca 9.3 Arachnids 9.4 Insects 9.5 Crustaceans 9.6 Animal transport 9.6.1 Remoras 9.6.2 Angler fish 9.6.3 Parasites 10 Changes between media 10.1 Fish 10.2 Marine mammals 10.3 Birds 11 Changes during the life-cycle 12 Function 12.1 Food procurement 13 Quantifying body and limb movement 14 Galleries 15 See also 16 References 17 Further reading 18 External links Animal locomotion Afrikaans العربية অসমীয়া Български Català Čeština Cymraeg Deutsch Eesti Español Esperanto فارسی Français ГӀалгӀай Gĩkũyũ 한국어 Bahasa Indonesia Italiano Kabɩyɛ ქართული Magyar Македонски Bahasa Melayu Монгол Nederlands 日本語 Português Romnă Русский Scots Suomi Svenska தமிழ் Türkçe Українська اردو Tiếng Việt 粵語中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Self-propulsion by an animal For the Eadweard Muybridge study, see Animal Locomotion. In ethology, animal locomotion is any of a variety of methods that animals use to move from one place to another. Some modes of locomotion are (initially) self-propelled, e.g., running, swimming, jumping, flying, hopping, soaring and gliding. There are also many animal species that depend on their environment for transportation, a type of mobility called passive locomotion, e.g., sailing (some jellyfish), kiting (spiders), rolling (some beetles and spiders) or riding other animals (phoresis). Animals move for a variety of reasons, such as to find food, a mate, a suitable microhabitat, or to escape predators. For many animals, the ability to move is essential for survival and, as a result, natural selection has shaped the locomotion methods and mechanisms used by moving organisms. For example, migratory animals that travel vast distances (such as the Arctic tern) typically have a locomotion mechanism that costs very little energy per unit distance, whereas non-migratory animals that must frequently move quickly to escape predators are likely to have energetically costly, but very fast, locomotion. The anatomical structures that animals use for movement, including cilia, legs, wings, arms, fins, or tails are sometimes referred to as locomotory organs or locomotory structures. Etymology [edit] The term "locomotion" is formed in English from Latin loco "from a place" (ablative of locus "place") + motio "motion, a moving". The movement of whole body is called locomotion Aquatic [edit] Main article: Aquatic locomotion Swimming [edit] Further information: Nekton and fish locomotion In water, staying afloat is possible using buoyancy. If an animal's body is less dense than water, it can stay afloat. This requires little energy to maintain a vertical position, but requires more energy for locomotion in the horizontal plane compared to less buoyant animals. The drag encountered in water is much greater than in air. Morphology is therefore important for efficient locomotion, which is in most cases essential for basic functions such as catching prey. A fusiform, torpedo-like body form is seen in many aquatic animals, though the mechanisms they use for locomotion are diverse. The primary means by which fish generate thrust is by oscillating the body from side-to-side, the resulting wave motion ending at a large tail fin. Finer control, such as for slow movements, is often achieved with thrust from pectoral fins (or front limbs in marine mammals). Some fish, e.g. the spotted ratfish (Hydrolagus colliei) and batiform fish (electric rays, sawfishes, guitarfishes, skates and stingrays) use their pectoral fins as the primary means of locomotion, sometimes termed labriform swimming. Marine mammals oscillate their body in an up-and-down (dorso-ventral) direction. Other animals, e.g. penguins, diving ducks, move underwater in a manner which has been termed "aquatic flying". Some fish propel themselves without a wave motion of the body, as in the slow-moving seahorses and Gymnotus. Other animals, such as cephalopods, use jet propulsion to travel fast, taking in water then squirting it back out in an explosive burst. Other swimming animals may rely predominantly on their limbs, much as humans do when swimming. Though life on land originated from the seas, terrestrial animals have returned to an aquatic lifestyle on several occasions, such as the fully aquatic cetaceans, now very distinct from their terrestrial ancestors. Dolphins sometimes ride on the bow waves created by boats or surf on naturally breaking waves. Benthic [edit] Benthic locomotion is movement by animals that live on, in, or near the bottom of aquatic environments. In the sea, many animals walk over the seabed. Echinoderms primarily use their tube feet to move about. The tube feet typically have a tip shaped like a suction pad that can create a vacuum through contraction of muscles. This, along with some stickiness from the secretion of mucus, provides adhesion. Waves of tube feet contractions and relaxations move along the adherent surface and the animal moves slowly along. Some sea urchins also use their spines for benthic locomotion. Crabs typically walk sideways (a behaviour that gives us the word crabwise). This is because of the articulation of the legs, which makes a sidelong gait more efficient. However, some crabs walk forwards or backwards, including raninids, Libinia emarginata and Mictyris platycheles. Some crabs, notably the Portunidae and Matutidae, are also capable of swimming, the Portunidae especially so as their last pair of walking legs are flattened into swimming paddles. A stomatopod, Nannosquilla decemspinosa, can escape by rolling itself into a self-propelled wheel and somersault backwards at a speed of 72 rpm. They can travel more than 2 m using this unusual method of locomotion. Aquatic surface [edit] Main article: Animal locomotion on the water surface Velella, the by-the-wind sailor, is a cnidarian with no means of propulsion other than sailing. A small rigid sail projects into the air and catches the wind. Velella sails always align along the direction of the wind where the sail may act as an aerofoil, so that the animals tend to sail downwind at a small angle to the wind. While larger animals such as ducks can move on water by floating, some small animals move across it without breaking through the surface. This surface locomotion takes advantage of the surface tension of water. Animals that move in such a way include the water strider. Water striders have legs that are hydrophobic, preventing them from interfering with the structure of water. Another form of locomotion (in which the surface layer is broken) is used by the basilisk lizard. Aerial [edit] See also: Aeroplankton Active flight [edit] Main articles: Flight and Flying and gliding animals Gravity is the primary obstacle to flight. Because it is impossible for any organism to have a density as low as that of air, flying animals must generate enough lift to ascend and remain airborne. One way to achieve this is with wings, which when moved through the air generate an upward lift force on the animal's body. Flying animals must be very light to achieve flight, the largest living flying animals being birds of around 20 kilograms. Other structural adaptations of flying animals include reduced and redistributed body weight, fusiform shape and powerful flight muscles; there may also be physiological adaptations. Active flight has independently evolved at least four times, in the insects, pterosaurs, birds, and bats. Insects were the first taxon to evolve flight, approximately 400 million years ago (mya), followed by pterosaurs approximately 220 mya, birds approximately 160 mya, then bats about 60 mya.[better source needed] Gliding [edit] Main articles: Flying and gliding animals and Aerial locomotion in marine animals Rather than active flight, some (semi-) arboreal animals reduce their rate of falling by gliding. Gliding is heavier-than-air flight without the use of thrust; the term "volplaning" also refers to this mode of flight in animals. This mode of flight involves flying a greater distance horizontally than vertically and therefore can be distinguished from a simple descent like a parachute. Gliding has evolved on more occasions than active flight. There are examples of gliding animals in several major taxonomic classes such as the invertebrates (e.g., gliding ants), reptiles (e.g., banded flying snake), amphibians (e.g., flying frog), mammals (e.g., sugar glider, squirrel glider). Some aquatic animals also regularly use gliding, for example, flying fish, octopus and squid. The flights of flying fish are typically around 50 meters (160 ft), though they can use updrafts at the leading edge of waves to cover distances of up to 400 m (1,300 ft). To glide upward out of the water, a flying fish moves its tail up to 70 times per second. Several oceanic squid, such as the Pacific flying squid, leap out of the water to escape predators, an adaptation similar to that of flying fish. Smaller squids fly in shoals, and have been observed to cover distances as long as 50 m. Small fins towards the back of the mantle help stabilize the motion of flight. They exit the water by expelling water out of their funnel, indeed some squid have been observed to continue jetting water while airborne providing thrust even after leaving the water. This may make flying squid the only animals with jet-propelled aerial locomotion. The neon flying squid has been observed to glide for distances over 30 m (100 ft), at speeds of up to 11.2 m/s (37 ft/s; 25 mph). Soaring [edit] Soaring birds can maintain flight without wing flapping, using rising air currents. Many gliding birds are able to "lock" their extended wings by means of a specialized tendon. Soaring birds may alternate glides with periods of soaring in rising air. Five principal types of lift are used: thermals, ridge lift, lee waves, convergences and dynamic soaring. Examples of soaring flight by birds are the use of: Thermals and convergences by raptors such as vultures Ridge lift by gulls near cliffs Wave lift by migrating birds Dynamic effects near the surface of the sea by albatrosses Ballooning [edit] Ballooning is a method of locomotion used by spiders. Certain silk-producing arthropods, mostly small or young spiders, secrete a special light-weight gossamer silk for ballooning, sometimes traveling great distances at high altitude. Terrestrial [edit] Main article: Terrestrial locomotion See also: Comparative foot morphology Forms of locomotion on land include walking, running, hopping or jumping, dragging and crawling or slithering. Here friction and buoyancy are no longer an issue, but a strong skeletal and muscular framework are required in most terrestrial animals for structural support. Each step also requires much energy to overcome inertia, and animals can store elastic potential energy in their tendons to help overcome this. Balance is also required for movement on land. Human infants learn to crawl first before they are able to stand on two feet, which requires good coordination as well as physical development. Humans are bipedal animals, standing on two feet and keeping one on the ground at all times while walking. When running, only one foot is on the ground at any one time at most, and both leave the ground briefly. At higher speeds momentum helps keep the body upright, so more energy can be used in movement. Jumping [edit] Further information: Jumping Jumping (saltation) can be distinguished from running, galloping, and other gaits where the entire body is temporarily airborne by the relatively long duration of the aerial phase and high angle of initial launch. Many terrestrial animals use jumping (including hopping or leaping) to escape predators or catch prey—however, relatively few animals use this as a primary mode of locomotion. Those that do include the kangaroo and other macropods, rabbit, hare, jerboa, hopping mouse, and kangaroo rat. Kangaroo rats often leap 2 m and reportedly up to 2.75 m at speeds up to almost 3 m/s (6.7 mph). They can quickly change their direction between jumps. The rapid locomotion of the banner-tailed kangaroo rat may minimize energy cost and predation risk. Its use of a "move-freeze" mode may also make it less conspicuous to nocturnal predators. Frogs are, relative to their size, the best jumpers of all vertebrates. The Australian rocket frog, Litoria nasuta, can leap over 2 metres (6 ft 7 in), more than fifty times its body length. Peristalsis and looping [edit] Other animals move in terrestrial habitats without the aid of legs. Earthworms crawl by a peristalsis, the same rhythmic contractions that propel food through the digestive tract. Leeches and geometer moth caterpillars move by looping or inching (measuring off a length with each movement), using their paired circular and longitudinal muscles (as for peristalsis) along with the ability to attach to a surface at both anterior and posterior ends. One end is attached, often the thicker end, and the other end, often thinner, is projected forward peristaltically until it touches down, as far as it can reach; then the first end is released, pulled forward, and reattached; and the cycle repeats. In the case of leeches, attachment is by a sucker at each end of the body. Sliding [edit] Due to its low coefficient of friction, ice provides the opportunity for other modes of locomotion. Penguins either waddle on their feet or slide on their bellies across the snow, a movement called tobogganing, which conserves energy while moving quickly. Some pinnipeds perform a similar behaviour called sledding. Climbing [edit] Some animals are specialized for moving on non-horizontal surfaces. One common habitat for such climbing animals is in trees; for example, the gibbon is specialized for arboreal movement, travelling rapidly by brachiation (see below). Others living on rock faces such as in mountains move on steep or even near-vertical surfaces by careful balancing and leaping. Perhaps the most exceptional are the various types of mountain-dwelling caprids (e.g., Barbary sheep, yak, ibex, rocky mountain goat, etc.), whose adaptations can include a soft rubbery pad between their hooves for grip, hooves with sharp keratin rims for lodging in small footholds, and prominent dew claws. Another case is the snow leopard, which being a predator of such caprids also has spectacular balance and leaping abilities, such as ability to leap up to 17 m (50 ft). Some light animals are able to climb up smooth sheer surfaces or hang upside down by adhesion using suckers. Many insects can do this, though much larger animals such as geckos can also perform similar feats. Walking and running [edit] Species have different numbers of legs resulting in large differences in locomotion. Modern birds, though classified as tetrapods, usually have only two functional legs, which some (e.g., ostrich, emu, kiwi) use as their primary, Bipedal, mode of locomotion. A few modern mammalian species are habitual bipeds, i.e., whose normal method of locomotion is two-legged. These include the macropods, kangaroo rats and mice, springhare, hopping mice, pangolins and homininan apes. Bipedalism is rarely found outside terrestrial animals—though at least two types of octopus walk bipedally on the sea floor using two of their arms, so they can use the remaining arms to camouflage themselves as a mat of algae or floating coconut. There are no three-legged animals—though some macropods, such as kangaroos, that alternate between resting their weight on their muscular tails and their two hind legs could be looked at as an example of tripedal locomotion in animals. Many familiar animals are quadrupedal, walking or running on four legs. A few birds use quadrupedal movement in some circumstances. For example, the shoebill sometimes uses its wings to right itself after lunging at prey. The newly hatched hoatzin bird has claws on its thumb and first finger enabling it to dexterously climb tree branches until its wings are strong enough for sustained flight. These claws are gone by the time the bird reaches adulthood. A relatively few animals use five limbs for locomotion. Prehensile quadrupeds may use their tail to assist in locomotion and when grazing, the kangaroos and other macropods use their tail to propel themselves forward with the four legs used to maintain balance. Insects generally walk with six legs—though some insects such as nymphalid butterflies do not use the front legs for walking. Arachnids have eight legs. Most arachnids lack extensor muscles in the distal joints of their appendages. Spiders and whipscorpions extend their limbs hydraulically using the pressure of their hemolymph. Solifuges and some harvestmen extend their knees by the use of highly elastic thickenings in the joint cuticle. Scorpions, pseudoscorpions and some harvestmen have evolved muscles that extend two leg joints (the femur-patella and patella-tibia joints) at once. The scorpion Hadrurus arizonensis walks by using two groups of legs (left 1, right 2, Left 3, Right 4 and Right 1, Left 2, Right 3, Left 4) in a reciprocating fashion. This alternating tetrapod coordination is used over all walking speeds. Centipedes and millipedes have many sets of legs that move in metachronal rhythm. Some echinoderms locomote using the many tube feet on the underside of their arms. Although the tube feet resemble suction cups in appearance, the gripping action is a function of adhesive chemicals rather than suction. Other chemicals and relaxation of the ampullae allow for release from the substrate. The tube feet latch on to surfaces and move in a wave, with one arm section attaching to the surface as another releases. Some multi-armed, fast-moving starfish such as the sunflower seastar (Pycnopodia helianthoides) pull themselves along with some of their arms while letting others trail behind. Other starfish turn up the tips of their arms while moving, which exposes the sensory tube feet and eyespot to external stimuli. Most starfish cannot move quickly, a typical speed being that of the leather star (Dermasterias imbricata), which can manage just 15 cm (6 in) in a minute. Some burrowing species from the genera Astropecten and Luidia have points rather than suckers on their long tube feet and are capable of much more rapid motion, "gliding" across the ocean floor. The sand star (Luidia foliolata) can travel at a speed of 2.8 m (9 ft 2 in) per minute. Sunflower starfish are quick, efficient hunters, moving at a speed of 1 m/min (3.3 ft/min) using 15,000 tube feet. Many animals temporarily change the number of legs they use for locomotion in different circumstances. For example, many quadrupedal animals switch to bipedalism to reach low-level browse on trees. The genus of Basiliscus are arboreal lizards that usually use quadrupedalism in the trees. When frightened, they can drop to water below and run across the surface on their hind limbs at about 1.5 m/s for a distance of approximately 4.5 m (15 ft) before they sink to all fours and swim. They can also sustain themselves on all fours while "water-walking" to increase the distance travelled above the surface by about 1.3 m. When cockroaches run rapidly, they rear up on their two hind legs like bipedal humans; this allows them to run at speeds up to 50 body lengths per second, equivalent to a "couple hundred miles per hour, if you scale up to the size of humans." When grazing, kangaroos use a form of pentapedalism (four legs plus the tail) but switch to hopping (bipedalism) when they wish to move at a greater speed. Bipedal ostrich Hexapedal stick-insect Octopedal locomotion by a spider Multi-legged millipede Powered cartwheeling [edit] The Moroccan flic-flac spider (Cebrennus rechenbergi) uses a series of rapid, acrobatic flic-flac movements of its legs similar to those used by gymnasts, to actively propel itself off the ground, allowing it to move both down and uphill, even at a 40 percent incline. This behaviour is different than other huntsman spiders, such as Carparachne aureoflava from the Namib Desert, which uses passive cartwheeling as a form of locomotion. The flic-flac spider can reach speeds of up to 2 m/s using forward or back flips to evade threats. Subterranean [edit] Some animals move through solids such as soil by burrowing using peristalsis, as in earthworms, or other methods. In loose solids such as sand some animals, such as the golden mole, marsupial mole, and the pink fairy armadillo, are able to move more rapidly, "swimming" through the loose substrate. Burrowing animals include moles, ground squirrels, naked mole-rats, tilefish, and mole crickets. Arboreal locomotion [edit] Main article: arboreal locomotion Arboreal locomotion is the locomotion of animals in trees. Some animals may only scale trees occasionally, while others are exclusively arboreal. These habitats pose numerous mechanical challenges to animals moving through them, leading to a variety of anatomical, behavioural and ecological consequences as well as variations throughout different species. Furthermore, many of these same principles may be applied to climbing without trees, such as on rock piles or mountains. The earliest known tetrapod with specializations that adapted it for climbing trees was Suminia, a synapsid of the late Permian, about 260 million years ago. Some invertebrate animals are exclusively arboreal in habitat, for example, the tree snail. Brachiation (from brachium, Latin for "arm") is a form of arboreal locomotion in which primates swing from tree limb to tree limb using only their arms. During brachiation, the body is alternately supported under each forelimb. This is the primary means of locomotion for the small gibbons and siamangs of southeast Asia. Some New World monkeys such as spider monkeys and muriquis are "semibrachiators" and move through the trees with a combination of leaping and brachiation. Some New World species also practice suspensory behaviors by using their prehensile tail, which acts as a fifth grasping hand. Pandas are known to swig their heads laterally as they ascend vertical surfaces astonishingly utilizing their head as a propulsive limb in an anatomical way that was thought to only be practiced by certain species of birds. Energetics [edit] Animal locomotion requires energy to overcome various forces including friction, drag, inertia and gravity, although the influence of these depends on the circumstances. In terrestrial environments, gravity must be overcome whereas the drag of air has little influence. In aqueous environments, friction (or drag) becomes the major energetic challenge with gravity being less of an influence. Remaining in the aqueous environment, animals with natural buoyancy expend little energy to maintain a vertical position in a water column. Others naturally sink, and must spend energy to remain afloat. Drag is also an energetic influence in flight, and the aerodynamically efficient body shapes of flying birds indicate how they have evolved to cope with this. Limbless organisms moving on land must energetically overcome surface friction, however, they do not usually need to expend significant energy to counteract gravity. Newton's third law of motion is widely used in the study of animal locomotion: if at rest, to move forwards an animal must push backwards against something. Terrestrial animals must push the solid ground, swimming and flying animals must push against a fluid (either water or air). The effect of forces during locomotion on the design of the skeletal system is also important, as is the interaction between locomotion and muscle physiology, in determining how the structures and effectors of locomotion enable or limit animal movement. The energetics of locomotion involves the energy expenditure by animals in moving. Energy consumed in locomotion is not available for other efforts, so animals typically have evolved to use the minimum energy possible during movement. However, in the case of certain behaviors, such as locomotion to escape a predator, performance (such as speed or maneuverability) is more crucial, and such movements may be energetically expensive. Furthermore, animals may use energetically expensive methods of locomotion when environmental conditions (such as being within a burrow) preclude other modes. The most common metric of energy use during locomotion is the net (also termed "incremental") cost of transport, defined as the amount of energy (e.g., Joules) needed above baseline metabolic rate to move a given distance. For aerobic locomotion, most animals have a nearly constant cost of transport—moving a given distance requires the same caloric expenditure, regardless of speed. This constancy is usually accomplished by changes in gait. The net cost of transport of swimming is lowest, followed by flight, with terrestrial limbed locomotion being the most expensive per unit distance. However, because of the speeds involved, flight requires the most energy per unit time. This does not mean that an animal that normally moves by running would be a more efficient swimmer; however, these comparisons assume an animal is specialized for that form of motion. Another consideration here is body mass—heavier animals, though using more total energy, require less energy per unit mass to move. Physiologists generally measure energy use by the amount of oxygen consumed, or the amount of carbon dioxide produced, in an animal's respiration. In terrestrial animals, the cost of transport is typically measured while they walk or run on a motorized treadmill, either wearing a mask to capture gas exchange or with the entire treadmill enclosed in a metabolic chamber. For small rodents, such as deer mice, the cost of transport has also been measured during voluntary wheel running. Energetics is important for explaining the evolution of foraging economic decisions in organisms; for example, a study of the African honey bee, Apis mellifera scutellata, has shown that honey bees may trade the high sucrose content of viscous nectar off for the energetic benefits of warmer, less concentrated nectar, which also reduces their consumption and flight time. Passive locomotion [edit] Passive locomotion in animals is a type of mobility in which the animal depends on their environment for transportation; such animals are vagile but not motile. Hydrozoans [edit] The Portuguese man o' war (Physalia physalis) lives at the surface of the ocean. The gas-filled bladder, or pneumatophore (sometimes called a "sail"), remains at the surface, while the remainder is submerged. Because the Portuguese man o' war has no means of propulsion, it is moved by a combination of winds, currents, and tides. The sail is equipped with a siphon. In the event of a surface attack, the sail can be deflated, allowing the organism to briefly submerge. Mollusca [edit] The violet sea-snail (Janthina janthina) uses a buoyant foam raft stabilized by amphiphilic mucins to float at the sea surface. Arachnids [edit] The wheel spider (Carparachne aureoflava) is a huntsman spider approximately 20 mm in size and native to the Namib Desert of Southern Africa. The spider escapes parasitic pompilid wasps by flipping onto its side and cartwheeling down sand dunes at speeds of up to 44 turns per second. If the spider is on a sloped dune, its rolling speed may be 1 metre per second. A spider (usually limited to individuals of a small species), or spiderling after hatching, climbs as high as it can, stands on raised legs with its abdomen pointed upwards ("tiptoeing"), and then releases several silk threads from its spinnerets into the air. These form a triangle-shaped parachute that carries the spider on updrafts of winds, where even the slightest breeze transports it. The Earth's static electric field may also provide lift in windless conditions. Insects [edit] The larva of Cicindela dorsalis, the eastern beach tiger beetle, is notable for its ability to leap into the air, loop its body into a rotating wheel and roll along the sand at a high speed using wind to propel itself. If the wind is strong enough, the larva can cover up to 60 metres (200 ft) in this manner. This remarkable ability may have evolved to help the larva escape predators such as the thynnid wasp Methocha. Members of the largest subfamily of cuckoo wasps, Chrysidinae, are generally kleptoparasites, laying their eggs in host nests, where their larvae consume the host egg or larva while it is still young. Chrysidines are distinguished from the members of other subfamilies in that most have flattened or concave lower abdomens and can curl into a defensive ball when attacked by a potential host, a process known as conglobation. Protected by hard chitin in this position, they are expelled from the nest without injury and can search for a less hostile host. Fleas can jump vertically up to 18 cm and horizontally up to 33 cm; however, although this form of locomotion is initiated by the flea, it has little control of the jump—they always jump in the same direction, with very little variation in the trajectory between individual jumps. Crustaceans [edit] Although stomatopods typically display the standard locomotion types as seen in true shrimp and lobsters, one species, Nannosquilla decemspinosa, has been observed flipping itself into a crude wheel. The species lives in shallow, sandy areas. At low tides, N. decemspinosa is often stranded by its short rear legs, which are sufficient for locomotion when the body is supported by water, but not on dry land. The mantis shrimp then performs a forward flip in an attempt to roll towards the next tide pool. N. decemspinosa has been observed to roll repeatedly for 2 m (6.6 ft), but they typically travel less than 1 m (3.3 ft). Again, the animal initiates the movement but has little control during its locomotion. Animal transport [edit] Main article: Phoresis Some animals change location because they are attached to, or reside on, another animal or moving structure. This is arguably more accurately termed "animal transport". Remoras [edit] Remoras are a family (Echeneidae) of ray-finned fish. They grow to 30–90 cm (0.98–2.95 ft) long, and their distinctive first dorsal fins take the form of a modified oval, sucker-like organ with slat-like structures that open and close to create suction and take a firm hold against the skin of larger marine animals. By sliding backward, the remora can increase the suction, or it can release itself by swimming forward. Remoras sometimes attach to small boats. They swim well on their own, with a sinuous, or curved, motion. When the remora reaches about 3 cm (1.2 in), the disc is fully formed and the remora can then attach to other animals. The remora's lower jaw projects beyond the upper, and the animal lacks a swim bladder. Some remoras associate primarily with specific host species. They are commonly found attached to sharks, manta rays, whales, turtles, and dugongs. Smaller remoras also fasten onto fish such as tuna and swordfish, and some small remoras travel in the mouths or gills of large manta rays, ocean sunfish, swordfish, and sailfish. The remora benefits by using the host as transport and protection, and also feeds on materials dropped by the host. Angler fish [edit] In some species of anglerfish, when a male finds a female, he bites into her skin, and releases an enzyme that digests the skin of his mouth and her body, fusing the pair down to the blood-vessel level. The male becomes dependent on the female host for survival by receiving nutrients via their shared circulatory system, and provides sperm to the female in return. After fusing, males increase in volume and become much larger relative to free-living males of the species. They live and remain reproductively functional as long as the female lives, and can take part in multiple spawnings. This extreme sexual dimorphism ensures, when the female is ready to spawn, she has a mate immediately available. Multiple males can be incorporated into a single individual female with up to eight males in some species, though some taxa appear to have a one male per female rule. Parasites [edit] Many parasites are transported by their hosts. For example, endoparasites such as tapeworms live in the alimentary tracts of other animals, and depend on the host's ability to move to distribute their eggs. Ectoparasites such as fleas can move around on the body of their host, but are transported much longer distances by the host's locomotion. Some ectoparasites such as lice can opportunistically hitch a ride on a fly (phoresis) and attempt to find a new host. Changes between media [edit] Some animals locomote between different media, e.g., from aquatic to aerial. This often requires different modes of locomotion in the different media and may require a distinct transitional locomotor behaviour. There are a large number of semi-aquatic animals (animals that spend part of their life cycle in water, or generally have part of their anatomy underwater). These represent the major taxa of mammals (e.g., beaver, otter, polar bear), birds (e.g., penguins, ducks), reptiles (e.g., anaconda, bog turtle, marine iguana) and amphibians (e.g., salamanders, frogs, newts). Fish [edit] Some fish use multiple modes of locomotion. Walking fish may swim freely or at other times "walk" along the ocean or river floor, but not on land (e.g., the flying gurnard—which does not actually fly—and batfishes of the family Ogcocephalidae). Amphibious fish, are fish that are able to leave water for extended periods of time. These fish use a range of terrestrial locomotory modes, such as lateral undulation, tripod-like walking (using paired fins and tail), and jumping. Many of these locomotory modes incorporate multiple combinations of pectoral, pelvic and tail fin movement. Examples include eels, mudskippers and the walking catfish. Flying fish can make powerful, self-propelled leaps out of water into air, where their long, wing-like fins enable gliding flight for considerable distances above the water's surface. This uncommon ability is a natural defence mechanism to evade predators. The flights of flying fish are typically around 50 m, though they can use updrafts at the leading edge of waves to cover distances of up to 400 m (1,300 ft). They can travel at speeds of more than 70 km/h (43 mph). Maximum altitude is 6 m (20 ft) above the surface of the sea. Some accounts have them landing on ships' decks. Marine mammals [edit] When swimming, several marine mammals such as dolphins, porpoises and pinnipeds, frequently leap above the water surface whilst maintaining horizontal locomotion. This is done for various reasons. When travelling, jumping can save dolphins and porpoises energy as there is less friction while in the air. This type of travel is known as "porpoising". Other reasons for dolphins and porpoises performing porpoising include orientation, social displays, fighting, non-verbal communication, entertainment and attempting to dislodge parasites. In pinnipeds, two types of porpoising have been identified. "High porpoising" is most often near (within 100 m) the shore and is often followed by minor course changes; this may help seals get their bearings on beaching or rafting sites. "Low porpoising" is typically observed relatively far (more than 100 m) from shore and often aborted in favour of anti-predator movements; this may be a way for seals to maximize sub-surface vigilance and thereby reduce their vulnerability to sharks Some whales raise their (entire) body vertically out of the water in a behaviour known as "breaching". Birds [edit] Some semi-aquatic birds use terrestrial locomotion, surface swimming, underwater swimming and flying (e.g., ducks, swans). Diving birds also use diving locomotion (e.g., dippers, auks). Some birds (e.g., ratites) have lost the primary locomotion of flight. The largest of these, ostriches, when being pursued by a predator, have been known to reach speeds over 70 km/h (43 mph), and can maintain a steady speed of 50 km/h (31 mph), which makes the ostrich the world's fastest two-legged animal: Ostriches can also locomote by swimming. Penguins either waddle on their feet or slide on their bellies across the snow, a movement called tobogganing, which conserves energy while moving quickly. They also jump with both feet together if they want to move more quickly or cross steep or rocky terrain. To get onto land, penguins sometimes propel themselves upwards at a great speed to leap out the water. Changes during the life-cycle [edit] An animal's mode of locomotion may change considerably during its life-cycle. Barnacles are exclusively marine and tend to live in shallow and tidal waters. They have two nektonic (active swimming) larval stages, but as adults, they are sessile (non-motile) suspension feeders. Frequently, adults are found attached to moving objects such as whales and ships, and are thereby transported (passive locomotion) around the oceans. Function [edit] Animals locomote for a variety of reasons, such as to find food, a mate, a suitable microhabitat, or to escape predators. Food procurement [edit] Animals use locomotion in a wide variety of ways to procure food. Terrestrial methods include ambush predation, social predation and grazing. Aquatic methods include filterfeeding, grazing, ram feeding, suction feeding, protrusion and pivot feeding. Other methods include parasitism and parasitoidism. Quantifying body and limb movement [edit] Main article: Study of animal locomotion The study of animal locomotion is a branch of biology that investigates and quantifies how animals move. It is an application of kinematics, used to understand how the movements of animal limbs relate to the motion of the whole animal, for instance when walking or flying. Galleries [edit] Swimming in major groups of formerly terrestrial animals Coypu (Rodentia) Frog (Anura) Sperm whales (Cetacea) Gentoo penguin (Aves) Marine iguana (Reptilia) Flight in major groups Australian Emperor dragonfly (Insecta) Magpie goose (Aves) Townsend's big-eared bat (Chiroptera) See also [edit] Animal migration Animal navigation Bird feet and legs Feather Joint Kinesis (biology) Microswimmer Movement of Animals (book) Role of skin in locomotion Sessile Taxis References [edit] ^ a b "Animal locomotion". Encyclopædia Britannica. Retrieved December 16, 2014. ^ Beckett, B. S. (1986). Biology: A Modern Introduction. 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Further reading [edit] McNeill Alexander, Robert. (2003) Principles of Animal Locomotion. Princeton University Press, Princeton, N.J. ISBN 0-691-08678-8 External links [edit] Media related to Animal locomotion at Wikimedia Commons Beetle Orientation Archived 2012-03-10 at the Wayback Machine Unified Physics Theory Explains Animals' Running, Flying And Swimming \| v t e Fins, limbs and wings \| \| Fins \| Aquatic locomotion Cephalopod fin Fish locomotion Fin and flipper locomotion Caudal fin Dorsal fin Fish fin Flipper Lobe-finned fish Ray-finned fish Pectoral fins Pelvic fin \| \| Limbs \| Limb development Limb morphology + digitigrade + plantigrade + unguligrade + uniped + biped + facultative biped + triped + quadruped Arthropod Cephalopod Tetrapod + dactyly Digit Webbed foot \| \| Wings \| Flying and gliding animals Bat wing Bird wing + keel + skeleton + feathers Insect wing Pterosaur wing Wingspan \| \| Evolution \| Evolution of fish Evolution of tetrapods Evolution of birds Origin of birds Origin of avian flight Evolution of cetaceans Comparative anatomy Convergent evolution Analogous structures Homologous structures \| \| Related \| Animal locomotion Gait Robot locomotion Samara Terrestrial locomotion Tradeoffs for locomotion in air and water Rotating locomotion Undulatory locomotion \| \| v t e Animal locomotion on land \| \| Gait class \| \| \| \| --- \| \| Legged \| Arboreal locomotion (Brachiation) Hand-walking Jumping Knuckle-walking Running Walking \| \| Legless \| Concertina movement Undulatory locomotion Rectilinear locomotion Rolling Sidewinding Other modes \| \| \| Anatomy \| Comparative foot morphology Arthropod leg \| \| \| --- \| \| Foot structure \| Digitigrade Plantigrade Unguligrade \| \| by limb \| Uniped Biped (Facultative) Triped Quadruped \| \| \| Specific \| Canine gait Horse gait Human gait \| \| Animal locomotion on the water surface Fish locomotion Volant animals \| \| v t e Physiology of muscles \| \| Exertion \| Exercise Movement + Eye movement + Gait + Locomotion \| \| Other \| \| \| \| Hand strength Muscle tone \| \| Muscle contraction + Isometric + Isotonic + Uterine contraction \| \| End-plate potential \| \| \| \| \| Authority control databases \| \| International \| \| \| National \| United States France BnF data Czech Republic Spain Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Animal locomotion Ethology Zoology Animals Hidden categories: CS1 maint: multiple names: authors list CS1: unfit URL All articles with incomplete citations Articles with incomplete citations from June 2023 CS1 maint: bot: original URL status unknown CS1 errors: periodical ignored CS1 maint: article number as page number Articles with short description Short description matches Wikidata All articles lacking reliable references Articles lacking reliable references from October 2016 Commons link from Wikidata Webarchive template wayback links Articles containing video clips Animal locomotion Add topic
12121	https://guides.library.tulsacc.edu/c.php?g=118452&p=771586	Skip to Main Content Math Help: Converting percents to decimals Refresh your competency in the various areas of Mathematics. Getting the Most Out of DIY Math Guide Integers Toggle Dropdown The number line Ordering negative numbers Number opposites Absolute value Adding and subtracting with negative numbers Multiplying with negative numbers Dividing with negative numbers Fractions Toggle Dropdown Understanding fractions Simplifying (reducing) fractions Equivalent fractions Comparing fractions Ordering fractions Mixed numbers and improper fractions Adding fractions Subtracting fractions Adding and subtracting mixed numbers Multiplying fractions Dividing fractions Multiplying and dividing mixed numbers Converting fractions to decimals Converting fractions to percents Converting between fractions, decimals and percents Ordering fractions, decimals and percents Decimals Toggle Dropdown Place value Comparing and ordering decimals Rounding to a place Significant figures Standard form and engineering notation Adding decimals Subtracting decimals Multiplying decimals Dividing decimals Converting decimals to fractions Converting decimals to percents Converting between decimals, fractions and percents Ordering decimals, fractions and percents Percents Toggle Dropdown Solving percent problems Percent increase and decrease Converting percents to fractions Converting percents to decimals Converting between percents, fractions and decimals Ordering percents, fractions and decimals Measurement Toggle Dropdown Metric units of measure Perimeter of squares, rectangles and triangles Circumference of circles Area of squares, rectangles and triangles Area of circles Surface area of cubes, cuboids and triangular prisms Surface area of cylinders Volume of cubes, cuboids and triangular prisms Volume of cylinders, cones, spheres and pyramids Statistics Toggle Dropdown Mean, median and mode Range, variation and standard deviation Algebra Toggle Dropdown Variables and expressions Order of operations Expanding expressions with brackets Exponent basics Radicals basics Evaluating expressions Evaluating a formula using substitution Solving one-step equations Solving two-step equations Solving multi-step equations Transposition Trigonometry Toggle Dropdown All about angles Similar triangles The Pythagorean Theorem Introduction to sine, cosine and tangent Find an unknown side length Find an unknown angle measure Free Mathematics Resources Calculators CONVERTING PERCENTS TO DECIMALS In this module, you can study learn how to convert a percent to a decimal number. Did you know? Changing a percentage to a decimal number is a simple as taking off the percent sign and moving the decimal point two places to the left! For example, 32.8% = .328, and 29% = .29! Study and practice Study converting percents to decimals Read about how to convert a percent to a decimal number at mathsisfun.com, then try a few problems on your own. Download a percent-to-decimal conversion worksheet Design and print a percent-to-decimal conversion worksheet at math-aids.com. (Under "Types of Conversion Problems", tick "Percent to Decimal.") << Previous: Converting percents to fractions Next: Converting between percents, fractions and decimals >> Last Updated: Aug 5, 2025 4:18 PM URL: Print Page Login to LibApps Subjects: Math & Engineering Metro Campus Library: 918.595.7172 \| Northeast Campus Library: 918.595.7501 \| Southeast Campus Library: 918.595.7701 \| West Campus Library: 918.595.8010 email: Library Website Help \| MyTCC \| © 2025 Tulsa Community College
12122	https://bookdown.org/robert_statmind/testyy2/set-theory.html	Chapter 3 Set Theory 3.1 Sets in Mathematics Intuitively, we understand the concept of sets. In dictionaries, you will find definitions like a group or collection of things that belong together or resemble one another. In mathematics, sets have a pretty precise definition: A set is a well-defined collection of objects. Each object in a set is called an element of the set. Two sets are equal if they have exactly the same elements in them. A set that contains no elements is called a null set or an empty set. If every element in Set A is also in Set B, then Set A is a subset of Set B. Example: if A is the set of first five charachters of the alphabet, then: A = {a,b,c,d,e} In R we can create this set as follows. We can decide on the name of the set. Here, we use SetA. The two characters “<” and “-” form an arrow pointing to the left. Remember that this arrow consists of two characters! We combine the characters on the right of the arrow, using the c() function We then assign this combination to SetA, on the lefthand side. Once we have defined SetA, we can check if certain elements, like “a” and “f” are elements of the set. SetA <- c("a","b","c","d","e") "a" %in% SetA ``` TRUE ``` "f" %in% ``` FALSE ``` Likewise, we can create a set of numbers, from 1 to 100. You can use arrows pointing to the right, as you see below! But it’s very uncommon to do so. Normally, we use the logic Object <- Created from, rather than Created from -> Object. So, we will do it once and never again, for illustration. c(1:100) -> SetN length(SetN) ``` 100 ``` 88 %in% # Is 88 an element of our set? [TRUE, it is!] ``` TRUE ``` 105 %in%# Is 105 an element of our set? [FALSE, it is not!] ``` FALSE ``` 3.2 Relations between Sets 3.2.1 Empty Set An empty set is a set without elements. SetEmpty <- NULL length(SetEmpty) # the length of an empty set is zero ``` 0 ``` "2" %in% # is 2 part of our empty set? ``` FALSE ``` 3.2.2 Identical Sets Apart from the sequence of elements, the two sets below are identical. All elements of A occur in B, and all elements of B are part of A. SetA <- c("1","3","4","6") SetB <- c("1","4","6","3") SetB %in% SetA ``` TRUE TRUE TRUE TRUE ``` %in% ``` TRUE TRUE TRUE TRUE ``` The overlap between two sets, is called the intersection. The intersection is graphically displayed below. All elements that are part of two sets combined (either A, B or both) form the union of the two sets. In a graph: The intersect of two sets is written as A∩B. The union of two sets is written as A∪B. A little trick to help you remember is that the ∪ symbol resembles the first character (“u”) of the word “union”. We can test the differences and similarities using the commands below. setdiff ``` character(0) ``` setequal ``` TRUE ``` intersect ``` "1" "3" "4" "6" ``` Let’s look at some other examples. The two sets below are not identical, and contain duplicates. From the result we see that duplicates are removed! The element “a” appears twice in SetA, but only once in the union of both sets. SetA <- c("a","a","b","c","d") SetB <- c("c","c","d","e") union(SetA, SetB) ``` "a" "b" "c" "d" "e" ``` There are various functions in R that help you detect unique and duplicate cases. Two of these functions are unique() and duplicated(). Applying unique() to a set, filters out any duplicates. The function duplicated() indicates, for each element, starting from the first element in the set, if it is the same as any of the preceding elements. Since the first element, by definition, has no predecessor, the indication has to be FALSE. If, like in SetA, the second element is identical to the first element, it will be TRUE. Going through the entire set, we generate a list of duplicates: elements with indicator TRUE. Note that duplication means that the same elements occurs twice or more. That is, triplication, quadruplication, and so on, are just subsets of duplication: elements appearing three times or more, are a subset of elements apperaring twice or more. We can use the result of duplicated() to index SetA. SetA has five elements (including duplicates). We can index these elements using square brackets [ ]. For example, SetA returns the first element of SetA. And SetA[3:5] returns the third to fifth elements of SetA. An alternative way for the latter, would be to use a list of TRUE and FALSE indicators, here c(F,F,T,T,T). Note that TRUE and FALSE, in this and many other cases, can be abbreviated as T and F. It is considered bad practice in programming. But you will encounter it time and again. 1 ``` "a" ``` 3: 5 ``` "b" "c" "d" ``` c# Equivalent to previous! ``` "b" "c" "d" ``` Since the duplicated() function produces a list of TRUE and FALSE values, we can use it to index SetA. However, duplicated() returns TRUE for duplicated elements. If we want to detect unique elements, we use the NOT operator (the exclamation mark, !) to change the TRUE indicators to FALSE, and vice versa. (SetA) ``` "a" "a" "b" "c" "d" ``` unique # Removes duplicates ``` "a" "b" "c" "d" ``` duplicated ``` FALSE TRUE FALSE FALSE FALSE ``` ! duplicated# Removes duplicates, in a slightly more complicated way ``` "a" "b" "c" "d" ``` Differences between sets can be checked using the setdiff() function. The function only compares two sets at the time, and the order is important, as you see below! setdiff# a en b occur in A, but not in B ``` "a" "b" ``` setdiff# e occurs in B, but not in A ``` "e" ``` 3.2.3 Advanced To challenge your skills, we can define the union of sets A and B as follows: Elements unique to A, plus: Elements unique to B, plus: Elements in the interesection of A and B. We can sort the combination of these three parts alphabetically, and check if indeed the result is equal to what we have defined as the union. The all() function can be used to see if all elements of one set, are contained by the other. sort c setdiff setdiff intersect ``` "a" "b" "c" "d" "e" ``` all sort c setdiff setdiff intersect == union ``` TRUE ``` An alternative to the above code, is to first create an object (say, combi) as the sorted elements of the combined parts. Then, we compare combi to the result of the union() function. It makes our code slightly more readable. Adding some comments (text after “#”), always helps making your code more readable. ``` union, as combination of unique and intersecting elements combi <- sort(c(setdiff(SetA, SetB), setdiff(SetB, SetA), intersect(SetB, SetA))) all(combi == union(SetA, SetB)) # check if combi equals the result of union() ``` ``` TRUE ``` Let’s see what happens if duplicated elements of one set, are checked against another set. All elements of set C (duplicated elements “a”) are part of D. One element of D, is not contained in A. SetC <- c("a","a"); SetD <- c("a","b") SetC %in% SetD ``` TRUE TRUE ``` all %in% ``` TRUE ``` %in% ``` TRUE FALSE ``` all %in% ``` FALSE ``` This is what we would expect: “a”, though a duplicated element in setC, occurs in SetD (where “a” only occurs once). 3.3 Sets of Numbers Sets can contain any type of element: characters, binary indicators (like TRUE or FALSE), numbers, and more. It can be tedious to create sets of characters or words, as you have to do a lot of typing. Sets of numbers are often easier, especially if it’s a line of numbers, say from 1, 2 .. to 99. You can use the length() function to return the number of elements (here: numbers). You can access specific elements using square brackets (indexing). Note that curved brackets () do not work, for indexing. Computers can be very picky. c - 10:+ 10# Putting the expression between brackets, prints the object in the console! ``` -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 ``` length ``` 21 ``` 3 ``` -8 ``` 3< 4 ``` TRUE ``` 3.4 Application of Set Theory, and Sets You may wonder why all of the above would be of interest to a data scientist like yourself. There are a least three reasons. Logical thinking First of all, dealing with data and data analysis will force you to think logically. Even when working with just one data set, say, a (partial) list of employees in your organization who have filled out survey questionnaire, requires you to first assess the quality of the list. How many employees have filled out the survey twice or more (and what do we do with duplicates, especially if responses by duplicate respondents are inconsistent)? Is the list of (unique) respondents a subset of employees? Is the response rate the same for all departments of the organization? And so on and so forth. Answers to such questions can be obtained using the tools discussed above, or any out of many alternative tools. But it is the thinking that counts! Comparing data sets One of the starting points in the data mining process, is data understanding. Often, your analysis is based on data from various sources. Before you start your analysis, you have to understand all sources in depth. Combining, or merging, data sets from various sources almost without exception reveals errors, inconsistencies and gaps. Applying the basics of set theory will help you in preparing high-quality, analyzable data sets. Or to put it negatively: data science follows the garbage-in-garbage-out principle! Probability theory In some of the modules, we will use probability theory. Concepts like conditional probablitity have close links to set theory. It helps if you’re familiar with the terminologies, and with notations like ∪ and ∩. 3.5 Exercises 3.5.1 Exercise 1 Given are the following sets: A = {2,4,6,8,10,12,14,16} B = {1,2,3,4,5,6,7} C = {3,6,9,12,15,18} A ∩ B = (Intersection of A and B) A ∩ C = B ∩ C = A ∩ B ∩ C = A ∪ B = (Union of A and B) A ∪ B ∪ C = 3.5.1.1 Solutions Let’s first create the sets A, B and C. We use the seq() function in R. You can find documentation on this function here. For a sequence of numbers that runs from 2 to 16, in steps of 2, like A, we type seq(2,16,2). B can be defined as seq(1,7,1), but c(1:7) is shorter. B <- 1:7 would work as well. A <- seq(2, 16, 2) B <- c(1:7) C <- seq(3, 18, 3) A; B; C # Prints the three sets ``` 2 4 6 8 10 12 14 16 ``` ``` 1 2 3 4 5 6 7 ``` ``` 3 6 9 12 15 18 ``` A ∩ B intersect# Putting the command between brackets, prints the result ``` 2 4 6 ``` A ∩ C intersect ``` 6 12 ``` B ∩ C intersect ``` 3 6 ``` A ∩ B ∩ C intersect(A, B, C) will give an error message, as intersect() only works on two sets! We can do it in 2 steps; first, we determine the intersection of A and B; followed by the intersection of that result with C! You can imagine that this is doable for three sets. But if we were to find the intersection between many sets, then the expression would get very lengthy! A better alternative is to use the Reduce() function. Often, you can find solutions to this kind of challenges by just googling. The Reduce() function, for example, we found on link intersect intersect ``` 6 ``` Reduce list ``` 6 ``` A ∪ B sort union ``` 1 2 3 4 5 6 7 8 10 12 14 16 ``` A ∪ B ∪ C sort Reduce list ``` 1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 ``` 3.5.1.2 Are you in for a challenge? Which elements are in intersections of sets, but not in all sets? intersect# Pairwise intersection A and B; stored as set AB ``` 2 4 6 ``` intersect# Id, for A and C ``` 6 12 ``` intersect# Id, for B and C ``` 3 6 ``` sort Reduce list# All pairwise intersection combined, stored as "pairs" ``` 2 3 4 6 12 ``` Reduce list # Elements present in the intersection of all three sets ``` 6 ``` sort setdiff# Elements in pairs, but not in trios ``` 2 3 4 12 ``` You see that elements 2, 3, 4 and 12 are in pairwise intersections, but not in the three-way intersection! Graphically, just to convince you: 3.5.2 Exercise 2 Are the following statements true or false? {2,4} ⊂ (A ∩ B) 6 ∈ (A ∩ B ∩ C) {6} ⊂ (A ∩ B) (A ∩ B) ⊂ (A ∪ B) 3.5.2.1 Solutions {2,4} ⊂ (A ∩ B) SetTest <- c(2,4) AB <- intersect(A,B) all(SetTest %in% AB) ``` TRUE ``` 6 ∈ (A ∩ B ∩ C) See below, under c. {6} ⊂ (A ∩ B) ``` In the formulation of c. SetTest <- c(6) trios ``` ``` 6 ``` all %in% # Remember we stored the intersect as "trios"! ``` TRUE ``` ``` Alternatively (as in b.), define 6 as a single element rather than a set of one element 6 %in% trios ``` ``` TRUE ``` ``` Note that b. and c. are basically the same! ``` (A ∩ B) ⊂ (A ∪ B) This is true in general. Elements in the intersection of A and B, are by definition part of both A and B. The intersection is therefore a subset of all elements in A or B! Just to train the formulation of this exercise in R: all intersect %in% union ``` TRUE ``` 3.5.3 Exercise 3 A = {1,2,3,4,5,6} B = {5,6} C = {1,2,5,6} D = {2,3,4} E = {2,3,4,5} Complete the statement with one of the symbols: ∈ (element of) ∉ (not an element of) ⊂ (subset of) ⊃ (superset; if A is subset of B, then B is superset of A) ∩ (intersection) ∪ B….C B….C = B B….C = C B….D = 0 (empty set) C….D = A D….E = D 4….C ∩ B D….E….A 3.5.3.1 Solutions Use reason to answer each of the questions! As an additional challenge, formulate the sets in R, and use any of the commands introduced in this chapter to check your answer! ⊂ (B is obviously a subset of C, as all elements of B are also in C) ∩ (the intersect of B and C, is equal to B, as B is a subset of B) ∪ (the combined elements of B and C, are equal to C, as C is a superset of B) ∩ (as B and D have no elements in common, the interesect is the empty set) ∪ (all elements of C and D combined, match A) ∩ or ⊂ ∉ (4 is not part of the intersection of B and C; it is not even part of the union of B and C) ⊂ (D is a subset of E, which in turn is a subset of A; you can conclude that therefore D is a subset of A) Working with data files (in data science) requires logical thinking. Set theory is a good exercise in logical thinking! 3.5.4 Exercise 4 Consider the following sets. A = {x \| x is an even number and x<20; x is a positive natural number} B = {x \| x is a multiple of 3 and x<20; x is a positive natural number} C = {x \| x is an odd numbber and x<20; x is a positive natural number} This looks cryptic. Set A, for example, reads like the set of numbers x conditioned by the following rules. x is: a positive natural number (1, 2, 3 to infinity) smaller than 20, and divisible by 2. We can enumerate these numbers easily: 2, 4, 6 .. up to 18. Set B then is 3, 6, 9 .. up to 18. And Set C is 1, 3, 5 .. up to 19. A ∩ B A ∩ B ∩ C A ∪ B ∪ C Again, create the sets in R and use the proper functions to get the solutions! 3.5.4.1 Solutions {6, 12, 18} 0 (empty set) {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19} seq 2 18 2 ``` 2 4 6 8 10 12 14 16 18 ``` seq 3 18 3 ``` 3 6 9 12 15 18 ``` seq 1 19 2 ``` 1 3 5 7 9 11 13 15 17 19 ``` intersect # Answer a ``` 6 12 18 ``` Reduce list # Answer B ``` numeric(0) ``` Reduce list # Answer C ``` 2 4 6 8 10 12 14 16 18 3 9 15 1 5 7 11 13 17 19 ```
12123	https://us.sofatutor.com/math/videos/the-definition-of-negative-exponents	Try sofatutor for 30 Days Do you already have an account? Log in: Try sofatutor for 30 Days Discover why over 1.6 MILLION students choose sofatutor! Math Middle School Exponential Notation and Properties of Integer Exponents The Definition of Negative Exponents The Definition of Negative Exponents Watch videos Start exercises Show worksheets Content The Definition of Negative Exponents Negative Exponents – Introduction Negative Exponents – Explanation The Laws of Exponents Understanding Negative Exponents with Expansion Negative Exponents – Summary Definition of Negative Exponents – Frequently Asked Questions Improve your grades nowwhile having fun Unlock this video in just a few steps, and benefit from all sofatutor content: Try it for 30 Days Rating Ø 5.0 / 1 ratings You must be logged in to be able to give a rating. Wow, thank you!Please rate us on Google too! We look forward to it! Go to Google The authors Chris S. Basics on the topic The Definition of Negative Exponents Negative Exponents – Introduction Negative exponents are an essential concept in algebra, often encountered in scientific notation, algebraic expressions, and real-world scenarios such as calculating decay rates or understanding exponential growth in finance. A negative exponent indicates repeated division, contrary to a positive exponent, which denotes repeated multiplication. A negative exponent expresses how many times the reciprocal of the base is multiplied by itself. For instance, $a^{-n}$ means $1$ divided by $a$ multiplied by itself $n$ times. Negative Exponents – Explanation The shift from positive to negative exponents represents a transition from multiplication to division in terms of how we handle the base number. With positive exponents, like $a^{3}$, we multiply the base ($a$) by itself a certain number of times (three times here). However, with negative exponents, such as $a^{-3}$, it indicates division. Instead of multiplying the base, we divide $1$ by the base multiple times. This reflects the inverse relationship - while positive exponents grow the number, negative exponents reduce it reciprocally. Example: Positive Exponent: $2^{3} = 2 \times 2 \times 2 = 8$ Negative Exponent: $2^{-3} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ Practice converting negative exponents to positive exponents. Convert the expression with a negative exponent into an expression with a positive exponent: $a^{-4}$ $\frac{1}{a^{4}}$ Simplify the expression with a negative exponent: $b^{-3}$ $\frac{1}{b^{3}}$ Rewrite the following with a positive exponent: $c^{-5}$ $\frac{1}{c^{5}}$ Change the negative exponent to a positive exponent: $d^{-2}$ $\frac{1}{d^{2}}$ The Laws of Exponents To understand the definition of negative exponents, it is important to understand the proofs of laws of exponents. Here, you will see all the important laws of exponents, including the law of negative exponents! \| Law \| Mathematical Representation \| Description \| --- \| Product of Powers \| $a^{m} \times a^{n} = a^{m+n}$ \| When multiplying like bases, add the exponents. \| \| Quotient of Powers \| $\frac{a^{m}}{a^{n}} = a^{m-n}$ \| When dividing like bases, subtract the exponents. \| \| Power of a Power \| $(a^{m})^{n} = a^{mn}$ \| When raising a power to a power, multiply the exponents. \| \| Zero Exponent \| $a^{0} = 1$ (where $a \neq 0$) \| Any non-zero number raised to the power of zero is one. \| \| Negative Exponent \| $a^{-n} = \frac{1}{a^{n}}$ \| A negative exponent indicates the reciprocal of the base raised to the positive exponent. \| Let’s practice applying these laws of exponents and specifically involving negative exponents. Product of Powers: Simplify $2^{-3} \times 2^{-2}$. Add the exponents: $2^{-3} \times 2^{-2} = 2^{(-3)+(-2)} = 2^{-5}$. Convert to a positive reciprocal: $2^{-5} = \frac{1}{2^{5}} = \frac{1}{32}$. Quotient of Powers: Simplify $\frac{3^{-4}}{3^{-2}}$. Subtract the exponents: $\frac{3^{-4}}{3^{-2}} = 3^{-4-(-2)} = 3^{-2}$. Convert to a positive reciprocal: $3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}$. Power of a Power: Simplify $(5^{-2})^{3}$. Multiply the exponents: $(5^{-2})^{3} = 5^{-2 \times 3} = 5^{-6}$. Convert to a positive reciprocal: $5^{-6} = \frac{1}{5^{6}}$. Try some on your own! Simplify $2^{-4} \times 2^{-3}$ $2^{-4} \times 2^{-3} = 2^{-7} = \frac{1}{2^{7}$ Simplify $\frac{4^{-3}}{4^{-1}}$ $\frac{4^{-3}}{4^{-1}} = 4^{-2} = \frac{1}{4^{2}}$ Simplify $(3^{-1})^{4}$ $(3^{-1})^{4} = 3^{-4} = \frac{1}{3^{4}}$ Simplify $5^{-2} \times 5^{-3}$ $5^{-2} \times 5^{-3} = 5^{-5} = \frac{1}{5^{5}}$ Simplify $\frac{6^{-4}}{6^{-2}}$ $\frac{6^{-4}}{6^{-2}} = 6^{-2} = \frac{1}{6^{2}}$ Simplify $(7^{-1})^{3}$ $(7^{-1})^{3} = 7^{-3} = \frac{1}{7^{3}}$ Understanding Negative Exponents with Expansion Expanding an expression written in exponential notation can help us better visualize how negative exponents work. Expand and simplify the expression: $\dfrac{5^{3}}{5^{7}}$ Expand Both Exponents: Numerator $5^{3}$ as $5 \times 5 \times 5$ Denominator $5^{7}$ as $5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5$ Divide Step by Step: Each $5$ in the numerator cancels out a $5$ in the denominator. Left with $\frac{1}{5 \times 5 \times 5 \times 5}$ (four $5$'s remain in the denominator). Result After Cancelation: $\dfrac{1}{5^{4}}$ Practice some expanding and working with negative exponents on your own Expand and simplify the expression: $\frac{2^{4}}{2^{6}}$ Expand both exponents: Numerator $2^{4}$ as $2 \times 2 \times 2 \times 2$, Denominator $2^{6}$ as $2 \times 2 \times 2 \times 2 \times 2 \times 2$. Divide step by step: Each $2$ in the numerator cancels out a $2$ in the denominator, leaving $\frac{1}{2 \times 2}$. Result after cancelation: $\frac{1}{2^{2}} = \frac{1}{4}$. Expand and simplify the expression: $\frac{3^{2}}{3^{5}}$ Expand both exponents: Numerator $3^{2}$ as $3 \times 3$, Denominator $3^{5}$ as $3 \times 3 \times 3 \times 3 \times 3$. Divide step by step: Each $3$ in the numerator cancels out a $3$ in the denominator, leaving $\frac{1}{3 \times 3 \times 3}$. Result after cancelation: $\frac{1}{3^{3}} = \frac{1}{27}$. Expand and simplify the expression: $\frac{4^{5}}{4^{7}}$ Expand both exponents: Numerator $4^{5}$ as $4 \times 4 \times 4 \times 4 \times 4$, Denominator $4^{7}$ as $4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4$. Divide step by step: Each $4$ in the numerator cancels out a $4$ in the denominator, leaving $\frac{1}{4 \cdot 4}$. Result after cancelation: $\frac{1}{4^{2}} = \frac{1}{16}$. Negative Exponents – Summary Key Learnings from this Text: Negative exponents represent reciprocal values and repeated division. Understanding and applying exponent rules is crucial for simplifying expressions. Expanding negative exponents helps to understand their meaning visually. Converting negative exponents to positive exponents simplifies the expression. Definition of Negative Exponents – Frequently Asked Questions What is a negative exponent? A negative exponent indicates the reciprocal of the base raised to the absolute value of the exponent. How do you simplify expressions with negative exponents? To simplify, convert the negative exponent into a positive one by taking the reciprocal of the base. Can you give an example of using negative exponents in real life? Yes, negative exponents are often used in scientific notation to represent very small numbers. What is the quotient of exponents rule? The quotient of exponents rule states that $\frac{a^{m}}{a^{n}} = a^{m-n}$. How do you multiply terms with exponents? When multiplying terms with the same base, add their exponents: $a^{m} \times a^{n} = a^{m+n}$. What does the zero exponent rule state? The zero exponent rule states that any non-zero base raised to the power of zero equals one: $a^{0} = 1$. Is there a rule for dividing terms with exponents? Yes, the rule is similar to multiplication: $\frac{a^{m}}{a^{n}} = a^{m-n}$. How do negative exponents relate to division? Negative exponents represent division, as they show how many times the reciprocal of the base is multiplied by itself. Can negative exponents be used in scientific notation? Yes, they are frequently used in scientific notation to express very small numbers. Transcript The Definition of Negative Exponents Wheezy, Macaroni, and Pipsqueak are starting to notice a disturbing trend. The iceberg where they love to play keeps breaking in half! It's a repeated-division catastrophe! Fortunately, Professor Penguin is on hand to explain using the definition of negative exponents. Let's take a closer look at how the iceberg is breaking apart. We started with one iceberg. Dividing this iceberg in half gives use two icebergs that are one-half the size of the original. But the icebergs continue to split in half. One-half of the original, divided by two, gives us one-fourth. One-fourth divided by two gives us one-eighth. So after 3 rounds of splitting in half, each mini-iceberg is one-eighth the size of the original. How small would each berg be after 6 rounds of splitting in half? We can answer this question by looking at exponents and division. You should already be familiar with the Law of Quotients of Powers, which tells us how to divide numbers written in exponential notation, when they both have the same base. For example, how would you simplify 2 to the 5th over 2 squared? When dividing numbers with the same base, we subtract the exponents. So this expression can be simplified to 2 cubed. But let's take a look at what happens if we have a larger exponent in our denominator than in our numerator. To simplify 2 to the fifth divided by 2 to the eighth we subtract the exponents and are left with 2 to the negative third power. But what does that mean? If we write our numerator and denominator in their factored forms and then cancel we can see that 1 over 2 to the third power is the same as 2 to the negative third power. In general, this is how we define negative exponents: x' to the negative 'n' equals 1 over 'x' to the 'n'. So while positive exponents represent repeated multiplication, negative exponents represent repeated division. This will come in handy, since we always want to write numbers without a negative exponent when simplifying. So let's go back to our iceberg. Originally, we had one whole iceberg. After one round of dividing in half, one mini-berg is half the size of the original, which we can write as 2 to the negative first power. After two rounds of dividing, that's 1 over 2 squared, or 2 to the negative second power. After 3 rounds it's one over 2 to the third, or 2 to the negative 3. We want to know how small are the mini-bergs after dividing by 2, six times. How could we write that using negative exponents? Well we're dividing by two over and over again, so that number is our base. And we're going to divide six times, so our exponent is negative 6. Using the definition of negative exponents we can write this as a fraction and see that each mini-berg is one sixty-fourth of the original. “Let's look at three quick examples to see how negative exponents apply to what you already know about the Laws of Exponents." Our first example uses the Product of Powers Law. Four to the negative 3rd times four to the negative 5th. We are multiplying numbers with the same base, so what should do with the exponents? Adding the exponents, gives us 4 to the negative eighth power. We don't want to leave this number with a negative exponent, so we use the definition of negative exponents to simplify to the fraction one over 4 to the 8th. Now an example of the quotient of Powers Law. Three to the negative 1st over three to negative the 7th. Here we are dividing numbers with the same base, so what should we do with the exponents? The Quotient of Powers Law tells us to subtract the exponents, so negative 1 minus negative 7 is equal to negative 1 plus 7. Leaving us with 3 to the 6th power. Our third and final example is a number raised to a power, then raised to a power again. What does the Powers of Powers Law tell us to do with the exponents in this case? A power raised to a power means we multiply exponents. Negative 3 times 2 is negative 6. Because we don't want to leave this number with a negative exponent, we rewrite eight to the negative 6th as one over 8 to the 6th power. Let's review. While positive exponents are how we represent repeated multiplication negative exponents represent repeated division. In general, we define a negative exponent as 'x' to the negative 'n' is equal to one over 'x' to the 'n'. And remember, the laws of exponents apply to all numbers written in exponential notation. This includes Product of Powers Quotient of Powers and Powers of Powers. Uh oh, it looks Professor Penguin got a bit carried away with splitting those icebergs in half. I think we're gonna need another iceberg! The Definition of Negative Exponents exercise Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video The Definition of Negative Exponents. Simplify the powers. Hints When we multiply exponents with the same base we use the product of powers rule. For example, $5^3\times5^2 = 5^{3+2} = 5^5$ We add the exponents. When we divide exponents with the same base we use the quotient of powers rule. For example, $\dfrac{7^6}{7^4} = 7^{6-4} = 7^2$ We subtract the exponents. When subtracting negatives we can use a number line. For example $-3 - (+2) = -3 - 2 = -5$ Solution $3^4\times3^2 = 3^{4+2} = 3^6$${}$ $3^7\times3^{-2} = 3^{7-2} = 3^5$${}$ $\dfrac{3^8}{3^4} = 3^{8-4} = 3^4$${}$ $\dfrac{3^{-2}}{3^4} = 3^{-2-+4} = 3^{-2-4} = 3^{-6}$ Simplify the negative exponent. Hints The rule we are using states that a negative power is the same as $1$ over the positive power. If a negative power is equal to $1$ over the positive power then $8^{-5} = \dfrac{1}{8^5}$ Solution $9^{-3} = \dfrac{1}{9^3}$ $9^{-2} = \dfrac{1}{9^2}$ $9^{-1} = \dfrac{1}{9}$ $9^{-4} = \dfrac{1}{9^4}$ ### Calculate the product of negative powers. Hints For the product of powers we add the powers when the base is the same. Example shown. If we have added the powers and we get a negative power. We know that a number to the negative power is equal to one over the number to the positive power. This should be done to simplify further. Solution $\mathbf{\frac{1}{3^6}}$ We use the product of powers rule, adding the powers Negative power is $1$ over the positive power Calculate the quotient of negative powers. Hints When subtracting negative numbers subtracting a negative means to add. For example $-7 - (-3) = -7 + 3 = -4$ For the quotient of powers we subtract the powers when the base is the same. Example shown. When we have subtracted the powers, we get a negative power. We know that a number to the negative power is equal to one over the number to the positive power. This should be done to simplify further. Solution $\dfrac{1}{5^4}$ We use the quotient of powers rule, subtracting the powers Subtracting a negative is the same as adding Negative power is $1$ over the positive power Calculate the quotient of negative powers. Hints To work this out we subtract the powers which gives us a negative power. We then use the definition of negative exponents. A number to a negative power is equal to $1$ over the number with a positive power. See the example. If we had $\dfrac{8^2}{8^7}$ we subtract the powers first. $2 - 7 = -5$ Our answer is $8^{-5} = \dfrac{1}{8^5}$. We can see the example where we use repeated division to show this. Solution Answer $\dfrac{1}{5^3}$ We subtract the powers first. $3 - 6 = -3$ Our answer is $5^{-3}$ By using repeated division shown = $\dfrac{1}{5^3}$. Calculate the power of negative powers. Hints A positive number $\times$ negative number $=$ negative answer. For example $8\times(-2) = -16$. For the power of powers we multiply the powers when the base is the same. Example shown. If we have multiplied the powers and the answer is a negative power. We know that a number to the negative power is equal to one over the number to the positive power. This should be done to simplify further. Solution We use the power of powers rule, multiplying the powers positive $\times$ negative $=$ negative Negative power is $1$ over the positive power More videos and learning texts for the topic Exponential Notation and Properties of Integer Exponents Exponential Notation Numbers Raised to the Zeroth Power Proofs of Laws of Exponents Product and Quotient of Powers Laws Numbers in Exponential Form Raised to a Power The Definition of Negative Exponents Exponents with Negative Bases Square Roots Combining Opposite Quantities to make 0 Company Our Team Prices Jobs Platform My Account Enter code Proven learning success Gamification How it works Tutorial Videos Exercises Sofahero Worksheets Interactive books Help FAQ Give Us Feedback Legal Terms & Conditions Privacy Policy Cookie Details Cookie Preferences Contact Us Do Not Sell My Personal Information Any questions? Contact us! sofatutor.com sofatutor.ch sofatutor.at sofatutor.com sofatutor.co.uk Any questions? 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12124	https://www.cuemath.com/calculators/degrees-to-radians-calculator/	Degrees to Radians calculator A degree is a unit to measure angles. We use degree in place of radians when we want to calculate the angle. What is Degrees to Radians calculator? 'Cuemath's Degrees to Radians calculator' is an online tool that helps to convert degrees to radians for a given value. Cuemath's online Degrees to Radians calculator helps you to convert degrees to radians in a few seconds. Note: Enter the input up to 5 digits. How to Use Degrees to Radians calculator? Please follow the steps below on how to use the calculator: How to Convert Degrees to Radians? Angles are measured in degrees. One revolution is divided into 360 equal parts and each part is called a degree. The radian is an S.I. unit that is used to measure angles. One radian is the angle made at the center of a circle by an arc whose length is equal to the radius of the circle. To convert an angle given in radians to degrees, we just need to multiply the angle by π / 180°. Angle in radians = π / 180° (Angle in degrees) Book a Free Trial Class Solved Example: Convert 45 degrees to radians. Solution: Angle in degree = π / 180° To convert 45 degrees to radians, 45 × π / 180° = 0.7763 radians Similarly, you can use the calculator to convert degrees to radians for the following:
12125	https://brainly.com/question/44226322	[FREE] Why is \pi radians equal to 180 degrees? - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +34,3k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +31,3k Ace exams faster, with practice that adapts to you Practice Worksheets +7,2k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Why is π radians equal to 180 degrees? 1 See answer Explain with Learning Companion NEW Asked by tylerkrobinson3668 • 12/10/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 6367285 people 6M 0.0 0 Upload your school material for a more relevant answer Pi radians is equal to 180 degrees because there are 360 degrees in a circle or one revolution. Explanation Pi radians is equal to 180 degrees because there are 360 degrees in a circle or one revolution. To understand this relationship, consider a circle. If we divide the circumference of a circle into 360 equal arcs, then each arc will correspond to one degree. Similarly, if we divide the circumference of a circle into 2π equal arcs (radians), then each arc will correspond to one radian. Since there are 2π radians in a circle, which is equal to 360 degrees, we can say that pi radians is equal to 180 degrees. For example, if A0 is a point on a CD that has made one complete revolution, every point on the CD is back at its original position. Hence, A0 = 2π rad = 360°. Answered by DominikBryan •24.7K answers•6.4M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 6367285 people 6M 0.0 0 College Physics 1e - OpenStax Physics - Boundless Physics - Paul Peter Urone, Roger Hinrichs Upload your school material for a more relevant answer π radians is equal to 180 degrees because a full circle contains 2 π radians, which corresponds to 360 degrees. By dividing, we find that π radians is equivalent to 180 degrees. This relationship is fundamental in understanding angle measurements in mathematics. Explanation The reason that π radians is equal to 180 degrees is rooted in the relationship between degrees and radians as units for measuring angles in a circle. Understanding a Circle: A full circle is traditionally divided into 360 degrees. This division is an arbitrary yet convenient way to measure angles. Radian Definition: A radian is defined based on the radius of a circle. Specifically, one radian is the angle formed when the length of the arc of a circle is equal to the radius of that circle. Total Radians in a Circle: Since a circle has a circumference of 2 π r, where r is the radius, the total number of radians in a complete circle is obtained by dividing the circumference by the radius: Total Radians=r 2 π r=2 π radians. Conversion to Degrees: Since there are 360 degrees in a circle and 2 π radians, we can set up the relationship: 2 π radians=360 degrees. Finding the Value of π: To find out how many degrees equal π radians, we just divide both sides of the equation by 2: π radians=2 360=180 degrees. Thus, π radians corresponds to 180 degrees, meaning one half of a circle in degrees translates to π radians in radians. This conversion is essential for various applications in trigonometry and calculus. Examples & Evidence For example, if an angle measures 90 degrees, it can be converted to radians as follows: 90 degrees=360 90×2 π=2 πradians. Similarly, a 45-degree angle equals 4 πradians, illustrating the conversion between the two units of measurement. The relationship between radians and degrees is defined by the fact that a complete revolution has both 2 π radians and 360 degrees. This conversion is widely accepted and taught in geometry and trigonometry courses. Thanks 0 0.0 (0 votes) Advertisement tylerkrobinson3668 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 8 What is the radian equivalent for 180°? 2 pi radians pi radians startfraction pi over 2 endfraction radians Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics The approximate antler length L (in inches) of a deer buck can be modeled by L=9 3 t+15 where t is the age in years of the buck. If a buck has an antler length of 36 inches, what is its age? Find three consecutive odd integers such that the sum of the largest and twice the smallest is 25. If x represents the smallest integer, then which equation could be used to solve the problem? A. 2 x+x+2=25 B. 2 x+2 x+4=25 C. 2 x+x+4=25 The greater of two consecutive integers is 7 greater than twice the smaller. Find the integers. Cholesterol levels (in m g/d L) were collected from a random sample of 24 patients two days after they had a heart attack. \| Cholesterol Levels (in mg/dL) \| \| 186 \| \| 224 \| \| 280 \| \| 236 \| \| 142 \| \| 226 \| \| 244 \| \| 282 \| \| 236 \| \| 220 \| \| 278 \| \| 272 \| \| 234 \| \| 276 \| \| 310 \| \| 288 \| \| 242 \| \| 280 \| \| 294 \| \| 282 \| \| 160 \| \| 288 \| \| 206 \| \| 266 \| For the data shown above, find the following. Do not round any of your answers. a) Find the 5-number summary: Evaluate 1 F 4+9 C 2 in base Sixteen. 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12126	https://pressbooks.bccampus.ca/introductorygeneralphysics1phys1107/chapter/1-7-significant-figures-originally-from-openstax-college-chemistry-1st-canadian-edition/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Chapter 1 The Nature of Science and Physics 1.7 Significant Figures (Originally from OpenStax College Chemistry 1st Canadian Edition) Learning Objectives Apply the concept of significant figures to limit a measurement to the proper number of digits. Recognize the number of significant figures in a given quantity. Limit mathematical results to the proper number of significant figures. If you use a calculator to evaluate the expression 337/217, you will get the following: 337217=1.55299539171… and so on for many more digits. Although this answer is correct, it is somewhat presumptuous. You start with two values that each have three digits, and the answer has twelve digits? That does not make much sense from a strict numerical point of view. Consider using a ruler to measure the width of an object, as shown in Figure 2.6 “Expressing Width”. The object is definitely more than 1 cm long, so we know that the first digit in our measurement is 1. We see by counting the tick marks on the ruler that the object is at least three ticks after the 1. If each tick represents 0.1 cm, then we know the object is at least 1.3 cm wide. But our ruler does not have any more ticks between the 0.3 and the 0.4 marks, so we can’t know exactly how much the next decimal place is. But with a practiced eye we can estimate it. Let us estimate it as about six-tenths of the way between the third and fourth tick marks, which estimates our hundredths place as 6, so we identify a measurement of 1.36 cm for the width of the object. What is the proper way to express the width of this object? Does it make any sense to try to report a thousandths place for the measurement? No, it doesn’t; we are not exactly sure of the hundredths place (after all, it was an estimate only), so it would be fruitless to estimate a thousandths place. Our best measurement, then, stops at the hundredths place, and we report 1.36 cm as proper measurement. This concept of reporting the proper number of digits in a measurement or a calculation is called significant figures. Significant figures (sometimes called significant digits) represent the limits of what values of a measurement or a calculation we are sure of. The convention for a measurement is that the quantity reported should be all known values and the first estimated value. The conventions for calculations are discussed as follows. Example 3 Use each diagram to report a measurement to the proper number of significant figures. Solution The arrow is between 4.0 and 5.0, so the measurement is at least 4.0. The arrow is between the third and fourth small tick marks, so it’s at least 0.3. We will have to estimate the last place. It looks like about one-third of the way across the space, so let us estimate the hundredths place as 3. Combining the digits, we have a measurement of 4.33 psi (psi stands for “pounds per square inch” and is a unit of pressure, like air in a tire). We say that the measurement is reported to three significant figures. The rectangle is at least 1.0 cm wide but certainly not 2.0 cm wide, so the first significant digit is 1. The rectangle’s width is past the second tick mark but not the third; if each tick mark represents 0.1, then the rectangle is at least 0.2 in the next significant digit. We have to estimate the next place because there are no markings to guide us. It appears to be about halfway between 0.2 and 0.3, so we will estimate the next place to be a 5. Thus, the measured width of the rectangle is 1.25 cm. Again, the measurement is reported to three significant figures. Check Your Understanding 1 What would be the reported width of this rectangle? In many cases, you will be given a measurement. How can you tell by looking what digits are significant? For example, the reported population of the United States is 306,000,000. Does that mean that it is exactly three hundred six million or is some estimation occurring? The following conventions dictate which numbers in a reported measurement are significant and which are not significant: Any nonzero digit is significant. Any zeros between nonzero digits (i.e., embedded zeros) are significant. Zeros at the end of a number without a decimal point (i.e., trailing zeros) are not significant; they serve only to put the significant digits in the correct positions. However, zeros at the end of any number with a decimal point are significant. Zeros at the beginning of a decimal number (i.e., leading zeros) are not significant; again, they serve only to put the significant digits in the correct positions. So, by these rules, the population figure of the United States has only three significant figures: the 3, the 6, and the zero between them. The remaining six zeros simply put the 306 in the millions position. Example 4 Give the number of significant figures in each measurement. 36.7 m 0.006606 s 2,002 kg 306,490,000 people Solution By rule 1, all nonzero digits are significant, so this measurement has three significant figures. By rule 4, the first three zeros are not significant, but by rule 2 the zero between the sixes is; therefore, this number has four significant figures. By rule 2, the two zeros between the twos are significant, so this measurement has four significant figures. The four trailing zeros in the number are not significant, but the other five numbers are, so this number has five significant figures. Check Your Understanding 2 Give the number of significant figures in each measurement. 0.000601 m 65.080 kg How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column. We drop the last digit—the 1—because it is not significant to the final answer. The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater and rounded down if the first dropped digit is less than 5. Example 5 Express the final answer to the proper number of significant figures. 101.2 + 18.702 = ? 202.88 − 1.013 = ? Solution If we use a calculator to add these two numbers, we would get 119.902. However, most calculators do not understand significant figures, and we need to limit the final answer to the tenths place. Thus, we drop the 02 and report a final answer of 119.9 (rounding down). A calculator would answer 201.867. However, we have to limit our final answer to the hundredths place. Because the first number being dropped is 7, which is greater than 7, we round up and report a final answer of 201.87. Check Your Understanding 3 Express the answer for 3.445 + 90.83 − 72.4 to the proper number of significant figures. If the operations being performed are multiplication or division, the rule is as follows: limit the answer to the number of significant figures that the data value with the least number of significant figures has. So if we are dividing 23 by 448, which have two and three significant figures each, we should limit the final reported answer to two significant figures (the lesser of two and three significant figures): 23448=0.051339286…=0.051 The same rounding rules apply in multiplication and division as they do in addition and subtraction. Example 6 Express the final answer to the proper number of significant figures. 76.4 × 180.4 = ? 934.9 ÷ 0.00455 = ? Solution The first number has three significant figures, while the second number has four significant figures. Therefore, we limit our final answer to three significant figures: 76.4 × 180.4 = 13,782.56 = 13,800. The first number has four significant figures, while the second number has three significant figures. Therefore we limit our final answer to three significant figures: 934.9 ÷ 0.00455 = 205,472.5275… = 205,000. Check Your Understanding 4 Express the final answer to the proper number of significant figures. 22.4 × 8.314 = ? 1.381 ÷ 6.02 = ? As you have probably realized by now, the biggest issue in determining the number of significant figures in a value is the zero. Is the zero significant or not? One way to unambiguously determine whether a zero is significant or not is to write a number in scientific notation. Scientific notation will include zeros in the coefficient of the number only if they are significant. Thus, the number 8.666 × 106 has four significant figures. However, the number 8.6660 × 106 has five significant figures. That last zero is significant; if it were not, it would not be written in the coefficient. So when in doubt about expressing the number of significant figures in a quantity, use scientific notation and include the number of zeros that are truly significant. Key Takeaways Significant figures in a quantity indicate the number of known values plus one place that is estimated. There are rules for which numbers in a quantity are significant and which are not significant. In calculations involving addition and subtraction, limit significant figures based on the rightmost place that all values have in common. In calculations involving multiplication and division, limit significant figures to the least number of significant figures in all the data values. Problems & Exercises Express each measurement to the correct number of significant figures. a) b) 2. Express each measurement to the correct number of significant figures. a) b) 3. How many significant figures do these numbers have? (a) 23 (b) 23.0 (c) 0.00023 (d) 0.0002302 How many significant figures do these numbers have? (a) 5.44 × 108 (b) 1.008 × 10−5 (c) 43.09 (d) 0.0000001381 How many significant figures do these numbers have? (a) 765,890 (b) 765,890.0 (c) 1.2000 × 105 (d) 0.0005060 6) How many significant figures do these numbers have? (a) 0.009 (b) 0.0000009 (c) 65,444 (d) 65,040 Compute and express each answer with the proper number of significant figures, rounding as necessary. (a) 56.0 + 3.44 = ? (b) 0.00665 + 1.004 = ? (c) 45.99 − 32.8 = ? (d) 45.99 − 32.8 + 75.02 = ? Compute and express each answer with the proper number of significant figures, rounding as necessary. (a) 1.005 + 17.88 = ? (b) 56,700 − 324 = ? (c) 405,007 − 123.3 = ? (d) 55.5 + 66.66 − 77.777 = ? Compute and express each answer with the proper number of significant figures, rounding as necessary. (a) 56.7 × 66.99 = ? (b) 1.000 ÷ 77 = ? (c) 1.000 ÷ 77.0 = ? (d) 6.022 × 1.89 = ? Compute and express each answer with the proper number of significant figures, rounding as necessary. (a) 0.000440 × 17.22 = ? (b) 203,000 ÷ 0.044 = ? (c) 67 × 85.0 × 0.0028 = ? (d) 999,999 ÷ 3,310 = ? Write the number 87,449 in scientific notation with four significant figures. Write the number 0.000066600 in scientific notation with five significant figures. Write the number 306,000,000 in scientific notation to the proper number of significant figures. Write the number 0.0000558 in scientific notation with two significant figures. Perform each calculation and limit each answer to three significant figures. (a) 67,883 × 0.004321 = ? (b) (9.67 × 103) × 0.0055087 = ? Perform each calculation and limit each answer to four significant figures. (a) 18,900 × 76.33 ÷ 0.00336 = ? (b) 0.77604 ÷ 76,003 × 8.888 = ? Solutions Check Your Understanding 1 0.63 cm Check Your Understanding 2 three significant figures five significant figures Check Your Understanding 3 Check Your Understanding 4 186 0.229 Problems & Exercises 1. (a) 375 psi (b) 1.30 cm 3. (a) two (b) three (c) two (d) four 5. (a) five (b) seven (c) five (d) four 7. (a) 59.4 (b) 1.011 (c) 13.2 (d) 88.2 9. (a) 3.80 × 103 (b) 0.013 (c) 0.0130 (d) 11.4 11. (a) 8.745 × 104 (b) 6.6600 × 10−5 13. (a) 293 (b) 53.3 License x-Douglas College Physics 1107 Fall 2019 Custom Textbook Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. Share This Book
12127	https://www.open.edu/openlearncreate/mod/oucontent/view.php?id=57302&printable=1	TI-AIE: Using a number line and the expression ‘Imagine if …’: positive and negative numbers: View as single page \| OLCreate Skip to main content Print Printable page generated Sunday, 28 September 2025, 10:52 PM Use 'Print preview' to check the number of pages and printer settings. Print functionality varies between browsers. Unless otherwise stated, copyright © 2025 The Open University, all rights reserved. Printable page generated Sunday, 28 September 2025, 10:52 PM ◀︎TI-AIE: Using a number line and the expression ‘Imagine if …’: positive and negative numbers TI-AIE: Using a number line and the expression ‘Imagine if …’: positive and negative numbers What this unit is about In this unit you will explore ways to encourage your students to think about the meaning of numbers and why the concept of ‘negative numbers’ was developed. Students first meet the minus sign when it is used to indicate the mathematical operation of subtraction; therefore, careful introduction to its use in negative numbers will be needed. Explaining that the sign is used differently and exploring why it is used for negative numbers will help your students to recognise and understand the similarities and differences in using this sign. Through the activities in this unit you will also think about developing the use of a number line to help your students visualise the movements indicated by positive and negative numbers. Actually making those movements themselves will further help the students understand what is meant by ‘positive’ and ‘negative’. The value of saying ‘Imagine if …’ to trigger imagination when teaching mathematics is also explored. What you can learn in this unit Some ideas to help your students understand the difference between positive and negative numbers. The role of saying ‘Imagine if …’ to trigger the imagination of your students when learning mathematics. How to use a number line to understand positive and negative numbers. This unit links to the teaching requirements of the NCF (2005) and NCFTE (2009) outlined in Resource 1. 1 The meaning of numbers Numbers were probably invented for the purpose of counting animals or other possessions. Numbering systems were originally thought to only have words for ‘one’, ‘two’ and ‘many’, as that was all that was needed. Further developments allowed herds to be counted, and as trade between people increased, the number system we use today was developed – including zero and negative numbers. Number names are almost always structured using a logical procedure so that they can express numbers that are to all intents and purposes, infinite. Numbers are used to represent: quantity, for answering questions like ‘How many?’ or ‘How far?’ relationships between numbers, for answering questions like ‘How many more?’ or ‘How many less?’ transformations in terms of quantities, for answering questions like ‘Imagine if Zuree owed Mary three rupees. She gave Mary one rupee. How much does she owe now?’, or ‘Manu won three marbles in the first match and lost five marbles in the second. How many marbles did he lose altogether?’ Pause for thought Reflect on how and where your students may have encountered negative numbers. They may, for example, have come across the idea that the temperature in an ice cream freezer is below zero. Where else might they have encountered these ideas? Zero is a number Zero plays a key role in understanding numbers. Mathematically, there are several uses or meanings attributed to zero that students have to work with. With one meaning the quantity ‘nothing’ is represented. This may mean ‘none’ as in ‘a team scored no goals in a football match’, or it may mean ‘no tens and no units’ as in the number 600. Zero is also used as a coordinate of an arbitrary point of reference or origin, for example (0, 0). From this point at least two mutually opposite directions can be considered. Understanding that zero has all these different meanings is important when teaching negative numbers. Negative numbers When a negative or minus sign is attached to a number as a prefix, it indicates the polarity of that number with respect to zero. Natural numbers are considered as positive numbers. Both positive and negative numbers have both magnitude and direction. Negative numbers can cause confusion between magnitude and order. For example, –4 is conventionally less than –1, despite –4 appearing to have a magnitude of more than –1. Pause for thought Think back to when you were learning negative numbers. Did it all seem straightforward? Try to express why negative numbers seemed straightforward to you (if they did). Maybe it was because the negative numbers fitted in with the ideas you already had about natural numbers and extended those ideas in a satisfactory way? Try to remember how you came to understand how to perform mathematical operations on negative numbers – did you learn rules by rote first? Think about some students in your classroom and the difficulties they have with natural numbers. Think about students you have taught and how they can get mixed up about when to apply the rule ‘two negatives make a positive’. How can your students be helped to understand negative numbers and not rely solely on remembering rules? 2 The need for negative numbers The activities in this unit will help you to develop your students’ understanding of why negative numbers are used and how useful they can be. They will also offer ideas about how to help your students understand how to work with negative numbers rather than just remembering rules. The first activity is designed to help students appreciate the need for negative numbers as part of the numbering system. Before attempting to use the activities in this unit with your students, it would be a good idea to complete all, or at least part, of the activities yourself. It would be even better if you could try them out with a colleague, as that will help you when you reflect on the experience. Trying them for yourself will mean you get insights into a learner’s experiences, which can, in turn, influence your teaching and your experiences as a teacher. Activity 1: Understanding the need for negative numbers Preparation This activity suggests three different ideas to help students appreciate the need for negative numbers as part of the number system. Using all the ideas, not necessarily in the same lesson, will give the students a wider exposure to thinking about negative numbers. To illustrate ‘positive’ and ‘negative’, try to find pictures of mountains and deep seas so that the ideas of ‘above’ and ‘below’ can be discussed, along with a zero that is at sea level. Can you think of other situations where positive and negative would be obvious or intuitive to the students? The activity Idea 1: Above and below sea level Draw a large picture, either on a large piece of paper on the wall or on the blackboard. Your picture should show the sea, mountains above the sea, and space below sea level. Use the pictures that you have collected from magazines or drawn. Suitable items would be a plane, an octopus, a whale, a boat, a house, a car, a fish, etc. Ask the students where they would place the items on your picture. Encourage them to say ‘above the sea level’ or ‘under the sea level’. When all of the items are stuck on, discuss how high the plane might be and how far under the sea the octopus might be, and so on. Introduce the minus sign to indicate ‘under the sea level’. Idea 2: Robot steps Make a space down the centre of the classroom, ensuring that all the students can see this pathway. Mark its centre with a chalk cross and ask a student to stand on the cross. Tell the class to imagine that the student is a robot who only moves forwards and backwards in a straight line. Use pieces of paper or chalk marks to number paces forward from the cross. Ask the robot to move to 2, then ask them to move two spaces back. Ask the students to say what number should be put on the cross – hopefully they will say zero. Ask others to give the robot instructions to move to a certain number and then back to a certain number. Now ask the robot to move to 3 and then move four spaces backwards. They have gone beyond zero! What number can be used to represent one step back from zero? Introduce other numbers beyond zero and ask the students to practise saying negative numbers by telling the robot where to move to. Idea 3: A game with benches Place as many benches as you can across the front of the room and divide the benches into individual seats with chalk lines. Write with chalk a zero on one of the seats (not at the end) and then number the other seats on the benches to the right of zero as 1, 2, 3 and so on. Ask the students how the seats to the left could be numbered. Suggest the negative sign if they do not think of it. Then play games that involve negative and positive numbers. For example: Stand a student behind a seat. The class call out the seat they want the student to move to, for example ‘5’ or ‘–2’ and so on. Sit a student on the seat and ask the class to say which seat the student should sit on. Encourage them to use just ‘3’ or ‘5’, and so on for numbers to the right, and ‘negative 2’, ‘negative 4’ and so on for numbers to the left of zero. Next, make the task more difficult. Sit a student on the seat labelled 5 and ask the class what ‘move’ has to be made to go to seat 2. This is more difficult because ‘negative 3’ can indicate a position relative to zero and can indicate the action of moving three to the left. Make sure you discuss these two meanings. Now ask the student to make a move and then ask what move would ‘undo’ that move. Use games like this as often as you can to help build confidence. You could stick the numbers to the wall rather than use chairs. In this way the students will use addition and subtraction of negative numbers naturally for the game. Video: Using local resources Case Study 1: Mrs Kapur reflects on using Activity 1 This is the account of a teacher who tried Activity 1 with her elementary students. I remember that my classes developed a dislike for negative numbers because there seemed to be so much to remember and it was easy to get mixed up. I decided to play some of the games in Activity 1 with them. They already knew about negative numbers so they were quick to say that the octopus would be at negative 8 metres. I drew the picture on paper on the wall with a scale marked positive and negative, and left it there after we had done this brief activity. In the morning many of the students arrived with pictures they had drawn so we put them in their correct places on the big picture and had another occasion to think about positive and negative numbers. Later in the term we played the bench game. They enjoyed this and, although they sometimes found it hard working out a move that is across the zero – from say 5 to –2 – they practised this a lot just because they wanted to keep playing. I definitely think making the actual moves themselves or instructing others to do so helped them to be able to visualise what was happening when we started to do the exercises on negative numbers in the textbook. To help make the step to using the textbook even easier, I think I will repeat some of these ideas and then also have discussions with the students about how we could record what we are doing in mathematical notation and write this on the blackboard. Hopefully they will then see how the actions relate to the mathematical notation and sums, and to what is asked in the textbooks. Reflecting on your teaching practice When you do such an exercise with your class, reflect afterwards on what went well and what went less well. Consider the questions that led to the students being interested and being able to progress, and those you needed to clarify. Such reflection always helps with finding a ‘script’ that helps you engage the students to find mathematics interesting and enjoyable. When you reflect on how the ideas in Activity 1 went with your class, make a note, as Mrs Kapur did, of some quite small things that made a difference. Pause for thought In the case study, Mrs Kapur said that she was thinking about repeating some of the activities and recording the outcomes on the blackboard using mathematical notation and sums. What do you feel might be the advantages of doing this after the students have a lot of experience of the activities and games? Now think about the following questions: How did the activities and games go with your class? What responses from students were unexpected? What did these tell you about their understanding of positive and negative numbers? What questions did you use to probe your students’ understanding? What points did you feel you had to reinforce? What will you do differently next time you use these activities in your teaching? 3 Using number lines to develop understanding of positive and negative numbers A number line such as Figure 1 is a geometrical idea that can be imagined as a set of points arranged in particular order on a straight line. A mathematical line has infinite length and is infinite in mutually opposite directions, but it is always centred at the origin, or zero. A number line can help students make sense of negative numbers and begin to understand adding and subtracting them. Figure 1 A number line. A number line can be so useful that it is a good idea to construct and display a large line divided into equally spaced intervals in any classroom where mathematics is learned, as in Figure 2. Figure 2 A blank number line Making the line in such a way that the numbers it represents can be written or attached separately will mean that it can be used for thinking about any part of the number system. Each division can then represent: units, tens or hundreds, etc. fractions or decimals, including very small decimals standard form – as well as many other mathematical ideas. Once students are used to seeing a number line on a wall or on their desks, they will be able to imagine the line in order to check their reasoning. The idea of a negative number only has existence in relation to positive numbers using zero as the origin. That is, a point is selected on a number line and zero is assigned to it so that one side of zero is positive and the other negative. In order to help students think in terms of opposites, it is conventional to use the right-hand side of a horizontal line to represent positive numbers and the left-hand side to represent negative numbers. However, it is also a good idea to also use a vertical line, where the numbers above zero are represented by positive numbers and the negative numbers are below zero. Whether using a horizontal or a vertical line, moving the point that is assigned to zero can help students understand that this is just a part of an infinite line, and that this part is being considered in lessons on negative numbers because zero is where things change. The following activity uses a number line drawn on a blackboard in a way that will help the students think about how to use negative numbers and how to add and subtract those numbers. The activity also uses the expression ‘Imagine if …’. This expression can help the students to use their imagination and not be limited by their belief that mathematics can only be ‘right’ or ‘wrong’. Knowing this is especially important in mathematical modelling (such as in word problems) where a model represents a perceived situation, which is not necessarily valid in all cases or may not even reflect a true real-life situation (Bruner, 1986). Activity 2: Learning from misconceptions and mistakes Part 1: How positive was that? Draw a number line on the blackboard going from –10 to 10. Ask the students to imagine positive things that can happen and ask them to imagine where they would put that on the number line. For example, ‘Someone gives me Rs. 10’ is a little positive; ‘Someone gives me Rs. 100’ is more positive. Then ask them to suggest negative things, for example, ‘My new outfit was splashed with mud when a rickshaw went past and I didn’t notice’, or ‘My cricket team lost a match’. Each time ask them to imagine where to put the idea on the number line, asking them to imagine ‘How positive do you feel?’ or ‘How negative was that?’ Part 2: The ‘happiness’model The ideas in Part 1 can then be extended to adding and subtracting negative numbers. Say to the class: I feel OK today; imagine I score 2 (pointing to the number line) on this happiness scale. Imagine if someone gave me nine sweets (a positive!), how would I feel then? Yes, I could move up 4 to 6. Now imagine if someone told me I had to stay after school (negative) how would I feel then? Yes, down 1, to 5. Imagine if you took away seven of my chocolates? How would I feel? Sadder? Yes, I need to go down 7, to –2. What if you told me I could go home early? … Adding something positive or taking away something negative improves the situation (go up the number line). Adding something negative or taking away something positive makes the situation worse (go down the number line). (Source: Part 2 was adapted from NRICH, undated.) Video: Talk for learning Case Study 2: Mrs Agarwal reflects on using Activity 2 I used the ideas in Activity 2 to explain positive and negative numbers to my class. Before I started I said, ‘I believe that adding and subtracting with negative numbers makes sense’. I wrote a big number line along the top of my blackboard. With the students, I brainstormed on ‘Things that are POSITIVE’ and ‘Things that are NEGATIVE’. We talked for quite a while about how you feel if someone gives you a positive thing, or if someone takes one away. We also talked about how you feel if someone gives you a negative thing, or if someone takes one away. Then we went on to use the happiness model. I went through getting sweets and losing my sweets, pointing to where I was on the happiness scale and then writing down the mathematical expression of what I was saying. I asked several of the students to tell their own story using the scale and to start with I wrote the sums as they were telling the story. I then asked the students to work in groups of three or four. They drew a number line in chalk on their desks then one told a story while another pointed to where they were on the number line and another wrote the addition and subtractions that they were doing. I have never seen so many smiles! Pause for thought How well did Activity 2 go with your class? What responses from students were unexpected? Why? Did you modify the task in any way? If so, what was your reasoning for this? What did you learn about your students’ understanding of positive and negative numbers? 4 Meanings of addition and subtraction processes Addition and subtraction are mutually inverse mathematical operations. For example: 5 + 1 = 6 6 – 1 = 5 6 – 5 = 1 According to some researchers (Linchevski and Williams, 1999; Bruno and Martinon, 1999) subtraction skills assist students to learn the concept of ‘negative numbers’. The processes of addition and subtraction become interchangeable in the case of integers. For example: 3 + 5 = 8 = 3 – (–5) The next activity aims to help your students focus on the thinking processes involved in calculating positive and negative numbers. Activity 3: Adding positive and negative numbers Part 1: Using the ‘counters’ model to understand addition and subtraction with negative numbers For this activity you will need a quantity of counters or pieces of card in two different colours. One colour should feature positive signs and the other colour should feature negative signs. An important part of this activity is getting students to talk about addition and subtraction with negative numbers and to explain their thinking. When you are planning your lesson you may want to have a look at Resource 2, ‘Talk for learning’. Tell your students that all of the following arrangements of counters add up to four. Figure 3 Various arrangements of counters that add up to four. Ask the students to suggest some other possibilities. Then ask them to explain how all these representations represent 4. Now use positive and negative counters to represent –2 in lots of different ways, starting with just two ‘negative’ counters. Be sure to ask the students to make suggestions themselves. Can they explain why all the different representations represent –2? Part 2: Using the ‘counters’ model for larger numbers, in small groups Put your students into small groups and tell them: Using counters or pieces of coloured paper, choose a number under ten and make at least four representations of that number using positive and negative counters. Record your representations as addition sums. Choose a negative number and do the same. If you have time, choose another number that you think will challenge you. Write out the sums for each of your representations. Swap with another group to mark them. If you can, go to the NRICH website to find several more ways to develop ideas for using positive and negative counters. (Source: Part 1 adapted from NRICH, undated.) Video: Using local resources Case Study 3: Mrs Nagaraju reflects on using Activity 3 I used the positive and negative counters with my class as they were having difficulty understanding how to work with negative numbers. I started off with some paper plates with positive and negative signs on them and got the students to hold the plates at the front of the class. They came up with some good ideas of combinations that made 4. It took longer for them to start making suggestions for negative 2 but they soon did. I wrote the total number of positives and the total number of negatives on the blackboard and then asked the students for the sign that meant put them together. They quickly suggested the ‘add’ sign. I put the students into groups of six for the next activity because there were nearly 60 in my class and they worked well together. Each group had ten pieces of paper in two colours and they wrote the positive and negative signs on for themselves. They made three different representations for each number they chose and I also made sure that they chose different numbers to the neighbouring groups. They wrote their addition sums on pieces of paper that we stuck on the wall so that everyone could see. I wanted to see if I could use the same ideas to help them understand what happens when you take away a negative, and of course you can! I got out the paper plates again and made 5 from 8 positives and 3 negatives. I asked what number we got when we took away 2 negatives and they could all tell me the answer was now 7. I wrote on the blackboard: 5 – (–2) = 7 For homework I asked them to look in their textbook for three similar examples like 5 – (–2) = 7 and to make a counter drawing of what these sums could look like. Pause for thought In the case study, Mrs Nagaraju asked a question to the whole class about which sign was needed to put the combinations together. The correct answer was given, but do you think she could be sure that this point was understood fully by all of the students? What other strategies might she have used to ensure that all of the students were involved in thinking about and discussing the answer? Now think about the following questions: When you carried out Activity 3, what questions did you use to probe your students’ understanding? Did you feel you had to intervene at any point? Were there any points that you feel some or all of your students need to revisit to secure their understanding? 5 Summary This unit has focused on negative numbers, understanding why they are needed and how they can be worked with. In reading this unit you have thought about how to enable your students to develop ways to visualise what negative numbers actually mean and ways to see addition and subtraction with negative numbers. You also used the expression ‘Imagine if...’ to trigger your students’ imagination. Students need rich and varied experiences of working with negative numbers if they are to see them as a normal part of using mathematics to represent the world. The activities in this unit have looked at many different ways to allow students to experience and understand negative numbers. You have also seen how reflecting on learning and how learning happens is important in becoming better at teaching. Resources Resource 1: NCF/NCFTE teaching requirements This unit links to the following teaching requirements of the NCF (2005) and NCFTE (2009) and will help you to meet those requirements: View students as active participants in their own learning and not as mere recipients of knowledge; how to encourage their capacity to construct knowledge; how to shift learning away from rote methods. Let students see mathematics as something to talk about, to communicate through, to discuss among themselves, to work together on. Resource 2: Talk for learning Why talk for learning is important Talk is a part of human development that helps us to think, learn and make sense of the world. People use language as a tool for developing reasoning, knowledge and understanding. Therefore, encouraging students to talk as part of their learning experiences will mean that their educational progress is enhanced. Talking about the ideas being learnt means that: those ideas are explored reasoning is developed and organised as such, students learn more. In a classroom there are different ways to use student talk, ranging from rote repetition to higher-order discussions. Traditionally, teacher talk was dominant and was more valued than students’ talk or knowledge. However, using talk for learning involves planning lessons so that students can talk more and learn more in a way that makes connections with their prior experience. It is much more than a question and answer session between the teacher and their students, in that the students’ own language, ideas, reasoning and interests are given more time. Most of us want to talk to someone about a difficult issue or in order to find out something, and teachers can build on this instinct with well-planned activities. Planning talk for learning activities in the classroom Planning talking activities is not just for literacy and vocabulary lessons; it is also part of planning mathematics and science work and other topics. It can be planned into whole class, pair or groupwork, outdoor activities, role play-based activities, writing, reading, practical investigations, and creative work. Even young students with limited literacy and numeracy skills can demonstrate higher-order thinking skills if the task is designed to build on their prior experience and is enjoyable. For example, students can make predictions about a story, an animal or a shape from photos, drawings or real objects. Students can list suggestions and possible solutions about problems to a puppet or character in a role play. Plan the lesson around what you want the students to learn and think about, as well as what type of talk you want students to develop. Some types of talk are exploratory, for example: ‘What could happen next?’, ‘Have we seen this before?’, ‘What could this be?’ or ‘Why do you think that is?’ Other types of talk are more analytical, for example weighing up ideas, evidence or suggestions. Try to make it interesting, enjoyable and possible for all students to participate in dialogue. Students need to be comfortable and feel safe in expressing views and exploring ideas without fear of ridicule or being made to feel they are getting it wrong. Building on students’ talk Talk for learning gives teachers opportunities to: listen to what students say appreciate and build on students’ ideas encourage the students to take it further. Not all responses have to be written or formally assessed, because developing ideas through talk is a valuable part of learning. You should use their experiences and ideas as much as possible to make their learning feel relevant. The best student talk is exploratory, which means that the students explore and challenge one another’s ideas so that they can become confident about their responses. Groups talking together should be encouraged not to just accept an answer, whoever gives it. You can model challenging thinking in a whole class setting through your use of probing questions like ‘Why?’, ‘How did you decide that?’ or ‘Can you see any problems with that solution?’ You can walk around the classroom listening to groups of students and extending their thinking by asking such questions. Your students will be encouraged if their talk, ideas and experiences are valued and appreciated. Praise your students for their behaviour when talking, listening carefully, questioning one another, and learning not to interrupt. Be aware of members of the class who are marginalised and think about how you can ensure that they are included. It may take some time to establish ways of working that allow all students to participate fully. Encourage students to ask questions themselves Develop a climate in your classroom where good challenging questions are asked and where students’ ideas are respected and praised. Students will not ask questions if they are afraid of how they will be received or if they think their ideas are not valued. Inviting students to ask the questions encourages them to show curiosity, asks them to think in a different way about their learning and helps you to understand their point of view. You could plan some regular group or pair work, or perhaps a ‘student question time’ so that students can raise queries or ask for clarification. You could: entitle a section of your lesson ‘Hands up if you have a question’ put a student in the hot-seat and encourage the other students to question that student as if they were a character, e.g. Pythagoras or Mirabai play a ‘Tell Me More’ game in pairs or small groups give students a question grid with who/what/where/when/why questions to practise basic enquiry give the students some data (such as the data available from the World Data Bank, e.g. the percentage of children in full-time education or exclusive breastfeeding rates for different countries), and ask them to think of questions you could ask about this data design a question wall listing the students’ questions of the week. You may be pleasantly surprised at the level of interest and thinking that you see when students are freer to ask and answer questions that come from them. As students learn how to communicate more clearly and accurately, they not only increase their oral and written vocabulary, but they also develop new knowledge and skills. Additional resources A newly developed maths portal by the Karnataka government: National Centre for Excellence in the Teaching of Mathematics: National STEM Centre: National Numeracy: BBC Bitesize: Khan Academy’s math section: NRICH: Art of Problem Solving’s resources page: Teachnology: Math Playground’s logic games: Maths is Fun: Coolmath4kids.com: National Council of Educational Research and Training’s textbooks for teaching mathematics and for teacher training of mathematics: AMT-01 Aspects of Teaching Primary School Mathematics, Block 1 (‘Aspects of Teaching Mathematics’), Block 2 (‘Numbers (I)’), Block 3 (‘Numbers (II)’): LMT-01 Learning Mathematics, Block 1 (‘Approaches to Learning’) Block 2 (‘Encouraging Learning in the Classroom’), Block 4 (‘On Spatial Learning’), Block 6 (‘Thinking Mathematically’): Manual of Mathematics Teaching Aids for Primary Schools, published by NCERT: Learning Curve and At Right Angles, periodicals about mathematics and its teaching: Textbooks developed by the Eklavya Foundation with activity-based teaching mathematics at the primary level: Central Board of Secondary Education’s books and support material (also including List of Hands-on Activities in Mathematics for Classes III to VIII) – select ‘CBSE publications’, then ‘Books and support material’: References Bruner, J. (1986) Actual Minds, Possible Worlds. Cambridge, MA: Harvard University Press. Bruno, A. and Martinon, A. (1999) ‘The teaching of numerical extensions: the case of negative numbers’, International Journal of Mathematical Education in Science and Technology, vol. 30, no. 6, pp. 789–809. Byers, V. and Herscovics, N. (1977) ‘Understanding school mathematics’, Mathematics Teaching, vol. 81, pp. 24–7. Egan, K. (1986) Teaching as Story Telling: An Alternative Approach to Teaching and Curriculum in the Elementary School. University of Chicago Press, Chicago. Fishbein, E. (1987) Intuition in Science and Mathematics: An Educational Approach. Dordrecht: Reidel. Linchevski, L. and Williams, J. (1999) ‘Using intuition from everyday life in “filling” the gap in children’s extension of their number concept to include the negative numbers’, Educational Studies in Mathematics, vol. 39, nos 1–3, pp. 131–47. National Council for Teacher Education (2009) National Curriculum Framework for Teacher Education (online). New Delhi: NCTE. Available from: publicnotice/ NCFTE_2010.pdf (accessed 5 February 2014). National Council of Educational Research and Training (2005) National Curriculum Framework (NCF). New Delhi: NCERT. NRICH (undated) ‘Making sense of positives and negatives: stage 3’ (online). Available from: (accessed 6 February 2014). Watson, A., Jones, K. and Pratt, D. (2013) Key Ideas in Teaching Mathematics. Oxford: Oxford University Press. Acknowledgements Except for third party materials and otherwise stated below, this content is made available under a Creative Commons Attribution-ShareAlike licence ( The material acknowledged below is Proprietary and used under licence for this project, and not subject to the Creative Commons Licence. This means that this material may only be used unadapted within the TESS-India project and not in any subsequent OER versions. This includes the use of the TESS-India, OU and UKAID logos. Grateful acknowledgement is made to the following sources for permission to reproduce the material in this unit: Activity 3, Part 1: adapted from ‘Making sense of positives and negatives’, © 1997–2014 University of Cambridge. Every effort has been made to contact copyright owners. If any have been inadvertently overlooked the publishers will be pleased to make the necessary arrangements at the first opportunity. Video (including video stills): thanks are extended to the teacher educators, headteachers, teachers and students across India who worked with The Open University in the productions. ◀︎TI-AIE: Using a number line and the expression ‘Imagine if …’: positive and negative numbers Unless otherwise stated, copyright © 2025 The Open University, all rights reserved.
12128	https://pubmed.ncbi.nlm.nih.gov/23790512/	Modern radiation therapy for Hodgkin lymphoma: field and dose guidelines from the international lymphoma radiation oncology group (ILROG) - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. 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Modern radiation therapy for Hodgkin lymphoma: field and dose guidelines from the international lymphoma radiation oncology group (ILROG) Lena Specht1,Joachim Yahalom2,Tim Illidge3,Anne Kiil Berthelsen4,Louis S Constine5,Hans Theodor Eich6,Theodore Girinsky7,Richard T Hoppe8,Peter Mauch9,N George Mikhaeel10,Andrea Ng9;ILROG Affiliations Expand Affiliations 1 Department of Oncology and Hematology, Rigshospitalet, University of Copenhagen, Denmark. Electronic address: lena.specht@regionh.dk. 2 Department of Radiation Oncology, Memorial Sloan-Kettering Cancer Center, New York, New York. 3 Institute of Cancer Sciences, University of Manchester, Manchester Academic Health Sciences Centre, Christie Hospital NHS Trust, Manchester, United Kingdom. 4 Department of Radiation Oncology and PET Centre, Rigshospitalet, University of Copenhagen, Denmark. 5 Department of Radiation Oncology and Pediatrics, James P. Wilmot Cancer Center, University of Rochester Medical Center, Rochester, New York. 6 Department of Radiation Oncology, University of Münster, Germany. 7 Department of Radiation Oncology, Institut Gustave-Roussy, Villejuif, France. 8 Department of Radiation Oncology, Stanford University, Stanford, California. 9 Department of Radiation Oncology, Brigham and Women's Hospital and Dana-Farber Cancer Institute, Harvard University, Boston, Massachusetts. 10 Department of Clinical Oncology and Radiotherapy, Guy's & St Thomas' NHS Foundation Trust, London, United Kingdom. PMID: 23790512 DOI: 10.1016/j.ijrobp.2013.05.005 Item in Clipboard Practice Guideline Modern radiation therapy for Hodgkin lymphoma: field and dose guidelines from the international lymphoma radiation oncology group (ILROG) Lena Specht et al. Int J Radiat Oncol Biol Phys.2014. Show details Display options Display options Format Int J Radiat Oncol Biol Phys Actions Search in PubMed Search in NLM Catalog Add to Search . 2014 Jul 15;89(4):854-62. doi: 10.1016/j.ijrobp.2013.05.005. Epub 2013 Jun 18. Authors Lena Specht1,Joachim Yahalom2,Tim Illidge3,Anne Kiil Berthelsen4,Louis S Constine5,Hans Theodor Eich6,Theodore Girinsky7,Richard T Hoppe8,Peter Mauch9,N George Mikhaeel10,Andrea Ng9;ILROG Affiliations 1 Department of Oncology and Hematology, Rigshospitalet, University of Copenhagen, Denmark. Electronic address: lena.specht@regionh.dk. 2 Department of Radiation Oncology, Memorial Sloan-Kettering Cancer Center, New York, New York. 3 Institute of Cancer Sciences, University of Manchester, Manchester Academic Health Sciences Centre, Christie Hospital NHS Trust, Manchester, United Kingdom. 4 Department of Radiation Oncology and PET Centre, Rigshospitalet, University of Copenhagen, Denmark. 5 Department of Radiation Oncology and Pediatrics, James P. Wilmot Cancer Center, University of Rochester Medical Center, Rochester, New York. 6 Department of Radiation Oncology, University of Münster, Germany. 7 Department of Radiation Oncology, Institut Gustave-Roussy, Villejuif, France. 8 Department of Radiation Oncology, Stanford University, Stanford, California. 9 Department of Radiation Oncology, Brigham and Women's Hospital and Dana-Farber Cancer Institute, Harvard University, Boston, Massachusetts. 10 Department of Clinical Oncology and Radiotherapy, Guy's & St Thomas' NHS Foundation Trust, London, United Kingdom. PMID: 23790512 DOI: 10.1016/j.ijrobp.2013.05.005 Item in Clipboard Cite Display options Display options Format Abstract Radiation therapy (RT) is the most effective single modality for local control of Hodgkin lymphoma (HL) and an important component of therapy for many patients. These guidelines have been developed to address the use of RT in HL in the modern era of combined modality treatment. The role of reduced volumes and doses is addressed, integrating modern imaging with 3-dimensional (3D) planning and advanced techniques of treatment delivery. The previously applied extended field (EF) and original involved field (IF) techniques, which treated larger volumes based on nodal stations, have now been replaced by the use of limited volumes, based solely on detectable nodal (and extranodal extension) involvement at presentation, using contrast-enhanced computed tomography, positron emission tomography/computed tomography, magnetic resonance imaging, or a combination of these techniques. The International Commission on Radiation Units and Measurements concepts of gross tumor volume, clinical target volume, internal target volume, and planning target volume are used for defining the targeted volumes. Newer treatment techniques, including intensity modulated radiation therapy, breath-hold, image guided radiation therapy, and 4-dimensional imaging, should be implemented when their use is expected to decrease significantly the risk for normal tissue damage while still achieving the primary goal of local tumor control. The highly conformal involved node radiation therapy (INRT), recently introduced for patients for whom optimal imaging is available, is explained. A new concept, involved site radiation therapy (ISRT), is introduced as the standard conformal therapy for the scenario, commonly encountered, wherein optimal imaging is not available. There is increasing evidence that RT doses used in the past are higher than necessary for disease control in this era of combined modality therapy. The use of INRT and of lower doses in early-stage HL is supported by available data. Although the use of ISRT has not yet been validated in a formal study, it is more conservative than INRT, accounting for suboptimal information and appropriately designed for safe local disease control. The goal of modern smaller field radiation therapy is to reduce both treatment volume and treatment dose while maintaining efficacy and minimizing acute and late sequelae. This review is a consensus of the International Lymphoma Radiation Oncology Group (ILROG) Steering Committee regarding the modern approach to RT in the treatment of HL, outlining a new concept of ISRT in which reduced treatment volumes are planned for the effective control of involved sites of HL. Nodal and extranodal non-Hodgkin lymphomas (NHL) are covered separately by ILROG guidelines. Copyright © 2014 Elsevier Inc. All rights reserved. PubMed Disclaimer Comment in Involved-Site Radiation Therapy for Early-Stage NLPHL.Patel CG, Ng AK.Patel CG, et al.Int J Radiat Oncol Biol Phys. 2020 May 1;107(1):21-22. doi: 10.1016/j.ijrobp.2019.09.025.Int J Radiat Oncol Biol Phys. 2020.PMID: 32277917 No abstract available. Similar articles Modern radiation therapy for nodal non-Hodgkin lymphoma-target definition and dose guidelines from the International Lymphoma Radiation Oncology Group.Illidge T, Specht L, Yahalom J, Aleman B, Berthelsen AK, Constine L, Dabaja B, Dharmarajan K, Ng A, Ricardi U, Wirth A; International Lymphoma Radiation Oncology Group.Illidge T, et al.Int J Radiat Oncol Biol Phys. 2014 May 1;89(1):49-58. doi: 10.1016/j.ijrobp.2014.01.006.Int J Radiat Oncol Biol Phys. 2014.PMID: 24725689 Radiotherapy for early mediastinal Hodgkin lymphoma according to the German Hodgkin Study Group (GHSG): the roles of intensity-modulated radiotherapy and involved-node radiotherapy.Koeck J, Abo-Madyan Y, Lohr F, Stieler F, Kriz J, Mueller RP, Wenz F, Eich HT.Koeck J, et al.Int J Radiat Oncol Biol Phys. 2012 May 1;83(1):268-76. doi: 10.1016/j.ijrobp.2011.05.054. Epub 2011 Nov 11.Int J Radiat Oncol Biol Phys. 2012.PMID: 22079733 Recommendations for the use of radiotherapy in nodal lymphoma.Hoskin PJ, Díez P, Williams M, Lucraft H, Bayne M; Participants of the Lymphoma Radiotherapy Group.Hoskin PJ, et al.Clin Oncol (R Coll Radiol). 2013 Jan;25(1):49-58. doi: 10.1016/j.clon.2012.07.011. Epub 2012 Aug 11.Clin Oncol (R Coll Radiol). 2013.PMID: 22889569 Does Radiation Have a Role in Advanced Stage Hodgkin's or Non-Hodgkin Lymphoma?Specht L.Specht L.Curr Treat Options Oncol. 2016 Jan;17(1):4. doi: 10.1007/s11864-015-0377-x.Curr Treat Options Oncol. 2016.PMID: 26739151 Review. A decade of comparative dose planning studies for early-stage Hodgkin lymphoma: what can we learn?Maraldo MV, Specht L.Maraldo MV, et al.Int J Radiat Oncol Biol Phys. 2014 Dec 1;90(5):1126-35. doi: 10.1016/j.ijrobp.2014.06.069.Int J Radiat Oncol Biol Phys. 2014.PMID: 25539371 Review. See all similar articles Cited by Long-Term Results of IFRT vs. ISRT in Infradiaphragmal Fields in Aggressive Non-Hodgkins's Lymphoma Patients-A Single Centre Experience.Galunic Bilic L, Santek F, Mitrovic Z, Basic-Kinda S, Dujmovic D, Vodanovic M, Mandac Smoljanovic I, Ostojic Kolonic S, Galunic Cicak R, Aurer I.Galunic Bilic L, et al.Cancers (Basel). 2024 Feb 2;16(3):649. doi: 10.3390/cancers16030649.Cancers (Basel). 2024.PMID: 38339400 Free PMC article. Past, Present, and Future of Radiation-Induced Cardiotoxicity: Refinements in Targeting, Surveillance, and Risk Stratification.Bergom C, Bradley JA, Ng AK, Samson P, Robinson C, Lopez-Mattei J, Mitchell JD.Bergom C, et al.JACC CardioOncol. 2021 Sep 21;3(3):343-359. doi: 10.1016/j.jaccao.2021.06.007. eCollection 2021 Sep.JACC CardioOncol. 2021.PMID: 34604796 Free PMC article.Review. Value of PET imaging for radiation therapy.Lapa C, Nestle U, Albert NL, Baues C, Beer A, Buck A, Budach V, Bütof R, Combs SE, Derlin T, Eiber M, Fendler WP, Furth C, Gani C, Gkika E, Grosu AL, Henkenberens C, Ilhan H, Löck S, Marnitz-Schulze S, Miederer M, Mix M, Nicolay NH, Niyazi M, Pöttgen C, Rödel CM, Schatka I, Schwarzenboeck SM, Todica AS, Weber W, Wegen S, Wiegel T, Zamboglou C, Zips D, Zöphel K, Zschaeck S, Thorwarth D, Troost EGC; Arbeitsgemeinschaft Nuklearmedizin und Strahlentherapie der DEGRO und DGN.Lapa C, et al.Strahlenther Onkol. 2021 Sep;197(9):1-23. doi: 10.1007/s00066-021-01812-2. Epub 2021 Jul 14.Strahlenther Onkol. 2021.PMID: 34259912 Review. Role of Radiation Therapy in the Treatment of Hodgkin Lymphoma.Gonzalez VJ.Gonzalez VJ.Curr Hematol Malig Rep. 2017 Jun;12(3):244-250. doi: 10.1007/s11899-017-0385-y.Curr Hematol Malig Rep. 2017.PMID: 28497317 Review. Congestive heart failure after anthracycline-containing treatment for Hodgkin lymphoma: A Swedish matched cohort study.Baech J, El-Galaly TC, Entrop JP, Glimelius I, Molin D, Godtfredsen SJ, Crowther MJ, Smedby KE, Eloranta S, Dietrich CE.Baech J, et al.EJHaem. 2024 Nov 13;5(6):1190-1200. doi: 10.1002/jha2.1048. eCollection 2024 Dec.EJHaem. 2024.PMID: 39691265 Free PMC article. 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12129	https://www.youtube.com/watch?v=QdYu_iv1CBM	Maximum Power Transfer Theorem with dependent source solved example \| Electrical Engineering Electrical and Electronics Engineering 77800 subscribers Description 12297 views Posted: 27 Jul 2022 Welcome to Electrical Engineering — your all-in-one platform to learn, practice, and master electrical engineering! 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Watch More ➤ Basic Electrical Engineering Capacitor Resistor Diode Thevenin’s Theorem Current Division Rule Operational Amplifiers Kirchhoff's Voltage Law (KVL) Kirchhoff’s Current Law (KCL) Series and parallel resistors \| Equivalent Resistance Impedance Combinations Impedance and Admittance Wye Delta Transformations Sinusoids Phasors Norton’s Theorem Nodal Analysis Source Transformation Mesh Analysis Impedance Electrical Circuits Superposition Theorem Mesh Analysis with Supermesh Two Port Network Electric Circuits and Networks (ECN) Maximum Power Transfer Theorem \| Electric Circuits Digital Techniques \| Digital Electronics Nodal Analysis with Supernode Alternating Current electricalengineering #electronics #electrical #engineering #math #education #learning #college #polytechnic #school #physics #electricity #mathematics #course #degree #diploma #semester #digital #theory #circuit #analysis #ElectricCircuits #math #science #networktheory #circuitanalysis #electriccircuits #tutorial #lectures Transcript: झाल फाइनली मैक्सिमम पावर ट्रांसपोर्ट टू रजिस्टर आर इन द सर्किट ऑफ कर दो एक सलूशन को हॉस्पिटल फाइन आर्ट एस छोटू फाइन आर्ट एस एच जो हेयर जिस पीआर कुच झाल कि यह फाइल मैक्सिमम पावर हिंदी रजिस्टर आर सुबह-सुबह और th9 में नोबेल टर्नऑफ अवधि इंडिपेंडेंस ऑफिस अनुसार वोल्टेज शोल्ड यू विल बी टर्न ऑफ एंड शॉर्ट सर्किट एंड तो फिर वैकेंसी डिपेंड छोड़ छोटू फाइन आर्ट ईच विल कंसीडर ऑल वर्ल्ड वोल्टेज और शेयर में त्रिशूल भी पॉजिटिव तहसील की नेगेटिव इस रिबन गोल्ड प्लेटेड डिश कारण हाई एक्स थोड़ी देर फॉर आर्ट एस विल बे इक्वल टू वन वर्ल्ड डिवाइडेड बाय 6a है ना व्यू टो फाइंड 620 वैल्यू आफ आर्ट एस इट्स ए डिसीजन ऑडियो एंड डिसित नोड को फील भी नोड आफ वन वोल्टेज सो लेट्स से विशेष ए रियल एज तो फिर इक्वल टू वन वर्ल्ड प् को लेकर डिसूजा रेफ्रेंस नोड थे लेटर दिस इज करंट आईएस वन विच करंट i2c कि सैंडविच करंट I3 रघु दीक्षित करंट iPhone तो अप्लाई के सीरियल अलार्म सेट नोड यह Vivo का नाम हिंदी अप्लाई फेशियल ऐड नोट वीओए फॉर एंटरिंग करंट एडिशन नोट इससे रोल करेंट्स आफ लिविंग बाय वन प्लस टू प्लस i3s पॉइंट्स फोन प्लस टू प्लस i3s 200 इक्वल टू Vivo - 0 को भी ओमाइनस हीरो बॉएड टैंक किलो हम छोटे के प्लस i2i तो इस विडियो - 01 यह Vivo - हीरो 14k ए प्लस i3s और नो आई थी इस विडियो - 151 - 120 के दो यह Vivo माइनस वन बाय तो 16 इज इक्वल टू जीरो दो कि लव यू टो फाइंड ए जॉब आफ रिलीफ फ्रॉम थिस इक्वेशन फॉर दिस विल बी रियो डी जेनेरो ए प्लस बी ओ बाय 4000 ए प्लस वीवो y20 माइनस वन बाय 22020 फॉर थिस विल बी 150 है वन बाय 1000 के दिन प्लस है वन बाय 4000 ए प्लस वन बाय है वन बाय तो एरो कि छोटी सी एल बी में 170 पॉइंट पांच से ज़रूरी पावर - 6 कि रियो ए माइनस वन बाय तो एरो का रिकार्ड 204 डायरेक्टली वे कैन राइट दिस इज इक्वल टू वन ट्वेंटी-20 फ्रॉम दिस अव्वल 251 बाय 20223 विल बी कि वन डिवाइडेड बाय 2202 आंसर थिस विल बी 0.2 छ कि हीरो पॉइंट टू 6 व्हाट इज द वैल्यू ऑफ वीवो है और सैफरन फॉर मोर प्रसेंस विल राइट फॉर बाय पागलवर्ल्ड कि नव यह Vivo फ्रॉम विच वे कैन फाइंड द वैल्यू ऑफ एक्स 612 इक्वल टू है इसलिए अप्लाइड के सीरियल ऐड 9वी ये है कि आप लाइव कि के सीरियल फैट नोड वेज कल शाम मिशन आफ ऑल थे इंटर इन करंट एंटरिंग करंट इस हाई एक्स प्लस i3s है हाई एक्स प्लस i3s इक्वल टू आईफोन एप थे वॉइस आफ 2003 वीडियोज वंडरिंग व्हेयर डिड नॉट फॉर थिस विल बी कि i386 प्लस ए प्लस 0 0 ए फ्री ऑडियो एप्स इक्वल टू अ हाइट फोन 234 आईएस ऐडेड पीस ए प्लस आई 39 i30 यह Vivo - 121 Twenty20 यह Vivo माइनस वन बाय 22k प्लस से ज़ ए फॉर वाइफ इन है और 220 इक्वल टू और - कि आई 49 ऑफिसर इन थिस आईएस फॉर थिस वन - 01 3000 कि वन - 01 30k लुट लुट 34 विल बी कि आई एक्स प्लस Vivo इस कि फॉर फिफ्टीन हैं सौंफ और y15 कि 415 माइनस वन डिवाइडेड बाय 2200 पॉइंट 03 2010 ए माइनस वन बाय 3000 कि रिक्वेस्ट 0 कि आई एक्स इंडिया विल बी ए फॉर्म आफ फिफ्टीन है - वंश डिवाइडेड मेज़ ट्वेंटी-20 इस है जीरो प्वाइंट 03 है फॉर 225 झाल ए माइनस वन बाय 3000 झाल 235 73.33 ए प्लस 733 यह पॉइंट 3310 टूटे इंडस्ट्री पावर - 6 इज इक्वल टू जीरो फोन 206 इज इक्वल टू - 733 लुटरे स्टोरी पावर - 6 एंपियर इज द वैल्यू ऑफ एक्स टो फाइंड आईटी हार्ड टो अचीव व्हाट इज इक्वल टू वन अधिसूचित की मैं वन अपऑन - आंसर 34540 6 - 136 3.63 होम दिस इज द वैल्यू ऑफ आर्ट एस माइनस वन 3.2 - 134 606 ॐ राधे राधे ॐ है - है नेक्स्ट टू हाउ टू फाइंड मी टीच ऊ ए व्यू टो फाइंड ए मैक्सिमम पावर ट्रांसपोर्ट व्ह रजिस्टर rs.30000 के चीज ग्रेट यू डेयर फॉर पीएम एक्स इज इक्वल टू डायरेक्टली ब्रिटेन से पीएम एक सी रिक्वेस्ट ऊ टो इंफिनिटी अ मैं सतीश राठौर फाइनल आंसर अ
12130	https://pmc.ncbi.nlm.nih.gov/articles/PMC5477786/	2016 AHA/ACC Guideline on the Management of Patients With Lower Extremity Peripheral Artery Disease: A Report of the American College of Cardiology/American Heart Association Task Force on Clinical Practice Guidelines - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. Learn more Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Advanced Search Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Circulation . Author manuscript; available in PMC: 2017 Sep 21. Published in final edited form as: Circulation. 2016 Nov 13;135(12):e726–e779. doi: 10.1161/CIR.0000000000000471 Search in PMC Search in PubMed View in NLM Catalog Add to search 2016 AHA/ACC Guideline on the Management of Patients With Lower Extremity Peripheral Artery Disease A Report of the American College of Cardiology/American Heart Association Task Force on Clinical Practice Guidelines Writing Committee Members, Marie D Gerhard-Herman Marie D Gerhard-Herman, MD, FACC, FAHA Chair Find articles by Marie D Gerhard-Herman , Heather L Gornik Heather L Gornik, MD, FACC, FAHA, FSVM Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information Vice Chair Find articles by Heather L Gornik , Coletta Barrett Coletta Barrett, RN †Functioning as the lay volunteer/patient representative Find articles by Coletta Barrett †, Neal R Barshes Neal R Barshes, MD, MPH ‡ACC/AHA Representative Find articles by Neal R Barshes ‡, Matthew A Corriere Matthew A Corriere, MD, MS, FAHA §Vascular and Endovascular Surgery Society Representative Find articles by Matthew A Corriere §, Douglas E Drachman Douglas E Drachman, MD, FACC, FSCAI Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information ‖Society for Cardiovascular Angiography and Interventions Representative Find articles by Douglas E Drachman ,‖, Lee A Fleisher Lee A Fleisher, MD, FACC, FAHA ¶ACC/AHA Task Force on Clinical Practice Guidelines Liaison Find articles by Lee A Fleisher ¶, Francis Gerry R Flowkes Francis Gerry R Flowkes, MD, FAHA Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information Inter-Society Consensus for the Management of Peripheral Arterial Disease Representative Find articles by Francis Gerry R Flowkes ,#, Naomi M Hamburg Naomi M Hamburg, MD, FACC, FAHA ‡ACC/AHA Representative Find articles by Naomi M Hamburg ‡, Scott Kinlay Scott Kinlay, MBBS, PhD, FACC, FAHA, FSVM, FSCAI Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information Society for Vascular Medicine Representative Find articles by Scott Kinlay ,, Robert Lookstein Robert Lookstein, MD, FAHA, FSIR Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information ‡ACC/AHA Representative Find articles by Robert Lookstein ,‡, Sanjay Misra Sanjay Misra, MD, FAHA, FSIR Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information ††Society of Interventional Radiology Representative Find articles by Sanjay Misra ,††, Leila Mureebe Leila Mureebe, MD, MPH, RPVI ‡‡Society for Clinical Vascular Surgery Representative Find articles by Leila Mureebe ‡‡, Jeffrey W Olin Jeffrey W Olin, DO, FACC, FAHA Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information ‡ACC/AHA Representative Find articles by Jeffrey W Olin ,‡, Rajan AG Patel Rajan AG Patel, MD, FACC, FAHA, FSCAI Inter-Society Consensus for the Management of Peripheral Arterial Disease Representative Find articles by Rajan AG Patel , Judith G Regensteiner Judith G Regensteiner, PhD, FAHA ‡ACC/AHA Representative Find articles by Judith G Regensteiner ‡, Andres Schanzer Andres Schanzer, MD Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information §§Society for Vascular Surgery Representative Find articles by Andres Schanzer ,§§, Mehdi H Shishehbor Mehdi H Shishehbor, DO, MPH, PhD, FACC, FAHA, FSCAI Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information ‡ACC/AHA Representative Find articles by Mehdi H Shishehbor ,‡, Kerry J Stewart Kerry J Stewart, EdD, FAHA, MAACVPR ‡ACC/AHA Representative ‖‖American Association of Cardiovascular and Pulmonary Rehabilitation Representative Find articles by Kerry J Stewart ‡,‖‖, Diane Treat-Jacobson Diane Treat-Jacobson, PhD, RN, FAHA ‡ACC/AHA Representative Find articles by Diane Treat-Jacobson ‡, M Eileen Walsh M Eileen Walsh, PhD, APN, RN-BC, FAHA ¶¶Society for Vascular Nursing RepresentativeChair, ACC/AHA Task Force on Clinical Practice Guidelines Find articles by M Eileen Walsh ¶¶, Jonathan L Halperin Jonathan L Halperin, MD, FACC, FAHA Chair Find articles by Jonathan L Halperin ; ACC/AHA Task Force on Clinical Practice Guidelines Author information Article notes Copyright and License information Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply; see Appendix 1 for recusal information †Functioning as the lay volunteer/patient representative ‡ACC/AHA Representative §Vascular and Endovascular Surgery Society Representative ‖Society for Cardiovascular Angiography and Interventions Representative ¶ACC/AHA Task Force on Clinical Practice Guidelines Liaison Inter-Society Consensus for the Management of Peripheral Arterial Disease Representative Society for Vascular Medicine Representative ††Society of Interventional Radiology Representative ‡‡Society for Clinical Vascular Surgery Representative §§Society for Vascular Surgery Representative ‖‖American Association of Cardiovascular and Pulmonary Rehabilitation Representative ¶¶Society for Vascular Nursing RepresentativeChair, ACC/AHA Task Force on Clinical Practice Guidelines Developed in Collaboration With the American Association of Cardiovascular and Pulmonar Rehabilitation, Inter-Society Consensus for the Management of Peripheral Arterial Disease, Society for Cardiovascular Angiography and Interventions, Society for Clinical Vascular Surgery, Society of Interventional Radiology, Society for Vascular Medicine, Society for Vascular Nursing, Society for Vascular Surgery, and Vascular and Endovascular Surgery Society Roles Marie D Gerhard-Herman: MD, FACC, FAHA, Chair Heather L Gornik: MD, FACC, FAHA, FSVM, Vice Chair Jonathan L Halperin: MD, FACC, FAHA, Chair Issue date 2017 Mar 21. Permissions: Multiple copies, modification, alteration, enhancement, and/or distribution of this document are not permitted without the express permission of the American Heart Association. Instructions for obtaining permission are located at A link to the “Copyright Permissions Request Form” appears on the right side of the page. PMC Copyright notice PMCID: PMC5477786 NIHMSID: NIHMS867255 PMID: 27840333 The publisher's version of this article is available at Circulation This article has been corrected. See Circulation. 2017 Mar 21;135(12):e791. Preamble Since 1980, the American College of Cardiology (ACC) and American Heart Association (AHA) have translated scientific evidence into clinical practice guidelines with recommendations to improve cardiovascular health. These guidelines, based on systematic methods to evaluate and classify evidence, provide a cornerstone of quality cardiovascular care. In response to reports from the Institute of Medicine1,2 and a mandate to evaluate new knowledge and maintain relevance at the point of care, the ACC/AHA Task Force on Clinical Practice Guidelines (Task Force) modified its methodology.3–5 The relationships among guidelines, data standards, appropriate use criteria, and performance measures are addressed elsewhere.5 Intended Use Practice guidelines provide recommendations applicable to patients with or at risk of developing cardiovascular disease. The focus is on medical practice in the United States, but guidelines developed in collaboration with other organizations may have a broader target. Although guidelines may be used to inform regulatory or payer decisions, the intent is to improve quality of care and align with patients' interests. Guidelines are intended to define practices meeting the needs of patients in most, but not all, circumstances, and should not replace clinical judgment. Guidelines are reviewed annually by the Task Force and are official policy of the ACC and AHA. Each guideline is considered current until it is updated, revised, or superseded by published addenda, statements of clarification, focused updates, or revised full-text guidelines. To ensure that guidelines remain current, new data are reviewed biannually to determine whether recommendations should be modified. In general, full revisions are posted in 5-year cycles.3–6 Modernization Processes have evolved to support the evolution of guidelines as “living documents” that can be dynamically updated. This process delineates a recommendation to address a specific clinical question, followed by concise text (ideally <250 words) and hyperlinked to supportive evidence. This approach accommodates time constraints on busy clinicians and facilitates easier access to recommendations via electronic search engines and other evolving technology. Evidence Review Writing committee members review the literature; weigh the quality of evidence for or against particular tests, treatments, or procedures; and estimate expected health outcomes. In developing recommendations, the writing committee uses evidence-based methodologies that are based on all available data.3–7 Literature searches focus on randomized controlled trials (RCTs) but also include registries, nonrandomized comparative and descriptive studies, case series, cohort studies, systematic reviews, and expert opinion. Only selected references are cited. The Task Force recognizes the need for objective, independent Evidence Review Committees (ERCs) that include methodologists, epidemiologists, clinicians, and biostatisticians who systematically survey, abstract, and assess the evidence to address systematic review questions posed in the PICOTS format (P=population, I=intervention, C=comparator, O=outcome, T=timing, S=setting).2,4–6 Practical considerations, including time and resource constraints, limit the ERCs to evidence that is relevant to key clinical questions and lends itself to systematic review and analysis that could affect the strength of corresponding recommendations. Guideline-Directed Management and Treatment The term “guideline-directed management and therapy” (GDMT) refers to care defined mainly by ACC/AHA Class I recommendations. For these and all recommended drug treatment regimens, the reader should confirm dosage with product insert material and carefully evaluate for contraindications and interactions. Recommendations are limited to treatments, drugs, and devices approved for clinical use in the United States. Class of Recommendation and Level of Evidence The Class of Recommendation (COR; ie, the strength of the recommendation) encompasses the anticipated magnitude and certainty of benefit in proportion to risk. The Level of Evidence (LOE) rates evidence supporting the effect of the intervention on the basis of the type, quality, quantity, and consistency of data from clinical trials and other reports (Table 1).3–5 Unless otherwise stated, recommendations are sequenced by COR and then by LOE. Where comparative data exist, preferred strategies take precedence. When >1 drug, strategy, or therapy exists within the same COR and LOE and no comparative data are available, options are listed alphabetically. Relationships With Industry and Other Entities The ACC and AHA sponsor the guidelines without commercial support, and members volunteer their time. The Task Force zealously avoids actual, potential, or perceived conflicts of interest that might arise through relationships with industry or other entities (RWI). All writing committee members and reviewers are required to disclose current industry relationships or personal interests, from 12 months before initiation of the writing effort. Management of RWI involves selecting a balanced writing committee and assuring that the chair and a majority of committee members have no relevant RWI (Appendix 1). Members are restricted with regard to writing or voting on sections to which their RWI apply. For transparency, members' comprehensive disclosure information is available online. Comprehensive disclosure information for the Task Force is also available online. The Task Force strives to avoid bias by selecting experts from a broad array of backgrounds representing different geographic regions, sexes, ethnicities, intellectual perspectives/biases, and scopes of clinical practice, and by inviting organizations and professional societies with related interests and expertise to participate as partners or collaborators. Individualizing Care in Patients With Associated Conditions and Comorbidities Managing patients with multiple conditions can be complex, especially when recommendations applicable to coexisting illnesses are discordant or interacting.8 The guidelines are intended to define practices meeting the needs of patients in most, but not all, circumstances. The recommendations should not replace clinical judgment. Clinical Implementation Management in accordance with guideline recommendations is effective only when followed. Adherence to recommendations can be enhanced by shared decision making between clinicians and patients, with patient engagement in selecting interventions on the basis of individual values, preferences, and associated conditions and comorbidities. Consequently, circumstances may arise in which deviations from these guidelines are appropriate. Keywords: AHA Scientific Statements, peripheral artery disease, claudication, critical limb ischemia, acute limb ischemia, antiplatelet agents, supervised exercise, endovascular procedures, bypass surgery, limb salvage, smoking cessation 1. Introduction 1.1. Methodology and Evidence Review The recommendations listed in this guideline are, whenever possible, evidence based. An initial extensive evidence review, which included literature derived from research involving human subjects, published in English, and indexed in MEDLINE (through PubMed), EMBASE, the Cochrane Library, the Agency for Healthcare Research and Quality, and other selected databases relevant to this guideline, was conducted from January through September 2015. Key search words included but were not limited to the following: acute limb ischemia, angioplasty, ankle-brachial index, anticoagulation, antiplatelet therapy, atypical leg symptoms, blood pressure lowering/hypertension, bypass graft/bypass grafting/surgical bypass, cilostazol, claudication/intermittent claudication, critical limb ischemia/severe limb ischemia, diabetes, diagnostic testing, endovascular therapy, exercise rehabilitation/exercise therapy/exercise training/supervised exercise, lower extremity/foot wound/ulcer, peripheral artery disease/peripheral arterial disease/peripheral vascular disease/lower extremity arterial disease, smoking/smoking cessation, statin, stenting, and vascular surgery. Additional relevant studies published through September 2016, during the guideline writing process, were also considered by the writing committee, and added to the evidence tables when appropriate. The final evidence tables included in theOnline Data Supplementsummarize the evidence utilized by the writing committee to formulate recommendations. Additionally, the writing committee reviewed documents related to lower extremity peripheral artery disease (PAD) previously published by the ACC and AHA.9,10 References selected and published in this document are representative and not all-inclusive. As stated in the Preamble, the ACC/AHA guideline methodology provides for commissioning an independent ERC to address systematic review questions (PI-COTS format) to inform recommendations developed by the writing committee. All other guideline recommendations (not based on the systematic review questions) were also subjected to an extensive evidence review process. For this guideline, the writing committee in conjunction with the Task Force and ERC Chair identified the following systematic review questions: 1) Is antiplatelet therapy beneficial for prevention of cardiovascular events in the patient with symptomatic or asymptomatic lower extremity PAD? 2) What is the effect of revascularization, compared with optimal medical therapy and exercise training, on functional outcome and quality of life (QoL) among patients with claudication? Each question has been the subject of recently published, systematic evidence reviews.11–13 The quality of these evidence reviews was appraised by the ACC/AHA methodologist and a vendor contracted to support this process (Doctor Evidence [Santa Monica, CA]). Few substantive randomized or nonrandomized studies had been published after the end date of the literature searches used for the existing evidence reviews, so the ERC concluded that no additional systematic review was necessary to address either of these critical questions. A third systematic review question was then identified: 3) Is one revascularization strategy (endovascular or surgical) associated with improved cardiovascular and limb-related outcomes in patients with critical limb ischemia (CLI)? This question had also been the subject of a high-quality systematic review that synthesized evidence from observational data and an RCT14; additional RCTs addressing this question are ongoing.15–17 The writing committee and the Task Force decided to expand the survey to include more relevant randomized and observational studies. Based on evaluation of this additional evidence the ERC decided that further systematic review was not needed to inform the writing committee on this question. Hence, the ERC and writing committee concluded that available systematic reviews could be used to inform the development of recommendations addressing each of the 3 systematic review questions specified above. The members of the Task Force and writing committee thank the members of the ERC that began this process and their willingness to participate in this volunteer effort. They include Aruna Pradhan, MD, MPH (ERC Chair); Natalie Evans, MD; Peter Henke, MD; Dharam J. Kumbhani, MD, SM, FACC; and Tamar Polonsky, MD. 1.2. Organization of the Writing Committee The writing committee consisted of clinicians, including noninvasive and interventional cardiologists, exercise physiologists, internists, interventional radiologists, vascular nurses, vascular medicine specialists, and vascular surgeons, as well as clinical researchers in the field of vascular disease, a nurse (in the role of patient representative), and members with experience in epidemiology and/or health services research. The writing committee included representatives from the ACC and AHA, American Association of Cardiovascular and Pulmonary Rehabilitation, Inter-Society Consensus for the Management of Peripheral Arterial Disease, Society for Cardiovascular Angiography and Interventions, Society for Clinical Vascular Surgery, Society of Interventional Radiology, Society for Vascular Medicine, Society for Vascular Nursing, Society for Vascular Surgery, and Vascular and Endovascular Surgery Society. 1.3. Document Review and Approval This document was reviewed by 2 official reviewers nominated by the ACC and AHA; 1 to 2 reviewers each from the American Association of Cardiovascular and Pulmonary Rehabilitation, Inter-Society Consensus for the Management of Peripheral Arterial Disease, Society for Cardiovascular Angiography and Interventions, Society for Clinical Vascular Surgery, Society of Interventional Radiology, Society for Vascular Medicine, Society for Vascular Nursing, Society for Vascular Surgery, and Vascular and Endovascular Surgery Society; and 16 additional individual content reviewers. Reviewers' RWI information was distributed to the writing committee and is published in this document (Appendix 2). Appendix 2. Reviewer Relationships With Industry and Other Entities (Comprehensive)—2016 AHA/ACC Guideline on the Management of Patients With Lower Extremity Peripheral Artery Disease (March 2016). \| Reviewer \| Representation \| Employment \| Consultant \| Speakers Bureau \| Ownership/Partnership /Principal \| Personal Research \| Institutional,Organizational, or Other FinancialBenefit \| Expert Witness \| :--- :--- :---: :--- \| Deepak L. Bhatt \| Official Reviewer—ACC Board of Trustees \| Brigham and Women's Hospital— Executive Director of Interventional Cardiovascular Programs; Harvard Medical School—Professor of Medicine \| Elsevier \| None \| None \| Amarin Amgen AstraZeneca Bristol-Myers Squibb Cardax† Eisai Ethicon FlowCo† Forest Laboratories Ischemix Mayo Clinic Medtronic Merck† Pfzer PLx Pharma† Regado Biosciences† Roche Sanof-aventis St. Jude Medical Takeda† The Medicines Company WebMD \| Belvoir Publications (Editor) Biotronik Boston Scientific Clinical Cardiology(Deputy Editor)† Harvard Clinical Research Institute HMP Communications (Editor) Duke Clinical Research Institute Journal of Invasive Cardiology (Editor) Medscape Cardiology Slack Publications (Editor) St. Jude Medical VA Healthcare System† \| None \| \| Mark A. Creager \| Official Reviewer—AHA \| Dartmouth-Hitchcock Medical Center— Director \| None \| None \| None \| None \| AHA (Past President)† \| None \| \| Philip Goodney \| Official Reviewer—AHA \| Dartmouth-Hitchcock—Associate Professor of Surgery and The Dartmouth Institute Director \| None \| None \| None \| NIH \| NIH \| None \| \| John S. Ikonomidis \| Official Reviewer—ACC/ AHA Task Force on Clinical Practice Guidelines \| Medical University of South Carolina—Chief \| None \| None \| None \| None \| None \| None \| \| Amy W. Pollak \| Official Reviewer—AHA \| Mayo Clinic—Cardiovascular Medicine Physician \| None \| None \| None \| None \| None \| None \| \| Michael D. White \| Official Reviewer—ACC Board of Governors \| Catholic Health Initiatives—Chief Academic Officer \| Anthera Pharmaceuticals† \| None \| None \| AstraZeneca† \| None \| None \| \| Ehrin J. Armstrong \| Organizational Reviewer—SVM \| University of Colorado—Director, Interventional Cardiology \| Abbott Medtronic Merck Spectranetics \| None \| None \| None \| None \| None \| \| Bernadette Aulivola \| Organizational Reviewer—VESS \| Loyola University medical Center, Stritch School of Medicine—Director, Division of Vascular Surgery and Endovascular Therapy; Associate Professor, Department of Surgery; Program Director, Vascular Surgery Fellowship; Medical Director, Vascular Noninvasive lab \| None \| None \| None \| None \| None \| None \| \| Alison Bailey \| Organizational Reviewer—AACVPR \| University of Tennessee Chattanooga—Cardiologist \| None \| None \| None \| CSL Behring \| AACVPR† ZOLL Medical \| None \| \| Todd Brown \| Organizational Reviewer—AACVPR \| University of Alabama at Birmingham—Associate Professor \| None \| None \| None \| Amgen Omthera† NIH \| None \| None \| \| Kristen Columbia \| Organizational Reviewer—SVN \| University of Maryland Baltimore Washington Medical Center, Maryland Vascular Center—Nurse practitioner \| None \| None \| None \| None \| None \| None \| \| Michael S. Conte \| Organizational Reviewer—SVS \| University of California San Francisco—Professor and Chief \| Cook Medical Medtronic \| None \| None \| Bard \| University of California Department of Surgery \| None \| \| Alik Farber \| Organizational Reviewer—SCVS \| Boston Medical Center—Chief, Division of Vascular Surgery \| Bard† \| None \| None \| None \| None \| None \| \| Robert Feezor \| Organizational Reviewer—VESS \| University of Florida—Associate Professor of Surgery, Division of Vascular Surgery and Endovascular Therapy \| Cook Medical Medtronic Terumo \| None \| None \| Cook Medical \| Cook Medical Novate \| Defendant, peripheral angioplasty, 2015 \| \| Dmitriy N. Feldman \| Organizational Reviewer—SCAI \| Weill Cornell Medical College, New York Presbyterian Hospital—Associate Professor of Medicine \| AstraZeneca \| Abbott Bristol-Myers Squibb† Daiichi-Sankyo Eli Lilly Medtronic Pfizer The Medicines Company \| None \| None \| Biotronic The Medicines Company \| None \| \| Jonathan Golledge \| Organizational Reviewer—TASC \| James Cook University—Professor, Department of Surgery, Head of Vascular Biology Unit \| None \| None \| None \| James Cook University \| None \| None \| \| Bruce H. Gray \| Organizational Reviewer—SCAI \| Greenville Health System—Director of Clinical Trials, Department of Surgery \| None \| Medtronic† \| None \| Abbott† W.L. Gore† \| NCDR† ACC† \| None \| \| William R. Hiatt \| Organizational Reviewer—TASC \| Colorado Prevention Center—Professor of Medicine \| None \| None \| None \| AstraZeneca Bayer CSI Kowa Kyushu University Merck Pluristem ReNeuron \| CPC Clinical Research NIH \| None \| \| Joseph Mills \| Organizational Reviewer—SVS \| Baylor College of Medicine—Professor and Chief, Division of Vascular surgery and Endovascular Therapy \| None \| None \| None \| None \| AnGes Bayer Cesca \| None \| \| Mohammad Reza Rajebi \| Organizational Reviewer—SIR \| University of Colorado Denver— Assistant Professor \| None \| None \| None \| None \| None \| None \| \| Mitchell J. Silver \| Organizational Reviewer—SVM \| McConnell Heart Hospital for Critical Limb Care—Director of Vascular Imaging \| Boston Scientific W.L. Gore Medtronic \| Bristol-Myers Squibb Pfzer \| Contego Medical \| None \| W.L. Gore Medtronic NIH \| None \| \| Lily Thomson \| Organizational Reviewer—SVN \| Hôpital St-Boniface Hospital—Clinical Research Coordinator, Vascular Surgery Nurse, Section of Vascular Surgery, Health Sciences Centre \| None \| None \| None \| None \| None \| None \| \| Sana M. Al-Khatib \| Content Reviewer—ACC/ AHA Task Force on Clinical Practice Guidelines \| Duke Clinical Research Institute—Associate Professor of Medicine \| None \| None \| None \| FDA NHLBI PCORI VA (DSMB) \| HRS (Board of Trustees)† Elsevier \| None \| \| Herbert Aronow \| Content Reviewer—ACC Peripheral Vascular Disease Member Section \| Rhode Island Hospital—Director of Cardiac Catheterization Laboratories \| None \| None \| None \| Silk Road Medical† Saint Luke's Health System The Medicines Company† \| Bard NIH PCORI† SVM† W.L. Gore \| \| \| Joshua A. Beckman \| Content Reviewer \| Vanderbilt University Medical Center— Director \| AstraZeneca Merck Sanof \| None \| EMX† JanaCare† \| Bristol-Myers Squibb Merck NIH \| Vascular Interventional Advances \| Defendant, venous thrombo-embolism, 2015 \| \| James C. Blankenship \| Content Reviewer \| Geisinger Medical Center—Staff Physician; Director, Cardiac Catheterization Laboratory \| None \| None \| None \| Abbott† AstraZeneca† Boston Scientific† GlaxoSmithKline† Hamilton Health Sciences† Medinal LTD† Orexigen Therapeutics† St. Jude Medical† Stentys† Takeda Pharmaceuticals† \| SCAI (Past President)† AMA† \| None \| \| Biykem Bozkurt \| Content Reviewer—ACC/AHA Task Force on Clinical Practice Guidelines \| Michael E. DeBakey VA Medical Center—The Mary and Gordon Cain Chair and Professor of Medicine \| None \| None \| None \| Novartis \| None \| None \| \| Joaquin E. Cigarroa \| Content Reviewer—ACC/AHA Task Force on Clinical Practice Guidelines \| Oregon Health and Science University—Clinical Professor of Medicine \| None \| None \| None \| None \| ACC/AHA† AHA† ASA† Catheterization and Cardiovascular Intervention† Portland Metro Area AHA(President)† SCAI Quality Interventional Council† NIH \| None \| \| Federico Gentile \| Content Reviewer—ACC/AHA Task Force on Clinical Practice Guidelines \| Centro Medico Diagnostico—Director, Cardiovascular Disease \| None \| None \| None \| None \| None \| None \| \| Anuj Gupta \| Content Reviewer—ACC Peripheral Vascular Disease Member Section \| University of Maryland—Assistant Professor of Medicine \| None \| None \| None \| Seimens Medtronic† \| Direct Flow Medical† Edwards† \| None \| \| John Jeb Hallett \| Content Reviewer \| Medical University of South Carolina—Clinical Professor of Surgery \| None \| None \| None \| None \| None \| None \| \| Alan Hirsch \| Content Reviewer \| University of Minnesota Medical School—Professor of Medicine, Epidemiology and Community Health, and Director Vascular Medicine Program \| Merck Novartis† \| None \| None \| Bayer Pluristem (PLX-PAD trial-PI)† AstraZeneca (EUCLID trial–PI)† Pluristem \| AHA† Tactile Medical \| None \| \| Mark A. Hlatky \| Content Reviewer—ACC/AHA Task Force on Clinical Practice Guidelines \| Stanford University School of Medicine—Professor of Health Research and Policy, Professor of Medicine \| Acumen Genentech \| None \| None \| Blue Cross/Blue Shield Center for Effectiveness Evaluation George Institute HeartFlow NHLBI Sanofi-aventis \| ACC (Associate Editor) \| None \| \| Michael R. Jaff \| Content Reviewer \| Newton-Wellesley Hospital; Harvard Medical School— Professor of Medicine \| AOPA Cardinal Health Covidien† Micell Vascular Therapies \| None \| MC10† Janacare† Northwind PQ Bypass Primacea SanoV Valiant Medical \| Abbott† Boston Scientific† Cordis† IC Sciences Medtronic† Novello \| CBSET Intersocietal Accreditation Commission SCAI† VIVA Physicians Group \| None \| \| José A. Joglar \| Content Reviewer—ACC/AHA Task Force on Clinical Practice Guidelines \| UT Southwestern Medical Center—Professor of Internal Medicine; Clinical Cardiac Electrophysiology—Fellowship Program Director \| None \| None \| None \| None \| None \| None \| \| Glenn N. Levine \| Content Reviewer—ACC/AHA Task Force on Clinical Practice Guidelines \| Baylor College of Medicine—Professor of Medicine; Director, Cardiac Care Unit \| None \| None \| None \| None \| None \| None \| \| Khusrow Niazi \| Content Reviewer—ACC Peripheral Vascular Disease Member Section \| Emory University Department of Medicine—Associate Professor of Medicine \| None \| Medtronic \| None \| Bard Impeto Terumo \| None \| Plaintiff, MI resulting in death, 2015 \| \| Paul D. Varosy \| Content Reviewer—Task Force on Performance Measures \| VA Eastern Colorado Health Care System—Associate Professor \| None \| None \| None \| VA Health Services Research and Development (PI) \| AHA (Guest Editor)† \| None \| \| Christopher J. White \| Content Reviewer \| Ochsner Clinical School, University of Queensland—Chairman, Department of Cardiology \| Neovasc \| None \| None \| AstraZeneca Pharmaceuticals NIH Neovasc Surmodics \| ACE (Board of Directors)† \| None \| Open in a new tab This table represents all relationships of reviewers with industry and other entities that were reported by authors, including those not deemed to be relevant to this document, at the time this document was under development. The table does not necessarily reflect relationships with industry at the time of publication. A person is deemed to have a significant interest in a business if the interest represents ownership of ≥5% of the voting stock or share of the business entity, or ownership of ≥$5000 of the fair market value of the business entity; or if funds received by the person from the business entity exceed 5% of the person's gross income for the previous year. Relationships that exist with no financial benefit are also included for the purpose of transparency. Relationships in this table are modest unless otherwise noted. Please refer to relationships-with-industry-policy for definitions of disclosure categories or additional information about the ACC/AHA Disclosure Policy for Writing Committees. Significant relationship. † No financial benefit. AACVPR indicates American Association of Cardiovascular and Pulmonary Rehabilitation; ACC, American College of Cardiology; ACE, Accreditation for Cardiovascular Excellence; AHA, American Heart Association; AMA, American Medical Association; DSMB, data and safety monitoring board; EUCLID, Effects of Ticagrelor and Clopidogrel in Patients with Peripheral Artery Disease; FDA, US Food and Drug Administration; HRS, Heart Rhythm Society; MI, myocardial infarction; NCDR, National Cardiovascular Data Registry; NIH, National Institutes of Health; NHLBI, National Heart, Lung, and Blood Institute; PCORI, Patient-Centered Outcomes Research Institute; PI, primary investigator; PLX-PAD, placental-derived adherent stromal cell; SCAI, Society for Cardiovascular Angiography and Interventions; SCVS, Society for Clinical Vascular Surgery; SIR, Society of Interventional Radiology; SVM, Society for Vascular Medicine; SVN, Society for Vascular Nursing; SVS, Society for Vascular Surgery; TASC, Trans-Atlantic Inter-Society Consensus for the Management of Peripheral Arterial Disease; VA, Veterans Affairs; VESS, Vascular and Endovascular Surgery Society; and VIVA, Vascular Intervention Advances. This document was approved for publication by the governing bodies of the ACC and the AHA and endorsed by the American Association of Cardiovascular and Pulmonary Rehabilitation, Inter-Society Consensus for the Management of Peripheral Arterial Disease, Society for Cardiovascular Angiography and Interventions, Society for Clinical Vascular Surgery, Society of Interventional Radiology, Society for Vascular Medicine, Society for Vascular Nursing, Society for Vascular Surgery, and Vascular and Endovascular Surgery Society. 1.4. Scope of Guideline Lower extremity PAD is a common cardiovascular disease that is estimated to affect approximately 8.5 million Americans above the age of 40 years and is associated with significant morbidity, mortality, and QoL impairment.18 It has been estimated that 202 million people worldwide have PAD.19 The purpose of this document is to provide a contemporary guideline for diagnosis and management of patients with lower extremity PAD. This document supersedes recommendations related to lower extremity PAD in the “ACC/AHA 2005 Guidelines for the Management of Patients With Peripheral Arterial Disease”9 and the “2011 ACCF/AHA Focused Update of the Guideline for the Management of Patients With Peripheral Artery Disease.”10 The scope of this guideline is limited to atherosclerotic disease of the lower extremity arteries (PAD) and includes disease of the aortoiliac, femoropopliteal, and infrapopliteal arterial segments. It does not address nonatherosclerotic causes of lower extremity arterial disease, such as vasculitis, fibromuscular dysplasia, physiological entrapment syndromes, cystic adventitial disease, and other entities. Future guidelines will address aneurysmal disease of the abdominal aorta and lower extremity arteries and diseases of the renal and mesenteric arteries. In developing the “2016 AHA/ACC Guideline on the Management of Patients With Lower Extremity Peripheral Artery Disease,” the writing committee reviewed the evidence to support recommendations in the relevant ACC/AHA guidelines noted in Table 2 and affirms the ongoing validity of the related recommendations, thus obviating the need to repeat existing guideline recommendations in the current guideline. Table 2 also contains a list of other statements that may be of interest to the reader. Table 3 includes definitions for PAD key terms used throughout the guideline. Table 2. Important Guideline Policy. \| Title \| Organization \| Publication Year (Reference) \| :---: \| ACC/AHA Guideline policy relevant to the management of lower extremity PAD \| \| Duration of dual antiplatelet therapy in patients with coronary artery disease \| ACC/AHA \| 201620 \| \| Perioperative cardiovascular evaluation and management of patients undergoing noncardiac surgery \| ACC/AHA \| 201421 \| \| Lifestyle management to reduce cardiovascular risk \| AHA/ACC \| 201322 \| \| Assessment of cardiovascular risk \| ACC/AHA \| 201323 \| \| Blood cholesterol to reduce atherosclerotic cardiovascular risk in adults \| ACC/AHA \| 201324 \| \| PAD (lower extremity, renal, mesenteric, and abdominal aortic) \| ACC/AHA \| 20059 and 201110 \| \| Secondary prevention and risk-reduction therapy for patients with coronary and other atherosclerotic vascular disease \| AHA/ACC \| 201125 \| \| Other related publications \| \| Atherosclerotic occlusive disease of the lower extremities guideline \| SVS \| 201526 \| \| Measurement and interpretation of the ankle-brachial index \| AHA \| 201227 \| \| Cardiac disease evaluation and management among kidney and liver transplantation candidates \| AHA/ACC \| 201228 \| \| Intensive glycemic control and the prevention of cardiovascular events \| ADA/ACC/AHA \| 200929 \| \| Influenza vaccination as secondary prevention for cardiovascular disease \| AHA/ACC \| 200630 \| \| Indications for renal arteriography at the time of coronary arteriography \| AHA/CLCD/CVRI/KCVD \| 200631 \| \| Seventh Report of the Joint National Committee on Prevention, Detection, Evaluation, and Treatment of High Blood Pressure (JNC 7) \| NHLBI \| 200332 \| Open in a new tab A revision to the current document is being prepared, with publication expected in 2017. The new title is expected to be “ACC/AHA/AAPA/ABC/ACPM/AGS/APhA/ASH/ASPC/NMA/PCNA Guideline for the Detection, Evaluation, Prevention and Management of High Blood Pressure.” AAPA indicates American Academy of Physician Assistants; ABC, Association of Black Cardiologists; ACC, American College of Cardiology; ACPM, American College of Preventive Medicine; ADA, American Diabetes Association; AGS, American Geriatrics Society; AHA, American Heart Association; APhA, American Pharmacists Association; ASH, American Society of Hypertension; ASPC, American Society for Preventive Cardiology; CLCD, Council on Clinical Cardiology; CVRI, Council on Cardiovascular Radiology and Intervention; KCVD, Council on the Kidney in Cardiovascular Disease; NHLBI, National Heart, Lung, and Blood Institute; NMA, National Medical Association; PAD, peripheral artery disease; PCNA, Preventive Cardiovascular Nurses Association; and SVS, Society for Vascular Surgery. Table 3. Definition of PAD Key Terms. \| Term \| Definition \| :--- \| \| Claudication \| Fatigue, discomfort, cramping, or pain of vascular origin in the muscles of the lower extremities that is consistently induced by exercise and consistently relieved by rest (within 10 min). \| \| Acute limb ischemia (ALI) \| Acute (<2 wk), severe hypoperfusion of the limb characterized by these features: pain, pallor, pulselessness, poikilothermia (cold), paresthesias, and paralysis. One of these categories of ALI is assigned (Section 10): 1. Viable—Limb is not immediately threatened; no sensory loss; no muscle weakness; audible arterial and venous Doppler. 2. Threatened—Mild-to-moderate sensory or motor loss; inaudible arterial Doppler; audible venous Doppler; may be further divided into IIa (marginally threatened) or IIb (immediately threatened). 3. Irreversible—Major tissue loss or permanent nerve damage inevitable; profound sensory loss, anesthetic; profound muscle weakness or paralysis (rigor); inaudible arterial and venous Doppler.33,34 \| \| Tissue loss \| Type of tissue loss: Minor—nonhealing ulcer, focal gangrene with diffuse pedal ischemia. Major—extending above transmetatarsal level; functional foot no longer salvageable.33 \| \| Critical limb ischemia (CLI) \| A condition characterized by chronic (≥2 wk) ischemic rest pain, nonhealing wound/ulcers, or gangrene in 1 or both legs attributable to objectively proven arterial occlusive disease. The diagnosis of CLI is a constellation of both symptoms and signs. Arterial disease can be proved objectively with ABI, TBI, TcPO 2, or skin perfusion pressure. Supplementary parameters, such as absolute ankle and toe pressures and pulse volume recordings, may also be used to assess for significant arterial occlusive disease. However, a very low ABI or TBI does not necessarily mean the patient has CLI. The term CLI implies chronicity and is to be distinguished from ALI.35 \| \| In-line blood flow \| Direct arterial flow to the foot, excluding collaterals. \| \| Functional status \| Patient's ability to perform normal daily activities required to meet basic needs, fulfill usual roles, and maintain health and well-being. Walking ability is a component of functional status. \| \| Nonviable limb \| Condition of extremity (or portion of extremity) in which loss of motor function, neurological function, and tissue integrity cannot be restored with treatment. \| \| Salvageable limb \| Condition of extremity with potential to secure viability and preserve motor function to the weight-bearing portion of the foot if treated. \| \| Structured exercise program \| Planned program that provides individualized recommendations for type, frequency, intensity, and duration of exercise. Program provides recommendations for exercise progression to assure that the body is consistently challenged to increase exercise intensity and levels as functional status improves over time. There are 2 types of structured exercise program for patients with PAD: 1. Supervised exercise program 2. Structured community- or home-based exercise program \| \| Supervised exercise program \| Structured exercise program that takes place in a hospital or outpatient facility in which intermittent walking exercise is used as the treatment modality. Program can be standalone or can be made available within a cardiac rehabilitation program. Program is directly supervised by qualified healthcare provider(s). Training is performed for a minimum of 30 to 45 min per session, in sessions performed at least 3 times/wk for a minimum of 12 wk.36–46 Patients may not initially achieve these targets, and a treatment goal is to progress to these levels over time. Training involves intermittent bouts of walking to moderate-to-maximum claudication, alternating with periods of rest. Warm-up and cool-down periods precede and follow each session of walking. \| \| Structured community- or home-based exercise program \| Structured exercise program that takes place in the personal setting of the patient rather than in a clinical setting.41,47–51 Program is self-directed with the guidance of healthcare providers who prescribe an exercise regimen similar to that of a supervised program. Patient counseling ensures that patients understand how to begin the program, how to maintain the program, and how to progress the difficulty of the walking (by increasing distance or speed). Program may incorporate behavioral change techniques, such as health coaching and/or use of activity monitors. \| \| Emergency versus urgent \| An emergency procedure is one in which life or limb is threatened if the patient is not in the operating room or interventional suite and/or where there is time for no or very limited clinical evaluation, typically within <6 h. An urgent procedure is one in which there may be time for a limited clinical evaluation, usually when life or limb is threatened if the patient is not in the operating room or interventional suite, typically between 6 and 24 h. \| \| Interdisciplinary care team \| A team of professionals representing different disciplines to assist in the evaluation and management of the patient with PAD. For the care of patients with CLI, the interdisciplinary care team should include individuals who are skilled in endovascular revascularization, surgical revascularization, wound healing therapies and foot surgery, and medical evaluation and care. Interdisciplinary care team members may include: Vascular medical and surgical specialists (ie, vascular medicine, vascular surgery, interventional radiology, interventional cardiology) Nurses Orthopedic surgeons and podiatrists Endocrinologists Internal medicine specialists Infectious disease specialists Radiology and vascular imaging specialists Physical medicine and rehabilitation clinicians Orthotics and prosthetics specialists Social workers Exercise physiologists Physical and occupational therapists Nutritionists/dieticians \| \| Cardiovascular ischemic events \| Acute coronary syndrome (acute MI, unstable angina), stroke, or cardiovascular death. \| \| Limb-related events \| Worsening claudication, new CLI, new lower extremity revascularization, or new ischemic amputation. \| Open in a new tab ABI indicates ankle-brachial index; ALI, acute limb ischemia; CLI, critical limb ischemia; MI, myocardial infarction; PAD, peripheral artery disease; TBI, toe-brachial index; and TcPO 2, transcutaneous oxygen pressure. 2. Clinical Assessment for PAD Evaluating the patient for PAD begins with the clinical history, review of symptoms, and physical examination. 2.1. History and Physical Examination: Recommendations Recommendations for History and Physical Examination CORLOERecommendations IB-NRPatients at increased risk of PAD (Table 4) should undergo a comprehensive medical history and a review of symptoms to assess for exertional leg symptoms, including claudication or other walking impairment, ischemic rest pain, and nonhealing wounds.52–57 See Online Data Supplement 1.The symptoms and signs of PAD are variable. Patients with PAD may experience the classic symptom of claudication or may present with advanced disease, including CLI. Studies have demonstrated that the majority of patients with confirmed PAD do not have typical claudication but have other non–joint-related limb symptoms or are asymptomatic.53,55 Atypical lower extremity symptoms related to PAD may include pain or discomfort that begins at rest but worsens with exertion, pain or discomfort that does not stop an individual from walking, and pain or discomfort that begins with exertion but is not alleviated within 10 minutes of rest.54 Patients with PAD who do not have typical claudication but have other leg symptoms, or who are asymptomatic, have been shown to have functional impairment comparable to patients with claudication.54 Thus, all patients at increased risk of PAD should be asked not only about claudication but also about other exertional non–joint-related limb symptoms and perceived walking impairment. IB-NRPatients at increased risk of PAD (Table 4) should undergo vascular examination, including palpation of lower extremity pulses (ie, femoral, popliteal, dorsalis pedis, and posterior tibial), auscultation for femoral bruits, and inspection of the legs and feet.56,58,59 See Online Data Supplements.A thorough lower extremity vascular examination and careful inspection of the legs and feet are important components of the clinical assessment for PAD. To perform a thorough examination, legs and feet are examined with lower garments (pants/skirt, shoes, and socks) removed. Examination findings suggestive of PAD are shown in Table 5. Lower extremity pulses should be assessed and rated as follows: 0, absent; 1, diminished; 2, normal; or 3, bounding. Reproducibility of pulse assessment is better for detection of normal versus absent pulse than for normal versus diminished pulse.56 Absence of the dorsalis pedis pulse is less accurate for diagnosis of PAD than is absence of the posterior tibial pulse because the dorsalis pedis pulse can be absent on examination in a significant percentage of healthy patients.56,58 The presence of multiple abnormal physical findings (ie, multiple pulse abnormalities, bruits) increases the likelihood of confirmed PAD.56,58,59 Abnormal physical findings, such as a pulse abnormality, require confirmation with the ankle-brachial index (ABI) to establish the diagnosis of PAD. Similarly, an entirely normal pulse examination and absence of bruits decreases the likelihood of confirmed PAD.56,58 The presence of nonhealing lower extremity wounds may be a sign of CLI. Findings of cool or discolored skin and delayed capillary refill are not reliable for PAD diagnosis.56 To confirm the diagnosis of PAD, abnormal physical examination findings must be confirmed with diagnostic testing (Section 3), generally with the ABI as the initial test. IB-NRPatients with PAD should undergo noninvasive blood pressure measurement in both arms at least once during the initial assessment.60–62 See Online Data Supplement 1.An inter-arm blood pressure difference of >15 to 20 mm Hg is abnormal and suggestive of subclavian (or innominate) artery stenosis. Patients with PAD are at increased risk of subclavian artery stenosis.60–62 Measuring blood pressure in both arms identifies the arm with the highest systolic pressure, a requirement for accurate measurement of the ABI.27 Identification of unequal blood pressures in the arms also allows for more accurate measurement of blood pressure in the treatment of hypertension (ie, blood pressure is taken at the arm with higher measurements). Although a difference in arm systolic pressures of >15 to 20 mm Hg suggests subclavian (or innominate) artery stenosis, in the absence of symptoms (eg, arm claudication or symptoms of vertebral artery steal), no further imaging or intervention is warranted. Open in a new tab Table 4. Patients at Increased Risk of PAD. Age ≥65 y Age 50–64 y, with risk factors for atherosclerosis (eg, diabetes mellitus, history of smoking, hyperlipidemia, hypertension) or family history of PAD63 Age <50 y, with diabetes mellitus and 1 additional risk factor for atherosclerosis Individuals with known atherosclerotic disease in another vascular bed (eg, coronary, carotid, subclavian, renal, mesenteric artery stenosis, or AAA) Open in a new tab AAA indicates abdominal aortic aneurysm; PAD, peripheral artery disease. Table 5. History and/or Physical Examination Findings Suggestive of PAD. History Claudication Other non–joint-related exertional lower extremity symptoms (not typical of claudication) Impaired walking function Ischemic rest pain Physical Examination Abnormal lower extremity pulse examination Vascular bruit Nonhealing lower extremity wound Lower extremity gangrene Other suggestive lower extremity physical findings (eg, elevation pallor/dependent rubor) Open in a new tab PAD indicates peripheral artery disease. 3. Diagnostic Testing for the Patient with Suspected Lower Extremity PAD (Claudication or CLI) 3.1. Resting ABI for Diagnosing PAD: Recommendations Recommendations for Resting ABI for Diagnosing PAD CORLOERecommendations IB-NRIn patients with history or physical examination findings suggestive of PAD (Table 5), the resting ABI, with or without segmental pressures and waveforms, is recommended to establish the diagnosis.64–69 See Online Data Supplement 4.The resting ABI is obtained by measuring systolic blood pressures at the arms (brachial arteries) and ankles (dorsalis pedis and posterior tibial arteries) in the supine position by using a Doppler device. The ABI of each leg is calculated by dividing the higher of the dorsalis pedis or posterior tibial pressure by the higher of the right or left arm blood pressure.27 In patients with a history or physical examination suggestive of PAD, the ABI has good validity as a first-line test in the diagnosis of PAD, as shown by vascular imaging, with sensitivities ranging from 68% to 84% and specificities from 84% to 99%.64–69 Segmental lower extremity blood pressures and Doppler or plethysmographic waveforms (pulse volume recordings) can be used to localize anatomic segments of disease (eg, aortoiliac, femoropopliteal, infrapopliteal).34,70,71 IC-LDResting ABI results should be reported as abnormal (ABI ≤0.90), borderline (ABI 0.91–0.99), normal (1.00–1.40), or noncompressible (ABI >1.40).27,67–69,72 See Online Data Supplement 4.Standardized reporting improves communication among healthcare providers. Calculated ABI values should be recorded to 2 decimal places. Patients with ABI ≤0.90 are diagnosed with PAD.67–69 Those with ABI 0.91 to 0.99 may possibly have PAD and should undergo exercise ABI, if the clinical suspicion of PAD is significant (Tables 4 and 5).73,74 Values >1.40 indicate that the arteries were not able to be compressed, which is more common among individuals with diabetes mellitus and/or advanced chronic kidney disease. In the setting of noncompressible ABI values, additional imaging can be used to diagnose PAD if the clinical suspicion is significant (Figures 1 and 2).72 These cutpoints for ABI interpretation have been previously proposed and represent a reasonable standardized categorization.27 IIaB-NRIn patients at increased risk of PAD (Table 4) but without history or physical examination findings suggestive of PAD (Table 5), measurement of the resting ABI is reasonable.54,55,75–97 See Online Data Supplements 3 and 4.The ABI test is noninvasive, is simple to perform, and has minimal risks, making it suitable for use in asymptomatic individuals. Previous studies have demonstrated a significant prevalence of abnormal resting ABI among asymptomatic patients with risk factors for PAD.55,79,95 A significant body of evidence demonstrates that patients with an abnormal ABI who are asymptomatic have poorer cardiovascular morbidity and mortality outcomes than do patients with normal ABI.79–87 While there is no conclusive evidence that aspirin treatment changes cardiovascular or limb outcomes in this population, in 1 cohort study of 5480 patients with asymptomatic PAD, statin treatment improved cardiovascular outcomes.75–78,96 There is also evidence that asymptomatic patients with a low resting ABI have a poorer functional status and a more rapid rate of functional decline than do patients with a normal ABI.54,88–92 Although physical activity has been shown to be associated with improvement in functional status in patients with asymptomatic PAD,93,94 the benefit of resting ABI testing to identify asymptomatic patients who are at increased risk of functional decline and may benefit from structured exercise programs remains to be determined. III: No BenefitB-NRIn patients not at increased risk of PAD (Table 4) and without history or physical examination findings suggestive of PAD (Table 5), the ABI is not recommended.95,98 See Online Data Supplement 4.The prevalence of PAD among individuals without risk factors for atherosclerosis and who are <50 years of age is low. Data from population-based cohort studies have demonstrated a low prevalence (approximately 1%) of abnormal resting ABI among individuals <50 years of age.95,98 In the NHANES (National Health and Nutrition Study), approximately 95% of participants with an abnormal resting ABI had at least 1 risk factor for atherosclerosis.95 The yield of ABI testing among younger, asymptomatic individuals without risk factors for atherosclerosis is low, and these patients should not be routinely tested for PAD.95,98 Open in a new tab Figure 1. Diagnostic Testing for Suspected PAD. Open in a new tab Colors correspond to Class of Recommendation in Table 1. ABI indicates ankle-brachial index; CLI, critical limb ischemia; CTA, computed tomography angiography; GDMT, guideline-directed management and therapy; MRA, magnetic resonance angiography; PAD, peripheral artery disease; and TBI, toe-brachial index. Figure 2. Diagnostic Testing for Suspected CLI. Open in a new tab Colors correspond to Class of Recommendation in Table 1. Order based on expert consensus. †TBI with waveforms, if not already performed. ABI indicates ankle-brachial index; CLI, critical limb ischemia; CTA, computed tomography angiography; MRA, magnetic resonance angiography; TcPO 2, transcutaneous oxygen pressure; and TBI, toe-brachial index. 3.2. Physiological Testing: Recommendations Recommendations for Physiological Testing CORLOERecommendations IB-NRToe-brachial index (TBI) should be measured to diagnose patients with suspected PAD when the ABI is greater than 1.40.72,99–102 See Online Data Supplement 5.TBI is a noninvasive test that is useful to evaluate for PAD in patents with noncompressible arteries, which cause an artificial elevation of the ABI.99,100,102,103 A TBI ≤0.70 is abnormal and diagnostic of PAD because the digital arteries are rarely noncompressible.99–102,104,105 Patients with longstanding diabetes mellitus72,101 or advanced chronic kidney disease106 have a high incidence of noncompressible arteries. Therefore, TBI assessment allows for the diagnosis of PAD in these patients with noncompressible arteries who have history or physical examination findings suggestive of PAD (Figure 1). IB-NRPatients with exertional non–joint-related leg symptoms and normal or borderline resting ABI (>0.90 and ≤1.40) should undergo exercise treadmill ABI testing to evaluate for PAD.71,74,107–110 See Online Data Supplement 5.Exercise treadmill ABI testing is important to objectively measure symptom limitations and diagnose PAD.71,74,107–110 It is useful in establishing the diagnosis of lower extremity PAD in the symptomatic patient when resting ABIs are normal or borderline and to differentiate claudication from pseudoclaudication in individuals with exertional leg symptoms. If the post-exercise treadmill ABI is normal, alternative causes of leg pain are considered (Table 6). If a treadmill is not available, the pedal plantarflexion ABI test is a reasonable alternative because the results correlate well with treadmill ABIs (Figure 1).111 IIaB-NRIn patients with PAD and an abnormal resting ABI (≤0.90), exercise treadmill ABI testing can be useful to objectively assess functional status.71,74,107–110 See Online Data Supplement 5.In patients with PAD, exercise treadmill ABI testing can objectively assess symptoms, measure change in ABI in response to exercise, and assess functional status71,74,107–110 (Figure 1). It can be useful to correlate exertional lower extremity symptoms to a decline in ABI after treadmill exercise. Exercise treadmill ABI testing can document the magnitude of symptom limitation in patients with PAD and provide objective data that can demonstrate the safety of exercise and help to individualize exercise prescriptions in patients with PAD before initiation of a formal program of structured exercise training. Exercise ABI may also be used to objectively measure the functional improvement obtained in response to claudication treatment (eg, structured exercise program or revascularization). Administration of a 6-minute walk test in a corridor is a reasonable alternative to treadmill ABI testing for assessment of functional status.54 IIaB-NRIn patients with normal (1.00–1.40) or borderline (0.91–0.99) ABI in the setting of nonhealing wounds or gangrene, it is reasonable to diagnose CLI by using TBI with waveforms, transcutaneous oxygen pressure (TcPO 2), or skin perfusion pressure (SPP).112–116 See Online Data Supplement 5.The toe pressure and TBI may be discordant with the ABI 0.90 to 1.40 in some patients with diabetes mellitus and a nonhealing wound (Figure 2).115,116 A TBI ≤0.70 is considered diagnostic of PAD.101,104,105 Doppler or plethysmographic waveforms taken at the toe supplement the toe pressure and TBI measurement and may be severely dampened in the setting of CLI. The likelihood of wound healing decreases with toe pressure <30 mm Hg.100 Perfusion assessment measures (ie, TBI with waveforms, TcPO 2, SPP) are obtained in a warm room to prevent arterial vasoconstriction in response to the cold. TcPO 2 measurements are performed with a standardized protocol and are taken at multiple sites.117 Correlation between TBI, TcPO 2, and SPP has been reported.113 TcPO 2>30 mm Hg has been used to predict ulcer healing.118 SPP ≥30 to 50 mm Hg is associated with increased likelihood of wound healing.113 If perfusion measures are normal or only mildly impaired, alternative causes of the nonhealing wounds are considered (Table 7). TcPO 2 and SPP can be used in angiosome-targeted assessment for revascularization.119 IIaB-NRIn patients with PAD with an abnormal ABI (≤0.90) or with noncompressible arteries (ABI >1.40 and TBI ≤0.70) in the setting of nonhealing wounds or gangrene, TBI with waveforms, TcPO 2, or SPP can be useful to evaluate local perfusion.112–116 See Online Data Supplement 5.Perfusion assessment measures (eg, TBI with waveforms, TcPO 2, SPP) can be useful when the ABI is only mildly reduced (eg, ABI 0.70–0.90) to determine whether factors other than PAD may be contributing to impaired wound healing (Figure 2). These perfusion assessment measures are obtained in a warm room to prevent arterial vasoconstriction in response to the cold. TcPO 2 measurements are performed with a standardized protocol and are taken at multiple sites.117 The likelihood of wound healing decreases with toe pressure <30 mm Hg.100 There is correlation between TBI, TcPO 2, and SPP. TcPO 2>30 mm Hg has been used to predict ulcer healing.118 SPP ≥30 to 50 mm Hg is associated with increased likelihood of wound healing.113 TcPO 2 and SPP can be used in angiosome-targeted assessment for revascularization.119 Additional perfusion assessment may also be useful for patients with nonhealing wounds or gangrene who have noncompressible arteries (ABI >1.40) but who have a diagnosis of PAD that is based on an abnormal TBI (ABI ≤0.70). Open in a new tab Table 6. Alternative Diagnoses for Leg Pain or Claudication With Normal Physiological Testing (Not PAD-Related). \| Condition \| Location \| Characteristic \| Effect of Exercise \| Effect of Rest \| Effect of Position \| Other Characteristics \| :--- :--- :--- \| Symptomatic Baker's cyst \| Behind knee, down calf \| Swelling, tenderness \| With exercise \| Also present at rest \| None \| Not intermittent \| \| Venous claudication \| Entire leg, worse in calf \| Tight, bursting pain \| After walking \| Subsides slowly \| Relief speeded by elevation \| History of iliofemoral deep vein thrombosis; edema; signs of venous stasis \| \| Chronic compartment syndrome \| Calf muscles \| Tight, bursting pain \| After much exercise (jogging) \| Subsides very slowly \| Relief with rest \| Typically heavy muscled athletes \| \| Spinal stenosis \| Often bilateral buttocks, posterior leg \| Pain and weakness \| May mimic claudication \| Variable relief but can take a long time to recover \| Relief by lumbar spine flexion \| Worse with standing and extending spine \| \| Nerve root compression \| Radiates down leg \| Sharp lancinating pain \| Induced by sitting, standing, or walking \| Often present at rest \| Improved by change in position \| History of back problems; worse with sitting; relief when supine or sitting \| \| Hip arthritis \| Lateral hip, thigh \| Aching discomfort \| After variable degree of exercise \| Not quickly relieved \| Improved when not weight bearing \| Symptoms variable; history of degenerative arthritis \| \| Foot/ankle arthritis \| Ankle, foot, arch \| Aching pain \| After variable degree of exercise \| Not quickly relieved \| May be relieved by not bearing weight \| Symptoms variable; may be related to activity level or present at rest \| Open in a new tab Modified from Norgren L et al.35 PAD indicates peripheral artery disease. Table 7. Alternative Diagnoses for Nonhealing Wounds With Normal Physiological Testing (Not PAD-Related). \| Condition \| Location \| Characteristics and Causes \| :--- \| Venous ulcer \| Distal leg, especially above medial mellolus \| Develops in regions of skin changes due to chronic venous disease and local venous hypertension Typically wet (ie, wound drainage) rather than dry lesion \| \| Distal small arterial occlusion (microangiopathy) \| Toes, foot, leg \| Diabetic microangiopathy End-stage renal disease Thromboangiitis obliterans (Buerger's) Sickle cell anemia Vasculitis (eg, Churg-Strauss, Henoch-Schonlein purpura, leukocytoclastic vasculitis, microscopic polyangiitis, polyarteritis nodosa) Scleroderma Cryoagglutination Embolic (eg, cholesterol emboli, thromboemboli, endocarditis) Thrombotic (eg, antiphospholipid antibody syndrome, Sneddon's syndrome, warfarin skin necrosis, disseminated intravascular coagulation, livedoid vasculitis, protein C or S deficiency, prolonged vasospasm) \| \| Local injury \| Toes, foot, leg \| Trauma Insect or animal bite Burn \| \| Medication related \| Toes, foot, leg \| Drug reactions (eg, erythema multiforme) Medication direct toxicity (eg, doxorubicin, hydroxyurea, some tyrosine kinase inhibitors) \| \| Neuropathic \| Pressure zones of foot \| Hyperkeratosis surrounds the ulcer Diabetes mellitus with peripheral neuropathy Peripheral neuropathy without diabetes mellitus Leprosy \| \| Autoimmune injury \| Toes, foot, leg \| With blisters (eg, pemphigoid, pemphigus, epidermolysis bullosa) Without blisters (eg, dermatomyositis, lupus, scleroderma) \| \| Infection \| Toes, foot, leg \| Bacterial (eg, pseudomonas, necrotizing streptococcus) Fungal (eg, blastomycosis, Madura foot, chromomycosis) Mycobacterial Parasitic (eg, Chagas, leishmaniasis) Viral (eg, herpes) \| \| Malignancy \| Toes, foot, leg \| Primary skin malignancy Metastatic malignancy Malignant transformation of ulcer \| \| Inflammatory \| Toes, foot, leg \| Necrobiosis lipoidica Pyoderma gangrenosum Granuloma annulare \| Open in a new tab PAD indicates peripheral artery disease. 3.3. Imaging for Anatomic Assessment: Recommendations Recommendations for Imaging for Anatomic Assessment CORLOERecommendations IB-NRDuplex ultrasound, computed tomography angiography (CTA), or magnetic resonance angiography (MRA) of the lower extremities is useful to diagnose anatomic location and severity of stenosis for patients with symptomatic PAD in whom revascularization is considered.118,120–122 See Online Data Supplement 6.For symptomatic patients in whom ABI/TBI confirms PAD and in whom revascularization is considered, additional imaging with duplex ultrasonography, CTA, or MRA is useful to develop an individualized treatment plan, including assistance in selection of vascular access sites, identification of significant lesions, and determination of the feasibility of and modality for invasive treatment. All 3 of these noninvasive imaging methods have good sensitivity and specificity as compared with invasive angiography.118,120–122 Renal function does not affect the safety of duplex ultrasonography, although duplex offers lower spatial resolution than CTA and MRA in the setting of arterial calcification. The tomographic data from CTA and MRA afford 3-dimensional reconstruction of the vessels examined. The iodinated contrast used in CTA confers risk of contrast-induced nephropathy and (rarely) severe allergic reaction123,124; CTA uses ionizing radiation. MRA does not use ionizing radiation; however, gadolinium contrast used frequently in MRA studies confers risk of nephrogenic systemic sclerosis for patients with advanced renal dysfunction and is therefore contraindicated in this population.125 The choice of the examination should be determined in an individualized approach to the anatomic assessment for each patient, including risk–benefit assessment of each study type. If these noninvasive tests are nondiagnostic, then invasive angiography may be required to delineate anatomy and plan revascularization. IC-EOInvasive angiography is useful for patients with CLI in whom revascularization is considered. N/A By definition, CLI results from extensive PAD that limits tissue perfusion. Because timely diagnosis and treatment are essential to preserve tissue viability in CLI, it is often most effective and expeditious to pursue invasive angiography with endovascular revascularization directly, without delay and potential risk of additional noninvasive imaging. IIaC-EOInvasive angiography is reasonable for patients with lifestyle-limiting claudication with an inadequate response to GDMT for whom revascularization is considered. N/A For patients with lifestyle-limiting claudication despite GDMT (including structured exercise therapy) for whom revascularization is being considered, proceeding directly to invasive angiography for anatomic assessment and to determine revascularization strategy is reasonable. In certain clinical settings, noninvasive imaging studies for anatomic assessment (ie, duplex ultrasound, CTA, or MRA) may not be available because of lack of local resources or expertise. In addition, there are clinical scenarios in which noninvasive studies for anatomic assessment may be perceived to confer greater risk to the patient than invasive angiography (eg, patient with advanced chronic kidney disease for whom contrast dose for invasive angiography would be lower than that required for CTA). III: HarmB-RInvasive and noninvasive angiography (ie, CTA, MRA) should not be performed for the anatomic assessment of patients with asymptomatic PAD.123,124,126 See Online Data Supplements 6 and 7.Angiography, either noninvasive or invasive, should not be performed for the anatomic assessment of patients with PAD without leg symptoms because delineation of anatomy will not change treatment for this population. This lack of benefit occurs in the setting of risk of contrast-induced nephropathy, patient discomfort, and allergic reactions.123,124,126 This recommendation does not address assessment of lower extremity aneurysmal disease or nonatherosclerotic causes of arterial disease, which is beyond the scope of this document. Open in a new tab 4. Screening for Atherosclerotic Disease in Other Vascular Beds for the Patient with PAD 4.1. Abdominal Aortic Aneurysm: Recommendation Recommendation for Abdominal Aortic Aneurysm CORLOERecommendation IIaB-NRA screening duplex ultrasound for abdominal aortic aneurysm (AAA) is reasonable in patients with symptomatic PAD.127–129 See Online Data Supplement 8.PAD has been recognized as a risk factor for AAA. In observational studies, the prevalence of AAA (aortic diameter ≥3 cm) was higher in patients with symptomatic PAD than in the general population127,129 and in a population of patients with atherosclerotic risk factors.128 The prevalence of AAA among patients with PAD increased with age, beginning in patients ≥55 years of age, and was highest in patients ≥75 years of age.129 There are no data on AAA screening in patients with asymptomatic PAD. This recommendation refers to screening patients with symptomatic PAD for AAA regardless of patient age, sex, smoking history, or family history of AAA. Recommendations for screening the general population with risk factors for AAA (based on age, sex, smoking history, and family history) have been previously published.9 Open in a new tab 4.2. Screening for Asymptomatic Atherosclerosis in Other Arterial Beds (Coronary, Carotid, and Renal Arteries) The prevalence of atherosclerosis in the coronary, carotid, and renal arteries is higher in patients with PAD than in those without pad.128,130–135 However, intensive atherosclerosis risk factor modification in patients with PAD is justified regardless of the presence of disease in other arterial beds. Thus, the only justification for screening for disease in other arterial beds is if revascularization results in a reduced risk of myocardial infarction (MI), stroke, or death, and this has never been shown. Currently, there is no evidence to demonstrate that screening all patients with PAD for asymptomatic atherosclerosis in other arterial beds improves clinical outcome. Intensive treatment of risk factors through GDMT is the principle method for preventing adverse cardiovascular ischemic events from asymptomatic disease in other arterial beds. 5. Medical Therapy for the Patient with PAD Patients with PAD should receive a comprehensive program of GDMT, including structured exercise and lifestyle modification, to reduce cardiovascular ischemic events and improve functional status. Smoking cessation is a vital component of care for patients with PAD who continue to smoke. A guideline-based program of pharmacotherapy to reduce cardiovascular ischemic events and limb-related events should be prescribed for each patient with PAD and is customized to individual risk factors, such as whether the patient also has diabetes mellitus. Previous studies have demonstrated that patients with PAD are less likely to receive GDMT than are patients with other forms of cardiovascular disease, including coronary artery disease (CAD).136–138 5.1. Antiplatelet Agents: Recommendations Recommendations for Antiplatelet Agents CORLOERecommendations IAAntiplatelet therapy with aspirin alone (range 75–325 mg per day) or clopidogrel alone (75 mg per day) is recommended to reduce MI, stroke, and vascular death in patients with symptomatic PAD.139–142 See Online Data Supplement 13.The effect of antiplatelet therapy on cardiovascular events has been systematically reviewed by the Antithrombotic Trialists' Collaboration.139 Of note, this meta-analysis included studies of antiplatelet agents other than aspirin or clopidogrel. Among patients with symptomatic PAD treated with antiplatelet therapy, there was a 22% odds reduction for cardiovascular events, including MI, stroke, or vascular death.139 Symptomatic patients with lower extremity PAD included both those with claudication and those with prior lower extremity revascularization. The Antithrombotic Trialists' Collaboration meta-analysis also compared the efficacy of different doses of aspirin.139 The proportional reduction in vascular events was 32% with 75 to 150 mg daily, 26% with 160 to 325 mg daily, and 19% with 500 to 1500 mg daily, whereas there was a significantly smaller (13%) reduction in cardiovascular events in patients being treated with <75 mg of aspirin per day.139 CLIPS (Critical Leg Ischaemia Prevention Study) demonstrated a benefit of aspirin (100 mg daily) compared with placebo in preventing vascular events, but the study was too small to derive meaningful conclusions.140 A meta-analysis of trials of aspirin (alone or in combination with dipyridamole) for prevention of cardiovascular events in patients with PAD found a non–statistically significant reduction in the primary endpoint of cardiovascular death, MI, and stroke and a statistically significant reduction in the secondary endpoint of nonfatal stroke with aspirin versus placebo.141 The CAPRIE (Clopidogrel Versus Aspirin in Patients at Risk of Ischemic Events) trial demonstrated a benefit of clopidogrel as compared with aspirin in cardiovascular risk reduction and bleeding events in a population of patients with symptomatic atherosclerotic vascular disease, including a subgroup of patients with symptomatic PAD.142 IIaC-EOIn asymptomatic patients with PAD (ABI ≤0.90), antiplatelet therapy is reasonable to reduce the risk of MI, stroke, or vascular death. See Online Data Supplement 13.Patients with PAD (ie, ABI ≤0.90) who do not have claudication may have leg symptoms atypical for claudication or may be too functionally limited to allow for adequate leg symptom assessment. Patients with PAD without claudication are at increased cardiovascular risk.79 Subgroup analysis in a trial evaluating asymptomatic patients did not show an effect of aspirin in patients with an abnormally low ABI (<0.80 or ≤0.90).76 However, the trial was not powered to analyze subgroups, and the uncertainty of the result does not rule out the possibility that aspirin could provide benefit in such patients, especially in those at increased risk of cardiovascular events. Another trial that included asymptomatic patients was too small to derive meaningful conclusions.140 IIbB-RIn asymptomatic patients with borderline ABI (0.91–0.99), the usefulness of antiplatelet therapy to reduce the risk of MI, stroke, or vascular death is uncertain.75,76 See Online Data Supplement 13.In asymptomatic patients with an abnormal or borderline ABI, 2 RCTs found that aspirin had no effect in reducing cardiovascular events75,76 and might increase bleeding.76 However, the trials were not powered to examine patients with borderline ABI separately. Given that cardiovascular risk is lower in patients with borderline ABI than in those with abnormal ABI,80 it would be unlikely that aspirin would have a meaningful effect in this subgroup when there was no evidence of an effect in the total trial populations. IIbB-RThe effectiveness of dual antiplatelet therapy (DAPT) (aspirin and clopidogrel) to reduce the risk of cardiovascular ischemic events in patients with symptomatic PAD is not well established.143,144 See Online Data Supplement 13.Based on findings from a subset of patients with PAD in the CHARISMA (Clopidogrel for High Atherothrombotic Risk and Ischemic Stabilization, Management, and Avoidance) trial, DAPT with aspirin plus clopidogrel may be considered for patients with PAD at particularly high risk of cardiovascular ischemic events who are not at high risk of bleeding.143,144 Currently, there are sparse data on newer P2Y 12 antagonists for PAD. There is uncertainty about the net benefit of long-term DAPT for patients with PAD—specifically the balance of risks of cardiovascular ischemic events versus major bleeding. Additional clinical trials are needed in the population with PAD. Refer to the DAPT guideline focused update for DAPT recommendations specifically for CAD.20 IIbC-LDDAPT (aspirin and clopidogrel) may be reasonable to reduce the risk of limb-related events in patients with symptomatic PAD after lower extremity revascularization.145–148 See Online Data Supplements 13 and 14.There are sparse data on DAPT after lower extremity revascularization. Still, DAPT is prescribed in up to 55% of patients after endovascular revascularization for CLI.146 One small RCT of aspirin or aspirin plus clopidogrel in patients undergoing endovascular revascularization demonstrated that patients with DAPT had fewer repeat revascularization procedures for clinical symptoms.145 A subsequent small RCT of aspirin plus placebo or aspirin plus clopidogrel in patients after endovascular revascularization also showed a decrease in the need for repeat revascularization at 6 months in patients receiving clopidogrel.147 An RCT of aspirin plus placebo or aspirin plus clopidogrel in patients who underwent below-knee bypass graft showed a decrease in limb-related events only in the prespecified subgroup of patients with prosthetic bypass grafts.148 Refer to the DAPT guideline focused update for DAPT recommendations specifically for CAD.20 IIbB-RThe overall clinical benefit of vorapaxar added to existing antiplatelet therapy in patients with symptomatic PAD is uncertain.149–152 See Online Data Supplement 13.This novel antagonist of protease-activated receptor-1 added to existing antiplatelet therapy reduced the risk of cardiovascular ischemic events in patients with atherosclerosis who were receiving standard therapy in an RCT.150,151 However, it also increased the risk of moderate or severe bleeding. Although the cardiovascular benefit was not demonstrated in the subgroup with symptomatic PAD, there was a reduction in limb-related events with vorapaxar, specifically in acute limb ischemia (ALI) and peripheral revascularization.149,152 More than half of ALI events in the PAD subset were due to thrombosis of lower extremity bypass grafts.149 Unfortunately, the benefit in limb events in patients with PAD was accompanied by an increased risk of bleeding.149,152 Therefore, the overall clinical benefit of vorapaxar in patients with PAD is uncertain. Open in a new tab 5.2. Statin Agents: Recommendation Recommendation for Statin Agents CORLOERecommendation IATreatment with a statin medication is indicated for all patients with PAD.96,153–157 See Online Data Supplements 15 and 16.Statin therapy improves both cardiovascular and limb outcomes in patients with PAD.157 In a subgroup of 6748 patients with PAD in the HPS (Heart Protection Study), simvastatin 40 mg daily reduced the rate of first major vascular event by 22% relative to placebo.155 In a multinational registry, statin use among patients with PAD reduced 4-year adverse limb-related events (ie, worsening claudication, new CLI, new lower extremity revascularization, new ischemic amputation) compared with no statin.153 Use of simvastatin in the HPS reduced relative risk of peripheral vascular events (including noncoronary revascularization, aneurysm repair, major amputation, or PAD death) compared with placebo.155 In Medicare patients undergoing lower extremity revascularization, 1-year limb salvage rates were improved among those receiving statin medication.154 In a multicenter RCT, use of atorvastatin 80 mg daily improved pain-free walking time and community-based walking at 12 months compared with placebo.156 In 1 cohort study of 5480 patients with asymptomatic PAD, statin treatment improved cardiovascular outcomes.96 Guidelines for dosing of statin medications have been previously published.24 Open in a new tab 5.3. Antihypertensive Agents: Recommendations Recommendations for Antihypertensive Agents CORLOERecommendations IAAntihypertensive therapy should be administered to patients with hypertension and PAD to reduce the risk of MI, stroke, heart failure, and cardiovascular death.158–162 See Online Data Supplements 17 and 18.Treatment of elevated blood pressure is indicated to lower the risk of cardiovascular events.162 Target blood pressure and selection of antihypertensive therapy should be consistent with current published guidelines for hypertension management. Concerns have been raised that antihypertensive therapy may reduce limb perfusion. However, multiple studies have demonstrated that blood pressure treatment, including the use of beta blockers, does not worsen claudication symptoms or impair functional status in patients with PAD.163–165 There is no evidence that one class of antihypertensive medication or strategy is superior for blood pressure lowering in PAD.158,166,167 An updated multisocietal guideline on the management of high blood pressure is anticipated in 2017. IIaAThe use of angiotensin-converting enzyme inhibitors or angiotensin-receptor blockers can be effective to reduce the risk of cardiovascular ischemic events in patients with PAD.161,168,169 See Online Data Supplement 17.The effect of ramipril versus placebo on cardiovascular events was studied in high-risk patients free of heart failure in the HOPE (Heart Outcomes Prevention Evaluation) trial.168,169 Patients were normotensive on average at the time of enrollment. In a subgroup of 4051 patients with PAD, ramipril reduced the risk of MI, stroke, or vascular death by 25%, similar to the efficacy in the entire study population.168,169 The efficacy was similar in patients with PAD with symptomatic disease and asymptomatic low ABI.168 ONTARGET (Ongoing Telmisartan Alone and in Combination With Ramipril Global Endpoint Trial) compared telmisartan, ramipril, and combination therapy in patients with cardiovascular disease, including PAD, and/or diabetes mellitus.161 All 3 treatments had similar cardiovascular event rates with higher rates of adverse events (including hypotension, syncope, and renal failure) in the combination-therapy group. The efficacy of telmisartan was similar in the subgroup of 3468 patients with PAD, which supports the use of angiotensin-receptor blockers as an alternative to angiotensin-converting enzyme inhibitors.161 The effect of angiotensin-receptor blockers in asymptomatic PAD has not been studied. Open in a new tab 5.4. Smoking Cessation: Recommendations Recommendations for Smoking Cessation CORLOERecommendations IAPatients with PAD who smoke cigarettes or use other forms of tobacco should be advised at every visit to quit.170–172 See Online Data Supplements 19 and 20.Tobacco use is a strong risk factor for the development and progression of PAD.173,174 Sparse evidence exists with regard to the association of novel tobacco product use, including electronic cigarettes, and PAD.175 Observational studies suggest that smoking cessation is associated with lower rates of cardiovascular ischemic events, limb-related events, bypass graft failure, amputation, and death in patients with PAD.172,176–178 Clinician advice increases quit rates, which supports simple provider-based measures as a component of smoking cessation programs.22,171,179 IAPatients with PAD who smoke cigarettes should be assisted in developing a plan for quitting that includes pharmacotherapy (ie, varenicline, bupropion, and/or nicotine replacement therapy) and/or referral to a smoking cessation program.170,180–182 See Online Data Supplements 19 and 20.Coordinated smoking cessation interventions that include nonpharmacological and pharmacological approaches have the greatest efficacy. An RCT of a follow-up program and smoking cessation medications provided to hospitalized patients, including those with PAD, demonstrated a modest increase in quit rates.181 In an RCT of patients with PAD specifically, a comprehensive smoking cessation program combining counseling and pharmacological agents increased the rates of smoking cessation to 21.3%, compared with 6.8% with standard advice.170 Three pharmacological approaches (ie, varenicline, bupropion, and nicotine replacement therapy) used alone or in combination all increase smoking cessation rates.179,180,182 Two meta-analyses of RCTs of smoking cessation medications showed no evidence of increased cardiovascular event rates with nicotine replacement, bupropion, or varenicline.183,184 Sparse data suggest that electronic cigarettes have no benefit on smoking cessation rates.179 IB-NRPatients with PAD should avoid exposure to environmental tobacco smoke at work, at home, and in public places.185,186 See Online Data Supplement 20.Passive smoke exposure has been associated with the development of PAD.186 Observational studies have shown lower cardiovascular and cerebrovascular event rates in the general population after enactment of smoke-free legislation.185 The effects of avoidance of passive smoke exposure on limb-related events are not known. Open in a new tab 5.5. Glycemic Control: Recommendations Recommendations for Glycemic Control CORLOERecommendations IC-EOManagement of diabetes mellitus in the patient with PAD should be coordinated between members of the healthcare team. N/A Diabetes mellitus is an important risk factor for the development of PAD.187 Furthermore, the presence of diabetes mellitus increases the risk of adverse outcomes among patients with PAD, including progression to CLI, amputation, and death.188,189 A comprehensive care plan for patients with PAD and diabetes mellitus is important and may include diet and weight management, pharmacotherapy for glycemic control and management of other cardiovascular risk factors, and foot care and ulcer prevention.25,190 Guidelines for glycemic control among patients with diabetes mellitus and atherosclerotic vascular disease have been previously published.25,29 Regular follow-up with and communication among the patient's healthcare providers, including vascular specialists and diabetes care providers (eg, primary care physicians, endocrinologists) constitute an important component of care for patients with PAD and diabetes mellitus. IIaB-NRGlycemic control can be beneficial for patients with CLI to reduce limb-related outcomes.191,192 See Online Data Supplement 22.In a cohort of 1974 participants with diabetes mellitus from the Strong Heart Study, compared with patients without PAD, patients with PAD and a Hg A1c level <6.5% had lower age-adjusted odds of major amputation compared to patients with PAD and hemoglobin A1c 6.5% to 9.5% and hemoglobin A1c >9.5%.188 Glycemic control is particularly important for patients with PAD and diabetes mellitus who have CLI. Single-center observational studies have demonstrated improved limb-related outcomes, including lower rates of major amputation and improved patency after infrapopliteal intervention, among patients with CLI who have more optimized glycemic control parameters compared with patients with inferior glycemic control.191,192 Open in a new tab 5.6. Oral Anticoagulation: Recommendations Recommendations for Oral Anticoagulation CORLOERecommendations IIbB-RThe usefulness of anticoagulation to improve patency after lower extremity autogenous vein or prosthetic bypass is uncertain.193–195 See Online Data Supplements 23 and 24.Two RCTs evaluating the effectiveness of oral anticoagulation (warfarin) in improving lower extremity bypass patency demonstrated improved patency among the subgroup of patients with autogenous vein bypass grafts.193,194 However, a Cochrane systematic review showed no patency benefit with the use of anticoagulation compared with antiplatelet therapy.195 All RCTs and observational studies evaluating the effect of anticoagulants on bypass patency demonstrated increased bleeding complications associated with anticoagulant use. One RCT evaluating the effectiveness of oral anticoagulation (warfarin) in addition to aspirin in improving lower extremity bypass patency demonstrated improved patency in a subgroup of patients with 6-mm polytetrafluoroethylene (known as PTFE) bypass graft.196 Randomization to anticoagulation plus aspirin was associated with increased risk of death and major hemorrhage versus aspirin alone. III: HarmAAnticoagulation should not be used to reduce the risk of cardiovascular ischemic events in patients with PAD.194,196–198 See Online Data Supplements 23 and 24RCTs and observational studies have uniformly demonstrated that oral anticoagulation therapy aimed at decreasing major cardiovascular ischemic events provided no benefit and resulted in increased morbidity.194,196–198 In the WAVE (Warfarin Antiplatelet Vascular Evaluation) trial of patients with atherosclerotic vascular disease, including PAD, there was no difference in cardiovascular ischemic events among patients randomized to oral anticoagulation and antiplatelet therapy versus antiplatelet therapy alone.198 In addition, there was an increase in bleeding endpoints including life-threatening and intracranial bleeding.198 One RCT demonstrated increased death rate among patients randomized to warfarin plus aspirin versus aspirin alone after lower extremity bypass grafting.196 Open in a new tab 5.7. Cilostazol: Recommendation Recommendation for Cilostazol CORLOERecommendation IACilostazol is an effective therapy to improve symptoms and increase walking distance in patients with claudication.199,200 See Online Data Supplement 25.In a Cochrane review including 15 double-blind RCTs with a total of 3718 participants, cilostazol was associated with improvement in claudication symptoms but no changes in cardiovascular deaths or QoL when compared with placebo.199 In 1 RCT, cilostazol was more effective than pentoxifylline or placebo.200 Side effects include headache, abnormal stool (diarrhea), dizziness, and palpitations. Cilostazol is contraindicated in patients with congestive heart failure.201 In 1 trial, 20% of patients discontinued cilostazol within 3 months.202 Open in a new tab 5.8. Pentoxifylline: Recommendation Recommendation for Pentoxifylline CORLOERecommendation III: No BenefitB-RPentoxifylline is not effective for treatment of claudication.200,203 See Online Data Supplement 26.In a Cochrane review of 24 studies with 3377 participants, there was large variability in study design and results between individual studies, and therefore the review's effectiveness was unclear.203 Pentoxifylline was shown to be generally well tolerated.203 In a multicenter RCT of pentoxifylline, cilostazol, or placebo for patients with moderate-to-severe claudication, there was no difference between pentoxifylline and placebo in the primary endpoint of maximal walking distance.200 Therefore, pentoxifylline is not recommended as treatment for claudication. Open in a new tab 5.9. Chelation Therapy: Recommendation Recommendation for Chelation Therapy CORLOERecommendation III: No BenefitB-RChelation therapy (eg, ethylenediaminetetraacetic acid) is not beneficial for treatment of claudication.204 See Online Data Supplement 27.In a Cochrane review of 5 studies with 260 participants, chelation therapy showed no significant difference in symptoms (maximal and pain-free walking distance) compared with placebo.204 Open in a new tab 5.10. Homocysteine Lowering: Recommendation Recommendation for Homocysteine Lowering CORLOERecommendation III: No BenefitB-RB-complex vitamin supplementation to lower homocysteine levels for prevention of cardiovascular events in patients with PAD is not recommended.205–207 See Online Data Supplements 28 and 29.Although patients with PAD have been shown to have increased plasma homocysteine levels compared with patients without PAD, there is no evidence that B-complex vitamin supplementation improves clinical outcomes in patients with PAD.207 The HOPE-2 trial randomized 5522 patients with atherosclerotic vascular disease, including symptomatic PAD, or diabetes mellitus with additional risk factors to receive folic acid/vitamin B6/vitamin B12 or placebo.205,206 Despite lowering of homocysteine levels in the vitamin supplementation arm, there was no improvement in the primary endpoint of cardiovascular death, MI, or stroke. Open in a new tab 5.11. Influenza Vaccination: Recommendation Recommendation for Influenza Vaccination CORLOERecommendation IC-EOPatients with PAD should have an annual influenza vaccination. See Online Data Supplements 30 and 31.Observational studies have demonstrated reduced cardiovascular event rates among patients with cardiovascular disease who have received an influenza vaccination.30 Two RCTs that enrolled patients with CAD demonstrated a benefit of an influenza vaccination on the prevention of cardiovascular events, particularly coronary ischemic events.208,209 Although these trials did not specifically enroll participants with PAD, a majority of patients with PAD also have CAD.30 On the basis of this evidence, an annual influenza vaccination is recommended as a component of medical therapy for patients with PAD. Open in a new tab 6. Structured Exercise Therapy: Recommendations Structured exercise therapy is an important element of care for the patient with PAD. Components of structured exercise programs for PAD are outlined in Table 8. Table 8. Structured Exercise Programs for PAD: Definitions. Supervised exercise program (COR I, LOE A) Program takes place in a hospital or outpatient facility. Program uses intermittent walking exercise as the treatment modality. Program can be standalone or within a cardiac rehabilitation program. Program is directly supervised by qualified healthcare provider(s). Training is performed for a minimum of 30–45 min/session; sessions are performed at least 3 times/wk for a minimum of 12 wk.36–46 Training involves intermittent bouts of walking to moderate-to-maximum claudication, alternating with periods of rest. Warm-up and cool-down periods precede and follow each session of walking. Structured community- or home-based exercise program (COR IIa, LOE A) Program takes place in the personal setting of the patient rather than in a clinical setting.41,47–51 Program is self-directed with guidance of healthcare providers. Healthcare providers prescribe an exercise regimen similar to that of a supervised program. Patient counseling ensures understanding of how to begin and maintain the program and how to progress the difficulty of the walking (by increasing distance or speed). Program may incorporate behavioral change techniques, such as health coaching or use of activity monitors. Open in a new tab COR indicates Class of Recommendation; LOE, Level of Evidence; and PAD, peripheral artery disease. Recommendations for Structured Exercise Therapy CORLOERecommendations IAIn patients with claudication, a supervised exercise program is recommended to improve functional status and QoL and to reduce leg symptoms.36–38,40–46,48,210,211 See Online Data Supplement 32.The data supporting the efficacy of supervised exercise training as an initial treatment for claudication continue to develop and remain convincing, building on many earlier RCTs.40–46,48,210,211 Trials with long-term follow-up from 18 months37,38 to 7 years36 have demonstrated a persistent benefit of supervised exercise in patients with claudication. Data also support a benefit of supervised exercise for patients with symptomatic PAD and diabetes mellitus.212 The risk–benefit ratio for supervised exercise in PAD is favorable, with an excellent safety profile in patients screened for absolute contraindications to exercise such as exercise-limiting cardiovascular disease, amputation or wheelchair confinement, and other major comorbidities that would preclude exercise.36,39,49,213–216 Despite the health benefits associated with supervised exercise in patients with PAD, initiating and maintaining a high level of adherence remain challenging. Frequent contact with patients both when performing exercise in the supervised setting and at home has been somewhat effective in promoting retention.37,38 IB-RA supervised exercise program should be discussed as a treatment option for claudication before possible revascularization.36–38 See Online Data Supplement 32.The CLEVER (Claudication: Exercise Versus Endoluminal Revascularization) trial randomized patients with symptomatic aortoiliac PAD and showed comparable benefits for supervised exercise and stent revascularization at 6 and 18 months, with each therapy being superior to optimal medical care.37,38 Overall, the safety profile for supervised exercise was excellent. An RCT that compared 7-year effectiveness of supervised exercise or endovascular revascularization in patients with stable claudication with iliac or femoropopliteal disease found no differences in improved walking and QoL outcomes.36 Although more secondary interventions occurred in the exercise group, the total number of interventions was greater in the endovascular revascularization group. Collectively, these studies provide strong support for offering patients a supervised exercise program for reducing claudication symptoms and for improving functional status and QoL. A 3-month RCT that compared percutaneous transluminal angioplasty (PTA), supervised exercise, and combined treatment for claudication found that both supervised exercise and PTA improved clinical and QoL outcomes, whereas PTA plus supervised exercise produced greater benefits than either therapy alone.217 The ERASE (Endovascular Revascularization and Supervised Exercise) study randomized participants with claudication to endovascular revascularization plus supervised exercise or supervised exercise alone. After 1 year, patients in both groups had significant improvements in walking distances and health-related QoL, with greater improvements in the combined-therapy group.218 Collectively, these studies support the continued provision of supervised exercise to patients with claudication, whether as a monotherapy or combined with revascularization. IIaAIn patients with PAD, a structured community- or home-based exercise program with behavioral change techniques can be beneficial to improve walking ability and functional status.49,88,94,213 See Online Data Supplement 32.Unstructured community-based or home-based walking programs that consist of providing general recommendations to patients with claudication to simply walk more are not efficacious.50 Studies supporting structured community- or home-based programs for patients with symptomatic PAD (claudication and/or leg symptoms atypical for claudication) are more recent than studies supporting supervised exercise programs, and have provided strong evidence in support of the community- or home-based approach.47,49,51,88,94,213 For example, the GOALS (Group Oriented Arterial Leg Study) trial94 included patients with confirmed PAD with and without claudication (atypical lower extremity symptoms or no symptoms) and showed increases in several parameters of functional status for both of these patient cohort subgroups, versus nonexercising controls, after 6 months,88 with improvement maintained at 12 months.94 As with supervised exercise programs, despite proven benefit, initiating and maintaining a high level of adherence to community- or home-based exercise programs remains challenging. Studies that have incorporated behavioral change techniques, such as health coaching and activity tracking used in supervised settings, appear to reduce attrition and promote higher levels of adherence, thereby improving functional and QoL outcomes, both short term and long term.49,88,94 IIaAIn patients with claudication, alternative strategies of exercise therapy, including upper-body ergometry, cycling, and pain-free or low-intensity walking that avoids moderate-to-maximum claudication while walking, can be beneficial to improve walking ability and functional status.39,215,219,220 See Online Data Supplements 32 and 33.Protocols for exercise therapy for PAD traditionally have recommended intermittent walking bouts to moderate or higher pain levels interspersed with short periods of rest. Although these protocols are efficacious, intolerance of pain may lead to poor exercise adherence. An increasing number of studies have shown that modalities of exercise that avoid claudication or walking performed at intensities that are pain free or produce only mild levels of claudication can achieve health benefits comparable to walking at moderate or higher levels of claudication pain.39,41,215,219–221 Open in a new tab 7. Minimizing Tissue Loss in Patients with PAD: Recommendations Recommendations for Minimizing Tissue Loss in Patients With PAD CORLOERecommendations IC-LDPatients with PAD and diabetes mellitus should be counseled about self–foot examination and healthy foot behaviors.222,223 See Online Data Supplement 34.Some RCTs have suggested that patient education may help reduce the incidence of serious foot ulcers and lower extremity amputations, but the quality of evidence supporting patient education is low.222 Educational efforts generally include teaching patients about healthy foot behaviors (eg, daily inspection of feet, wearing of shoes and socks; avoidance of barefoot walking), the selection of proper footwear, and the importance of seeking medical attention for new foot problems.223 Educational efforts are especially important for patients with PAD who have diabetes mellitus with peripheral neuropathy. IC-LDIn patients with PAD, prompt diagnosis and treatment of foot infection are recommended to avoid amputation.224–228 See Online Data Supplement 34.Foot infections (infection of any of the structures distal to the malleoli) may include cellulitis, abscess, fasciitis, tenosynovitis, septic joint space infection, and osteomyelitis. Studies have investigated the accuracy of physical findings for identification of infection and determining infection severity and risk of amputation.224–226 Because of the consequences associated with untreated foot infection—especially in the presence of PAD—clinicians should maintain a high index of suspicion.228 It is also recognized that the presence of diabetes mellitus with peripheral neuropathy and PAD may make the presentation of foot infection more subtle than in patients without these problems. Foot infection should be suspected if the patient presents with local pain or tenderness; periwound erythema; periwound edema, induration or fluctuance; pretibial edema; any discharge (especially purulent); foul odor; visible bone or a wound that probes-to-bone; or signs of a systemic inflammatory response (including temperature >38°C or <36°C, heart rate >90/min, respiratory rate >20/min or Paco 2<32 mm Hg, white blood cell count >12 000 or <4000/mcL or >10% immature forms).226 Probe-to-bone test is moderately predictive for osteomyelitis but is not pathognomonic.227 IIaC-LDIn patients with PAD and signs of foot infection, prompt referral to an interdisciplinary care team (Table 9) can be beneficial.228–230 See Online Data Supplement 34.The EuroDIALE (European Study Group on Diabetes and the Lower Extremity) study demonstrated that the presence of both PAD and foot infection conferred a nearly 3-fold higher risk of leg amputation than either infection or PAD alone.228 The treatment of deep soft-tissue infection typically requires prompt surgical drainage; vascular imaging and expeditious revascularization generally follow. Experienced clinical teams have reported very good outcomes when this is performed in a coordinated and timely fashion.229,230 Previous groups have described various combinations of functions of interdisciplinary care teams (See Online Data Supplement 34a for a complete list of functions). See Section 9.2 for recommendations related to the role of the interdisciplinary care team in wound healing therapies for CLI. IIaC-EOIt is reasonable to counsel patients with PAD without diabetes mellitus about self–foot examination and healthy foot behaviors. N/A Although there are limited data to support patient education about self–foot examination and foot care for patients with diabetes mellitus, there are no data that have evaluated this practice in a population of patients with PAD but without diabetes mellitus. Nonetheless, this is a very low-risk intervention with potential for benefit. Educational efforts generally include teaching patients about healthy foot behaviors (eg, daily inspection of feet; foot care and hygiene, including appropriate toenail cutting strategies; avoidance of barefoot walking), the selection of appropriately fitting shoes, and the importance of seeking medical attention for new foot problems.223 IIaC-EOBiannual foot examination by a clinician is reasonable for patients with PAD and diabetes mellitus. N/A A history of foot ulcers, foot infections, or amputation identifies patients with a very high (>10%) yearly incidence of recurrent ulcers.231 Examination includes a visual inspection for foot ulcers (full-thickness epithelial defects) and structural (bony) deformities, monofilament testing for sensory neuropathy, and palpation for pedal pulses. Open in a new tab Table 9. Interdisciplinary Care Team for PAD. A team of professionals representing different disciplines to assist in the evaluation and management of the patient with PAD. For the care of patients with CLI, the interdisciplinary care team should include individuals who are skilled in endovascular revascularization, surgical revascularization, wound healing therapies and foot surgery, and medical evaluation and care. Interdisciplinary care team members may include: Vascular medical and surgical specialists (ie, vascular medicine, vascular surgery, interventional radiology, interventional cardiology) Nurses Orthopedic surgeons and podiatrists Endocrinologists Internal medicine specialists Infectious disease specialists Radiology and vascular imaging specialists Physical medicine and rehabilitation clinicians Orthotics and prosthetics specialists Social workers Exercise physiologists Physical and occupational therapists Nutritionists/dieticians Open in a new tab CLI indicates critical limb ischemia; and PAD, peripheral artery disease. 8. Revascularization for Claudication An individualized approach to revascularization for claudication is recommended for each patient to optimize outcome. Revascularization is but one component of care for the patient with claudication, as each patient should have a customized care plan that also includes medical therapy (Section 5), structured exercise therapy (Section 6), and care to minimize tissue loss (Section 7). If a strategy of revascularization for claudication is undertaken, the revascularization strategy should be evidence based and can include endovascular revascularization, surgery, or both. Because of the variability of ischemic limb symptoms and impact of these symptoms on functional status and QoL, patients should be selected for revascularization on the basis of severity of their symptoms. Factors to consider include a significant disability as assessed by the patient, adequacy of response to medical and structured exercise therapy, status of comorbid conditions, and a favorable risk–benefit ratio. Patient preferences and goals of care are important considerations in the evaluation for revascularization. The revascularization strategy should have a reasonable likelihood of providing durable relief of symptoms. A general recommendation for revascularization as a treatment option for claudication is provided below followed by specific recommendations for endovascular (Section 8.1.1) and surgical (Section 8.1.2) procedures if a revascularization strategy is undertaken. 8.1. Revascularization for Claudication: Recommendation Recommendation for Revascularization for Claudication CORLOERecommendation IIaARevascularization is a reasonable treatment option for the patient with lifestyle-limiting claudication with an inadequate response to GDMT.12,37,38,232,233 See Online Data Supplements 35 and 36.A minority of patients with claudication (estimated at <10% to 15% over 5 years or more) will progress to CLI.234–237 Therefore, the role of revascularization in claudication is improvement in claudication symptoms and functional status, and consequently in QoL, rather than limb salvage. Revascularization is reasonable when the patient who is being treated with GDMT (including structured exercise therapy) presents with persistent lifestyle-limiting claudication.12,37,38,232,233 Lifestyle-limiting claudication is defined by the patient rather than by any test. It includes impairment of activities of daily living and/ or vocational and/or recreational activities due to claudication. There should be clear discussion with the patient about expected risks and benefits of revascularization, as well as discussion of the durability of proposed procedures. Open in a new tab 8.1.1. Endovascular Revascularization for Claudication: Recommendations Endovascular techniques to treat claudication include balloon dilation (angioplasty), stents, and atherectomy. These techniques continue to involve and now include covered stents, drug-eluting stents (DES), cutting balloons, and drug-coated balloons. The technique chosen for endovascular treatment is related to lesion characteristics (eg, anatomic location, lesion length, degree of calcification) and operator experience. Assessment of the appropriateness of specific endovascular techniques for specific lesions for the treatment of claudication is beyond the scope of this document. Revascularization is performed on lesions that are deemed to be hemodynamically significant, and stenoses selected for endovascular treatment should have a reasonable likelihood of limiting perfusion to the distal limb. Stenoses of 50% to 75% diameter by angiography may not be hemodynamically significant, and resting or provoked intravascular pressure measurements may be used to determine whether lesions are significant.238,239 Multiple RCTs have compared endovascular procedures to various combinations of medical treatment with or without supervised or unsupervised exercise programs.12,37,38,217,232,233,240–251 These trials have used different endpoints and enrolled patients with anatomic disease distribution at different levels. Recommendations for Endovascular Revascularization for Claudication CORLOERecommendations IAEndovascular procedures are effective as a revascularization option for patients with lifestyle-limiting claudication and hemodynamically significant aortoiliac occlusive disease.12,37,38,232,240,242,246 See Online Data Supplements 35 and 36.Two separate systematic analyses that included RCTs that enrolled patients with aortoiliac disease reported that endovascular treatment of claudication improved walking parameters and QoL.11,12,233 The CLEVER trial enrolled only patients with aortoiliac disease and compared endovascular therapy to supervised exercise therapy and to medications alone.37,38 At 6-month follow-up, both the endovascular therapy and supervised exercise groups had improved peak walking time compared with medication alone, with a greater improvement in the supervised exercise group.37 By 18 months, there was no significant difference between the endovascular therapy and supervised exercise groups, with a sustained benefit versus medication alone.38 Other RCTs that included patients with aortoiliac disease have shown QoL, as assessed by questionnaires and time to onset of claudication, may be superior with endovascular treatment in combination with a medical and an exercise treatment plan, compared versus medical treatment alone.232,233,246 The ERASE trial randomized patients with claudication and aortoiliac (as well as femoropopliteal) disease to endovascular revascularization plus supervised exercise or supervised exercise alone. After 1 year, patients in both groups had significant improvements in walking distances and health-related QoL, with greater improvements in the combined-therapy group.218 The long-term comparative efficacy of endovascular revascularization versus supervised exercise therapy and medical therapy compared to supervised exercise therapy and medical therapy without revascularization for aortoiliac disease is unknown. IIaB-REndovascular procedures are reasonable as a revascularization option for patients with lifestyle-limiting claudication and hemodynamically significant femoropopliteal disease.217,232,243–245,250,251 See Online Data Supplement 35.Multiple RCTs have demonstrated short-term efficacy with endovascular treatment of femoropopliteal disease for claudication versus supervised exercise training or medical therapy, with benefit that diminishes by 1 year.217,232,240–246,250,251 Two separate systematic reviews that included RCTs that enrolled patients with femoropopliteal disease, reported that endovascular treatment of claudication improved walking parameters and QoL.11,12,233 The durability of endovascular treatment for claudication is directly related to vessel patency. Long-term patency is greater in the iliac artery than in the femoropopliteal segment. Furthermore, durability is diminished with greater lesion length, occlusion rather than stenosis, the presence of multiple and diffuse lesions, poor-quality runoff, diabetes mellitus, chronic kidney disease, renal failure, and smoking.252–255 The choice of endovascular therapy as a revascularization approach for claudication due to femoropopliteal disease therefore should include a discussion of outcomes, addressing the risk of restenosis and repeat intervention, particularly for lesions with poor likelihood of long-term durability. IIbC-LDThe usefulness of endovascular procedures as a revascularization option for patients with claudication due to isolated infrapopliteal artery disease is unknown.256–258 See Online Data Supplement 35.Isolated infrapopliteal disease is unlikely to cause claudication. Incidence of in-stent restenosis is high and long-term benefit lacking with bare-metal stenting of the infrapopliteal arteries.256 Studies that have enrolled patients with claudication as well as CLI have demonstrated a benefit of DES versus bare-metal stents or versus drug-coated balloons for revascularization of infrapopliteal lesions.257,258 However, these differences were mainly for patency and restenosis endpoints, and neither of these studies included patient-oriented outcomes, such as walking function or QoL parameters. Additional efficacy data on the use of infrapopliteal drug-coated balloon or DES for the treatment of claudication are likely to be published in the near future. III: HarmB-NREndovascular procedures should not be performed in patients with PAD solely to prevent progression to CLI.234–237,259–261 See Online Data Supplements 36 and 38.There are no data to support a practice paradigm of performing endovascular procedures on patients with PAD for the purpose of preventing progression of claudication symptoms to CLI. Reported rates of amputation or progression to CLI from prospective cohort studies of patients with claudication are <10% to 15% over 5 years or more, and increased mortality rate associated with claudication is usually the result of cardiovascular events rather than limb-related events.234–237,262 Similarly, there are no data to support revascularization in patients with asymptomatic PAD. Procedural risks include bleeding, renal failure from contrast-induced nephropathy, and the possibility of adverse limb outcomes.259–261 Therefore, the known risks of endovascular procedures outweigh any hypothetical benefit of preventing progression from asymptomatic PAD or claudication to CLI. Open in a new tab 8.1.2. Surgical Revascularization for Claudication: Recommendations Recommendations for Surgical Revascularization for Claudication CORLOERecommendations IAWhen surgical revascularization is performed, bypass to the popliteal artery with autogenous vein is recommended in preference to prosthetic graft material.263–271 See Online Data Supplements 37 and 38.The superficial femoral and proximal popliteal arteries are the most common anatomic sites of stenosis or occlusion among individuals with claudication. Femoral-popliteal bypass is therefore one of the most common surgical procedures for claudication and may be performed under general or regional anesthesia. The type of conduit and site of popliteal artery anastomosis (above versus below knee) are major determinants of outcomes associated with femoral-popliteal bypass. Systematic reviews and meta-analyses have identified a clear and consistent primary patency benefit for autogenous vein versus to prosthetic grafts for popliteal artery bypass.270,271 Prosthetic grafts to the popliteal artery above the knee have reduced patency rates and increased rates of repeat intervention.263,266,269,272 Sparse evidence suggests a long-term patency advantage for Dacron over polytetrafluoroethylene (known as PTFE) graft for above-knee bypass,270 although this finding has not been consistently demonstrated in all RCTs.266,273,274 IIaB-NRSurgical procedures are reasonable as a revascularization option for patients with lifestyle-limiting claudication with inadequate response to GDMT, acceptable perioperative risk, and technical factors suggesting advantages over endovascular procedures.232,265,275–277 See Online Data Supplements 37 and 38.Systematic reviews have concluded that surgical procedures are an effective treatment for claudication and have a positive impact on QoL and walking parameters but have identified sparse evidence supporting the effectiveness of surgery compared with other treatments.11,233,278,279 Although symptom and patency outcomes for surgical interventions may be superior versus less invasive endovascular treatments for specific patients, surgical interventions are also associated with greater risk of adverse perioperative events.280–286 Treatment selection should therefore be individualized on the basis of the patient's goals, perioperative risk, and anticipated benefit. Surgical procedures for claudication are usually reserved for individuals who a) do not derive adequate benefit from nonsurgical therapy, b) have arterial anatomy favorable to obtaining a durable result with surgery, and c) have acceptable risk of perioperative adverse events. Acceptable risk is defined by the individual patient and provider on the basis of symptom severity, comorbid conditions, and appropriate GDMT risk evaluation. Guidelines for the evaluation and management of patients undergoing noncardiac surgery, including vascular surgical procedures, have been previously published.21 III: HarmB-RFemoral-tibial artery bypasses with prosthetic graft material should not be used for the treatment of claudication.287–289 See Online Data Supplement 37.Bypasses to the tibial arteries with prosthetic material for treatment of claudication should be avoided because of very high rates of graft failure and amputation.287–289 III: HarmB-NRSurgical procedures should not be performed in patients with PAD solely to prevent progression to CLI.234–237,262 See Online Data Supplements 37 and 38.Claudication does not commonly progress to CLI. Reported rates of amputation or progression to CLI from prospective cohort studies of patients with claudication are <10% to 15% for 5 years or more, and increased mortality rate associated with claudication is usually the result of cardiovascular events rather than limb-related events.234–237,262 Surgical intervention should not be performed primarily to prevent disease progression, given the risk of adverse perioperative events without potential for significant benefit. Similarly, there are no data to support surgical revascularization in patients with asymptomatic PAD to prevent progression to CLI. Open in a new tab 9. Management of CLI Patients with CLI are at increased risk of amputation and major cardiovascular ischemic events. Care of the patient with CLI includes evaluation for revascularization and wound healing therapies, with the objective to minimize tissue loss, completely heal wounds, and preserve a functional foot. Medical therapy to prevent cardiovascular ischemic events is also an important component of care for the patient with CLI (Section 5). 9.1. Revascularization for CLI: Recommendations Recommendation for Revascularization for CLI CORLOERecommendation IB-NRIn patients with CLI, revascularization should be performed when possible to minimize tissue loss.290 See Online Data Supplement 39.Patients with CLI are at high risk of major cardiovascular ischemic events, as well as nonhealing wounds and major amputation. In a systematic review of 13 studies of patients with CLI who did not receive revascularization, which included patients enrolled in medical and angiogenic therapy trials, there was a 22% all-cause mortality rate and a 22% rate of major amputation at a median follow-up of 12 months.290 The goal of surgical or endovascular revascularization is to provide in-line blood flow to the foot through at least 1 patent artery, which will help decrease ischemic pain and allow healing of any wounds, while preserving a functional limb. Multiple RCTs comparing contemporary surgical and endovascular treatment for patients with CLI are ongoing.15–17 Revascularization is not warranted in the setting of a nonviable limb. IC-EOAn evaluation for revascularization options should be performed by an interdisciplinary care team (Table 9) before amputation in the patient with CLI. N/A Patients with CLI should be evaluated by an interdisciplinary care team. Before amputation, evaluation generally includes imaging for assessment of revascularization options (eg, duplex ultrasound, CTA, MRA, or catheter-based angiogram). The objective of this strategy is to minimize tissue loss and preserve a functional limb with revascularization. Open in a new tab 9.1.1. Endovascular Revascularization for CLI: Recommendations Recommendations for Endovascular Revascularization for CLI CORLOERecommendations IB-REndovascular procedures are recommended to establish in-line blood flow to the foot in patients with nonhealing wounds or gangrene.292,293 See Online Data Supplement 39.The technique chosen for endovascular treatment of CLI is related to anatomic location of lesions, lesion characteristics, and operator experience. Revascularization is performed on hemodynamically significant stenoses that are likely to be limiting blood flow to the limb. For stenoses of 50% to 75%, where the hemodynamic significance is unclear, intravascular pressure measurements may be used to determine hemodynamic significance.294 The BASIL (Bypass versus Angioplasty in Severe Ischemia of the Leg) RCT demonstrated that endovascular revascularization is an effective option for patients with CLI as compared with open surgery.292,293 The primary endpoint of amputation-free survival was the same in the endovascular and surgical arms. Of note, the endovascular arm used only PTA.292,293 Multiple RCTs comparing contemporary surgical and endovascular treatment for patients with CLI are ongoing.15–17Table 10 addresses factors that may prompt an endovascular versus surgical approach to the patient with CLI. IIaC-LDA staged approach to endovascular procedures is reasonable in patients with ischemic rest pain.295,296 N/A For patients with multilevel disease who suffer from ischemic rest pain, in-flow lesions are generally addressed first.295,296 Depending on procedural characteristics, including contrast volume used, radiation exposure, and procedure time, out-flow lesions can be addressed in the same setting or at a later time if symptoms persist. This strategy for ischemic rest pain is distinct from the strategy recommended for CLI in the patient with a nonhealing wound or gangrene. In that scenario, restoration of direct in-line flow to the foot is essential for wound healing. IIaB-REvaluation of lesion characteristics can be useful in selecting the endovascular approach for CLI.297,298 See Online Data Supplement 39.The lesion characteristics to consider include length, anatomic location, and extent of occlusive disease. For example, if an adequate angioplasty result can be achieved with PTA alone for short (<10 cm) stenoses in the femoropopliteal segment, then stent placement is not necessary.297,298 Presence of thrombosis or calcification at the lesion site will also affect the endovascular approach. In general, the advantages of DES and drug-coated balloons over PTA alone or bare-metal stents are more consistent in the femoropopliteal segment than for infrapopliteal interventions.257,258,299–309 However, these differences are mainly for patency, restenosis, and repeat-revascularization endpoints. Most studies were underpowered or did not examine other patient-oriented outcomes, such as amputation or wound healing in CLI. Endovascular techniques continue to evolve rapidly, and there has been limited literature comparing techniques with regard to clinically significant outcomes, such as amputation or wound healing. IIbB-NRUse of angiosome-directed endovascular therapy may be reasonable for patients with CLI and nonhealing wounds or gangrene.310–319 See Online Data Supplements 39 and 40.During the past decade, the goal of care with regard to endovascular therapy for the treatment of nonhealing wounds due to CLI has been establishment of direct in-line blood flow to the affected limb. The angiosome concept has also been described in the literature in relation to the treatment of nonhealing wounds. Angiosome-directed treatment entails establishing direct blood flow to the infrapopliteal artery directly responsible for perfusing the region of the leg or foot with the nonhealing wound. Multiple retrospective studies and 1 small nonrandomized prospective study assessing the efficacy of this concept have been published.119,310–321 Meta-analyses of these studies found improved wound healing and limb salvage with angiosome-guided therapy but cautioned that the quality of the evidence was low.322,323 Although the angiosome concept is theoretically satisfying, randomized data comparing the establishment of in-line flow versus angiosome-guided therapy have yet to be published. Furthermore, there is no evidence yet to demonstrate the potential benefit of treating additional infrapopliteal arteries once in-line flow has been established in one artery, regardless of angiosome. Important considerations with regard to angiosome-guided therapy include the potential for longer procedural times, more contrast exposure, and more technically complex procedures. The impact of all these factors needs to be weighed against the likelihood of a technically successful procedure providing hypothetical added benefit over the establishment of in-line blood flow. Open in a new tab Table 10. Therapy for CLI: Findings That Prompt Consideration of Surgical or Endovascular Revascularization. \| Findings That Favor Consideration of Surgical Revascularization \| Examples \| :--- \| \| Factors associated with technical failure or poor durability with endovascular treatment \| Lesion involving common femoral artery, including origin of deep femoral artery \| \| Long segment lesion involving the below-knee popliteal and/or infrapopliteal arteries in a patient with suitable single-segment autogenous vein conduit \| \| Diffuse multilevel disease that would require endovascular revascularization at multiple anatomic levels \| \| Small-diameter target artery proximal to site of stenosis or densely calcified lesion at location of endovascular treatment \| \| Endovascular treatment likely to preclude or complicate subsequent achievement of in-line blood flow through surgical revascularization \| Single-vessel runoff distal to ankle \| \| Findings That Favor Consideration of Endovascular Revascularization \| Examples \| \| The presence of patient comorbidities may place patients at increased risk of perioperative complications from surgical revascularization. In these patients, an endovascular-first approach should be used regardless of anatomy \| Patient comorbidities, including coronary ischemia, cardiomyopathy, congestive heart failure, severe lung disease, and chronic kidney disease \| \| Patients with rest pain and disease at multiple levels may undergo a staged approach as part of endovascular-first approach \| In-flow disease can be addressed first, and out-flow disease can be addressed in a staged manner, when required, if clinical factors or patient safety prevent addressing all diseased segments at one setting \| \| Patients without suitable autologous vein for bypass grafts \| Some patients have had veins harvested for previous coronary artery bypass surgery and do not have adequate remaining veins for use as conduits. Similarly, patients may not have undergone prior saphenous vein harvest, but available vein is of inadequate diameter \| Open in a new tab CLI indicates critical limb ischemia. 9.1.2. Surgical Revascularization for CLI: Recommendations Recommendations for Surgical Revascularization for CLI CORLOERecommendations IAWhen surgery is performed for CLI, bypass to the popliteal or infrapopliteal arteries (ie, tibial, pedal) should be constructed with suitable autogenous vein.263,266,269,272 See Online Data Supplement 37.Many large RCTs have demonstrated that bypasses above the knee should be autogenous vein either reversed or in situ vein.263,266,269,272 There are large single-center trials showing the efficacy of autogenous vein to distal tibial vessels.324,325 In addition, composite sequential femoropopliteal-tibial bypass and bypass to an isolated popliteal arterial segment that has collateral out flow to the foot are both acceptable methods of revascularization and should be considered when no other form of bypass with adequate autogenous conduit is possible.326,327 IC-LDSurgical procedures are recommended to establish in-line blood flow to the foot in patients with nonhealing wounds or gangrene.328–330 See Online Data Supplement 42.In patients presenting with nonhealing ulcers or gangrene, surgical procedures should be performed to establish in-line blood flow to the foot.328–330Table 10 addresses factors that may prompt a surgical approach to the patient with CLI. IIaB-NRIn patients with CLI for whom endovascular revascularization has failed and a suitable autogenous vein is not available, prosthetic material can be effective for bypass to the below-knee popliteal and tibial arteries.331–333 See Online Data Supplement 42.There are studies demonstrating that patients for whom endovascular treatment for CLI has failed can be treated successfully with autogenous vein bypass graft332,333 or prosthetic material.331 Although autogenous vein is the preferred conduit for surgical revascularization, prosthetic conduit is a secondary option for patients with CLI without suitable saphenous vein who require surgical revascularization. IIaC-LDA staged approach to surgical procedures is reasonable in patients with ischemic rest pain.334–336 N/A It is reasonable to perform a staged approach to revascularization in patients with ischemic rest pain with multilevel disease. For example, aortoiliac (inflow) disease may be treated first with endovascular treatment or by surgical reconstruction, depending on lesion characteristics, patient comorbidities, and patient preference.337,338 Combined percutaneous and surgical revascularization may require separate interventions, typically with the most proximal procedure performed first. Open in a new tab 9.2. Wound Healing Therapies for CLI: Recommendations Recommendations for Wound Healing Therapies for CLI CORLOERecommendations IB-NRAn interdisciplinary care team should evaluate and provide comprehensive care for patients with CLI and tissue loss to achieve complete wound healing and a functional foot.229,339–341 See Online Data Supplement 44.The management of patients with CLI and nonhealing wounds should include coordinated efforts for both revascularization and wound healing, because the risk of limb-threatening infections remains until complete wound healing is achieved. The structure and activities of interdisciplinary care teams for CLI may vary according to several factors, including the local availability of resources. Previous groups have described various combinations of activities of this team, which are in addition to revascularization and include functions such as wound care, infection management, orthotics, and prosthetics (see Online Data Supplement 34a for a complete list of functions). Coordination of these activities and some degree of organized team structure are recommended, as opposed to ad hoc or unstructured referrals among various specialty clinicians not involved in interdisciplinary care. Ambulatory patients with PAD and nonhealing foot ulcers should be considered for efforts to prevent amputation. The components of this effort may include revascularization, offloading, treatment of infection, and wound care. The long-term outcome of the limb is excellent when complete wound healing can be achieved.339Revascularization should be coordinated with the efforts of clinicians who manage foot infections, provide offloading, and achieve complete wound healing, either through medical therapy, surgical options, or a combination thereof. Coordinated and timely interdisciplinary care can achieve excellent limb outcomes for patients with PAD and nonhealing foot wounds.229,339–341 IC-LDIn patients with CLI, wound care after revascularization should be performed with the goal of complete wound healing.339 See Online Data Supplement 44.A comprehensive plan for treatment of CLI must include a plan for achieving an intact skin surface on a functional foot. One study demonstrated a limb salvage rate of 100% at 3 years in a cohort of patients with CLI who achieved complete wound healing with endovascular revascularization and dedicated wound care.339 Before revascularization, the interdisciplinary care team should devise a plan to achieve the goal of complete wound healing. After successful revascularization, most patients with gangrene of the foot are evaluated for minor amputation with staged/delayed primary closure or surgical reconstruction when feasible.342–344 Negative-pressure wound therapy dressings are helpful to achieve wound healing after revascularization and minor (ie, digit or partial foot) amputation when primary or delayed secondary closure is not feasible.345,346 Spontaneous amputation, or autoamputation, of gangrenous digits should be reserved for palliation in patients without options for revascularization.345,347,348 Other evidence-based guidelines relevant to those with nonhealing foot wounds following revascularization cover the full spectrum of diabetic foot problems349 or separately consider the management of infection,225,350 offloading,351 and wound care.352 To date, there are no RCTs or high-quality studies that have focused on wound healing adjuncts in limbs with severe PAD (eg, topical cytokine ointments, skin substitutes, cell-based therapies intended to optimize wound healing). IIbB-NRIn patients with CLI, intermittent pneumatic compression (arterial pump) devices may be considered to augment wound healing and/or ameliorate severe ischemic rest pain.353 See Online Data Supplement 44.A systematic review of studies that used intermittent pneumatic compression devices specifically designed to augment arterial perfusion of the lower extremities suggests that these may provide modest clinical benefit (specifically, decreased amputation rates and improved QoL) in patients with CLI who were ineligible for revascularization.353 The potential benefit appears to outweigh the low risk associated with the use of these devices. IIbC-LDIn patients with CLI, the effectiveness of hyperbaric oxygen therapy for wound healing is unknown.354 See Online Data Supplement 44.The literature evaluating the utility of hyperbaric oxygen therapy has focused on patients without severe PAD and has not demonstrated a long-term benefit on wound healing or improving amputation-free survival when compared with sham treatment.355 There are no published studies evaluating the role of hyperbaric oxygen therapy for patients with nonreconstructible PAD. One small RCT that focused on patients with foot ulcers and PAD (ABI <0.80 or TBI <0.70) for whom no revascularization was planned demonstrated a significant decrease in ulcer area at 6 weeks, but no significant differences in ulcer size at 6 months, complete ulcer healing at 6 weeks or 6 months, and major or minor amputations.354 Further research on the utility of hyperbaric oxygen therapy in this context is needed. III: No BenefitB-RProstanoids are not indicated in patients with CLI.356 See Online Data Supplement 43.A systematic review and meta-analysis concluded that RCTs have not demonstrated meaningful long-term clinical benefit from the administration of prostanoids to patients with CLI attributable to nonreconstructible PAD.356 Open in a new tab 10. Management of ALI ALI is one of the most treatable and potentially devastating presentations of PAD. Timely recognition of arterial occlusion as the cause of an ischemic, cold, painful leg is crucial to successful treatment. The writing committee has used a standard definition of ALI in which symptom duration is <2 weeks (Table 3).33,34 Category I refers to viable limbs that are not immediately threatened. Category II refers to threatened limbs. Category IIa limbs are marginally threatened and salvageable, if promptly treated. Category IIb are immediately threatened limbs that require immediate revascularization if salvage is to be accomplished. Category III are irreversibly damaged limbs, in which case resultant major tissue loss or permanent nerve damage is inevitable.34 10.1. Clinical Presentation of ALI: Recommendations Recommendations for Clinical Presentation of ALI CORLOERecommendations IC-EOPatients with ALI should be emergently evaluated by a clinician with sufficient experience to assess limb viability and implement appropriate therapy. N/A Patients with ALI should be rapidly evaluated by a vascular specialist if one is available. Depending on local clinical expertise, the vascular specialist may be a vascular surgeon, interventional radiologist, cardiologist, or a general surgeon with specialized training and experience in treating PAD. If such expertise is not locally or rapidly available, there should be strong consideration of transfer of the patient to a facility with such resources. The more advanced the degree of ischemia, the more rapidly the communication (including communication about potential patient transfer) needs to occur. IC-LDIn patients with suspected ALI, initial clinical evaluation should rapidly assess limb viability and potential for salvage and does not require imaging.357–361 See Online Data Supplements 45 and 46.ALI is a medical emergency and must be recognized rapidly. The time constraint is due to the period that skeletal muscle will tolerate ischemia—roughly 4 to 6 hours.362 A rapid assessment of limb viability and ability to restore arterial blood flow should be performed by a clinician able to either complete the revascularization or triage the patient.358 Lower extremity symptoms in ALI can include both pain and loss of function. The longer these symptoms are present, the less likely the possibility of limb salvage.360,361 Clinical assessment must include symptom duration, pain intensity, and motor and sensory deficit severity to distinguish a threatened from a nonviable extremity (Figure 3). The bedside assessment should include arterial and venous examination with a handheld continuous-wave Doppler because of the inaccuracy of pulse palpation.34 The loss of dopplerable arterial signal indicates that the limb is threatened. The absence of both arterial and venous Doppler signal indicates that the limb may be irreversibly damaged (nonsalvageable). Comorbidities should be investigated and managed aggressively, but this must not delay therapy. Even in the setting of rapid and effective revascularization, the 1-year morbidity and mortality rates associated with ALI are high.360,363 Open in a new tab Figure 3. Diagnosis and Management of ALI. Open in a new tab 33,34, Colors correspond to Class of Recommendation in Table 1. ALI indicates acute limb ischemia. 10.2. Medical Therapy for ALI: Recommendations Recommendation for ALI Medical Therapy CORLOERecommendation IC-EOIn patients with ALI, systemic anticoagulation with heparin should be administered unless contraindicated. N/A Heparin (generally intravenous unfractionated heparin) is given to all patients acutely.35,364 This can stop thrombus propagation and may provide an anti-inflammatory effect that lessens the ischemia. Patients who have received heparin before the onset of ALI and have a decrease in platelet count may have heparin-induced thrombocytopenia.365,366 In this situation, a direct thrombin inhibitor is given, rather than heparin, if heparin-induced thrombocytopenia with thrombosis is suspected. Open in a new tab 10.3. Revascularization for ALI: Recommendations Recommendations for Revascularization for ALI CORLOERecommendations IC-LDIn patients with ALI, the revascularization strategy should be determined by local resources and patient factors (eg, etiology and degree of ischemia).367–369 See Online Data Supplement 47.For marginally or immediately threatened limbs (Category IIa and IIb ALI [Figure 3]), revascularization should be performed emergently (within 6 hours). For viable limbs (Category I ALI [Figure 3]), revascularization should be performed an on urgent basis (within 6–24 hours). The revascularization strategy can range from catheter-directed thrombolysis to surgical thromboembolectomy. Available facilities and clinical expertise are factors that should be considered when determining the revascularization strategy. The technique that will provide the most rapid restoration of arterial flow with the least risk to the patient should be selected. For example, catheter-directed thrombolysis can provide rapid restoration of arterial flow to a viable or marginally threatened limb, particularly in the setting of recent occlusion, thrombosis of synthetic grafts, and stent thrombosis.367 If this is not available locally, surgical options for timely revascularization should be considered, along with the feasibility of timely transfer to a facility with the necessary expertise. IACatheter-based thrombolysis is effective for patients with ALI and a salvageable limb.367–371 See Online Data Supplement 47.Assessment of the comparative effectiveness of catheter-based thrombolysis versus open surgery is complicated by variable definitions of ALI in this literature. Four RCTs comparing catheter-based thrombolysis to surgery,367,369–371 as well as a meta-analysis,368 have demonstrated similar limb salvage rates between the 2 approaches but better survival with catheter-based therapy. The survival advantage of catheter-based therapy may be at least in part attributable to multiple comorbidities found among the population of patients who present with ALI. Increased comorbidities are likely to contribute to increased perioperative risk. Several of the RCTs included patients with relatively chronic ischemia. Acuity and severity are both factors in the decision to consider thrombolysis.367,369–371 IC-LDAmputation should be performed as the first procedure in patients with a nonsalvageable limb.372,373 See Online Data Supplement 48.For patients with Category III ALI (Figure 3), amputation should be performed as the index procedure. Prolonged duration of ischemia is the most common factor in patients requiring amputation for treatment of ALI. The risks associated with reconstruction outweigh the potential benefit in a limb that is already insensate or immobile because of prolonged ischemia. Patients who have an insensate and immobile limb in the setting of prolonged ischemia (>6 to 8 hours) are unlikely to have potential for limb salvage.34,362 In addition, in this setting the reperfusion and circulation of ischemic metabolites can result in multiorgan failure and cardiovascular collapse. However, if pain can be controlled and there is no evidence of infection, amputation may be deferred if this meets with the patient's goals. IC-LDPatients with ALI should be monitored and treated (eg, fasciotomy) for compartment syndrome after revascularization.372,373 See Online Data Supplement 48.The lower extremity muscles reside in compartments, surrounded by fascia and bones. Reperfusion to ischemic muscles can cause cellular edema, resulting in increased compartment pressure. When compartment pressure is >30 mm Hg, there is capillary and venule compression that leads to malperfusion of the muscle; this is compartment syndrome. Fasciotomy is indicated when the compartment pressure increases. Measurement of intracompartment pressure is not always easily accessible. In such cases, evaluation for fasciotomy is prompted by development of increased pain, tense muscle, or nerve injury. Fasciotomy should be considered for patients with Category IIb ischemia for whom the time to revascularization is >4 hours. IIaB-NRIn patients with ALI with a salvageable limb, percutaneous mechanical thrombectomy can be useful as adjunctive therapy to thrombolysis.374–378 See Online Data Supplements 49 and 50.Multiple nonrandomized studies have suggested that percutaneous mechanical thrombectomy in combination with pharmacological therapy can be beneficial in the treatment of threatened limbs.374–378 IIaC-LDIn patients with ALI due to embolism and with a salvageable limb, surgical thromboembolectomy can be effective.379–381 See Online Data Supplements 49 and 50.Patients with arterial embolism and an absent pulse ipsilateral to the ischemic limb can be treated by exposure of an artery in the affected limb and balloon-catheter thromboembolectomy. These patients may benefit from adjunctive intraoperative fibrinolytics. In the event that thromboembolectomy does not restore arterial flow, bypass can be performed.381–383 IIbC-LDThe usefulness of ultrasound-accelerated catheter-based thrombolysis for patients with ALI with a salvageable limb is unknown.384–386 See Online Data Supplements 47 and 50.The use of ultrasound-accelerated catheter delivery of thrombolytic agents has been published in case series384 and retrospective analyses.385 However, the single RCT comparing this technique to standard catheter-based thrombolytic therapy failed to demonstrate a difference in outcomes, including bleeding, despite a lower total amount of lytic delivered.386 Open in a new tab 10.4. Diagnostic Evaluation of the Cause of ALI: Recommendations Recommendations for Diagnostic Evaluation of the Cause of ALI CORLOERecommendations IC-EOIn the patient with ALI, a comprehensive history should be obtained to determine the cause of thrombosis and/or embolization. N/A In addition to identifying a known history of PAD, the history should focus on uncovering clinical evidence of other conditions that can result in ALI through either embolic or thrombotic mechanisms. These conditions include atrial fibrillation, left ventricular thrombus, aortic dissection, trauma, hypercoagulable state, and presence of a limb artery bypass graft. The clinical history should identify the presence or absence of a history of MI, symptoms and signs of left ventricular dysfunction resulting in congestive heart failure, or possible endocarditis. The history should evaluate for possibility of deep vein thrombosis with intracardiac shunt (eg, patent foramen ovale or other that may result in paradoxical arterial embolism), hypercoagulable state, and family history of thrombosis. IIaC-EOIn the patient with a history of ALI, testing for a cardiovascular cause of thromboembolism can be useful. N/A Treatment of ALI should not be delayed for testing for the underlying cause of the limb ischemia. Delay from symptom onset to revascularization is a major determinant of outcome.360,361 The evaluation of a cardiovascular cause of ALI is most useful in the patient without underlying PAD. Evaluation for cardiovascular cause includes electrocardiogram or additional heart rhythm monitoring to detect atrial fibrillation, electrocardiogram to detect evidence of MI, and echocardiography to further determine whether there is a cardiac etiology for thromboembolism, such as valvular vegetation, left atrial or left ventricular thrombus, or intracardiac shunt.387 Open in a new tab 11. Longitudinal Follow-Up: Recommendations PAD is a lifelong chronic medical condition. Ongoing care focuses on cardiovascular risk reduction with medical therapy, optimizing functional status with structured exercise and, when indicated, revascularization. Recommendations for Longitudinal Follow-Up CORLOERecommendations IC-EOPatients with PAD should be followed up with periodic clinical evaluation, including assessment of cardiovascular risk factors, limb symptoms, and functional status. N/A A comprehensive care plan for patients with PAD includes periodic clinical evaluation by a healthcare provider with experience in the care of vascular patients. Clinical evaluation should include assessment of cardiovascular risk factors, assessment of adherence to medical therapy, and re-evaluation of smoking cessation efforts. Comprehensive lifestyle modification, including heart-healthy nutrition, is encouraged.22 Patients with PAD should also undergo periodic assessment of limb symptoms, functional status, and their ability to participate in vocational and recreational activities. Ongoing participation in a structured exercise program should be facilitated. Foot examination and patient counseling about healthy foot behaviors in PAD are addressed in Section 7. IC-EOPatients with PAD who have undergone lower extremity revascularization (surgical and/or endovascular) should be followed up with periodic clinical evaluation and ABI measurement. N/A In addition to the clinical evaluation of cardiovascular risk factors, functional status, and adherence to medical therapy and smoking cessation, patients with PAD who have previously undergone lower extremity revascularization (surgical and/or endovascular) require additional ongoing assessment and care. Follow-up visits after revascularization should include reassessment of the patient's limb symptoms and interval change in functional status, as well as participation in a structured exercise program. Pulse examination and ABI are included in the assessment. A change in ABI of 0.15 is considered clinically significant.388 IIaB-RDuplex ultrasound can be beneficial for routine surveillance of infrainguinal, autogenous vein bypass grafts in patients with PAD.389–395 See Online Data Supplements 51 and 52.A general surveillance schedule may be at 4 to 6 weeks, 6 months, and 12 months in the first year and yearly thereafter. It is important that testing frequency is individualized to the patient, type of arterial bypass, and any prior duplex scan findings. Duplex graft surveillance focuses on the identification of high-grade stenosis (eg, peak systolic velocity >300 cm/s and peak systolic velocity ratio across the stenosis >3.5) or impending graft failure (eg, PSV <40 cm/s).392,395 Detection of a graft stenosis prompts the consideration of further revascularization to treat the stenosis and maintain graft patency. Duplex may detect significant stenoses that may not be detected by a decline in ABI.394 Although case series have demonstrated high rates of primary assisted patency with a duplex ultrasound-surveillance strategy, RCTs of duplex surveillance versus clinical surveillance with the ABI have demonstrated mixed results in terms of a benefit on patency and limb outcomes.391,393,396 IIaC-LDDuplex ultrasound is reasonable for routine surveillance after endovascular procedures in patients with PAD.397–399 See Online Data Supplement 52.Studies have developed duplex ultrasound diagnostic criteria for diagnosing restenosis at the site of endovascular revascularization. Diagnostic criteria need to be customized to the location (eg, iliac or superficial femoral artery) and type of intervention (eg, angioplasty, uncovered stent, or covered stent). The optimal timing for surveillance after endovascular procedures is unclear.397–399 There are limited outcome data on routine duplex surveillance versus clinical surveillance plus the ABI after endovascular revascularization.397–399 The value of duplex ultrasound may be greater in cases with higher rates of restenosis, such as after interventions to treat very long lesions or occlusions.400 IIbB-RThe effectiveness of duplex ultrasound for routine surveillance of infrainguinal prosthetic bypass grafts in patients with PAD is uncertain.393,401–403 See Online Data Supplements 51 and 52.Duplex ultrasound of prosthetic bypass grafts may be used to characterize mid-graft velocity, because low velocities can predict impending graft failure.401–403 Outcome studies of duplex surveillance of prosthetic grafts have not shown consistent benefit.393,401–403 One RCT of duplex versus clinical surveillance with the ABI for femoropopliteal grafts did not show a benefit of duplex on outcome in the subset of patients with prosthetic grafts, though there was a benefit of duplex surveillance for vein bypass grafts.393 Open in a new tab 12. Evidence Gaps and Future Research Directions In performing the evidence review and in developing the present guidelines, the writing committee identified the following critical evidence gaps and future directions for PAD-related research: Basic science and translational studies to better understand the vascular biology of endovascular therapies and bypass grafting and to develop new methods for preventing restenosis after revascularization. Determination of risk factors for progression from asymptomatic PAD to symptomatic disease, including CLI. RCTs needed to determine the value of using the ABI to identify asymptomatic patients with PAD for therapies to reduce cardiovascular risk (eg, antiplatelet agents, statins, and other therapies). Advancement in PAD diagnostics, such as technologies for simplified yet highly accurate measurement of the ABI and tools for more reliable noninvasive perfusion assessment in CLI. Comparative-effectiveness studies to determine the optimal antiplatelet therapy (drug or drugs and dosage) for prevention of cardiovascular and limb-related events in patients with PAD. Development of additional medical therapies for claudication–an area of unmet medical need with a currently limited research pipeline.404 Studies to investigate the role of dietary intervention, in addition to statin therapy, to improve outcome and modify the natural history of PAD. Additional research to identify the best community-or home-based exercise programs for patients with PAD to maximize functional status and improve QoL, as well as the role of such exercise programs before or in addition to revascularization. Development and validation of improved clinical classification systems for PAD that incorporate symptoms, anatomic factors, and patient-specific risk factors and can be used to predict clinical outcome and optimize treatment approach. An example of a recently developed classification system is the Society for Vascular Surgery limb classification system, based on wound, ischemia, and foot infection (WIfI), which has been validated in different populations and may permit more meaningful prognosis in patients with CLI.405–409 Comparative- and cost-effectiveness studies of the different endovascular technologies for treatment of claudication and CLI, including drug-coated balloons and DES. Studies should include patient-centered end-points, such as functional parameters, time to wound healing, and QoL, in addition to standard patency-focused outcomes. These studies could then be incorporated into value-based clinical algorithms for approach to revascularization for claudication and CLI. Additional studies to demonstrate the impact of multisocietal registries on clinical outcomes and appropriate use. At present, these include the Vascular Quality Initiative (VQI), the National Cardiovascular Data Registry Peripheral Vascular Intervention Registry™ (PVI Registry™), and the National Radiology Data Registry for Interventional Radiology (NRDR). These registries provide an opportunity to obtain “real-world” data on surgical and endovascular procedures for PAD and to improve quality by providing feedback to participating centers. Future efforts should incorporate these registries into interventional RCTs and postmarketing studies of PAD-related devices. 13. Advocacy Priorities The writing committee identified 3 priorities for multi-societal advocacy initiatives to improve health care for patients with PAD. First, the writing committee supports the availability of the ABI as the initial diagnostic test to establish the diagnosis of PAD in patients with history or physical examination findings suggestive of PAD (Table 5). Although the ABI test is generally reimbursed by third-party payers for patients with classic claudication or lower extremity wounds, payers may not provide reimbursement for the ABI with other findings suggestive of PAD, such as lower extremity pulse abnormalities or femoral bruits. The writing committee affirms the importance of confirming the diagnosis of PAD in such patients to allow for GDMT as delineated in this document. Second, the writing committee supports the vital importance of insuring access to supervised exercise programs for patients with PAD. Although extensive high-quality evidence supports supervised exercise programs to improve functional status and QoL, only a minority of patients with PAD participate in such programs because of lack of reimbursement by third-party payers. Third, the writing committee recognizes the need for incorporation of patient-centered outcomes into the process of regulatory approval of new medical therapies and revascularization technologies. For revascularization technologies, regulatory approval is driven primarily by data on angiographic efficacy (ie, target lesion patency) and safety endpoints. The nature of the functional limitation associated with PAD warrants the incorporation of patient-centered outcomes, such as functional parameters and QoL, into the efficacy outcomes for the approval process. Supplementary Material Evidence Table 1. Nonrandomized Trials, Observational Studies, and/or Registries of History for Clinical Assessment for PAD–Section 2.1. Evidence Table 2. Nonrandomized Trials, Observational Studies, and/or Registries of Physical Examination for Clinical Assessment for PAD–Section 2.1. Evidence Table 3. RCTs of Resting ABI for Diagnosing PAD–Section 3.1. Evidence Table 4. Nonrandomized Trials, Observational Studies, and/or Registries of Resting ABI for Diagnosing PAD–Section 3.1. Evidence Table 5. Nonrandomized Trials, Observational Studies, and/or Registries of Physiological Testing–Section 3.2. Evidence Table 6. Nonrandomized Trials, Observational Studies, and/or Registries of Imaging for Anatomic Assessment (Ultrasound, CTA, MRA, Angiography)–Section 3.3. Evidence Table 7. RCTs of Imaging for Anatomic Assessment (Ultrasound, CTA, MRA, Angiography)–Section 3.3. Evidence Table 8. Nonrandomized Trials, Observational Studies, and/or Registries for Abdominal Aortic Aneurysm–Section 4.1. Evidence Table 9. Nonrandomized Trials, Observational Studies, and/or Registries of Coronary Artery Disease Screening in PAD–Section 4.2. Evidence Table 10. RCTs for CAD Screening in PAD–Section 4.2. Evidence Table 11. Nonrandomized Trials, Observational Studies, and/or Registries of Screening in Carotid Artery Disease–Section 4.3. Evidence Table 12. Nonrandomized Trials, Observational Studies, and/or Registries for Renal Artery Disease–Section 4.4. Evidence Table 13. RCTs Evaluating Antiplatelet Agents– Section 5.1. Evidence Table 14. Nonrandomized Trials, Observational Studies, and/or Registries of Antiplatelet Agents–Section 5.2. Evidence Table 15. Randomized Trials Comparing Statin Agents–Section 5.2. Evidence Table 16. Nonrandomized Trials, Observational Studies, and/or Registries of Statin Agents–Section 5.2. Evidence Table 17. RCTs for Antihypertensive Agents– Section 5.3. Evidence Table 18. Nonrandomized Trials, Observational Studies, and/or Registries of Antihypertensive Agents–Section 5.3. Evidence Table 19. RCTs for Smoking Cessation–Section 5.4. Evidence Table 20. Nonrandomized Trials, Observational Studies, and/or Registries of Smoking Cessation–Section 5.4. Evidence Table 21. RCTs Evaluating Glycemic Control in Patients with PAD and Diabetes Mellitus–Section 5.5. Evidence Table 22. Nonrandomized Trials, Observational Studies, and/or Registries of Glycemic Control–Section 5.5. Evidence Table 23. RCTs Evaluating Oral Anticoagulation–Section 5.6. Evidence Table 24. Nonrandomized Trials, Observational Studies, and/or Registries of Oral Anticoagulation–Section 5.6. Evidence Table 25. RCTs and Observational Studies of Cilostazol–Section 5.7. Evidence Table 26. Nonrandomized Trials, Observational Studies, and/or Registries of Pentoxifylline–Section 5.8. Evidence Table 27. Systematic Review of Chelation Therapy–Section 5.9. Evidence Table 28. Nonrandomized Trials, Observational Studies, and/or Registries of Homocysteine Lowering Therapy for Lower Extremity PAD in Patients with Diabetes Mellitus–Section 5.10.1. Evidence Table 29. RCTs Comparing Additional Medical Therapies of Homocysteine Lowering Therapy for Lower Extremity PAD–Section 5.10.1. Evidence Table 30. RCTs for Influenza Vaccination–Section 5.10.2. Evidence Table 31. Nonrandomized Trials for Influenza Vaccination–Section 5.10.2. Evidence Table 32. RCTs for Exercise Therapy–Section 6. Evidence Table 33. Nonrandomized Trials, Observational Studies, and/or Registries for Exercise Therapy–Section 6. Evidence Table 34. Nonrandomized Trials and Observational Studies of Minimizing Tissue Loss in Patients with PAD–Section 7. Data Supplement 34a. Functions of a Multidisciplinary Foot Care / Amputation Prevention Team–Section 7. Evidence Table 35. RCTs Comparing Endovascular Treatment and Endovascular Versus Noninvasive Treatment of Claudication–Section 8.1. Evidence Table 36. Nonrandomized Trials, Observational Studies, and/or Registries of Endovascular and Endovascular Versus Noninvasive Treatment of Claudication–Section 8.1. Evidence Table 37. RCTs Evaluating Surgical Treatment for Claudication–Section 8.1.2. Evidence Table 38. Nonrandomized Trials, Observational Studies, and/or Registries of Surgical Treatment for Claudication–Section 8.1.2. Evidence Table 39. RCTs Comparing Endovascular Revascularization for Chronic CLI–Section 8.2. Evidence Table 40. Nonrandomized Trials, Observational Studies, and/or Registries of Endovascular Revascularization for Chromic CLI–Section 8.2.1. Evidence Table 41. RCTs of Surgical Revascularization for Chronic CLI–Section 8.2. Evidence Table 42. Nonrandomized Trials, Observational Studies, and/or Registries for Surgical Revascularization for Chronic CLI–Section 8.2. Evidence Table 43. RCT Comparing Prostanoids for End-Stage Peripheral Artery Disease–Section 8.2.3. Evidence Table 44. Nonrandomized Trials, Observational Studies, and/or Registries for Would Healing Therapies for CLI–Section 8.2.3. Evidence Table 45. Nonrandomized Trials, Observational Studies, and/or Registries of Acute Limb Ischemia–Section 9.1. Evidence Table 46. Nonrandomized Trials, Observational studies, and/or Registries Comparing Evaluating Noninvasive Testing and Angiography for ALI–Section 9.1. Evidence Table 47. RCTs of Revascularization Strategy for ALI–Section 9.2.2. Evidence Table 48. Nonrandomized Trials, Observational Studies, and/or Registries of Clinical Presentation of ALI–Section 9.2.2. Evidence Table 49. Nonrandomized Trials, Observational Studies, and/or Registries of Diagnostic Evaluation of the Cause of ALI–Section 9.2.2. Evidence Table 50. Nonrandomized Trials, Observational Studies, and/or Registries of Revascularization Strategy for ALI–Section 9.2.2. Evidence Table 51. RCTs for Longitudinal Follow-Up–Section 10. Evidence Table 52. Nonrandomized Trials, Observational Studies, and/or Registries for Longitudinal Follow-Up–Section 10. References. NIHMS867255-supplement-supplement_1.pdf (2.2MB, pdf) Table 1. ACC/AHA Recommendation System: Applying Class of Recommendation and Level of Evidence to Clinical Strategies, Interventions, Treatments, or Diagnostic Testing in Patient Care (Updated August 2015). \| CLASS (STRENGTH) OF RECOMMENDATION \| \| CLASS 1 (STRONG) \| Benefit >>> Risk \| \| Suggested phrases for writing recommendations: ■ Is recommended ■ Is indicated/useful/effective/beneficial ■ Should be performed/administered/other ■ Comparative-Effectiveness Phrases†: ○ Treatment/strategy A is recommended/indicated in preference to treatment B ○ Treatment A should be chosen over treatment B \| \| CLASS IIa (MODERATE) \| Benefit >> Risk \| \| Suggested phrases for writing recommendations; ■ Is reasonable ■ Can be useful/effective/beneficial ■ Comparative-Effectiveness Phrases†: ○ Treatment/strategy A is probably recommended/indicated in preference to treatment B ○ It is reasonable to choose treatment A over treatment B \| \| CLASS IIb (WEAK) \| Benefit ≥ Risk \| \| Suggested phrases for writing recommendations: ■ May/might be reasonable ■ May/might be considered ■ Usefulness/effectiveness is unknown/unclear/uncertain or not well established \| \| CLASS III: No Benefit (MODERATE) (Generally, LOE A or B use only) \| Benefit = Risk \| \| Suggested phrases for writing recommendations: ■ Is not recommended ■ Is not indicated/useful/effective/beneficial ■ Should not be performed/administered/other \| \| CLASS III: Harm (STRONG) \| Risk > Benefit \| \| Suggested phrases for writing recommendations: ■ Potentially harmful ■ Causes harm ■ Associated with excess morbidity/mortality ■ Should not be performed/administered/other \| \| LEVEL (QUALITY) OF EVIDENCE‡ \| \| LEVEL A \| \| ■ High-quality evidence‡ from more than 1 RCT ■ Meta-analyses of high-quality RCTs ■ One or more RCTs corroborated by high-quality registry studies \| \| LEVEL B-R \| (Randomized) \| \| ■ Moderate-quality evidence‡ from 1 or more RCTs ■ Meta-analyses of moderate-quality RCTs \| \| LEVEL B-NR \| (Nonrandomized) \| \| ■ Moderate-quality evidence‡ from 1 or more well-designed, well-executed nonrandomized studies, observational studies, or registry studies ■ Meta-analyses of such studies \| \| LEVEL C-LD \| (Limited Data) \| \| ■ Randomized or nonrandomized observational or registry studies with limitations of design or execution ■ Meta-analyses of such studies ■ Physiological or mechanistic studies in human subjects \| \| LEVEL C-EO \| (Expert Opinion) \| \| Consensus of expert opinion based on clinical experience \| Open in a new tab COR and LOE are determined independently (any COR may be paired with any LOE), A recommendation with LOE C does not imply that the recommendation is weak. Many important clinical questions addressed in guidelines do not lend themselves to clinical trials. Although RCTs are unavailable, there may be a very dear clinical consensus that a particular test or therapy is useful or effective. The outcome or result of the intervention should be specified (an improved clinical outcome or increased diagnostic accuracy or incremental prognostic information). † For comparative-effectiveness recommendations (COR I and lla; LOE A and B only), studies that support the use of comparator verbs should involve direct comparisons of the treatments or strategies being evaluated. ‡ The method of assessing quality is evolving, including the application of standardized, widely used, and preferably validated evidence grading tools; and for systematic reviews, the incorporation of an Evidence Review Committee. COR indicates Class of Recommendation; EO, expert opinion; LD, limited data; LOE, Level of Evidence; NR, nonrandomized; R, randomized; and RCT randomized controlled trial. Appendix 1. Author Relationships With Industry and Other Entities (Relevant)—2016 AHA/ACC Guideline on the Management of Patients With Lower Extremity Peripheral Artery Disease (March 2014). \| Committee Member \| Employment \| Consultant \| Speakers Bureau \| Ownership/Partnership /Principal \| Personal Research \| Institutional, Organizational, or Other Financial Benefit \| Expert Witness \| Voting Recusals by Section \| :--- :---: :---: :---: \| Marie D. Gerhard-Herman, Chair \| Harvard Medical School—Associate Professor \| None \| None \| None \| None \| None \| None \| None \| \| Heather L. Gornik, Vice Chair \| Cleveland Clinic Foundation, Cardiovascular Medicine—Medical Director, Noninvasive Vascular Laboratory \| None \| None \| Summit Doppler Systems Zin Medical \| Astra Zeneca Theravasc \| None \| None \| 3.1, 3.2, 5.1–5.3, and 5.6. \| \| Coletta Barrett \| Our Lady of the Lake Regional Medical Center—Vice President \| None \| None \| None \| None \| None \| None \| None \| \| Neal R. Barshes \| Baylor College of Medicine, Division of Vascular Surgery and Endovascular Therapy Michael E. DeBakey Department of Surgery—Assistant Professor \| None \| None \| None \| None \| None \| None \| None \| \| Matthew A. Corriere \| University of Michigan—Frankel Professor of Cardiovascular Surgery, Associate Professor of Surgery \| None \| None \| None \| None \| None \| None \| None \| \| Douglas E. Drachman \| Massachusetts General Hospital—Training Director \| Abbott Vascular St. Jude Medical \| None \| None \| Atrium Medical Bard Lutonix \| None \| None \| 4, 8.1.1–9.1.2, and 10.2.2. \| \| Lee A. Fleisher \| University of Pennsylvania Health System Department of Anesthesiology and Critical Care—Chair \| None \| None \| None \| None \| None \| None \| None \| \| Francis Gerry R. Fowkes \| University of Edinburgh—Emeritus Professor of Epidemiology \| AstraZeneca† Bayer Merck \| None \| None \| None \| None \| None \| 5.1–5.3, 5.6, 5.10, 7, and 9.2. \| \| Naomi M. Hamburg \| Boston University School of Medicine, Cardiovascular Medicine Section—Associate Professor of Medicine \| None \| None \| None \| None \| None \| None \| None \| \| Scott Kinlay \| VA Boston Healthcare System—Associate Chief, Cardiology Director, Cardiac Catheterization Laboratory & Vascular Medicine \| None \| None \| None \| Medtronic† The Medicines Company† \| None \| None \| 4, 5.6, 8.1.1, 9.1.1, 10.2.1, and 10.2.2. \| \| Robert Lookstein \| Mount Sinai Medical Center—Chief, Interventional Radiology; Professor of Radiology and Surgery; Vice Chair, Department of Radiology \| Boston Scientific Medrad Interventional Possis The Medicines Company \| Cordis‡ \| None \| Shockwave (DSMB) \| None \| None \| 4, 5.6, 8.1.1, 9.1.1, 10.2.1, and 10.2.2. \| \| Sanjay Misra \| Mayo Clinic, Division of Vascular and Interventional Radiology—Professor; Department of Radiology— Interventional Radiologist \| None \| None \| None \| Johnson & Johnson (DSMB) \| None \| None \| 4, 7, 8, and 10.2.2. \| \| Leila Mureebe \| Duke University Medical Center—Associate Professor of Surgery, Division of Vascular Surgery \| None \| None \| None \| None \| None \| None \| None \| \| Jeffrey W. Olin \| Ichan School of Medicine at Mount Sinai, Zena and Michael A. Wiener Cardiovascular Institute and Marie-Josée and Henry R. Kravis Center for Cardiovascular Health—Professor of Medicine, Cardiology; Director, Vascular Medicine \| AstraZeneca Merck Novartis Plurestem \| None \| Northwind† \| AstraZeneca† \| None \| None \| 5.1–5.3, 5.6, 5.10, and 12. \| \| Rajan A.G. Patel \| John Ochsner Heart & Vascular Center, Ochsner Clinical School, University of Queensland School of Medicine— Senior Lecturer \| None \| None \| None \| None \| None \| None \| None \| \| Judith G. Regensteiner \| University of Colorado, Health Sciences Center, Division of Cardiology—Associate Professor of Medicine \| None \| None \| None \| None \| None \| None \| None \| \| Andres Schanzer \| University of Massachusetts Medical School—Professor of Surgery and Quantitative Health Sciences; Program Director, Vascular Surgery Residency \| Cook Medical \| None \| None \| None \| None \| None \| 4, 8.1.1, 9.1.1, and 10.2.2. \| \| Mehdi H. Shishehbor \| Cleveland Clinic, Interventional Cardiology and Vascular Medicine— Director, Endovascular Services \| Boston Scientific‡ Medtronic‡ \| None \| None \| None \| Atrium Medical AstraZeneca† \| None \| 4, 8.1.1– 9.1.2, and 10.2.2. \| \| Kerry J. Stewart \| Johns Hopkins University, School of Medicine; Johns Hopkins Bayview Medical Center— Professor of Medicine; Director, Clinical and Research Exercise Physiology \| None \| None \| None \| None \| None \| None \| None \| \| Diane Treat-Jacobson \| University of Minnesota, School of Nursing— Professor \| None \| None \| None \| None \| None \| None \| None \| \| M. Eileen Walsh \| University of Toledo, College of Nursing— Professor \| None \| None \| None \| None \| None \| None \| None \| Open in a new tab This table represents the relationships of committee members with industry and other entities that were determined to be relevant to this document. These relationships were reviewed and updated in conjunction with all meetings and/or conference calls of the writing committee during the document development process. The table does not necessarily reflect relationships with industry at the time of publication. A person is deemed to have a significant interest in a business if the interest represents ownership of ≥5% of the voting stock or share of the business entity, or ownership of ≥$5000 of the fair market value of the business entity; or if funds received by the person from the business entity exceed 5% of the person's gross income for the previous year. Relationships that exist with no financial benefit are also included for the purpose of transparency. Relationships in this table are modest unless otherwise noted. According to the ACC/AHA, a person has a relevant relationship IF: a) the relationship or interest relates to the same or similar subject matter, intellectual property or asset, topic, or issue addressed in the document; or b) the company/entity (with whom the relationship exists) makes a drug, drug class, or device addressed in the document, or makes a competing drug or device addressed in the document; or c) the person or a member of the person's household, has a reasonable potential for financial, professional, or other personal gain or loss as a result of the issues/content addressed in the document. Writing committee members are required to recuse themselves from voting on sections to which their specific relationships with industry and other entities may apply. † Significant relationship. ‡ No financial benefit. ACC indicates American College of Cardiology; AHA, American Heart Association; DSMB, data safety monitoring board; and VA, Veterans Affairs. Appendix 3. Abbreviations. AAA = abdominal aortic aneurysm ABI = ankle-brachial index ALI = acute limb ischemia CAD = coronary artery disease CLI = critical limb ischemia CTA = computed tomography angiography DAPT = dual antiplatelet therapy DES = drug-eluting stent(s) GDMT = guideline-directed management and therapy MI = myocardial infarction MRA = magnetic resonance angiography PAD = peripheral artery disease PTA = percutaneous transluminal angioplasty RCT = randomized controlled trial SPP = skin perfusion pressure TBI = toe-brachial index TcPO 2 = transcutaneous oxygen pressure QoL = quality of life Open in a new tab The American Heart Association requests that this document be cited as follows: Gerhard-Herman MD, Gornik HL, Barrett C, Barshes NR, Corriere MA, Drachman DE, Fleisher LA, Fowkes FGR, Hamburg NM, Kinlay S, Lookstein R, Misra S, Mureebe L, Olin JW, Patel RAG, Regensteiner JG, Schanzer A, Shishehbor MH, Stewart KJ, Treat-Jacobson D, Walsh ME. 2016 AHA/ACC guideline on the management of patients with lower extremity peripheral artery disease: a report of the American College of Cardiology/American Heart Association Task Force on Clinical Practice Guidelines. ACC/AHA Task Force Members Jonathan L. Halperin, MD, FACC, FAHA, Chair; Glenn N. Levine, MD, FACC, FAHA, Chair-Elect; Sana M. Al-Khatib, MD, MHS,FACC, FAHA; Kim K. Birtcher, PharmD, MS, AACC; Biykem Bozkurt, MD, PhD, FACC, FAHA; Ralph G. Brindis, MD, MPH, MACC; Joaquin E. Cigarroa, MD, FACC; Lesley H. Curtis, PhD, FAHA; Lee A. Fleisher, MD, FACC, FAHA; Federico Gentile, MD, FACC; Samuel Gidding, MD, FAHA; Mark A. Hlatky, MD, FACC; John Ikonomidis, MD, PhD, FAHA; José Joglar, MD, FACC, FAHA; Susan J. Pressler, PhD, RN, FAHA; Duminda N. Wijeysundera, MD, PhD Presidents and Staff American College of Cardiology: Richard A. Chazal, MD, FACC, President Shalom Jacobovitz, Chief Executive Officer William J. Oetgen, MD, MBA, FACC, Executive Vice President, Science, Education, Quality, and Publishing Amelia Scholtz, PhD, Publications Manager, Science, Education, Quality, and Publishing American College of Cardiology/American Heart Association: Katherine Sheehan, PhD, Director, Guideline Strategy and Operations Lisa Bradfield, CAE, Director, Guideline Methodology and Policy Abdul R. Abdullah, MD, Associate Science and Medicine Advisor Allison Rabinowitz, MPH, Project Manager, Clinical Practice Guidelines American Heart Association: Steven R. Houser, PhD, FAHA, President Nancy Brown, Chief Executive Officer Rose Marie Robertson, MD, FAHA, Chief Science and Medicine Officer Gayle R. Whitman, PhD, RN, FAHA, FAAN, Senior Vice President, Office of Science Operations Comilla Sasson, MD, PhD, FACEP, Vice President, Science and Medicine Jody Hundley, Production Manager, Scientific Publications, Office of Science Operations The Data Supplement is available with this article at Copies: This document is available on the World Wide Web sites of the American College of Cardiology (www.acc.org) and the American Heart Association (professional.heart.org). A copy of the document is available at org/statements by selecting either the “By Topic” link or the “By Publication Date” link. To purchase additional reprints, call 843-216-2533 or kelle.ramsay@wolterskluwer.com. Expert peer review of AHA Scientific Statements is conducted by the AHA Office of Science Operations. For more on AHA statements and guidelines development, visit and select the “Policies and Development” link. 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Nonrandomized Trials, Observational Studies, and/or Registries of Physiological Testing–Section 3.2. Evidence Table 6. Nonrandomized Trials, Observational Studies, and/or Registries of Imaging for Anatomic Assessment (Ultrasound, CTA, MRA, Angiography)–Section 3.3. Evidence Table 7. RCTs of Imaging for Anatomic Assessment (Ultrasound, CTA, MRA, Angiography)–Section 3.3. Evidence Table 8. Nonrandomized Trials, Observational Studies, and/or Registries for Abdominal Aortic Aneurysm–Section 4.1. Evidence Table 9. Nonrandomized Trials, Observational Studies, and/or Registries of Coronary Artery Disease Screening in PAD–Section 4.2. Evidence Table 10. RCTs for CAD Screening in PAD–Section 4.2. Evidence Table 11. Nonrandomized Trials, Observational Studies, and/or Registries of Screening in Carotid Artery Disease–Section 4.3. Evidence Table 12. Nonrandomized Trials, Observational Studies, and/or Registries for Renal Artery Disease–Section 4.4. Evidence Table 13. RCTs Evaluating Antiplatelet Agents– Section 5.1. Evidence Table 14. Nonrandomized Trials, Observational Studies, and/or Registries of Antiplatelet Agents–Section 5.2. Evidence Table 15. Randomized Trials Comparing Statin Agents–Section 5.2. Evidence Table 16. Nonrandomized Trials, Observational Studies, and/or Registries of Statin Agents–Section 5.2. Evidence Table 17. RCTs for Antihypertensive Agents– Section 5.3. Evidence Table 18. Nonrandomized Trials, Observational Studies, and/or Registries of Antihypertensive Agents–Section 5.3. Evidence Table 19. RCTs for Smoking Cessation–Section 5.4. Evidence Table 20. Nonrandomized Trials, Observational Studies, and/or Registries of Smoking Cessation–Section 5.4. Evidence Table 21. RCTs Evaluating Glycemic Control in Patients with PAD and Diabetes Mellitus–Section 5.5. Evidence Table 22. Nonrandomized Trials, Observational Studies, and/or Registries of Glycemic Control–Section 5.5. Evidence Table 23. RCTs Evaluating Oral Anticoagulation–Section 5.6. Evidence Table 24. Nonrandomized Trials, Observational Studies, and/or Registries of Oral Anticoagulation–Section 5.6. Evidence Table 25. RCTs and Observational Studies of Cilostazol–Section 5.7. Evidence Table 26. Nonrandomized Trials, Observational Studies, and/or Registries of Pentoxifylline–Section 5.8. Evidence Table 27. Systematic Review of Chelation Therapy–Section 5.9. Evidence Table 28. Nonrandomized Trials, Observational Studies, and/or Registries of Homocysteine Lowering Therapy for Lower Extremity PAD in Patients with Diabetes Mellitus–Section 5.10.1. Evidence Table 29. RCTs Comparing Additional Medical Therapies of Homocysteine Lowering Therapy for Lower Extremity PAD–Section 5.10.1. Evidence Table 30. RCTs for Influenza Vaccination–Section 5.10.2. Evidence Table 31. Nonrandomized Trials for Influenza Vaccination–Section 5.10.2. Evidence Table 32. RCTs for Exercise Therapy–Section 6. Evidence Table 33. Nonrandomized Trials, Observational Studies, and/or Registries for Exercise Therapy–Section 6. Evidence Table 34. Nonrandomized Trials and Observational Studies of Minimizing Tissue Loss in Patients with PAD–Section 7. Data Supplement 34a. Functions of a Multidisciplinary Foot Care / Amputation Prevention Team–Section 7. Evidence Table 35. RCTs Comparing Endovascular Treatment and Endovascular Versus Noninvasive Treatment of Claudication–Section 8.1. Evidence Table 36. Nonrandomized Trials, Observational Studies, and/or Registries of Endovascular and Endovascular Versus Noninvasive Treatment of Claudication–Section 8.1. Evidence Table 37. RCTs Evaluating Surgical Treatment for Claudication–Section 8.1.2. Evidence Table 38. Nonrandomized Trials, Observational Studies, and/or Registries of Surgical Treatment for Claudication–Section 8.1.2. Evidence Table 39. RCTs Comparing Endovascular Revascularization for Chronic CLI–Section 8.2. Evidence Table 40. Nonrandomized Trials, Observational Studies, and/or Registries of Endovascular Revascularization for Chromic CLI–Section 8.2.1. Evidence Table 41. RCTs of Surgical Revascularization for Chronic CLI–Section 8.2. Evidence Table 42. Nonrandomized Trials, Observational Studies, and/or Registries for Surgical Revascularization for Chronic CLI–Section 8.2. Evidence Table 43. RCT Comparing Prostanoids for End-Stage Peripheral Artery Disease–Section 8.2.3. Evidence Table 44. Nonrandomized Trials, Observational Studies, and/or Registries for Would Healing Therapies for CLI–Section 8.2.3. Evidence Table 45. Nonrandomized Trials, Observational Studies, and/or Registries of Acute Limb Ischemia–Section 9.1. Evidence Table 46. Nonrandomized Trials, Observational studies, and/or Registries Comparing Evaluating Noninvasive Testing and Angiography for ALI–Section 9.1. Evidence Table 47. RCTs of Revascularization Strategy for ALI–Section 9.2.2. Evidence Table 48. Nonrandomized Trials, Observational Studies, and/or Registries of Clinical Presentation of ALI–Section 9.2.2. Evidence Table 49. Nonrandomized Trials, Observational Studies, and/or Registries of Diagnostic Evaluation of the Cause of ALI–Section 9.2.2. Evidence Table 50. Nonrandomized Trials, Observational Studies, and/or Registries of Revascularization Strategy for ALI–Section 9.2.2. Evidence Table 51. RCTs for Longitudinal Follow-Up–Section 10. Evidence Table 52. Nonrandomized Trials, Observational Studies, and/or Registries for Longitudinal Follow-Up–Section 10. References. NIHMS867255-supplement-supplement_1.pdf (2.2MB, pdf) ACTIONS View on publisher site PDF (1014.2 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Preamble 1. Introduction 2. Clinical Assessment for PAD 3. Diagnostic Testing for the Patient with Suspected Lower Extremity PAD (Claudication or CLI) 4. Screening for Atherosclerotic Disease in Other Vascular Beds for the Patient with PAD 5. Medical Therapy for the Patient with PAD 6. Structured Exercise Therapy: Recommendations 7. Minimizing Tissue Loss in Patients with PAD: Recommendations 8. Revascularization for Claudication 9. Management of CLI 10. Management of ALI 11. Longitudinal Follow-Up: Recommendations 12. Evidence Gaps and Future Research Directions 13. Advocacy Priorities Supplementary Material ACC/AHA Task Force Members Presidents and Staff References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
12131	https://brainly.in/question/60350706	Find the value of a for which p^n = (a × 5)^n ends with digit 0. - Brainly.in Skip to main content Ask Question Log in Join for free For parents For teachers Honor code Textbook Solutions Brainly App riya15061 19.05.2024 Math Primary School answered Find the value of a for which p^n = (a × 5)^n ends with digit 0. 2 See answers See what the community says and unlock a badge. 0:00 / 0:15 Read More Answer No one rated this answer yet — why not be the first? 😎 ay4409290 ay4409290 Answer: If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5 It can be observed that 2 is not in the prime factorisation of 5n. Hence, for any value of n, 5n will not be divisible by 2. Therefore, 5n cannot end with the digit 0 for any natural number n. Explore all similar answers Thanks 0 rating answer section Answer rating 3.7 (6 votes) Find Math textbook solutions? See all Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 Mathematics 619 solutions NCERT Class 8 Mathematics 815 solutions NCERT Class 7 Mathematics 916 solutions NCERT Class 10 Mathematics 721 solutions NCERT Class 6 Mathematics 1230 solutions Xam Idea Mathematics 10 2278 solutions ML Aggarwal - Understanding Mathematics - Class 8 2090 solutions R S Aggarwal - Mathematics Class 8 1964 solutions R D Sharma - Mathematics 9 2199 solutions R S Aggarwal - Mathematics Class 7 2222 solutions SEE ALL Advertisement Answer No one rated this answer yet — why not be the first? 😎 gokulat55 gokulat55 thanks for the help ☺️ its really helpful Explore all similar answers Thanks 0 rating answer section Answer rating 0.0 (0 votes) Advertisement Still have questions? Find more answers Ask your question New questions in Math using basic geometric shapes create traditional indian rt form like warli painting 3. 5a-(3b-2a + 4c) simplify each of the following algebraic expression by removing gropping symbols if the area of a square plot is 1600sq.m , what is the side length of square plot Line ab parallel line CD line PQ is transferor major of angle BRT is equal 105 degree find angle RTC angle dtq angle ART the median of 23 35 15 8 x 42 and 28 is x. which of the following cannot be the value of x PreviousNext Advertisement Ask your question Free help with homework Why join Brainly? ask questions about your assignment get answers with explanations find similar questions I want a free account Company Careers Advertise with us Terms of Use Copyright Policy Privacy Policy Cookie Preferences Help Signup Help Center Safety Center Responsible Disclosure Agreement Get the Brainly App ⬈(opens in a new tab)⬈(opens in a new tab) Brainly.in We're in the know (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
12132	https://www.collinsdictionary.com/us/dictionary/english/delay	English Spanish Portuguese Hindi Chinese Korean Japanese More English Italiano Português 한국어 简体中文 Deutsch Español हिंदी 日本語 English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese Definitions Summary Synonyms Sentences Pronunciation Collocations Conjugations Grammar Credits × Definition of 'delay' COBUILD frequency band delay (dɪleɪ ) Word forms: plural, 3rd person singular present tense delays, present participle delaying, past tense, past participle delayed 1. transitive verb/intransitive verb If you delay doing something, you do not do it immediately or at the planned or expected time, but you leave it until later. For sentimental reasons I wanted to delay my departure until June. They had delayed having children, for the usual reason, to establish their careers. 2. transitive verb To delay someone or something means to make them late or to slow them down. Can you delay him in some way? Various setbacks and problems delayed production. 3. intransitive verb If you delay, you deliberately take longer than necessary to do something. If he delayed any longer, the sun would be up. Synonyms: linger, lag, loiter, dawdle More Synonyms of delay 4. variable noun If there is a delay, something does not happen until later than planned or expected. They claimed that such a delay wouldn't hurt anyone. 5. uncountable noun Delay is a failure to do something immediately or in the required or usual time. We'll send you a quote without delay. Synonyms: dawdling, lingering, loitering, procrastination More Synonyms of delay More Synonyms of delay Collins COBUILD Advanced Learner’s Dictionary. Copyright © HarperCollins Publishers American English pronunciation ! It seems that your browser is blocking this video content. To access it, add this site to the exceptions or modify your security settings, then refresh this page. British English pronunciation ! It seems that your browser is blocking this video content. To access it, add this site to the exceptions or modify your security settings, then refresh this page. You may also like English Quiz ConfusablesSynonyms of 'delay'Language Lover's BlogFrench Translation of 'delay'Translate your textPronunciation PlaylistsWord of the day: 'curtain call'Spanish Translation of 'delay'English GrammarCollins AppsEnglish Quiz ConfusablesSynonyms of 'delay'Language Lover's BlogFrench Translation of 'delay'Translate your textPronunciation PlaylistsWord of the day: 'curtain call'Spanish Translation of 'delay'English GrammarCollins AppsEnglish Quiz ConfusablesSynonyms of 'delay'Language Lover's Blog COBUILD frequency band delay in American English (diˈleɪ ; dɪˈleɪ ) verb transitive 1. to put off to a future time; postpone 2. to make late; slow up; detain verb intransitive 3. to stop for a while; linger noun 4. a delaying or being delayed 5. the period of time for which something is delayed Webster’s New World College Dictionary, 4th Edition. Copyright © 2010 by Houghton Mifflin Harcourt. All rights reserved. Derived forms delayer (deˈlayer) noun Word origin ME delaien < OFr delaier < de-, intens. + laier, to leave, let, altered (? after conjugation of faire) < laissier < L laxare: see relax COBUILD frequency band delay in American English (dɪˈlei) transitive verb 1. to put off to a later time; defer; postpone The pilot delayed the flight until the weather cleared 2. to impede the process or progress of; retard; hinder The dense fog delayed the plane's landing intransitive verb 3. to put off action; linger; loiter He delayed until it was too late noun 4. the act of delaying; procrastination; loitering 5. an instance of being delayed There were many delays during the train trip 6. the period or amount of time during which something is delayed The ballet performance began after a half-hour delay SYNONYMS 1. See defer1. 2. slow, detain. 3. procrastinate, tarry. 4. tarrying, dawdling. 5. deferment, postponement, respite. Most material © 2005, 1997, 1991 by Penguin Random House LLC. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd Derived forms delayable adjective delayer noun delayingly adverb Word origin [1225–75; ME delaien (v.), delai(e) (n.) ‹ OF delaier (v.), delai (n.)] COBUILD frequency band delay in British English (dɪˈleɪ ) verb 1. (transitive) to put off to a later time; defer 2. (transitive) to slow up, hinder, or cause to be late; detain 3. (intransitive) to be irresolute or put off doing something; procrastinate 4. (intransitive) to linger; dawdle noun 5. the act or an instance of delaying or being delayed 6. the interval between one event and another; lull; interlude Collins English Dictionary. Copyright © HarperCollins Publishers Derived forms delayer (deˈlayer) noun Word origin C13: from Old French delaier, from des- off + laier, variant of laissier to leave, from Latin laxāre to loosen, from laxus slack, lax1 Examples of 'delay' in a sentence delay These examples have been automatically selected and may contain sensitive content that does not reflect the opinions or policies of Collins, or its parent company HarperCollins. We welcome feedback: report an example sentence to the Collins team. Read more… These disruptions have caused inevitable financial information preparation delays. Wall Street Journal (2020) Those delays have held back potential sales. Wall Street Journal (2023) One of the most common reasons for delays in discharging people is problems organising social care. The Guardian (2018) Then work was delayed by the death of his own wife. The Guardian (2022) They can not afford the delay, obstruction and cost of immigration and customs controls. The Guardian (2016) The governor emphasised yesterday that economic pain would be delayed rather than cancelled. Times, Sunday Times (2016) If you suffered a delay this summer and are due compensation, do not give up. The Sun (2016) Before take-off, the pilot said a technical fault would delay their departure. Times, Sunday Times (2016) There may be a temptation to bolster security by delaying their departure. Times, Sunday Times (2006) But the Government cannot delay acting. The Sun (2011) Related word partners delay delay a departure delay a hearing delay a plan delay a process delay a project delay a trial delay a vote delay development delay implementation delay indefinitely delay legislation delay payment delay progress delay publication delay recovery delay repeatedly delay retirement delay surgery delay the introduction of delay the onset of delay the return of delay the start delay treatment delayed gratification delays occur delivery delay developmental delay experience a delay force a delay lengthy delay long delay passenger delay production delay rain delay reduce a delay regulatory delay repeated delays significant delay slight delay suffer delay traffic delay travel delays unexpected delay unnecessary delay weather delay Trends of delay View usage over: Source: Google Books Ngram Viewer In other languages delay British English: delay /dɪˈleɪ/ NOUN If there is a delay, something does not happen until later than planned or expected. This caused delays and disruption to flights. American English: delay /dɪˈleɪ/ Arabic: تَأْخِير Brazilian Portuguese: atraso Chinese: 拖延 Croatian: odgoda Czech: zpoždění Danish: forsinkelse Dutch: vertraging European Spanish: retraso Finnish: viivytys French: retard German: Verspätung Greek: καθυστέρηση Italian: ritardo Japanese: 遅れ Korean: 지연 Norwegian: forsinkelse Polish: opóźnienie European Portuguese: atraso Romanian: întrziere Russian: задержка Spanish: retraso Swedish: försening Thai: ความล่าช้า Turkish: gecikme Ukrainian: затримка Vietnamese: sự trì hoãn British English: delay /dɪˈleɪ/ VERB If you delay doing something, you do not do it until a later time. He intends to delay his departure until next Sunday. American English: delay /dɪˈleɪ/ Arabic: يَتَأَخَّرُ Brazilian Portuguese: atrasar Chinese: 拖延 Croatian: odgoditi Czech: odkládat na později Danish: forsinke Dutch: vertragen European Spanish: retrasar Finnish: viivyttää French: retarder German: verzögern Greek: καθυστερώ Italian: ritardare Japanese: 遅れる Korean: 지연하다 Norwegian: forsinke Polish: opóźnić European Portuguese: atrasar Romanian: a amna Russian: отложить Spanish: retrasar Swedish: försena Thai: เลื่อนออกไป Turkish: gecikmek Ukrainian: відкладати Vietnamese: trì hoãn Translate your text for free Browse alphabetically delay Delaware Bay Delaware Jargon Delawarean delay delay a date delay a decision delay a departure All ENGLISH words that begin with 'D' Related terms of delay avoid delay brief delay delay line long delay rain delay View more related words ## Wordle Helper ## Scrabble Tools Quick word challenge Quiz Review Question: 1 - Score: 0 / 5 tide or tied? Which version is correct? He tied the dog securely to the railing.He tide the dog securely to the railing. place or plaice? Which version is correct? The best place to catch fish on a canal is close to a lock.The best plaice to catch fish on a canal is close to a lock. muscles or mussels? Which version is correct? Exercise will tone up your stomach mussels.Exercise will tone up your stomach muscles. censor or sensor? Which version is correct? Television companies tend to sensor bad language in feature films.Television companies tend to censor bad language in feature films. read or reed? Drag the correct answer into the box. read reed She was standing in front of a bed on the far side of the lake. Your score: New collocations added to dictionary Collocations are words that are often used together and are brilliant at providing natural sounding language for your speech and writing. 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12133	https://artofproblemsolving.com/wiki/index.php/Divisibility_rules?srsltid=AfmBOorh1yaCjems96oKR-lRp6fEVp_zPFTCOhHnFc_5L1fUE4hgWYBd	Art of Problem Solving Divisibility rules - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Divisibility rules Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Divisibility rules These divisibility rules help determine when positive integers are divisible by particular other integers. All of these rules apply for base-10only -- other bases have their own, different versions of these rules. Contents [hide] 1 Divisibility Videos 2 Basics 2.1 Divisibility Rule for 2 and Powers of 2 2.2 Divisibility Rule for 3 and 9 2.3 Divisibility Rule for 5 and Powers of 5 2.4 Divisibility Rule for 7 2.5 Divisibility Rule for 10 and Powers of 10 2.6 Divisibility Rule for 11 2.7 General Rule for Composites 2.7.1 Example 3 Advanced 3.1 General Rule for Primes 3.2 Divisibility Rule for 13 3.3 Divisibility Rule for 17 3.4 Divisibility Rule for 19 3.5 Divisibility Rule for 29 3.6 Divisibility Rule for 49 4 Special 4.1 Mod-preserving tests 4.1.1 Mod-preserving for 7 4.1.2 Mod-preserving for 13 4.2 Block tests 4.2.1 Small blocks -- 101 and 1001 4.2.2 Bigger blocks -- 10001 and 10000001 4.2.3 Type 2 blocks -- 111 and 11111 5 Problems 6 Resources 6.1 Books 6.2 Classes 7 See also Divisibility Videos Basics Divisibility Rule for 2 and Powers of 2 A number is divisible by if and only if the last digits of the number are divisible by . Thus, in particular, a number is divisible by 2 if and only if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8. Proof Divisibility Rule for 3 and 9 A number is divisible by 3 or 9 if and only if the sum of its digits is divisible by 3 or 9, respectively. Note that this does not work for higher powers of 3. For instance, the sum of the digits of 1899 is divisible by 27, but 1899 is not itself divisible by 27. Proof Divisibility Rule for 5 and Powers of 5 A number is divisible by if and only if the last digits are divisible by that power of 5. Proof Divisibility Rule for 7 Rule 1: Partition into 3 digit numbers from the right (). The alternating sum () is divisible by 7 if and only if is divisible by 7. Proof Rule 2: Truncate the last digit of , double that digit, and subtract it from the rest of the number (or vice-versa). is divisible by 7 if and only if the result is divisible by 7. Proof Rule 3: "Tail-End divisibility." Note. This only tells you if it is divisible and NOT the remainder. Take a number say 12345. Look at the last digit and add or subtract a multiple of 7 to make it zero. In this case we get 12380 or 12310 (both are acceptable; I am using the former). Lop off the ending 0's and repeat. 1238 - 28 ==> 1210 ==> 121 - 21 ==> 100 ==> 1 NOPE. Works in general with numbers that are relatively prime to the base (and works GREAT in binary). Here's one that works. 12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY! Tiny extension to tell you the remainder: Count how many zeroes you lop off and mod 6. Multiply mod 7 with the corresponding number Zeroes (mod 6) Number to multiply by 0 1 1 3 2 2 3 6 4 4 5 5 And that's the remainder. Divisibility Rule for 10 and Powers of 10 If a number is power of 10, define it as a power of 10. The exponent is the number of zeros that should be at the end of a number for it to be divisible by that power of 10. Example: A number needs to have 6 zeroes at the end of it to be divisible by 1,000,000 because . Divisibility Rule for 11 A number is divisible by 11 if the alternating sum of the digits is divisible by 11. Proof General Rule for Composites A number is divisible by , where the prime factorization of is , if the number is divisible by each of . Example For the example, we will check if 55682168544 is divisible by 36. The prime factorization of 36 to be . Thus we must check for divisibility by 4 and 9 to see if it's divisible by 36. Since the last two digits, 44, of the number is divisible by 4, so is the entire number. To check for divisibility by 9, we look to see if the sum of the digits is divisible by 9. The sum of the digits is 54 which is divisible by 9. Thus, the number is divisible by both 4 and 9 and must be divisible by 36. Advanced General Rule for Primes For every prime number other than 2 and 5, there exists a rule similar to rule 2 for divisibility by 7. For a general prime , there exists some number such that an integer is divisible by if and only if truncating the last digit, multiplying it by and subtracting it from the remaining number gives us a result divisible by . Divisibility rule 2 for 7 says that for , . The divisibility rule for 11 is equivalent to choosing . The divisibility rule for 3 is equivalent to choosing . These rules can also be found under the appropriate conditions in number bases other than 10. Also note that these rules exist in two forms: if is replaced by then subtraction may be replaced with addition. We see one instance of this in the divisibility rule for 13: we could multiply by 9 and subtract rather than multiplying by 4 and adding. is any number so that Divisibility Rule for 13 Rule 1: Truncate the last digit, multiply it by 4 and add it to the rest of the number. The result is divisible by 13 if and only if the original number was divisble by 13. This process can be repeated for large numbers, as with the second divisibility rule for 7. Proof Rule 2: Partition into 3 digit numbers from the right (). The alternating sum () is divisible by 13 if and only if is divisible by 13. Proof Rule 3: Works for . Let . If is odd add 39 to . Round up to the nearest multiple of 80, call the result . Find . Check: Is . Proof Divisibility Rule for 17 Truncate the last digit, multiply it by 5 and subtract from the remaining leading number. The number is divisible if and only if the result is divisible. The process can be repeated for any number. Proof Divisibility Rule for 19 Truncate the last digit, multiply it by 2 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers. Proof Divisibility Rule for 29 Truncate the last digit, multiply it by 3 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers. Proof Divisibility Rule for 49 Why 49? For taking pesky out of a root. Useful below 4900. Round up to a multiple of 50, call it , and subtract the original number, call this . If , it is divisible by 49. Examples: Round up: . Difference: . ? Yes! Round up: . Difference: . ? No! Round up: . Difference: . ? Yes! Extension to work for all numbers. Floor divide by 4950, multiply by 50, and add to before calculating Proof Special Mod-preserving tests These tests allow you take the modulo operation easily. Mod-preserving for 7 Multiply the first digit by 3 and add it to the rest. Mod-preserving for 13 Multiply the first digit by 3 and subtract it from the rest Block tests As a bonus, these are also mod-preserving Small blocks -- 101 and 1001 The divisibility for 101 test is simple: Alternate adding and subtracting blocks of two digits starting from the end two, which are added. Ex. 1102314 by 101 01 + 10 - 23 + 14 ← last block is always two digits and positive =0 so 1102314 is divisible by 101 The divisibility for 1001 is the same, but with blocks of three. (Starting with the end three, this time) The 1001 test also works for all it's divisors. The most useful are 7, 11, and 13. Bigger blocks -- 10001 and 10000001 10001 has block size length 4, and factors nicely into 73137. 1000001 has block size 6, and factors into 175882353. 5882353 isn't much use, but 17 is, when we're testing a large number. Type 2 blocks -- 111 and 11111 A different type of test can be yielded from adding all the blocks, but again starting with the end. 111 has a block length of three, and factors into 37 and 3. 11111 has a length of five, and factors to 41 and 271. 1111111, with a length of seven, can provide a test for 239 and 4649, if you ever need it. Problems Practice Problems on Alcumus Divisibility (Prealgebra) 2000 AMC 8 Problems/Problem 11 2006 AMC 10B Problems/Problem 25 Resources Books The AoPS Introduction to Number Theory by Mathew Crawford. The Art of Problem Solving by Sandor Lehoczky and Richard Rusczyk. Classes AoPS Introduction to Number Theory Course See also Number theory Modular arithmetic Math books Mathematics competitions Retrieved from " Category: Divisibility Rules Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
12134	https://people.tamu.edu/~yvorobets/MATH433-2023A/Lect3-09web.pdf	MATH 433 Applied Algebra Lecture 34: Zeros of polynomials (continued). Greatest common divisor of polynomials. Zeros of polynomials Deﬁnition. An element α ∈R of a ring R is called a zero (or root) of a polynomial f ∈R[x] if f (α) = 0. Theorem Let F be a ﬁeld. Then α ∈F is a zero of f ∈F[x] if and only if the polynomial f (x) is divisible by x −α. Idea of the proof: The remainder after division of f (x) by x −α is f (α). Problem. Find the remainder after division of f (x) = x100 by g(x) = x2 + x −2. We have x100 = (x2 + x −2)q(x) + r(x), where r(x) = ax + b for some a, b ∈R. The polynomial g has zeros 1 and −2. Evaluating both sides at x = 1 and x = −2, we obtain f (1) = r(1) and f (−2) = r(−2). This gives rise to a system of linear equations: a + b = 1, −2a + b = 2100. It has a unique solution: a = (1 −2100)/3, b = (2100 + 2)/3. Rational roots Theorem Let f (x) = cnxn + cn−1xn−1 + · · · + c1x + c0 be a polynomial with integer coeﬃcients and cn, c0 ̸= 0. Assume that f has a rational root α = p/q, where the fraction is in lowest terms. Then p divides c0 and q divides cn. Proof: By assumption, cn p q n + cn−1 p q n−1 + · · · + c1 p q + c0 = 0. Multiplying both sides of this equality by qn, we obtain cnpn + cn−1pn−1q + · · · + c1pqn−1 + c0qn = 0. It follows that c0qn is divisible by p while cnpn is divisible by q. Since the fraction p/q is in lowest terms, we have gcd(p, q) = 1. This implies that, in fact, c0 is divisible by p and cn is divisible by q. Corollary If cn = 1 then any rational root of the polynomial f is, in fact, an integer. Example. f (x) = x3 + 6x2 + 11x + 6. Since all coeﬃcients are integers and the leading coeﬃcient is 1, all rational roots of f (if any) are integers. Moreover, the only possible integer roots of f are divisors of the constant term: ±1, ±2, ±3, ±6. Notice that there are no positive roots as all coeﬃcients are positive. We obtain that f (−1) = 0, f (−2) = 0, and f (−3) = 0. First we divide f (x) by x + 1: x3 + 6x2 + 11x + 6 = (x + 1)(x2 + 5x + 6). Then we divide x2 + 5x + 6 by x + 2: x2 + 5x + 6 = (x + 2)(x + 3). Thus f (x) = (x + 1)(x + 2)(x + 3). Greatest common divisor of polynomials Deﬁnition. Given non-zero polynomials f , g ∈F[x], a greatest common divisor gcd(f , g) is a polynomial over the ﬁeld F such that (i) gcd(f , g) divides f and g, and (ii) if any p ∈F[x] divides both f and g, then it divides gcd(f , g) as well. Theorem (Bezout) The polynomial gcd(f , g) exists and is unique up to a scalar multiple. Moreover, it is a non-zero polynomial of the least degree that can be represented as uf + vg, where u, v ∈F[x]. Theorem The polynomial gcd(f , g) exists and is unique up to a scalar multiple. Moreover, it is a non-zero polynomial of the least degree that can be represented as uf + vg, where u, v ∈F[x]. Proof: Let S denote the set of all polynomials of the form uf + vg, where u, v ∈F[x]. The set S contains non-zero polynomials, say, f and g. Let d(x) be any such polynomial of the least possible degree. It is easy to show that the remainder after division of any polynomial h ∈S by d belongs to S as well. By the choice of d, that remainder must be zero. Hence d divides every polynomial in S. In particular, d is a common divisor of f and g. Further, if any p(x) ∈F[x] divides both f and g, then it also divides every element of S. In particular, it divides d. Thus d = gcd(f , g). Now assume d1 is another greatest common divisor of f and g. By deﬁnition, d1 divides d and d divides d1. This is only possible if d and d1 are scalar multiples of each other.
12135	https://simple.wikipedia.org/wiki/Regular_polygon	Regular polygon - Simple English Wikipedia, the free encyclopedia Jump to content [x] Main menu Main menu move to sidebar hide Getting around Main page Simple start Simple talk New changes Show any page Help Contact us About Wikipedia Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Give to Wikipedia Create account Log in [x] Personal tools Give to Wikipedia Create account Log in [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 Examples 2 Properties Regular polygon [x] 52 languages العربية অসমীয়া Azərbaycanca বাংলা Български Чӑвашла Čeština Cymraeg Deutsch Eesti Ελληνικά English Español Esperanto Euskara فارسی Français Galego 한국어 Հայերեն हिन्दी Ido Italiano עברית ქართული Latviešu Lietuvių Magyar Македонски Монгол Nederlands नेपाली 日本語 Norsk bokmål Oʻzbekcha / ўзбекча ភាសាខ្មែរ Polski Português Română Русский Slovenščina Ślůnski کوردی Suomi Tagalog தமிழ் Українська اردو Tiếng Việt 粵語中文 ⵜⴰⵎⴰⵣⵉⵖⵜ ⵜⴰⵏⴰⵡⴰⵢⵜ Change links Page Talk [x] English Read Change Change source View history [x] Tools Tools move to sidebar hide Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sandbox Expand all Edit interlanguage links Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikidata item From Simple English Wikipedia, the free encyclopedia \| Regular heptadecagon \| \| A regular heptadecagon \| \| Type \| Regular polygon \| \| Edges and vertices \| 17 \| \| Schläfli symbol \| {17} \| \| Coxeter diagram \| \| \| Symmetry group \| Dihedral (D 17), order 2×17 \| \| Internal angle (degrees) \| ≈158.82° \| \| Dual polygon \| Self \| \| Properties \| Convex, cyclic, equilateral, isogonal, isotoxal \| A regular polygon is a shape that can be drawn on a flat surface. It has sides that are all the same length, and its angles are all the same. In other words, a polygon which is both equilateral (which means all its sides have the same length) and equiangular (which means that all its angles are the same) is a regular polygon. It always has the same number of edges and points. A polygon is also convex if there aren't any two points within it that can't be connected with a straight line, where the entire straight line is also in it. An example of a concave shape is a five-pointed star. It's concave because you can connect two of the points of a star with a straight line, and most of the straight line doesn't go through the star. If a five-pointed star is both equilateral and equiangular, it is also a regular polygon, but it isn't a convex one. All regular concave polygons are stars. Examples [change \| change source] The simplest examples of regular convex polygons are: equilateral triangle square pentagon hexagon heptagon octagon enneagon decagon Polygons with more sides look more like circles. A polygon with an infinite number of sides is called an apeirogon. Polygons can be described in text by a special symbol called a Schläfli symbol. This symbol is written as curly brackets, with some numbers inside it, like {3}. {3} is a triangle. The number inside can also be a fraction. A Schläfli symbol of {5/2} is a five-pointed star. Properties [change \| change source] Most mathematical objects have things that can be said about them that are always true. These are called properties. All polygons (convex and star polygons) can be rotated in place by some angle and still look the same—in other words, they're all rotationally symmetric. For example, a square could be rotated a quarter turn, and it would look the same as before. Another way of seeing this property is that any polygon can have a circle drawn around it, touching all its points, and the polygon's points are evenly spaced around the circle. A polygon can also have a circle drawn inside it, touching all its edges (but not overlapping), and the places where the circle touches the polygon are also evenly spaced. The centers of the two circles drawn are in the same place. This proves that a regular polygon is rotationally symmetric. The circle outside the polygon is called circumscribed, and the circle inside the polygon is called inscribed. Any polygon where a circle circumscribes all its points is called a cyclic polygon. All regular polygons are cyclic, but not all cyclic polygons are regular. \| Collapse v t e Fundamental convex regular and uniform polytopes in dimensions 2–10 \| \| Family \| A n \| B n \| I 2(p) / D n \| E 6 / E 7 / E 8 / F 4 / G 2 \| H n \| \| Regular polygon \| Triangle \| Square \| p-gon \| Hexagon \| Pentagon \| \| Uniform polyhedron \| Tetrahedron \| Octahedron • Cube \| Demicube \| \| Dodecahedron • Icosahedron \| \| Uniform 4-polytope \| 5-cell \| 16-cell • Tesseract \| Demitesseract \| 24-cell \| 120-cell • 600-cell \| \| Uniform 5-polytope \| 5-simplex \| 5-orthoplex • 5-cube \| 5-demicube \| \| \| \| Uniform 6-polytope \| 6-simplex \| 6-orthoplex • 6-cube \| 6-demicube \| 1 22 • 2 21 \| \| \| Uniform 7-polytope \| 7-simplex \| 7-orthoplex • 7-cube \| 7-demicube \| 1 32 • 2 31 • 3 21 \| \| \| Uniform 8-polytope \| 8-simplex \| 8-orthoplex • 8-cube \| 8-demicube \| 1 42 • 2 41 • 4 21 \| \| \| Uniform 9-polytope \| 9-simplex \| 9-orthoplex • 9-cube \| 9-demicube \| \| \| \| Uniform 10-polytope \| 10-simplex \| 10-orthoplex • 10-cube \| 10-demicube \| \| \| \| Uniform n-polytope \| n-simplex \| n-orthoplex • n-cube \| n-demicube \| 1 k2 • 2 k1 • k 21 \| n-pentagonal polytope \| \| Topics: Polytope families • Regular polytope • List of regular polytopes and compounds \| Retrieved from " Category: Polygons This page was last changed on 1 February 2024, at 20:04. Text is available under the Creative Commons Attribution-ShareAlike License and the GFDL; additional terms may apply. See Terms of Use for details. Privacy policy About Wikipedia Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Regular polygon 52 languagesAdd topic
12136	https://math.stackexchange.com/questions/2345511/cosets-of-mathbbz-n-under-addition-and-resulting-structure-of-mathbbz	abstract algebra - Cosets of $\mathbb{Z}_n$ (under addition) and resulting structure of $\mathbb{Z}/\mathbb{Z}_n$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Cosets of Z n Z n (under addition) and resulting structure of Z/Z n Z/Z n Ask Question Asked 8 years, 2 months ago Modified8 years, 2 months ago Viewed 334 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. It is well known (and asked countless times as I gathered while searching an answer for this question) that Z/n Z Z/n Z≅≅Z n Z n. However, what I am trying to do is to infer the structure of Z/Z n Z/Z n (if this is even a valid structure that is, as I am not even sure that Z n Z n is a valid normal subgroup of Z Z under normal addition: details below). Obviously taking a quotient group is not the same thing as division as we are used to (as in multiplying by the multiplicative inverse), but the separation (partition) into "equal" groups of the original supergroup does resemble similar concepts. I was wondering if this structure will also be isomorphic to n Z n Z as if we simply switched the denominator like in regular division. I simply cannot comprehend how to partition Z Z by the cosets of Z n Z n. And this ultimately is my problem. If I take regular (nonmodular addition) clearly if I take the coset where I add 1 to, say, Z 5 Z 5, this overlaps with Z 5 Z 5, and does not parition Z Z. But then I also consider how nonmodular addition will not make Z 5 Z 5 a well defined binary structure in the first place. So if we consider just modular addition we arrive that it is its only coset. So did I answer my own question? The only reason that we can't have cosets of Z n Z n in the context of a supergroup Z Z is because of the fact that their operations are incompatible, and would not even make sense, as the incompatibility would not make it a proper subgroup. I am curious as to the relationship here as it has never been brought up in any introductory algebra class. abstract-algebra group-theory finite-groups normal-subgroups quotient-group Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jul 3, 2017 at 21:32 Charlie TianCharlie Tian 176 1 1 silver badge 11 11 bronze badges 5 I'm not sure I understand the question, but it seems as if you are having trouble with the idea of a quotient group. A quotient group of G G is not a subgroup of G G, and thus the notion of Z/Z n Z/Z n does not make sense. If Z n Z n was a subgroup of Z Z, then Z Z would have torsion elements, which it does not!TomGrubb –TomGrubb 2017-07-03 21:36:44 +00:00 Commented Jul 3, 2017 at 21:36 Yes I figured that I answered my own question, because they are incompatible. And I never once said the quotient group was a subgroup of G, I was referring the group we were taking quotients of. However, I believe that there is not a way to describe Z n Z n as a normal subgroup of Z Z, and thus not a valid thing to take quotients of.Charlie Tian –Charlie Tian 2017-07-03 21:38:45 +00:00 Commented Jul 3, 2017 at 21:38 also not sure what a torsion element is either Charlie Tian –Charlie Tian 2017-07-03 21:39:02 +00:00 Commented Jul 3, 2017 at 21:39 1 @CharlieTian: A torsion element is one whose product with a non-zero element equals zero, e.g., in Z 6,2,3,4 Z 6,2,3,4 are torsion, since 2(3)=3(2)=4(3)=0(M o d 6)2(3)=3(2)=4(3)=0(M o d 6)MSIS –MSIS 2017-07-03 22:24:33 +00:00 Commented Jul 3, 2017 at 22:24 Would you define a torsion as nonzero? Because then 0 is also a torsion element. This sounds very similar to the concept of zero divisors Charlie Tian –Charlie Tian 2017-07-06 15:02:47 +00:00 Commented Jul 6, 2017 at 15:02 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. I think you are having notation issues. The notation Z/n Z Z/n Z always means the integers modulo n n. The notation Z n Z n is sometimes used sometimes for the integers modulo n n, but sometimes it is not. See the section of the wikipedia on notation. However the way you are using it means the integers modulo n n. Same object, different notation. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 3, 2017 at 21:41 ybermanyberman 1,974 11 11 silver badges 20 20 bronze badges 1 Yes this is what I mean, and the textbook I have followed (Fraleigh) uses this frequently as the integers mod n n Charlie Tian –Charlie Tian 2017-07-03 21:57:35 +00:00 Commented Jul 3, 2017 at 21:57 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. 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12137	https://www.virtuallearningacademy.net/LessonDisplay/Lesson8477/MATHALGIU35Geometric_Sequences.pdf	GEOMETRIC SEQUENCES Unit Overview An important mathematical skill is discovering patterns. In this unit, you will investigate different types of patterns represented in geometric sequences. You will also discern the difference between an arithmetic sequence and a geometric sequence. Geometric Sequences Watch this video: Example 2: Increasing Geometric Sequence--Population (02:30) An example of a geometric sequence 5 10 20 40 80 A geometric sequence is one in which each number is multiplied by a constant ratio to get the next number in the sequence. In the example above, notice that each term is multiplied by 2 to get the next term. The following is another example of a Geometric Sequence that still follows the pattern of multiplying by 2. 2 3 4 5 2, 2 , 2 , 2 , 2 2, 4, 8, 16, 32 = Another way to look at this sequence is: 1 2 1 3 2 4 3 5 4 2, 2, 2, 2, 2 a a a a a a a a a = = ⋅ = ⋅ = ⋅ = ⋅ … In each case, the next term in the sequence is the product of the previous term and the constant 2. For a geometric sequence, the number that multiplies the present term to give the next result is called the common ratio. The effect of the common ratio is perhaps better seen in the next example. 2 3 4 1 1 2 3 4 4, 4 3, 4 3 , 4 3 , 4 3 4, 12, 36, 108, 324, 972 , 3, 3, 3, 3 a a a a a ⋅ ⋅ ⋅ ⋅ = = ⋅ ⋅ ⋅ ⋅ In this example, the common ratio is 3 and is denoted r = 3. × 2 × 2 × 2 × 2 Geometric Sequence: A Geometric Sequence is a sequence where each term after the first term, 1 a , is the product of the preceding term and the common ratio, r, where r ≠ 0 or 1. The terms of a geometric sequence can be represented by; 1 2 1 3 2 4 3 5 4 , , , , . . . a a a r a a r a a r a a r = ⋅ = ⋅ = ⋅ = ⋅ This formula for the general term is as follows: Example #1: Find the next three terms in the following geometric sequence. 5, 15, 45, … Step #1: Determine the common ratio between terms. 15 3 5 45 3 5 = = Step #2: Multiply the last term by the common ratio and then again and as often as required. 45 3 135 135 3 405 405 3 1215 ⋅ = ⋅ = ⋅ = Therefore, the next three terms are 135, 405, 1215. nth Term of a Geometric Sequence The nth term, of a geometric sequence whose first term is and whose ratio is r is given by the explicit formula , where n 1 Example #2: Find the first five terms of the geometric sequence for which 1 1 576 and 2 a r = = − . 1 2 3 4 5 576 1 576 288 2 1 288 144 2 1 144 72 2 1 72 36 2 a a a a a = = ⋅− = − = − ⋅− = = ⋅− = − = − ⋅− = The first five terms of the sequence are 576, –288, 144, –72, 36. Click on the link to watch the video "Geometric sequence or progression" or click on the video. Example #3: Find the 6th term of the geometric sequence for which 1 2 and 3. a r = = − Substitute the values for n and r into the formula for the nth term of an geometric sequence along with 1 2 a = . 1 1 5 6 6 6 (2)( 3) (2)( 243) 486 n n a a r a a a − = = − = − = − The 6th term of the sequence is –486. Stop! Go to Questions #1-7 about this section, then return to continue on to the next section. Example #4: Find the explicit formula for the sequence 2, 4, 8, 16, 32, … First, find the common ratio (r). In this example, 4/2 = 2, 8/4 = 2, 16/8 = 2 and 32/16 = 2 so 2 is our common ratio. Using our explicit formula , 1 1 n n a a r − = , we will plug in 2 for r and 2 for a1 (the first term). This gives the formula 2 2 2n n a − = × . Using our exponent rules, we could simplify this further. Remember that when we have like bases, we can add exponents. If we think of this as 1 2 2 2n n a − = × , we can add the exponents to get 1 2 1 2 2 n n n a + − − = = . Notice that this is an exponential function. If graphed, we would get this: Geometric sequences are exponential functions. Example #5: Using the explicit formula 1 5 4n n a − = × , state a1 and r. Then, find the first 4 terms of the sequence. First, using the formula, you can easily identify a1 and r. a1 = 5 based on its position in the formula. r = 4, again based on its position in the formula. Now, use the formula to find the first 4 terms. We already know the first term (a1) is 5. 2nd term: 2 1 1 5 4 5 4 20 n a − = × = × = 3rd term: 3 1 2 5 4 5 4 5 16 80 n a − = × = × = × = 4th term: 4 1 3 5 4 5 4 5 64 320 n a − = × = × = × = Example #6: Use the table of values to write a function for the geometric sequence. First, find the common ratio, r. 5 5 1 25 5 5 125 5 5 = − − − = − = − − We see that –5 is the common ratio. Next, identify the first term a1. The first term is –1 so a1 = –1. Now, since we have 2 negative numbers, make sure to be careful and not make mistakes with those negatives! We will use our explicit formula, 1 1 n n a a r − = , but since we are writing in function notation, we will write it as 1 1 ( ) n f n a r − = . 1 ( ) 1( 5) n f n − = −− Notice that the –5 is in parenthesis. This is because –5 is r. All of r is taken to the exponent. Without including the negative in parenthesis, we are making r a positive 5. Also, we are not taking (–1) × (–5). Exponents first, and this will take care of the negative 5. Whatever you get after working the exponent is what will be multiplied by –1. Term # Value 1 –1 2 5 3 –25 4 125 5 –625 To make sure you can use this function properly, Try this! Using the function just found, find the 6th term of the geometric sequence. "Click here" to check the answer. f (6) = (–1) × (–5)6 – 1 f (6) = (–1) × (–5)5 f (6) = (–1)(–3125) f (6) = 3125 Stop! Go to Questions #8-12 about this section, then return to continue on to the next section. Arithmetic or Geometric? In the last unit, you learned about arithmetic sequences. In this unit, you have learned about geometric sequences. If it is not specified, how do you know which type of sequence it is? How do you know which rule to use to create the explicit formula? Remember that arithmetic sequences had a common difference. A number that you added or subtracted to find the next term. In a geometric sequence, you multiply by a common ratio to find the next term. When given problems that aren’t specified, you must discern if you have a common difference or a common ratio. For the next 4 problems, identify each sequence as arithmetic, geometric, or neither. If the sequence is arithmetic state the common difference. If the sequence is geometric, state the common ratio. Identify each sequence. 3, –3, 3, –3, …. "Click here" to check the answer. Geometric, r = –1 Identify each sequence. 1, 4, 9, 16, … "Click here" to check the answer. Neither, the same number is neither added nor multiplied each time to get the next term. Identify each sequence. 25, 50, 75, 100, … "Click here" to check the answer. Arithmetic, d = 25 Identify each sequence. 2, 1, 0.5, 0.25 "Click here" to check the answer. Geometric, r = ½ or 0.5 Stop! Go to Questions #13-17 about this section, then return to continue on to the next section. Word Problems on Geometric Sequences Example#1: A culture of bacteria doubles every 3 hours. If there are 200 bacteria present at the beginning, how many bacteria will there be after 24 hours? Step #1: Determine whether the situation represents an arithmetic or geometric sequence. Starting with 200 and doubling produces the following sequence. 200, 400, 800,… The sequence 200, 400, 800, … is a geometric sequence. Step #2: Identify the variables. If the bacteria doubles every 3 hours, it will double 8 times in a 24 hour period. So, we are looking for the 8th term of the sequence. 1 200 2, 8 a r n = = = Step #3: Substitute and evaluate. Use the formula for a geometric sequence. 1 1 7 8 8 200(2) 25,600 n n a a r a a − = = = There are 25,600 bacteria present after 24 hours. Example #2: Mrs. Readworthy is serious when she says she wants her students to study. Each week, she requires them to study 5 times the number of spelling words required the week before. If she starts off the year requiring only 1 spelling word, how many spelling words will students be studying by week 4? From the problem, we discern that a1 = 1 (1 word the first week). We also see the common ratio is 5. From this, we write the formula 1 1 n n a a r − = and fill in what we know. 1 1 5n n a − = × . We are solving for week 4, so 4 1 3 4 1 5 1 5 1 125 125 a − = × = × = × = . Mrs. Readworthy will require 125 spelling words on week 4. Stop! Go to Questions #18-19 about this section, then return to continue on to the next section. The Fibonacci Sequence There is one special sequence worth noting since we are discussing sequences. That is the Fibonacci Sequence. It is a special pattern found by adding the previous 2 numbers in the sequence. a1 = 1 and a2 = 1. By adding the first 2 terms, a3 = 2. Continuing the pattern 1, 1, 2, 3, 5, 8, 13, … Notice that the pattern is to add the previous two terms to get the next term. Watch this video about the Fibonacci Sequence: Fibonacci Sequence (03:50) Watch this video to see where the Fibonacci Sequence is seen in nature: The Fibonacci Sequence in Nature (03:00) What type of sequence is the Fibonacci Sequence? Is it arithmetic? geometric? Do we add the same number each time to get the next term? No. Do we multiply by the same number each time to get the next term? No. We always add the previous two terms. So, this sequence is not arithmetic or geometric. We can represent it with an explicit formula, but that is beyond the scope of this lesson. However, we can represent it with a recursive formula. Remember, recursive means that it is dependent on knowing the previous term(s). For this sequence, we could write, f (n + 1) = f (n) + f (n – 1) for n ≥ 1. We must also state that f (0) = f (1) = 1. This means that the first two terms are 1. Then the formula basically says that the next term (n + 1) equals the last two terms added together. If you’d like to learn more about this fascinating sequence, take it further and research it. There are many wonderful websites and books on this topic. Stop! Go to Questions #20-30 to complete this unit.
12138	https://www.nycmathteam.org/wp-content/uploads/2020/11/Divisibility_and_Bases_Lesson-2.pdf	Divisibility and Bases Lesson Theo Schiminovich November 2020 1 Divisibility 1.1 Divisibility Rules What are the divisibility rules for: • 2? • 3? • 4? • 5? • 6? How can this be applied to other numbers? • 8? • 9? • 10? • 11? 1.2 Exercises 1. Is 1748391 divisible by 9? 2. Is 3953851 divisible by 11? 1.3 Proving Divisibility Rules • Prove the divisibility rule for 2. • Prove the divisibility rule for 4. • Prove the divisibility rule for 8. • Prove the divisibility rule for 3. • Prove the divisibility rule for 9. • Prove the divisibility rule for 11. 1.4 Problems 1. Find the largest 3-digit number that is divisible by 22 such that the sum of the units digit and the tens digit is 11. 2. If the ﬁve-digit number ABCDE is the 4th power of a whole number and A + C + E = B + D, ﬁnd C. 1 2 Bases Base: The number of digits in a number system Base 10 is what’s typically used, which has 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A number AnAn−1...A2A1A0 in base 10 has a value An ∗10n + An−1 ∗10n−1 + ... + A2 ∗102 + A1 ∗101 + A0. Likewise, a number AnAn−1...A2A1A0 in base b has a value An ∗bn +An−1 ∗bn−1 +...+A2 ∗b2 +A1 ∗b1 +A0. When a number N is in a certain base b, it is written as Nb. When writing in a base system which has more than 10 digits, the additional digits will become letters. For example, the digits in base 16 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F 2.1 How to convert a number c from base a to base 10 • Consider a number c = AnAn−1...A2A1A0 in base a. • Write it as c = An ∗an + An−1 ∗an−1 + ... + A2 ∗a2 + A1 ∗a1 + A0. • Multiply and add (using base 10 rules) until you simplify it down to one number. That number is c in base 10. 2.2 How to convert a number c from base 10 to base b • Find the largest power of b that is smaller than your number c. Call that power bn. • Subtract bn from c until you get a value less than bn. Put the number of times you subtracted bn in the bn’s place. • Repeat for bn−1, bn−2, ...b1, 1, putting the number of times you subtracted in the correct place. 2.3 How to convert a number c from base a to base b To do this, you can convert the number from base a to base 10, and then from base 10 to base b, using the two sections above. 2.4 Exercises 1. Convert 8310 to base 8. 2. Convert 10010 to base 2. 3. Convert 2257 to base 9. 4. Convert DF16 to base 6. 2.5 Problems 1. If N is the base 4 equivalent of 36110, ﬁnd the square root of N in base 4. 2. Find all solutions to the equation 72b = 2(27b) or ﬁnd that there is no solution. 3. Find all solutions to the equation 73b = 2(37b) or ﬁnd that there is no solution. 4. ABC and CBA are, respectively, the base 9 and base 7 numerals for the same positive integers. Express this integer in base 10. 5. The three-digit base x numeral 7y3 is twice the three-digit base x numeral 3y7. Express x + y as a base 10 numeral. 2 3 Operations in Diﬀerent Bases 3.1 Addition and Subtraction Addition and subtraction are the same, except instead of carrying multiples of 10, you carry multiples of b. 3.2 Problems 1. 1011011012 + 1010000102 (in base 2) 2. DEF16 + ABC16 (in base 16) 3. 1A4C16 −DFE16 (in base 16) 3.3 Multiplication When using the multiplication algorithm: • In the ﬁrst step, when multiplying digits, ﬁnd the result in base b, then drop down the one’s digit and carry over the ”b’s digit”. • In the second step, the addition is done as mentioned earlier. 3.4 Problems 1. 919 ∗919 (in base 19) 2. 101112 ∗10102 (in base 2) 3. 578 ∗678 (in base 8) 4 Decimals in Diﬀerent Bases 4.1 Converting from Base 10 to Base b (non-repeating decimals) Way 1 • Find the largest power of b that is smaller than your number c. Call that power bn. • Subtract bn from c until you get a value less than bn. Put the number of times you subtracted bn in the bn’s place. • Repeat for bn−1, bn−2, ...b1, 1, b−1, b−2, ..., putting the number of times you subtracted in the correct place. Stop when some subtraction gives you a result of 0. Way 2 • Multiply c by b repeatedly until you get a whole number. Call the number of times you have multiplied k, and your new number d. Now you know c = d bk . You can start by converting d to base b. • Find the largest power of b that is smaller than your number d. Call that power bn. • Subtract bn from d until you get a value less than bn. Put the number of times you subtracted bn in the bn’s place. • Repeat for bn−1, bn−2, ...b1, 1, putting the number of times you subtracted in the correct place. • Now, divide d by bk to get your result c by shifting the decimal place to the left k times. Note: These methods will only work if c is a non-repeating decimal in base b. An extreme example: 1 49 = 0.02040816326530612244897959183673469387755110 = 0.017. 3 4.2 Problems 1. Express 9.2510 in base 2. 2. Express 3.6562510 in base 2. 3. Express 11.¯ 510 in base 3. 5 AMC/AIME Problems 1. The base-ten representation for 19! is 121, 6T5, 100, 40M, 832, H00, where T, M, and H denote digits that are not given. What is T + M + H? (2019 AMC 10B) (A=3, B=8, C=12, D=14, E=17) 2. The number n can be written in base 14 as abc, can be written in base 15 as acb, and can be written in base 6 as acac, where a > 0. Find the base-10 representation of n. (2018 AIME I) 3. A positive integer N has base-eleven representation abc and base-eight representation 1bca, where a, b, and c represent (not necessarily distinct) digits. Find the least such N expressed in base ten. (2020 AIME I) 4
12139	https://www.wyzant.com/resources/answers/415097/three_vertices_of_parallelogram_wxyz_are_w_5_2_x_2_4_and_z_7_3_find_the_coordinates_of_vertex_y	WYZANT TUTORING Ethan H. three vertices of parallelogram WXYZ are W (-5,2) X (2,4) and Z (-7,-3). find the coordinates of vertex Y 1 Expert Answer Carl M. answered • 11/02/17 Pre-Algebra, Algebra 1, Algebra 2 and Geometry Tutoring Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. RELATED TOPICS RELATED QUESTIONS what is a bisector geometry Answers · 5 what is an equation equal of a line parallel to y=2/3x-4 and goes through the point (6,7) Answers · 9 what are some angles that can be named with one vertex? Answers · 2 name of a 2 demension figure described below Answers · 8 how to find the distance of the incenter of an equlateral triangle to mid center of each side? Answers · 7 RECOMMENDED TUTORS Cindy M. Nicholas P. Graciele O. find an online tutor Download our free app A link to the app was sent to your phone. Get to know us Learn with us Work with us Download our free app Let’s keep in touch Need more help? Learn more about how it works Tutors by Subject Tutors by Location IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Education.com 35,000 worksheets, games, and lesson plans TPT Marketplace for millions of educator-created resources Vocabulary.com Adaptive learning for English vocabulary ABCya Fun educational games for kids SpanishDictionary.com Spanish-English dictionary, translator, and learning Inglés.com Diccionario inglés-español, traductor y sitio de aprendizaje Emmersion Fast and accurate language certification
12140	https://www.investopedia.com/terms/r/residual-value.asp	Skip to content Trade Please fill out this field. Top Stories Even With Fed Cuts, This One Move Can Protect Your Savings Fed Cut or Not, Here’s How Much You Lose by Keeping Savings at the Biggest Banks Grab a CD Rate up to 4.60% Before the Fed Acts Despite Fed Moves, Earn 5% and $300 With This Bank Account Now Table of Contents Table of Contents What Is Residual Value? How to Calculate Residual Value Examples Using Residual Value The Bottom Line Residual Value Explained, With Calculation and Examples By Adam Hayes Full Bio Adam Hayes, Ph.D., CFA, is a financial writer with 15+ years Wall Street experience as a derivatives trader. Besides his extensive derivative trading expertise, Adam is an expert in economics and behavioral finance. Adam received his master's in economics from The New School for Social Research and his Ph.D. from the University of Wisconsin-Madison in sociology. He is a CFA charterholder as well as holding FINRA Series 7, 55 & 63 licenses. He currently researches and teaches economic sociology and the social studies of finance at the University of Lucerne in Switzerland.Adam's new book, "Irrational Together: The Social Forces That Invisibly Shape Our Economic Behavior" (University of Chicago Press) is a must-read at the intersection of behavioral economics and sociology that reshapes how we think about the social underpinnings of our financial choices. Learn about our editorial policies Updated May 13, 2025 Fact checked by Vikki Velasquez Fact checked by Vikki Velasquez Full Bio Vikki Velasquez is a researcher and writer who has managed, coordinated, and directed various community and nonprofit organizations. She has conducted in-depth research on social and economic issues and has also revised and edited educational materials for the Greater Richmond area. Learn about our editorial policies Definition Residual value is the estimated worth of an asset at the end of its useful life or lease term. Residual value is what's left of an asset's worth after you're done using it. In finance or accounting, this concept is crucial for determining depreciation schedules, lease payments, and investment decisions. Also known as salvage value or scrap value, residual value helps businesses and investors understand how much is retained after an asset's primary use period ends. Key Takeaways Residual value is an asset's estimated worth at the end of its useful life. Depreciation calculations and lease terms are based on residual value. To determine residual value, you need to look at several factors—from what's happening in the market to how quickly technology is becoming outdated and what's trending in your industry. Different depreciation methods can significantly impact how residual value affects financial statements and tax calculations. The difference between estimated residual value and actual resale value has significant financial implications, particularly in leasing arrangements. What Is Residual Value? Residual value is what you expect to get back when you're ready to sell or dispose of something—whether it's a piece of equipment, a vehicle, or a building. The idea is rooted in the understanding that assets lose value over time because of wear and tear, technological obsolescence, or market saturation. From an accounting perspective, it affects companies' depreciation and amortization expenses. In this way, it influences capital budgeting decisions by helping determine the total cost of ownership and whether to buy versus lease. For leased assets, the residual value then helps figure out the terms of leases. A higher residual value generally results in lower regular payments because the lessee is expected to pay for the asset’s usage minus its estimated remaining worth. Here is the most basic formula for residual value: Residual Value = Original Cost – (Annual Depreciation Expense × Useful Life) - Disposal Cost This formula assumes straight-line depreciation. But in practice, residual value calculations can be more complex given the following: Market conditions and demand for used assets Technological obsolescence rates Historical price trends for similar assets Maintenance costs and asset condition Industry-specific factors Tip Different industries have their own ways of figuring out residual value. For example, the auto industry considers everything from how well certain brands hold their value to what's been happening in the used car market.Equipment manufacturers do something similar but focus more on how their machines are used and maintained. How to Calculate Residual Value Here's how to figure out residual value. While different industries might tweak the process a bit, the underlying principles are largely the same. Step 1: Determine Initial Cost First, add up everything you spent to get the asset up and running. This isn't just the purchase price—don't forget to include things like installation costs as well. Example:A company purchases a piece of machinery for $50,000, plus a $1,000 installation fee, totaling $51,000. Step 2: Determine the Asset’s Useful Life Estimate the duration (typically in years) when the asset will be operational and productive for the business. The useful life of an asset is based on manufacturer guidelines, as well as industry standards and historical data. Example:The machine is expected to last 10 years. Step 3: Forecast the Salvage Value Estimate the amount the asset will be worth at the end of its useful life. This forecast should consider market conditions, technological obsolescence (are there newer, better versions available?), and expected wear and tear. Salvage value is usually given as a percentage of the asset’s original cost. Example Option A (Percentage Method):Industry data suggests that similar machines have a residual value of about 20% remaining after 10 years. Residual Value = Original Cost × Residual Percentage Residual Value = $50,000 × 0.20 = $10,000 Example Option B (Market Comparison):Based on market research and transaction data, the anticipated resale value for this type of machine after 10 years is projected at $8,500. In practice, companies may combine approaches to refine their estimates. In our example, let's take the average of both A and B: ($10,000 + 8,500) / 2 = $9,250 Step 4: Adjust for Disposal Costs Subtract any costs directly associated with disposing of or selling the asset. Example:If selling the machine would incur a $250 removal fee to haul it away, then you'd get the following: Residual Value = Forecasted Salvage Value – Disposal Costs Residual Value = $9,250 – $250 = $9,000 Step 5: Subtract the Residual Value from the Acquisition Cost The company purchased the asset for a total basis of $51,000 and is projected to net $9,000 upon its eventual disposal, resulting in a total cost to the business of $42,000. Examples Using Residual Value Let's see how this works in practice for different assets: Depreciation of a Machine Let's revisit the example above, where a company purchases a machine for $50,000, paying $1,000 for installation. The machine has a useful life of 10 years and is expected to have a residual value of $10,000 at the end of its life. Using the straight-line depreciation method, we get the following: Annual Depreciation = (Purchase Price of Asset - Salvage Value) / Estimated Useful Life ($51,000 – $10,000) / 10 = $4,100 per year Vehicle Leasing When a consumer leases a vehicle, the leasing company estimates the car’s residual value at the beginning of the lease. Suppose a car is leased with an original cost of $30,000, and the residual value is set at 50% after a three-year lease term, which is $15,000. This is the residual value used in the lease calculation. However, when the lease expires, market conditions may differ: Estimated residual value: $15,000 (used in the lease contract) Actual market (resale) value: Upon lease-end, the car may actually be sold for $13,000 fair market value because of market downturns or higher-than-expected wear. As you can see, guessing what something will be worth in the future isn't always spot-on. That's why setting residual values can be a bit tricky. Important In lease contracts, the residual value is used to set the buyout price at the end of the lease term. The Bottom Line Whether buying new equipment, setting up a lease, or planning for depreciation, calculating residual value is worth the effort. Keep an eye on your estimates and adjust them when needed. This way, your financial statements stay accurate. Remember, the goal isn't to make a perfect prediction (that's almost impossible) but to make a solid estimate that helps you plan. Whether leasing a car, buying manufacturing equipment, or managing a fleet of vehicles, understanding residual value puts you in a better position to make sound financial decisions. Article Sources Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy. Oliver Wyman, LLC. "A Better Approach To Residual Value." EquipmentWatch by Fusable. "Residual Value." The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace. Open an Account Before the Fed Decision on Sept. 17 Read more Business Corporate Finance Financial Analysis Partner Links The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace. Open an Account Before the Fed Decision on Sept. 17 Related Articles Total Enterprise Valuation (TEV): Definition, Calculation, and Uses What Is CAPM Formula in Excel? 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12141	https://www.profedeele.es/gramatica/verbos/indicativo/preterito-indefinido/	Recibe nuestras novedades por correo electrónico: Actividades Gramática Verbos Tiempos verbales Indicativo Pretérito indefinido Pretérito indefinido Filtrar Categoría Formato Temáticas Nivel B1 B2 La Fama - Rosalía & The Weeknd Actividades para trabajar el vocabulario de la fama y el contraste de pasados. Pretérito imperfectoPretérito indefinidoEspañaCanción B1 B2 Más allá del arcoíris Unidad Didáctica Interactiva sobre algunos conceptos básicos LGTBIQ+ y personajes famosos del mundo hispano que pertenecen al colectivo Pretérito imperfectoPretérito indefinidoEspañaArgentinaChileCubaMéxicoEducación sexualUnidad didácticaKahoot A2 B1 Mujeres con historia Descubre la historia de mujeres hispanas mientras practicas el pretérito indefinido con actividades interactivas. Ideal para niveles A2 y B1. Pretérito indefinidoCulturaValoresFeminismoUnidad didáctica B1 La rana que quería ser una rana auténtica, de Augusto Monterroso Unidad didáctica interactiva para trabajar con una fábula de Augusto Monterroso los pasados y los marcadores temporales. Nivel B1. NarrativaPretérito imperfectoPretérito indefinidoMarcadores del discursoGuatemalaUnidad didáctica A2 B1 La leyenda de san Jorge y el dragón Unidad didáctica interactiva con la leyenda de San Jorge y el dragón para trabajar el contraste de pasados. Nivel A2/B1. Fiestas y tradicionesPretérito imperfectoPretérito indefinidoEspañaUnidad didáctica B1 Contraste de pasados: imperfecto e indefinido Material para aprender el contraste de pasados según la gramática cognitiva. Pretérito imperfectoPretérito indefinidoPresentación B1 Caperucita Roja y otras versiones del cuento para trabajar los pasados Actividad para trabajar el contraste de pasados (pretérito imperfecto - pretérito indefinido). Pretérito imperfectoPretérito indefinidoActividad breve Boletín / Newsletter de ProfeDeELE Recibe gratis por correo electrónico nuestras novedades. "" señala los campos obligatorios Niveles A1 A2 B1 B2 C1 C2 Categorías Alfabetización Actualidad Adverbios de frecuencia Gramática Vocabulario Funciones Exámenes Ortografía Cultura Formatos Actividad breve Presentación Lecturas Ponte al día Unidad didáctica Test Canción Pódcast Escape Room Kahoot Temáticas Países hispanos Español terapia Valores Fines específicos ELE para niños Recibe gratis por correo electrónico nuestras novedades: Suscribirse Accede a los recursos premium de ProfedeELE Actividades interactivas Para profesores y alumnos Acceder a los ejercicios Descargables y formación Solo para profesores Ir a la membresía Redirección a mas.profedeele.es
12142	https://www.cut-the-knot.org/s_4x4_2.shtml	Second 4x4 Coloring problem, Solution Typesetting math: 100% Site... What's new Content page Front page Index page About Privacy policy Help with math Subjects... Arithmetic Algebra Geometry Probability Trigonometry Visual illusions Articles... Cut the knot! What is what? Inventor's paradox Math as language Problem solving Collections... Outline mathematics Book reviews Interactive activities Did you know? Eye opener Analogue gadgets Proofs in mathematics Things impossible Index/Glossary Simple math... Fast Arithmetic Tips Stories for young Word problems Games and puzzles Our logo Make an identity Elementary geometry Second 4x4 Coloring Problem, Solution We are concerned with the problem of getting on black-white pattern from another on a 4×4 board by certain moves related to the Merlin Magic Squares game. The game lives in the 16-dimensional boolean space. The previous variant of the problem had only 13 moves and, for this reason, there were patterns not obtainable from other patterns. To complicate matters, we intoduced additional moves. Here's the whole list: four 3×3-square moves nine 2×2-square moves four 1-row moves four 1-column moves two 4-diagonal moves four 3-diagonal moves four 2-diagonal moves four 1-corner-square moves with the total of 35 different moves in a 16-dimensional game! Interestingly, the answer to the extended problem is still in the negative. Now we have To understand why all this abundance of moves is not sufficient to attain every possible coloring consider the coloring on the right. The region of interest consists of the red squares. For a given coloring, let N b and N w be the numbers of black and white squares, respectively, in the red region. Both (N b mod 2) and (N w mod 2) can easily be shown to be invariant under any of the legal moves. In other words, N b will be even after any of the legal moves if and only if it was even before that move. The same is true for N w. Since for the all-white coloring N w=0 mod 2, it is impossible, starting with this coloring, to obtain any coloring for which N w=1 mod 2. The argument can be framed as suggesting a counterexample: i.e., an example of a board that could not be obtained by the stipulated moves from, say, the above one. Here's one such example Martin Gardner discusses the problem (with no 2×2- and 3×3-square moves) in his Puzzles from Other Worlds, #30. \|Merlin Magic Square\|\|Front page\|\|Up\| Copyright © 1996-2018 Alexander Bogomolny 73255410
12143	https://d296n67kxwq0ge.cloudfront.net/attachments/5126/assets/originals.pdf?1529905502	100+ SAT Words to Study on the Go abolish to put an end to Abraham Lincoln is perhaps best known for his efforts to abolish slavery. abrupt sudden or curt Although she was trying to be nice, her response was still abrupt . agitation anxiety; the fast stirring of a liquid The night before the big game, I was in a state of agitation . alteration change from a previous norm After we missed the bus the third time, our homeroom decided that serious alterations in the schedule were needed. ambiguous open to more than one interpretation; unclear My teacher's instructions about the paper were ambiguous ; nobody knew what to write. ambivalent having mixed feelings Due to his ambiguous instructions, I had ambivalent feelings about my teacher despite his warm manner. arcane difficult to understand I fell asleep moments after I opened the arcane and academic book. 100+ SAT Words to Study on the Go by aromatic having a pleasant smell Smelling aromatic oils can be a good way to reduce stress. assumption an idea accepted as fact without proof My mother made the assumption that I wanted to got to State, but she hadn't talked to me first. begrudging to envy; to give reluctantly I begrudged my sister her new laptop, since I hadn't had a new one in three years. belligerent war-like, inclined to fight It is hard not to provoke my cat, who is belligerent with all other animals. bias inclination for or against a group of people or a particular outcome The scientific study seemed compelling at first, but eventually scientists found that it showed numerous biases . characterize describe distinctive features; to be typical of I shouldn't have been surprised that Jess was so spontaneous, since that was characteristic of her. condescension disdain My father loves to deliver lectures at the dinner table, not realizing how condescending they can make him sound. consequently as a result I got a C on the midterm; consequently , my final grade will not be higher than a B. conserve protect The importance of conserving our natural environment has become increasingly clear. magoosh.com 2100+ SAT Words to Study on the Go by contentious controversial Rebecca would have become student body president if some of her proposals hadn't been so contentious . conventional aligned with general beliefs Jake, who won the election, had much more conventional beliefs, in line with what most students thought. convey express I tried to convey my disappointment to my parents without telling them directly. corroborate confirm; support Lucy said that she'd been in the house all day, and her sister corroborated this statement. corrupt dishonest for personal gain Politicians are known for being corrupt, but I think some of them must have more altruistic goals. counterargument an argument opposing an idea set forth elsewhere While I thought I made a strong case, my debate opponent's counterarguments were just too good. curtail cut short My neighbor had been talking for two hours and eventually, I had to curtail her. deplete use up I was only two blocks from home when I realized that I had completely depleted my car's gas supply. dismay disappointment and distress My father was dismayed that I wouldn't attend his alma mater, but I'm happy with my decision. magoosh.com 3100+ SAT Words to Study on the Go by ebullient highly enthusiastic Luke laughs a lot at everything; he's always been ebullient . eloquent well-spoken Very few presidents have been known for their eloquence , even though they give many speeches. emerging beginning; new The emerging music scene had few followers at the beginning but soon became very popular. empathetic feeling sympathy When I volunteered at the hospital, it was hard not to be empathetic to the patients there. engagement participation; an appointment One component of our grades would be based on our engagement in class discussions. enigmatic mysterious The enigmatic guy seemed compelling at first, but once the mystery was gone, so was my interest. entrenched solidly established It can be difficult to get rid of deep-rooted, entrenched beliefs, since we often take them for granted. enumerate to list I didn't think I was late that much, but my mother enumerated many occasions on which I had been. ephemeral short-lived My interest in the boy band was embarrassing but thankfully ephemeral . magoosh.com 4100+ SAT Words to Study on the Go by equivocal vague My mother's answer to my question about how she'd met my father was equivocal , and I could tell she was embarrassed by the story. esoteric known only to a select group I thought I should have known the book they were discussing, but later I discovered it was esoteric . exertion effort Running a marathon would take so much exertion that I just don't feel up to it. exhilarating wildly exciting The day I got my college acceptance letters was exhilarating . exonerate free from blame My aunt thought I'd taken the car until she saw me in the living room; I was then exonerated . fastidious nit-picky, fussing over details There are some typos in my essay; I just hope the teacher isn't too fastidious . fluctuate to change irregularly My grades fluctuated for a while when I was a freshman, but they're consistent now. foreshadow to allude to coming events Looking back, my early interest in dancing foreshadowed my life in my teens. fundamentally centrally Fundamentally , I don't believe that certain human rights are up for debate. magoosh.com 5100+ SAT Words to Study on the Go by garner gather Before submitting my applications, I had to garner several letters of recommendation. garrulous talkative My grade would have been better but I was too garrulous in class; my best friend sat beside me and we could never shut up. gregarious flocking, sociable Sometimes I wish I were more gregarious , but I prefer having one or two close friends instead. hasten to do something quickly Realizing the deadline was the next day, I hastened to put together my application. hypothetical based on a hypothesis; theoretical Hypothetically , I'd love to travel the world, but I don't have the time or money right now. imperative absolutely necessary The principal insisted that taking an interest in our classes was imperative to our success. indifferent expressing no opinions on a matter The school newspaper reporter wanted to interview me about the issue, but I wasn't a good source because I was too indifferent . indigenous native to a certain area Planting indigenous plants is one way to begin setting up a carbon sink. indiscriminate without consideration There were so many random people at the party that it felt like Tom must have handed out invitations indiscriminately . magoosh.com 6100+ SAT Words to Study on the Go by indispensable vital I've found that keeping my planner up to date is an indispensable practice. indistinct unclear Her voice was indistinct and so I missed most of what she said. infrastructure the organizational structures needed for the operation of a society or project One major challenge to foreign aid is distributing it, since infrastructure —including roads and local organizations to help with distribution—can be lacking in remote areas. keen eager; incisive; to wail He had a keen wit, but sometimes his observations hit too close to home. magnanimous kind-hearted, likely to forgive I'd really insulted her, but luckily she was magnanimous and we were soon friends again. malevolence hostility I don't think that political campaigns have to be characterized by malevolence , even though many politicians see their opponents as enemies. melodramatic exaggerated I wasn't sure how worried I should be, since his stories were always a little melodramatic . menacing inspiring fear The grey sky was menacing , and I was sure it would rain later. modification change I always revise my essays because I find that modifications are usually necessary to make my ideas clear. magoosh.com 7100+ SAT Words to Study on the Go by naïve innocent; likely to believe anything I used to think it was easy to get into a top-10 school, but I was a little naïve . neglect fail to care for I've never been able to keep a plant alive, mostly because I tend to neglect them. null invalid; related to zero We spent years researching the problem only to find that our hypothesis was null . obsolete no longer useful My dad thought I could use his old word processor without realizing that the technology was totally obsolete . omnipotence having unlimited power David acted like his position as editor gave him omnipotence and was always a little mean to his "subordinates." opaque not transparent I really prefer my tights to be opaque , because I hate how shiny the transparent ones can be. oppress to keep in a state of hardship While we thought our allowances were low, it wasn't as though we were being oppressed . ornate highly decorated My sister loves having a really ornate holiday season, but I think it's better to be more subdued. pantheon the group of gods of a people I didn't know about the Greek pantheon until I read about legends containing Zeus. magoosh.com 8100+ SAT Words to Study on the Go by pending awaiting decision Even after three days, I could see that my status hadn't been decided and was still listed as " pending ." preclude to prevent from happening To preclude my tendency to procrastinate, I set aside the same hour every day to study. profuse excessive She thanked me so profusely that it was embarrassing and, after a minute, I had to ask her to stop. regression return to an earlier state My uncle saw my low grade as a regression , but I just saw it as a step in the learning process rather than a set-back. reinforce strengthen; back up I studied for the SAT in different ways to reinforce my knowledge. render to provide; to make I was on the fence about which school to attend until one offered me a full scholarship, rendering my indecision a thing of the past. renounce to declare the abandonment of something formally After the club refused to let Julie in, I renounced my membership in it due to the unfairness of the decision. repeal to revoke Our school tried to pass a strict dress code but soon repealed it when students ignored the guidelines. repose rest When the competition was over, I was exhausted and in desperate need of repose . magoosh.com 9100+ SAT Words to Study on the Go by reproach express disapproval My mother's behavior is always perfect, completely beyond reproach . restorative possessing characteristics that allow it to return health or well-being I felt so much better after the restorative massage. reticent not saying much My cousin can be reticent sometimes, but I know his mind is going a mile a minute. revere respect deeply I wouldn't say I revere my teacher, but I do like her a lot. sampling a representative group It turned out that the three people we'd used weren't a representative sampling of the whole group. scope the extent to which something is relevant The comments on my essay were all good except for one which called the evidence "out of scope ." secession formally withdrawing from membership The South's secession from the United States in the 19th century was unprecedented. selfless unselfish My sister is entirely selfless and always helps me with anything I ask for. simulate imitate The program was meant to simulate what it was like to be in space, but I didn't feel like I was there at all. magoosh.com 10 100+ SAT Words to Study on the Go by soporific causing sleep The arcane book was so boring that I found it soporific . spawn produce, often in terms of offspring Her idea, while later disproven, spawned many more hypotheses that changed the face of science. spectacle a visually impressive performance The play used red and black scenery covered in glitter, making the performance into more of a spectacle than a simple play. stimulate build interest in The simulation was not at all realistic and failed to stimulate my interest in the subject. subsequent following While I didn't do well on my first paper, I used my teacher's advice on the subsequent essays and got better grades. supremacy predominance; the state of being in control of all others Our principal was so kind that nobody ever challenged her supremacy in the school. synchronized occurring at the same time Our answers were synchronized , as though we'd practiced speaking together. tenacious not giving in easily I knew that to get into the school of my dreams, I'd have to work long and hard; I'd have to be tenacious . undermine make less effective Her argument seemed compelling, but when I left her house I thought of six examples that would undermine it. magoosh.com 11 100+ SAT Words to Study on the Go by urbane refined, of the city Going to college in New York seemed like a great idea; I thought I'd come back more sophisticated and urbane . venerable deserving of respect My grandfather didn't speak much, but he was wise when he did, making him a venerable figure in our family. verbose using too many words If you used all these words in the same sentence, you would definitely be verbose . vitality being strong and energetic I was exhausted and weak, but a nap soon restored my vitality . warrant justification; to justify Andrew didn't think the punishment was warranted and argued against it successfully. yield to produce; to give way The apple crop was great this year and yielded way more than we thought it would. Happy studying! magoosh.com 12
12144	https://webbook.nist.gov/cgi/cbook.cgi?ID=C64175&Mask=4&Type=ANTOINE&Plot=on	Jump to content National Institute of Standards and Technology NIST Chemistry WebBook, SRD 69 Home Search Name Formula IUPAC identifier CAS number More options NIST Data SRD Program Science Data Portal Office of Data and Informatics About FAQ Credits More documentation Ethanol Formula: C2H6O Molecular weight: 46.0684 IUPAC Standard InChI: InChI=1S/C2H6O/c1-2-3/h3H,2H2,1H3 IUPAC Standard InChIKey: LFQSCWFLJHTTHZ-UHFFFAOYSA-N CAS Registry Number: 64-17-5 Chemical structure: This structure is also available as a 2d Mol file or as a computed 3d SD file View 3d structure (requires JavaScript / HTML 5) Isotopologues: [2H6]ethanol Ethanol-d1 C2H3D3O Ethanol-d5 Ethanol-1,1-d2 Other names: Ethyl alcohol; Alcohol; Alcohol anhydrous; Algrain; Anhydrol; Denatured ethanol; Ethyl hydrate; Ethyl hydroxide; Jaysol; Jaysol S; Methylcarbinol; SD Alchol 23-hydrogen; Tecsol; C2H5OH; Absolute ethanol; Cologne spirit; Fermentation alcohol; Grain alcohol; Molasses alcohol; Potato alcohol; Aethanol; Aethylalkohol; Alcohol, dehydrated; Alcool ethylique; Alcool etilico; Alkohol; Cologne spirits; Denatured alcohol CD-10; Denatured alcohol CD-5; Denatured alcohol CD-5a; Denatured alcohol SD-1; Denatured alcohol SD-13a; Denatured alcohol SD-17; Denatured alcohol SD-23a; Denatured alcohol SD-28; Denatured alcohol SD-3a; Denatured alcohol SD-30; Denatured alcohol SD-39b; Denatured alcohol SD-39c; Denatured alcohol SD-40m; Etanolo; Ethanol 200 proof; Ethyl alc; Etylowy alkohol; EtOH; NCI-C03134; Spirits of wine; Spirt; Alkoholu etylowego; Ethyl alcohol anhydrous; SD alcohol 23-hydrogen; UN 1170; Tecsol C; Alcare Hand Degermer; Absolute alcohol; Denatured alcohol; Ethanol, silent spirit; Ethylol; Punctilious ethyl alcohol; SD 3A Permanent link for this species. Use this link for bookmarking this species for future reference. Information on this page: Antoine Equation Parameters References Notes Other data available: Gas phase thermochemistry data Condensed phase thermochemistry data Phase change data Reaction thermochemistry data: reactions 1 to 50, reactions 51 to 77 Henry's Law data Gas phase ion energetics data Ion clustering data IR Spectrum Mass spectrum (electron ionization) Gas Chromatography Data at other public NIST sites: Gas Phase Kinetics Database X-ray Photoelectron Spectroscopy Database, version 5.0 Options: Switch to calorie-based units Data at NIST subscription sites: NIST / TRC Web Thermo Tables, "lite" edition (thermophysical and thermochemical data) NIST / TRC Web Thermo Tables, professional edition (thermophysical and thermochemical data) NIST subscription sites provide data under the NIST Standard Reference Data Program, but require an annual fee to access. The purpose of the fee is to recover costs associated with the development of data collections included in such sites. Your institution may already be a subscriber. Follow the links above to find out more about the data in these sites and their terms of usage. Antoine Equation Parameters Go To: Top, References, Notes Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved. log10(P) = A − (B / (T + C)) P = vapor pressure (bar) T = temperature (K) \| Temperature (K) \| A \| B \| C \| Reference \| Comment \| --- --- --- \| \| 364.8 to 513.91 \| 4.92531 \| 1432.526 \| -61.819 \| Ambrose, Sprake, et al., 1975 \| Coefficents calculated by NIST from author's data. \| \| 292.77 to 366.63 \| 5.24677 \| 1598.673 \| -46.424 \| Ambrose and Sprake, 1970 \| Coefficents calculated by NIST from author's data. \| \| 273. to 351.70 \| 5.37229 \| 1670.409 \| -40.191 \| Kretschmer and Wiebe, 1949 \| Coefficents calculated by NIST from author's data. \| Notice: A plot could not be displayed here becuse plotting data currently requires a JavaScript and HTML 5 canvas enabled browser. Plot Help Software credits The interactive plot requires a browser with JavaScript and HTML 5 canvas support. Select a region with data to zoom. Select a region with no data or click the mouse on the plot to revert to the orginal display. Moving the mouse pointer over or near a point will display the coordinates of the point. The number of digits shown do not reflect the uncertainty of the value. On newer browsers the following keyboard commands are supported: \| \| \| --- \| \| Key \| Action \| \| L \| Toggle log-log display if possible. \| \| S \| Toggle semilog (log Y) display if possible. \| \| K \| Move / hide legend (key). Repeated presses of this key will posiiton the legend in the following sequence: upper left, upper right, lower right, lower left, hidden. \| Click in or tab to the plot area to enable keyboard commands. The following components were used in generating the plot: jQuery jQuery UI Flot Plugins for Flot: Resize (distributed with Flot) Selection (distributed with Flot) Axis labels Additonal code used was developed at NIST: plot-data.js. Use or mention of technologies or programs in this web site is not intended to imply recommendation or endorsement by the National Institute of Standards and Technology, nor is it intended to imply that these items are necessarily the best available for the purpose. References Go To: Top, Antoine Equation Parameters, Notes Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved. Ambrose, Sprake, et al., 1975 Ambrose, D.; Sprake, C.H.S.; Townsend, R., Thermodynamic Properties of Organic Oxygen Compounds. XXXVII. Vapour Pressures of Methanol, Ethanol, Pentan-1-ol, and Octan-1-ol from the Normal Boiling Temperature to the Critical Temperature, J. Chem. Thermodyn., 1975, 7, 2, 185-190, . [all data] Ambrose and Sprake, 1970 Ambrose, D.; Sprake, C.H.S., Thermodynamic properties of organic oxygen compounds XXV. Vapour pressures and normal boiling temperatures of aliphatic alcohols, The Journal of Chemical Thermodynamics, 1970, 2, 5, 631-645, . [all data] Kretschmer and Wiebe, 1949 Kretschmer, Carl B.; Wiebe, Richard., Liquid-Vapor Equilibrium of Ethanol--Toluene Solutions, J. Am. Chem. Soc., 1949, 71, 5, 1793-1797, . [all data] Notes Go To: Top, Antoine Equation Parameters, References Data from NIST Standard Reference Database 69: NIST Chemistry WebBook The National Institute of Standards and Technology (NIST) uses its best efforts to deliver a high quality copy of the Database and to verify that the data contained therein have been selected on the basis of sound scientific judgment. However, NIST makes no warranties to that effect, and NIST shall not be liable for any damage that may result from errors or omissions in the Database. Customer support for NIST Standard Reference Data products.
12145	http://www.geom.uiuc.edu/~doty/degree.html	Network Tutorial Looking at the Degree of the Vertex and Graph Degree of Verticies To analize a graph it is important to look at the degree of a vertex. One way to find the degree is to count the number of edges which has that vertx as an endpoint. An easy way to do this is to draw a circle around the vertex and count the number of edges that cross the circle. \| \| \| --- \| \| Vertex \| Degree \| \| M \| 3 \| \| A \| 5 \| \| T \| 3 \| \| H \| 5 \| To find the degree of a graph, figure out all of the vertex degrees. The degree of the graph will be its largest vertex degree. The degree of the network is 5. Once you know the degree of the verticies we can tell if the graph is a traversable by lookin at odd and even vertecies. First lets look how you tell if a vertex is even or odd. Graphs/networks Even / Odd degree Traversible Euler Circuit Eurler Path Summary \| \| \| \| --- \| Main Page \| Glossary \| Activities \| \| Bridge Problem \| Applications \| Other Links \|
12146	https://www.conjuga.app/spanish%20conjugation/2025/04/03/spanish-verb-conjugation-verbo-Querer-conditional-simple_past/	Verbo Querer: Guía de Conjugación en Pasado Simple Condicional para Principiantes \| Conjuga \| Follow French Verbs English Verbs Spanish Verbs Apps Spanish Conjugation Verbo Querer: Guía de Conjugación en Pasado Simple Condicional para Principiantes El verbo español querer (to want) es uno de los verbos más utilizados, y lo mejor de todo es que es regular. En esta guía, aprenderás cómo conjugar querer en los tiempos más importantes para que puedas comenzar a usarlo con confianza en conversaciones cotidianas. Conjugación del “Querer” Tiempo Presente: Yo quiero – I want Tú quieres – You want (informal) Él/Ella/Usted quiere – He/She/You (formal) wants Nosotros/Nosotras queremos – We want Vosotros/Vosotras queréis – You all want (informal plural, Spain) Ellos/Ellas/Ustedes quieren – They/You all want Pretérito (Pasado Simple): Yo quise – I wanted Tú quisiste – You wanted Él/Ella/Usted quiso – He/She/You (formal) wanted Nosotros/Nosotras quisimos – We wanted Vosotros/Vosotras quisisteis – You all wanted Ellos/Ellas/Ustedes quisieron – They/You all wanted Imperfecto: Yo quería – I used to want Tú querías – You used to want Él/Ella/Usted quería – He/She/You used to want Nosotros/Nosotras queríamos – We used to want Vosotros/Vosotras queríais – You all used to want Ellos/Ellas/Ustedes querían – They/You all used to want Futuro: Yo querré – I will want Tú querrás – You will want Él/Ella/Usted querrá – He/She/You will want Nosotros/Nosotras querremos – We will want Vosotros/Vosotras querréis – You all will want Ellos/Ellas/Ustedes querrán – They/You all will want Ejemplos de uso: Presente: “Yo quiero aprender español para viajar a México.” → I want to learn Spanish to travel to Mexico. 2. Pretérito: “Ayer quisiste comer pizza para la cena.” → Yesterday, you wanted to eat pizza for dinner. 3. Imperfecto: “Cuando era niño, quería ser astronauta.” → When I was a child, I used to want to be an astronaut. 4. Futuro: “Mañana querré ir al cine.” → Tomorrow I will want to go to the cinema. Aprender a conjugar querer hará que tu español suene más natural y fluido. ¡Practica regularmente y mejora tus habilidades de comunicación! Acelera tu aprendizaje de español con nuestra [Aplicación de Con « Verb 'to saw': Past Perfect Conditional Conjugation Guide for Beginners Verbo Seguir: Guía de Conjugación en el Condicional Simple Pasado para Principiantes » ### Download on the App Store! Categories Explore all post categories. Latest Posts ###### Rendezvous: Mastering the Verb 'Se Rendre' and Its Famous French Expression Sep 18, 2025 ###### Verbo Llegar: Guía de Conjugación en Indicativo Pluscuamperfecto para Principiantes Sep 13, 2025 ###### Verbe To hinder: Guide de Conjugaison du Subjonctif Passé pour les Débutants Sep 13, 2025 Exclusive Deals Await Join Now! Sign up now for a chance to win free premium access to our app forever! Conjugation Privacy Policy Contact Download Copyright © Conjuga All rights reserved.
12147	https://fq.math.ca/Papers1/49-4/Sha.pdf	ON THE CYCLE STRUCTURE OF REPEATED EXPONENTIATION MODULO A PRIME POWER MIN SHA Abstract. We obtain some results about the repeated exponentiation modulo a prime power from the viewpoint of arithmetic dynamical systems. In particular, we extend two asymptotic formulas about periodic points and tails in the case of modulo a prime to the case of modulo a prime power. 1. Introduction For a positive integer M, denote by Z/MZ the residue ring of Z modulo M and (Z/MZ)∗ the unit group. For an integer k ≥2, we consider the following endomorphism of (Z/MZ)∗, f : (Z/MZ)∗→(Z/MZ)∗, x →xk. For any initial value x ∈(Z/MZ)∗, we repeat the action of f, then we get a sequence x0 = x, xn = xk n−1, n = 1, 2, 3, . . . . This sequence is known as the power generator of pseudorandom numbers. Studying such sequences in the cases that M is a prime or a product of two distinct primes, is of independent interest and is also important for several cryptographic applications, see [1, 6]. From the viewpoint of cryptography, there are numerous results about these sequences, see the papers mentioned in , more recently see and its references. If we view (Z/MZ)∗as a vertex set and draw a directed edge from a to b if f(a) = b, then we get a digraph. There are also many results in this direction, see and the papers mentioned there, more recently see [8, 9, 10, 11]. As in , in this article we will study (Z/MZ)∗under the action of f from the viewpoint of arithmetic dynamical systems, where M is a prime power. Speciﬁcally we will extend two asymptotic formulas in to the case of modulo a prime power. It is easy to see that for any initial value x ∈(Z/MZ)∗the corresponding sequence becomes eventually periodic, that is, for some positive integer sk,M(x) and tail tk,M(x) < sk,M(x), the elements x0 = x, x1, . . ., xsk,M(x)−1 are pairwise distinct and xsk,M(x) = xtk,M (x). So we can deﬁne a tail function tk,M on (Z/MZ)∗. The sequence xtk,M (x), . . . , xsk,M(x)−1, ordered up to a cyclic shift, is called a cycle. The cycle length is ck,M(x) = sk,M(x) −tk,M(x). The elements in the cycle are called periodic points and their periods are ck,M(x). So we can deﬁne a cycle length function ck,M on (Z/MZ)∗. In particular, [4, 5] gave lower bounds for the largest period. We denote by Pr(k, M) and P(k, M), respectively the number of periodic points with period r and the number of periodic points in (Z/MZ)∗. Also, we denote by Cr(k, M) and C(k, M), respectively the number of cycles with length r and the number of cycles in (Z/MZ)∗. We denote the average values of ck,M(x) and tk,M(x) over all x ∈(Z/MZ)∗by c(k, M) and t(k, M), 340 VOLUME 49, NUMBER 4 ON THE CYCLE STRUCTURE OF EXPONENTIATION MODULO A PRIME POWER respectively, c(k, M) = 1 ϕ(M) X x∈(Z/MZ)∗ ck,M(x), t(k, M) = 1 ϕ(M) X x∈(Z/MZ)∗ tk,M(x), where ϕ is the Euler totient function. When M is an odd prime power, we will derive explicit formulas for Pr(k, M) and Cr(k, M) by the results in , and we will also derive explicit formulas for c(k, M) and t(k, M) which generalize those in . For two integers r, m ≥1, we call lim X→∞ 1 π(X) X p≤X Pr(k, pm) the asymptotic mean number of periodic points with period r in (Z/pmZ)∗for diﬀerent choices of prime p, and we denote it by APr(k, m). Similarly, we can deﬁne the asymptotic mean number for cycles with length r and denote it by ACr(k, m). We will derive explicit formulas for APr(k, m) and ACr(k, m). For an integer m ≥1, following , we study the average values of P(k, pm) and t(k, pm) over all primes p ≤N, S0(k, m, N) = 1 π(N) X p≤N P(k, pm), S(k, m, N) = 1 π(N) X p≤N t(k, pm). where, as usual, π(N) is the number of primes p ≤N. Following the method in , we will get asymptotic formulas for S0(k, m, N) and S(k, m, N). 2. Preparations For two integers l and n, we denote their greatest common divisor by gcd(l, n). For a positive integer n, we denote by τ(n) the number of its positive divisors. Theorem 4.9 in tells us that lim X→∞ 1 π(X) X p≤X gcd(p −1, n) = τ(n). (2.1) For two integers m ≥1 and n ≥2, we denote the largest prime divisor of n by q. Then we have lim X→∞ 1 π(X) X p≤X gcd(pm−1(p −1), n) = lim X→∞ 1 π(X) X q<p≤X gcd(pm−1(p −1), n) (2.2) = lim X→∞ 1 π(X) X q<p≤X gcd(p −1, n) = τ(n). Notice that if p is an odd prime, gcd(pm −pm−1, n) is the number of solutions of the equation xn = 1 in (Z/pmZ)∗. NOVEMBER 2011 341 THE FIBONACCI QUARTERLY Given two integers a and n with gcd(a, n) = 1, following the method in the proof of formula (2) in , we can get X p≤X p≡a (mod n) pm = Xm+1 (m + 1)ϕ(n)lnX + O(Xm+1ln−2X). (2.3) Then we have X p≤X p≡a (mod n) pm−1(p −1) = Xm+1 (m + 1)ϕ(n)lnX + O(Xm+1ln−2X). (2.4) Following the same method in the proof of formula (4) in , we have X p≤X p≡a (mod n) pm−1(p −1) = O Xm+1 n + Xm . (2.5) 3. Main Results For two integers d and n satisfying gcd(d, n) = 1, we denote the multiplicative order of n modulo d by orddn. For an integer n and a prime p, we denote vp(n) the exact power of p dividing n. Let µ be the M¨ obius function. For a real number a, we denote ⌈a⌉the least integer which is not less than a. Write k = pn1 1 pn2 2 · · · pns s ≥2, where p1, . . . , ps are distinct primes, p1 < p2 < · · · < ps and n1, . . . , ns ≥1. Let m be a ﬁxed positive integer. Proposition 3.1. Let p be an odd prime and r be a positive integer. Write pm −pm−1 = pr1 1 · · · prs s · ρ, where r1, . . . , rs ≥0 are integers and gcd(p1 . . . ps, ρ) = 1. We have (1) Cr(k, pm) = 1 r P d\|r µ(d) gcd(pm −pm−1, kr/d −1). (2) Pr(k, pm) = P d\|r µ(d) gcd(pm −pm−1, kr/d −1). (3) P(k, pm) = ρ. (4) C(k, pm) = P d\|ρ ϕ(d) orddk. (5) For any x ∈(Z/pmZ)∗, denote ordpmx by ordx, ck,pm(x) = ordgcd(ordx,ρ)k. (6) c(k, pm) = 1 ρ P d\|ρ ϕ(d)orddk. (7) For any x ∈(Z/pmZ)∗, denote ordpmx by ordx, tk,pm(x) = max vp1(ordx) n1 , vp2(ordx) n2 , . . . , vps(ordx) ns . (8) t(k, pm) = 1 pr1 1 ···prs s P d\|pr1 1 ···prs s ϕ(d) max nl vp1(d) n1 m , . . . , l vps(d) ns mo . Proof. (1) and (2) By M¨ obius inversion formula and Theorem 5.6 in . (3) A special case of Corollary 3 in . (4) By Theorem 2 and Theorem 3 in . (5) By Lemma 3 and Theorem 2 in . 342 VOLUME 49, NUMBER 4 ON THE CYCLE STRUCTURE OF EXPONENTIATION MODULO A PRIME POWER (6) Denote pr1 1 · · · prs s by w, from (5), we have c(k, pm) = 1 pm −pm−1 X x∈(Z/pmZ)∗ ck,pm(x) = 1 pm −pm−1 X d\|ρ X n\|w ϕ(dn)orddk = 1 pm −pm−1 X n\|w ϕ(n) X d\|ρ ϕ(d)orddk = 1 ρ X d\|ρ ϕ(d)orddk. (7) Let wx be the factor of ordx such that ordx wx is the largest factor relatively prime to k. By Lemma 3 in , we have tk,pm(x) is the least non-negative integer l such that wx\|kl. In other words, tk,pm(x) is the least non-negative integer l such that vpi(ord x) ≤lni, for any 1 ≤i ≤s. Then we get the desired result. (8) Notice that for any x ∈(Z/pmZ)∗, ordx\|(pm −pm−1), and there are ϕ(ordx) elements with the order ordx. By (7), we have t(k, pm) = 1 pm −pm−1 X d\|(pm−pm−1) ϕ(d) max vp1(d) n1 , vp2(d) n2 , . . . , vps(d) ns . Furthermore, we have t(k, pm) = 1 pm −pm−1 X d\|pr1 1 ...prs s ρ ϕ(d) max vp1(d) n1 , . . . , vps(d) ns = 1 pm −pm−1 r1 X i1=0 · · · rs X is=0 X d\|ρ ϕ(pi1 1 · · · pis s d) max i1 n1 , . . . , is ns = 1 pm −pm−1 X d\|ρ ϕ(d) r1 X i1=0 . . . rs X is=0 ϕ(pi1 1 · · · pis s ) max i1 n1 , . . . , is ns = 1 pr1 1 · · · prs s X d\|pr1 1 ···prs s ϕ(d) max vp1(d) n1 , . . . , vps(d) ns . □ Remark 3.2. If we put k = 2 and m = 1, then the formulas (3), (4), (6), and (8) correspond to Theorem 6 in . Remark 3.3. Since the conclusions in and are about the general case of modulo a positive integer, it is easy to get similar formulas for the case of p = 2. Proposition 3.4. Let r be a positive integer, we have APr(k, m) = X d\|r µ(d)τ(kr/d −1), (3.1) ACr(k, m) = 1 r X d\|r µ(d)τ(kr/d −1). (3.2) Proof. Combining (2.2) and Proposition 3.1 (1) and (2), we can get the desired formulas. □ NOVEMBER 2011 343 THE FIBONACCI QUARTERLY In the following, we denote by Ωthe set of positive S-units with S = {p1, . . . , ps}. Here a positive S-unit means a positive integer whose prime divisors all belong to S. Proposition 3.5. We have lim N→∞ S0(k, m, N) N m = 1 m + 1 s Y i=1 p2 i p2 i −1 −1 ! . Proof. Let Q = p1p2 . . . ps and denote by UQ the set of integers u, 1 ≤u ≤Q, and gcd(u, Q) = 1. For each odd prime p, let ρp be the largest divisor of pm −pm−1 coprime to p1p2 . . . ps. It is easy to see lim N→∞ S0(k, m, N) N m = lim N→∞ 1 N mπ(N) X ps ps, then vpi(pm −pm−1) = vpi(p −1) for any 1 ≤i ≤s. Hence, following the method in Theorem 2 of , we have lim N→∞ S0(k, m, N) N m = lim N→∞ 1 N mπ(N) X q∈Ω q−1 X u∈UQ X p≤N p≡qu+1 (mod qQ) (pm −pm−1). Following the method in Theorem 2 of , we have lim N→∞ S0(k, m, N) N m = 1 m + 1 X q∈Ω 1 q2 . Moreover, we have X q∈Ω 1 q2 = ∞ X i1,...,is=0 1 (pi1 1 · · · pis s )2 −1 = ∞ X i1=0 1 p2i1 1 · · · ∞ X is=0 1 p2is s −1 = s Y i=1 p2 i p2 i −1 −1. Hence, we get the desired result. □ Corollary 3.6. We have 1 k2(m + 1) < lim N→∞ S0(k, m, N) N m < 2s −1 m + 1. Proof. Notice that for any prime p, we have 1 + p−2 < p2 p2 −1 = 1 + 1 p2 −1 < 2. □ Given q = pr1 1 · · · prs s ∈Ω, we denote ψ(q) = 1 q X d\|q ϕ(d) max vp1(d) n1 , . . . , vps(d) ns . 344 VOLUME 49, NUMBER 4 ON THE CYCLE STRUCTURE OF EXPONENTIATION MODULO A PRIME POWER Proposition 3.7. We have lim N→∞S(k, m, N) = X q∈Ω ψ(q) q . Proof. Given q = pr1 1 · · · prs s ∈Ω. Suppose r1 ≥1. We want to estimate 1 q P d\|q ϕ(d) l vp1(d) n1 m . For simplicity, we replace p1, r1, and n1 by p, r, and n, respectively. By the division algorithm, we write r = ln + d with 0 ≤d < n. We have 1 q X d\|q ϕ(d) vp(d) n = 1 pr X d\|pr ϕ(d) vp(d) n = p −1 pr r X i=1 pi−1 i n = p −1 pr   n X i=1 pi−1 + 2n X i=n+1 2pi−1 + · · · + ln X i=(l−1)n+1 lpi−1 + ln+d X i=ln+1 (l + 1)pi−1   = pn −1 pr h 1 + 2pn + · · · + lp(l−1)ni + (l + 1)pln(pd −1) pr = lpln pr − pln −1 pr(pn −1) + (l + 1)pln(pd −1) pr ≤l + (l + 1) ≤3r. Hence, we have ψ(q) ≤1 q X d\|q ϕ(d) vp1(d) n1 + · · · + vps(d) ns ≤3(r1 + · · · + rs) (3.3) ≤ 3 ln 2 ln q = O(lnq). Similarly to Proposition 3.5, by Proposition 3.1 (8), we have lim N→∞S(k, m, N) = lim N→∞ 1 π(N) X q∈Ω ψ(q) X u∈UQ X p≤N p≡qu+1 (mod qQ) 1. Then following the method in Theorem 2 of , we can get the desired result. □ Corollary 3.8. We have 1 k < lim N→∞S(k, m, N) < 5√p1 · · · √ps (√p1 −1) · · · (√ps −1). NOVEMBER 2011 345 THE FIBONACCI QUARTERLY Proof. On one hand we have X q∈Ω ψ(q) q > X i1≥n1,··· ,is≥ns ϕ(pi1 1 · · · pis s ) (pi1 1 · · · pis s )2 = (p1 −1) · · · (ps −1) p1 · · · ps ∞ X i1≥n1 1 pi1 1 · · · ∞ X is≥ns 1 pis s = 1 k. On the other hand, by (3.3) we have ψ(q) < 5 ln q, then we have X q∈Ω ψ(q) q < X q∈Ω 5 ln q q < 5 X q∈Ω 1 √q = 5 X i1=0,··· ,is=0 1 q pi1 1 · · · pis s = 5√p1 · · · √ps (√p1 −1) · · · (√ps −1). □ 4. Remarks on the General Case In this section, we will give some remarks on the case of modulo a positive integer. We can deduce formulas for Cr(k, M) and Pr(k, M) directly from Theorem 5.6 in . Corollary 3 in has given a formula for P(k, M). We can also derive a formula for C(K, M) directly by applying Theorem 2 and Theorem 3 in . Following the same methods, we can easily determine the cycle length function ck,M(x) and the tail function tk,M(x) on (Z/MZ)∗, then we can get formulas for c(k, M) and t(k, M). In fact, and can tell us more information about the properties of repeated expo-nentiation modulo a positive integer. 5. Acknowledgment We would like to thank Prof. I. E. Shparlinski for suggesting this problem and for his helpful advice. References L. Blum, M. Blum, and M. 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Nilsson, Cycles of monomials and perturbated monomial p-adic dynamical systems, Ann. Math. Blaise Pascal, 7.1 (2000), 37–63. L. Somer and M. Kˇ r´ ıˇ zek, Structure of digraphs associated with quadratic congruences with composite moduli, Discrete Math., 306 (2006), 2174–2185. L. Somer and M. Kˇ r´ ıˇ zek, On semiregular digraphs of the congruence xk ≡ y (mod n), Com-ment. Math. Univ. Carolin., 48 (2007), 41–58. L. Somer and M. Kˇ r´ ıˇ zek, On symmetric digraphs of the congruence xk ≡y (mod n), Discrete Math., 309 (2009), 1999–2009. T. Vasiga and J. Shallit, On the iteration of certain quadratic maps over GF(p), Discrete Math., 277 (2004), 219–240. B. Wilson, Power digraphs modulo n, The Fibonacci Quarterly, 36.3 (1998), 229–239. MSC2010: 37P35, 11K45, 11B50 Institut de Mathematiques de Bordeaux, Universite Bordeaux 1, 33405 Talence Cedex, France E-mail address: shamin2010@gmail.com NOVEMBER 2011 347
12148	https://climate.nenu.edu.cn/content/context2/4_1_1.htm	\| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| == \| \| 首页 \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| 课程计划 \| \| 课程内容 \| \| 学习资源 \| \| 学习交流 \| \| \| \| \| \| 课程简介 \| \| 学习目标 \| \| \| \| 引论 \| \| 气候系统的辐射过程与能量平衡 \| \| 气候系统的水分运动 \| \| 气候系统的大气运动 \| \| 气候系统的天气系统统 \| \| 气候的形成过程 \| \| 全球气候带和气候型分布的地理规律 \| \| 气候变化和人类活动对气候的影响 \| \| \| \| 文献资料库 \| \| 视频库 \| \| 新书阅览室 \| \| 云　图　库 \| \| 天气图标 \| \| \| \| 文稿上传 \| \| 留言板 \| \| 联系我们 \| \| 实践导航 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| \| \| \| 第四节大气温度随时间的变化 \| \| \| 一、气温的周期性变化 \| \| \| 地表从太阳辐射得到大量热量，同时又以长波辐射、显热和潜热的形式将部分热量传输给大气，从而失去热量。从长时间平均看，热量得失总和应该平衡，因此地面的平均温度维持不变。但在某一段时间内，可能得多于失，地面有热量累积而升温，从而导致支出增加，趋于新的平衡。反之，当失多于得时，地面将伴随着降温过程。由于在这种热量收支平衡过程中，太阳辐射处于主导地位，因此随着日夜、冬夏的交替，地面的温度也会相应地出现日变化和年变化，且变化的幅度与纬度、天气及地表性质等影响热量平衡的控制因子有关。此外地面温度的变化也会通过非绝热因子传递给大气，大气温度也会相应出现变化。一、气温的周期性变化（一）气温的日变化大气边界层的温度主要受地表面增热与冷却作用的影响而发生变化。例如白天当地表面吸收了太阳辐射能而逐渐增热，通过辐射、分子运动、湍流及对流运动和潜热输送等方式将热量传递给边界层大气，使大气温度随之升高；夜间地表面因放射长波辐射而冷却，使边界层大气温度也随之降低。因而引起边界层大气温度的日变化。而地表面对大气边界层温度的影响是与地表面的性质（森林、草原、沙漠、不同类型的土壤等）有关的。广阔洋面上的冷暖洋流也影响洋面上空的大气。此外，大气中的水平运动与垂直运动都会引起局地气温的变化。例如暖平流移来时，会使局地上空的气温升高。冷平流移来时则会使局地上空的气温下降。大气中的垂直运动使得垂直方向上热量分布趋于一致。当地表面受热时，垂直交换作用使地表面增热现象减弱。当地表面冷却时，交换作用使降温现象减小。近地层气温日变化的特征是：在一日内有一个最高值，一般出现在午后14时左右，一个最低值，一般出现在日出前后（图2•30）。一天中气温的最高值与最低值之差，称为气温日较差，其大小反映气温日变化的程度。一天中正午太阳辐射最强，但最高气温却出现在午后两点钟左右。这是因为大气的热量主要来源于地面。地面一方面吸收太阳的短波辐射而得热，一方面又向大气输送热量而失热。若净得热量，则温度升高。若净失热量，则温度降低。这就是说地温的高低并不直接决定于地面当时吸收太阳辐射的多少，而决定于地面储存热量的多少。从图2•30中看出，早晨日出以后随着太阳辐射的增强，地面净得热量，温度升高。此时地面放出的热量随着温度升高而增强，大气吸收了地面放出的热量，气温也跟着上升。到了正午太阳辐射达到最强。正午以后，地面太阳辐射强度虽然开始减弱，但得到的热量比失去的热量还是多些，地面储存的热量仍在增加，所以地温继续升高，长波辐射继续加强，气温也随着不断升高。到午后一定时间，地面得到的热量因太阳辐射的进一步减弱而少于失去的热量，这时地温开始下降。地温的最高值就出现在地面热量由储存转为损失，地温由上升转为下降的时刻。这个时刻通常在午后13时左右。由于地面的热量传递给空气需要一定的时间，所以最高气温出现在午后14时左右。随后气温便逐渐下降，一直下降到清晨日出之前地面储存的热量减至最少为止。所以最低气温出现在清晨日出前后，而不是在半夜。气温日变化的另一特征是日较差的大小与纬度、季节和其它自然地理条件有关。日较差最大的地区在副热带，向两极减小。热带地区的平均日较差约为12℃，温带约为8—9℃，极圈内为3—4℃。日较差夏季大于冬季，但最大值并不出现在夏至日。这是因为气温日较差不仅与白天的最高温度值有关，还取决于夜间的最低温度值。夏至日，中午太阳高度角虽最高，但夜间持续时间短，地表面来不及剧烈降温而冷却，最低温度不够低。所以，中纬度地区日较差最大值出现在初夏，最小值出现在冬季。海洋上日较差小于大陆。盆地和谷地由于坡度及空气很少流动之故，白天增热与夜间冷却都较大，日较差大。而小山峰等凸出地形区，地表面对气温影响不大，日较差小。气温日较差还与地面的特性和天气情况等有关。例如沙漠地区日较差很大。潮湿地区日较差较小。就天气情况来说，如果有云层存在，则白天地面得到的太阳辐射少，最高气温比晴天低。而在夜间，云层覆盖又不易使地面热量散失，最低气温反而比晴天高。所以阴天的气温日较差比晴天小（图2•31）。由此可见，在任何地点，每一天的气温日变化，既有一定的规律性，又不是前一天气温日变化的简单重复，而是要考虑上述诸因素的综合影响。气温日变化的极值出现时间随离地面的高度增大而后延，振幅随离地高度的增大而减小。冬季约在0.5km高度处日振动已不明显，但夏季日振动可扩展到1.5km到2km高度处。（二）气温的年变化气温的年变化和日变化在某些方面有着共同的特点，如地球上绝大部分地区，在一年中月平均气温有一个最高值和一个最低值。由于地面储存热量的原因，使气温最高和最低值出现的时间，不是在太阳辐射最强和最弱的一天（北半球夏至和冬至），也不是在太阳辐射最强和最弱一天所在的月份（北半球6月和12月），而是比这一时段要落后1—2个月。大体而论，海洋上落后较多，陆地上落后较少。沿海落后较多，内陆落后较少。就北半球来说，中、高纬度内陆的气温以7月为最高，1月为最低。海洋上的气温以8月为最高，2月为最低。一年中月平均气温的最高值与最低值之差，称为气温年较差。气温年较差的大小与纬度、海陆分布等因素有关。赤道附近，昼夜长短几乎相等，最热月和最冷月热量收支相差不大，气温年较差很小；愈到高纬度地区，冬夏区分明显，气温的年较差就很大。例如我国的西沙群岛（16°50＇N）气温年较差只有6℃，上海（31°N）为25℃，海拉尔（49°13＇N）达到46.7℃。图2•32给出了不同纬度气温年变化的情况。低纬度地区气温年较差很小，高纬度地区气温年较差可达40—50℃。如以同一纬度的海陆相比，大陆区域冬夏两季热量收支的差值比海洋大，所以陆上气温年较差比海洋大得多。在一般情况下，温带海洋上年较差为11℃，大陆上年较差可达到20—60℃。根据温度年较差的大小及最高、最低值出现的时间，可将气温的年变化按纬度分为四种类型。 1.赤道型它的特征是一年中有两个最高值，分别出现在春分和秋分以后，因赤道地区春秋分时中午太阳位于天顶。两个最低值出现在冬至与夏至以后，此时中午太阳高度角是一年中的最小值。这里的年较差很小，在海洋上只有1℃左右，大陆上也只有5—10℃左右。这是因为该地区一年内太阳辐射能的收入量变化很小之故。 2.热带型其特征是一年中有一个最高（在夏至以后）和一个最低（在冬至以后），年较差不大（但大于赤道型），海洋上一般为5℃，在陆地上约为20℃左右。 3.温带型一年中也有一个最高值，出现在夏至后的7月。一个最低值出现在冬至以后的1月。其年较差较大，并且随纬度的增加而增大。海洋上年较差为10—15℃，内陆一般达40—50℃，最大可达60℃。另外，海洋上极值出现的时间比大陆延后，最高值出现在8月，最低值出现在2月。 4.极地型一年中也是一次最高值和一次最低值，冬季长而冷，夏季短而暖，年较差很大是其特征。这里特别要指出的是，随着纬度的增高，气温日较差减小而年较差却增大。这主要是由于高纬度地区，太阳辐射强度的日变化比低纬度地区小，即纬度高的地区，在一天内太阳高度角的变化比纬度低的地区小，而太阳辐射的年变化在高纬地区比低纬地区大的缘故。 [设本页为书签] \| \| \| \| \| \| \| \| \| \| \| \| \| \|
12149	https://nrich.maths.org/problems/which-scripts	Which Scripts? \| NRICH Skip to main content Problem-Solving Schools can now access the Hub! Contact us if you haven't received login details Main navigation Teachersexpand_more Early years Primary Secondary Post-16 Professional development Studentsexpand_more Primary Secondary Post-16 Parentsexpand_more Early years Primary Secondary Post-16 Problem-Solving Schoolsexpand_more What is the Problem-Solving Schools initiative? Becoming a Problem-Solving School Charter Hub Resources and PD Events About NRICHexpand_more About us Impact stories Support us Our funders Contact us search menu search close Search NRICH search Or search by topic Number and algebra Properties of numbers Place value and the number system Calculations and numerical methods Fractions, decimals, percentages, ratio and proportion Patterns, sequences and structure Coordinates, functions and graphs Algebraic expressions, equations and formulae Geometry and measure Measuring and calculating with units Angles, polygons, and geometrical proof 3D geometry, shape and space Transformations and constructions Pythagoras and trigonometry Vectors and matrices Probability and statistics Handling, processing and representing data Probability Working mathematically Thinking mathematically Mathematical mindsets Advanced mathematics Calculus Decision mathematics and combinatorics Advanced probability and statistics Mechanics For younger learners Early years foundation stage Which scripts? There are six numbers written in five different scripts. Can you sort out which is which? Age 7 to 11 Challenge level Primary Number - Number and Place Value Understanding Place Value Exploring and noticingWorking systematicallyConjecturing and generalisingVisualising and representingReasoning, convincing and proving Being curiousBeing resourcefulBeing resilientBeing collaborative Problem Getting Started Student Solutions Teachers' Resources Problem Which Scripts? printable sheet There are six numbers written in five different scripts. Can you sort out which is which? Write $51$ in each script. Image Here is an interactive version for you to try out your ideas. [Thank you to the SMILE Centre for permission to use this puzzle.] Getting Started You might find it helpful to print outthis sheetof the numbers. Which numbers do you know? Can you see any similarities between any of the numbers? Which numbers are the 'shortest' and the 'longest'? Student Solutions Well done to everybody who worked out which numbers belong to which scripts! Poppy from Acomb First School in the UK shared this strategy with us: First of all I found the numbers that I recognised. Then I found the numbers that looked a little bit like the numbers that I recognised and looked for one-digit numbers and three-digit numbers because I knew that one-digit numbers would be 2 and that three-digit numbers would be 100. Then I grouped the numbers into 5 groups of similar scripts. To work out the trickier numbers, I was able to work out the 1 from the 100 and the 2 from the 2. Then I could work out the 13 and the 25. Once I knew the 13 and 25, I could find the 58 from the 5 and then the 83 from the 3 and the 8. Good ideas, Poppy - lots of children found it easiest to start by looking for the digit 2. James from Co-op Academy New Islington, Manchester in the UK started by sorting the numbers into the scripts based on which numbers looked similar: I cut the number grid into squares so I could move them around. I found out what the six numbers in English were first because I know what they look like. Then, I started to look for ones that looked similar. I put them in rows and columns for the the same number and the same scripts to be organised. Next, I looked for numbers in other scripts that looked very similar and matched them up with the correct number. Because I knew 2 it was easier to work the rest of them out because after you've found 2 you can work out 25. Once you've worked out 25 you can work out 58. Once you've worked out 58 you can work out 83. After you've worked out 83 you can work out 13. For the 100s you can use the 1 from 13 and there are three digits or two zeroes or two dots on some of them. This is a good step-by-step method, James! The numbers we use in English are also the same numbers that are used in many other languages. I wonder if anyone knows the name for these numbers? We received a lot of solutions from the children at St. Helen's School in Abbotsham, England. Amelia-May and Frances explained their strategy, which was similar to James's: First we looked for all the 2s (which were kind of easy), then found the 25s because we knew what the 2 would look like. After that we got the 58s because we knew what the 5 looked like. Then we found that it had an 8 like the 83 so we moved on to that looking for the 83s. We realised that the 83 linked to 13 because of the 3. So we found all of the 13s. Last but not least we did the 100! Image Thank you as well to Edgar, Will, Amber, Grace, Kacie, Fraser, Oliver,Myles, Albie, Charlie and Lucy from St. Helen's School who also sent in some similar solutions. Gabe and Muhammad from Wembrook Primary also used similar reasoning, and explained their thinking very clearly: Image Image Dhruv from The Glasgow Academy in the UK used their prior knowledge of two of the scripts to solve this problem: First, I separated these numbers into different groups based on their writing pattern. Secondly, I knew numbers that were in English and Hindi because I am from India. Thirdly, I arranged the numbers in ascending order. Finally, for the first row in Chinese I took a guess for the first number and then linked it to find the other numbers. Example: For the Chinese group I took a guess that the two lines were the number 2 and then found the same two lines in another number and so on. Image Then I arranged the numbers in ascending order. Image I followed the same strategy for the two scripts which were unknown to me. Well explained, Dhruv - this looks similar to some of the other strategies above, but the pictures make it really clear how you got from one solution to the next. I wonder if anybody has worked out what the other scripts might be? Thank you as well to the following children for sending in their ideas about which numbers belonged to which scripts: Zoe from Canada;Chloe, Meriam,Sophia, Milan, Josh,Thomas,Hogan,Henry and Harry from Banstead Prep; Blossom,Gabe, Kinel and Lilah from Onchan Primary School on the Isle of Man; Freddie; and the children at Ganit Kreeda in Vicharvatika,India. The second part of this problem involved writing the number 51 in each script. Only the next four groups of children sent in a correct solution to this, as lots of children made a mistake with writing 51 in the Chinese script. Isobelle, Edie and James from Richmond Methodist School in the UK explained: We took the 5 and the 1 from each script to make 51 in the different scripts. I wonder if you had actually had to use a slightly different strategy with the Chinese script? Sophie from Glenfall in the UK explained: The trickiest was the Chinese style numbers because they actually used a 'plus' symbol which wasn't a digit. Elliott from Richmond Methodist School in the UK had an idea about what the plus symbol might mean: The second step is to write 51 in all of the scripts. You just need to take the 5 and 1 from all the scripts except script E (the Chinese script). I saw that a + equals x10. If one of the symbols has lines going up, one line is 1, two are 2 and 3 are 3. So adding a plus after a symbol multiplies the previous symbol by 10. Good ideas! I wonder how this works with the number 13 as there isn't a digit before the plus symbol? Junior Maths Club at Caulfield Grammar School in Wheelers Hill, Australia sent in these full solutions: Image Thank you all for sharing your ideas with us. Teachers' Resources Which Scripts? Which Scripts? printable sheet There are six numbers written in five different scripts. Can you sort out which is which? Write $51$ in each script. Image [Thank you to the SMILE Centre for permission to use this puzzle.] Why do this problem? This problem consolidates understanding of place value in a demanding but intriguing context. In order to tackle the problem, learners will have to organise and sort the information given. We hope they are curious enough to keep going, even when it gets tricky! Possible approach Show the image of the numbers to the group and ask them to talk to a partner about what they notice. Gather some suggestions and explain what the image shows, if this has not already come up in discussion. Invite learners to suggest ways of beginning the problem and then set them off in pairs to work together, usingthis copyof the image and providing squared paper. As they work, encourage them to develop a good way to record their findings. In the plenary, it might be helpful for you to enlargethis sheetand cut out the numbers so they can be moved around on the board. (If these were laminated, they would make a useful set of cards to be used again.) You could invite pairs of children to explain how they reached their conclusions and recorded the results. This could lead into a discussion of the place value system (compared with, for example, Roman numerals). Key questions Which numbers do you know? Can you see any similarities between any of the numbers? Which numbers are the 'shortest' and the 'longest'? Possible extension Learners could write hints which might help others work on the task without giving away the solution. You could also encourage children to find out the name of each script. Another idea would be to include Roman Numeral versions of the numbers: XIII, II, LVIII, XXV, LXXXIII, C.This sheet includes six cards which could be printed off to accompany the original numbers. Alternatively, challenge children to create calculations and their answers in one of the scripts. Possible support Some children may find it useful to cut out the individual numbers so they can be sorted more easily. Below you can see some pictures of children at Lancasterian Primary School in Haringey working on this task: Image Image Image Footer Sign up to our newsletter Technical help Accessibility statement Contact us Terms and conditions Links to the NRICH Twitter account Links to the NRICH Facebook account Links to the NRICH Bluesky account NRICH is part of the family of activities in the Millennium Mathematics Project.
12150	https://unacademy.com/content/nda/study-material/mathematics/probability-definition-of-probability-classical-and-statistical/	Access free live classes and tests on the app Login Join for Free Profile Settings Refer your friends Sign out Terms & conditions • Privacy policy About • Careers • Blog © 2023 Sorting Hat Technologies Pvt Ltd NDA » NDA Study Material » Mathematics » Probability-Definition of Probability Classical and Statistical Probability-Definition of Probability Classical and Statistical Share Probability is the branch of mathematics that deals with the possibility of any outcome of an event or experiment. There are different types of probabilities. Classical probability states the possible outcome of any event in a classic manner. On the other hand, statistical probability involves the laws governing random events and their data collection, analysis, interpretation, and display. In other words, we can say that in a classical probability, the possible outcomes are at equal odds. For example, when we roll a dice, there are 50-50% chances of getting an even and odd number. Similarly, when we toss a coin, there are equal chances of getting either a head or a tail. Therefore, we can conclude that classical probability is the simplest and easiest to understand probability type. Classical Probability As stated above, the classical probability is the probability of any event in a classic manner, i.e. equal happening of events. When we roll a dice, the probability of each number is equal to 1/6 of the equal. This is an example of a classical probability. The Formula of Classical Probability The formula of classical probability is as follows: P(A)= f/N Where P(A)= classical probability f= Frequency or the number of favourable outcomes N= Number of total possible outcomes. Examples of Classical Probability in Daily Life Example 1 Suppose you have a multiple-choice format exam of mathematics tomorrow. There will be four options: A, B, C, and D. You know that any of these options can be correct. Thus, by the formula of probability, we can say that the probability of getting a correct answer in each case is ¼, i.e., 25% for each option. Example 2 We usually toss a coin while deciding the teams for batting or bowling. The coin has two sides, i.e., a head and a tail. Thus, there are equal chances of getting a head and a tail. Therefore, it is an example of classical probability with 50% chance of each outcome. Statistical Probability In statistics, we collect the data in a specific form and represent it in an order to get the result. Similarly, if we talk about statistics in probability, it deals with governing events, collecting data of events, and also its representation in a specific manner for better understanding. Let’s take an example of a coin. If we toss a coin four times, the outcomes will vary. It can either be 50-50 or all head or all tails or maybe 3 heads-1 tails and vice versa. But by tossing the same coin 400 times, we shall get the heads and tails in approx. equal ratios. We can collect the data and analyse it to calculate the probable outcome of it. How to Calculate Statistical Probability? We can calculate statistical probability just like other probability questions. We need the number of favourable outcomes and total outcomes for the calculation. By dividing the number of favourable outcomes by the number of total outcomes, we can get the statistical probability of that event. The statistical probability will also involve representing it in a certain way (like a frequency table or a graph) for better analysis. Probability(Event) = Favorable Outcomes/Total Outcomes = x/n Other Types of Probability There are three other types of probabilities. These are as follows: Empirical Probability: Empirical probability is the type of experimental probability that evaluates outcomes based on conducting experiments. For instance, if you roll a weighted dice without knowing the side having weight, you’ll get the idea of the probability of each time (outcome) by rolling that dice a number of times and determining the proportion of times the dice gives that desired outcome. That outcome will then be the probability. Subjective Probability: Subjective probability deals with one’s own belief of the happening or not happening of a certain event. For example, while watching a cricket match, you believe that the probability of winning your favourite team is the highest. However, the fans of the other team might think the opposite. Therefore, subjective probability is completely based on a person’s belief. Axiomatic Probability: While calculating axiomatic probability, we must follow certain rules or axioms specified by Kolmogorov. By these rules, we determine whether the event will happen or not. These three rules are as follows: The first point states that the least possibility or probability of happening an event is 0. Similarly, the highest probability is 1. Every certain event (an event that must occur) has the probability 1. Two mutually exclusive events will never occur simultaneously. However, we can say that only one of them will happen. For example, any place will either have a hot or cold climate at a time (not both). Solved Questions on Classical and Statistical Probabilities Nina wants a head to win the toss. What is the possibility of winning for Nina? Solution: Number of total possible outcomes= 2 Number of favourable outcomes= 1 Tossing a coin is an example of classical probability. Therefore, we can use the formula of classical probability: P(A)= f/N = ½ Aman needs an even number to escape the snake while playing the game of snakes and ladders. What is the probability of winning for Aman? Solution: Number of total possible outcomes= 6 Number of favourable outcomes= 3 P(A)= f/N = 3/6 =1/2 Conclusion Classical probability states the possible outcome of any event in a classic manner, whereas statistical probability is the statistical representation of any random even. In classical probability, all the outcomes have equal odds of happening. For example, rolling a dice or tossing a coin. The formula of classical probability is as follows: P(A)= f/N; where, P(A)= classical probability, f= frequency or the number of favourable outcomes and N= Number of total possible outcomes. There are three more types of probabilities: empirical, subjective and axiomatic probabilities. Frequently asked questions Get answers to the most common queries related to the NDA Examination Preparation. What is the relation between probability and statistics? Ans. Probability and statistics are both branches of mathematics that deal with the outcomes of any event and...Read full What is the formula of classical probability? Ans. The formula of classical probability is as follows: P (A)= ...Read full State an example of classical probability? Ans. When you have a multiple-choice form of exam, there are four options: A, B, C, and D. You know that any of t...Read full What is the difference between subjective and empirical probability? Ans. Empirical probability is the type of experimental probability that evaluates outcomes on the basis of co...Read full Ans. Probability and statistics are both branches of mathematics that deal with the outcomes of any event and its representation. Statistics in probability deals with governing events, collecting data of events and also its representation in a specific manner for better understanding. Ans. The formula of classical probability is as follows: P (A)= f/N Ans. When you have a multiple-choice form of exam, there are four options: A, B, C, and D. You know that any of these options can be correct. Thus, by the formula of probability, we can say that the probability of getting a correct answer in each case is ¼, that is, 25% for each option. Ans. Empirical probability is the type of experimental probability that evaluates outcomes on the basis of conducting experiments. For instance, if you roll a weighted dice without knowing the side having weight, you can get the idea of the probability of each time (outcome) by rolling that dice a number of times and determining the proportion of times the dice gives that desired outcome. That outcome will be the probability. On the other hand, subjective probability deals with one’s own belief of the happening or not happening of a certain event. For example, while watching a cricket match, you believe that the probability of winning your favourite team is the highest. Share via
12151	https://webster-wotd.livejournal.com/1782787.html	contentious Main Top Interesting Checklist ☕🧦🍂 Help webster_wotd Follow us: Follow us on Twitter Applications Download Huawei RuStore COMPANY About News Help PRODUCTS "Share" button COMMUNITY Frank CHOOSE LANGUAGE English ▾ English Deutsch Dansk español Français Italiano Русский Українська Беларуская 日本語 Português Esperanto עברית Nederlands Magyar Gaeilge íslenska suomi Ελληνικά Norsk bokmål Svenska polski 简体中文 Latviešu Türkçe Bahasa Melayu हिन्दी Brazilian Portuguese Chinese Traditional Lietuvių Norsk nynorsk Privacy Policy User Agreement Recommendation technologies Help LiveJournal — v.880 Main Top Interesting Checklist ☕🧦🍂 Help Follow us: Follow us on Twitter Applications Download Huawei RuStore COMPANY About News Help PRODUCTS "Share" button COMMUNITY Frank CHOOSE LANGUAGE English ▾ English Deutsch Dansk español Français Italiano Русский Українська Беларуская 日本語 Português Esperanto עברית Nederlands Magyar Gaeilge íslenska suomi Ελληνικά Norsk bokmål Svenska polski 简体中文 Latviešu Türkçe Bahasa Melayu हिन्दी Brazilian Portuguese Chinese Traditional Lietuvių Norsk nynorsk Privacy Policy User Agreement Recommendation technologies Help LiveJournal — v.880 LiveJournal Find more Communities RSS Reader Shop Help Search Log in webster_wotd Log in Join free Join English (en) English (en) Русский (ru) Українська (uk) Français (fr) Português (pt) español (es) Deutsch (de) Italiano (it) Беларуская (be) Log in No account?Create an account [x] Remember me Forgot password Log in Log in QR code No account?Create an account By logging in to LiveJournal using a third-party service you accept LiveJournal's User agreement webster_wotd Subscribe August 5 2025, 01:00 0 00 0 0 0 0 00 0 contentious Merriam-Webster's Word of the Day for August 5, 2025 is: contentious• \kun-TEN-shuss\ •adjective Contentious describes something that is likely to cause people to argue or disagree or that involves a lot of arguing. When used of a person,contentious describes someone likely or willing to argue. // I think it’s wise to avoid such a contentious topic at a dinner party. // After a contentious debate, members of the committee finally voted to approve the funding. // The dispute involves one of the region’s most contentious leaders. See the entry > Examples: “Next up will be Peter Shaffer’s ‘Amadeus,’ which opened in 1979 and won the Tony for best play in 1981 with Ian McKellen winning lead actor honors. ... The story is a fictional account of the contentious relationship between Wolfgang Amadeus Mozart and his rival, Antonio Salieri, the court composer of the Austrian emperor.” — Jessica Gelt,The Los Angeles Times, 12 June 2025 Did you know? If everyone has abone to picknow and then, contentious types have entire skeletons. While English has plenty of words for people prone to fighting—combativeandbelligerentamong them—contentious implies a fondness for arguing that others find particularly tedious or wearying. Thankfully, even the most contentious cranks and crabs among us have no cause to quibble over the history of the word contentious, as its origins are very clear:contentious comes (by way of Middle French) from the Latin adjective contentiōsus, meaning “persistent, obstinate, argumentative, or quarrelsome.” 0 0Subscribe0 0 0 0 0 00 0 LJ Video No comments yet Post a new comment )_ Your IP address will be recorded For a line break, press Shift + Enter Press Enter to send comment Applications Download Huawei RuStore Follow us: Follow us on Twitter LiveJournal COMPANY About News Help PRODUCTS "Share" button COMMUNITY Frank CHOOSE LANGUAGE English ▾ English Deutsch Dansk español Français Italiano Русский Українська Беларуская 日本語 Português Esperanto עברית Nederlands Magyar Gaeilge íslenska suomi Ελληνικά Norsk bokmål Svenska polski 简体中文 Latviešu Türkçe Bahasa Melayu हिन्दी Brazilian Portuguese Chinese Traditional Lietuvių Norsk nynorsk v.880 Privacy Policy User Agreement Recommendation technologies Help
12152	https://www.dummies.com/article/academics-the-arts/math/geometry/the-properties-of-trapezoids-and-isosceles-trapezoids-188119/	Home Academics & The Arts Articles Math Articles Geometry Articles The Properties of Trapezoids and Isosceles Trapezoids By Mark Ryan Updated 2021-07-09 13:59:53 From the book Geometry For Dummies Share Geometry For Dummies Explore Book Buy Now Subscribe on Perlego Geometry For Dummies Explore Book Buy Now Subscribe on Perlego A trapezoid is a quadrilateral (a shape with four sides) with exactly one pair of parallel sides (the parallel sides are called bases). The following figure shows a trapezoid to the left, and an isosceles trapezoid on the right. The properties of the trapezoid are as follows: The bases are parallel by definition. Each lower base angle is supplementary to the upper base angle on the same side. The properties of the isosceles trapezoid are as follows: The properties of a trapezoid apply by definition (parallel bases). The legs are congruent by definition. The lower base angles are congruent. The upper base angles are congruent. Any lower base angle is supplementary to any upper base angle. The diagonals are congruent. The supplementary angles might be the hardest property to spot in the diagrams above. Because of the parallel sides, consecutive angles are same-side interior angles and are thus supplementary. (All the special quadrilaterals except the kite, by the way, contain consecutive supplementary angles.) Here’s an isosceles trapezoid proof for you: Statement 1: Reason for statement 1: Given. Statement 2: Reason for statement 2: The legs of an isosceles trapezoid are congruent. Statement 3: Reason for statement 3: The upper base angles of an isosceles trapezoid are congruent. Statement 4: Reason for statement 4: Reflexive Property. Statement 5: Reason for statement 5: Side-Angle-Side, or SAS (2, 3, 4) Statement 6: Reason for statement 6: CPCTC (Corresponding Parts of Congruent Triangles are Congruent). Statement 7: Reason for statement 7:If angles are congruent, then so are sides. About This Article This article is from the book: Geometry For Dummies About the book author: Mark Ryan has more than three decades’ experience as a calculus teacher and tutor. He has a gift for mathematics and a gift for explaining it in plain English. He tutors students in all junior high and high school math courses as well as math test prep, and he’s the founder of The Math Center on Chicago’s North Shore. Ryan is the author of Calculus For Dummies, Calculus Essentials For Dummies, Geometry For Dummies, and several other math books. This article can be found in the category: Geometry
12153	https://brightchamps.com/en-us/math/calculus/derivative-of-power-of-x	Table Of Contents What is the Derivative of Power of x? Derivative of Power of x Formula Proofs of the Derivative of Power of x Higher-Order Derivatives of Power of x Special Cases: Common Mistakes and How to Avoid Them in Derivatives of Power of x Examples Using the Derivative of Power of x FAQs on the Derivative of Power of x Important Glossaries for the Derivative of Power of x Explore More calculus Table Of Contents Table Of Contents Summarize this article: ChatGPT Perplexity Last updated on August 5, 2025 Derivative of Power of x We use the derivative of x^n, which is nx^(n-1), as a fundamental tool to understand how polynomial functions change in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of x^n in detail. What is the Derivative of Power of x? We now understand the derivative of x^n. It is commonly represented as d/dx (x^n) or (x^n)', and its value is nx^(n-1). The function x^n has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Power Rule: A rule for differentiating functions of the form x^n. Polynomial Functions: Functions like x^n, where n is a non-negative integer. Derivative of Power of x Formula The derivative of x^n can be denoted as d/dx (x^n) or (x^n)'. The formula we use to differentiate x^n is: d/dx (x^n) = nx^(n-1) This formula applies to all x and for any real number n. Proofs of the Derivative of Power of x We can derive the derivative of x^n using proofs. To show this, we will use the properties of limits along with the rules of differentiation. There are several methods we use to prove this, such as: We will now demonstrate that the differentiation of x^n results in nx^(n-1) using the above-mentioned methods: By First Principle The derivative of x^n can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of x^n using the first principle, we will consider f(x) = x^n. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = x^n, we write f(x + h) = (x + h)^n. Substituting these into equation (1), f'(x) = limₕ→₀ [(x + h)^n - x^n] / h Using the Binomial Theorem, expand (x + h)^n: = limₕ→₀ [x^n + nx^(n-1)h + ... + h^n - x^n] / h = limₕ→₀ [nx^(n-1)h + ... + h^n] / h = limₕ→₀ [nx^(n-1) + ... + h^(n-1)] Using limit properties, all terms containing h vanish as h approaches 0, leaving: f'(x) = nx^(n-1) Hence, proved. Using Binomial Theorem To prove the differentiation of x^n using the binomial theorem, We use the expansion: (x + h)^n = x^n + nx^(n-1)h + ... + h^n By focusing on the terms that involve h, when we divide by h and take the limit as h approaches 0, we find that only the first-order term, nx^(n-1), remains, yielding the derivative. Higher-Order Derivatives of Power of x When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like x^n. For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues. For the nth Derivative of x^n, we generally use f^(n)(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives). Special Cases: When n = 0, the derivative is undefined because x^0 = 1, which is a constant with a derivative of 0. When n = 1, the derivative of x^1 = 1x^(1-1), which is 1. Common Mistakes and How to Avoid Them in Derivatives of Power of x Students frequently make mistakes when differentiating x^n. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them: Students frequently make mistakes when differentiating x^n. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them: Mistake 1 Not applying the power rule correctly Not applying the power rule correctly Students may forget to apply the power rule correctly, which can lead to incorrect results. They often skip the step of multiplying by the exponent and directly arrive at the result. Ensure that each step follows the rule: d/dx (x^n) = nx^(n-1). Students may forget to apply the power rule correctly, which can lead to incorrect results. They often skip the step of multiplying by the exponent and directly arrive at the result. Ensure that each step follows the rule: d/dx (x^n) = nx^(n-1). Mistake 2 Forgetting to adjust the exponent Forgetting to adjust the exponent They might not remember to subtract 1 from the exponent after differentiating. Keep in mind that you should always reduce the exponent by 1 when using the power rule. They might not remember to subtract 1 from the exponent after differentiating. Keep in mind that you should always reduce the exponent by 1 when using the power rule. Mistake 3 Misapplying the power rule to non-polynomial functions Misapplying the power rule to non-polynomial functions While differentiating functions that are not of the form x^n, students misapply the power rule. For example: Incorrectly trying to apply the power rule to e^x or ln(x). Recognize the types of functions where the power rule is applicable and ensure its proper usage. While differentiating functions that are not of the form x^n, students misapply the power rule. For example: Incorrectly trying to apply the power rule to e^x or ln(x). Recognize the types of functions where the power rule is applicable and ensure its proper usage. Mistake 4 Not simplifying expressions Not simplifying expressions There is a common mistake where students do not simplify the resulting expression after differentiation. For example, they might leave expressions in a complicated form instead of simplifying them. Always simplify your result to its simplest form. There is a common mistake where students do not simplify the resulting expression after differentiation. For example, they might leave expressions in a complicated form instead of simplifying them. Always simplify your result to its simplest form. Mistake 5 Ignoring constants Ignoring constants Students often forget to multiply constants with the derivative. For example, they incorrectly write d/dx (5x^3) = 3x^2. Always remember to multiply the constant by the derivative of the power: d/dx (5x^3) = 15x^2. Students often forget to multiply constants with the derivative. For example, they incorrectly write d/dx (5x^3) = 3x^2. Always remember to multiply the constant by the derivative of the power: d/dx (5x^3) = 15x^2. Examples Using the Derivative of Power of x Problem 1 Calculate the derivative of x^3 e^x Here, we have f(x) = x^3 e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x^3 and v = e^x. Let’s differentiate each term, u′ = d/dx (x^3) = 3x^2 v′ = d/dx (e^x) = e^x Substituting into the given equation, f'(x) = (3x^2) (e^x) + (x^3) (e^x) Let’s simplify terms to get the final answer, f'(x) = 3x^2 e^x + x^3 e^x f'(x) = e^x (3x^2 + x^3) Thus, the derivative of the specified function is e^x (3x^2 + x^3). Here, we have f(x) = x^3 e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x^3 and v = e^x. Let’s differentiate each term, u′ = d/dx (x^3) = 3x^2 v′ = d/dx (e^x) = e^x Substituting into the given equation, f'(x) = (3x^2) (e^x) + (x^3) (e^x) Let’s simplify terms to get the final answer, f'(x) = 3x^2 e^x + x^3 e^x f'(x) = e^x (3x^2 + x^3) Thus, the derivative of the specified function is e^x (3x^2 + x^3). Explanation We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result. We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result. Problem 2 A new rocket is being launched, and its height is given by h(x) = x^3 + 5x^2 meters, where x is time in seconds. Find the rate of change of height when x = 2 seconds. We have h(x) = x^3 + 5x^2 (height of the rocket)...(1) Now, we will differentiate the equation (1) Take the derivative: h'(x) = 3x^2 + 10x Given x = 2 (substitute this into the derivative) h'(2) = 3(2)^2 + 10(2) = 34 + 20 = 12 + 20 = 32 Hence, the rate of change of height when x = 2 seconds is 32 meters/second. We have h(x) = x^3 + 5x^2 (height of the rocket)...(1) Now, we will differentiate the equation (1) Take the derivative: h'(x) = 3x^2 + 10x Given x = 2 (substitute this into the derivative) h'(2) = 3(2)^2 + 10(2) = 34 + 20 = 12 + 20 = 32 Hence, the rate of change of height when x = 2 seconds is 32 meters/second. Explanation We find the rate of change of height at x = 2 seconds as 32 meters/second, meaning the rocket's height is increasing at this rate at that moment in time. We find the rate of change of height at x = 2 seconds as 32 meters/second, meaning the rocket's height is increasing at this rate at that moment in time. Problem 3 Derive the second derivative of the function y = x^4. The first step is to find the first derivative, dy/dx = 4x^3...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4x^3] = 4 3x^2 = 12x^2 Therefore, the second derivative of the function y = x^4 is 12x^2. The first step is to find the first derivative, dy/dx = 4x^3...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4x^3] = 4 3x^2 = 12x^2 Therefore, the second derivative of the function y = x^4 is 12x^2. Explanation We use the step-by-step process, starting with the first derivative. We apply the power rule again to find the second derivative, resulting in 12x^2. We use the step-by-step process, starting with the first derivative. We apply the power rule again to find the second derivative, resulting in 12x^2. Problem 4 Prove: d/dx (x^2 + x^3) = 2x + 3x^2. Let’s apply the power rule: Consider y = x^2 + x^3. To differentiate, we take each term separately: dy/dx = d/dx (x^2) + d/dx (x^3) = 2x + 3x^2 Hence proved. Let’s apply the power rule: Consider y = x^2 + x^3. To differentiate, we take each term separately: dy/dx = d/dx (x^2) + d/dx (x^3) = 2x + 3x^2 Hence proved. Explanation In this step-by-step process, we differentiate each term separately using the power rule. We then add the derivatives to derive the equation. In this step-by-step process, we differentiate each term separately using the power rule. We then add the derivatives to derive the equation. Problem 5 Solve: d/dx (x^2/x) To differentiate the function, we simplify first: x^2/x = x Now take the derivative: d/dx (x) = 1 Therefore, d/dx (x^2/x) = 1. To differentiate the function, we simplify first: x^2/x = x Now take the derivative: d/dx (x) = 1 Therefore, d/dx (x^2/x) = 1. Explanation In this process, we simplify the given function first and then differentiate the simplified form, resulting in the derivative of 1. In this process, we simplify the given function first and then differentiate the simplified form, resulting in the derivative of 1. FAQs on the Derivative of Power of x 1.Find the derivative of x^n. Using the power rule for x^n, we get: d/dx (x^n) = nx^(n-1) (simplified) Using the power rule for x^n, we get: d/dx (x^n) = nx^(n-1) (simplified) 2.Can we use the derivative of x^n in real life? Yes, we can use the derivative of x^n in real life in calculating the rate of change in various scientific and engineering contexts, such as velocity and acceleration. Yes, we can use the derivative of x^n in real life in calculating the rate of change in various scientific and engineering contexts, such as velocity and acceleration. 3.Is it possible to take the derivative of x^n at x = 0? Yes, it is possible to take the derivative of x^n at x = 0, and it is typically straightforward unless n is negative, where x^n might be undefined at x = 0. Yes, it is possible to take the derivative of x^n at x = 0, and it is typically straightforward unless n is negative, where x^n might be undefined at x = 0. 4.What rule is used to differentiate x^n ln(x)? We use the product rule to differentiate x^n ln(x): d/dx (x^n ln(x)) = (d/dx (x^n)) ln(x) + x^n (d/dx (ln(x))) We use the product rule to differentiate x^n ln(x): d/dx (x^n ln(x)) = (d/dx (x^n)) ln(x) + x^n (d/dx (ln(x))) 5.Are the derivatives of x^n and e^x the same? No, they are different. The derivative of x^n is nx^(n-1), while the derivative of e^x is simply e^x. No, they are different. The derivative of x^n is nx^(n-1), while the derivative of e^x is simply e^x. Important Glossaries for the Derivative of Power of x Explore More calculus Important Math Links IconPrevious to Derivative of Power of x Important Math Links IconNext to Derivative of Power of x Jaskaran Singh Saluja About the Author Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles. Fun Fact : He loves to play the quiz with kids through algebra to make kids love it.
12154	https://www.ncbi.nlm.nih.gov/books/NBK538197/	Dandy-Walker Malformation - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Dandy-Walker Malformation Edgar A. Zamora; Joe M. Das; Tahani Ahmad. Author Information and Affiliations Authors Edgar A. Zamora 1; Joe M. Das 2; Tahani Ahmad 3. Affiliations 1 Montefiore Medical Center 2 Imperial College Healthcare NHS Trust, London 3 IWK Health Center Last Update: November 12, 2023. Go to: Continuing Education Activity Dandy-Walker malformation or syndrome is a rare congenital neurological anomaly that affects the development of the cerebellum, the region of the brain responsible for motor coordination and balance. This posterior fossa anomaly is characterized by agenesis or hypoplasia of the vermis and cystic enlargement of the fourth ventricle, causing upward displacement of the tentorium and torcula. Most patients have hydrocephalus at the time of diagnosis. Dandy-Walker malformation is the most common posterior fossa malformation, and it typically occurs sporadically.The syndrome can manifest with a wide spectrum of neurological and developmental symptoms, making timely recognition and management crucial for improving patient outcomes.Treatment consists of treating the manifestations and associated comorbidities. This activity reviews the presentation of Dandy-Walker malformation and highlights the role of the interprofessional team in its management. Objectives: Identify the etiology of Dandy-Walker malformation. Screen patients with suspected Dandy-Walker malformation efficiently, employing appropriate diagnostic tools and considering comorbidities. Implement evidence-based treatment strategies for managing patients with Dandy-Walker malformation. Collaborate with an interprofessional healthcare team to deliver comprehensive care and address the complex needs of patients with Dandy-Walker malformations. Access free multiple choice questions on this topic. Go to: Introduction Dandy-Walker malformation (DWM) or syndrome is a posterior fossa anomaly characterized by: Agenesis or hypoplasia of the vermis Cystic enlargement of the fourth ventricle with communication to a large cystic dilated posterior fossa Upward displacement of tentorium and torcula (torcular-lambdoid inversion) Enlargement of the posterior fossa The first description of DWM dates back to 1887 by Sutton. Dandy and Blackfan described this deformity in 1914, and Taggart and Walker expanded on it in 1942. Bender first described the condition as DWM in 1954. The "Dandy-Walker complex" and "Dandy-Walker spectrum" are 2 of the most often used radiological entities proposed by Barkovich et al. in 1989. Most patients have hydrocephalus at the time of diagnosis. DWM is the most common posterior fossa malformation, and it typically occurs sporadically. Keeping the development of cerebellar vermis as the standard, the Dandy-Walker spectrum can be divided into: Dandy-Walker malformation Dandy-Walker variant Simple posterior fossa cistern widening Dandy-Walker variant is characterized by cerebellar vermian hypoplasia, cystic fourth ventricular dilatation, and normal posterior fossa volume. This terminology is currently going out of favor as it generates a lot of confusion. Rather, individual physical abnormalities should be mentioned. Many patients remain clinically asymptomatic for years, while others may present with a variety of comorbidities leading to earlier diagnosis.Treatment is generally focused on alleviating hydrocephalus and posterior fossa symptoms, often including surgical interventions like ventriculoperitoneal and cystoperitoneal shunting. Go to: Etiology Historically, DWM was believed to be caused by atresia of the Luschka and Magendie foramina, leading to enlargement of the fourth ventricle and vermian hypoplasia. However, recent evidence suggests that DWM results from developmental abnormalities affecting the roof of the rhombencephalon, leading to variable degrees of vermian hypoplasia and cystic enlargement. This complex malformation may be initiated by 2 different pathophysiological mechanisms: the arrest of vermian development and the failure of fourth ventricle foramina fenestration, leading to an enlarged Blake’s pouch and causing compression of the vermis. DWM may be isolated or associated with chromosomal abnormalities, Mendelian disorders, syndromic malformations, congenital infections, and various other comorbidities. Central nervous system (CNS) disorders related to DWM include malformations of cortical development, holoprosencephaly, dysgenesis of the corpus callosum, and neural tube defects. Rare mutations have been described in some genes including FOXC1 (in locus 6p25.3), ZIC1, and ZIC4 (in locus 3q24), FGF17, LAMC1, and NID1. Go to: Epidemiology DWM and related variants have a prevalence of 1 in 350,000 live births in the United States. These malformations account for approximately 1% to 4% of hydrocephalus cases.Most DWM patients present with signs and symptoms from increased intracranial pressure, most commonly related to hydrocephalus and the posterior fossa cyst. For this reason, therapy generally aims to control intracranial pressure, usually through surgery. Some institutions have an estimated mortality rate of 12% to 50%.Treatment of hydrocephalus with shunting substantially improves mortality.Fetal mortality directly correlates with the presence of extra-CNS abnormalities. Go to: Pathophysiology The majority of cases are sporadic. However, some may result from chromosomal aneuploidy, Mendelian disorders, and environmental exposures, including congenital rubella and fetal alcohol exposure.This condition underscores the intricate interplay between genetic and environmental factors in brain development. Several conditions have been reported to be associated with DWM: Mendelian disorders: Warburg syndrome Coffin-Siris syndrome Fraser cryptophthalmos Joubert–Boltshauser syndrome Meckel–Gruber syndrome Aicardi syndrome Neurofibromatosis type 1 Chromosomal aberrations: Duplication syndromes, including those involving 5q; 8p; 8q; 17q Trisomies 9, 13, and 18 Triploidy Deletion at 6(p24–p25) 9ph+ heteromorphism Environmental agents: Prenatal exposure to rubella,cytomegalovirus,toxoplasmosis, coumadin, alcohol Maternal diabetes Sporadic syndromes: Klippel–Feil syndrome Goldenhar syndrome Cornelia de Lange syndrome Concept of the Midline as a Developmental Field (Opitz and Gilbert) An insult to this field can result in pleiotropic sequelae on midline structures- the central nervous system (CNS),heart,palate, midface,vertebrae, genitalia, and sacrococcygeal region. DWM should be considered a complex developmental disturbance of the midline of marked genetic and etiological heterogeneity. Embryology The cerebellar vermis develops from top to bottom at 17 to 18 weeks gestation. Due to vermian developmental arrest or compression of the vermis brought on by a failure of fourth ventricle foramina fenestration and consequent growing of Blake's pouch, the roof of the rhombencephalon does not form correctly. Go to: History and Physical The clinical presentation is nonspecific, subject to multiple factors, including the severity of hydrocephalus, intracranial hypertension, and associated comorbidities.Most patients will present in their first year of life with signs and symptoms of increased intracranial pressure.The most common manifestation is macrocephaly, affecting 90% to 100% of patients during their first months of life. The syndromic form of DWM may also have malformations of the heart, face, limbs, and gastrointestinal or genitourinary system that could draw initial medical attention.Almost 75% of cases will have hydrocephalus by the age of 3 months, and eventually, almost 90% of the patients will have hydrocephalus. A case series of 42 patients with DWM showed hydrocephaly in all patients at the time of diagnosis, vermian hypoplasia in 88%, and cerebellar hypoplasia in 59%. Go to: Evaluation Ultrasound is typically the first imaging modality in assessing the fetal brain. CNS structures evaluated through this modality should include head size and shape, choroid plexus, thalami, cerebellum, cavum septum pellucidum, lateral ventricles,nuchal fold, cisterna magna, and spine. Imaging may also demonstrate other CNS abnormalities that commonly correlate with DWM. Measurement of the cisterna magna forms part of the ultrasound evaluation of the fetal brain. A case series showed an upper limit of normal of 10 mm and a mean size of 5 plus or minus 3 mm at 15 or more gestational weeks. A prominent cisterna magna during prenatal fetal ultrasound evaluation may raise concern for congenital posterior fossa abnormalities. However, as an isolated finding, a prominent cisterna magna is unlikely to be clinically significant if the patient presents no other abnormalities. Radiological Criteria for Diagnosis The classical radiological triad of DWM includes hypoplastic vermis, fourth ventricle enlargement, and torcular elevation. However,Klein et al. modified the classification by Barkovich et al. in 2003 to identify the radiological criteria for the diagnosis of DWM: Large, median posterior fossa cyst widely communicating with the fourth ventricle (seeImage.Occipital Meningocele With Dandy-Walker Cyst) Absence of the lower portion of the vermis at different degrees (lower 3/4, lower half, lower 1/4) Hypoplasia, anterior rotation, and upward displacement of the remnant of the vermis Absence or flattening of the angle of the fastigium Large bossing posterior fossa with an elevation of the torcula Anterolateral displacement of normal or hypoplastic cerebellar hemispheres Like ultrasound, magnetic resonance imaging (MRI) is more helpful after the 20th week of gestation, and it may aid in evaluating CNS malformations not satisfactorily described by ultrasound and search for commonly associated abnormalities. Prenatal diagnosis by ultrasound is possible after the 18th week of gestation when the cerebellar vermis has completely developed. Diagnosis may be confirmed by magnetic resonance.For patients with the Dandy-Walker variant (DWV), discrepancies may be found in prenatal and postnatal imaging based on the variation of vermian development. Magnetic resonance will also help distinguish DWM from other posterior fossa lesions (see Image.Dandy-Walker Malformation). Karyotype and postnatal imaging should be offered for all patients with prenatal imaging findings consistent with DWM to confirm findings and search for other possibly associated abnormalities. Measurement of the brainstem-vermian (BV) angle and brainstem-tentorium (BT) angle substantially help in the differential diagnosis of patients with an increased size of the cisterna magna. The BV angle increases with the severity of the condition. Angles <18 degrees are normal, while 18 to 30 degrees suggest Blake's pouch malformation. The BT angle ranges typically between 21 to 44 degrees. Both angles >45 degrees are strongly suggestive of DWM. Associated CNS abnormalities include occipital encephalocele,polymicrogyria,agenesis, or dysgenesis of the corpus callosum, and heterotopia. Go to: Treatment / Management Treatment consists of treating the manifestations and associated comorbidities. Most patients present with signs and symptoms from increased intracranial pressure, most commonly related to hydrocephalus and posterior fossa cyst. For this reason, therapy generally aims to control intracranial pressure, usually through surgery. Surgical treatment may include ventriculoperitoneal (VP) or cystoperitoneal (CP) shunts. Other patients may be candidates for endoscopic procedures, including endoscopic third ventriculostomy (ETV). Go to: Differential Diagnosis Differential diagnosis includes posterior fossa abnormalities that may be associated with hydrocephalus sometimes, which commonly may present in a similar way to DWM; these include retrocerebellar arachnoid cysts, cystic hygroma, Blake's pouch cyst, mega cisterna magna, and vermian hypoplasia. Additionally, several syndromes correlate with DWM, such as Aase-Smith, cerebro-oculo-muscular syndrome, Coffin-Siris, Cornelia de Lange, and Aicardi. The other cystic malformations of the posterior fossa include persistent Blake's pouch, mega cisterna magna, and arachnoid cyst. The non-cystic malformations include Joubert syndrome, rhombencephalosynapsis, tectocerebellar dysraphia, and neocerebellar dysgenesis. Retrocerebellar arachnoid cysts may be large enough to cause compression of cerebellar hemispheres and the fourth ventricle. Blake's cyst is a retrocerebellar fluid collection with a medial line communication to the fourth ventricle. Mega cisterna manga refers to a fluid collection located posteroinferior to a normally developed cerebellum. Several comorbidities may also correlate with DWM, including syndromic and non-syndromic, CNS and non-CNS anomalies, chromosomal abnormalities, cardiovascular conditions, mental illness, and severe intellectual disability. Go to: Prognosis The prognosis for patients with DWM is highly variable and dependent on several factors, including the severity of the malformation, the timeliness of diagnosis and intervention, and the presence of associated complications. In milder cases, where the cerebellar vermis is less affected and with minimal intracranial pressure, individuals may experience fewer neurological deficits and lead relatively normal lives with appropriate management and support. However, in more severe cases with significant cerebellar and posterior fossa abnormalities, patients are at a higher risk of experiencing developmental delays, intellectual disabilities, motor coordination issues, and hydrocephalus. Fifty percent of children with untreated hydrocephalus die before age 3, and only approximately 20% to 23% will reach adult life. Most of the patients with untreated hydrocephalus who reach adult life will have motor, visual, and auditory deficits. The diameter of the fetal lateral ventricle measured by obstetric ultrasound may have substantial prognostic value. As described in a previous study, lateral ventricles measuring between 11 mm to 15 mm correlate with a 21% risk of developmental delay. If the diameter measures >15 mm, the risk of developmental delay is above 50%. Most case series of patients with nontumoral hydrocephalus present an overall risk for epilepsy of approximately 30%. Functional outcome is subject to several factors, which include other structural brain abnormalities, extra-CNS manifestations, epilepsy, motor, visual, or hearing impairment, and other congenital abnormalities. Go to: Complications Complications in DWM can encompass a broad spectrum of neurological and developmental challenges. One of the primary complications is hydrocephalus, leading to increased intracranial pressure. This increased pressure can lead to symptoms such as headaches, vomiting, and cognitive impairments, making timely intervention essential.The most common complications associated with shunting are infection and shunt malfunction. Patients with DWM may experience various motor and coordination deficits due to the cerebellar abnormalities, affecting their daily functioning and quality of life. Cognitive and intellectual disabilities, speech and language impairments, and behavioral problems are common as well, adding to the complexity of managing this condition. Furthermore, the associated structural anomalies may result in additional complications, including the risk of syringomyelia or other spinal cord abnormalities. Sixteen percent of patients with isolated DWM have chromosomal abnormalities.DWM may be associated with malformations of the face, limbs, heart, and genitourinary or gastrointestinal system. Go to: Deterrence and Patient Education The cerebellar vermis development varies between individuals and usually completes its formation late in the first half of pregnancy. Some fetuses may achieve vermis development around gestational week 18. Therefore, the diagnosis of DWM based on imaging performed before weeks 16 to 18 of development may be premature and erroneous. Additionally, the cisterna magna has not reached its final anatomy during the first half of pregnancy. A relatively wide opening in the cerebellomedullary cistern may not indicate vermian dysgenesis at that stage of development. Therefore, the cerebellum should be reevaluated at 20 to 22 weeks gestation to rule out vermian abnormalities. Karyotype and comprehensive fetal ultrasound should be offered for patients with features of the Dandy-Walker complex. In the absence of other lesions in the posterior fossa, an isolated prominent cisterna magna may not be clinically significant. Go to: Enhancing Healthcare Team Outcomes In managing patients with DWM, a multidisciplinary healthcare team comprised of physicians, advanced practitioners, nurses, pharmacists, and other health professionals plays a pivotal role in enhancing patient-centered care, outcomes, patient safety, and team performance. Pediatric healthcare providers should be familiar with the broad spectrum of congenital posterior fossa abnormalities to provide an accurate diagnosis, optimal therapy, and genetic counseling.Additionally, posterior fossa imaging findings will change based on the gestational age; as such, findings suggestive of disease in earlier weeks should be reevaluated at weeks 20 to 22 gestation. Skillful expertise, honed through continuous education and training, is essential for correctly diagnosing and treating this complex neurological condition. A well-structured strategy involving the collaboration of various professionals ensures that each aspect of the patient's care plan is addressed comprehensively. Timely consultation should be obtained from the pediatric neurosurgeon/neurologist. Ethical considerations, such as informed consent and respect for patient autonomy, guide decision-making and promote trust within the interprofessional team. Responsibilities are distributed efficiently, with clear roles and accountabilities to optimize patient care. Effective interprofessional communication fosters the exchange of critical information and the seamless coordination of care, helping to minimize errors and improve patient safety. Care coordination, facilitated by healthcare professionals, enables a holistic approach to patient management, offering the best chance for positive outcomes in the challenging realm of DWM. Go to: Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Figure Dandy-Walker Malformation Image courtesy S Bhimji, MD Figure Occipital Meningocele With Dandy-Walker Cyst. A transverse image through the fetal head shows a large cystic structure outside the calvaria in the occipital region (arrows). A large cystic structure is also seen in the posterior fossa, consistent with (more...) Go to: References 1. 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The Most Common Comorbidities in Dandy-Walker Syndrome Patients: A Systematic Review of Case Reports. J Child Neurol. 2017 Sep;32(10):886-902. [PubMed: 28635420] 10. Osenbach RK, Menezes AH. Diagnosis and management of the Dandy-Walker malformation: 30 years of experience. Pediatr Neurosurg. 1992;18(4):179-89. [PubMed: 1472430] 11. Robinson AJ. Inferior vermian hypoplasia--preconception, misconception. Ultrasound Obstet Gynecol. 2014 Feb;43(2):123-36. [PubMed: 24497418] 12. Bokhari I, Rehman L, Hassan S, Hashim MS. Dandy-Walker Malformation: A Clinical and Surgical Outcome Analysis. J Coll Physicians Surg Pak. 2015 Jun;25(6):431-3. [PubMed: 26100996] 13. Nyberg DA, Cyr DR, Mack LA, Fitzsimmons J, Hickok D, Mahony BS. The Dandy-Walker malformation prenatal sonographic diagnosis and its clinical significance. J Ultrasound Med. 1988 Feb;7(2):65-71. [PubMed: 3279226] 14. Shah SP, Gohil JA, Kesavapisharady K, Easwer HV. Revisiting Dandy-Walker Malformation with Associated Neurofibromatosis. Asian J Neurosurg. 2021 Oct-Dec;16(4):850-853. [PMC free article: PMC8751521] [PubMed: 35071091] 15. Ecker JL, Shipp TD, Bromley B, Benacerraf B. The sonographic diagnosis of Dandy-Walker and Dandy-Walker variant: associated findings and outcomes. Prenat Diagn. 2000 Apr;20(4):328-32. [PubMed: 10740206] 16. Mahony BS, Callen PW, Filly RA, Hoddick WK. The fetal cisterna magna. Radiology. 1984 Dec;153(3):773-6. [PubMed: 6387792] 17. Beaman MM, Guidugli L, Hammer M, Barrows C, Gregor A, Lee S, Deak KL, McDonald MT, Jensen C, Zaki MS, Masri AT, Hobbs CA, Gleeson JG, Cohen JL. Novel association of Dandy-Walker malformation with CAPN15 variants expands the phenotype of oculogastrointestinal neurodevelopmental syndrome. Am J Med Genet A. 2023 Nov;191(11):2757-2767. [PMC free article: PMC11141336] [PubMed: 37596828] 18. Klein O, Pierre-Kahn A, Boddaert N, Parisot D, Brunelle F. Dandy-Walker malformation: prenatal diagnosis and prognosis. Childs Nerv Syst. 2003 Aug;19(7-8):484-9. [PubMed: 12879343] 19. Kline-Fath BM, Calvo-Garcia MA. Prenatal imaging of congenital malformations of the brain. Semin Ultrasound CT MR. 2011 Jun;32(3):167-88. [PubMed: 21596274] 20. Harper T, Fordham LA, Wolfe HM. The fetal dandy walker complex: associated anomalies, perinatal outcome and postnatal imaging. Fetal Diagn Ther. 2007;22(4):277-81. [PubMed: 17369695] 21. Yu F, Fu L, Xu C, Wang J, Huang X, Zhang G, Zhang H. Prenatal Magnetic Resonance Imaging helps Discover Cerebellar Dysplasia or Malformations in Foetuses. Curr Med Imaging. 2023 Oct 31; [PubMed: 37921151] 22. Volpe P, Contro E, De Musso F, Ghi T, Farina A, Tempesta A, Volpe G, Rizzo N, Pilu G. Brainstem-vermis and brainstem-tentorium angles allow accurate categorization of fetal upward rotation of cerebellar vermis. Ultrasound Obstet Gynecol. 2012 Jun;39(6):632-5. [PubMed: 22253138] 23. Mohanty A, Biswas A, Satish S, Praharaj SS, Sastry KV. Treatment options for Dandy-Walker malformation. J Neurosurg. 2006 Nov;105(5 Suppl):348-56. [PubMed: 17328256] 24. Chumas P, Tyagi A, Livingston J. Hydrocephalus--what's new? Arch Dis Child Fetal Neonatal Ed. 2001 Nov;85(3):F149-54. [PMC free article: PMC1721342] [PubMed: 11668153] 25. Yashon D, Jane JA, Sugar O. The course of severe untreated infantile hydrocephalus. Prognostic significance of the cerebral mantle. J Neurosurg. 1965 Nov;23(5):509-16. [PubMed: 5295268] 26. Donders J, Canady AI, Rourke BP. Psychometric intelligence after infantile hydrocephalus. A critical review and reinterpretation. Childs Nerv Syst. 1990 May;6(3):148-54. [PubMed: 2357712] 27. Dennis M, Fitz CR, Netley CT, Sugar J, Harwood-Nash DC, Hendrick EB, Hoffman HJ, Humphreys RP. The intelligence of hydrocephalic children. Arch Neurol. 1981 Oct;38(10):607-15. [PubMed: 6975094] 28. Grinberg I, Northrup H, Ardinger H, Prasad C, Dobyns WB, Millen KJ. Heterozygous deletion of the linked genes ZIC1 and ZIC4 is involved in Dandy-Walker malformation. Nat Genet. 2004 Oct;36(10):1053-5. [PubMed: 15338008] 29. Bromley B, Nadel AS, Pauker S, Estroff JA, Benacerraf BR. Closure of the cerebellar vermis: evaluation with second trimester US. Radiology. 1994 Dec;193(3):761-3. [PubMed: 7972820] Disclosure:Edgar Zamora declares no relevant financial relationships with ineligible companies. Disclosure:Joe Das declares no relevant financial relationships with ineligible companies. Disclosure:Tahani Ahmad declares no relevant financial relationships with ineligible companies. Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Deterrence and Patient Education Enhancing Healthcare Team Outcomes Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK538197 PMID: 30855785 Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Deterrence and Patient Education Enhancing Healthcare Team Outcomes Review Questions References Related information PMCPubMed Central citations PubMedLinks to PubMed Similar articles in PubMed Review [Dandy-Walker malformation].[Radiologe. 2018]Review [Dandy-Walker malformation].Reith W, Haussmann A. Radiologe. 2018 Jul; 58(7):629-635. Review Neuroimaging of Dandy-Walker malformation: new concepts.[Top Magn Reson Imaging. 2011]Review Neuroimaging of Dandy-Walker malformation: new concepts.Correa GG, Amaral LF, Vedolin LM. Top Magn Reson Imaging. 2011 Dec; 22(6):303-12. Dandy-Walker malformation: prenatal diagnosis and prognosis.[Childs Nerv Syst. 2003]Dandy-Walker malformation: prenatal diagnosis and prognosis.Klein O, Pierre-Kahn A, Boddaert N, Parisot D, Brunelle F. Childs Nerv Syst. 2003 Aug; 19(7-8):484-9. Epub 2003 Jul 16. Dandy-Walker malformation: a rare association with hypoparathyroidism.[Pediatr Neurol. 2010]Dandy-Walker malformation: a rare association with hypoparathyroidism.Coban D, Akin MA, Kurtoglu S, Oktem S, Yikilmaz A. Pediatr Neurol. 2010 Dec; 43(6):439-41. Posterior fossa and vermian morphometry in the characterization of fetal cerebellar abnormalities: a prospective three-dimensional ultrasound study.[Ultrasound Obstet Gynecol. 2006]Posterior fossa and vermian morphometry in the characterization of fetal cerebellar abnormalities: a prospective three-dimensional ultrasound study.Paladini D, Volpe P. Ultrasound Obstet Gynecol. 2006 May; 27(5):482-9. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Dandy-Walker Malformation - StatPearlsDandy-Walker Malformation - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. Cite this Page Close Zamora EA, Das JM, Ahmad T. Dandy-Walker Malformation. 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12155	http://www.fen.bilkent.edu.tr/~franz/ag05/ag-07.pdf	Chapter 7 Projective Transformations 7.1 Aﬃne Transformations In aﬃne geometry, aﬃne transformations (translations, rotations, . . . ) play a central role; by deﬁnition, an aﬃne transformation is an invertible linear map A 2K − →A 2K followed by a translation, that is, a map (x, y) 7− →(x′, y′), where x′ = ax + by + c, y′ = dx + ey + f, and ad −bc ̸= 0. Note that aﬃne transformations form a group under composition of maps. Proposition 7.1.1. Let P1, P2, P3 be non-collinear points in the aﬃne plane. Then there is a unique aﬃne transformation that sends P1 to (0, 0), P2 to (1, 0), and P3 to (0, 1). Proof. We only sketch the proof. Write Pi = (xi, yi); then we get a linear system of 6 equations in 6 unknowns, and since the Pi are not collinear, the corresponding system has nonzero determinant and thus a unique solution. 7.2 Projective Transformations Now let us deﬁne projective transformations. An invertible 3 × 3-matrix A = (aij) ∈M3(K) acts on the projective plane P2K via A([x : y : z]) = [x′ : y′ : z′], where (x′, y′, z′) = (x, y, z)   a11 a12 a13 a21 a22 a23 a31 a32 a33  . This is well deﬁned, since A([λx : λy : λz]) = [λx′ : λy′ : λz′], so rescaling is harmless. Note that we write A(P) for the point whose coordinates are computed by pA, where p is a vector (x, y, z) corresponding to P = [x : y : z]. There are, however, matrices in GL3(K) that have no eﬀect on points in the projective plane: the diagonal matrix diag(λ, λ, λ) (this is the matrix with 33 aij = 0 except for a11 = a22 = a33 = λ) for nonzero λ ∈K ﬁxes every [x : y : z] ∈P2K. The group of all diagonal matrices with entry λ ∈K× is isomorphic to K×, and we can make the projective general linear group PGL3(K) = GL3(K)/K× act on the projective plane. Its elements are 3 × 3-matrices with nonzero determinant, and two such matrices are considered to be equal if they diﬀer by a nonzero factor λ ∈K×. Some Abstract Nonsense This is a very special case of some fairly general observation. Assume that a group G acts on a set X (this means that there is a map G×X − →X : (g, x) 7− → gx such that 1x = x and g(g′x) = (gg′)x). For any x ∈X, there is a group Stab(x) = {g ∈G : gx = x}, the stabilizer. Now consider the intersection H of all these stabilizers. Then H is normal in G: in fact, for h ∈H and g ∈G we have (g−1hg)x = g−1h(gx) = g−1gx = x, since h ﬁxes everything (in particular gx), and therefore g−1hg ∈H. Back to Projective Transformation Lemma 7.2.1. Let A be a projective transformation represented by a a nonsin-gular 3 × 3-matrix A = (aij). Then the following assertions are equivalent: 1. The restriction of A to A2 = {(x : y : 1) ∈P2} is an aﬃne transformation; 2. a13 = a23 = 0; 3. A ﬁxes the line z = 0 at inﬁnity. Proof. 1 ⇐ ⇒2: We have [x : y : 1]A = [x′ : y′ : z′] with z′ = a13x+a23y +a33. If A induces an aﬃne transformation, then we must have z′ ̸= 0 for all x, y ∈K, and this implies a13 = a23 = 0. Note that we automatically have a33 ̸= 0, since det A ̸= 0. Thus we can rescale A to get a33 = 1. Conversely, if a13 = a23 = 0 and a33 = 1, then A([x : y : 1]) = [x′ : y′ : 1], where x′ = a11x + a21y + a31 and y′ = a12x + a22y + a32. This is an aﬃne transformation. 2 ⇐ ⇒3: If a13 = a23 = 0, then A([x : y : 0]) = [x′ : y′ : 0], hence the line z = 0 is preserved. Conversely, if A([x : y : 0]) = [x′ : y′ : 0] for all x, y ∈K, then a31 = a32 = 0. This result shows that we have a lot more choice in the projective world; as an example, we have Proposition 7.2.2. Let Pi = [xi : yi : zi] (i = 1, 2, 3, 4) be four points in the projective plane, no three of which are collinear. Then there is a unique projective transformation sending the standard frame, namely [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1] and [1 : 1 : 1], to P1, P2, P3 and P4, respectively. 34 Proof. The transformation deﬁned by A = (aij) ∈PGL3(K) will map [1 : 0 : 0] to P1 if and only if there is some α1 ∈K× with α1(x1, y1, z1) = (1, 0, 0)A = (a11, a12, a13). This determines the ﬁrst row of A up to some nonzero factor. Similarly, the second and the third rows are determined up to nonzero factors α2, α3 ∈K× by the second and third condition. Thus the rows of A are given by α1p1, α2p2 and α3p3, where the pi are vectors corresponding to the Pi. Now P4 will be the image of [1 : 1 : 1] if and only if α4p4 = α1p1 + α2p2 + α3p3 (rescaling allows us to assume that α4 = 1). Now this is a linear system of three equations in three unknowns; since the vectors p1, p2, p3 are linearly independent, there is a unique solution (α1, α2, α3). Since p4 is independent of any two out of p1, p2, p3, the numbers αi are all nonzero; this implies that the matrix with rows αipi (i = 1, 2, 3) is invertible, hence A deﬁnes a projective transformation. Finally, A is unique except for the rescaling α4 = 1, hence is unique as an element of PGL3(K). This result has a number of important corollaries: Corollary 7.2.3. Let Pi and Qi (i = 1, 2, 3, 4) denote two sets of four points in the projective plane such that no three Pi and no three Qi are collinear. Then there is a projective transformation sending Pi to Qi for i = 1, 2, 3, 4. Proof. Let A denote the projective transformation that sends the standard frame to the Pi; let B denote the transformation that does the same with the Qi. Then A ◦B−1 is the projective transformation we are looking for. Projective transformations A act on projective planes and therefore on plane algebraic curves CF : F(X, Y, Z) = 0; the image of C under A is some curve CG : G(U, V, W) = 0. How can we compute G from F? Given a point [x : y : z] ∈CF (K), we must have G(A(P)) = 0, and this is accomplished by G = F ◦A−1. Here is an example. Take F(X, Y, Z) = Y Z −X2 and the transformation [u : v : w] = [x : y : z]A = [x + y : y : z]. For getting G, we solve for x, y, z, that is, put [x : y : z] = [u : v : w]A−1 and then plug the result into F: [x : y : z] = [u −v : y : z], hence G(U, V, W) = F(U −V, V, W) = V W −(U −V )2. Thus we get G by evaluating F at (X, Y, Z)A−1, that is, G = F ◦A−1. This ensures that a point [x : y : z] on CF will get mapped by A to a point [u : v : w] = [x : y : z]A on CG. Proposition 7.2.4. Projective transformations preserve the degree of curves. Proof. Projective transformations map a monomial XiY iZk of degree m = i + j+k either to 0 or to another homogeneous polynomial of degree m. If f(X, Y, Z) is transformed by some transformation T into the zero polynomial, then the inverse transformation maps the zero polynomial into f, which is nonsense. 35 Finally, let us talk a little bit about singular points. We have F = G ◦A, hence the chain rule implies that the derivative of F is the derivative of G with respect to the new variables multiplied by the derivative of the linear map (u, v, w) = (x, y, z)A, which is the matrix A itself. In symbols: ∂F ∂X , ∂F ∂Y , ∂F ∂Z = ∂G ∂U , ∂G ∂V , ∂G ∂W · A. Now a point on CF is singular if and only if all three derivatives vanish at some point P = [x : y : z]. Since the matrix A is nonsingular, this happens if and only if the point [u : v : w] = [x : y : z]A is singular. Proposition 7.2.5. Projective transformations preserve singularities. With some more work it can also be shown that projective transformations also preserve multiplicities, tangents, ﬂexes etc. 7.3 Projective Conics Observe that this means that projective transformations map lines into lines and conics into conics. Aﬃne transformations preserve the line at inﬁnity, hence cannot map a (real) circle (no point at inﬁnity) into a hyperbola (two points at inﬁnity). Projective transformations can do this: the projective circle has equation X2 + Y 2 −Z2 = 0; the projective transformation X = Y ′, Y = Z′, Z = X′ transforms this into Y ′2 −X′2 + Z′2 = 0, which, after dehomogenizing with respect to Z′, is just the hyperbola x2 −y2 = 1. What happened here is that Y = Z′ has moved the two points with Y = 0 to inﬁnity. Similarly, the hyperbola XY −Z2 = 1 can be transformed into a parabola via X = Y ′, Y = Z′, Z = X′: after dehomogenizing we get y = x2. The hyperbola had two points [1 : 0 : 0] and [0 : 1 : 0] at inﬁnity; the ﬁrst one was moved to the point [0 : 1 : 0] at inﬁnity, the second one to [0 : 0 : 1], which is the origin in the aﬃne plane. As a matter of fact it can easily be proved that, over the complex numbers (or any algebraically closed ﬁeld of characteristic ̸= 2), there is only one nondegenerate conic up to projective transformations. Note that f(X, Y, Z) = XY Z−XY 2 is transformed into the zero polynomial by the singular transformation X = X′, Y = X′, Z = X′. Let us call two conics projectively equivalent if there is a projective trans-formation mapping one to the other. Proposition 7.3.1. Any nondegenerate projective conic deﬁned over some ﬁeld K with at least one K-rational point is projectively equivalent to the conic XY + Y Z + ZX = 0. (7.1) More exactly, given a nondegenerate conic C and three points on C, there is a unique projective transformation mapping C to (7.1) and the three points to [1 : 0 : 0], [0 : 1 : 0] and [0 : 0 : 1], respectively. 36 Proof. Take any three points on a conic (it has one K-rational point, hence a parametrization gives all of them; there are inﬁnitely many over inﬁnite ﬁelds and exactly q +1 over ﬁnite ﬁelds with q elements. Now observe that q +1 ≥3). Then there is a projective transformation mapping them into [1 : 0 : 0], [0 : 1 : 0] and [0 : 0 : 1], respectively (note that the three points on the conic are not collinear since the conic is degenerate). If the transformed conic has the equation aX2 + bXY + cY 2 + dY Z + eZX + fZ2 = 0, then we immediately see that a = c = f = 0: bXY + dY Z + eZX = 0. Moreover, bde ̸= 0 since otherwise the conic is degenerate: if, for example, b = 0, then the equation 0 = dY Z + ZX = Z(dY + eX) describes a pair of lines, which is a degenerate conic. Using the transformation X = dX′, Y = eY ′, Z = bZ′, this becomes (7.1). If there are two such maps A, B, then B ◦A−1 maps the standard conic onto itself and preserves the three points of the standard frame. It is then easily seen that B ◦A−1 must be the identity map in PGL3(K). This result allows us to simplify computational proofs of a number of theo-rems in projective geometry. As an example, we prove Pascal’s Theorem (1640); its analog for degenerate conics is due to Pappus of Alexandria (ca. 320). For its proof, we use a little Lemma 7.3.2. A point P ∈P2K diﬀerent from [0 : 0 : 1] is on the conic (7.1) if and only if there is some r ∈K such that P = [r : 1 −r : r(r −1)]. Proof. The equation of the conic is (x+y)z = −xy. If x+y = 0, then x = y = 0 and thus P = [0 : 0 : 1]. Therefore we can rescale the coordinates such that x + y = 1. Write x = r; then y = 1 −r and z = −xy/(x + y) = r(r −1). Conversely, every point [r : 1 −r : r(r −1)] is easily seen to be on the conic. Theorem 7.3.3 (Pascal’s Theorem). Let ABCDEF be a hexagon inscribed in a nondegenerate conic. Then the points of intersection X = AE ∩BF, Y = BD ∩CE and Z = AD ∩CF are collinear. Proof. Since projective transformations preserve lines, conics, and points of in-tersection, we may assume that the conic has the form (7.1) and that A = [1 : 0 : 0], B = [0 : 1 : 0] and C = [0 : 0 : 1]. Now let D = [d : 1 −d : d(d −1), E = [e : 1−e : e(e−1)] and F = [f : 1−f : f(f −1)] and observe that def ̸= 0. Now we see AE : ey + z = 0, BF : (1 −f)x + z = 0, X = [e : 1 −f : e(f −1)] BD : (1 −d)x + z = 0, CE : (e −1)x + ey = 0, Y = [e : 1 −e : e(d −1)], CF : (f −1)x + fy = 0, AD : dy + z = 0, Z = [f : 1 −f : d(f −1)]. 37 Now three points are collinear in P2K if and only if the determinant whose columns are the coordinates of these points is 0. A standard calculation shows that this is the case. 38
12156	https://dictionary.cambridge.org/us/dictionary/english-portuguese/awkwardness	AWKWARDNESS definition \| Cambridge Dictionary Dictionary Translate Grammar Thesaurus +Plus Cambridge Dictionary +Plus Games Shop Cambridge Dictionary +Plus My profile +Plus help Log out {{userName}} Cambridge Dictionary +Plus My profile +Plus help Log out Log in / Sign up English (US) English–Portuguese {{word}} {{#beta}} Beta{{/beta}} Translation of awkwardness – English-Portuguese dictionary awkwardness noun[U] uk Your browser doesn't support HTML5 audio /ˈɔː.kwəd.nəs/us Your browser doesn't support HTML5 audio /ˈɑː.kwɚd.nəs/ EMBARRASSMENT Add to word listAdd to word list embarrassment, or a situation that is difficult and not relaxed constrangimento In spite of the divorce, there was no awkwardness between them - in fact they seemed very much at ease. Apesar do divórcio, não havia constrangimento entre eles; na verdade, eles pareciam muito à vontade. The internet removes the awkwardness of a face-to-faceinteraction. More examples Fewer examples Once we had overcome an initial awkwardness we had an interestingconversation. the awkwardness of a first meeting I was aware of the potential for awkwardness and embarrassment in arriving late to the meeting. Despite his shyness and occasionalsocial awkwardness, friendsknew he was fiercelyloyal. DIFFICULTY the situation of being difficult to use, do, or deal with estranheza, dificuldade The awkwardness of the shape makes the deskdifficult to lift. A estranheza da forma torna a mesa difícil de levantar. the awkwardness of making complexarrangements to suit so many people More examples Fewer examples There was some awkwardness in the process. The strikescreated some awkwardness for peoplecommuting to work. The design of the handlebars only adds to the bike's awkwardness. OF MOVEMENT a quality of not being natural, relaxed, or attractive desconforto He outgrew his physical awkwardness and became a leading man in schoolplays. Ele superou o desconforto em relação a seu físico e se tornou protagonista nas peças escolares. The world of high fashion has embracedugliness or awkwardness as a new kind of beauty. More examples Fewer examples I was getting cramp because of the awkwardness of the position. At 16, she was very tall and people made fun of her youthful awkwardness. UNHELPFUL UK the fact of being deliberatelyunhelpful constrangimento Is he really unable to help or is it just awkwardness? Ele é realmente incapaz de ajudar ou é só constrangimento? See awkward More examples Fewer examples the awkwardness of the frontdeskstaff I've had enough of her awkwardness every time I ask her to do something. As a colleague he was known for his uncooperative awkwardness. (Translation of awkwardness from the Cambridge English-Portuguese Dictionary © Cambridge University Press) Translation of awkwardness \| GLOBAL English–Portuguese Dictionary awkwardness noun[uncountable] /ˈɔkwərdnɪs/ estranheza[feminine]falta de jeito[feminine] the awkwardness of his movements a falta de jeito de seus movimentos (Translation of awkwardness from the GLOBAL English-Portuguese Dictionary © 2021 K Dictionaries Ltd) C1 Translations of awkwardness in Chinese (Traditional) 局促不安，尴尬, 尷尬, 笨拙… See more in Chinese (Simplified) 局促不安，尴尬, 尴尬, 笨拙… See more in Spanish vergüenza, incomodidad, lo incómodo… See more in more languages in French in Japanese in Turkish in Dutch in Czech in Danish in Indonesian in Thai in Vietnamese in Polish in Swedish in Malay in German in Norwegian in Ukrainian côté [masculine] embarrassant, maladresse [feminine], gêne [feminine]… See more ぎこちなくさ… See more hantallık, beceriksizlik… See more onhandigheid… See more neobratnost, trapná situace… See more kejtethed… See more sikap kikuk… See more ความอึดอัดใจ… See more sự vụng về… See more niezręczność, niezdarność… See more tafatthet… See more kekekokan… See more die Ungeschicklichkeit… See more klossethet [masculine], klossethet… See more незграбність, ніяковість… See more Need a translator? Get a quick, free translation! Translator tool What is the pronunciation of awkwardness? See the definition of awkwardness in the English dictionary Browse awfulness awhile awkward awkwardly awkwardness awl awning awoke awoken Word of the Day Victoria sponge UK Your browser doesn't support HTML5 audio /vɪkˌtɔː.ri.ə ˈspʌndʒ/ US Your browser doesn't support HTML5 audio /vɪkˌtɔːr.i.ə ˈspʌndʒ/ a soft cake made with eggs, sugar, flour, and a type of fat such as butter. It is made in two layers with jam or cream, or both, between them About this Blog Calm and collected (The language of staying calm in a crisis) September 24, 2025 Read More New Words lawnmower poetry September 29, 2025 More new words has been added to list To top Contents English–PortugueseGLOBAL English–PortugueseTranslations © Cambridge University Press & Assessment 2025 Learn LearnLearnNew WordsHelpIn PrintWord of the Year 2021Word of the Year 2022Word of the Year 2023Word of the Year 2024 Develop DevelopDevelopDictionary APIDouble-Click LookupSearch WidgetsLicense Data About AboutAboutAccessibilityCambridge EnglishCambridge University Press & AssessmentCookies SettingsCookies and PrivacyCorpusTerms of Use © Cambridge University Press & Assessment 2025 Cambridge Dictionary +Plus My profile +Plus help Log out Dictionary Definitions Clear explanations of natural written and spoken English English Learner’s Dictionary Essential British English Essential American English Translations Click on the arrows to change the translation direction. 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12157	https://en.wikipedia.org/wiki/Lightning	Jump to content Lightning Afrikaans አማርኛ अंगिका العربية ܐܪܡܝܐ Armãneashti Asturianu Aymar aru Azərbaycanca Basa Bali বাংলা 閩南語 / Bn-lm-gí Башҡортса Беларуская Беларуская (тарашкевіца) भोजपुरी Bikol Central Български Bosanski Català Чӑвашла Cebuano Čeština Chavacano de Zamboanga ChiShona Corsu Cymraeg Dagbanli Dansk Deutsch Eesti Ελληνικά Emiliàn e rumagnòl Español Esperanto Estremeñu Euskara فارسی Français Frysk Gaeilge Gaelg Galego 贛語客家語 / Hak-k-ngî 한국어 Hawaiʻi Հայերեն हिन्दी Hrvatski Bahasa Hulontalo Ido Ilokano Bahasa Indonesia ᐃᓄᒃᑎᑐᑦ / inuktitut Iñupiatun Íslenska Italiano עברית Jawa ಕನ್ನಡ ქართული کٲشُر Kiswahili Kreyòl ayisyen Kurdî Кыргызча ລາວ Latina Latviešu Lëtzebuergesch Lietuvių Lingála Livvinkarjala Lombard Magyar Македонски Malagasy മലയാളം Māori मराठी Bahasa Melayu Minangkabau 閩東語 / Mìng-dĕ̤ng-ngṳ̄ မြန်မာဘာသာ Nāhuatl Nederlands Nedersaksies नेपाली 日本語 Norsk bokmål Norsk nynorsk Nouormand Occitan ଓଡ଼ିଆ Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ پنجابی ភាសាខ្មែរ Picard Polski Português Romnă Runa Simi Русский Саха тыла संस्कृतम् Scots Shqip Sicilianu සිංහල Simple English SiSwati Slovenčina Slovenščina Soomaaliga کوردی Српски / srpski Srpskohrvatski / српскохрватски Sunda Suomi Svenska Tagalog தமிழ் Taqbaylit Татарча / tatarça Tayal తెలుగు ไทย Тоҷикӣ Türkçe Українська اردو ئۇيغۇرچە / Uyghurche Vahcuengh Vèneto Tiếng Việt Walon Wayuunaiki 文言 Winaray 吴语 Xitsonga ייִדיש Yorùbá 粵語 Zazaki Žemaitėška 中文 Edit links From Wikipedia, the free encyclopedia Weather phenomenon involving electrostatic discharge For other uses, see Lightning (disambiguation). Not to be confused with lighting or thunder. "Lightening" redirects here. For other uses, see wikt:lightening. \| \| \| Part of a series on \| \| Weather \| \| Temperate and polar seasons Winter Spring Summer Autumn \| \| Tropical seasons Dry season + Harmattan Wet season \| \| Storms Cloud + Cumulonimbus cloud + Arcus cloud Downburst + Microburst + Heat burst + Derecho Lightning + Volcanic lightning Thunderstorm + Air-mass thunderstorm + Thundersnow + Dry thunderstorm Mesocyclone + Supercell Tornado + Anticyclonic tornado + Landspout + Waterspout Dust devil Fire whirl Anticyclone Cyclone Polar low Extratropical cyclone + European windstorm + Nor'easter Subtropical cyclone Tropical cyclone + Atlantic hurricane + Typhoon Storm surge Dust storm + Simoom + Haboob Monsoon + Amihan Gale Sirocco Firestorm Winter storm + Ice storm + Blizzard + Ground blizzard + Snow squall \| \| Precipitation Drizzle + Freezing Graupel Hail + Megacryometeor Ice pellets Diamond dust Rain + Freezing Cloudburst Snow + Rain and snow mixed + Snow grains + Snow roller + Slush \| \| Topics Air pollution Atmosphere + Chemistry + Convection + Physics + River Climate Cloud + Physics Fog + Mist + Season Cold wave Heat wave Jet stream Meteorology Severe weather + List + Extreme + Severe weather terminology - Canada - Japan - United States Weather forecasting Weather modification \| \| Glossaries Meteorology Climate change Tornado terms Tropical cyclone terms \| \| Weather portal \| \| v t e \| \| \| \| Part of a series of \| \| Geophysics \| \| \| \| Computational Exploration History Outline \| \| Electricity Ionosphere Polar wind Thunderstorms Lightning \| \| Fluid dynamics Atmospheric science Magnetohydrodynamics Oceanography Turbulence \| \| Geodynamics Climate Earth's mantle Exoplanetology Geochemistry Glaciology Planetary science Plate tectonics Tectonics Volcanism \| \| Gravity Geodesy Geoid Physical geodesy \| \| Magnetism Earth's magnetic field Geomagnetic reversal Magnetosphere Paleomagnetism Solar wind \| \| Waves Seismology Spectroscopy Vibration \| \| Geophysicists Aki Alfven Anderson Benioff Bowie Dziewonski Forbes Eotvos Gilbert Gutenberg Heiskanen Hotine von Humboldt Jeffreys Kanamori Love Matthews McKenzie Mercalli Molodenskii Munk Press Richter Turcotte Van Allen Vanicek Vening Meinesz Wegener Wilson \| \| v t e \| Lightning is a natural phenomenon consisting of electrostatic discharges occurring through the atmosphere between two electrically charged regions. One or both regions are within the atmosphere, with the second region sometimes occurring on the ground. Following the lightning, the regions become partially or wholly electrically neutralized. Lightning involves a near-instantaneous release of energy on a scale averaging between 200 megajoules and 7 gigajoules. The air around the lightning flash rapidly heats to temperatures of about 30,000 °C (54,000 °F). There is an emission of electromagnetic radiation across a wide range of wavelengths, some visible as a bright flash. Lightning also causes thunder, a sound from the shock wave which develops as heated gases in the vicinity of the discharge experience a sudden increase in pressure. The most common occurrence of a lightning event is known as a thunderstorm, though they can also commonly occur in other types of energetic weather systems, such as volcanic eruptions. Lightning influences the global atmospheric electrical circuit and atmospheric chemistry and is a natural ignition source of wildfires. Lightning is considered an Essential Climate Variable by the World Meteorological Organization, and its scientific study is called fulminology. Forms Three primary forms of lightning are distinguished by where they occur: Intra-cloud (IC) or in-cloud – Within a single thundercloud Cloud-to-cloud (CC) or inter-cloud – Between two clouds Cloud-to-ground (CG) – Between a cloud and the ground, in which case it is referred to as a lightning strike. Many other observational variants are recognized, including: volcanic lightning, which can occur during volcanic eruptions; "heat lightning", which can be seen from a great distance but not heard; dry lightning, which can cause forest fires; and ball lightning, which is rarely observed scientifically. The most direct effects of lightning on humans occur as a result of cloud-to-ground lightning, even though intra-cloud and cloud-to-cloud are more common. Intra-cloud and cloud-to-cloud lightning indirectly affect humans through their influence on atmospheric chemistry. There are variations of each type, such as "positive" versus "negative" CG flashes, that have different physical characteristics common to each which can be measured. Cloud to ground (CG) Cloud-to-ground (CG) lightning is a lightning discharge between a thundercloud and the ground. It is initiated by a stepped leader moving down from the cloud, which is met by a streamer moving up from the ground. CG is the least common, but best understood of all types of lightning. It is easier to study scientifically because it terminates on a physical object, namely the ground, and lends itself to being measured by instruments on the ground. Of the three primary types of lightning, it poses the greatest threat to life and property, since it terminates on the ground or "strikes". The overall discharge, termed a flash, is composed of a number of processes such as preliminary breakdown, stepped leaders, connecting leaders, return strokes, dart leaders, and subsequent return strokes. The conductivity of the electrical ground, be it soil, fresh water, or salt water, may affect the lightning discharge rate and thus visible characteristics. Positive and negative lightning Cloud-to-ground (CG) lightning is either positive or negative, as defined by the direction of the conventional electric current between cloud and ground. Most CG lightning is negative, meaning that a negative charge is transferred (electrons flow) downwards to ground along the lightning channel (the conventional current flows from the ground up to the cloud). The reverse happens in a positive CG flash, where electrons travel upward along the lightning channel, while also a positive charge is transferred downward to the ground (the conventional current travels from cloud to ground). Positive lightning is less common than negative lightning and on average makes up less than 5% of all lightning strikes. There are a number of mechanisms theorized to result in the formation of positive lightning. These are mainly based on movement or intensification of charge centres in the cloud. Such changes in cloud charging may come about as a result of variations in vertical wind shear or precipitation, or dissipation of the storm. Positive flashes may also result from certain behaviour of in-cloud discharges, e.g. breaking off or branching from existing flashes. Positive lightning strikes tend to be much more intense than their negative counterparts. An average bolt of negative lightning creates an electric current of 30,000 amperes (30 kA), transferring a total 15 C (coulombs) of electric charge and 1 gigajoule of energy. Large bolts of positive lightning can create up to 120 kA and transfer 350 C. The average positive ground flash has roughly double the peak current of a typical negative flash, and can produce peak currents up to 400 kA and charges of several hundred coulombs. Furthermore, positive ground flashes with high peak currents are commonly followed by long continuing currents, a correlation not seen in negative ground flashes. As a result of their greater power, positive lightning strikes are considerably more dangerous than negative strikes.[citation needed] Positive lightning produces both higher peak currents and longer continuing currents, making them capable of heating surfaces to much higher levels which increases the likelihood of a fire being ignited. The long distances positive lightning can propagate through clear air explains why they are known as "bolts from the blue", giving no warning to observers. Positive lightning has also been shown to trigger the occurrence of upward lightning flashes from the tops of tall structures and is largely responsible for the initiation of sprites several tens of kilometers above ground level. Positive lightning tends to occur more frequently in winter storms, as with thundersnow, during intense tornadoes and in the dissipation stage of a thunderstorm. Huge quantities of extremely low frequency (ELF) and very low frequency (VLF) radio waves are also generated. Contrary to popular belief, positive lightning flashes do not necessarily originate from the anvil or the upper positive charge region and strike a rain-free area outside of the thunderstorm. This belief is based on the outdated idea that lightning leaders are unipolar and originate from their respective charge region.[citation needed] Despite the popular misconception that flashes originating from the anvil are positive, because they seem to originate from the positive charge region, observations have shown that these are in fact negative flashes. They begin as IC flashes within the cloud, the negative leader then exits the cloud from the positive charge region before propagating through clear air and striking the ground some distance away. Cloud to cloud (CC) and intra-cloud (IC) Lightning discharges may occur between areas of cloud without contacting the ground. When it occurs between two separate clouds, it is known as cloud-to-cloud (CC) or inter-cloud lightning; when it occurs between areas of differing electric potential within a single cloud, it is known as intra-cloud (IC) lightning. IC lightning is the most frequently occurring type. IC lightning most commonly occurs between the upper anvil portion and lower reaches of a given thunderstorm. This lightning can sometimes be observed at great distances at night as so-called "sheet lightning". In such instances, the observer may see only a flash of light without hearing any thunder. Another term used for cloud–cloud or cloud–cloud–ground lightning is "Anvil Crawler", due to the habit of charge, typically originating beneath or within the anvil and scrambling through the upper cloud layers of a thunderstorm, often generating dramatic multiple branch strokes. These are usually seen as a thunderstorm passes over the observer or begins to decay. The most vivid crawler behavior occurs in well developed thunderstorms that feature extensive rear anvil shearing. Branching of cloud to cloud lightning, New Delhi, India. Multiple paths of cloud-to-cloud lightning, Swifts Creek, Australia. Intra-clouds lightning over the Baltic Sea. Cloud-to-cloud lightning, Albury, Australia Formation The processes involved in lightning formation fall into the following categories: Large-scale atmospheric phenomena in which charge separation can occur (e.g. storm) Microscopic and macroscopic processes that result in charge separation Establishment of an electric field Discharge through a lightning channel Atmospheric phenomena in which lightning occurs Lightning primarily occurs when warm air is mixed with colder air masses, resulting in atmospheric disturbances necessary for polarizing the atmosphere. The disturbances result in storms, and when those storms also result in lightning and thunder, they are called a thunderstorm. Lightning can also occur during dust storms, forest fires, tornadoes, volcanic eruptions, and even in the cold of winter, where the lightning is known as thundersnow. Hurricanes typically generate some lightning, mainly in the rainbands as much as 160 km (99 mi) from the center. Intense forest fires, such as those seen in the 2019–20 Australian bushfire season, can create their own weather systems that can produce lightning (also called Fire Lightning) and other weather phenomena. Intense heat from a fire causes air to rapidly rise within the smoke plume, causing the formation of pyrocumulonimbus clouds. Cooler air is drawn in by this turbulent, rising air, helping to cool the plume. The rising plume is further cooled by the lower atmospheric pressure at high altitude, allowing the moisture in it to condense into cloud. Pyrocumulonimbus clouds form in an unstable atmosphere. These weather systems can produce dry lightning, fire tornadoes, intense winds, and dirty hail. As well as the thermodynamic and dynamic conditions of the atmosphere, aerosol (e.g. dust or smoke) composition is thought to influence the frequency of lightning flashes in a storm. A specific example of this is that relatively high lightning frequency is seen along ship tracks. Airplane contrails have also been observed to influence lightning to a small degree. The water vapor-dense contrails of airplanes may provide a lower resistance pathway through the atmosphere having some influence upon the establishment of an ionic pathway for a lightning flash to follow. Rocket exhaust plumes provided a pathway for lightning when it was witnessed striking the Apollo 12 rocket shortly after takeoff. Thermonuclear explosions, by providing extra material for electrical conduction and a very turbulent localized atmosphere, have been seen triggering lightning flashes within the mushroom cloud. In addition, intense gamma radiation from large nuclear explosions may develop intensely charged regions in the surrounding air through Compton scattering. The intensely charged space charge regions create multiple clear-air lightning discharges shortly after the device detonates. Some high energy cosmic rays produced by supernovas as well as solar particles from the solar wind, enter the atmosphere and electrify the air, which may create pathways for lightning channels. Charge separation Charge separation in thunderstorms The details of the charging process are still being studied by scientists, but there is general agreement on some of the basic concepts of thunderstorm charge separation, also known as electrification. Electrification can be by the triboelectric effect leading to electron or ion transfer between colliding bodies. The main charging area in a thunderstorm occurs in the central part of the storm where air is moving upward rapidly (updraft) and temperatures range from −15 to −25 °C (5 to −13 °F); see Figure 1. In that area, the combination of temperature and rapid upward air movement produces a mixture of super-cooled cloud droplets (small water droplets below freezing), small ice crystals, and graupel (soft hail). The updraft carries the super-cooled cloud droplets and very small ice crystals upward. At the same time, the graupel, which is considerably larger and denser, tends to fall or be suspended in the rising air. The differences in the movement of the cloud particles cause collisions to occur. When the rising ice crystals collide with graupel, the ice crystals become positively charged and the graupel becomes negatively charged; see Figure 2. The updraft carries the positively charged ice crystals upward toward the top of the storm cloud. The larger and denser graupel is either suspended in the middle of the thunderstorm cloud or falls toward the lower part of the storm. Typically, the upper part of the thunderstorm cloud becomes positively charged while the middle to lower part of the thunderstorm cloud becomes negatively charged. The above process of charge separation as a result of cloud particle collisions is normally referred to as the non-inductive charging mechanism. The upward motions within the storm and winds at higher levels in the atmosphere tend to cause the small ice crystals (and positive charge) in the upper part of the thunderstorm cloud to spread out horizontally some distance from the thunderstorm cloud base. This part of the thunderstorm cloud is called the anvil. While this is the main charging process for the thunderstorm cloud, some of these charges can be redistributed by air movements within the storm (updrafts and downdrafts). In addition, there is a small but important positive charge buildup near the bottom of the thunderstorm cloud due to the precipitation and warmer temperatures. The positive-negative-positive charge regions commonly occur in mature thunderstorms, and referred to as the tripolar charge structure. There are also other charging processes that may play a role in thunderstorms, but are generally thought to be less important. An inductive charging mechanism has been studied, and would arise from the polarization of cloud droplets in the presence of the fair-weather electric field. It has also been stated that uncharged, colliding water-drops can become charged because of charge transfer between them (as aqueous ions) in an electric field as would exist in a thunderstorm. The induced separation of charge in pure liquid water has been known since the 1840s as has the electrification of pure liquid water by the triboelectric effect. William Thomson (Lord Kelvin) demonstrated that charge separation in water occurs in the usual electric fields at the Earth's surface and developed a continuous electric field measuring device using that knowledge. The physical separation of charge into different regions using liquid water was demonstrated by Kelvin with the Kelvin water dropper. The most likely charge-carrying species were considered to be the aqueous hydrogen ion and the aqueous hydroxide ion. An electron is not stable in liquid water concerning a hydroxide ion plus dissolved hydrogen for the time scales involved in thunderstorms. The electrical charging of solid water ice has also been considered. The charged species were again considered to be the hydrogen ion and the hydroxide ion. Establishing an electric field Main article: Thunderstorm In order for an electrostatic discharge to occur, two preconditions are necessary: first, a sufficiently high potential difference between two regions of space must exist, and second, a high-resistance medium must obstruct the free, unimpeded equalization of the opposite charges. The atmosphere provides the electrical insulation, or barrier, that prevents free equalization between charged regions of opposite polarity. Meanwhile, a thunderstorm can provide the charge separation and aggregation in certain regions of the cloud. When the local electric field exceeds the dielectric strength of damp air (about 3 MV/m), electrical discharge results in a strike, often followed by commensurate discharges branching from the same path. Mechanisms that cause the charges to build up to lightning are still a matter of scientific investigation. A 2016 study confirmed dielectric breakdown is involved. Lightning may be caused by the circulation of warm moisture-filled air through electric fields. Ice or water particles then accumulate charge as in a Van de Graaff generator. As a thundercloud moves over the surface of the Earth, an equal electric charge, but of opposite polarity, is induced on the Earth's surface underneath the cloud. The induced positive surface charge, when measured against a fixed point, will be small as the thundercloud approaches, increasing as the center of the storm arrives and dropping as the thundercloud passes. The referential value of the induced surface charge could be roughly represented as a bell curve. The oppositely charged regions create an electric field within the air between them. This electric field varies in relation to the strength of the surface charge on the base of the thundercloud – the greater the accumulated charge, the higher the electrical field. Electrical discharge as flashes and strikes The charge carrier in lightning is mainly electrons in a plasma. The process of going from charge as ions (positive hydrogen ion and negative hydroxide ion) associated with liquid water or solid water to charge as electrons associated with lightning must involve some form of electro-chemistry, that is, the oxidation and/or the reduction of chemical species. The best-studied and understood form of lightning is cloud to ground (CG) lightning. Although more common, intra-cloud (IC) and cloud-to-cloud (CC) flashes are difficult to study because there are no fixed points to monitor inside the clouds. Also, because of the very low probability of lightning striking the same point repeatedly and consistently, scientific inquiry is difficult even in areas of high CG frequency. Lightning leaders In a process not well understood, a bidirectional channel of ionized air, called a "leader", is initiated between oppositely-charged regions in a thundercloud. Leaders are electrically conductive channels of ionized gas that propagate through, or are otherwise attracted to, regions with a charge opposite of that of the leader tip. The negative end of the bidirectional leader fills a positive charge region, also called a well, inside the cloud while the positive end fills a negative charge well. Leaders often split, forming branches in a tree-like pattern. In addition, negative and some positive leaders travel in a discontinuous fashion, in a process called "stepping". The resulting jerky movement of the leaders can be readily observed in slow-motion videos of lightning flashes. It is possible for one end of the leader to fill the oppositely-charged well entirely while the other end is still active. When this happens, the leader end which filled the well may propagate outside of the thundercloud and result in either a cloud-to-air flash or a cloud-to-ground flash. In a typical cloud-to-ground flash, a bidirectional leader initiates between the main negative and lower positive charge regions in a thundercloud. The weaker positive charge region is filled quickly by the negative leader which then propagates toward the inductively-charged ground. The positively and negatively charged leaders proceed in opposite directions, positive upwards within the cloud and negative towards the earth. Both ionic channels proceed, in their respective directions, in a number of successive spurts. Each leader "pools" ions at the leading tips, shooting out one or more new leaders, momentarily pooling again to concentrate charged ions, then shooting out another leader. The negative leader continues to propagate and split as it heads downward, often speeding up as it gets closer to the Earth's surface. About 90% of ionic channel lengths between "pools" are approximately 45 m (148 ft) in length. The establishment of the ionic channel takes a comparatively long amount of time (hundreds of milliseconds) in comparison to the resulting discharge, which occurs within a few dozen microseconds. The electric current needed to establish the channel, measured in the tens or hundreds of amperes, is dwarfed by subsequent currents during the actual discharge. Initiation of the lightning leader is not well understood. The electric field strength within the thundercloud is not typically large enough to initiate this process by itself. Many hypotheses have been proposed. One hypothesis postulates that showers of relativistic electrons are created by cosmic rays and are then accelerated to higher velocities via a process called runaway breakdown. As these relativistic electrons collide and ionize neutral air molecules, they initiate leader formation. Another hypothesis involves locally enhanced electric fields being formed near elongated water droplets or ice crystals. Percolation theory, especially for the case of biased percolation, [clarification needed] describes random connectivity phenomena, which produce an evolution of connected structures similar to that of lightning strikes. A streamer avalanche model has recently been favored by observational data taken by LOFAR during storms. Upward streamers When a stepped leader approaches the ground, the presence of opposite charges on the ground enhances the strength of the electric field. The electric field is strongest on grounded objects whose tops are closest to the base of the thundercloud, such as trees and tall buildings. If the electric field is strong enough, a positively charged ionic channel, called a positive or upward streamer, can develop from these points. This was first theorized by Heinz Kasemir. As negatively charged leaders approach, increasing the localized electric field strength, grounded objects already experiencing corona discharge will exceed a threshold and form upward streamers. Attachment Once a downward leader connects to an available upward leader, a process referred to as attachment, a low-resistance path is formed and discharge may occur. Photographs have been taken in which unattached streamers are clearly visible. The unattached downward leaders are also visible in branched lightning, none of which are connected to the earth, although it may appear they are. High-speed videos can show the attachment process in progress. Discharge – Return stroke "Return stroke" redirects here. For other uses, see Return stroke (disambiguation). Once a conductive channel bridges the air gap between the negative charge excess in the cloud and the positive surface charge excess below, there is a large drop in resistance across the lightning channel. Electrons accelerate rapidly as a result in a zone beginning at the point of attachment, which expands across the entire leader network at up to one third of the speed of light. This is the "return stroke" and it is the most luminous and noticeable part of the lightning discharge. A large electric charge flows along the plasma channel, from the cloud to the ground, neutralising the positive ground charge as electrons flow away from the strike point to the surrounding area. This huge surge of current creates large radial voltage differences along the surface of the ground. Called step potentials,[citation needed] they are responsible for more injuries and deaths in groups of people or of other animals than the strike itself. Electricity takes every path available to it. Such step potentials will often cause current to flow through one leg and out another, electrocuting an unlucky human or animal standing near the point where the lightning strikes. The electric current of the return stroke averages 30 kiloamperes for a typical negative CG flash, often referred to as "negative CG" lightning. In some cases, a ground-to-cloud (GC) lightning flash may originate from a positively charged region on the ground below a storm. These discharges normally originate from the tops of very tall structures, such as communications antennas. The rate at which the return stroke current travels has been found to be around 100,000 km/s (one-third of the speed of light). A typical cloud-to-ground lightning flash culminates in the formation of an electrically conducting plasma channel through the air in excess of 5 km (3.1 mi) tall, from within the cloud to the ground's surface. The massive flow of electric current occurring during the return stroke combined with the rate at which it occurs (measured in microseconds) rapidly superheats the completed leader channel, forming a highly electrically conductive plasma channel. The core temperature of the plasma during the return stroke may exceed 27,800 °C (50,000 °F), causing it to radiate with a brilliant, blue-white color. Once the electric current stops flowing, the channel cools and dissipates over tens or hundreds of milliseconds, often disappearing as fragmented patches of glowing gas. The nearly instantaneous heating during the return stroke causes the air to expand explosively, producing a powerful shock wave which is heard as thunder. Discharge – Re-strike High-speed videos (examined frame-by-frame) show that most negative CG lightning flashes are made up of 3 or 4 individual strokes, though there may be as many as 30. Each re-strike is separated by a relatively large amount of time, typically 40 to 50 milliseconds, as other charged regions in the cloud are discharged in subsequent strokes. Re-strikes often cause a noticeable "strobe light" effect. To understand why multiple return strokes utilize the same lightning channel, one needs to understand the behavior of positive leaders, which a typical ground flash effectively becomes following the negative leader's connection with the ground. Positive leaders decay more rapidly than negative leaders do. For reasons not well understood, bidirectional leaders tend to initiate on the tips of the decayed positive leaders in which the negative end attempts to re-ionize the leader network. These leaders, also called recoil leaders, usually decay shortly after their formation. When they do manage to make contact with a conductive portion of the main leader network, a return stroke-like process occurs and a dart leader travels across all or a portion of the length of the original leader. The dart leaders making connections with the ground are what cause a majority of subsequent return strokes. Each successive stroke is preceded by intermediate dart leader strokes that have a faster rise time but lower amplitude than the initial return stroke. Each subsequent stroke usually re-uses the discharge channel taken by the previous one, but the channel may be offset from its previous position as wind displaces the hot channel. Since recoil and dart leader processes do not occur on negative leaders, subsequent return strokes very seldom utilize the same channel on positive ground flashes which are explained later in the article. Discharge – Transient currents during flash The electric current within a typical negative CG lightning discharge rises very quickly to its peak value in 1–10 microseconds, then decays more slowly over 50–200 microseconds. The transient nature of the current within a lightning flash results in several phenomena that need to be addressed in the effective protection of ground-based structures. Rapidly changing (alternating) currents tend to travel on the surface of a conductor, in what is called the skin effect, unlike direct currents, which "flow-through" the entire conductor like water through a hose. Hence, conductors used in the protection of facilities tend to be multi-stranded, with small wires woven together. This increases the total bundle surface area in inverse proportion to the individual strand radius, for a fixed total cross-sectional area. The rapidly changing currents also create electromagnetic pulses (EMPs) that radiate outward from the ionic channel. This is a characteristic of all electrical discharges. The radiated pulses rapidly weaken as their distance from the origin increases. However, if they pass over conductive elements such as power lines, communication lines, or metallic pipes, they may induce a current which travels outward to its termination. The surge current is inversely related to the surge impedance: the higher in impedance, the lower the current. This is the surge that, more often than not, results in the destruction of delicate electronics, electrical appliances, or electric motors. Devices known as surge protectors (SPD) or transient voltage surge suppressors (TVSS) attached in parallel with these lines can detect the lightning flash's transient irregular current, and, through alteration of its physical properties, route the spike to an attached earthing ground, thereby protecting the equipment from damage. Distribution, frequency and properties Main article: Distribution of lightning Global monitoring indicates that lightning on Earth occurs at an average frequency of approximately 44 (± 5) times per second, equating to nearly 1.4 billion flashes per year. Median duration is 0.52 seconds made up from a number of much shorter flashes (strokes) of around 60 to 70 microseconds. Occurrences are distributed unevenly across the planet with about 70% being over land in the tropics where atmospheric convection is the greatest. Many factors affect the frequency, distribution, strength and physical properties of a typical lightning flash in a particular region of the world. These factors include ground elevation, latitude, prevailing wind currents, relative humidity, and proximity to warm and cold bodies of water.[citation needed] Lightning is usually produced by cumulonimbus clouds, which have bases that are typically 1–2 km (0.62–1.24 mi) above the ground and tops up to 15 km (9.3 mi) in height. In general, CG lightning flashes account for only 25% of all total lightning flashes worldwide. The proportions of intra-cloud, cloud-to-cloud, and cloud-to-ground lightning may also vary by season at latitude. In the tropics, where the freezing level is generally higher in the atmosphere, only 10% of lightning flashes are CG. At the latitude of Norway (around 60° North latitude), where the freezing elevation is lower, 50% of lightning is CG. The place on Earth where lightning occurs most often is over Lake Maracaibo, wherein the Catatumbo lightning phenomenon produces 250 bolts of lightning a day. This activity occurs on average, 297 days a year. The second most lightning density is near the village of Kifuka in the mountains of the eastern Democratic Republic of the Congo, where the elevation is around 975 m (3,200 ft). On average, this region receives 158 lightning strikes per square kilometre per year (410/sq mi/yr). Other lightning hotspots include Singapore and Lightning Alley in Central Florida. Researchers at the University of Florida found that the final one-dimensional speeds of 10 flashes observed were between 1.0×105 and 1.4×106 m/s, with an average of 4.4×105 m/s. Megaflashes An unusually long lightning bolt is sometimes called a megaflash. The American Meteorological Society defines a megaflash as "a continuous mesoscale lightning flash with a horizontal path length of approximately 100 km or greater." Because of their long distance the duration of a megaflash is also long, typically more than 5 seconds. Megaflashes are caused by expansive electrified clouds that discharge slowly; these do not occur in ordinary thunderstorms, only in mesoscale convective systems. As of 2022, megaflashes have been observed only in the Great Plains of North America and the Río de la Plata Basin of South America. Though this is, in part, because only part of the world is observed with the kind of satellite instrument necessary to detect megaflashes. The launch of the Meteosat Third Generation satellite in 2022 increased that coverage to now include Africa and Europe. According to the World Meteorological Organization, on April 29, 2020, a bolt 768 km (477.2 mi) long was observed in the southern U.S.—sixty km (37 mi) longer than the previous distance record (southern Brazil, October 31, 2018). A single flash in Uruguay and northern Argentina on June 18, 2020, lasted for 17.1 seconds—0.37 seconds longer than the previous record (March 4, 2019, also in northern Argentina).In 2025, scientists discovered the longest observed flash occurred in October 2017 across Texas and Kansas, measuring 829 km (515 mi). The flash, which was detected by NOAA’s Geostationary Lightning Mapper, touched ground in five states. Extraterrestrial Lightning has been observed within the atmospheres of planets other than Earth, such as Jupiter, Saturn, and probably Uranus and Neptune. Lightning on Jupiter is far more energetic than on Earth, despite seeming to be generated via the same mechanism. Recently, a new type of lightning was detected on Jupiter, thought to originate from "mushballs" including ammonia. On Saturn, lightning, initially referred to as "Saturn Electrostatic Discharge", was discovered by the Voyager 1 mission. Lightning on Venus has been a controversial subject after decades of study. During the Soviet Venera and U.S. Pioneer missions of the 1970s and 1980s, signals suggesting lightning may be present in the upper atmosphere were detected. The short Cassini–Huygens mission fly-by of Venus in 1999 detected no signs of lightning, but radio pulses recorded by the spacecraft Venus Express (which began orbiting Venus in April 2006) may originate from lightning on Venus. Effects A lightning strike can unleash a variety of effects, some temporary, including very brief emission of light, sound and electromagnetic radiation, and some long-lasting, such as death, damage, and atmospheric and environmental changes. Injury, damage and destruction Main article: Lightning strike The immense amount of energy transferred in a lightning strike can have potentially devastating effect in a multitude of areas. To nature Objects struck by lightning experience heat and magnetic forces of great magnitude. Consequently: The heat created by lightning currents travelling through a tree may vaporize its sap, causing a steam explosion that rips off bark or even bursts the trunk. Similarly water in a fractured rock may be rapidly heated such that it splits further apart. A struck tree may catch fire, or a forest fire may be started. See also fire lightning below. As lightning travels through sandy soil, the soil surrounding the plasma channel may melt, forming tubular structures called fulgurites. To man-made structures and their contents Buildings or tall structures hit by lightning may be damaged as the lightning seeks unimpeded paths to the ground. By safely conducting a lightning strike to the ground, a lightning protection system, usually incorporating at least one lightning rod, can greatly reduce the probability of severe property damage. Surge protection devices (SPDs) can additionally or alternatively be used to help protect electrical installations from lightning induced electrical surges that risk damaging or destroying electrical equipment or starting a fire. Electrical fires obviously threaten not only structures but all assets, personal possessions, and living beings (people, pets and livestock) within. What, if any, protection system a building or structure requires is determined through a risk assessment. Threats to structures come not only from direct strikes to the structure itself, but also from direct or indirect strikes to connected electrically conductive services (electrical power lines; communication lines; water/gas pipes), or even to the surrounding area from which a surge may reach a service connection as it spreads out into the ground. To aircraft Aircraft are highly susceptible to being struck due to their metallic fuselages, but lightning strikes are generally not dangerous to them. Due to the conductive properties of aluminium alloy, the fuselage acts as a Faraday cage. Present day aircraft are built to be safe from a lightning strike and passengers will generally not even know that it has happened. However, there have been suspicions that lightning strikes can ignite fuel vapor and cause explosion,[citation needed] and nearby lightning can momentarily blind the pilot and cause permanent errors in magnetic compasses. To living beings Although 90 percent of people struck by lightning survive, humans and other animals struck by lightning may suffer severe injury due to internal organ and nervous system damage. Noise (thunder) Main article: Thunder Because the electrostatic discharge of terrestrial lightning superheats the air to plasma temperatures along the length of the discharge channel in a short duration, kinetic theory dictates gaseous molecules undergo a rapid increase in pressure and thus expand outward from the lightning creating a shock wave audible as thunder. Since the sound waves propagate not from a single point source but along the length of the lightning's path, the sound origin's varying distances from the observer can generate a rolling or rumbling effect. Perception of the sonic characteristics is further complicated by factors such as the irregular and possibly branching geometry of the lightning channel, by acoustic echoing from terrain, and by the usually multiple-stroke characteristic of the lightning strike. Thunder is heard as a rolling, gradually dissipating rumble because the sound from different portions of a long stroke arrives at slightly different times. Lightning at a sufficient distance may be seen and not heard; there is data that a lightning storm can be seen at over 160 km (100 miles) whereas the thunder travels about 32 km (20 miles). Anecdotally, there are many examples of people describing a 'storm directly overhead' or 'all-around' and yet 'no thunder'. Since thunderclouds can be up to 20 km (12 miles) high, lightning occurring high up in the cloud may appear close but is actually too far away to produce noticeable thunder. The distance approximation trick Light travels at about 300,000,000 m/s (980,000,000 ft/s), while sound only travels through air at about 343 m/s (1,130 ft/s). An observer can approximate the distance to the strike by timing the interval between the visible lightning and the audible thunder it generates. A lightning flash preceding its thunder by one second would be approximately 343 m (0.213 miles) away; thus a delay of three seconds would indicate a distance of about 1 km (0.62 miles); while a flash preceding thunder by five seconds would indicate a distance of roughly 1 mile (1.6 km). Consequently, a lightning strike observed at a very close distance will be accompanied by a sudden clap of thunder, with almost no perceptible time lapse, possibly accompanied by the smell of ozone (O3). Electromagnetic radiation and interference Electromagnetic waves are emitted in a variety of wavelengths, most obviously that of visible light – the big bright flash. This emitted radiation results partly from black-body radiation due to the temperature increase caused by electrical resistance of the air, and partly for other reasons that are still being actively researched. Radio frequency radiation Further information: Radio atmospheric, Whistler, and Schumann resonances Lightning discharges generate radio-frequency electromagnetic waves which can be received thousands of kilometers from their source. The discharge by itself is relatively simple short-lived dipole source that creates a single electromagnetic pulse with a duration of about 1 ms and a wide spectral density. In the absence in the nearby environment of materials with magnetic or electrical interaction properties, at a large distances in a far field zone, the electromagnetic wave will be proportional to the second derivation of the discharge current. This is what happens with high-altitude discharges or discharges over areas of a dry land. In other cases, the surrounding environment will change the shape of the source signal by absorbing some of its spectrum and converting it into a heat or re-transmitting it back as modified electromagnetic waves. High-energy radiation Further information: Terrestrial gamma-ray flash The production of X-rays by a bolt of lightning was predicted as early as 1925 by C.T.R. Wilson, but no evidence was found until 2001/2002, when researchers at the New Mexico Institute of Mining and Technology detected X-ray emissions from an induced lightning strike along a grounded wire trailed behind a rocket shot into a storm cloud. In the same year, University of Florida and Florida Tech researchers used an array of electric field and X-ray detectors at a lightning research facility in North Florida to confirm that natural lightning makes X-rays in large quantities during the propagation of stepped leaders. The cause of the X-ray emissions is still a matter for research, as the temperature of lightning is too low to account for the X-rays observed. A number of observations by space-based telescopes have revealed even higher energy gamma ray emissions, the so-called terrestrial gamma-ray flashes (TGFs). These observations pose a challenge to current theories of lightning, especially with the recent discovery of the clear signatures of antimatter produced in lightning. Recent research has shown that secondary species, produced by these TGFs, such as electrons, positrons, neutrons or protons, can gain energies of up to several tens of MeV. Environmental changes More permanent or longer-lasting environmental changes include the following. Atmospheric chemistry The very high temperatures generated by lightning lead to significant local increases in ozone and oxides of nitrogen. Each lightning flash in temperate and sub-tropical areas produces 7 kg of NOx on average. In the troposphere the effect of lightning can increase NOx by 90% and ozone by 30%. Ground fertilisation Lightning serves an important role in the nitrogen cycle by oxidizing diatomic nitrogen in the air into nitrates which are deposited by rain and can fertilize the growth of plants and other organisms. Induced permanent magnetism The movement of electrical charges produces a magnetic field (see electromagnetism). The intense currents of a lightning discharge create a fleeting but very strong magnetic field. Where the lightning current path passes through rock, soil, or metal these materials can become permanently magnetized. This effect is known as lightning-induced remanent magnetism, or LIRM. These currents follow the least resistive path, often horizontally near the surface but sometimes vertically, where faults, ore bodies, or ground water offers a less resistive path. One theory suggests that lodestones, natural magnets encountered in ancient times, were created in this manner. Lightning-induced magnetic anomalies can be mapped in the ground, and analysis of magnetized materials can confirm lightning was the source of the magnetization and provide an estimate of the peak current of the lightning discharge. Magnetic hallucinations Research at the University of Innsbruck has calculated that magnetic fields generated by plasma may induce hallucinations in subjects located within 200 m (660 ft) of a severe lightning storm, like what happened in Transcranial magnetic stimulation (TMS). Detection and monitoring Main article: Lightning detection The earliest detector invented to warn of the approach of a thunderstorm was the lightning bell. Benjamin Franklin installed one such device in his house. The detector was based on an electrostatic device called the 'electric chimes' invented by Andrew Gordon in 1742. Lightning discharges generate a wide range of electromagnetic radiations, including radio-frequency pulses. The Earth-ionosphere waveguide traps electromagnetic VLF- and ELF waves. Electromagnetic pulses transmitted by lightning strikes propagate within that waveguide. The waveguide is dispersive, which means that their group velocity depends on frequency. The difference of the group time delay of a lightning pulse at adjacent frequencies is proportional to the distance between transmitter and receiver. Together with direction-finding methods, this allows locating lightning strikes up to distances of 10,000 km from their origin. Moreover, the eigenfrequencies of the Earth-ionospheric waveguide, the Schumann resonances at about 7.5 Hz, are used to determine the global thunderstorm activity. A number of countries have installed nationwide lightning detector networks. The United States federal government has constructed a nationwide grid of such lightning detectors, allowing lightning discharges to be tracked in real time throughout the continental U.S. The EUCLID detection network is a combination of several national networks across Europe. Other examples of nations with lightning detection networks are India and Brazil. There are a range of global detection networks, which vary in their commercial and academic principles. Blitzortung (a private global detection system that consists of over 500 detection stations owned and operated by hobbyists/volunteers) provides near real-time lightning maps. The World Wide Lightning Location Network (WWLLN) is an academic led detection system. The Vaisala GLD360 network is a private enterprise. In addition to ground-based lightning detection, several instruments aboard satellites have been constructed to observe lightning distribution. Some of the first satellite-based observations were made in the late 1970s. The global and tropical long-term climatology of lightning has been observed by the Optical Transient Detector (OTD), aboard the OrbView-1 satellite launched on April 3, 1995, and the subsequent Lightning Imaging Sensor (LIS) aboard TRMM launched on November 28, 1997. In addition, the ISS carried a LIS instrument for three years from March 2017. Starting in 2016, the National Oceanic and Atmospheric Administration launched Geostationary Operational Environmental Satellite–R Series (GOES-R) weather satellites outfitted with Geostationary Lightning Mapper (GLM) instruments which are near-infrared optical transient detectors that can detect the momentary changes in an optical scene, indicating the presence of lightning. The lightning detection data can be converted into a real-time map of lightning activity across the Western Hemisphere; this mapping technique has been implemented by the United States National Weather Service. At the end of 2022, EUMETSAT launched the Lightning Imager (MTG-LI) on board the Meteosat Third Generation. This complements NOAA's GLM as MTG-LI will observe Europe and Africa. Artificial triggering Rocket-triggered : Lightning can be "triggered" by launching specially designed rockets trailing spools of wire into thunderstorms. The wire unwinds as the rocket ascends, creating an elevated ground that can attract descending leaders. If a leader attaches, the wire provides a low-resistance pathway for a lightning flash to occur. The wire is vaporized by the return current flow, creating a straight lightning plasma channel in its place. This method allows for scientific research of lightning to occur under a more controlled and predictable manner. The International Center for Lightning Research and Testing (ICLRT) at Camp Blanding, Florida typically uses rocket triggered lightning in their research studies. Laser-triggered : Since the 1970s, researchers have attempted to trigger lightning strikes by means of infrared or ultraviolet lasers, which create a channel of ionized gas through which the lightning would be conducted to ground. Such triggering of lightning is intended to protect rocket launching pads, electric power facilities, and other sensitive targets. : In New Mexico, U.S., scientists tested a new terawatt laser which provoked lightning. Scientists fired ultra-fast pulses from an extremely powerful laser thus sending several terawatts into the clouds to call down electrical discharges in storm clouds over the region. The laser beams sent from the laser make channels of ionized molecules known as filaments. Before the lightning strikes earth, the filaments lead electricity through the clouds, playing the role of lightning rods. Researchers generated filaments that lived a period too short to trigger a real lightning strike. Nevertheless, a boost in electrical activity within the clouds was registered. According to the French and German scientists who ran the experiment, the fast pulses sent from the laser will be able to provoke lightning strikes on demand. Statistical analysis showed that their laser pulses indeed enhanced the electrical activity in the thundercloud where it was aimed—in effect they generated small local discharges located at the position of the plasma channels. Impact of climate change and air pollution It is difficult to accurately predict changes in lightning due to climate change because it is challenging to simulate cloud physics variables that predict lightning (such as convection and cloud ice) in climate models. A large share of the world's lightning occurs over Africa. While there are regional variations in how climate change affects lightning across the continent, one study predicts a small increase in the total amount of lightning across the continent with warming. More specifically, the total number of lightning days per year is predicted to decrease, while more cloud ice and stronger convection leads to more lightning strikes occurring on days when lightning does occur. Lightning is much less common near the North and South Poles than in other regions. However, observations are beginning to show that lightning in the Arctic is increasing. and models suggest that climate change will continue to increase the frequency of lightning in the Arctic in future. The ratio of Arctic summertime lightning strikes has increased from 2010 to 2020 compared to the total lightning strikes in the world, indicating that the region is becoming more influenced by lightning. Lightning activity is increased by particulate emissions (a form of air pollution). However, this only occurs up to a point (aerosol optical depth = 0.3). Once this threshold is crossed, lightning is then suppressed by further increases in particulates. When lightning occurs, it generates rapid heating causing nitrogen and oxygen molecules in the atmosphere to break apart. This process leads to the formation of nitrogen oxides (NOx), which can subsequently result in the production of ozone, a greenhouse gas when occurring in the troposphere. However, lightning NOx also leads to increased amounts of hydroxyl (OH) and hydroperoxyl (HO2) radicals. These reactive molecules initiate chemical reactions that break down greenhouse gases like methane, effectively cleaning the atmosphere. Lightning and climate change feedbacks As lightning is influenced by climate change, there is a corresponding change to the lightning's influence on climate. These changes can lead to further climate change, thus creating a climate change feedback. Lightning leads to the production of tropospheric ozone and destruction of methane, both greenhouse gases and air pollutants. Therefore, the net impact of lightning on climate depends on the balance between this warming and cooling effect of the gases' effects on atmospheric chemistry. Predictions of this feedback can vary, resulting in either no change (net zero feedback), or a warming effect (positive feedback), depending on the method used to predict lightning. Lightning is the major natural cause of wildfire, estimated to cause 10% of forest fires worldwide. Wildfire can contribute to climate change. Because wildfires emit greenhouse gases, and also affect vegetation cover (which affects how much sunlight is reflected), a lightning-wildfire feedback is possible. Multiple studies suggest there could be an increase in Boreal and Arctic lightning-ignited fires in response to climate change. There is evidence that arctic lightning wildfire feedback may also influence vegetation and permafrost cover. The impact of lightning on fires in the tropics remains uncertain. In culture The first known photograph of lightning is from 1847, by Thomas Martin Easterly. The first surviving photograph is from 1882, by William Nicholson Jennings. Religion and mythology Further information: Lightning in religion In many cultures, lightning has been viewed as a sign or part of a deity or a deity in and of itself. These include the Greek god Zeus, the Aztec god Tlaloc, the Mayan God K, Slavic mythology's Perun, the Baltic Pērkons/Perkūnas, Thor in Norse mythology, Ukko in Finnish mythology, the Hindu god Indra, the Yoruba god Sango, Illapa in Inca mythology and the Shinto god Raijin. The ancient Etruscans produced guides to divining the future based on the omens supposedly displayed by thunder or lightning. Such use of thunder and lightning in divination is also known as ceraunoscopy, a kind of aeromancy. In the traditional religion of the African Bantu tribes, lightning is a sign of the ire of the gods. Scriptures in Judaism, Islam and Christianity also ascribe supernatural importance to lightning. In popular culture Although sometimes used figuratively, the idea that lightning never strikes the same place twice is a common myth. In fact, lightning can, and often does, strike the same place more than once. Lightning in a thunderstorm is more likely to strike objects and spots that are more prominent or conductive. For instance, lightning strikes the Empire State Building in New York City on average 23 times per year. In French and Italian, the expression for "Love at first sight" is coup de foudre and colpo di fulmine, respectively, which literally translated means "lightning strike". Political and military culture The bolt of lightning in heraldry is called a thunderbolt. This symbol usually represents power and speed. Some political parties use lightning flashes as a symbol of power, such as the People's Action Party in Singapore, the British Union of Fascists during the 1930s, and the National States' Rights Party in the United States during the 1950s. The Schutzstaffel, the paramilitary wing of the Nazi Party, used the Sig rune in their logo which symbolizes lightning. The German word Blitzkrieg, which means "lightning war", was a major offensive strategy of the German army during World War II. The lightning bolt is a common insignia for military communications units. A lightning bolt is also the NATO symbol for a signal asset. See also Environment portal Weather portal Lightning strike Volcanic lightning Paleolightning Harvesting lightning energy Keraunography Keraunomedicine – medical study of lightning casualties Lichtenberg figure Lightning injury Lightning-prediction system Roy Sullivan – Sullivan is recognized by Guinness World Records as the person struck by lightning more recorded times than any other human St. Elmo's fire Upper-atmospheric lightning Vela satellites – satellites which could record lightning superbolts ^ Maggio, Christopher R.; Marshall, Thomas C.; Stolzenburg, Maribeth (2009). 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Retrieved August 9, 2023. ^ Picture of John Kaspar of the National States Rights Party speaking in front of the party's lightning bolt flag (the flag was red, white, and blue) Archived February 3, 2013, at the Wayback Machine. Mauryk2.com (November 6, 2010). Retrieved on April 9, 2013. Rakov, Vladimir A.; Uman, Martin A. (2003). Lightning: Physics and effects. Cambridge, England: Cambridge University Press. ISBN 978-0521583275. Uman, Martin A. (1986). All About Lightning. Dover Publications, Inc. pp. 103–110. ISBN 978-0-486-25237-7. This article incorporates public domain material from Understanding Lightning: Thunderstorm Electrification. National Oceanic and Atmospheric Administration. Further reading Anders, André (2003). "Tracking Down the Origin of Arc Plasma Science I. Early Pulsed and Oscillating Discharges". IEEE Transactions on Plasma Science. 31 (4): 1052–1059. Bibcode:2003ITPS...31.1052A. doi:10.1109/TPS.2003.815476. OSTI 823201. S2CID 46204216. This is also available at Anders, A. (2003). "Energy Citations Database (ECD)" (PDF). IEEE Transactions on Plasma Science. 31 (5): 1052–1059. Bibcode:2003ITPS...31.1052A. doi:10.1109/TPS.2003.815476. OSTI 823201. S2CID 46204216. Retrieved September 5, 2008. Cooray, Vernon (2014). An Introduction to Lightning. Springer Verlag. doi:10.1007/978-94-017-8938-7. ISBN 978-94-017-8937-0. S2CID 127691542. Field, P. R.; W. H. Hand; G. Cappelluti; et al. (November 2010). "Hail Threat Standardisation" (PDF). European Aviation Safety Agency. Research Project EASA.2008/5. Archived from the original (PDF) on December 7, 2013. Gosline, Anna (May 2005). "Thunderbolts from space". New Scientist. 186 (2498): 30–34. Sample, in .PDF form, consisting of the book through page 20. "Effects of Lightning". The Mirror of Literature, Amusement, and Instruction. Vol. 12, no. 323. Columbia College, New York. July 19, 1828 – via Project Gutenberg. Early lightning research. External links "Lightning". Encyclopædia Britannica. Vol. 16 (11th ed.). 1911. p. 673. 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12158	https://math.stackexchange.com/questions/1351994/three-variable-second-degree-diophantine-equation	elementary number theory - Three variable, second degree diophantine equation - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Three variable, second degree diophantine equation Ask Question Asked 10 years, 2 months ago Modified7 years, 2 months ago Viewed 1k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am trying to solve this diophantine equation: x 2+y x+y 2=z 2 x 2+y x+y 2=z 2 In other words, I am trying to find integers x x and y y such that x 2+y x+y 2 x 2+y x+y 2 is a perfect square. So far, the only methods to solve quadratic diophantine equations I am familiar with are Pythagorean triples and Pell equations. The y x y x term has a coefficient of one, so I can't complete the square to reduce it to a Pell equation somehow, and I am wondering if there are other methods to solve this kind of equation. Some insight would be highly appreciated. elementary-number-theory diophantine-equations Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jul 7, 2015 at 3:25 AshuraAshura asked Jul 6, 2015 at 23:28 AshuraAshura 185 11 11 bronze badges 2 en.wikipedia.org/wiki/…Will Jagy –Will Jagy 2015-07-07 18:28:25 +00:00 Commented Jul 7, 2015 at 18:28 en.wikipedia.org/wiki/Eisenstein_tripleWill Jagy –Will Jagy 2015-07-07 18:31:06 +00:00 Commented Jul 7, 2015 at 18:31 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. We can generate solutions by a procedure similar to the one used to generate Pythagorean triples. Namely, it is equivalent to find rational points on the ellipse a 2+a b+b 2=1 a 2+a b+b 2=1. We do this by choosing a rational slope m m, taking the line through (1,−1)(1,−1) with slope m m, and finding the other intersection of this line with the ellipse. This procedure gives the solution a=−m 2+2 m m 2+m+1 a=−m 2+2 m m 2+m+1, b=2 m+1 m 2+m+1 b=2 m+1 m 2+m+1. Now, if m=p q m=p q, we get a solution a=−p 2+2 p q p 2+p q+q 2 a=−p 2+2 p q p 2+p q+q 2, b=2 p q+q 2 p 2+p q+q 2 b=2 p q+q 2 p 2+p q+q 2. Therefore, x=−(p 2+2 p q)x=−(p 2+2 p q), y=2 p q+q 2 y=2 p q+q 2, z=p 2+p q+q 2 z=p 2+p q+q 2 gives a solution to the original equation; and every solution in integers is a rational multiple of one of these solutions (since in reverse, the line through (1,−1)(1,−1) and (x z,y z)(x z,y z) has rational slope). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jul 18, 2018 at 17:44 Daniel ScheplerDaniel Schepler 23.1k 1 1 gold badge 25 25 silver badges 48 48 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical. Write the formula can someone come in handy. the equation: Y 2+a X Y+X 2=Z 2 Y 2+a X Y+X 2=Z 2 Has a solution: X=a s 2−2 p s X=a s 2−2 p s Y=p 2−s 2 Y=p 2−s 2 Z=p 2−a p s+s 2 Z=p 2−a p s+s 2 more: X=(4 a+3 a 2)s 2−2(2+a)p s−p 2 X=(4 a+3 a 2)s 2−2(2+a)p s−p 2 Y=(a 3−8 a−8)s 2+2(a 2−2)p s+a p 2 Y=(a 3−8 a−8)s 2+2(a 2−2)p s+a p 2 Z=(2 a 3+a 2−8 a−8)s 2+2(a 2−2)p s−p 2 Z=(2 a 3+a 2−8 a−8)s 2+2(a 2−2)p s−p 2 more: X=(a+4)p 2−2 p s X=(a+4)p 2−2 p s Y=3 p 2−4 p s+s 2 Y=3 p 2−4 p s+s 2 Z=(2 a+5)p 2−(a+4)p s+s 2 Z=(2 a+5)p 2−(a+4)p s+s 2 more: X=8 s 2−4 p s X=8 s 2−4 p s Y=p 2−(4−2 a)p s+a(a−4)s 2 Y=p 2−(4−2 a)p s+a(a−4)s 2 Z=−p 2+4 p s+(a 2−8)s 2 Z=−p 2+4 p s+(a 2−8)s 2 For the particular case: Y 2+X Y+X 2=Z 2 Y 2+X Y+X 2=Z 2 You can draw more formulas. X=3 s 2+2 p s X=3 s 2+2 p s Y=p 2+2 p s Y=p 2+2 p s Z=p 2+3 p s+3 s 2 Z=p 2+3 p s+3 s 2 more: X=3 s 2+2 p s−p 2 X=3 s 2+2 p s−p 2 Y=p 2+2 p s−3 s 2 Y=p 2+2 p s−3 s 2 Z=p 2+3 s 2 Z=p 2+3 s 2 In the equation: X 2+a X Y+b Y 2=Z 2 X 2+a X Y+b Y 2=Z 2 there is always a solution and one of them is quite simple. X=s 2−b p 2 X=s 2−b p 2 Y=a p 2+2 p s Y=a p 2+2 p s Z=b p 2+a p s+s 2 Z=b p 2+a p s+s 2 p,s p,s - integers asked us. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 7, 2015 at 4:31 individindivid 4,435 1 1 gold badge 16 16 silver badges 26 26 bronze badges 4 I was looking for something like this! Can all solutions be generated from each case? or should one use all of them together to generate all solutions?Ashura –Ashura 2015-07-07 04:34:19 +00:00 Commented Jul 7, 2015 at 4:34 @Ashura formally, we need to use all the formulas. But it is enough to use only General. Some formulas are mapped to each other.individ –individ 2015-07-07 04:38:13 +00:00 Commented Jul 7, 2015 at 4:38 Can you possibly direct me to some resource with more information about this? Just out of curiousity. Thanks!Ashura –Ashura 2015-07-07 04:47:19 +00:00 Commented Jul 7, 2015 at 4:47 @Ashura these formulas are a special case of a more General formulas. For example of such. math.stackexchange.com/questions/738446/…math.stackexchange.com/questions/794510/… Better in my blog to see. Many decisions have collected. artofproblemsolving.com/community/c3046individ –individ 2015-07-07 04:52:50 +00:00 Commented Jul 7, 2015 at 4:52 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Let z z be a fixed integer. We have x 2+y 2+x y=z 2 x 2+y 2+x y=z 2 is equivalent to (x+y)2−x y=z 2(x+y)2−x y=z 2 and (x−y)2+3 x y=z 2.(x−y)2+3 x y=z 2. Therefore, (x+y)2=z 2+x y≥0(x+y)2=z 2+x y≥0 and (x−y)2=z 2−3 x y≥0.(x−y)2=z 2−3 x y≥0. Then, we have z 2+x y z 2+x y and z 2−3 x y z 2−3 x y are squares. So, \|x+y\|=z 2+x y−−−−−−√\|x+y\|=z 2+x y and \|x−y\|=z 2−3 x y−−−−−−−√.\|x−y\|=z 2−3 x y. This implies that x+y=z 2+x y−−−−−−√x+y=z 2+x y or x+y=−z 2+x y−−−−−−√x+y=−z 2+x y and x−y=z 2−3 x y−−−−−−−√x−y=z 2−3 x y or x−y=−z 2−3 x y−−−−−−−√.x−y=−z 2−3 x y. You use after that the basic tools of calculus to find x x and y y separatly. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 7, 2015 at 0:19 Khadija MbarkiKhadija Mbarki 631 3 3 silver badges 13 13 bronze badges 1 I like what you did there, but it doesn't look like I can use those formulae to systematically generate all solutions to the equation above. Unless I'm missing something.Ashura –Ashura 2015-07-07 04:29:57 +00:00 Commented Jul 7, 2015 at 4:29 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions elementary-number-theory diophantine-equations See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 8Solutions to a x 2+b y 2=c z 2 a x 2+b y 2=c z 2 3a 2−a b+b 2=c 2 a 2−a b+b 2=c 2. Show prime factors of c c are of the form 6 k+1 6 k+1 0How to find multiple solutions for 3 variable, 2 degree Diophantine equation? 1Curves triangular numbers. 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12159	https://surgeryreference.aofoundation.org/orthopedic-trauma/adult-trauma/metacarpals/basic-technique/lag-screw-fixation-distal	Lag-screw fixation Login Authors of section Authors Fabio A Suarez, Aida Garcia Executive Editor Simon Lambert Open all credits Share Lag-screw fixation Select a chapter 1/5 – General considerations 2/5 – Approach 3/5 – Reduction 4/5 – Fixation 5/5 – Final assessment 1. General considerations Partial articular fractures require anatomical reduction and may be fixed with lag screws if the size of the fragment allows. The reduction can be assisted with arthroscopy if skill and equipment are available. The joint may collapse if there is impaction or comminution while the lag screw is tightened. In this case, plate fixation should be considered for chondral support. This fracture type may be associated with metacarpophalangeal (MCP) joint dislocation. In this case, the dislocation must be manipulated, and any interposed soft-tissue structures removed. Note: Axial traction may permit soft tissue to be interposed. This obstructs relocation and should therefore be avoided. Percutaneous vs open reduction and fixation Percutaneous reduction and fixation may be performed. The advantages are: Shorter operation time Less soft-tissue damage Faster mobilization This treatment option needs some skills and experience and special reduction forceps to avoid impingement of swollen soft tissue (atraumatic technique). These reduction forceps allow to hold the reduction at the planned screw placement, and drilling and screw insertion. There is no need for additional K-wire fixation or reduction forceps placement. If a percutaneous reduction is not achievable, the treatment can be changed to open surgery. 2. Approach For this procedure, a dorsal approach to the MCP jointcan be used. 3. Reduction Closed reduction Reduce the fracture indirectly by manual traction. The articular reduction can be confirmed with arthroscopy. Special reduction forceps designed for percutaneous fixation may be used to hold the reduction. Open reduction For more accurate reduction, use small pointed reduction forceps gently to manipulate the fracture. Application of excessive force can result in fragmentation. Confirm reduction with an image intensifier. Note: Anatomical reduction is important to prevent chronic instability or posttraumatic degenerative joint disease. Preliminary K-wire fixation Preliminarily fix the fragments by inserting a K-wire. Be careful to place it so it will not conflict with later screw placement. Stability evaluation Confirm reduction with an image intensifier and check the joint stability by flexion and extension. This should show congruent movement compared with the adjacent joints. 4. Fixation Planning for screw insertion Each lag screw must be inserted perpendicularly to the fracture plane. Do not insert screws too close to the fracture apex or the subchondral bone. A minimal distance from the fracture line, equal to the screw head diameter, must be observed. The screw insertion should avoid conflicts with the MCP ligaments. The screw length needs to be adequate for the screw to penetrate and purchase in the opposite (trans) cortex. Screw size selection The exact size of the diameter of the screws used will be determined by the fragment size and the fracture configuration. The various gliding and thread hole drill sizes for different screws are illustrated here. Pitfall: countersinking in the metaphysis Avoid countersinking in the metaphyseal regions, as the cortex is very thin and may be damaged. Screw insertion Insert the screw closest to the articular surface first. Cortical lag-screw insertion Alternate tightening of the two lag screws helps to avoid tilting the fragment and applies even compression forces across the fracture surface. 5. Final assessment Confirm anatomical reduction and fixation with an image intensifier. 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12160	https://math.stackexchange.com/questions/485264/agreed-upon-domain-of-inverse-trig-functions	trigonometry - Agreed upon domain of inverse trig functions - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Agreed upon domain of inverse trig functions Ask Question Asked 12 years ago Modified8 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Why do we choose [−π/2,π/2][−π/2,π/2] as the domain restriction for sin(x)sin⁡(x) when determining arcsin(x)arcsin⁡(x), clearly other restrictions are possible, which also yield a 1-1 function (like [π/2,3 π/2][π/2,3 π/2]). No pre-calculus book I have seen goes beyond saying that this choice is completely arbitrary. However, I have noticed that every domain restriction for inverse trig functions contains the whole first quadrant in its entirety (with the exception of x values where there are vertical asymptopes). I think these domain restrictions we choose are the only possible choices to get a 1-1 function on a domain that includes "effectively" the whole first quadrant. Is this the reason why we choose these domain restrictions? Are there other reasons why these domain restrictions are more natural than others that would yield a 1-1 function? trigonometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Sep 5, 2013 at 22:35 Applied Squared MathematicianApplied Squared Mathematician 835 1 1 gold badge 6 6 silver badges 23 23 bronze badges 5 Well, the interval is symmetric. Also [3 π/2,5 π/2][3 π/2,5 π/2] yields the same arguments as [−π/2,π/2][−π/2,π/2].AlexR –AlexR 2013-09-05 22:47:31 +00:00 Commented Sep 5, 2013 at 22:47 I'm not sure what you mean Alex; is an interval being symmetric a standard term? I've never heard this word used in this way. You could also make similar types of statements when adding 2 π 2 π to the endpoints of any interval.Applied Squared Mathematician –Applied Squared Mathematician 2013-09-05 23:32:07 +00:00 Commented Sep 5, 2013 at 23:32 The domain should include the "normal" angles between 0 0 and 90∘90∘, and be connected. Doesn't leave much choice for arcsin arcsin!André Nicolas –André Nicolas 2013-09-05 23:45:44 +00:00 Commented Sep 5, 2013 at 23:45 "Symmetric", when applied to an interval in R R, or more generally a region in R n R n, means that for every point x x in the region, the point −x−x is also in the region.G Tony Jacobs –G Tony Jacobs 2013-09-05 23:58:38 +00:00 Commented Sep 5, 2013 at 23:58 Oh I see, I thought Alex's second sentence was some how related to his definition of symmetry now I see that these were two separate statements. Thanks for the clarification.Applied Squared Mathematician –Applied Squared Mathematician 2013-09-06 20:14:59 +00:00 Commented Sep 6, 2013 at 20:14 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. It's nice to have a domain that includes 0 0. The domain of a trig function is the range of the corresponding inverse function, and it's nice to have 0 0 in the range. Think of applications. If I'm building something, and I want to set two pieces at some angle, I want an angle between 0 0 and π π, because that's what I can measure with my tools. That's one reason that it's nice to use intervals either centered on 0 0, or with 0 0 as an endpoint. Also, in calculus, we like to represent functions as power series, and these things are a lot easier to write down if we can center them at x=0 x=0. Of course, this isn't an explanation that would have much place in one of those Precalculus books that you mention, but it would seem strange to define the ranges of inverse trig functions to be one thing in Precal and something else in Calculus. There may be other reasons, but I think these two are compelling. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 11, 2016 at 14:57 answered Sep 5, 2013 at 23:00 G Tony JacobsG Tony Jacobs 32.1k 4 4 gold badges 58 58 silver badges 115 115 bronze badges 3 This is a great point, but for say cos(x) [−π,0][−π,0] would also satisfy your condition that 0 should be included, although it would be an odd choice to have a completely non positive interval. I just pulled out my old calculus text book and there is also no explanation in there for the domain restriction (besides the standard 1-1 statements). I would up-vote this post, but unfortunately I am a newbie who doesn't have enough rep to do that.Applied Squared Mathematician –Applied Squared Mathematician 2013-09-05 23:24:19 +00:00 Commented Sep 5, 2013 at 23:24 Yeah, values of cos−1 cos−1 are angles, and insofar as angles are physical quantities, it's natural for them to be non-negative. I don't know why they don't explain this in more textbooks.G Tony Jacobs –G Tony Jacobs 2013-09-05 23:27:17 +00:00 Commented Sep 5, 2013 at 23:27 I have the same doubt about the domains of inverse trigonometric functions. Unfortunatelly I would like a more objective answer but it seems that it is a conveniente convention.Cleto Pereira –Cleto Pereira 2019-06-27 20:16:36 +00:00 Commented Jun 27, 2019 at 20:16 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions trigonometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 0Clarification about domains of inverse trigonometric functions 0What is so special about the principal value branch of inverse trigonometric functions. Related 3Intuition around why domain of x of arcsine and arccosine is [-1;1] for "real result" & domain for arctangent is all real numbers 1Do I need to use different trig functions in different quadrants? 4when to use restrictions (domain and range) on trig functions 3Walk me through some basic trig algebra? 0Inverse Trig Functions Domain Restrictions 1How does one define restrictions on x x or θ θ when simplifying trigonometric identities? 2Why is not f(x)=arcsin(2 x 1−x 2−−−−−√)f(x)=arcsin⁡(2 x 1−x 2) a one to one function? 1Convert pair of parametric trig equations to y=f(x)y=f(x) form Hot Network Questions Fundamentally Speaking, is Western Mindfulness a Zazen or Insight Meditation Based Practice? Do we need the author's permission for reference Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation Are there any legal strategies to modify the nature and rules of the UN Security Council in the face of one, or more SC vetos? 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12161	https://brainly.com/question/187754	[FREE] List all of the prime numbers less than 20 in numerical order. - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +84,6k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +20,5k Ace exams faster, with practice that adapts to you Practice Worksheets +5,9k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified List all of the prime numbers less than 20 in numerical order. 2 See answers Explain with Learning Companion NEW Asked by Barakah • 11/16/2014 0:06 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 226646060 people 226M 5.0 10 Upload your school material for a more relevant answer a prime number is a number only divisible by itself and 1 (this excludes decimals and fractions) surprise, 1 is not prime because it must be divisible by itself and 1 and the factors of 1 are 1 and 1 so 1 is not prime 2 is prime because it is the first even counting number and since all even numbers after 2 are divisble by 2 you can cross out all odd numbers after 2 3 is prime because it is the first odd counting number that satisfies the condition 4 is not because it is even 5 is prime 6 is not because it is even 7 is prime 8 is not because it is even 9 is not 10 is not because it is even 11 is prime 12 is not because it is even 13 is prime 14 is not because it is even 15 is not 16 is not because it is even 17 is prime 18 is not because it is even 19 is prime 20 is not because it is even so the numbers are 2,3,5,7,11,13,17,19 Answered by apologiabiology •9.2K answers•226.6M people helped Thanks 10 5.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 226646060 people 226M 5.0 10 The Basics of General, Organic, and Biological Chemistry - David W Ball How History is Made: A Student’s Guide to Reading, Writing, and Thinking in the Discipline - Stephanie Cole, Kimberly Breuer, Scott W. Palmer Corpus Linguistics: A Guide to the Methodology - Anatol Stefanowitsch Upload your school material for a more relevant answer The prime numbers less than 20 are: 2, 3, 5, 7, 11, 13, 17, and 19. A prime number is defined as a number that is greater than 1 and is only divisible by 1 and itself. Therefore, any composite number or number less than 2 is not included in this list. Explanation To find all the prime numbers less than 20, we first need to understand what a prime number is. A prime number is a whole number greater than 1 that is not divisible by any other numbers except for 1 and itself. Let's list the numbers from 1 to 19 and determine which of them are prime by checking their divisibility: 1 is not a prime number because it has only one divisor (itself). 2 is prime because it can only be divided by 1 and 2. 3 is prime as it can only be divided by 1 and 3. 4 is not prime because it can be divided by 1, 2, and 4 (it has more than two divisors). 5 is prime (only divisible by 1 and 5). 6 is not prime (divisible by 1, 2, 3, and 6). 7 is prime (only divisible by 1 and 7). 8 is not prime (divisible by 1, 2, 4, and 8). 9 is not prime (divisible by 1, 3, and 9). 10 is not prime (divisible by 1, 2, 5, and 10). 11 is prime (only divisible by 1 and 11). 12 is not prime (divisible by 1, 2, 3, 4, 6, and 12). 13 is prime (only divisible by 1 and 13). 14 is not prime (divisible by 1, 2, 7, and 14). 15 is not prime (divisible by 1, 3, 5, and 15). 16 is not prime (divisible by 1, 2, 4, 8, and 16). 17 is prime (only divisible by 1 and 17). 18 is not prime (divisible by 1, 2, 3, 6, 9, and 18). 19 is prime (only divisible by 1 and 19). After evaluating all the numbers, the prime numbers less than 20 are: 2, 3, 5, 7, 11, 13, 17, 19. Examples & Evidence For example, the number 11 is prime because it cannot be divided evenly by any whole number other than 1 and 11 itself. In contrast, the number 12 is not prime because it can be divided by 1, 2, 3, 4, 6, and 12, having more than two divisors. According to the definition of prime numbers, any number that has exactly two distinct positive divisors qualifies as prime. This is well-established in mathematics and can be verified by attempting to divide each number by integers greater than 1 and less than the number itself. Thanks 10 5.0 (1 vote) Advertisement Community Answer This answer helped 363646379 people 363M 3.5 36 19, 17, 13, 11, 7, 5, 3, 2 Answered by AL2006 •15.1K answers•363.6M people helped Thanks 36 3.5 (11 votes) 2 Advertisement ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them? Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216 Choose an equivalent expression for 1 0 6÷1 0 4. A. 1 0 2 B. 1 0 3 C. 1 0 10 D. 1 0 24 How would you write 1 2−3 using a positive exponent? A. 1 2 3 B. 1 2 0 C. 1 1 2 3 D. 1 2 3 1 Is this equation correct? 6 3⋅7 3=4 2 3 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
12162	https://oercommons.org/courseware/lesson/87594/overview	Preview Please log in to save materials. Log in Report Details Resource Library Author: : Anna McCollum, ALka Sharma, Jillian Gorrell, Amanda Spangler, Madonna Kemp Subject: : Agriculture, Biology Material Type: : Diagram/Illustration, Textbook Level: : High School Tags: : - Plant Science Log in to add tags to this item. License: : Creative Commons Attribution Non-Commercial Share Alike Language: : English Media Formats: : Graphics/Photos Show More Show Less Version History PDF 2.2 Chloroplast Download View PDF 2.3 Light Dependent Reaction Download View PDF 2.4 Light Independent Reaction Download View PDF 2.5 C4 & CAM Pathways as Means of Reducing Photorespiration Download View PDF 2.6 How Environmental Factors Affect Photosynthesis Download View PDF 2.7 The Energy Cycle Download View PDF 2_Photosynthesis Download View Statewide Dual Credit Introduction to Plant Science Plant Function Plant Form Plant Function Plant Reproduction and Propagation Soil, Mediums, and Plant Nutrition Plant Classification and Use Nursery Production Controlled Environment Production Plant Injuries and Their Control/Integrated Pest Management Impact of Plants and Horticulture on People Photosynthesis Cellular Respiration Photosynthesis Water Movement in Xylem Phloem Translocation Plant Hormones Photosynthesis Overview An in-depth overview of photosynthesis, detailing the structure and function of chloroplasts, the light-dependent and light-independent reactions, and the adaptations of plants to various environmental conditions.Also explains the importance of photosynthesis in the energy cycle and its role in supporting life on Earth. Did you have an idea for improving this content? We’d love your input. Introduction Learning Objectives Explain the structure of chloroplast. Describe the steps in the light-dependent reaction of photosynthesis. Describe the steps in the light-independent reaction of photosynthesis. Differentiate between C3, C4, and CAM photosynthesis. List the effects of high-intensity and low-intensity light on photosynthesis. Explain how plants adapt to changes in solar intensity. List the drought and flood adaptations. Explain how temperature, CO2 concentration, and air movement affect photosynthesis. Key Terms 3-phosphoglycerate - the first compound formed after CO2 assimilation by Rubisco Accessary pigment - light-absorbing pigments other than chlorophyll a C3 plants - plants in which the first product of CO2 fixation by Rubisco is a 3 carbon compound C4 plants -plants in which the first product of CO2 fixation is a 4 carbon compound CAM plants -plants that temporally separate light-dependent and light-independent reactions of photosynthesis Calvin cycle -light-independent reactions of photosynthesis that convert carbon dioxide from the atmosphere into carbohydrates using the energy and reducing power of ATP and NADPH Carbon fixation - a process of converting inorganic CO2 gas into organic compounds Carotenoid -photosynthetic pigment (yellow-orange-red) that functions to dispose of excess energy Chlorophyll - Two kinds of chlorophyll, athatabsorb violet-blue and red light and consequently have a bluish-green color; the only pigment molecule that performs the photochemistry by getting excited and losing an electron to the electron transport chain; chlorophyll b is an accessory pigment that absorbs blue and red-orange light and consequently has a yellowish-green tint. Chloroplast -organelle in which photosynthesis takes place Cyclic photophosphorylation - ATP production by cyclic movement of electron through photosystem I Drought avoidance - adaptations in a plant that allow it to avoid drought conditions Drought tolerance - adaptations in a plant that allow it to survive under drought conditions Light-dependent reaction - thefirst stage of photosynthesis where certain wavelengths of the visible light are absorbed to form two energy-carrying molecules (ATP and NADPH) Light independent reaction -thesecond stage of photosynthesis, through which carbon dioxide is used to build carbohydrate molecules using energy from ATP and NADPH Light intensity -number of photons falling on a unit area of the leaf surface in unit time NADPH - thehigh-energy molecule Non-cyclic photophosphorylation -ATP and NADPH production by the movement of electrons through photosystem I and photosystem II Photolysis - the splitting of a water molecule in presence of sunlight Photosynthesis - the process by which autotrophs use sunlight, water, and CO2 to produce sugars Photosystem I -integral pigment and protein complex in thylakoid membranes that uses light energy to transport electrons from plastocyanin to NADP+ (which becomes reduced to NADPH in the process) Photosystem II -integral protein and pigment complex in thylakoid membranes that transports electrons from water to the electron transport chain; oxygen is a product of PSII Pigment - amolecule that is capable of absorbing certain wavelengths of light and reflecting others (which accounts for its color) secondary Pigment - same as an accessory pigment The metabolic processes in all organisms—from bacteria to humans—require energy. To get this energy, many organisms access stored energy by eating, that is, by ingesting other organisms. But where does the stored energy in food originate? All of this energy can be traced back to photosynthesis. Photosynthesisis essential to all life on earth; both plants and animals depend on it. It is the only biological process that can capture the energy that originates from sunlight and converts it into chemical compounds (carbohydrates) that every organism uses to power its metabolism. It is also a source of oxygen necessary for many living organisms. In brief, the energy of sunlight is “captured” to energize electrons, whose energy is then stored in the covalent bonds of sugar molecules. How long-lasting and stable are those covalent bonds? The energy extracted today by the burning of coal and petroleum products represents sunlight energy captured and stored by photosynthesis 350 to 200 million years ago during the Carboniferous Period. Plants, algae, and a group of bacteria called cyanobacteria are the only organisms capable of performing photosynthesis (figure 2.2.2). Because they use light to manufacture their own food, they are called photoautotrophs (literally, “self-feeders using light”). Other organisms—such as animals, fungi, and most other bacteria—are termed heterotrophs (“other feeders”), because they must rely on the sugars produced by photosynthetic organisms for their energy needs. A third very interesting group of bacteria synthesize sugars not by using sunlight’s energy but by extracting energy from inorganic chemical compounds. For this reason, they are referred to as chemoautotrophs. The importance of photosynthesis is not just that it can capture sunlight’s energy. In contrast, photosynthesis is vital because it evolved as a way to store the energy from solar radiation (the “photo-” part) to energy in the carbon-carbon bonds of carbohydrate molecules (the “-synthesis” part). Those carbohydrates are the energy source that heterotrophs use to power the synthesis of ATP via respiration. Therefore, photosynthesis powers 99 percent of Earth’s ecosystems. When a top predator, such as a wolf, preys on a deer, the wolf is at the end of an energy path that went from nuclear reactions on the surface of the sun, to visible light, to photosynthesis, to vegetation, to deer, and finally to the wolf. Photosynthesis is a multi-step process that requires specific wavelengths of visible sunlight, carbon dioxide (which is low in energy), and water as substrates (figure 2.2.3). After the process is complete, it releases oxygen and produces glyceraldehyde-3-phosphate (G3P), as well as simple carbohydrate molecules (high in energy) that can then be converted into glucose, sucrose, or any of dozens of other sugar molecules. These sugar molecules contain energy and the energized carbon that all living things need to survive. The following is the chemical equation for photosynthesis (figure 2.2.4): Although the equation in Figure 2.2.4. looks simple, the many steps that take place during photosynthesis are quite complex. Before learning the details of how photoautotrophs turn sunlight into food, it is important to become familiar with the structures involved. Access for free at PDF 2_Photosynthesis Download View Chloroplast In plants, photosynthesis generally takes place in leaves, which consist of several layers of cells. The process of photosynthesis occurs in a middle layer called the mesophyll. The gas exchange of carbon dioxide and oxygen occurs through small, regulated openings called stomata (singular: stoma), which also play roles in the regulation of gas exchange and water balance. The stomata are typically located on the underside of the leaf, which helps to minimize water loss due to high temperatures on the upper surface of the leaf. Each stoma is flanked by guard cells that regulate the opening and closing of the stomata by swelling or shrinking in response to osmotic changes. In all autotrophic eukaryotes, photosynthesis takes place inside an organelle called a chloroplast. For plants, chloroplast-containing cells exist mostly in the mesophyll. Chloroplasts have a double membrane envelope (composed of an outer membrane and an inner membrane) and are ancestrally derived from ancient free-living cyanobacteria. Within the chloroplast are stacked, disc-shaped structures called thylakoids. Embedded in the thylakoid membrane is chlorophyll, a pigment (a molecule that absorbs light) responsible for the initial interaction between light and plant material, and numerous proteins that make up the electron transport chain. The thylakoid membrane encloses an internal space called the thylakoid lumen. As shown in Figure 2.2.5, a stack of thylakoids is called a granum, and the liquid-filled space surrounding the granum is called stroma or “bed” (not to be confused with stoma or “mouth,” an opening on the leaf epidermis). The Two Parts of Photosynthesis Photosynthesis takes place in two sequential stages: light dependent reactions and light independent reactions. In light-dependent reactions, energy from sunlight is absorbed by chlorophyll and that energy is converted into stored chemical energy. In light-independentreactions, the chemical energy harvested during the light-dependent reactions drives the assembly of sugar molecules from carbon dioxide. Therefore, although the light-independent reactions do not use light as a reactant, they require the products of the light-dependent reactions to function. In addition, however, several enzymes of the light-independent reactions are activated by light. The light-dependent reactions utilize certain molecules to temporarily store the energy: these are referred to as energy carriers. The energy carriers that move energy from light-dependent reactions to light-independent reactions can be thought of as “full” because they are rich in energy. After the energy is released, the “empty” energy carriers return to the light-dependent reaction to obtain more energy. Figure 2.2.6 illustrates the components inside the chloroplast where light-dependent and light-independent reactions take place. Learn more about photosynthesis. Access for free at PDF 2.2 Chloroplast Download View Light Dependent Reaction How can light energy be used to make food? When a person turns on a lamp, electrical energy becomes light energy. Like all other forms of kinetic energy, light can travel, change its form, and be harnessed to do work. In the case of photosynthesis, light energy is converted into chemical energy, which photoautotrophs use to build basic carbohydrate molecules (Figure 2.2.7). However, autotrophs only use a few specific wavelengths of sunlight. What Is Light Energy? The sun emits an enormous amount of electromagnetic radiation (solar energy in a spectrum from very short gamma rays to very long radio waves). How solar energy travels is described as waves. Scientists can determine the amount of energy of a wave by measuring its wavelength(represented by the Greek symbol lambda λ)—the distance between consecutive crest points of a wave (crest to crest or from trough to trough (figure 2.2.8). Thefrequency of a wavelength (represented by the Greek symbol nu n) is the number of crests or troughs passing a fixed point in unit time. Thus, for any electromagnetic wave, wavelength times frequency is equal to the speed of light (represented as C, equal to 3.0 X 108 m/s). C = λn Shorter wavelengths have more energy than longer wavelengths. The longer the wavelength, the less energy it carries. This may seem illogical but think of it in terms of a piece of moving heavy rope. It takes little effort by a person to move a rope in long, wide waves. To make a rope move in short, tight waves, a person would need to apply significantly more energy. Visible light constitutes only one of many types of electromagnetic radiation emitted from the sun and other stars. Scientists differentiate the various types of radiant energy from the sun within the electromagnetic spectrum. The electromagnetic spectrum is the range of all possible frequencies of radiation (Figure 2.2.9). The difference between wavelengths relates to the amount of energy carried by them. Electromagnetic radiations also exist as particles, called photons. All photons carry a definite amount of energy called quantum based on the wavelength. Photons are like packets of energy of an electromagnetic wave. Thus, frequency times Planck’s constant gives us the value of energy a photon is carrying. E = hn Where, h represents Planck’s constant (6.626 X 10-34 J s), n represents the frequency of the light. You can visualize photons as packet of sauces that you get at any fast-food restaurant, the spiciness depends on the kind of sauces that you picked (mild, medium, or hot). Each type of electromagnetic radiation travels at a particular wavelength. The electromagnetic spectrum (Figure 2.2.9) shows several types of electromagnetic radiation originating from the sun, including X-rays and ultraviolet (UV) rays. The higher-energy waves can penetrate tissues and damage cells and DNA, which explains why both X-rays and UV rays can be harmful to all living organisms. Absorption of Light Light energy initiates the process of photosynthesis when pigments absorb specific wavelengths of visible light. Organic pigments, of the chloroplast thylakoid, have a narrow range of energy levels that they can absorb. Energy levels lower than those represented by red light are insufficient to raise an orbital electron to an excited (quantum) state. Energy levels higher than those in blue light will physically tear the molecules apart, in a process called bleaching. Our retinal pigments can only “see” (absorb) wavelengths between 700 nm and 400 nm of light, a spectrum that is therefore called visible light. For the same reasons, pigment molecules in plants, absorb only light in the wavelength range of 700 nm to 400 nm; plant physiologists refer to this range for plants as photosynthetically active radiation (Figure 2.2.10). The visible light is seen by humans as white light exists in a rainbow of colors. Certain objects, such as a prism or a drop of water, disperse white light to reveal the colors to the human eye. The visible light portion of the electromagnetic spectrum shows the rainbow of colors, with violet and blue having shorter wavelengths, and therefore higher energy. At the other end of the spectrum toward red, the wavelengths are longer and have lower energy (Figure 2.2.11). Understanding Pigments Different kinds of pigments exist, and each absorbs only specific wavelengths (colors) of visible light. Pigments reflect or transmit the wavelengths they cannot absorb, making them appear a mixture of the reflected or transmitted light colors. Chlorophylls and carotenoids are the two major classes of photosynthetic pigments found in plants and algae; each class has multiple types of pigment molecules. There are five major chlorophylls: a, b, c, and d, as well as a related molecule found in prokaryotes called bacteriochlorophyll. Chlorophyll a and chlorophyll b are found in the chloroplast of higher plants. With dozens of different forms, carotenoids are a much larger group of pigments. The carotenoids found in fruit—such as the red of tomato (lycopene), the yellow of corn seeds (zeaxanthin), or the orange of an orange peel (β-carotene)—are used as advertisements to attract seed dispersers. In photosynthesis, carotenoids function as photosynthetic pigments that are very efficient molecules for the disposal of excess energy. When a leaf is exposed to full sun, the light-dependent reactions are required to process an enormous amount of energy; if that energy is not handled properly, it can do significant damage. Therefore, many carotenoids reside in the thylakoid membrane, absorb excess energy, and safely dissipate that energy as heat. Each type of pigment can be identified by the specific pattern of wavelengths it absorbs from visible light: This is termed the absorption spectrum. The graph in Figure 2.2.12 shows the absorption spectra for chlorophyll a, chlorophyll b, and a type of carotenoid pigment called β-carotene (which absorbs blue and green light). Notice how each pigment has a distinct set of peaks and troughs, revealing a highly specific pattern of absorption. Chlorophyll a absorbs wavelengths from either end of the visible spectrum (blue and red), but not green. Because green is reflected or transmitted, chlorophyll appears green. Carotenoids absorb in the short-wavelength blue region and reflect the longer yellow, red, and orange wavelengths. Many photosynthetic organisms have a mixture of pigments, and by using these pigments, the organism can absorb energy from a wider range of wavelengths. Not all photosynthetic organisms have full access to sunlight. Some organisms grow underwater where light intensity and quality decrease and change with depth. Other organisms grow in competition for light. Plants on the rainforest floor must be able to absorb any bit of light that comes through; this is because the taller trees absorb most of the sunlight and scatter the remaining solar radiation (Figure 2.2.13). When studying a photosynthetic organism, scientists can determine the types of pigments present by generating absorption spectra. An instrument called a spectrophotometer can differentiate which wavelengths of light a substance can absorb. A spectrophotometer measure transmitted light and compute from it the absorption. By extracting pigments from leaves and placing these samples into a spectrophotometer, scientists can identify which wavelengths of light an organism can absorb. Additional methods for the identification of plant pigments include various types of chromatography that separate the pigments by their relative affinities to solid and mobile phases. How Light-Dependent Reactions Work The overall function of light-dependent reactions is to convert solar energy into chemical energy in the form of NADPH and ATP. This chemical energy supports light-independent reactions and fuels the assembly of sugar molecules. The light-dependent reactions are depicted in Figure 2.2.14. Protein complexes and pigment molecules work together to produce NADPH and ATP. The numbering of the photosystems is derived from the order in which they were discovered, not in the order of the transfer of electrons. The actual step that converts light energy into chemical energy takes place in a multiprotein complex called a photosystem, two types of which are found embedded in the thylakoid membrane: photosystem II (PSII) and photosystem I (PSI) (Figure 2.2.15). The two complexes differ on the basis of what they oxidize (that is, the source of the low-energy electron supply) and what they reduce (the place to which they deliver their energized electrons). Both photosystems have the same basic structure; a number of antenna proteins to which the chlorophyll molecules are bound surrounding the reaction center where the photochemistry takes place. Each photosystem is serviced by the light-harvesting complex, which passes energy from sunlight to the reaction center; it consists of multiple antenna proteins that contain a mixture of 300 to 400 chlorophyll a and b molecules, as well as other pigments like carotenoids. The absorption of a single photon or distinct quantity or “packet” of light by any of the chlorophylls pushes that molecule into an excited state. In short, light energy has now been captured by biological molecules but is not stored in any useful form yet. The energy is transferred from chlorophyll to chlorophyll until eventually (after about a millionth of a second), it is delivered to the reaction center. Up to this point, only energy has been transferred between molecules, not electrons. The reaction center contains a pair of chlorophyll a that have a special property. Those two chlorophylls can undergo oxidation upon excitation; they can actually give up an electron in a process that is called a photoact. It is at this step in the reaction center, during photosynthesis, that light energy is converted into an excited electron. All of the subsequent steps involve getting that electron onto the energy carrier NADPH for delivery to the Calvin cycle where the electron is deposited onto carbon for long-term storage in the form of carbohydrate. PSII and PSI are two major components of the photosynthetic electron transport chain, which also includes the cytochrome complex. The cytochrome complex, an enzyme composed of two protein complexes, transfers the electrons from the carrier molecule plastoquinone (Pq) to the protein plastocyanin (Pc), thus enabling both the transfer of protons across the thylakoid membrane and the transfer of electrons from PSII to PSI. The reaction center of PSII (called P680) delivers its high-energy electrons, one at a time, to the primary electron acceptor, and through the electron transport chain (Pq to cytochrome complex to plastocyanin) to PSI. P680’s missing electron is replaced by extracting a low-energy electron from water; thus, water is “split” during this stage of photosynthesis, and PSII is re-reduced after every photoact. Splitting one H2O molecule releases two electrons, two hydrogen atoms, and one atom of oxygen. However, splitting two molecules is required to form one molecule of diatomic O2 gas. About 10 percent of the oxygen is used by mitochondria in the leaf to support oxidative phosphorylation. The remainder escapes to the atmosphere where it is used by aerobic organisms to support respiration. As electrons move through the proteins that reside between PSII and PSI, they lose energy. This energy is used to move hydrogen atoms from the stromal side of the membrane to the thylakoid lumen. Those hydrogen atoms, plus the ones produced by splitting water, accumulate in the thylakoid lumen and will be used to synthesize ATP in a later step. Because the electrons have lost energy prior to their arrival at PSI, they must be re-energized by PSI, hence, another photon is absorbed by the PSI antenna. That energy is relayed to the PSI reaction center (called P700). P700 is oxidized and sends a high-energy electron to NADP+ to form NADPH. Thus, PSII captures the energy to create proton gradients to make ATP, and PSI captures the energy to reduce NADP+ into NADPH. The two photosystems partly work in concert to guarantee that the production of NADPH will roughly equal the production of ATP. Other mechanisms exist to fine-tune that ratio to exactly match the chloroplast’s constantly changing energy needs. Generating an Energy Carrier: ATP As in the intermembrane space of the mitochondria during cellular respiration, the buildup of hydrogen ions inside the thylakoid lumen creates a concentration gradient. The passive diffusion of hydrogen ions from high concentration (in the thylakoid lumen) to low concentration (in the stroma) is harnessed to create ATP, just as in the electron transport chain of cellular respiration. The ions build up energy because of diffusion and because they all have the same electrical charge, repelling each other. To release this energy, hydrogen ions will rush through any opening, similar to water jetting through a hole in a dam. In the thylakoid, that opening is a passage through a specialized protein channel called the ATP synthase. The energy released by the hydrogen ion stream allows ATP synthase to attach a third phosphate group to ADP, which forms a molecule of ATP (Figure 2.2.15). The flow of hydrogen ions through ATP synthase is called chemiosmosis because the ions move from an area of high to an area of low concentration through a semi-permeable structure of the thylakoid. View the process of photosynthesis within a leaf. Access for free at PDF 2.3 Light Dependent Reaction Download View Light Independent Reaction After the energy from the sun is converted into chemical energy and temporarily stored in ATP and NADPH molecules, the cell has the fuel needed to build carbohydrate molecules for long-term energy storage. The products of the light-dependent reactions, ATP and NADPH—have lifespans in the range of millionths of seconds, whereas the products of the light-independent reactions (carbohydrates and other forms of reduced carbon) can survive almost indefinitely. The carbohydrate molecules made will have a backbone of carbon atoms. But where does the carbon come from? It comes from carbon dioxide—the gas that is a waste product of respiration in microbes, fungi, plants, and animals. In plants, carbon dioxide (CO2) enters the leaves through stomata, where it diffuses over short distances through intercellular spaces until it reaches the mesophyll cells. Once in the mesophyll cells, CO2 diffuses into the stroma of the chloroplast—the site of light-independent reactions of photosynthesis. These reactions actually have several names associated with them. One of those names, the Calvin cycle, is used to honor the man who discovered it, as well as because these reactions function as a cycle. Others call it the Calvin-Benson cycle to include the name of another scientist involved in its discovery. The most outdated name is “dark reaction,” which was used because light is not directly required (figure 2.2.16). The term dark reaction can be misleading because it implies incorrectly that the reaction only occurs at night or is independent of light, which is why most scientists and instructors no longer use it. The light-independent reactions of the Calvin cycle can be organized into three basic stages: fixation, reduction, and regeneration. Stage 1: Fixation In the stroma, in addition to CO2,two other components are present to initiate the light-independent reactions: an enzyme called ribulose-1,5-bisphosphate carboxylase/oxygenase (RuBisCO) and three molecules of ribulose bisphosphate (RuBP), as shown in Figure 2.2.17. RuBP has five atoms of carbon, flanked by two phosphates. RuBisCO catalyzes a reaction between CO2 and RuBP. Each RuBP molecule combines with one CO2 molecule producing one molecule of 1, 3-bisphosphoglycerate. This molecule splits into two molecules of 3-phosphoglyceric acid (3-PGA) or 3-phosphoglycerate. PGA has three carbons and one phosphate. Each turn of the cycle involves only one RuBP and one carbon dioxide and forms two molecules of 3-PGA. The number of carbon atoms remains the same, as the atoms move to form new bonds during the reactions: 3 C atoms from 3CO2 + 15 C atoms from 3RuBP = 18 C atoms in 6 molecules of 3-PGA This process is called carbon fixation because CO2 is “fixed” from an inorganic form into organic molecules. Since the first intermediate formed is a 3-C compound, plants that produce this compound are also calledC3 plants About 85% of the plant species on the planet are C3 plants; some examples are rice, wheat, soybeans, and all trees. Stage 2: Reduction ATP and NADPH are used to convert the six molecules of 3-PGA into six molecules of a chemical called glyceraldehyde 3-phosphate (G3P). That is a reduction reaction because it involves the gain of electrons by 3-PGA. (Recall that a reduction is the gain of an electron by an atom or molecule.) Six molecules of both ATP and NADPH are used. For ATP, energy is released with the loss of the terminal phosphate atom, converting it into ADP; for NADPH, both energy and a hydrogen atom are lost, converting it into NADP+. Both of these molecules return to the nearby light-dependent reactions to be reused and re-energized. Stage 3: Regeneration Interestingly, at this point, only one of the G3P molecules leaves the Calvin cycle and is sent to the cytoplasm to contribute to the formation of other compounds needed by the plant. Because the G3P exported from the chloroplast has three carbon atoms, it takes three “turns” of the Calvin cycle to fix enough net carbon to export one G3P. But each turn makes two G3Ps, thus three turns make six G3Ps. One is exported while the remaining five G3P molecules remain in the cycle and are used to regenerate RuBP, which enables the system to prepare for more CO2 to be fixed. Three more molecules of ATP are used in these regeneration reactions. The process of photosynthesis has a theoretical efficiency of 30% (i.e., the maximum amount of chemical energy output would be only 30% of the solar energy input), but the efficiency is much lower in reality. It is only about 3% on cloudy days. Why is so much solar energy lost? There are a number of factors contributing to this energy loss, and one metabolic pathway that contributes to this low efficiency is photorespiration. During photorespiration, the key photosynthetic enzyme Rubisco (ribulose-1,5-bisphosphate carboxylase oxygenase) uses O2 as a substrate instead of CO2. This process uses up a considerable amount of energy without making sugars (Figure 2.2.18). When a plant has its stomata open (when CO2 is diffusing in while O2 and water are diffusing out), photorespiration is minimized because Rubisco has a higher affinity for CO2 than for O2 when air temperatures are below 30°C (86°F). However, when a plant closes its stomata during times of water stress and O2 from respiration builds up inside the cell, the rate of photorespiration increases because O2 is now more abundant inside the mesophyll. So, there is a tradeoff. Plants can leave the stomata open and risk drying out, or they can close the stomata, thereby reducing the uptake of CO2, and decreasing the efficiency of photosynthesis. In addition, Rubisco has a higher affinity for O2 when temperatures increase, which means that C3 plants use more energy (ATP) for photorespiration at higher temperatures. Access for free at PDF 2.4 Light Independent Reaction Download View C4 & CAM Pathways as Means of Reducing Photorespiration The C4 and CAM pathways for fixing CO2 are two adaptations that improve the efficiency of photosynthesis, by ensuring that Rubisco encounters high CO2 concentrations and thus reduces photorespiration. These two photosynthetic adaptations for fixing CO2 have evolved independently several times in species that evolved from wet and dry, but typically warm climates. Why have these mechanisms evolved independently so many times? Plants that minimize photorespiration may have a significant competitive advantage because a considerable amount of energy (in the form of ATP) is lost in plants during photorespiration. In many environments, plants that use solar energy more efficiently should out-compete those which are less efficient. C4 Pathway Many angiosperms have developed adaptations that minimize the losses to photorespiration. They all use a supplementary method of CO2 uptake which initially forms a four-carbon molecule compared to the two three-carbon molecules that are initially formed in the C3 pathway. Hence, these plants are called C4 plants. Note that C4 plants will eventually conduct the light-independent reactions (C3 pathway), but they form a four-carbon molecule first. C4 plants have structural changes in their leaf anatomy (Kranz anatomy) so that synthesizing the four-carbon sugar (the C4 pathway) and resuming the light-independent reactions (C3 pathways) are separated in different parts of the leaf with RuBisCO sequestered in bundle sheath cells, where the CO2 level is high and the O2 level low. After entering through the stomata, CO2 diffuses into a mesophyll cell (Figure 2.2.19; Figure 2.2.20). Being close to the leaf surface, these cells are exposed to high levels of O2, but they have no RuBisCO so cannot start photorespiration (nor the light-independent reactions). How does this work? Atmospheric CO2 is fixed in the mesophyll cells as a simple 4-carbon organic acid (malate) by an enzyme that has no affinity for O2. Malate is then transported to the bundle sheath cells. Inside the bundle sheath, malate is oxidized to a 3-C organic acid, and in the process, 1 molecule of CO2 is produced from every malate molecule (Figure 2.2.21). The CO2 is then fixed by Rubisco into sugars, via the Calvin cycle, exactly as in C3 photosynthesis. There is an additional cost of two ATPs associated with moving the three-carbon “ferry” molecule from the bundle sheath cell back to the mesophyll to pick up another molecule of atmospheric CO2. Since the spatial separation in bundle-sheath cells minimizes O2 concentrations in the locations where Rubisco is located, photorespiration is minimized (Figure 2.2.21). This arrangement of cells reduces photorespiration and increases the efficiency of photosynthesis for C4 plants. In addition, C4 plants require about half as much water as a C3 plant. The reason C4 plants require less water is that the physical shape of the stomata and leaf structure of C4 plants helps reduce water loss by developing a large CO2 concentration gradient between the outside of the leaf (400 ppm) and the mesophyll cells (10 ppm). The large CO2 concentration gradient reduces water loss via transpiration through the stomata. These C4 plants are well adapted to (and likely to be found in) habitats with high daytime temperatures and intense sunlight. Because they use the C4pathway to prevent photorespiration, they do not have to open their stomata to the same extent as C3plants and can thus conserve water. Some examples are crabgrass, corn (maize), sugarcane, and sorghum. Although comprising only ~3% of the angiosperms by species, C4 plants are responsible for ~25% of all the photosynthesis on land. CAM Many plants such as cacti and pineapples, which are adapted to arid environments, use different energy and water-saving pathway called crassulacean acid metabolism (CAM). This name comes from the family of plants (Crassulaceae) in which scientists first discovered the pathway. Instead of spatially separating the light-dependent reactions and the use of CO2 in the Calvin cycle, CAM plants separate these processes temporally (Figure 2.2 22). At night, CAM plants open their stomata, and an enzyme in the mesophyll cells fixes the CO2 as an organic acid; then, they store the organic acid in vacuoles until morning. During the day the light-dependent reactions supply the ATP and NADPH necessary for the Calvin cycle to function, and the CO2 is released from those organic acids in order to make sugars. Plant species using CAM photosynthesis are the most water-efficient of all; the stomata are only open at night when humidity is typically higher, and the temperatures are much cooler (which serves to lower the diffusive gradient driving water loss from leaves). The CAM pathway is primarily an adaptation to water-limited environments; the fact that this pathway also stops photorespiration is an added benefit. Overall, C3, C4 and CAM plants all use the Calvin cycle to make sugars from CO2. Table 2.2.1, list the various ways in which plants fix CO2 and the advantages and disadvantages associated with the various mechanisms and the habitats where plants are found. Table 2.2.1. Characteristics of C3, C4, and CAM methods of fixing CO2 \| \| C3 plant \| C4 plant \| CAM Plant \| \| cost \| photorespiration \| The cost in terms of ATP associated with fixing carbon is double. Carbon fixation is less efficient under cold conditions. \| Reduced amount of fixed carbon, stomata only open at night \| \| benefits \| Carbon fixation without using ATP \| Reduced photorespiration and ability to fix Carbon under high temperatures and reduced water loss \| Reduced photorespiration and reduced water loss \| \| Separation of light-dependent reactions and carbon fixation \| None, all of these reactions occur in the same cells \| Spatial, these two sets of reactions occur in different cells \| Temporal, these two sets of reactions occur at different times of day \| View this explanation about how C4 plants can start with an alternative enzyme (PEP carboxylase) and CAM plants can alter the timing of their Calvin cycle processes to avoid/minimize O2 use and water loss. Watch an animation of photosynthesis and the Calvin cycle. PDF 2.5 C4 & CAM Pathways as Means of Reducing Photorespiration Download View How Environmental Factors Affect Photosynthesis In the previous section, we learn how light has both wave and particle properties as well as how photons are utilized in photosynthesis. Plants grow in diverse climates and under ever-changing environmental conditions. The rate of photosynthesis also changes with changing environmental conditions. This change in photosynthetic rate directly affects crop yield. In this section, we will briefly discuss how light intensity, the concentration of CO2, high and low temperatures, water availability, and air movement affect photosynthesis. Light Intensity In photosynthesis, irradiance or intensity can be measured as the number of photons falling on a unit area of the leaf surface in unit time. We already know that the energy of a photon depends on the wavelength. Photosynthesis-irradiance curve (figure 2.2.23) depicts how with increasing light intensity, the rate of photosynthesis continues to increase (left portion of the graph) until it reaches a saturation point (middle portion showing flat line or plateau) and then drops (right portion of the graph). This drop in the rate of photosynthesis is due to many factors, such as CO2 concentration, temperatures change, and water conditions. When exposure to high-intensity light causes damage to photosystem II, it is called photoinhibition (Figure 2.2.24). A few mechanisms that plants use to prevent photoinhibition are: Paraheliotropism: For efficient photosynthesis, plants orient their leaves perpendicular to the incident light. To reduce the damage from photoinhibition, some plant that exhibits paraheliotropism, orient their leaves parallel to the plane of incident light. Chloroplast movement: cytoplasmic streaming or cyclosis has been observed in algae, mosses as well as in the leaves of higher plants. Elements of the cytoskeleton (Unit 1, lesson 1, The cell) participate in cytoplasmic streaming and move organelles such as chloroplast around the plant cell. Cytoplasmic streaming leads to changes in the location and orientation of chloroplast. Repositioning of chloroplast blocks the over-exposure to high-intensity light and prevents photoinhibition. See cytoplasmic streaming in elodea leaves. Xanthophyll cycle: If too much light is absorbed, the pH gradient developed across the thylakoid membranes becomes greater. This is sensed by a protein, PsbS, and through subsequent conformational changes transmitted through the light-harvesting antennae, the excess light energy must be dissipated as thermal energy. Molecules called xanthophylls (synthesized from carotenes - vitamin A precursors) such as zeaxanthin are important in excess energy dissipation. These molecules appear to cause excited-state chlorophyll (singlet-like excited state dioxygen) to become de-excited (Figure 2.2.25). Without the xanthophylls, the excited state chlorophyll could deexcite by transfer of energy to ground state triplet dioxygen, promoting it to the singlet, reactive state, which through electron acquisition, could also be converted to superoxide. These reactive oxygen species (ROS) can lead to oxidative damage to proteins, lipids, and nucleic acids, alteration in gene transcription, and even programmed cell death. Carotenoids can also act as ROS scavengers. Hence both heat dissipation and inhibition of the formation of ROS (by such molecules as vitamin E) are both mechanisms of defense against excessive solar energy. Sun and shade leave: Plants show variation in the size and thickness of their leaves. Sun leaves are smaller and thicker than shade leaves. Thicker cuticle, more palisade parenchyma cells or layers, and more vascular tissues are observed in the sun leaves. This makes sun leaves more efficient in capturing high-intensity light while preventing water loss via stomata due to transpiration. Shade leaves are larger. This increased surface area in shade leaves increases the absorption of light (Figure 2.2.26). CO2 concentration The partial pressure of CO2 is identical in the atmosphere. C4 plants evolved anatomical structures called Kranz anatomy. (figure 2.2.19; unit 2 lesson 2 section 5) to concentrate CO2 while CAM plants separate CO2 fixation temporally. The rate of photosynthesis increases with increasing CO2 concentration only if stomata remain open. The opening and closing of stomata depend on the ratio of the rate of water loss by transpiration to the rate of CO2 fixation. Many factors affect this ratio. Under high CO2 concentrations, the components of light-independent reactions (RuBisCo enzyme, regeneration of Ribulose-1, 5-bisphosphate) are saturated and the rate of photosynthesis cannot be increased any further. The availability of ATP and NADPH from light-dependent reactions also limits the rate of CO2 fixation under high CO2 conditions. Photorespiration is likely to occur under low CO2 concentrations inside the leaf. Wind speed When winds are calm, the rate of exchange of CO2 and water vapor via stomata is slower than when there is a breeze or a gust. The rate of exchange of gases depends on the thickness of the boundary layer, a layer of stagnant air around the leaf. A breeze or a gust disrupts the boundary layer and helps in increasing the rate of gas exchange thus enhancing the rate of diffusion of CO2 into the leaves. The increased rate of CO2 diffusion supports a higher rate of photosynthesis. Similarly, on humid days, the rate of exchange of water vapor and CO2 goes down decreasing the rate of photosynthesis. Temperature Low and high temperatures are relative to the range of temperature where a plant is growing. The temperature range for photosynthesis is wide, 59°F to 104°F. Rate of photosynthesis increases as the temperature rises, but only up to a certain level. Plants grow in different climates and under a wide range of temperatures and are well adapted to the temperature fluctuations in the area. The probability of photorespiration increases with high temperatures. This is because, at high temperatures, RuBisCO has a lower affinity for CO2. Lower temperatures bring down the activity of multiple enzymes involved in photosynthesis as well as transport in phloem and xylem. Water availability As we already know, plants grow in different climates and are well adapted to the environmental conditions of those regions. Low water conditions or drought conditions disrupt the water potential of the cells and thus affect all functions including photosynthesis. Remember, CO2 exchange occurs via stomata. To do photosynthesis plants must open stomata and this can cause further aggravate the water stress. Depending on the species and climate, plants show many adaptations to avoid water loss, such as sunken stomata, thick waxy cuticle, fewer stomata, deep roots, water storage in roots or other parts of the plant, reduced growth, smaller or reduced leaf to decrease the surface area for transpiration or in some cases complete loss of leaves, for example, spines on cacti are reduced leaves. Some plants called xerophytes are naturally tolerant to drought and can survive under extreme drought conditions. Draught-tolerant plants continue to grow and function under severe drought conditions. Whereas plants that germinate, grow, and reproduce only when water is available are called drought-avoiding plants. Flooding also reduces the rate of photosynthesis. Roots are non-photosynthetic and depend on cellular respiration to support all functions. When submerged under water for long periods roots face oxygen scarcity. Many plants develop spaces or channels within tissue to store diffused air from aerial parts of the plant. For example, water lilies develop aerenchyma (figure 2.2.27). Many plants develop adventitious roots, roots growing from stems or other parts of plants above the water level. Intertidal zones are areas of seashore where land is submerged during high tide and exposed during low tide. Mangroves are well adapted to overcome the effects of flooding due to high tides. Mangroves grow aerial roots called pneumatophores (Figure 2.2.28) that grow upwards and stick out of the water surface. Numerous lenticels cover the surface of these roots to aid in O2 absorption. Lenticels (Figure 2.2.29) are openings in the bark, made up of cells that are permeable to O2 due to a lack of suberin deposition. Some plants, such as rice and cottonwood germinate as well as grow well under flooded conditions. PDF 2.6 How Environmental Factors Affect Photosynthesis Download View The Energy Cycle Whether the organism is a bacterium, plant, or animal, all living things access energy by breaking down carbohydrates and other carbon-rich organic molecules. But if plants make carbohydrate molecules, why would they need to break them down, especially when it has been shown that the gas organisms release as a “waste product” (CO2) acts as a substrate for the formation of more food in photosynthesis? Remember, living things need the energy to perform life functions. In addition, an organism can either make its own food or eat another organism—either way, the food still needs to be broken down. Finally, in the process of breaking down food, called cellular respiration, heterotrophs release needed energy and produce “waste” in the form of CO2 gas. However, in nature, there is no such thing as “waste.” Every single atom of matter and energy is conserved and recycled over and over infinitely. Substances change the form or move from one type of molecule to another, but their constituent atoms never disappear (Figure 8.2.30). In reality, CO2 is no more a form of waste than oxygen is wasteful to photosynthesis. Both are byproducts of reactions that move on to other reactions. Photosynthesis absorbs light energy to build carbohydrates in chloroplasts, and aerobic cellular respiration releases energy by using oxygen to metabolize carbohydrates in the cytoplasm and mitochondria. Both processes use electron transport chains to capture the energy necessary to drive other reactions. These two powerhouse processes, photosynthesis and cellular respiration (Figure 2.2.31), function in biological, cyclical harmony to allow organisms to access life-sustaining energy that originates millions of miles away in a burning star humans call the sun. Everyday Connection: Photosynthesis at the Grocery Store Major grocery stores in the United States are organized into departments, such as dairy, meats, produce, bread, cereals, and so forth. Each aisle (figure 2.2.32) contains hundreds, if not thousands, of different products for customers to buy and consume. Although there is a large variety, each item ultimately can be linked back to photosynthesis. Meats and dairy link because the animals were fed plant-based foods The bread, cereals, and pasta come largely from starchy grains, which are the seeds of photosynthesis in plants. What about desserts and drinks? All of these products contain sugar – sucrose in a plant product, a disaccharide, a carbohydrate molecule, which is built directly from photosynthesis. Moreover, many items are less obviously derived from plants: paper goods are generally plant products and many plastics (abundant as products and packaging) are derived from “algae” (unicellular photosynthesizing protozoans and cyanobacteria). Virtually every specie and flavoring in the spice aisle was produced by a plant as a leaf, root, bark, flower, fruit, or stem. Ultimately, photosynthesis connects to every meal and every food a person consumes. Access for free at PDF 2.7 The Energy Cycle Download View Attributions Title image: Plastids; Kristian Peters -- Fabelfroh, CC BY-SA 3.0 , via Wikimedia Commons Bear, Robert; Rintoul, David; Snyder, Bruce; Smith-Caldas, Martha; Herren, Christopher; and Horne, Eva, "Principles of Biology" (2016). Open Access Textbooks. 1. Biology 2e By Mary Ann Clark, Matthew Douglas, Jung Choi. OpenStax is licensed under Creative Commons Attribution License v4.0 Introduction to Organismal Biology at is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Botany (Ha, Morrow, and Algiers) is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Melissa Ha, Maria Morrow, & Kammy Algiers. Bear, Robert; Rintoul, David; Snyder, Bruce; Smith-Caldas, Martha; Herren, Christopher; and Horne, Eva, "Principles of Biology" (2016). Open Access Textbooks. 1. Glossary absorption spectrum -range of wavelengths of electromagnetic radiation absorbed by a given substance antenna protein -pigment molecule that directly absorbs light and transfers the energy absorbed to other pigment molecules Calvin cycle -light-independent reactions of photosynthesis that convert carbon dioxide from the atmosphere into carbohydrates using the energy and reducing power of ATP and NADPH carbon fixation -process of converting inorganic CO2 gas into organic compounds carotenoid -photosynthetic pigment (yellow-orange-red) that functions to dispose of excess energy chemoautotroph -organism that can build organic molecules using energy derived from inorganic chemicals instead of sunlight chlorophyll a -form of chlorophyll that absorbs violet-blue and red light and consequently has a bluish-green color; the only pigment molecule that performs the photochemistry by getting excited and losing an electron to the electron transport chain chlorophyll b -accessory pigment that absorbs blue and red-orange light and consequently has a yellowish-green tint chloroplast -organelle in which photosynthesis takes place cytochrome complex -group of reversibly oxidizable and reducible proteins that forms part of the electron transport chain between photosystem II and photosystem I electromagnetic spectrum -range of all possible frequencies of radiation electron transport chain -group of proteins between PSII and PSI that pass energized electrons and use the energy released by the electrons to move hydrogen ions against their concentration gradient into the thylakoid lumen granum -stack of thylakoids located inside a chloroplast heterotroph -organism that consumes organic substances or other organisms for food light harvesting complex -complex that passes energy from sunlight to the reaction center in each photosystem; consists of multiple antenna proteins that contain a mixture of 300 to 400 chlorophyll a and b molecules, as well as other pigments like carotenoids light-dependent reaction -first stage of photosynthesis where certain wavelengths of the visible light are absorbed to form two energy-carrying molecules (ATP and NADPH) light-independent reaction -second stage of photosynthesis, through which carbon dioxide is used to build carbohydrate molecules using energy from ATP and NADPH mesophyll -middle layer of chlorophyll-rich cells in a leaf P680 -reaction center of photosystem II P700 -reaction center of photosystem I photoact -ejection of an electron from a reaction center using the energy of an absorbed photon photoautotroph -organism capable of producing its own organic compounds from sunlight photon -distinct quantity or “packet” of light energy photosystem -group of proteins, chlorophyll, and other pigments that are used in the light-dependent reactions of photosynthesis to absorb light energy and convert it into chemical energy photosystem I -integral pigment and protein complex in thylakoid membranes that uses light energy to transport electrons from plastocyanin to NADP+ (which becomes reduced to NADPH in the process) photosystem II -integral protein and pigment complex in thylakoid membranes that transports electrons from water to the electron transport chain; oxygen is a product of PSII pigment -molecule that is capable of absorbing certain wavelengths of light and reflecting others (which accounts for its color) primary electron acceptor -pigment or other organic molecule in the reaction center that accepts an energized electron from the reaction center reaction center -complex of chlorophyll molecules and other organic molecules that is assembled around a special pair of chlorophyll molecules and a primary electron acceptor, capable of undergoing oxidation and reduction reduction -gain of electron(s) by an atom or molecule spectrophotometer -instrument that can measure transmitted light and compute the absorption stoma -opening that regulates gas exchange and water evaporation between leaves and the environment, typically situated on the underside of leaves stroma -fluid-filled space surrounding the grana inside a chloroplast where the light-independent reactions of photosynthesis take place thylakoid -disc-shaped, membrane-bound structure inside a chloroplast where the light-dependent reactions of photosynthesis take place; stacks of thylakoids are called grana thylakoid lumen -aqueous space bound by a thylakoid membrane where protons accumulate during light-driven electron transport wavelength -distance between consecutive points of equal position (two crests or two troughs) of a wave in a graphic representation; inversely proportional to the energy of the radiation Version History Remix published on Jan 27, 2025 by Abel Rosado: Photosynthesis `
12163	https://www.media4math.com/library/definition-direct-variation	Definition--Direct Variation \| Media4Math Skip to main content Enable accessibility for low vision Open the accessibility menu Skip to content Home Search Subscribe ContentNewest ResourcesLesson Plan LibraryContent ShowcaseAnimated Math Clip ArtSAT Test PrepMath Fluency CentersMath Visual GlossaryStudent Tutorials Standards AlignmentsStandards Alignments (K-12)Standards Reports Getting StartedEducatorsTutorsParents Partners Shop@Media4Math Persistent Menu Library Classroom Log in Register Discover more MATH math Math Mathematical Mathematics educators educational Education Register to SaveSubscribe to DownloadPreviewAdd to Slideshow (Subscribers Only) If you are a subscriber, please log in. Algebra>>Linear Functions and Equations>>Applications of Linear Functions Resource Related Resources Display Title Definition--Direct Variation Direct Variation Topic Linear Functions Definition Direct variation describes a linear relationship between two variables where one variable is a constant multiple of the other, expressed as y = kx, where k is the constant of variation. Description Direct variation is a fundamental concept in linear functions, illustrating how one variable changes proportionally with another. The constant of variation, 𝑘 k, represents the rate of change. In real-world scenarios, direct variation can model relationships such as speed and distance, where distance traveled varies directly with time at a constant speed. Understanding this concept is crucial in fields like physics and engineering. In education, direct variation helps students grasp proportional relationships and their graphical representations. It lays the foundation for more complex functions and is essential for solving real-life problems involving proportionality. For a complete collection of terms related to linear functions and equations click on this link: Linear Functions and Equations Collection Direct Variation Learn about direct variations in this video tutorial. The video was uploaded on 10/22/2022. You can view the video here. The video lasts for 1 minute and 16 seconds. Video Transcript A direct variation is a special type of linear function whose graph passes through the origin. The equation of a direct variation is of the form y = k•x, where k is the constant of variation. It also happens to be the slope of the line. This is an example of a direct variation where k is greater than 0. This is an example of a direct variation where k is less than 0. Direct variations can be used to model many real world phenomena. For example, pushing or pulling a spring a certain distance is a way to measure the force on it. The equation F = k•x is used to measure force on the spring based on how much it is stretched or compressed. The value k is called the spring constant. The direct variation between force and distance can be used to measure weight. Scales for measuring weight use springs. With the scale on the left, standing on it compresses the spring. With the scale on the right, putting produce on it stretches the spring. In both cases the amount the spring is compressed or stretched is a direct indicator of the weight. There are many examples of direct variations in the real world. Brief Review of Linear Functions What Is a Function? A function is a one-to-one mapping of input values (the independent variable) to output values (the dependent variable). Click on this link to see a quick tutorial on what a function is. This slide show goes over the following key points: For every input value (x), there is a unique output value, f(x). Functions can be represented as equations, tables, and graphs. A function machine is a useful visual representation of the input/output nature of functions. Dependent/Independent Variables. When one variable depends on another, then it is the dependent variable. For example, the faster your speed, the farther you travel. Suppose that speed is represented by the variable s and the distance traveled is represented by the variable d. Here’s how to describe the relationship between s and d: The faster the speed, the more distance traveled. Distance is dependent on speed. Distance is a function of speed. d = f(s) When studying functions, make sure you are comfortable telling the difference between the independent variable and dependent variable. Get comfortable using function notation. To learn more about function notation, click on this link. Domain and Range. A function shows the relationship between two variables, the independent variable and the dependent variable. The domain is the allowed values for the independent variable. The range is the allowed values for the dependent variable. The domain and range influence what the graph of the function looks like. For a detailed review of what domain and range are, click on this link to learn more. You’ll see definitions of the terms domain and range, as well as examples of how to find the domain and range for given functions. Multiple Representations of Functions.We mentioned previously that functions can be represented in different ways. In fact, any function can be represented by an equation, usually f(x) equal to some expression; a table; or a graph. For a detailed review of multiple representations of functions, click on this link, to see a slide show that includes examples of these multiple representations. Slope-Intercept Form The most important form of a linear function is the slope-intercept form. Given the slope, m, and the y-intercept, b, for a linear function, you can easily construct the equation and graph of the linear function. To see examples of graphing linear functions in slope-intercept form, click on this link. This slide show also includes a video tutorial. Point-Slope Form Another method of finding the slope-intercept form involves using the point-slope form. In this case you are given the slope, m, of the line and one set of coordinates, (x, y), on the line. This is what the point-slope form looks like. To see examples of deriving the linear function using the in point-slope form, click on this link. This slide show tutorial walks you through the process and provides several worked-out examples. Slope Formula First, let’s review the basics of slope. This is the slope formula: Given two coordinates, the slope of the line connecting the two points is found using the slope formula. The key to understanding this is the slope-intercept form for parallel and perpendicular lines. Review these definitions: Basically, lines that are parallel have the same slope. Lines that are perpendicular have slopes that are negative reciprocals. Look at the following examples. To see examples of finding the equations of parallel and perpendicular lines, click on the following links: Parallel Lines Perpendicular Lines Loading… An error occurred while loading related resources. No related resources found. \| Common Core Standards \| CCSS.MATH.CONTENT.8.EE.B.6 \| \| Grade Range \| 6 - 12 \| \| Curriculum Nodes \| Algebra •Linear Functions and Equations •Applications of Linear Functions \| \| Copyright Year \| 2013 \| \| Keywords \| defnitions, glossary term, direct variation \| The Media4Math Definitions Library A Visual Glossary for Your Students Vocabulary is an important part of the math curriculum. In fact, many students struggle with math concepts because they lack the mastery of key vocabulary. Textbook instruction or examples often rely on these key terms and without a proper grounding in the relevant vocabulary, students will continue to struggle. With that in mind, Media4Math has developed an extensive glossary of key math terms. Each definition is a downloadable image that can easily be incorporated into a lesson plan. Furthermore, each definition includes a clear explanation and a contextual example of the term. To see the complete collection of these terms, click on this link. Math vocabulary doesn't consist of isolated terms. In fact, for any given concept there are clusters of vocabulary terms that students need to learn in order to better understand the concept. The Media4Math glossary consists of clusters of such terms. This is a summary of these clusters. Click on each link to see that collection of terms and definitions. 3D Geometry Coordinate Systems Fraction Concepts Linear Systems Polygon Concepts Rationals and Radicals Triangle Concepts Angle Concepts Early Elementary Definitions Functions Concepts Linear Functions Concepts Polynomial Concepts Ratios, Proportions, and Percents Trigonometry Charts and Graphs Equation Concepts Geometry Math Properties Primes and Composites Sequences and Series Variables and Unknowns Circles Exponential Concepts Geometric Theorems Measures of Central Tendency Quadratics Concepts Slope Closure Property Factors and Multiples Inequalities Place Value Concepts Quadrilateral Concepts Statistics and Probability Instructional Suggestions Here are some idea for how to use this library of vocabulary terms: Creating Connections As you introduce a new topic, for example Slope, go to the corresponding collection of definitions by linking on one of the collections above. Each definition includes an example of the term. Have students research one or more of these terms. Have a group of students research these terms and begin making connections. The idea is to encourage students to start using these terms as they begin discussing the main concept. To continue this example, let's look at the collection of terms under slope. Clicking on the link reveals that there are 17 terms under the category of slope. As you can see, this is more than just a simple definition of a single term. As students analyze these definitions, they begin to see common terms: ratio, rise over run, change in coordinates, and so forth. Working in teams, students can begin to build connections among these terms. Encourage them make connections among these related terms, creating a graphic similar to this: With an activity like this, students begin to use math vocabulary but, more important, tie it to math concepts. Creating Word Games Once students are familiar with the collection of terms have them create word search or crossword puzzles using these terms. There are many free online tools for creating such puzzles. In addition, Media4Math has a collection of word games and vocabulary puzzles to allow students to further practice their vocabulary skills. To see the current collection of puzzles see the links below: Word Search Puzzles Crossword Puzzles The Illustrated Math Dictionary Subscribers to Media4Math also get access to The Illustrated Math Dictionary. This ebook brings together math definitions and related multimedia resources. This ebook brings together the glossary terms for concepts like Linear Functions, Quadratic Functions, and Polynomial functions. Each term has an audio component, along with related resources.The Illustrated Math Dictionary is more than just a vocabulary tool. Use it for instruction or review. Additional Resources The What Works Clearinghouse has a number of Practice Guides that focus on evidence-based practices that will help struggling students. In particular, the Practice Guide entitled, Assisting Students Struggling with Mathematics: Intervention in the Elementary Grades, emphasizes the importance of math vocabulary. In particular, make a not of the following point: Mathematical language is academic language that conveys mathematical ideas.This includes vocabulary, terminology, and language structures used when thinking about, talking about, and writing about mathematics. Mathematical language conveys a more precise understanding of mathematics than the conversational or informal language used every day to communicate with others. Footer menu About Privacy Policy Blog Contact Us Accessibility Advertise with Us! Follow © Media4Math. All rights reserved
12164	https://math.stackexchange.com/questions/2970197/is-the-function-fx-ex-cos-1-x-is-uniformly-continuous-on-a-0-1	calculus - Is the function $f(x) = e^x \cos (1/x)$ is uniformly continuous on $A = (0,1)$. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Is the function f(x)=e x cos(1/x)f(x)=e x cos⁡(1/x) is uniformly continuous on A=(0,1)A=(0,1). [duplicate] Ask Question Asked 6 years, 11 months ago Modified6 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. This question already has an answer here: Determine whether the function f(x)=e x cos(1/x)f(x)=e x cos⁡(1/x) is uniformly continuous on A=(0,1)A=(0,1). (1 answer) Closed 6 years ago. The question and its answer is given below: Determine whether t the function f(x)=e x cos(1/x)f(x)=e x cos⁡(1/x) is uniformly continuous on A=(0,1)A=(0,1). but I am wondering why the n n is chosen like this, and if it should be N N? and why 1/n π 1/n π is ≤δ?≤δ? could anyone explain this for me please? calculus real-analysis analysis uniform-continuity Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Oct 25, 2018 at 6:24 IntuitionIntuition asked Oct 25, 2018 at 6:23 IntuitionIntuition 3,123 1 1 gold badge 14 14 silver badges 56 56 bronze badges 4 No it is not a duplicate @LordSharktheUnknown my questions are different in both posts.Intuition –Intuition 2018-10-25 06:25:56 +00:00 Commented Oct 25, 2018 at 6:25 That only works if n n is an odd positive integer.Angina Seng –Angina Seng 2018-10-25 06:26:16 +00:00 Commented Oct 25, 2018 at 6:26 Thank you for your answer .... but this is not an answer to my questions.@LordSharktheUnknown .... or at least I can not see that it is answer Intuition –Intuition 2018-10-25 06:28:32 +00:00 Commented Oct 25, 2018 at 6:28 1 It demonstrates that the "answer" you reproduce is incorrect. The "answers at the back of the book" are usually produced as an afterthought, with little care. Take them with a pinch of salt!Angina Seng –Angina Seng 2018-10-25 06:33:38 +00:00 Commented Oct 25, 2018 at 6:33 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. There is a mistake in the proof. If you take n=[1 δ π]+1 n=[1 δ π]+1 the proof works fine because n>1 δ π n>1 δ π in this case , so 1 n π<δ 1 n π<δ. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 25, 2018 at 6:34 Kavi Rama MurthyKavi Rama Murthy 361k 20 20 gold badges 102 102 silver badges 195 195 bronze badges 12 1 You also need to ensure that n n is odd.Angina Seng –Angina Seng 2018-10-25 06:44:39 +00:00 Commented Oct 25, 2018 at 6:44 1 @hopefully The above proof says that e x cos(1/x)e x cos⁡(1/x) has NO limit as x→0+x→0+ and therefore it can not be extended to a continuous function on [0,1][0,1].Robert Z –Robert Z 2018-10-25 06:44:55 +00:00 Commented Oct 25, 2018 at 6:44 1 @hopefully The choice of the points x x and y y is quite natural and it comes from properties of cos cos. Once you choose these points you get \|x−y\|=1 6 n π\|x−y\|=1 6 n π. To make this less than δ δ you need n>1 6 π δ n>1 6 π δ. That is, you need an integer that exceeds 1 6 π δ 1 6 π δ. Natural choice would be the smallest integer greater than this number. This is what they have did except that they use 1/6<1 1/6<1.Kavi Rama Murthy –Kavi Rama Murthy 2018-10-25 06:45:39 +00:00 Commented Oct 25, 2018 at 6:45 1 @hopefully if n n is even then cos(3 n π)=1 cos⁡(3 n π)=1.Robert Z –Robert Z 2018-10-25 06:52:19 +00:00 Commented Oct 25, 2018 at 6:52 1 I don't know why you are bringing in unnecessary things into the argument. To prove that the function is not uniformly continuous they are assuming it is uniformly continuous, applying the defintion, and arriving at a contradiction. Why bother about extending the function to [0,1][0,1]?Kavi Rama Murthy –Kavi Rama Murthy 2018-10-25 07:18:51 +00:00 Commented Oct 25, 2018 at 7:18 \|Show 7 more comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus real-analysis analysis uniform-continuity See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 1Determine whether the function f(x)=e x cos(1/x)f(x)=e x cos⁡(1/x) is uniformly continuous on A=(0,1)A=(0,1). Related 5Is function y=tan x y=tan⁡x uniformly continuous in the open interval (0,π/2)?(0,π/2)? 7Is any differentiable function f:(0,1)→[0,1]f:(0,1)→[0,1] is uniformly continuous 1Determine whether the function f(x)=e x cos(1/x)f(x)=e x cos⁡(1/x) is uniformly continuous on A=(0,1)A=(0,1). 0[Determine whether the function f(x)=x−−√f(x)=x is uniformly continuous on A=[1,+∞)A=1,+∞). 21/x 2 1/x 2 is not uniformly continuous on (0,1)(0,1) 2Physical Interpretation of uniformly continuous function. 4Is f(x)=1/x f(x)=1/x over [0.1,1][0.1,1] uniformly continuous but 1/x 1/x on (0,1)(0,1) is not? 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12165	https://www.quora.com/Why-are-A-B-coordinates-of-the-normal-vector-of-line-Ax-By-C-0	Something went wrong. Wait a moment and try again. The Vector Equation of Lines Coordinate Systems 3D Geometry Vectors (mathematics) Linear Equations Normal Equations Vectors, Kinamatics 5 Why are (A, B) coordinates of the normal vector of line Ax+By+C=0? Richard Goldstone PhD in Mathematics, The Graduate Center, CUNY (Graduated 1995) · Author has 1.8K answers and 3.9M answer views · 4y First, please note that we are speaking of “A” normal vector, not “THE” normal vector. There are infinitely many normal vectors to a given line. Solving for y in the equation Ax+By+C=0 gives y=−(AB)x−CB. In vector format, this says [xy]=[x−(AB)x−CB]=x[1−AB]+[0−CB]. This is the vector equation for a line with direction vector [1−AB]. Since a First, please note that we are speaking of “A” normal vector, not “THE” normal vector. There are infinitely many normal vectors to a given line. Solving for y in the equation Ax+By+C=0 gives y=−(AB)x−CB. In vector format, this says [xy]=[x−(AB)x−CB]=x[1−AB]+[0−CB]. This is the vector equation for a line with direction vector [1−AB]. Since any scalar multiple of a direction vector of a line is a direction vector for the line, we can multiply the above vector by B and instead take the direction vector for the line to be d=[B−A]. If we set n=[AB], it is easily checked that n⋅d=0, so n is a normal vector to the given line. Related questions A line ax + by + cz + … + k = 0 and has a normal vector [a, b, c…] formed by coefficients. But can a normal vector determine a line's coefficients? If two vectors are equal (have the same magnitude and direction) then when representing them on the coordinates they will sure have the same x and y components and thus they will be represented with a single vector right? What is the unit normal vector to the surface xy^ {2} +2yz=8 at the point (3,-2,1)? How can we convert a unit vector into a normal vector in a cylindrical coordinate system? How do you find the normal vector of a line in an XY coordinate system? Dale Gray MA in Mathematics, The University of Texas at Arlington (Graduated 1974) · Author has 763 answers and 1.5M answer views · 4y See the picture below. The displacement vector from the x-intercept to the y-intercept is Δ→r=CA^i−CB^j. Let →N=A^i+B^j. Then, →N⋅Δ→r=ACA−BCB=0. Therefore, ^N is perpendicular to Δ^r. See the picture below. The displacement vector from the x-intercept to the y-intercept is Δ→r=CA^i−CB^j. Let →N=A^i+B^j. Then, →N⋅Δ→r=ACA−BCB=0. Therefore, ^N is perpendicular to Δ^r. Pavel Juranek Technician, Programmer, Analyst, Consultant · Author has 3.7K answers and 1.3M answer views · Updated 4y Have two points P,Q and direction vector u = Q - P Then point M=[x,y] of line can be written M = P + t . u In coordinates x = Px + t . ux, then t = (x - Px) / ux y = Py + t . uy, then t = (y - Py) / uy uy (x - Px) = ux (y - Py) … from comparing of t values uy x - ux y + (ux Py - uy Px) = 0 can be written A x + B y + C = 0 (-B, A) is direction vector u (A, B) is some vector n because u.n = 0 then n is normal vector OTHER APPROACH have two vectors C = (A, B) … position vector given by coefficients M = (x, y) … position vector of some point of line. Draw line given by equation C . M = d ( Have two points P,Q and direction vector u = Q - P Then point M=[x,y] of line can be written M = P + t . u In coordinates x = Px + t . ux, then t = (x - Px) / ux y = Py + t . uy, then t = (y - Py) / uy uy (x - Px) = ux (y - Py) … from comparing of t values uy x - ux y + (ux Py - uy Px) = 0 can be written A x + B y + C = 0 (-B, A) is direction vector u (A, B) is some vector n because u.n = 0 then n is normal vector OTHER APPROACH have two vectors C = (A, B) … position vector given by coefficients M = (x, y) … position vector of some point of line. Draw line given by equation C . M = d (scalar product equal to line equation) where d is distance of libe from the origin. Then it si clear that C is perpendicular to line. Dean Rubine Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994) · Author has 10.6K answers and 23.6M answer views · 4y Ax+By=−C We see (B,−A) is the direction of the line, because if we translate an (x,y) on the line by (B,−A) we’re still on the line: The vector is perpendicular to the direction of the line as indicated by the zero dot product. is perpendicular to the direction of the line; that’s the definition of a normal to a line. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. 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Three vectors satisfy a.b = b.c = c.a = -1 and a + b + c = 0. What is the magnitude of vector a, b , and c? Given a = (1, 1, 1), c = (0, 1, -1), a. b = 3 and a × b = c, where a, b, and c are vectors, what is vector b? If three vectors vector A, vector B, and vector C satisfy the relation vector A.vectorB=0 and vectorA.vectorC=0, then the vector A is parallel to vector B×vector C. Is it true or false? Assistant Bot · Sep 7 For the line given by Ax + By + C = 0, the vector (A, B) is a normal (perpendicular) vector because of how the line equation encodes orthogonality between direction vectors and position vectors on the line. Concise derivation and explanation: Geometric meaning of the equation The equation Ax + By + C = 0 can be written as the dot product of two vectors: A·x + B·y + C = 0 ⇒ (A, B) · (x, y) + C = 0. For any point (x, y) on the line, (A, B) · (x, y) is constant (equal to −C). Thus differences between any two points on the line are orthogonal to (A, B). Orthogonality via difference of two For the line given by Ax + By + C = 0, the vector (A, B) is a normal (perpendicular) vector because of how the line equation encodes orthogonality between direction vectors and position vectors on the line. Concise derivation and explanation: Geometric meaning of the equation The equation Ax + By + C = 0 can be written as the dot product of two vectors: A·x + B·y + C = 0 ⇒ (A, B) · (x, y) + C = 0. For any point (x, y) on the line, (A, B) · (x, y) is constant (equal to −C). Thus differences between any two points on the line are orthogonal to (A, B). Orthogonality via difference of two points Take two points P1 = (x1, y1) and P2 = (x2, y2) on the line. Both satisfy Ax + By + C = 0, so (A, B)·(x1, y1) + C = 0 and (A, B)·(x2, y2) + C = 0. Subtract: (A, B)·(x2 − x1, y2 − y1) = 0. The vector v = (x2 − x1, y2 − y1) is any direction vector along the line; its dot product with (A, B) is zero, so (A, B) is perpendicular to every direction vector of the line. Therefore (A, B) is a normal vector. Alternative differential viewpoint Solve for y (if B ≠ 0): y = (−A/B)x + (−C/B). Slope = −A/B. A direction vector is (1, −A/B) which is proportional to (B, −A). Dotting (A, B) with (B, −A) gives AB + B(−A) = 0, confirming perpendicularity. Intuition The coefficients A and B measure how the line changes with x and y in the linear form Ax + By. Those coefficients define a vector whose action (dot product) on position vectors is constant along the line; constancy implies orthogonality to any change along the line. Conclusion: The pair (A, B) appears as the normal vector because the line equation is exactly the condition that the dot product of (A, B) with any point on the line equals the constant −C, which forces (A, B) to be orthogonal to every direction tangent to the line. Dr-Michael-W-Ecker 45 years as a university mathematics professor/ Ph.D. in mathematics · Author has 1.9K answers and 2.1M answer views · 3y Related Why is it not possible for both (a) and (b) to equal zero in the form of ax+by+c=0? If both a and b were zero, then you would not have an equation with any variables at all, so there is no meaning then to solutions or graphing. If a = 0, we get an equation y = constant, whose graph is a horizontal line. If b = 0, we get an equation x - constant, whose graph is a vertical line. If neither is zero, then the graph is an oblique (slanted) line. Reddy Studied at University of North Carolina at Chapel Hill · Author has 267 answers and 371.5K answer views · 6y Related Why is vector (a,b) perpendicular to line ax+by=c? Vector (a, b) is a units away from the origin horizontally and b units away vertically. That means the slope of the vector is rise/run = b/a. Line ax + by = c has a slope of -a/b. You can verify this by simply subtracting ax from both sides and dividing by b, which will resemble y = mx + b. Because b/a and -a/b are opposite reciprocals, vector (a, b) is perpendicular to line ax + by = c. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Upvoted by Mike Drake , B.S. Physics & Mathematics, Saint Joseph's University, Philadelphia · Author has 6.8K answers and 52.8M answer views · 4y Related How can the equation of a line which is equation of degree one in x and y can be reduced to Ax+By+C=0 then if the two lines are perpendicular then AA'+BB'=0? This sounds interesting! I have never seen this before. Let’s investigate! I always like line equations in the form y = mx + c rather than ax + by + c = 0 If two lines are perpendicular then their gradients will multiply to – 1 So here are the equations of two general perpendicular lines… This sounds interesting! I have never seen this before. Let’s investigate! I always like line equations in the form y = mx + c rather than ax + by + c = 0 If two lines are perpendicular then their gradients will multiply to – 1 So here are the equations of two general perpendicular lines… Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 3y Related Does the equation ax+by+c=0 represent a plane or a straight line? Firstly, a LINE is straight. We don’t need to say “a straight line”. Secondly, I will use the other form of ax + by + c which is y = mx + c just to make my examples simpler. In fact I will choose m = 1 and c = 2. If we are working in TWO DIMENSIONS then y = x + 2 is a line as below… However, if we are working in THREE DIMENSIONS y = x + 2 is a PLANE where the third axis called z is not restricted so z can be any number as below… I have also added the LINE y = x + 2 for clarity. Firstly, a LINE is straight. We don’t need to say “a straight line”. Secondly, I will use the other form of ax + by + c which is y = mx + c just to make my examples simpler. In fact I will choose m = 1 and c = 2. If we are working in TWO DIMENSIONS then y = x + 2 is a line as below… However, if we are working in THREE DIMENSIONS y = x + 2 is a PLANE where the third axis called z is not restricted so z can be any number as below… I have also added the LINE y = x + 2 for clarity. Promoted by Spokeo Spokeo - People Search \| Dating Safety Tool Dating Safety and Cheater Buster Tool · Apr 16 Is there a way to check if someone has a dating profile? Originally Answered: Is there a way to check if someone has a dating profile? 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Dating Safety Check – Review your date’s background to help keep you safe. Donald Hartig PhD in Mathematics, University of California, Santa Barbara (Graduated 1970) · Author has 7.4K answers and 2.8M answer views · 6y Related Why is vector (a,b) perpendicular to line ax+by=c? Just as the standard equation of a plane through a point math\,[/math]in [math]\mathbb{R}^3\,[/math]having normal math\,[/math]is [math]\hspace{5ex}a(x-x_0)+b(y-y_0)+c(z-z_0)=0\,,[/math] the equation of a line through a point math\,[/math] in [math]\mathbb{R}^2\,[/math]with normal math\,[/math]is [math]\hspace{9ex}a(x-x_0)+b(y-y_0)=0\,.[/math] Samiul Abid B.Sc in Electrical and Electronics Engineering, Bangladesh University of Engineering and Technology (Graduated 2018) · Author has 263 answers and 420.1K answer views · 5y Related How do I normalize the general equation of a line Ax + By + C = 0 into the form A^2 + B^2 = 1? Before asking a question on Quora, you must first have the least knowledge that will make you confirmed that what you are asking on Quora is a valid question. Let A=5 and B=3. Then how can you make A^2+B^2= 1? I think, your question should be: “How can we write the general equation of a straight line ax+by+c=0 in the form x cos alpha + y sin alpha =p?” [Don’t worry, we are all human beings and it’s said that- “To err is human”. No problem. But always try to understand your mistakes and try to make them correct. It will be a good gesture.] Now, coming to the answer. Look at the following figure: Before asking a question on Quora, you must first have the least knowledge that will make you confirmed that what you are asking on Quora is a valid question. Let A=5 and B=3. Then how can you make A^2+B^2= 1? I think, your question should be: “How can we write the general equation of a straight line ax+by+c=0 in the form x cos alpha + y sin alpha =p?” [Don’t worry, we are all human beings and it’s said that- “To err is human”. No problem. But always try to understand your mistakes and try to make them correct. It will be a good gesture.] Now, coming to the answer. Look at the following figure: Let r be the x-intercept and s be the y-intercept of the straight line ax+by+c=0. That is, A(r,0) and B(0,s) are two points on the straight line. So, OA=\|r\|[absolute value of r has been taken. It us because, OA is a distance from O to A and distance is always ‘+’ve, but r itself can be negative, so, OA is not directly equal to r, rather OA=\|r\|.] Similarly, OB=\|s\|. Let O(0,0) be the origin. Let \|p\| is the distance of the straight line ax+by+c=0 from the origin(O), i.e, OD is perpendicular to AB and OD=\|p\|. Now, as r is the x-intercept and s is the y-intercept of the straight line ax+by+c=0, we can write the equation of the same straight line also in the following way: x/r + y/s =1……….(1) Let alpha(greek letter ‘alpha' as you can see from the figure above) be the angle between OD and OA. Now, from the figure, from triangle OAD, we can write, OD/OA=cos alpha. So, OA=OD/cos alpha Or, r=\|p\|/cos alpha[since, we have drawn both points A and B on the 1st quadrant, so, we can write, OA=r and OB=s in stead of OA=\|r\| and OB=\|s\| respectively] Since, angle DOA=alpha and angle AOB=90 (degree), so, <DOB=<AOB-<DOA=90-alpha. Now, in triangle BOD, we can write, OD/OB=cos <DOB Or, OB=OD/cos <DOB Or, s=\|p\|/cos (90-alpha) Or, s=\|p\|/sin alpha Now, putting r=\|p\|/cos alpha and s=\|p\|/sin alpha in equation (1), we get, x/(\|p\|/cos alpha) + y/(\|p\|/sin alpha)=1 Or, x cos alpha + y sin alpha=\|p\| More generally, x cos alpha + y sin alpha=p.……..(2) This is another form of writing the same straight line ax+by+c=0 as we are going to discover below. Now, the general form of a straight line is: Or, ax+by=-c Or, x. a/sqrt(a^2+b^2) + y. b/sqrt(a^2+b^2)= -c/sqrt(a^2+b^2)…………(3) [read ‘sqrt(g)’ as ‘square root of g’] Now, if we take a/sqrt(a^2+b^2)=cos alpha, then sin alpha=sqrt(1-(a/sqrt(a^2+b^2))^2) = b/sqrt(a^2+b^2). So, b/sqrt(a^2+b^2)=sin alpha. Let -c/sqrt(a^2+b^2)=p which can be negative if c is positive. So, if c>0, p<0 and vice versa. Now, putting a/sqrt(a^2+b^2)=cos alpha, b/sqrt(a^2+b^2)=sin alpha and -c/sqrt(a^2+b^2)=p, we can write eqn. (3) as: x cos alpha + y sin alpha = p. I think, you understood what the mistake in your question was and what should be the answer and also how to formulate it. Thank you for the question. Kurt Behnke PhD in Mathematics & Theoretical Physics, University of Hamburg (Graduated 1981) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and David Vanderschel , PhD Mathematics & Physics, Rice (1970) · Author has 8.3K answers and 14.2M answer views · 8y Related A line ax + by + cz + … + k = 0 and has a normal vector [a, b, c…] formed by coefficients. But can a normal vector determine a line's coefficients? That’s not a line, it is a hyperplane in n-space (whatever your coefficients are). By the equation, it is clear that [a,b,c, …] (except the trailing k) is a normal vector to the hyperplane: the scalar product of any vector sitting on the hyperplane and the normal vector is zero, almost by definition. So obviously the correspondence is there, except the constant vector k, which is the shift away from 0, and is not determined by the normal vector. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views · 5y Related If the line ax+by+c=0 is a normal to the curve xy=1, then what is the relation between a, b, and c? Related questions A line ax + by + cz + … + k = 0 and has a normal vector [a, b, c…] formed by coefficients. But can a normal vector determine a line's coefficients? If two vectors are equal (have the same magnitude and direction) then when representing them on the coordinates they will sure have the same x and y components and thus they will be represented with a single vector right? What is the unit normal vector to the surface xy^ {2} +2yz=8 at the point (3,-2,1)? How can we convert a unit vector into a normal vector in a cylindrical coordinate system? How do you find the normal vector of a line in an XY coordinate system? If vector A.B=0 & vector [A×B] =0, then what will be the angle between vector B & C? The points and have coordinates and respectively. What are the vectors and ? Three vectors satisfy a.b = b.c = c.a = -1 and a + b + c = 0. What is the magnitude of vector a, b , and c? Given a = (1, 1, 1), c = (0, 1, -1), a. b = 3 and a × b = c, where a, b, and c are vectors, what is vector b? If three vectors vector A, vector B, and vector C satisfy the relation vector A.vectorB=0 and vectorA.vectorC=0, then the vector A is parallel to vector B×vector C. Is it true or false? How can I explain different forms of in a straight line? The pyramid is defined by the coordinates A (0,0,0), B (1,0,0),C (0,1,0) and D (0,0,1) is rotated 90 degrees about the line L that has direction vector? What are the vector, parametric, and symmetric equations for the line through the point (1,0,-3) and parallel to the vector 2i-4j+5k.? How do I find a vector equation with parameter t for the line of intersection of the planes x + y + z = 4 and x + z=0? For the vector a= (-3,4,0) and b= (2,1,4), what is the vector projection of a and b? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
12166	https://www.jogcr.com/article_697326.html	Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer Register Login Journal of Obstetrics, Gynecology and Cancer Research ===================================================== Iranian Society of Gynecology Oncology (IRSGO) Home Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer [x] [x] Home [x] Browse Current Issue By Issue By Author By Subject Author Index Keyword Index [x] Journal Info About Journal Aims and Scope Editorial Board Indexing and Abstracting Related Links FAQ Journal Metrics News [x] For Authors Submission Instruction call for papers The Article's Request Form Publication Ethics Publishing Charges [x] Submit Manuscript [x] For reviewers and editors Reviewers Section Duties of Editors Peer Review Process [x] Contact Us Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer Document Type : Original Research Article Authors Zinatossadat Bouzari1 Tara Mohammadi2 Mohammad Ranaei3 Karimollah Hajian-Taliki1 Azita Ghanbarpour1 1 Infertility and Reproductive Health Research Center, Health Research Institute and Department of Obstetrics and Gynecology, Babol University of Medical Sciences, Babol, Iran 2 Student Committee Research, Babol University of Medical Sciences, Babol, Iran 3 Clinical Research Development Unit of Rouhani Hospital, Babol University of Medical Sciences, Babol, Iran Abstract Background & Objective: Gestational diabetes mellitus (GDM) is also defined as a metabolic disease associated with relative insulin resistance during pregnancy, and elevated circulating insulin may increase the risk of EH and EC development. This study aimed to investigate the association between GDM and the incidence of EH and EC. Materials & Methods: We conducted a retrospective case-control study, including 300 women with abnormal uterine bleeding (AUB) referred to Ayatollah Rouhani Hospital in Babol. Cases (n=152) were patients with HC and EC based on medical records, and the controls (n=148) were individuals without HC and EC. The groups were compared according to demographic information, GDM or diabetes mellitus (DM) history, and body mass index (BMI). The Chi-square, independent t-test, and logistic regression analyses were performed to compare groups. Results: Of 300 women studied, 72 people (24.1%) had a GDM history, and 64 people had a diabetes mellitus history. There was a significant difference between the incidence of EC and EH with GDM (P < /i>=0.001). Both GDM and DM were associated with the increased EC (OR: 17.98, 95% CI: 6.73-48.08, and OR: 1.84, 95% CI: 1.26-2.68, respectively). GDM was also associated with the increased risk of EH (OR: 6.68, 95% CI: 2.77-16.10), whereas diabetes mellitus had not a significant role in the increased risk of EH (P < /i>=0.14). Conclusion: This study indicated that a GDM history is significantly associated with HC and EC. Therefore, to prevent and control these two complications in the future, management and monitoring of diabetes during pregnancy should be considered. Keywords Gestational Diabetes Endometrial Cancer Endometrial hyperplasia Insulin resistance Main Subjects Obstetrics and Gynecology ##### Full Text Introduction E ndometrial cancer (EC) is the second most prevalent gynecological malignancy affecting females. Endometrial hyperplasia (EH) is an uncontrolled proliferation of endometrial glands, resulting in EC's precursor lesions. EC is classified into two types based on histopathological observations, metabolic features, and clinical manifestations (1). Type I EC accounts for 70%-80% of cases and typically represents endometrioid histology. This type is commonly described as a low-grade tumor, which is estrogen-dependent and is often associated with simple and complex atypical hyperplasia (2-4). Based on epidemiological findings, multiple risk factors are associated with EC, including increased body mass index (BMI), anovulation, infertility, estrogen replacement therapy, a family history of EC, aging, and diabetes mellitus (5-7). Insulin resistance, observed in diseases like type-2 diabetes, obesity, and polycystic ovary syndrome (PCOS), is an important risk factor for EC (8, 9). Extensive studies have been conducted on the role of insulin resistance as an influential factor in EC, most of which indicate a significant relationship between diabetes and the incidence of this cancer (10, 11). Gestational diabetes mellitus (GDM), a disease with relative insulin resistance during pregnancy, is also a risk factor in developing type-2 diabetes (12, 13). GDM is estimated to affect 3%-10% of pregnant women depending on some parameters, such as the population studied, dietary patterns, and lifestyle. There are limited studies on the GDM role in EH and EC risk, which have been associated with conflicting results (14-16). A study by Wartko et al. in 2017 found that EH and EC had a significant relationship with GDM after matching data for gender/ethnicity and maternal age during pregnancy (14). Another study reported an association between GDM and EH/EC in obese women, suggesting the possible role of augmented circulating insulin in obese women with GDM in EC development (17). Given the fact that GDM may be a fast marker for insulinopathy, we investigated the relationship between GDM, EH, and EC incidence. Materials and Methods This retrospective case-control study was performed on the medical records of 335 women with abnormal uterine bleeding (AUB) who were referred to Ayatollah Rouhani Hospital in Babol, Iran. Among them, 35 cases were excluded from the study due to infertility and abortion. Finally, 300 people were reviewed, and their demographic information, clinical features, and the history of gestational diabetes mellitus (GDM) and diabetes mellitus (DM) were collected in a checklist. The cases included all patients diagnosed with hyperplasia Cancer (HC) and endometrial cancer (EC) based on the pathology reports, and the controls were individuals without HC and EC (Figure 1). The inclusion criteria entailed a history of at least one delivery after the 29 th week of pregnancy. The exclusion criteria included a diagnosis of prenatal EH, DM before 29 th weeks, young age, the lack of parity, hysterectomy at delivery, stillbirth, the lack of information about GDM, and insufficient information. All data were obtained from medical records, including age (years), menarche age, weight, height, BMI, parity, pharmacologic treatment, the presence of DM or GDM, underlying diseases, menstrual cycle pattern, taking oral contraceptive pill (OCP), a family history of DM or GDM, the number of pregnancies, current hormonal replacement therapy (HRT), infertility, a history of abortion, insulin and tamoxifen use, and duration of AUB. Case and control groups were matched in terms of hormone consumption, age (±3 years), and parity. Based on the information obtained from the patients' medical records, the case and control groups were compared in terms of GDM and DM history. It should be noted that the definition of GDM in this center was based on the latest edition of the guidelines provided by the American Diabetes Association (ADA) and was recorded in the patient records. Statistical Analysis All results were analyzed using the SPSS software version 18 (IBM, Chicago, IL., USA). The normality of data distribution was evaluated based on the Kolmogorov-Smirnov test. Quantitative variables were expressed as mean±standard deviation (SD), and qualitative variables were expressed as numbers and percentages. The chi-Square and analysis of variance (ANOVA) were used to analyze the qualitative and quantitative variables, respectively. P<0.05 was considered significant. Moreover, logistic regression analysis was used to test the odds ratio (OR) of the risk of GDM and DM on EH and EC with a 95% confidence interval. Figure 1. Schematic view of the study group selection. EC: endometrial cancer; EH: endometrial hyperplasia. Results Demographic Characteristics Out of 300 participants, 88 patients had EH and 64 had EC. We observed that 148 participants did not have these two lesions in the control group. Demographic information for all three groups is given in Table 1. Our findings showed that the mean age of the EC group was 56.6±9.7 years. Moreover, the number of pregnancies, deliveries, live births, and stillbirths was significantly higher in patients with EC than in the other groups (P=0.001) Patients with EH had a higher BMI compared to other groups (P=0.023). The mean age of menarche in normal people was higher than in the EH and EC groups (P=0.023). There was no significant difference between different groups in terms of the mean number of abortions (Table 1). Table 1.Comparison of demographic information and reproductive behaviors Groups VariablesEH (n=88) Mean±SDEC (n=64) Mean±SDControls (n=148) Mean±SDP-value Age (years)47.8± 8.1 56.6± 9.7 51.08± 9.31 0.001 BMI (kg/m 2)34.2±20.5 32.42± 7.05 29.83± 4.7 0.023 Gravity3.37± 1.97 4.6± 2.25 3.83± 2.04 0.002 Parity2.94± 1.81 4.2± 2.05 3.59± 3.05 0.01 Number of live births2.85± 1.69 3.92± 1.8 3.25± 1.74 0.001 Number of abortions0.44 ± 0.7 0.39± 0.79 0.42 ± 0.78 0.93 Number of stillbirths0.1 ± 0.34 0.32 ± 0.73 0.19 ± 0.53 0.039 Menarche age11.52± 1.68 12.12± 1.81 12.13± 1.71 0.023 Frequency Distribution of the Clinical Features of the Participants Among 300 subjects, 227 (75.9%) had no GDM history, of which 138 patients (93.9%) belonged to the normal group. The current study results showed that 72 patients (24.1%) had a GDM history, and GDM prevalence in the group with EC was significantly higher than in others (P=0.001). In addition, 236 cases (78.7%) had no diabetes history, while 64 patients (21.3%) had diabetes. This rate was significantly higher in patients with EC compared to others (P=0.001). We found that 20 (22.7%) of women with EH, 13 (20.6%) of cases with EC, and 16 (10.8%) of normal individuals had a familial diabetes history (P=0.034). Furthermore, hyperplasia and EC had statistically significant relationships with age (P=0.001), gravity (P=0.001), history of underlying disease (P=0.016), age of GDM (P=0.001), and time of GDM (P=0.001). On the other hand, hyperplasia and EC had no significant relationship with a history of insulin use, OCP consumption, infertility, AUB, PCOS, and tamoxifen use in the study population. The difference in the duration of oral diabetes medication use was statistically significant between three groups of EH, EC, and healthy controls (P=0.004). The EH, EC, and controls distribution and frequency are presented in Table 3 based on diabetes duration, in which the three groups were significantly associated (P=0.001). The highest frequency of the EC group was in people who had diabetes for more than 10 years (14.1%). Moreover, EH, EC, and controls had a significant relationship with underlying diseases (P=0.001). Patients with EC (53.2%) had underlying diseases for a more extended period than other groups. We found that EH, EC, and controls did not have a significant relationship with infertility, insulin use, HRT, and OCP in the study population. DM and GDM Increase the Risk of EC and EH Compared to the Control Group According to the information recorded in Table 2, the unadjusted and adjusted ORs of EC in people with a history of DM compared to healthy individuals were 3.67 (CI 95%: 1.86-7.22; P= 0.001) and 1.84 (CI 95%: 1.26-2.68; P=0.001), respectively. The unadjusted and adjusted ORs of EC risk in people with a GDM history compared to the healthy individuals were 14.4 (CI 95%: 6.25-33.1; P=0.001) and 17.98 (CI 95%: 6.73-48.08; P=0.001), respectively. Therefore, the presence of a history of GDM and DM significantly increased EC development (Table 2). According to Table 2, the unadjusted and adjusted ORs in people with DM compared to healthy individuals were 1.37 (CI 95%: 0.68-2.75; P=0.37) and 1.36 (CI 95%: 0.9-2.05; P=0.014), respectively. These results suggest that DM did not significantly affect the risk of developing EH. In contrast, the data presented in Table 2 show that GDM significantly increased EH risk (at least six-fold). The unadjusted and adjusted ORs in GDM compared to the healthy individuals were 8.76 (CI 95%: 3.92-19.5; P=0.001) and 6.68 (CI 95%: 2.77-16.1; P=0.001), respectively. Table 2.The unadjusted and adjusted odds ratio of history of diabetes mellitus and GDM at risk of EH and EC compared with the normal group Independent variableUnadjusted odds ratio (CI 95%)p-valueAdjusted odds ratio (CI 95%)p-value EC groupDiabetes mellitus 3.67 (1.86-7.22)0.001 1.84 (1.26-2.68)0.001 GDM 14.4 (6.25-33.1)0.0001 17.98 (6.73-48.08)0.001 EH groupDiabetes mellitus 1.37 (0.68-2.75)0.37 1.36 (0.9-2.05)0.14 GDM 8.76 (3.92-19.5)0.001 6.68 (2.77-16.1)0.001 Relationship of DM and GDM History with EH and EC According to Table 3, 64 (21.3%) people in the study population had a DM history (type 1 and type 2 diabetes), and 236 (78.7%) individuals had no DM history. Out of 64 people with DM, 42 cases belonged to the case groups (EH and EC individuals), and 22 people were in the control group (healthy individuals). The incidence of EH and EC in the people with DM were 26.6% and 39.1%, respectively. In addition, 72 (24.1%) subjects had a positive history of GDM, and 227 (75.9%) did not have a GDM history. The incidence of EH and EC in people with a positive history of GDM was 44.4% and 43.1%, respectively. According to Table 4, 49 (16.3%) people in the study population had a positive family history of type1 and type2 DM, and 250 (83.7%) individuals did not have a DM family history. Among 49 people with a family history of diabetes, 33 cases were in the case group (EH and EC groups), and 16 people were in the control group. The EH and EC incidence in people with a family history of diabetes was 40.8% and 26.5%, respectively Comparison of Hyperplasia and Endometrial Cancer Incidence According to BMI and GDM History According to Table 5, out of 72 people with a history of GDM, 55 (76.3%) were overweight (BMI˃25), and 17 (23.7%) were average weight. The number of people with EH and EC who were overweight and had a history of GDM was 26 and 24, respectively. There was a statistically significant relationship between obesity and the incidence of EH and EC in people with a history of GDM (P=0.001). The findings of the present study demonstrated that 47.3% of overweight people and 35.3% of normal-weight people had EH. Furthermore, 43.6% of overweight and obese and 41.2% of normal-weight people had EC. The incidence of EH and EC in overweight and obese people without a history of GDM was 31.1% and 16%, respectively (P=0.001). Table 3.Comparison of the history of GDM and diabetes mellitus with the incidence of EH and EC in the subjects GroupsHistory of GDM N (%)History of diabetes mellitus N (%)P-value Positive Negative Positive Negative 0.001 EH group32 (44.4)56 (24.7)17 (26.6)71 (30.1) EC group31 (43.1)33 (14.5)25 (39.1)39 (16.5) Control group9 (12.5)138 (60.8)22 (34.4)126 (53.4) Table 4.Comparison of the diabetes mellitus family history with the EH and EC incidence in the subjects GroupsFamily history of diabetes N (%)P-value Positive Negative 0.034 EH group20 (40.8)68 (27.2) EC group13 (26.5)50 (20) Control group16 (32.7)132 (52.8) Table 5.Comparison of the incidence of EH and EC according to BMI and history of GDM GroupsPositive history of GDMNegative history of GDMP-value BMI < 25 kg/m2 BMI˃ 25 kg/m2 BMI < 25 kg/m2 BMI ˃ 25 kg/m2 0.001 EH group6 (5.3%)26 (47.3%)18 (16.8%)37 (31.1%) EC group7 (41.2%)24 (43.6 %)14 (13.1%)19 (16%) Control group4 (23.5%)5 (9.1%)75 (70.1%)63 (52.9%) Discussion In the current research, we observed that a history of GDM was significantly associated with EH and EC, and history of DM was significantly associated with EC. The association between EH/EC and GDM and between EC and DM was also statistically significant after adjusting for age, parity, abortions, menarche age, menopause, BMI, HRT, and OCP history. Histologically, EH refers to the increased proliferation of the endometrial glands and is the most common EC cause, mainly in postmenopausal women (18). An essential hypothesis involved in the etiology of EH is the exposure of endometrial tissue to large amounts of estrogen along with the deficiency of progesterone, which augments the mitogenic activity of endometrial cells (3, 19). After hormone replacement therapy, progesterone protects the endometrium tissue against the mitogenic effects of estrogen in women. Consequently, factors that reduce progesterone are also associated with a higher risk of EC (20). EC's most important known risk factors are a family history of EC, anovulatory cycles, estrogen therapy, and certain diseases in which estrogen levels are elevated in the body, such as estrogen-producing ovarian tumors and PCOS (20, 21). In addition, some metabolic factors, such as obesity, insulin resistance, and hyperinsulinemia, may also be associated with EC development, as insulin might play a vital role in altering the blood levels of estrogen and androgen and stimulating the proliferation of endometrial cells (1, 22, 23). GDM is a metabolic disorder with relative insulin resistance during pregnancy, a known risk factor for developing type 2 diabetes (12, 14). It occurs when the function of insulin receptors is impaired, or they become insensitive and resistant to this hormone, which can eventually lead to elevated blood glucose levels and gestational diabetes (24). A study by Peng et al. in 2019 revealed that women with a history of GDM were significantly more likely to develop laryngeal, lung, kidney, breast, and thyroid cancers (25). Limited studies have been performed on the relationship between GDM and the risk of EH and EC. Wartko et al. showed a significant relationship between EH/EC and GDM after adjustment for gender, ethnicity, and maternal age during pregnancy (14). A systematic review and meta-analysis by Saed et al. suggested that GDM increases EC risk by 1.72 times (26). The data of our study were in line with the results of previous studies. However, in this study, we investigated the role of GDM in the incidence of EH and EC in women. Our results indicated that 24.1% and 21.3% of participants had a positive history of GDM and DM, respectively. The history of DM in the group with EC (39.1%) was also significantly higher than in other groups. According to our study results, DM and GDM were considered the risk factors for EC and increased the chance of developing EC by 1.84 (CI 95%: 1.26-2.68) and 17.98 (CI 95%: 6.73-48.08) times, respectively. Moreover, the current investigation showed that DM did not significantly affect the chances of developing EH, while GDM increased the risk of EH by almost seven-fold. Wartko et al. also reported the significant association of EH and EC with a history of GDM in obese women (OR=1.31). They revealed that GDM could be associated with an increased risk of EH and EC by elevating insulin levels, which may synergistically raise estrogen levels in the body (17). The results of Lucenteforte et al. demonstrated that EC risk in people with diabetes and obese diabetic women was 1.7 and 1.5 times higher than in non-diabetic women, respectively (27). The findings of these studies are consistent with our research. We evaluated the association of EH and EC with BMI. The incidence of EH and EC in obese women with a history of GDM was 91.9%, while this rate in normal-weight women with a history of GDM was 76.5%. As a result, EH and EC incidence were observed to have a significant relationship with BMI and GDM history. A possible hypothesis that could justify and confirm our findings and the results of other studies is the role of insulin in stimulating endometrial cell growth (1). Type 2 DM and GDM are the disorders of insulinopathy in which circulating insulin levels increase (28, 29). Higher circulating insulin might stimulate the proliferation of endometrial stromal cells in several ways: (1) inducing cell proliferation by insulin binding to its receptors on the surface of endometrial cells (30), reducing the levels of the globulin-binding hormone that increase the active estrogens serum levels (31), decreasing the levels of insulin-like growth factor-1 protein (IGFBP-1) and raising the levels of IGF-1, which act as mitogen agents and cause abnormal endometrial cell proliferation (32), and activating mitogen-activated protein kinase and phosphoinositide 3-kinase pathways downstream of insulin receptors on the surface of endometrial cells, which are involved in the pathogenesis of ECs (33). The small number of patients and the lack of histological data for EH and EC groups were the limitations of the present study. However, physical activity status before pregnancy, GDM treatment, pre- and post-pregnancy obesity status, smoking history, the progression of type 2 diabetes after pregnancy, and insulin resistance status during pregnancy were not evaluated in this study. Conclusion In this case series, among 25 women with confirmed According to the results of this study, the history of diabetes mellitus and GDM are significantly effective in the incidence of EC. Therefore, women with a history of GDM are considered a high-risk group and should be managed and followed up. This study's findings can be used as a fundamental study to conduct more prospective studies on EH and EC's etiology and pathogenesis. It is recommended that prospective studies be performed with a larger sample volume and multicenter to obtain more reliable results. Author's Contributions ZB, AGH, MR, TM, and KH conceived the research question, designed the protocol, and were involved in the literature search, study selection, and data extraction. ZB, AGH, MR, TM, and KH contributed to data acquisition, analysis, and interpretation. ZB and KH created the tables, figures, and ZB, AGH, and TM contributed to both the draft and final versions of the manuscript.Contributed to study design/conduct/analysis: ZB, AGH, MR, TM and KH Acknowledgments We would like to thank the clinical research development unit of Rouhani Hospital of Babol. Conflicts of Interest The authors report no conflicts of interest. The authors alone are responsible for the content and writing of this article. Ethical Approval Our study proposal was approved by the ethics committee of the Babol University of Medical Sciences, Babol, Iran. Ethics number: IR.MUBABOL.HRI.REC.1397.268 Funding/Support This study was financially supported by the Babol University of Medical Sciences, Babol, Iran. Informed Consent Written informed consent was obtained from all participants after explaining the objectives of the study. ##### References Sidorkiewicz I, Jóźwik M, Niemira M, Krętowski A. Insulin Resistance and Endometrial Cancer: Emerging Role for microRNA. Cancers. 2020;12(9):2559. [DOI:10.3390/cancers12092559] [PMID] [PMCID] Setiawan VW, Yang HP, Pike MC, McCann SE, Yu H, Xiang Y-B, et al. Type I and II endometrial cancers: have they different risk factors? J Clin Oncol. 2013;31(20):2607. [DOI:10.1200/JCO.2012.48.2596] [PMID] [PMCID] Sanderson PA, Critchley HO, Williams AR, Arends MJ, Saunders PT. New concepts for an old problem: the diagnosis of endometrial hyperplasia. Hum Reprod Update. 2017;23(2):232-54. [DOI:10.1093/humupd/dmw042] [PMID] [PMCID] Abazari O, Shafaei Z, Divsalar A, Eslami-Moghadam M, Ghalandari B, Saboury AA. 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Oncol Lett. 2014;8(5):1993-9. [DOI:10.3892/ol.2014.2466] [PMID] [PMCID] Tian W, Teng F, Gao J, Gao C, Liu G, Zhang Y, et al. Estrogen and insulin synergistically promote endometrial cancer progression via crosstalk between their receptor signaling pathways. Cancer Biol Med. 2019;16(1):55. [DOI:10.20892/j.issn.2095-3941.2018.0157] [PMID] [PMCID] ) Volume 7, Issue 5 - Serial Number 28 July and August 2022 Pages 414-421 Files XML PDF 685.84 K History Receive Date: 21 November 2021 Revise Date: 02 February 2022 Accept Date: 05 February 2022 Publish Date: 07 July 2022 Share "Email") "Print") How to cite RIS EndNote Mendeley BibTeX APA MLA HARVARD CHICAGO VANCOUVER Statistics Article View: 496 PDF Download: 154 × APA Bouzari, Z. , Mohammadi, T. , Ranaei, M. , Hajian-Taliki, K. and Ghanbarpour, A. (2022). Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer. Journal of Obstetrics, Gynecology and Cancer Research, 7(5), 414-421. doi: 10.30699/jogcr.7.5.414 × MLA Bouzari, Z. , , Mohammadi, T. , , Ranaei, M. , , Hajian-Taliki, K. , and Ghanbarpour, A. . "Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer", Journal of Obstetrics, Gynecology and Cancer Research, 7, 5, 2022, 414-421. doi: 10.30699/jogcr.7.5.414 × HARVARD Bouzari, Z., Mohammadi, T., Ranaei, M., Hajian-Taliki, K., Ghanbarpour, A. (2022). 'Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer', Journal of Obstetrics, Gynecology and Cancer Research, 7(5), pp. 414-421. doi: 10.30699/jogcr.7.5.414 × CHICAGO Z. Bouzari , T. Mohammadi , M. Ranaei , K. Hajian-Taliki and A. Ghanbarpour, "Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer," Journal of Obstetrics, Gynecology and Cancer Research, 7 5 (2022): 414-421, doi: 10.30699/jogcr.7.5.414 × VANCOUVER Bouzari, Z., Mohammadi, T., Ranaei, M., Hajian-Taliki, K., Ghanbarpour, A. Association Between Gestational Diabetes History with Endometrial Hyperplasia and Cancer. 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12167	https://passnownow.com/classwork-series-and-exercises-mathematics-jss3-word-problems/	Classwork Series and Exercises {Mathematics- JSS3}: Word Problems - Passnownow Skip to content Passnownow Rated 4.8/5 by parents & students Home Class JSS1 Class Notes JSS2 Class Notes JSS3 Class Notes SSS1 Class Notes SSS2 Class Notes SSS3 Class Notes Past Questions JSSCE SSCE UTME Educational Resources Teachers Resources Post Secondary Education & Career Counseling Blog About About Passnownow Contact us Plans & Payments Home Class JSS1 Class Notes JSS2 Class Notes JSS3 Class Notes SSS1 Class Notes SSS2 Class Notes SSS3 Class Notes Past Questions JSSCE SSCE UTME Educational Resources Teachers Resources Post Secondary Education & Career Counseling Blog About About Passnownow Contact us Plans & Payments Log In Register Classwork Series and Exercises {Mathematics- JSS3}: Word Problems Oluwayemi Alabi November 29, 2013 One Comment Word Problems Word problems are mathematical problems which are written in words. This means that to be able to solve the problems, we must understand the words that are often used in word problems. Some of these words are; Sum:The result we get when we add numbers together e.g. the sum of 2 and 3 = 2 + 3 = 5 Difference:The result of subtracting one number from another. It can be positive or negative Positive difference:This is when we subtract the smaller number from the bigger number. Example: The positive difference between 2 & 5 = 5 – 2 = + 3 Negative difference:We get this by subtracting the bigger number from the smaller number. Example: The negative difference between 2 & 5 = 2 – 5 = – 3 Product:The result of multiplication e.g. Product of 4 and 6 = 4 x 6 = 24 Consecutive:Numbers that follow each other e.g.21, 22, 23, 24 are consecutive numbers Consecutive odd numbers:Odd numbers that follow each other e.g. 21, 23, 25, 27… Consecutive even numbers: Even numbers that follow each other e.g. 22, 24, 26, 28… NOTE: The difference between consecutive numbers is 1 but the difference between consecutive odd/consecutive even numbers is 2. Sum and Difference We have said earlier that sum of a set of numbers is the result of adding the numbers together and the difference is what we get when we subtract one number from another. Now let us take some examples to help our understanding of the topic. Example 1: Find the sum of 23, 27 and 33 Solution: Sum of numbers = 23 + 27 + 33 = 83 Example 2: Find the positive difference between – 7 and – 12 Solution: Positive difference = Higher – Lower Since – 7 is the higher number, positive difference will be – 7 – (– 12) = – 7 + 12 = +5 Example 3: The sum of three consecutive numbers is 63. Find the numbers. Solution: When we have word problems where one or more of the quantities is unknown, we represent that quantity with a letter So let the numbers be “t”, (t + 1) and (t + 1+1 = t +2) from difference between consecutive numbers Therefore t + t + 1 + t + 2 = 63; 3t + 3 = 63; 3t = 60; t = 20 So the numbers are 20, 21 and 22 Example 4: The difference between – 3 and a number is 8. Find the two possible values for the number. Solution: Let the number be y; we are not told if it is positive or negative difference so we solve for both Positive difference = y – (– 3) = 8; y + 3 = 8; y = 5 Negative difference= – 3 – y = 8 If we gather like terms; – 3 – 8 = y; y = – 11 TRY THESE: The sum of four consecutive odd numbers is 80. Find the numbers [fusion_builder_container hundred_percent=”yes” overflow=”visible”][fusion_builder_row][fusion_builder_column type=”1_1″ background_position=”left top” background_color=”” border_size=”” border_color=”” border_style=”solid” spacing=”yes” background_image=”” background_repeat=”no-repeat” padding=”” margin_top=”0px” margin_bottom=”0px” class=”” id=”” animation_type=”” animation_speed=”0.3″ animation_direction=”left” hide_on_mobile=”no” center_content=”no” min_height=”none”][17, 19, 21 & 23] The difference between 12.6 and a number is 5.4. Find the two possible values of the number [7.2, 18.0] Product When we multiply 2 or more numbers, the result obtained is called the product of the numbers. Example 1: Find the product of 2, 3 and 9 Solution: Product of 2, 3 and 9 = 2 x 3 x 9 = 54 Example 2: Find the product of 3 3/8 and 3 5/9 Solution: To multiply mixed fractions, it is important to convert them to improper fraction 3 3/8 = 27/8 and 3 5/9 =32/9 Therefore, the product of 3 3/8 and 3 5/9= 27/8 x 32/9 = 12 Example 3: The product of two numbers is 54. If one of the numbers is 27, find the other Solution: Let the number be K. K x 27 = 54 so K = 54/27 = 2 Example 4: Find the product of –2, –5 and +9 Solution: Product = (–2) x (–5) x (+9) = +90 The final sign is + because the two negative (–) signs cancel out themselves TRY THESE: The product of three numbers is 0.084. If two of the numbers are 0.7 and 0.2, find the third number [ANS: 0.6] Find the product of 12, 0.6 and 6¼. [ANS: 45] Combined Operations When we have the combination of addition, subtraction and multiplication in a word problem, the best thing to do is to take the operations one after the other following the BODMAS order of arithmetic operations. Example 1: Find the sum of the product of 5/9 and 3/5 and the product of 5/9 and 3/20 Solution:We want to find (5/9 x 3/5) + (5/9 x 3/20) (5/9 x 3/5) = 15/45 = 1/3 while (5/9 x 3/20) = 15/180) = 1/12 So (5/9 x 3/5) + (5/9 x 3/20) = ⅓ + 1/12 = 5/12 Example 2: Find the product of the difference between 2 and 7 and the sum of 2 and 7. Solution: Product of (difference between 2 and 7) and (sum of 2 and 7) = (7 – 2) x (7 + 2) = 5 x 9 = 45 TRY THESE: Find the difference between the sum of 16 and 17 and the product of 8 and 4. [ANS: 1] Find the sum of the product of 9 and 5 and the product of 10 and 6. [ANS: 105] Find the product of 6 and the sum of 2½ and 4 ½ [ANS: 42] Expressions with Fractions When we have expressions with fractions follow the simple steps below, it will help you Step I: Interpret and write out the expressions Step II: Simplify expressions in brackets (or the sub-expressions) Example 1:Divide 52 by the sum of 1 and the product of 5 and 6 Solution: 52 ÷ [1 + (5 x6)] = 52 ÷ (1 + 30) = 52 ÷ 31 = 1 21/31 Example 2: Subtract 16 from the product of 4 and 10; then divide the result by 6 Solution:[(4 x 10) – 16] ÷ 6 = (40 – 16) ÷ 6 = 24 ÷ 6 = 4 NOTE: The use of brackets is to make our work clearer and to make the solutions neater and more distinct. It also shows that we understand our expressions and we can interpret them correctly. TRY THESE: Divide 48 by the sum of 4 and 8 [ANS: 4] Find one-seventh of the sum of 19 and the product of 4 and 11. [ANS: 9] Expressions with equations When we want to solve word problems that have equations it is important to do the following. Step I: Write out the equations Step II: Use LCM to clear out the fractions (if fractions are involved) Step III: Carefully carry out the actions and take note of the negative sign if present. Example 1: 13 was subtracted from the product of 4 and a certain number. The result is equal to the sum of 5 and the original number. Find the number Solution: Let the unknown number be a. So (a x 4) – 13 = a + 5 ====> 4a – 13 = a + 5 Gather like terms; 4a – a = 5 + 13 3a = 18; a = 6 Example 2: Find the number that when ¾ of it is added to 3½, the sum is the same as when 2/3 of it is added to 6½ Solution: Let the number be f. ¾f + 3½ = 6½ + ⅔f Convert the mixed fractions to improper fractions so that we can use LCM ¾f + 7/2 = 13/2 + ⅔f; LCM of all denominators is 12 so we would multiply though by 12 9f + 42 = 78 + 8f ====> 9f – 8f = 78 – 42; f = 36 TRY THIS: I add 12 to a number and then double their sum. The result is one and a half times what I get when I double the original number and add 12. Find the number. [ANS: 6] Tests and Exercises The sum of the square roots of 9 and 25 is… (a) 15 (b) 5 (c) 3 (d) 8 Guideline: Square root of 9 = 3; Square root of 25 = 5 Sum of the square roots = 3 + 5 = 8 Answer: 8 Find the number which when divided by 0.7 gives 0.4 (a) 0.28 (b) 1.28 (c) 1.18 (d) 28 Guideline: Let the number be y Therefore, y ÷ 0.7 = 0.4; multiply both sides by 0.7 0.7 x (y/0.7) = 0.4 x 0.7; y = 0.28 Answer: 0.28 If a certain number is doubled, then the result divided by 7, the final result is 2. What is the number? (a) 5 (b) 6 (c) 7 (d) 8 Guideline: Let the number be P (P x 2) ÷ 7 = 2; 2P/7 = 2 Multiply both sides by 7====> 2P = 14; then divide the result by 2 P = 7 Answer: 7 Find one-sixth of the positive difference between 36 and 63 (a) – 4.5 (b) +3.6 (c) + 4.5 (d) +7 Guideline: Positive difference between 36 & 63 = 63 – 36 = 27 One-sixth of the difference = 27 ÷ 6 = +4.5 Answer: C The sum of three consecutive even numbers is 72. The highest of the three numbers is (a) 18 (b) 22 (c) 24 (d) 26 Guideline: Let the smallest number by m, so the next numbers will be (m+2) and (m+2+2) = (m + 4) Therefore m + m+2 + m +4 = 72 3m + 6 = 80; 3m = 72 – 6; 3m = 66; m = 66/3 = 22 If m = 22, then the biggest number m + 4 = 22 + 4 = 26 Answer: 26 For more on classwork notes, click here [/fusion_builder_column][/fusion_builder_row][/fusion_builder_container] 1 thought on “Classwork Series and Exercises {Mathematics- JSS3}: Word Problems” Chidinma November 19, 2017 at 8:56 pm The sum of two numbers is 38. When 8 is added to twice one of the number, the result is 5 times the other numbers find the number. Reply Leave a Comment Cancel Reply Your email address will not be published.Required fields are marked Type here.. Name Email Website [x] Save my name, email, and website in this browser for the next time I comment. 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12168	https://faculty.fiu.edu/~mebela/chm3410_chapter9.pdf	Chemical equilibrium Spontaneous chemical reactions The direction of spontaneous change at constant temperature and pressure – towards lower values of the Gibbs energy. We locate the equilibrium composition of the reaction mixture by calculating its Gibbs energy and identifying the composition that corresponds to minimum G. The reaction Gibbs energy Consider the equilibrium AB. Suppose an infinitesimal amount dζ of A turns into B: dnA = -dζ dnB = +dζ ζ - the extent of reaction, has the dimensions of amount of substance and is reported in moles. When the extent of reaction changes by a finite amount Δζ, the amount of A present changes from nA,0 to nA,0 - Δζ and the amount of B changes from nB,0 to nB,0 + Δζ. The reaction Gibbs energy, ΔrG – the slope of the graph of the Gibbs energy plotted against the extent of reaction: € ΔrG = ∂G ∂ζ       p,T dG = µAdnA + µBdnB = -µAdζ + µBdζ = (µB - µA)dζ € ∂G ∂ζ       p,T = µB −µ A ΔrG = µB - µA ΔrG – can be interpreted as the difference between the chemical potentials of the reactants and products at the composition of the reaction mixture. Because chemical potential varies with the composition, the slope of G changes as the reaction proceeds. Because the reaction runs in the direction of decreasing G (down the slope), the reaction A → B is spontaneous when µB < µA, whereas the reverse reaction is spontaneous µB > µA. The slope is zero and the reaction is spontaneous in neither direction when ΔrG = 0 and µB = µA. If we can find the composition of the reaction mixture that ensures µB = µA, then we can identify the composition of the reaction mixture at equilibrium. Exergonic and endergonic reactions At constant temperature and pressure: If ΔrG < 0, the forward reaction is spontaneous – exergonic (work-producing). If ΔrG > 0, the reverse reaction is spontaneous – endergonic (work-consuming). If ΔrG = 0, the reaction is at equilibrium. The description of equilibrium Perfect gas equilibria When A and B are perfect gases, ΔrG = µB - µA = (µB ∅ + RT ln pB) - (µA ∅ + RT ln pA) = = ΔrG∅ + RT ln(pB/pA) ΔrG = ΔrG∅ + RT ln(Q) Q = pB/pA Q – a reaction quotient. The standard reaction Gibbs energy is defined as a difference in the standard Gibbs energies of formation, so in practice we calculate ΔrG∅ as ΔrG∅ = ΔfG∅(B) – ΔfG∅(A) At equilibrium, ΔrG = 0 0 = ΔrG∅ + RT ln(K) RT ln(K) = -ΔrG∅ K = (pB/pA)equilibrium This is a link between tables of thermodynamic data and K. In molecular terms, the minimum in the Gibbs energy, which corresponds to ΔrG = 0, stems from the Gibbs energy of mixing of the two gases. Consider a hypothetical reaction in which A molecules change into B molecules without mingling together. The Gibbs energy of the system changes from G∅(A) to G∅(B) in proportion to the amount of B that had been formed and the slope of the plot of G against the extent of reaction is a constant and equal to ΔrG∅ at all stages of the reaction. There is no minimum in the graph (except of pure B). However, in fact, the newly produced B molecules mix with the surviving A molecules: ΔmixG = nRT(xAlnxA + xBlnxB) This expression makes a U-shaped contribution to the total change in Gibbs energy and there is now a minimum in the Gibbs energy, and its position corresponds to the equilibrium composition of the reaction mixture. When ΔrG∅ > 0, the equilibrium constant K < 1. Therefore, at equilibrium the partial pressure of A exceeds that of B, which means that the reactant A is favored at equilibrium. When ΔrG∅ < 0, the equilibrium constant K > 1, so at equilibrium the partial pressure of B exceeds that of A – the product B is favored at equilibrium. The general case of a reaction Consider the reaction 2 A + B → 3 C + D We can rewrite as 0 = 3 C + D – 2 A – B € 0 = νJJ J ∑ νJ – stoichiometric numbers: νA = -2, νB = -1, νC = 3, νD = 1 A stoichiometric number is positive for products and negative for reactants. We define ζ so that if it changes by Δζ, then the change in the amount of any species J is νJΔζ. To express the equation N2(g) + 3 H2(g) → 2 NH3(g) we rearrange it to 0 = 2 NH3(g) – N2(g) – 3 H2(g) ν(N2) = -1, ν(H2) = -3, ν(NH3) = +2 If initially there is 10 mol N2 present, then when the extent of reaction changes from ζ = 0 to ζ =1 (Δζ = +1 mol), the amount of N2 changes from 10 mol to 9 mol. All the N2 has been consumed when ζ = 10. When Δζ = +1 mol, the amount of H2 changes by -3×(1 mol) = -3 mol and the amount of NH3 changes by +2×(1 mol) = +2 mol. We will show that the Gibbs energy of reaction can always be written ΔrG = ΔrG∅ + RT ln(Q) with the standard Gibbs energy calculated from € ΔrGΘ = νΔ f Products ∑ GΘ − νΔ f Reactants ∑ GΘ € ΔrGΘ = νJΔ f J ∑ GΘ J ( ) The quotient: Q = (activities of products)/( activities of reactants) € Q = aJ νJ J ∏ Because reactants have negative stoichiometric numbers, they automatically appear as the denominator. For pure solids and liquids, the activity is 1, so they make no contribution to Q even when they appear in the chemical equation. Consider the reaction 2 A + 3 B → C + 2 D We can rewrite as 0 = 3 C + D – 2 A – B € 0 = νJJ J ∑ νA = -2, νB = -3, νC = +1, νD = +2 € Q = aCaD 2 aA 2aB 3 Justification: When the reaction advances by dζ, dnJ = νJdζ € dG = µJdnJ J ∑ = ν JµJ J ∑       dζ € ΔrG = ∂G ∂ζ       p,T = ν JµJ J ∑ µJ = µJ ∅ + RT ln aJ € ΔrG = νJµJ Θ J ∑ ΔrGΘ 6 7 4 8 4 + RT lnaJ νJ J ∑ = ΔrGΘ + RT lnaJ νJ J ∑ € = ΔrGΘ + RT ln aJ ν J J ∏ Q 6 7 8 = ΔrGΘ + RT lnQ At equilibrium, the slope of G is zero, ΔrG = 0, € K = aJ ν J J ∏       equilibrium For K we use the values of activities at equilibrium and for Q we use their values at the specific stage of reaction. An equilibrium constant K expressed in terms of activities (or fugacities for gases) – a thermodynamic equilibrium constant. Because activities are dimensionless numbers, the thermodynamic equilibrium constant is also dimensionless. In elementary applications, the activities are often replaced by the numerical values of molalities or molar concentrations, and fugacities are replaced by partial pressures. In either case, the resulting expressions are only approximations. The approximation is particularly severe for electrolyte solutions – in them the activity coefficients differ from 1 even in very dilute solutions. ΔrG = 0: RTln K = -ΔrG∅ – an exact and highly important thermodynamic relation, enables us to predict the equilibrium constant of any reaction from tables of thermodynamic data and to predict the equilibrium composition of the reaction mixture. Example 1. Calculating an equilibrium constant Calculate the equilibrium constant for the ammonia synthesis reaction, N2(g) + 3 H2(g) → 2 NH3(g) at 298 K and show how K is related to the partial pressures of the species at equilibrium when the overall pressure is low enough for the gases to be treated as perfect. We calculate the standard Gibbs energy and convert it to the equilibrium constant. Because the gases are perfect, we replace each fugacity by a partial pressure. -ΔrG∅ = 2ΔfG∅(NH3,g) – {ΔfG∅(N2,g) + 3ΔfG∅(H2,g)} = 2ΔfG∅(NH3,g) = 2 × (-16.5 kJ mol-1) ln K = -2 × (-16.5 kJ mol-1) /{(8.3145 J K-1 mol-1)×(298 K) K = aNH3 2 aN2aH2 3 = fNH3 pΘ ( ) 2 fN2 pΘ ( ) fH2 pΘ ( ) 3 = fNH3 2 pΘ2 fN2 fH2 3 ≈pNH3 2 pΘ2 pN2 pH2 3 K = 6.1×105 the thermodynamically exact result Example 2. Estimating the degree of dissociation at equilibrium The standard Gibbs energy of reaction for the decomposition H2O(g) → H2(g) + 1/2 O2(g) is +118.0 kJ mol-1 at 2300 K. What is the degree of dissociation of H2O at 2300 K and 1.00 bar. The equilibrium constant is obtained from the standard Gibbs energy of reaction, so the task is to relate the degree of dissociation, α, to K and then find its numerical value. We proceed by expressing the equilibrium composition in terms of α and solve for α in terms of K. As the standard Gibbs energy of reaction is large and positive, we expect K and α to be small. ln K = (+118.0×103 J mol-1) /{(8.3145 J K-1 mol-1)×(2300 K) K = 2.08×10-3 The equilibrium composition can be expressed in terms of α by drawing up the following table: H2O H2 O2 equilibrium table Initial amount n 0 0 The equilibrium constant Change to reach equilibrium -αn +αn +(1/2)αn is therefore Amount at equilibrium (1 - α)n αn (1/2)αn Mole fraction € 1−α 1+ 1 2 ( )α € α 1+ 1 2 ( )α € 1 2 ( )α 1+ 1 2 ( )α Partial pressure € 1−α ( )p 1+ 1 2 ( )α € αp 1+ 1 2 ( )α € 1 2 ( )αp 1+ 1 2 ( )α € K = pH2 pO2 1/2 pH2O = α3/2p1/2 1−α ( ) 2+α ( ) 1/2 For α << 1, € K ≈α3/2p1/2 21/2 € α ≈21/2K ( ) 2/3 = 0.0205 About 2% of water has decomposed. The relation between equilibrium constants We need to express the thermodynamic equilibrium constant in terms of the mole fractions, xJ, or molalities, bJ, of the species. For that, we need to know the activity coefficients and then use aJ = γjxJ or aJ = γjbJ/b∅ For an equilibrium A + B  C + D € K = aCaD aAaB = γCγ D γ Aγ B bCbD bAbB = KγKb The activity coefficients must be evaluated at the equilibrium composition, which may involve a complicated calculation, because the latter is known only if the equilibrium composition is already known. In elementary applications, and to begin the iterative calculation of the concentrations in a real example, we can assume that the activity coefficients are all close to 1, Kγ = 1. Then K ≈ Kb and equilibria are discussed in terms of molalities (or molar concentrations). Equilibria in biological systems For biological systems it is appropriate the biological standard state, pH = 7. The relation between the thermodynamic and biological standard Gibbs energies of reaction for the reaction A + ν H+(aq) → P is ΔrG⊕ = ΔrG∅ + 7νRT ln 10 There is no difference between the two standard values if hydrogen ions are not involved in the reaction. Consider the reaction NADH(aq) + H+(aq) → NAD+(aq) + H2(g) at 37°C, for which ΔrG∅ = -21.8 kJ mol-1. NADH is the reduced form of nicotinamide adenine dinucleotide and NAD+ is its oxidized form the molecules play an important role in the later stages of the respiratory process. 7 ln 10 = 16.1 ΔrG⊕ = -21.8 kJ mol-1+16.1×(8.3145×10-3 kJ K-1 mol-1)(310 K) = +19.7 kJ mol-1 Here, the biological standard value is opposite in sign to the thermodynamic standard value. The response of equilibria to the conditions Equilibria respond to changes in pressure, temperature, and concentrations of reactants and products. The equilibrium constant is not affected by the presence of a catalyst or an enzyme. Catalysts increase the rate at which equilibrium is attained but do not affect its position. How equilibria respond to pressure The equilibrium constant depends on the value of ΔrG∅, which is defined at a single, standard pressure. Therefore, the value of ΔrG∅ and hence K is independent of the pressure at which the equilibrium is actually established: € ∂K ∂p ( )T = 0 This does not necessarily mean that the equilibrium composition is independent of pressure. We distinguish between two ways to apply pressure: 1) The pressure within a reaction vessel can be increased by injecting an inert gas into it. However, so long as the gases are perfect, the addition of gas leaves all the partial pressures of the reacting gases unchanged. The addition of an inert gas leaves the molar concentrations of the original gases unchanged, as they continue to occupy the same volume – no effect on the equilibrium composition of the system. 2) The pressure may be increased by confining the gases to a smaller volume. Then the partial pressures are changed. The equilibrium constant itself is independent of pressure. However, compression can adjust the individual partial pressures of the reactants and products in such a way that, although each one changes, their ratio (as it appears in the equilibrium constant) remains the same. Consider the perfect gas equilibrium A  2 B, € K = pB 2 pA pΘ – remains constant only if an increase in pA cancels and increase in the square of pB. This relative steep increase in pA compared to pB will occur if the equilibrium composition shifts in favor of A at the expense of B. The number of A molecules will increase as the volume of the container decreases. The increase in the number of molecules A and the corresponding decrease in the number of B molecules due to compression – a special case of the Le Chatelier’s principle: When a system at equilibrium is subjected to a disturbance, the composition of the system adjusts so as to minimize the effect of the disturbance. If a system at equilibrium is compressed, then the reaction will adjust so as to minimize the increase in pressure. This it can do by a reduction in the number of molecules in the gas phase, which implies a shift A  2 B. The quantitative treatment of this effect shows that the extent of dissociation, α, of A into B is € α = 1 1+ 4 p/K       1/2 Here p is to be understood as p/p∅. Suppose that there is an amount n of A present initially (and no B). At equilibrium the amount of A is (1 - α)n and the amount of B is 2αn. The mole fractions present at equilibrium are € xA = 1−α ( )n 1−α ( )n+2αn = 1−α 1+α € xB = 2α 1+α € K = pB 2 pA = xB 2 p2 xA p = 4α2p 1−α2 The result shows that, even though K is independent of pressure, the amount of A and B do depend on pressure. As p is increased, α decreases, in accord with Le Chatelier’s principle. To predict the effect of an increase in pressure on the composition of the ammonia synthesis at equilibrium, note that the number of gas molecules decreases from 4 to 2 – an increase in pressure will favor the product. € K = pNH3 2 pN2 pH2 3 = xNH3 2 p2 xN2xH2 3 p4 = Kx p2 Kx – the part of the equilibrium constant that contains the equili-brium mole fractions of reactants and products. Doubling the pressure must increase Kx by a factor of 4 to preserve the value of K. The effect of temperature The equilibrium constant of a reaction changes when the temperature is changed. According to Le Chatelier’s principle, we can expect a reaction to respond to a lowering of temperature by releasing heat and to respond to an increase of temperature by absorbing heat. The equilibrium constant of an endothermic reaction increases with temperature: increased temperature favors the products. The equilibrium constant of an exothermic reaction decreases with temperature: increased temperature favors the reactants. The van’t Hoff equation (the van’t Hoff isochore) (a) € d lnK dT = ΔrH Θ RT 2 (b) € d lnK d 1/T ( ) = −ΔrH Θ R Derivation: (a): € lnK = −ΔrGΘ RT € d lnK dT = −1 R d ΔrGΘ /T ( ) dT = −1 R −ΔrH Θ T 2       = ΔrH Θ RT 2 We differentiated ln K and use the Gibbs-Helmholtz equation here. (b): € d 1/T ( ) dT = −1 T 2 € dT = −T 2d 1/T ( ) € d lnK d 1/T ( ) = −ΔrH Θ R The equation shows that d lnK / dT < 0 (and dK/dT < 0) for exothermic reaction under standard conditions (ΔrH∅ < 0). A negative slope means that ln K (and K itself) decreases as T rises – the equilibrium shifts away from products. The opposite occurs in the case of endothermic reactions. ΔrG∅ = ΔrH∅ - TΔrS∅ -ΔrG∅/T = -ΔrH∅/T + ΔrS∅ When the reaction is exothermic, -ΔrH∅/T corresponds to a positive change of entropy in the surroundings and favors the formation of products. When the temperature is raised, -ΔrH∅/T decreases, and the increasing entropy of the surroundings has a less important role – the equilibrium lies less to the right. When the reaction is endothermic, the principal factor is the increasing entropy of the reaction system. The importance of the unfavorable change of entropy of the surroundings is reduced if the temperature is raised and the reaction is able to shift towards products. Example 3. Measuring a reaction enthalpy The data below show the temperature dependence of the equilibrium constant of the reaction Ag2CO3(s)  Ag2O(s) + CO3(g) Calculate the standard reaction enthalpy of decomposition. T/K 350 400 450 500 K 3.98×10-4 1.41×10-2 1.86×10-1 1.48 Provided the reaction enthalpy can be assumed to be independent of temperature, a plot of –ln K against 1/T should be a straight line of slope ΔrH∅/R. T/K 350 400 450 500 (103 K)/T 2.86 2.50 2.22 2.0 -ln K 7.83 4.26 1.68 -0.39 The slope of the graph is +9.6×103, so ΔrH∅ = 9.6×103 × R = +80 kJ mol-1 This is a noncalometric method to get ΔrH∅. The value of K at different temperatures To find the value of the equilibrium constant at a temperature T2 in terms of its value K1 at another temperature T1, we integrate the equation between these two temperatures: € lnK2 −lnK1 = −1 R ΔrH Θd 1/T ( ) 1/T1 1/T2 ∫ If we suppose that ΔrH∅ varies only slightly with temperature, € lnK2 −lnK1 = −ΔrH Θ R 1 T2 −1 T 1       To estimate the equilibrium constant for the synthesis of ammonia at 500 K from its value at 298 K (6.1×105) we use the standard reaction enthalpy: ΔrH∅ =2ΔfH∅(NH3,g) and assume that its value is constant over the range of temperatures: ln K2 = ln(6.1×105) – (-92.2×103 J mol-1 / 8.3145 J K-1 mol-1) ×{1/(500 K) – 1/(298 K)} = -1.71 K2 = 0.18 Knowledge of the temperature dependence of the equilibrium constant for a reaction can be useful in the design of laboratory and industrial processes: synthetic chemists can improve the yield of a reaction by changing the temperature of the reaction mixture, etc.
12169	https://www.youtube.com/watch?v=FKJsR_K5ysE	Find TENSION and REACTION FORCE on a Cable Hung Beam \| Statics for Engineers INTEGRAL PHYSICS 51700 subscribers 303 likes Description 36193 views Posted: 20 Jan 2022 I am not sponsored by Sharpie... yet! A beam, supported at one end by an angled cable, and at the other by a pin is loaded with a 50lb block at one end. Calculate the tension force in the supporting cable as well as the force by the pin on the beam. This statics problem uses the sum of all torques around a point as well as the sum of all forces in both the horizontal and vertical axes to solve for the external forces on the beam. This problem comes up in physics and engineering courses such as PLTW POE as well as tests such as the AP Physics C Mechanics test and JEE. 27 comments Transcript: [Music] all right youtube today we're going to take a look at a block which is hanging from the end of a beam now the beam is held up partially by a wall or pin on a wall and partially by a cable and in this problem we're going to work out both the tension in the cable as well as how hard this little pin is pushing on the left side of this beam now there's a couple ways we can do this problem but what i find is the simplest is to go through and start this problem by looking at what might happen if we're to simply cut the string you see if we were to cut the string everything would just fall downward or rotate around this pin and it's that rotation that the cable is preventing so we're going to start by looking at the sum of all torques around this pin so in order for this beam and this block to remain static we need the sum of all torques around this point as well as any other point to equal zero so we're going to start by looking at how each force produces a torque around this point now there's actually four forces in this problem the first being the force by the wall the next being the tension of the cable and the last two being the weight or force by gravity on each of these objects now you'll remember torque is given by the equation fr sine theta so all we're going to do is take a look at each force to see how it contributes to the total torque acting around this pin so starting at the force by the wall there's going to be a force by the wall acting at a distance from the pin of zero that means the force by the wall is producing no torque around this point moving to our 10 newton force acting in the middle of this beam there's going to be 10 newtons acting at a radius of 1 meter and realize this force is straight down so it's going to be at a right angle to the radius vector so that means we're going to be dealing with an angle of 90 degrees between the force factor and the radius vector sine of 90 works out to be next we have our 50 newton force acting at a radius of 2 meters and again this force is acting straight down so it's at a right angle to the beam and last we have this tension by the string tension is gonna be some unknown value t it's acting at a radius of two meters and there's gonna be some angle between this cable and the beam and we're just going to say that angle is 30 degrees and this isn't something that you'd magically figure out just by looking at the picture this is something that i should have written down when i made the whole drawing but i forgot sorry youtube so looking at this entire function the only thing we don't know here is t the tension in the cable and we find the tension in the cable is 110 newtons so now that we know the total tension in the cable we need to turn our attention to the force by the wall the issue being we can't use torque to solve for the force at the wall here so what we're going to need to do is look at our other conditions for static equilibrium that is looking at the sum of all forces within any axis i'm going to start by looking at the sum of all forces in the y axis now if we want to keep this beam from moving vertically the sum of all forces in the y-axis needs to equal zero now it's clear we have two forces down but we also have two forces which are acting at least somewhat upward and we need to be careful with our positives and negatives when looking at the sum of all forces here so i'm going to go ahead and say the up so we're going to say up and to the right are the positive direction in this problem so looking first at the force by the wall the force by the wall is not entirely upward it's up and at an angle so it has a vertical component we're going to call that the force by the wall in the y-axis so there's that force by the wall and the y-axis plus this 10 newton force which is acting downward so it's going to be 10 newtons in the negative direction there's this 50 newton force downward this can be 50 newtons in the negative direction and last there's going to be the tension in this string which is acting partially upward just like this force was broken up to into its components this force can be broken up into its components so we'll call that vertical component of tension t y now realize this vertical component of tension is something we can solve for because we know the magnitude of the tension force that's 110 newtons as well as this angle we can solve for that vertical component and that's going to allow us to solve for the vertical component of force by the wall now realize this isn't the total force by the wall this is simply the vertical component of this force which is up and to the right so in order to solve for the total force by the wall we're going to need to solve for this horizontal component of this force by the wall which we'll call the force by the wall in the x direction now realize just like the sum of all forces in the y axis added up to zero the sum of all forces in the x-axis is also going to add up to zero now there's only two forces acting horizontally on this beam the force by the wall and the tension force these other two forces are directed straight down so to the right we're going to have the force by the wall in the x direction and to the left we're going to have the horizontal component of the tension we'll call that tx t in the x direction is to the left so it's negative and just like we did in the y axis we're going to be able to expand out t and the x axis and we find the force by the wall in the x direction is 95.2 newtons so to solve for the total force by the wall we're simply going to combine these two forces using the pythagorean theorem and we find the force by the wall is 95.3 newtons now since we're here we're also going to solve for the direction of this force by the wall and that's going to again involve the two components of the force by the wall except rather than using the pythagorean theorem like we did here we're going to take a look at the inverse tangent of these two in order to solve for the angle and we find the angle of the force by the wall relative to the horizontal is three degrees so this is how we go through and solve for the forces acting on this beam which is suspended by a cable i hope you found this useful and on that note that's all for now [Music] you
12170	https://www.teacherspayteachers.com/browse/free?search=linear%20equations%20puzzles	Log InSign Up Cart is empty Total: View Wish ListView Cart Linear Equations Puzzles 52+results Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Filters Grade Elementary 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education Subject English language arts Math Algebra Algebra 2 Applied math Arithmetic Basic operations Geometry Graphing Math test prep Measurement Mental math Precalculus Other (math) For all subjects Price Format Easel Easel Activities Google Apps Microsoft Microsoft PowerPoint Microsoft Word PDF Resource type Teacher tools Homeschool curricula Unit plans Printables Hands-on activities Activities Centers Internet activities Bell ringers Games Instruction Handouts Scaffolded notes Student assessment Assessment Critical thinking and problem solving Student practice Worksheets Flash cards Homework Task cards Standard Theme Seasonal Winter Holiday Halloween Audience Homeschool Staff & administrators Supports Special education Algebra Fun, Linear Equations Logic Puzzle, Created by Prickly Pear Puzzles Linear Equations Logic Puzzle! FREEBIE! This fun, challenging logic puzzle encourages students to combine their knowledge of positive and negative slope, and y intercepts. Students are given 9 unique linear equations and use logic & critical thinking skills to stack them in order based on the logic clues. Deductive reasoning is a must! :) Includes➪• List of linear equations to be used! All equations are y = mx + b form • Thought box for kids to show their thinking or take notes • Clear, easy 8th - 11th Algebra, Algebra 2 CCSS 8.EE.B.6 , 8.EE.C.8 , HSF-BF.B.3 +1 FREE Rated 4.79 out of 5, based on 34 reviews 4.8 (34) X and Y Intercepts of Linear Equations Puzzle Created by Mandy's Math World This FREEBIE sampler puzzle over X and Y Intercepts of Linear Equations is one of the five activities included in my Linear Function Review Party Stations resource. You'll find a PDF puzzle printable included with answer key. A digital version of this puzzle is also available in Google Slides. The product includes directions for how to add this resource along with the answer key to your Google Drive. 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Students will be given a 9 by 9 grid and worksheets with 45 problems, divided uniformly into 5 levels.. Each problem comes with a puzzle piece. The solutions to the problems tell students where in the grid to copy the figure in that puzzle piece. The end result will be a picture of a unicorn. The levels are as follows: Level 1: One-step equations Level 2: Two-step equations Level 3: Distributive property and combining like terms Level 8th - 11th Algebra, Math FREE Rated 4.71 out of 5, based on 7 reviews 4.7 (7) Writing Linear Equations in Slope-Intercept Form Puzzle Created by Math On The Spot Students will practice writing equations given in the forms Ax+By=C and Ax+By+C=0 in slope-intercept form. Two Puzzles Included: Level I Puzzle (Basic): 9 equations given to rewrite in slope-intercept form, “borders” of the final triangle are left blank. Level II Puzzle(Intermediate): 9 equations given to rewrite in slope-intercept form, “borders” of the final triangle include equations that do not have answers to match. Answer Key included! 7th - 11th Algebra FREE Rated 5 out of 5, based on 4 reviews 5.0 (4) Solving Linear Equations Puzzle Created by Robyn Holloway This puzzle is designed to be used in an algebra or geometry class. It consists of a page of puzzle details, the student puzzle page, and a key. The student puzzle page is a 4 by 4 grid containing linear equations and solutions. It includes equations requiring one-step, two-steps, distributive property, and dealing with variables on both sides. The goal is to put the sixteen squares together to form a block letter I by matching the equations and solutions. I usually have students complete the p 7th - 10th Algebra, Geometry FREE Rated 4.92 out of 5, based on 13 reviews 4.9 (13) Math Puzzle Worksheet Solving One Step Linear Equation (English Fact Series) Created by littleworld Math puzzle worksheet is a great way to make math practice become more fun. This is one of the English-Fact-Series Math-Puzzle-Worksheet about Solving Linear Equation. On each worksheet, students need to solve linear equation and trying to solve the puzzle about English fact at the same time. Something about English that they might not know yet. 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Great for sub (sub 7th - 10th Algebra, Math, Other (Math) FREE Rated 5 out of 5, based on 3 reviews 5.0 (3) LINEAR EQUATIONS IN ONE VARIABLE (math puzzles) - EASY Created by NEO SIMANA The student must complete five different tasks (printable PDF). Tasks are designed to repeat an entire chapter, and require logic, insight, and knowledge. Students will practice solving multi-step equations, including those with distribution and variables on both sides, equivalent equations, tasks of creating equations. Total assignments are designed to engage the student and stimulate the thought process. This file is level C – Easy. Also Included in this resource: Instructional assessmentAnswe 7th - 10th Algebra, Math CCSS 7.EE.B.3 , 7.EE.B.4 , 7.EE.B.4a +4 FREE Rated 4.5 out of 5, based on 2 reviews 4.5 (2) Jigsaw Puzzle #37 Frog - Solving Linear Equations (10 Qs) Created by Jigsaw Math Jigsaw Puzzle #37 Frog - Solving Linear Equations (10 Qs)Solving multi-step equationsOne solution, no solution, & IMSHow to JigsawCut out the puzzle pieces on pg 3Solve the math problems on pg 1 and 2Match each problem with the correct letter from the answer bankUse the problem #s and corresponding letter choice to assemble the puzzleGlue the completed puzzle on a separate sheet of paperGet creative by decorating your completed puzzle Google Doc AccessAccess to a copy of a google doc can be foun 8th Math, Other (Math) CCSS 8.EE.C.7 , 8.EE.C.7a , 8.EE.C.7b Also included in: Jigsaw Puzzles - 8th Grade Bundle FREE Rated 5 out of 5, based on 1 reviews 5.0 (1) Solving Circle and Linear System of Equations Puzzle Activity Created by Newton's Solutions Looking for a fun and engaging puzzle activity to help students practice solving a circle and linear system of equations? This easy to check activity is perfect for students to gain confidence when finding the solutions to a system of equations by providing repeated exposure of this skill. Just cut out the squares and have students match up a system with a solution. When correct, a 3x3 square will be formed. This puzzle is a great way for students to self-check their work! This activity pro 8th - 11th Algebra 2, Math CCSS HSA-REI.C.7 FREE Algebra 2 - Solving Linear Equations Matching Puzzle with Biblical Integration Created by Lambkins199 In this Matching Puzzle, students will learn about Peter's vision of clean and unclean animals and how God led him to witness to a Roman soldier and his family who were saved and baptized as they solve linear equations. Students will match the problem with the correct corresponding answer. 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If you find this resource useful, please leave a comment and rate it below! Don't forget to follow me! 7th - 10th Algebra, Arithmetic, Math CCSS HSA-REI.A.1 , HSA-REI.B.3 FREE Rated 4.81 out of 5, based on 31 reviews 4.8 (31) Solving Systems of Linear Equations using Elimination Digital Activity Created by Hello Algebra In this self-checking activity, students will work to solve systems of linear equations using elimination. As questions are answered correctly, pieces of the mystery puzzle will begin to appear, just like a pixel art activity! This resources comes with two activities. In one activity, students will solve question only focusing on addition and subtraction. In the second activity, students will solve questions where you must multiply one of the equations before using addition/subtraction to so 8th - 11th Algebra, Basic Operations, Math CCSS HSA-REI.C.6 Also included in: Algebra 1 Digital Mystery Puzzle Pixel Activities FREE Rated 4.94 out of 5, based on 17 reviews 4.9 (17) Solving Multi-Step Equations DIGITAL Sudoku Puzzle Activity Created by Amusing Algebra With this engaging digital activity, your students will enjoy solving math problems to solve the Sudoku puzzle! No prep and self checking, this activity will help your students practice solving multi-step linear equations. Great to use for practice, homework, review, or sub plans. When students fill in the answer to each problem, this answer is automatically filled into a corresponding Sudoku box. When finished solving all of the problems, students must solve the rest of the Sudoku puzzle. St 8th - 10th Algebra, Algebra 2, Math CCSS HSA-REI.B.3 Also included in: Solving Linear Equations DIGITAL Activity Bundle - Distance Learning FREE Rated 4.54 out of 5, based on 24 reviews 4.5 (24) Systems of Equations Puzzle Math Activity Created by Beyond the Worksheet with Lindsay Gould This puzzle has students match up a linear system, an equation that shows the system in a substituted format and the solution to the system. Puzzle must be pre-cut as the format it is in currently is the answer key! This puzzle can be found in my Algebra 1 Linear Systems Unit, Algebra 1 Linear Systems Digital Notebook and in Unit 5 - Linear Systems of Equations of my Pre-Algebra Curriculum!Licensing Terms :By purchasing this product, you own a license for one teacher only for personal use 7th - 8th Math Also included in: 8th Grade Math Curriculum Supplemental Activities Bundle FREE Rated 4.81 out of 5, based on 16 reviews 4.8 (16) Solving Two Variable Systems Sudoku Puzzle Created by Amusing Algebra With this engaging activity, your students will enjoy solving math problems to solve the Sudoku puzzle! No prep and ready to print, this activity will help your students practice solving linear systems of equations. Great to use for practice, homework, review, or sub plans. What's included:✅ Printable PDF activity containing 12 problems ✅ 3 problems are special cases ✅ Sudoku puzzle easy in difficulty ✅ Teacher answer key Looking for a digital self-checking version to use? Click here !⭐ You m 8th - 10th Algebra, Algebra 2, Math CCSS HSA-REI.C.5 , HSA-REI.C.6 Also included in: Solving Two Variable Systems Activity Bundle FREE Rated 4.71 out of 5, based on 14 reviews 4.7 (14) Systems of Linear Equations Task Cards and Review Activity - Sample Created by Teach With Fergy This 16 System of Linear Equations Task Card resource is unique. It can be utilized in a clue gathering, puzzle solving, highly active and engaging activity and/or as an alternative to worksheet review for Linear Systems. If you'd like to receive 50% off your next TeachWithFergy purchase, (includes your entire shopping cart) please CLICK HERE. A new page will open but don't worry, you won't lose this one. This is a portion of my Systems of Linear Equations Task Card Activity, please click HER 7th - 12th, Adult Education, Higher Education Algebra, Math, Other (Math) FREE Rated 4.77 out of 5, based on 14 reviews 4.8 (14) SYSTEMS OF EQUATIONS: Vocabulary Activities (Matching Cards & Crossword Puzzle) Created by Math Byrd This set of Vocabulary Activities covers vocabulary on Linear Inequalities. The set includes 2 activities, one is a vocabulary-definition matching card sort of 12 pairs and the other is a crossword puzzle of 12 vocabulary terms. The matching card sort can be used multiple times throughout the unit as a quick review of vocabulary within small groups or pairs. The cards can also be used as a great way to randomly pair students, just hand out the cards and ask them to find their matching pair! Voca 8th - 11th Algebra, Algebra 2, Math Also included in: Common Core Algebra 1 VOCABULARY ACTIVITIES - BUNDLE PRICE! FREE Rated 5 out of 5, based on 3 reviews 5.0 (3) Finding Missing Angels Puzzle #1 - vertical, supplementary, corresponding..... Created by Joshua Peretti All too often mathematical skills are taught in isolation. This of course, is the source of one of our greatest laments as teachers: “They forget things so quickly! I just taught them _____ last month!” To counter this natural tendency, the activities I create are designed to prompt students to draw from skills learned in multiple lessons over an extended period of time. In this activity, students will need to find the measures of multiple angles by drawing on their equation solving fluency and 7th - 12th Algebra, Geometry, Measurement FREE Rated 4.67 out of 5, based on 3 reviews 4.7 (3) Scavenger Hunt: Solving Systems of Linear Equations (Graphing with Calculator) Created by Prachi Parihar This scavenger hunt is a good way to get kids up and moving around the room while they are learning to use their calculators for solving systems of equations. There are 14 problems that they need to solve to complete the puzzle. For each system that they solve, the answer will lead them to the next system and also give them a letter. At the end, they will uncover a secret phrase. The first 15 pages are part of the scavenger hunt (and are in order so you have the solution). The last few pages are 9th - 10th Algebra, Algebra 2, Graphing FREE Rated 5 out of 5, based on 2 reviews 5.0 (2) GRAPHING LINES X-INTERCEPT and Y-INTERCEPT LINEAR EQUATIONS Created by iLoveToTeachKids Allow your students to GET OUT OF THE BOX while practicing USING X- and Y-INTERCEPTS TO GRAPH A LINE. Students will solve a puzzle while working through the problems. This high-interest engaging activity will provide students with opportunities to practice their Algebra skills. For this activity students are to graph the equation of a line given the x- and y-intercepts. Your students will need a ruler. Instruct students to extend the lines to the end of the graph. They will then identif 6th - 12th Algebra, Math, Other (Math) FREE Rated 5 out of 5, based on 5 reviews 5.0 (5) Equations Crossword Puzzle Created by Mrs. RP's Resources Engage your students with this fun and challenging algebra word search! This activity is perfect for review, exit tickets, or a quick warm-up. Students will search for 12 key math vocabulary words related to equations, making it an excellent way to reinforce concepts while keeping learning enjoyable. Included in this resource:✅ 1 Student Word Search (with word bank) ✅ 1 Answer Key for easy grading/checking ✅ 12 Algebra Vocabulary Words: expression, polynomial, equality, formula, variable, const 7th - 11th Algebra, Arithmetic FREE Practicing Linear Equations Riddle Created by Turners Teachable Resources In this activity, students can practicing analyzing the slope and y-intercepts of linear functions. Students select the correct answer and input letters into the puzzle at the bottom in order to solve the riddle. I have used this as both a homework and a post-unit assessment in my Algebra class. Students enjoyed it as it keeps them active while still practicing old skills. I think it is important to provide students with a variety of different assessment formats so that the stress of a typical 7th - 10th Algebra, Graphing, Math CCSS , 8.F.B.4 , HSF-IF.C.7a +1 Also included in: Algebraic Functions Activities Bundle FREE Rated 5 out of 5, based on 1 reviews 5.0 (1) Showing 1-24of 52+ results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Who we are We're hiring Press Blog Gift Cards Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Get our weekly newsletter with free resources, updates, and special offers. 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12171	https://math.stackexchange.com/questions/2382920/geometry-assuming-the-altitude-of-a-triangle-cuts-base-in-half	geometry: assuming the altitude of a triangle cuts base in half - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more geometry: assuming the altitude of a triangle cuts base in half Ask Question Asked 8 years, 1 month ago Modified8 years, 1 month ago Viewed 713 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. A friend of mine posted this GRE geometry question and I gave it a look: (below) - I saw another question posted here which is similar to this but doesn't quite answer what I am asking. The question is: is quantity A greater than quantity B, quantity B greater than quantity A, are they equal or is there not enough information to deduce this. And the only piece of information is the what's given below the figure: XY = YZ. The answer given is that they are equal. From what I know if the areas are to be equal then their altitudes must be the same. WY can only be the altitude if XW = WZ (isosceles triangle). But this isn't stated. Can we deduce this from similarity principles? If so, how? If not, can we assume that the two triangles will have the same height because they share a common vertex? geometry triangles Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 4, 2017 at 22:55 Xander Henderson♦ 32.8k 25 25 gold badges 73 73 silver badges 122 122 bronze badges asked Aug 4, 2017 at 22:47 dnclemdnclem 239 1 1 silver badge 6 6 bronze badges 1 1 W Y W Y does not need to be an altitude (i.e. perpendicular to X Z X Z) but if X Y=Y Z X Y=Y Z then W Y W Y is a median. Since the triangles X Y W X Y W and Y Z W Y Z W have the same perpendicular height and have equal bases, they must have the same area Henry –Henry 2017-08-04 23:01:10 +00:00 Commented Aug 4, 2017 at 23:01 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. If it is given that X Y=Y Z X Y=Y Z, then the two triangles have the same height and the same base, so they must have equal areas, because the area of a triangle is half the product of its base and height. The common height is simply the length of the perpendicular from W W to X Z X Z, which may or may not coincide with W Y W Y. Note that in such questions, the figures are not necessarily drawn to scale, so you can only use the information provided by the words. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 4, 2017 at 23:05 answered Aug 4, 2017 at 22:52 user468525 user468525 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. The shared side W Y W Y is not necessarily the altitude of either triangle, because we don't know if it is perpendicular to the base X Z X Z. However, the triangles do have a shared altitude -- it just might not be shown in the diagram. Consider the sketch below: In this diagram X Y X Y=Y Z Y Z and the dotted line is the altitude of both triangles. The altitude lies inside one triangle and outside the other, but is perpendicular to both of the two equal bases. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2017 at 23:09 mweissmweiss 24.6k 3 3 gold badges 50 50 silver badges 93 93 bronze badges 3 A lot of good answers here! I'm selecting this one as the answer because of the diagram. I guess the trap I fell into was assuming WY was the altitude, and then trying to prove that assumption.dnclem –dnclem 2017-08-04 23:13:27 +00:00 Commented Aug 4, 2017 at 23:13 BTW what program did you draw this diagram in?dnclem –dnclem 2017-08-05 05:30:03 +00:00 Commented Aug 5, 2017 at 5:30 1 @david Geogebra.mweiss –mweiss 2017-08-06 01:58:55 +00:00 Commented Aug 6, 2017 at 1:58 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Let Y Y be the midpoint of X Z X Z. Consider a line X E X E perpendicular to W Y W Y, and a line Z F Z F perpendicular to W Y W Y. Since these lines are both perpendicular to W Y W Y, they are parallel to each other. Hence ∠E X Y≅∠F Z Y∠E X Y≅∠F Z Y. Moreover, ∠X Y E≅∠Z Y F∠X Y E≅∠Z Y F, and X Y≅Y Z X Y≅Y Z by hypothesis. Hence by the angle-side-angle congruence relation, we have △Y X E≅△Y Z F△Y X E≅△Y Z F. In particular, X E≅Z F X E≅Z F, where these segments are the altitudes of the triangles Δ W Y X Δ W Y X and Δ W Y Z Δ W Y Z, respectively. As the triangles share a common base W Y W Y, and they have the same altitude, their areas are equal. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 4, 2017 at 23:20 answered Aug 4, 2017 at 22:53 Xander Henderson♦Xander Henderson 32.8k 25 25 gold badges 73 73 silver badges 122 122 bronze badges 3 Actually sorry that was an assumption I made. I am not sure if WY is an altitude (it's not stated).dnclem –dnclem 2017-08-04 23:04:58 +00:00 Commented Aug 4, 2017 at 23:04 It might be helpful to edit the title of your question to reflect that...Xander Henderson –Xander Henderson♦ 2017-08-04 23:08:37 +00:00 Commented Aug 4, 2017 at 23:08 To be fair I said "WY can only be the altitude if XW = WZ (isosceles triangle)", I didn't state that WY was indeed the altitude. But sorry for the ambiguity regardless.dnclem –dnclem 2017-08-04 23:14:20 +00:00 Commented Aug 4, 2017 at 23:14 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry triangles See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 0Is it safe to assume that the altitude of a triangle always cuts the base in half 5Constructing a triangle given three concurrent cevians? 5Area of Triangle when 2 Sides and No Angle Known 2Integer-sided triangles with equal perimeters and equal areas 6The hardest geometry question with "a triangle" and "a circle" - Circle intersecting triangle equally in 5 parts 5In the given triangle with two intersecting cevians, find the area of the shaded quadrilateral. 0Two overlapping triangles △A B C△A B C, △D B E△D B E. 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12172	https://www.cuemath.com/ncert-solutions/what-happens-to-the-value-of-a-fraction-if-the-denominator-of-the-fraction-is-decreased-while-numerator-is-kept-unchanged/	What happens to the value of a fraction if the denominator of the fraction is decreased while numerator is kept unchanged Grade KG 1st 2nd 3rd 4th 5th 6th 7th 8th Algebra 1 Algebra 2 Geometry Pre-Calculus Calculus Pricing Events About Us Get Started Grade KG 1st 2nd 3rd 4th 5th 6th 7th 8th Algebra 1 Algebra 2 Geometry Pre-Calculus Calculus Pricing Events About Us What happens to the value of a fraction if the denominator of the fraction is decreased while numerator is kept unchanged Solution: In fractions, if the denominator part of the fraction is decreased the value of a fraction is increased while the numerator remains the same. For example, 4/5 is a fraction with 4 as numerator and 5 as denominator. Let us decrease the denominator by 2 then the fraction we get is 4/3. This proves 4/5 < 4/3. ✦ Try This:Evaluate 4/8 × 7/8 To solve the above problem you can follow the fractions and decimals rules to get the required answer. ☛ Also Check:NCERT Solutions for Class 7 Maths Chapter 2 NCERT Exemplar Class 7 Maths Chapter 2 Exercise Problem 56 What happens to the value of a fraction if the denominator of the fraction is decreased while numerator is kept unchanged Summary: The denominator of the fraction is decreased the value of a fraction is increased while the numerator remains unchanged ☛ Related Questions: Which letter comes 2/5 of the way among A and J If 2/3 of a number is 10, then what is 1.75 times of that number In a class of 40 students, 1/5 of the total number of students like to eat rice only, 2/5 of the tot . . . . Explore math program Math worksheets and visual curriculum Get Started FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 ABOUT US Our Mission Our Journey Our Team MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math Terms and ConditionsPrivacy Policy
12173	https://www.stuvia.com/en-us/search/us/evolution-making-sense-of-life-2nd-edition/evolution-making-sense-of-life-2nd-edition	Evolution making sense of life 2nd edition Notes - Evolution making sense of life 2nd edition - Stuvia US LoginSell Where do you study Your language The Netherlands United Kingdom Germany Spain France Belgium South Africa Canada United States Other country Nederlands English English Deutsch English Español English Français English Nederlands Français English English English Français English Español English Español Indonesian Choose currency € EUR $ USD £ GBP R ZAR Save You searched for: Evolution making sense of life 2nd edition ×Evolution making sense of life 2nd edition Your school or university Improve your search results. Select your educational institution and subject so that we can show you the most relevant documents and help you in the best way possible. Ok, I understand! 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12174	https://www.youtube.com/watch?v=Mb5-3nQPBII	Auxin Discovery \| Frits Went Experiment \| Part 4 \| Lec. 16 \| Plant Physiology Biology With Wk 5240 subscribers 20 likes Description 487 views Posted: 14 Jun 2024 The discovery of auxin, a key plant hormone that regulates growth and development, is credited to the experiments conducted by the Dutch botanist Frits Went in the 1920s. Specifically, Went performed a series of experiments studying phototropism - the growth response of plants towards light. He found that when the tip of a plant's shoot was covered, the plant would still bend towards the light, indicating that the response was being mediated by a chemical signal transported from the shoot tip. Went then conducted further experiments where he placed agar blocks at the tips of decapitated plants. He found that the agar blocks were able to restore the phototropic response, suggesting that the shoot tip was producing a growth-promoting substance that was being transported downwards. Through additional experiments, Went was able to isolate and identify this substance as indoleacetic acid (IAA), which we now know as the plant hormone auxin. Went's pioneering work laid the foundation for our understanding of auxin and its central role in regulating plant growth and development. The "Went experiment" on phototropism and auxin transport is considered a landmark discovery in the history of plant physiology. auxindiscovery discoveryofauxin fritswentexperiment F.W.wentexperimsnt planthormones biologywithwk 2 comments Transcript: अस्सलाम वालेकुम डियर व्यूवर्स मैं हूं वाहिद वजीर वेलकम टू माय परफॉर्म की थी चार्लेस डार्विन ने 18811 परफॉर्म की थी पीटर बॉय एंड जनसन ने 1913 में और थर्ड परफॉर्म की थी फिट्स पेंट ने 1926 में प्रीवियस टू लेक्चर में भी मैं यह बता चुका हूं ये जो थ्री एक्सपेरिमेंट है न हार्मोन को डिस्कवर करने के लिए परफॉर्म नहीं किए गए थे बल्कि ये परफॉर्म किए गए थे टू स्टडी फोटोट्रॉपिज्म इन प्लांट प्लांट में फोटोट्रॉपिज्म को स्टडी करने के लिए ये थ्री एक्सपेरिमेंट परफॉर्म किए गए थे प्लांट पार्ट्स की जो मूवमेंट होती है लाइट की तरफ या प्लांट पार्ट्स की जो बेंडिंग होती है लाइट की तरफ उसको स्टडी करने के लिए यह थ्री एक्सपेरिमेंट परफॉर्म किए गए थे तो चार्लेस डारन एक्सपेरिमेंट एंड पीटर बॉय एंड जेमसन एक्सपेरिमेंट इनटू एक्सपेरिमेंट को मैं प्रीवियस टू लेक्चर में डिटेल के साथ डिस्कस कर चुका हूं तो इस वीडियो में हम डिस्कस करने वाले हैं फ्रीट्स बेंट एक्सपेरिमेंट तो हम टमेंट एक्सपेरिमेंट की तरफ आते हैं लेकिन इससे पहले ये नोट कर ले यहां पर इस वीडियो में हम ये टर्म कोइल बारबार यूज करेंगे तो पहले हम कोइल की बात करते हैं कि कोइल है क्या फॉर एग्जांपल य एक सीड है तो ड्यूरिंग जर्मिनेशन जब कोई सीड जर्मिनेट हो जाता है आप लोगों ने देखा होगा उससे इस तरह का स्ट्रक्चर जो है वो डिवेलप हो जाता है तो इसको प्लोम कहा जाता है और ये फ्यूचर शूड होता है यह बाद में शूट में डिवेलप हो जाता है या इसको यंग शूट भी कहा जाता है तो यह जो पलमल होता है इसके ऊपर इस तरह का प्रोटेक्टिव कवर मौजूद होता है शीप मौजूद होता है तो इसको कोइल कहा जाता है कोइल का फंक्शन यह होता है कि य पलमल को प्रोटेक्ट करता है जो यंग शूड होता है उसको प्रोटेक्ट करता है और यह सिर्फ आप लोगों को मिलेगा सीडलिंग ऑफ मनोका प्लांट में डाकड प्लांट्स में आप लोगों को कोले पटाल नहीं मिलेगा अब हम आते हैं फ्रीट स्वेंड एक्सपेरिमेंट की तरफ तो फ्रिट्स वेंट एक्सपेरिमेंट को कोपला कर्वेचर टेस्ट भी कहा जाता है एविना कर्वेचर टेस्ट भी कहा जाता है फ्रीट्स वेंट ने अपनी एक्सपेरिमेंट में कोलि पटाल ऑफ ओट प्लांट को यूज किया था ओट प्लांट का बोटेनिकल नेम है एविना सेवा या एविना सेवा फ्रीट्स वेंट ने अपनी एक्सपेरिमेंट में यूनिडायरेक्शनल लाइट को यूज किया था यूनिडायरेक्शनल लाइट वो लाइट होता है जो एक ही साइड से होता है सिंगल डायरेक्शन से होता है मैकेनिज्म की बात करते हैं तो फर्स्ट स्टेप में फ्रीट्स वेंट ने टू कोलि पटाइन लिए थे यहां पर आप लोग देख सकते हैं एक कोले पटाल है और ये दूसरा है और साथ साथ टू एगर ब्लॉक लिए थे ये एक एगर ब्लक है और ये दूसरा है तो फ्रीट स्व ने क्या किया दोनों कोले पटाल से टिप्स को कट किए इस तरह और ये जो टीप है इसको एक अगर ब्लॉक के ऊपर रखें इस तरह इसको मैं रिमूव करता हूं फिर क्या हुआ यह जो टिप्स है इसमें जो ग्रत प्रमोटिंग केमिकल थे वो इस एगर ब्लॉक में डिफ्यूज हो गए इस तरह फिर फ्रीट ने क्या किया यह जो एगर ब्लक है जिसमें ग्रोथ प्रमोटिंग केमिकल डिफ्यूज हो गए थे इसको एक कोइल के ऊपर रखा और य जो एगर बलक है जिसम ग्रत प्रमोटिंग केमिकल मौजूद नहीं है इसको दूसरे कोइल के ऊपर रखा फिर क्या हुआ जब फ्रीट्स ने थोड़ी देर बाद देखा तो ये जो कोइल था ये स्ट्रेट ग्रो हुआ था स्ट्रेट य क्या है एगर ब्लक है और य जो कोइल था इसकी ग्रत में कोई चेंज नहीं आ था सेकंड स्टेप में फट ने क्या किया था अगेन एक कोइल से को कट किया और ये जो टीप था इसको एर ब्लक के ऊपर इस तरह रखा तो यहां पर भी टीप में जो ग्र प्रमोटिंग केमिकल थे व इस एगर ब्लक में डिफ्यूज हो गए अब य पर यह जो एगर ब्लक है इसको कोइल के ऊपर रखा लेकिन य पर जो एगर ब्लक रखा था वो ए सिमेट्री रखा था अनइक्वली रखा था यानी हाफ साइड पर रखा था और हाफ साइड जो है कोइल का व वैसे छोड़ दिया था व एक्सपोज किया था य पर आप लोग देख सकते हैं तो फिर यह जोल था इसको यूनि डायरेक्शनल लाइट के लिए एक्सपोज किया जब फ ने थोड़ी देर बाद देखा कुछ घंटे बाद देखा तो यह जो कोइल था ये लाइट की तरफ बेंड हुआ था यहां पर आप लोग देख सकते हैं लाइट की तरफ बेंड हुआ था यानी इसने फोटोट्रॉपिज्म शो किया था तो अपनी एक्सपेरिमेंट को मध्य नजर रखते हुए फ्रीट स्वेंड ने यह कहा कि टीप ऑफ पलपटाइन में ग्रत प्रमोटिंग केमिकल मौजूद है जो कि प्लांट में या कोलप में फोटोट्रॉपिज्म को कॉज करती हैं अब यहां पर ये नोट कर ले यहां पर इसलिए कोलिप टाइल स्टेट ग्रो हुआ था क्योंकि यहां पर जो एगर ब्लॉक रखा था कोलिप टायर के ऊपर वो सिमेटिक रखा था इक्वली रखा था तो यहां पर जो डिस्ट्रीब्यूशन हुई थी ग्रोथ प्रमोटिंग केमिकल की वो इक्वली हुई थी वो इक्वल डिस्ट्रीब्यूशन हुई थी और यहां पर कोई सनलाइट भी मौजूद नहीं था या यूनि डायरेक्शन लाइट सोर्स भी मौजूद नहीं था तो इसलिए यहां पर ये स्ट्रेट ग्रो हुआ य पर इसलिए लाइट की तरफ ब हुआ क्योंकि यहां पर जो एगर ब्लक रखा था वो एमेटिक रखा था एक ही साइड पर रखा था हाफ साइड पर रखा था और हाफ साइड को वसे छोड़ दिया था तो इसमें जो ग्रोथ प्रमोटिंग केमिकल डिफ्यूज हुए थे यहां पर व सिर्फ कोइल के इस साइड में डिफ्यूज हो गए जो शड साइड था यानी जो लाइट के लिए एक्सपोज नहीं किया था तो इस साइड में हो गए थे और ये जो साइड था नेटेड साइड जो लाइट के लिए एक्सपोज किया था इसमें ग्रु प्रमोटिंग केमिकल एकलेट नहीं हुए थे तो इस साइड में एकलेट हुए थे तो य पर जो सेल थे उसम लंगे काज हो गई थी गथ प्रमोटिंग केमिकल की वजह से जिसके नतीजे में यह लाइट की तरफ बेंड हुआ और साथ साथ य पर य भी नोट कर ले यह जो टर्म हैन इसको फर द फर्स्ट टाइम फ्रीट्स बेंट ने यूज किया था
12175	https://www.quora.com/How-can-I-prove-mathematically-that-the-smallest-distance-from-the-origin-to-the-circle-shown-is-the-distance-from-the-origin-to-the-centre-minus-the-radius	Something went wrong. Wait a moment and try again. Proofs (mathematics) Centre Circle Distance Formula Coordinate Plane Radius and Diameter Geometry in Mathematics Distance Geometry 5 How can I prove mathematically that the smallest distance from the origin to the circle shown is the distance from the origin to the centre, minus the radius? David Joyce Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Author has 9.9K answers and 68.4M answer views · Updated 10y Theorem. The smallest distance from a point outside a circle to a point on the circle is equal to the distance from the point to the center of the circle minus the radius of the circle. Proof a la Euclid. Let A be the center of the circle and B the point outside the circle. Connect the points A and B with a straight line [segment] AB. Let C be the point of intersection of the circle with AB. Then AC + CB = AB. Let D be any other point on the circle. By the triangle inequality, Euclid's I.20, AB < AD + BD. So AC + CB < AD + BD. But AC and AD are equal being radii of the circle. Therefore, C Theorem. The smallest distance from a point outside a circle to a point on the circle is equal to the distance from the point to the center of the circle minus the radius of the circle. Proof a la Euclid. Let A be the center of the circle and B the point outside the circle. Connect the points A and B with a straight line [segment] AB. Let C be the point of intersection of the circle with AB. Then AC + CB = AB. Let D be any other point on the circle. By the triangle inequality, Euclid's I.20, AB < AD + BD. So AC + CB < AD + BD. But AC and AD are equal being radii of the circle. Therefore, CB < BD. Thus, C is the closest point on the circle to B. And the distance from C to B is equal to the distance from B to A minus the radius of the circle. Q.E.D. Note that when Euclid used the term "line" he meant what we mean by line segment. Swapnam Bajpai M. Tech. in Computer Science, Indian Institute of Technology, Bombay (IITB) (Graduated 2019) · 10y This is a proof by contradiction. Let the proposed line segment meet the circle at any other point than the one shown in the picture above. Join that point to the centre. You get two line segments, the one joining the origin to the circumference, and the one joining the circumference to the centre, namely the radius. Call the sum S = x + r where r is the radius, and x is the value we need. Now by a basic law of mathematics, a straight line is the shortest distance between any two points. So consider the problem of minimizing the sum S, the distance between the origin and the centre of the circ This is a proof by contradiction. Let the proposed line segment meet the circle at any other point than the one shown in the picture above. Join that point to the centre. You get two line segments, the one joining the origin to the circumference, and the one joining the circumference to the centre, namely the radius. Call the sum S = x + r where r is the radius, and x is the value we need. Now by a basic law of mathematics, a straight line is the shortest distance between any two points. So consider the problem of minimizing the sum S, the distance between the origin and the centre of the circle. It occurs when the two crooked lines reduce to a single straight line. Thus S is minimized in the situation showed in the figure. And since r is common to all the cases, the straight line case also minimizes x, contradicting our assumption that taking some another point on the circle and making two separate line segments would serve the purpose. Hence the given situation indeed is the right answer, and x = S - r, i.e the distance from the origin to the centre minus the radius. Samuel Chester 10y The proof here really is about identities in geometry, namely, that the shortest distance from 2 points in 2d, is the length of the line connecting both points. Let o be the origin Let c be the center of the circle The first thing is an identity. The shortest distance in 2d from o to c is the length of the line segment connecting both points. Also, let b be the point on the straight line from o to c that lies on the circle's circumference, if it exists. Therefore, since b is collinear with o and c: oc = ob + bc And since the distance from b to c is the radius r, by the definition of a circle: oc The proof here really is about identities in geometry, namely, that the shortest distance from 2 points in 2d, is the length of the line connecting both points. Let o be the origin Let c be the center of the circle The first thing is an identity. The shortest distance in 2d from o to c is the length of the line segment connecting both points. Also, let b be the point on the straight line from o to c that lies on the circle's circumference, if it exists. Therefore, since b is collinear with o and c: oc = ob + bc And since the distance from b to c is the radius r, by the definition of a circle: oc = ob + r So: ob = oc - r If the origin is within the circumference of the circle, ie, oc < r, then b doesn't exist according to this definition, and, in any case, the question doesn't really apply. Related questions What is the distance of any point of a circle from its centre called? What is the average distance from the origin to the edge of offset circle? If a point is on the circle, how is its distance from the center related to the radius of the circle? What is the area of a circle if its longest distance is 10 cm? The distance around a circle is 6.28 dm. What is the radius? Danya Rose Mathematician and general student of life, with a few years of Tai Chi Chuan. · Author has 2K answers and 7M answer views · 10y The circle looks like a unit-radius circle with centre (−1,1), which means it has Cartesian equation (x+1)2+(y−1)2=1. For the purpose of this question, consider only the region where −1≤x≤0 and 0≤y≤1. This is for obvious reasons that if the circle is divided into quadrants this is the quadrant clearly closest to the origin. Now the circle can be parameterised by x=−1+cost, y=1+sint (if it were any other radius that would appear as a coefficient of the trig functions), and distance from the origin is given by d = \sqrt{x^2 + y^2} = \sqrt{1 - 2\cos t + \cos^2 t + 1 + The circle looks like a unit-radius circle with centre (−1,1), which means it has Cartesian equation (x+1)2+(y−1)2=1. For the purpose of this question, consider only the region where −1≤x≤0 and 0≤y≤1. This is for obvious reasons that if the circle is divided into quadrants this is the quadrant clearly closest to the origin. Now the circle can be parameterised by x=−1+cost, y=1+sint (if it were any other radius that would appear as a coefficient of the trig functions), and distance from the origin is given by d=√x2+y2=√1−2cost+cos2t+1+2sint+sin2t=√3+2(sint−cost). We want to find for what value of t d is minimised. We can immediately restrict t to the interval [−π2,0 because of the parameterisation we've chosen and the observation in the second paragraph. As an optimisation problem we can use calculus: the distance is a smooth function of the parameter t, so wherever that function has a global or local minimum is a candidate. The only places it could have global minima that are not also local minima are at t=−π2,0, the endpoints of the domain, but clearly the distances there are 1. So we're only looking for local minima, or places where \frac{\mathrm{d}d{\mathrm{d}t} = 0 with "upwards" curvature. To make the differentiation process simpler I'm going to square d and use implicit differentiation: d2=3+2sint−2cost becomes 2d\frac{\mathrm{d}d{\mathrm{d}t} = 2\cos t + 2\sin t. In turn this becomes \frac{\mathrm{d}d{\mathrm{d}t} = \frac{1}{d}\left(\cos t + \sin t\right). We want to find where this is zero. Since we know that d>0 (just look at it), we can restrict ourselves to just cost+sint=0, or cost=−sint. There are a few different ways you can find this value of t exactly, but the value is t=−π4. That is exactly the line joining the centre of the circle to the origin of the plane. Sub in that value for t in the equation for distance and you can get what that minimum distance is, and it is as you have been asked to show: √2−1. Sponsored by Grubhub For Merchants Ready to expand your customer reach? With millions of customers, Grubhub is the platform to grow your business. Assistant Bot · 1y To prove that the smallest distance from the origin to a circle is the distance from the origin to the center of the circle minus the radius, we can use some basic concepts from geometry. Definitions Let the center of the circle be denoted as C, and let the radius of the circle be r. The position of the origin is denoted as O. Steps to Prove Distance from the Origin to the Center: Let d be the distance from the origin O to the center C of the circle. This distance can be calculated using the distance formula if the coordinates of O and C are known. Understanding the Circle: The circle ca To prove that the smallest distance from the origin to a circle is the distance from the origin to the center of the circle minus the radius, we can use some basic concepts from geometry. Definitions Let the center of the circle be denoted as C, and let the radius of the circle be r. The position of the origin is denoted as O. Steps to Prove Distance from the Origin to the Center: Let d be the distance from the origin O to the center C of the circle. This distance can be calculated using the distance formula if the coordinates of O and C are known. Understanding the Circle: The circle can be defined as the set of all points P such that the distance from C to P is equal to r. In mathematical terms, this can be expressed as: \|CP\|=r Here, CP is the distance from the center C to any point P on the circle. Finding the Smallest Distance: The smallest distance from the origin O to any point P on the circle occurs when the line segment connecting O to C is extended towards the circle. The point P that is closest to the origin lies directly along the line connecting O to C and is located at a distance r away from the center C. Calculating the Smallest Distance: The closest point P on the circle can be visualized as being directly in line with the origin and the center: \|OP\|=\|OC\|−\|CP\| Since \|CP\|=r, we have: \|OP\|=\|OC\|−r Thus, the smallest distance from the origin to the circle is given by: d−r where d=\|OC\| is the distance from the origin to the center. Conclusion The smallest distance from the origin O to the circle is indeed the distance from the origin to the center C minus the radius r of the circle: Smallest Distance=\|OC\|−r This completes the proof. Harun Šiljak thinks he's an applied mathematician · Author has 1.3K answers and 5.4M answer views · 10y Proof without numbers would go along these lines: The point(s) on the circle making the minimum distance cannot be outside of the right angle arc formed by the tangent axis points, as a ray connecting the origin would have to intersect another point on the mentioned arc, therefore suggesting a shorter distance exists, which is a contradiction. Assuming that, due to symmetry, there may exist two points with minimum distance, symmetrical w.r.t. the line joining origin and center, let us connect those two points with a line segment. The axis of symmetry of that segment crosses the arc providing a s Proof without numbers would go along these lines: The point(s) on the circle making the minimum distance cannot be outside of the right angle arc formed by the tangent axis points, as a ray connecting the origin would have to intersect another point on the mentioned arc, therefore suggesting a shorter distance exists, which is a contradiction. Assuming that, due to symmetry, there may exist two points with minimum distance, symmetrical w.r.t. the line joining origin and center, let us connect those two points with a line segment. The axis of symmetry of that segment crosses the arc providing a shorter distance, so it is a contradiction. The only possibility is the degenerate case of the point lying on the line connecting the origin and the center. Related questions A chord of a circle of length 6 cm and is at a distance of 4 cm from the centre, can you find the radius of the circle? What is the circumference of an imaginary circle with radius r centered at origin O? Without knowing about pi, how could I prove that the area of a circle is half the circumference times the radius? I have an area of a circle. How can I get the radius of the circle? Which point is on the circle centered at the origin with a radius of 5 units distance formula? Joshua Dawes Maths tutor during high school and university · Author has 625 answers and 716.5K answer views · 10y There are a few steps, so bear with me :-) (I'm assuming you have the equation of the circle already.) First, construct the equation of the line between the center of the circle and the origin. It's a line that passes through two points that we know, so we can get the equation from those two points. Then solve the simultaneous equation you get when you set the height of the circle equal to the height of the line (I.e find the intersections of the line and circle). This will give you two points (x,y) when you solve it, on the near and far sides of the circle. Then calculate the distance between There are a few steps, so bear with me :-) (I'm assuming you have the equation of the circle already.) First, construct the equation of the line between the center of the circle and the origin. It's a line that passes through two points that we know, so we can get the equation from those two points. Then solve the simultaneous equation you get when you set the height of the circle equal to the height of the line (I.e find the intersections of the line and circle). This will give you two points (x,y) when you solve it, on the near and far sides of the circle. Then calculate the distance between each point above and the origin, and choose the shorter distance. Finally, calculate the distance to the center of the circle from the origin, subtract the radius, and compare it with the distance from before. 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Zhihao Ding Works at GE Global Research · 10y First, you can ignore the axis. We are essentially finding the smallest distance from a point to any point of a circle. Our goal is to prove this distance equals to the distance between this point and the center of the circle subtracted by the radius. You can use vector to represent the problem and minimize the distance. For example, use O to represent the point, use R to represent the center of the circle and use X to represent any point at the circle. Then we have OX = OR + RX (these are all vectors) We are minimizing \|OX\| (the length of OX) We also observe that \|OR\| is fixed. Also \|RX\| is fixed First, you can ignore the axis. We are essentially finding the smallest distance from a point to any point of a circle. Our goal is to prove this distance equals to the distance between this point and the center of the circle subtracted by the radius. You can use vector to represent the problem and minimize the distance. For example, use O to represent the point, use R to represent the center of the circle and use X to represent any point at the circle. Then we have OX = OR + RX (these are all vectors) We are minimizing \|OX\| (the length of OX) We also observe that \|OR\| is fixed. Also \|RX\| is fixed, which is just the radius. So \|OX\|^2 = \|OR\|^2 + \|OX\|^2 - 2\|OR\|\|OX\|cos We find when cos = 1, \|OX\|^2 is minimized. In that case, OR and OX are in the same line. That's it. Tony Nguyen Knows Vietnamese · Author has 6.1K answers and 2.9M answer views · 10y Let r = radius Total distance from origin to centre is hypotenuse of right angled triangle with legs r long. Use the Pythagorean theorem to find it. To find the distance from the circle to the origin, you minus the hypotenuse with r. Let r = radius Total distance from origin to centre is hypotenuse of right angled triangle with legs r long. Use the Pythagorean theorem to find it. To find the distance from the circle to the origin, you minus the hypotenuse with r. Frank Lovell Physical-organic chemist (UKY '75), GE Retired Senior Professional; Critical Rationalist; agnostic atheist; little-l libertarian Republican. · Author has 615 answers and 759.4K answer views · 10y Simply plot another circle on the chart whose center is the graph's 0,0 origin whose circumference just touches the larger circle's circumference. From that it should be immediately apparent that the smallest distance from the larger circle's circumference to the graph's 0,0 origin is the length of the line from the circle's center to the graph's 0,0 origin minus the larger circle's radius (you can envision this second circle and discern its clear implication without really needing to draw anything on the graph above). Others have answered giving rigorous proofs, but what I like about plane ge Simply plot another circle on the chart whose center is the graph's 0,0 origin whose circumference just touches the larger circle's circumference. From that it should be immediately apparent that the smallest distance from the larger circle's circumference to the graph's 0,0 origin is the length of the line from the circle's center to the graph's 0,0 origin minus the larger circle's radius (you can envision this second circle and discern its clear implication without really needing to draw anything on the graph above). Others have answered giving rigorous proofs, but what I like about plane geometry is that many graphical proofs can be made unequivocally clear without using equations (which in my experience helps mathematically non-savvy folks appreciate math). Related questions What is the distance of any point of a circle from its centre called? What is the average distance from the origin to the edge of offset circle? If a point is on the circle, how is its distance from the center related to the radius of the circle? What is the area of a circle if its longest distance is 10 cm? The distance around a circle is 6.28 dm. What is the radius? A chord of a circle of length 6 cm and is at a distance of 4 cm from the centre, can you find the radius of the circle? What is the circumference of an imaginary circle with radius r centered at origin O? Without knowing about pi, how could I prove that the area of a circle is half the circumference times the radius? I have an area of a circle. How can I get the radius of the circle? Which point is on the circle centered at the origin with a radius of 5 units distance formula? What is the distance and displacement along the diameter, radius and quarter of a circle? What is the radius of a circle if two points on its circumference are the same distance from the center? What is the distance of the chord from the centre of the circle if the length of the chord and radius of the circle is 5cm and 8cm? What is the focus of a point whose distance from the vertex of a square of side 2 and the centre origin is greater than the distance from the origin? What is the line which is at a distance equal to the radius of a circle from its centre called? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
12176	https://openstax.org/books/contemporary-mathematics/pages/3-10-arithmetic-sequences	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in Contemporary Mathematics 3.10 Arithmetic Sequences Contemporary Mathematics 3.10 Arithmetic Sequences Search for key terms or text. Learning Objectives After completing this section, you should be able to: Identify arithmetic sequences. Find a given term in an arithmetic sequence. Find the nth term of an arithmetic sequence. Find the sum of a finite arithmetic sequence. Use arithmetic sequences to solve real- world applications As we saw in the previous section, we are adding about 2.5 quintillion bytes of data per day to the Internet. If there are 550 quintillion bytes of data today, then there will be 552.5 quintillion bytes tomorrow, and 555 quintillion bytes in 2 days. This is an example of an arithmetic sequence. There are many situations where this concept of fixed increases comes into play, such as raises or table arrangements. Identifying Arithmetic Sequences A sequence of numbers is just that, a list of numbers in order. It can be a short list, such as the number of points earned on each assignment in a class, such as {10, 10, 8, 9, 10, 6, 10}. Or it can be a longer list, even infinitely long, such as the list of prime numbers. For example, here’s a sequence of numbers, specifically, the squares of the first 12 natural numbers. {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} Each value in the sequence is called a term. Terms in the list are often referred to by their location in the sequence, as in the nth term. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, the first term of the sequence is 1, the fourth term is 16, and so on. In the sequence of assignment scores {10, 10, 8, 9, 10, 6, 10}, the first term is 10 and the third term is 8 (Figure 3.47). Figure 3.47 Sequence showing first, second, and fifth terms The notation we use with sequences is a letter, which represents a term in the sequence, and a subscript, which indicates what place the term is in the sequence. For the sequence {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}, we will use the letter a as a value in the sequence, and so a5 would be the term in the sequence at the fifth position. That number is 25, so we can write a5=25. In this section, we focus on a special kind of sequence, one referred to as an arithmetic sequence. Arithmetic sequences have terms that increase by a fixed number or decrease by a fixed number, called the constant difference (denoted by d), provided that value is not 0. This means the next term is always the previous term plus or minus a specified, constant value. Another way to say this is that the difference between any consecutive terms of the sequence is always the same value. To see a constant difference, look at the following sequence: {7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87}. Figure 3.48 illustrates that each term of the sequence is the previous term plus 8. Eight is the constant difference here. Figure 3.48 Sequence of numbers with 8 added to each term Example 3.132 Identifying Arithmetic Sequences Determine if the following sequences are arithmetic sequences. Explain your reasoning. {4,7,10,13,16,19,22,25,...} {20,40,80,160,320,640} {7,1,−5,−11,−17,−23,−29,−34,−40} Solution In the sequence {4,7,10,13,16,19,22,25,...}, every term is the previous term plus 3. The ellipsis indicates that the pattern continues, which means keep adding 3 to the previous term to get the new term. Therefore, this is an infinite arithmetic sequence. In the sequence {20,40,80,160,320,640}, terms increase by various amounts, for instance from term 1 to term 2, the sequence increases by 20, but from term 2 to term 3 the sequence increases by 40. So, this is not an arithmetic sequence. In the sequence {7,1,−5,−11,−17,−23,−29,−34,−40}, every term is the previous term minus 6, so this is an arithmetic sequence. Your Turn 3.132 Determine if the following sequences are arithmetic sequences. Explain your reasoning. 1. {7.6,5.4,3.2,1.0,−1.2,−3.4,−5.6,−7.8,−10.0} 2. {14,16,18,22,28,40,32,0} 3. {14,20,26,32,38,44,50,56,62,68,74,80,...} Arithmetic sequences can be expressed with a formula. When we know the first term of an arithmetic sequence, which we label a1, and we know the constant difference, which is denoted d, we can find any other term of the arithmetic sequence. The formula for the ith term of an arithmetic sequence is ai=a1+d×(i−1). FORMULA If we have an arithmetic sequence with first term a1 and constant difference d, then the ith term of the arithmetic sequence is ai=a1+d×(i−1). Let’s examine the formula with this arithmetic sequence: {4,7,10,13,16,19,22,25,...}. In this sequence a1=4 and d=3. The table below shows the values calculated. \| i, Place in Sequence \| ai, ith Term \| Value of Term \| Term Written as a1+3×(i−1) \| \| 1 \| a1 \| 4 \| 4+3×0 \| \| 2 \| a2 \| 7 \| 4+3×1 \| \| 3 \| a3 \| 10 \| 4+3×2 \| \| 4 \| a4 \| 13 \| 4+3×3 \| \| 5 \| a5 \| 16 \| 4+3×4 \| \| i \| ai \| \| 4+3×(i−1) \| We can see how the ith term can be directly calculated. In this sequence, the formula is a1+3×(i−1) where the first term, a1, is 4 and the constant difference d is 3. We can then determine the 47th term of this sequence: a47=4+3×(47−1)=4+3×46=4+138=142. Example 3.133 Calculating a Term in an Arithmetic Sequence Identify a1 and d for the following arithmetic sequence. Use this information to determine the 60th term. {18,31,44,57,70,83,...} Solution Inspecting the sequence shows that a1=18 and d=13. We use those values in the formula, with i=60. a60=a1+d×(i−1)=18+13×(60−1)=18+13×59=18+767=785 Your Turn 3.133 1. Identify a1 and d for the following arithmetic sequence. Use this information to determine the 86th term. {4.5,8.1,11.7,15.3,18.9,22.5,26.1,...} Video Arithmetic Sequences If we know two terms of the sequence, it is possible to determine the general form of an arithmetic sequence, ai=a1+d×(i−1). FORMULA If we have the ith term of an arithmetic sequence, ai, and the jth term of the sequence, aj, then the constant difference is d=aj−aij−i and the first term of the sequence is a1=ai−d(i−1). Example 3.134 Determining First Term and Constant Difference Using Two Terms A sequence is known to be arithmetic. Two of its terms are a7=56 and a19=104. Use that information to find the constant difference, the first term, and then the 50th term of the sequence. Solution To find the constant difference, use d=aj−aij−i. The location of the terms is given by the subscript of the two a terms, i=7 and j=19. So, the constant difference can be calculated as such: d=104−5619−7=4812=4 . The constant difference of 4 is then used to find a1. a1=ai−d(i−1)=a7−4(7−1)=56−4×6=32 . So d=4 and a1=32. With this information, the 50th term can be found. a50=a1+d×(i−1)=32+4×(50−1)=32+4×49=32+196=228 . The 50th term is a50=228. Your Turn 3.134 1. A sequence is known to be arithmetic. Two of the terms are a14=41 and a38=161 . Use that information to find the constant difference and the first term. Then determine the 151st term of the sequence. Video Finding the First Term and Constant Difference for an Arithmetic Sequence Finding the Sum of a Finite Arithmetic Sequence Sometimes we want to determine the sum of the numbers of a finite arithmetic sequence. The formula for this is fairly straightforward. FORMULA The sum of the first n terms of a finite arithmetic sequence, written sn, with first and last term a1 and an, respectively, is sn=n(a1+an2). Example 3.135 Finding the Sum of a Finite Arithmetic Sequence What is the sum of the first 60 terms of an arithmetic sequence with a1=4.5 and d=2.5? Solution The formula requires the first and last terms of the sequence. The first term is given, a1=4.5. The 60th term is needed. Using the formula a1=ai+d(i−1) provides the value for the 60th term. a60=4.5+2.5(60−1)=4.5+2.5×59=4.5+147.5=152 . Applying the formula sn=n(a1+an2) provides the sum of the first 60 terms. s60=60(4.5+1522)=60×156.52=4,695 . The sum of the first 60 terms is 4,695. Your Turn 3.135 1. What is the sum of the first 101 terms of an arithmetic sequence with a1=13 and d=2.25? Video Finding the Sum of a Finite Arithmetic Sequence Using Arithmetic Sequences to Solve Real-World Applications Applications of arithmetic sequences occur any time some quantity increases by a fixed amount at each step. For instance, suppose someone practices chess each week and increases the amount of time they study each week. The first week the person practices for 3 hours, and vows to practice 30 more minutes each week. Since the amount of time practicing increases by a fixed number each week, this would qualify as an arithmetic sequence. Example 3.136 Applying an Arithmetic Sequence Jordan has just watched The Queen’s Gambit and decided to hone their skills in chess. To really improve at the game, Jordan decides to practice for 3 hours the first week, and increase their time spent practicing by 30 minutes each week. How many hours will Jordan practice chess in week 20? Solution Jordan’s practice scheme is an arithmetic sequence, as it increases by a fixed amount each week. The first week there are 3 hours of practice. This means a1=3. Jordan increases the time spent practicing by 30 minutes, or half an hour, each week. This means d=0.5. Using those values, and that we want to know the amount of time Jordan will study in week 20, we determine the time in week 20 using ai=a1+d×(i−1). a20=3+0.5×(20−1)=3+0.5×19=3+9.5=12.5 So, Jordan will practice 12.5 hours in week 20. Your Turn 3.136 1. Christina decides to save money for after graduation. Christina starts by setting aside $10. Each week, Christina increases the amount she saves by $5. How much money will Christina save in week 52? Example 3.137 Finding the Sum of a Finite Arithmetic Sequence Let’s check back in on Jordan. Recall, Jordan had just watched The Queen’s Gambit and decided to hone their skills, practicing for 3 hours the first week, and increasing the time spent practicing by 30 minutes each week. How many hours total will Jordan have practiced chess after 30 weeks of practice? Solution To calculate the total amount of time that Jordan practiced, we need to use sn=n(a1+an2). The formula requires the first and last terms of the sequence. Since Jordan practiced 3 hours in the first week, the first term is a1=3. Because we want the total practice time after 30 weeks, we need the 30th term. Because the constant difference is d=0.5, the 30th term is a30=3+0.5(30−1)=3+0.5×29=3+14.5=17.5. Applying the formula sn=n(a1+an2) provides the sum of the first 30 terms. s30=30(3+17.52)=60×20.52=615 . This means that Jordan practiced a total of 615 hours after 30 weeks. Your Turn 3.137 1. In a theater, the first row has 24 seats. Each row after that has 2 more seats. How many total seats are there if there are 40 rows of seat in the theater? Who Knew? The Fibonacci Sequence Not all sequences are arithmetic. One special sequence is the Fibonacci sequence, which is the sequence that has as its first two terms 1 and 1. Every term thereafter is the sum of the previous two terms. The first nine terms of the Fibonacci sequence are 1, 1, 2, 3, 5, 8, 13, 21, and 34. This sequence is found in nature, architecture, and even music! In nature, the Fibonacci sequence describes the spirals of sunflower seeds, certain galaxy spirals, and flower petals. In music, the band Tool used the Fibonacci sequence in the song “Lateralus.” The Fibonacci sequence even relates to architecture, as it is closely related to the golden ratio. Video Fibonacci Sequence and “Lateralus” Check Your Understanding 52. Is the following an arithmetic sequence? Explain. {3, 6, 9, 15, 25, 39, 90} 53. What is the 7th term of the following sequence? {1, 5, 7, 100, 4, -17, 8, 100, 19, 7.6, 345} 54. In an arithmetic sequence, the first term is 10 and the constant difference is 4.5. What is the 135th term? 55. If the eighth term of an arithmetic sequence is 35 and the 40th term is 131, what is the constant difference and the first term of the sequence? 56. What is the sum of the first 100 terms of the arithmetic sequence with first term 4 and constant difference 7? 57. A new marketing firm began with 30 people in its survey group. The firm adds 4 people per day. How many people will be in their survey group after 100 days? Section 3.10 Exercises For the following exercises, determine if the sequence is an arithmetic sequence. 1 . {3,7,11,15,25,100,...} 2 . {27,24,21,18,15,12,9,...} 3 . {6,−1,−8,−15,−23,−31,−39,...} 4 . {−5,4,13,22,31,40,49,58,67,...} 5 . {14,19,24,29,34,50,60} 6 . {3.9,2.3,0.7,−0.9,−2.5,−4.1,−5.7,...} 7 . {4,−8,12,−16,20,−24,28,−32,...} 8 . {1,2,3,5,8,13,21,34,55,...} For the following exercises, the sequences given are arithmetic sequences. Determine the constant difference for each sequence. Verify that each term is the previous term plus the constant difference. 9 . {18,68,118,168,218,268,...} 10 . {13,35,57,79,101,123,145,167,...} 11 . {14,11,8,5,2,−1,−4,...} 12 . {4.5,1.9,−0.7,−3.3,−5.9,...} 13 . {−27,−13,1,15,29,43,57,71,...} 14 . {3.8,10.6,17.4,24.2,31,37.8,44.6,...} For the following exercises, the first term and the constant difference of an arithmetic sequence is given. Using that information, determine the indicated term of the sequence. 15 . a1=12 , d=11, find a20 . 16 . b1=5, d=8, find b38 . 17 . c1=48 , d=−7, find c50 . 18 . a1=110, d=−16, find a27 . 19 . t1=15.3, d=4.2, find t17 . 20 . b1=23.8, d=11.7, find b120 . 21 . b1=27.45, d=−3.67, find b40. 22 . a1=67.4, d=−12.3, find a200 . For the following exercises, two terms of an arithmetic sequence are given. Using that information, identify the first term and the constant difference. 23 . a5=27, a15=77 24 . b10=47, b25=137 25 . a9=38, a45=189.2 26 . a6=43, a41=−377 27 . a4=−12.3, a54=−106.5 28 . a12=45.9, a60=−563.7 For the following exercises, the first term and the constant difference is given for an arithmetic sequence. Use that information to find the sum of the first n terms of the sequence, sn. 29 . a1=15, d=7 , calculate s10 30 . a1=2, d=13, calculate s20 . 31 . a1=105, d=0.3, calculate s15 . 32 . a1=56.2, d=1.1, calculate a35 . 33 . a1=450, d=−20, calculate s20. 34 . a1=1400, d=−35, calculate s40 . For the following exercises, apply your knowledge of arithmetic sequences to these real-world scenarios. 35 . A collection is taken up to support a family in need. The initial amount in the collection is $135. Everyone places $20 in the collection. When the 35th person puts their $20 in the collection, how much is present in the collection? 36 . There are 50 songs on a playlist. Every minute, 3 more songs are added to the playlist. How many songs are on the playlist after 40 minutes have passed? 37 . One genre on Netflix has 1,000 shows. Every week, 20 shows are added to that genre. After 15 weeks, how many shows are in that genre? 38 . A new local band has 10 people come to their first show. News of the band spreads afterwards. Each week, 4 more people attend their show than the previous week. After 50 weeks, how many people are at their show? 39 . The Jester Comic book store is going out of business and is taking in no new inventory. Its inventory is currently 13,563 titles. Each day after, they sell or give away 250 titles. After 15 days, how many titles are left? 40 . Jasmyn has decided to train for a marathon. In week one, Jasmyn runs 5 miles. Each week, Jasmyn increased the running distance by 2 miles. How many miles will Jasmyn run in week 13 of the training schedule? 41 . A 42-gallon bathtub sits with 14 gallons in it. The faucet is turned on and is now being filled at the rate of 2.2 gallons per minute, but is draining slowly, at 1.8 gallons per minute. After 20 minutes, how many gallons are in the tub? 42 . A trained diver is 250 feet deep. The diver is nearly out of air and needs to surface. However, the diver can only comfortably ascend 30 feet per minute. How deep is the diver after ascending for 5 minutes? 43 . Jaclyn, an investor, begins a start-up to revitalize homes in South Bend, Indiana. She begins with $10,000, making her investor 1. Each investor that joins will invest $500 more than the previous investor. How much does the 50th investor invest in the project? With that 50th investor, what is the total amount invested in the project? 44 . Jasmyn has decided to train for a marathon. In week one, Jasmyn runs 5 miles. Each week, Jasmyn increased the running distance by 2 miles. After training for 14 weeks, how many total miles will Jasmyn have run? 45 . The base of a pyramidal structure has 144 blocks. Each level above has 5 fewer blocks than the previous level. How many total blocks are there if the pyramidal structure has 25 levels? 46 . As part of a deal, a friend tells you they will give you $10 on day 1, $20 on day 2, $30 on day 3, for all 30 days of a month. At the end of that month, what is the total amount your friend has given you? Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. 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It refers to the capacity of the resisting forces to prevent the wall from rotating with respect to the most bottom left corner of the base for the action of the overturning forces. Those two load groups (resisting and overturning) are divided as follows: Resisting loads: Retaining wall self-weight Active soil over the heel weight Surcharge load Overturning loads: Active soil lateral pressure Lateral pressure that results from the presence of the surcharge That said, the calculation process will be detailed in the following: Input data: Stem Height: 3.124 m Width: 0.305 m Offset: 0.686 m Base Width: 2.210 m Thickness: 0.381 m Active and Passive Soil Unit weight: 18.85 kN/m 3 Friction Angle: 35 degrees Substructure Soil Unit weight: 18.85 kN/m 3 Friction Angle: 35 degrees Soil-Concrete Friction Coefficient: 0.55 Allowable bearing pressure: 143.641 kPa Soil Layers: Active: 3.505 m Passive: 0.975 m Substructure: 0.792 m Surcharge Load Value: -17.237 kN/m Vertical loads: All the vertical loads to which the concrete cantilever retaining wall is subjected are shown in the following picture: It is worth to mention here, that the weight (vertical load) and the moment associated with the portion of passive soil are neglected as it might be removed or eroded and it is a conservative assumption. W s t e m=γ c o n c r e t e⋅(s t e m h e i g h t⋅s t e m w i d t h)=23.58 k N/m 3⋅3.124 m⋅0.305 m W s t e m=γ c o n c r e t e⋅(s t e m h e i g h t⋅s t e m w i d t h)=23.58 k N/m 3⋅3.124 m⋅0.305 m W s t e m=22.467 k N/m W s t e m=22.467 k N/m W b a s e=γ c o n c r e t e⋅(b a s e t h i c k n e s s⋅b a s e w i d t h)=23.58 k N/m 3⋅0.381 m⋅2.210 m W b a s e=γ c o n c r e t e⋅(b a s e t h i c k n e s s⋅b a s e w i d t h)=23.58 k N/m 3⋅0.381 m⋅2.210 m W b a s e=18.855 k N/m W b a s e=18.855 k N/m W a c t i v e=γ s o i l,a c t i v e⋅(s t e m h e i g h t⋅(b a s e w i d t h−s t e m o f f s e t−s t e m w i d t h))W a c t i v e=γ s o i l,a c t i v e⋅(s t e m h e i g h t⋅(b a s e w i d t h−s t e m o f f s e t−s t e m w i d t h)) W a c t i v e=18.85 k N/m 3⋅3.124 m⋅(2.210−0.686−0.305)m W a c t i v e=18.85 k N/m 3⋅3.124 m⋅(2.210−0.686−0.305)m W a c t i v e=71.784 k N/m W a c t i v e=71.784 k N/m W s u r c h a r g e=s u r c h a r g e v a l u e⋅((b a s e w i d t h−s t e m o f f s e t−s t e m w i d t h)W s u r c h a r g e=s u r c h a r g e v a l u e⋅((b a s e w i d t h−s t e m o f f s e t−s t e m w i d t h) W s u r c h a r g e=17.237 k N/m⋅(2.210−0.686−0.305)m W s u r c h a r g e=17.237 k N/m⋅(2.210−0.686−0.305)m W s u r c h a r g e=21.012 k N/m W s u r c h a r g e=21.012 k N/m Retaining Wall Calculator Restoring moment: The restoring moment is the one in charge of preventing the wall from rotating with respect to the most bottom-left corner of the base. For calculating it, it is required to perform a moment summation with respect to the mentioned point of all the vertical loads: M s t e m=W s t e m⋅d s t e m=22.467 k N/m⋅0.839 m=18.839 k N m/m M s t e m=W s t e m⋅d s t e m=22.467 k N/m⋅0.839 m=18.839 k N m/m M b a s e=W b a s e⋅d b a s e=18.855 k N/m⋅1.105 m=21.939 k N m/m M b a s e=W b a s e⋅d b a s e=18.855 k N/m⋅1.105 m=21.939 k N m/m M a c t i v e=W a c t i v e⋅d a c t i v e=71.784 k N/m⋅1.601 m=114.89 k N m/m M a c t i v e=W a c t i v e⋅d a c t i v e=71.784 k N/m⋅1.601 m=114.89 k N m/m M s u r c h a r g e=W s u r c h a r g e⋅d s u r c h a r g e=21.012 k N/m⋅1.601 m=33.630 k N m/m M s u r c h a r g e=W s u r c h a r g e⋅d s u r c h a r g e=21.012 k N/m⋅1.601 m=33.630 k N m/m Σ M R=M s t e m+M b a s e+M a c t i v e+M s u r c h a r g e Σ M R=M s t e m+M b a s e+M a c t i v e+M s u r c h a r g e Σ M R=18.839+21.939+114.89+33.630 Σ M R=18.839+21.939+114.89+33.630 Σ M R=189.298 k N m/m Σ M R=189.298 k N m/m Horizontal loads: All the horizontal loads to which the concrete cantilever retaining wall is subjected are shown in the following picture: In order to calculate the lateral earth pressure due to the retained soil active pressure and the surcharge resultant lateral pressure, it is necessary to calculate the Rankine active earth-pressure coefficient: K a=1−sin(γ s o i l,a c t i v e)1+sin(γ s o i l,a c t i v e)K a=1−sin⁡(γ s o i l,a c t i v e)1+sin⁡(γ s o i l,a c t i v e) K a=1−sin(35 º)1+sin(35 º)=0.271 K a=1−sin⁡(35 º)1+sin⁡(35 º)=0.271 With that result, it is now possible to calculate the horizontal load resulting from the lateral active pressure that the retained soil exerts: H a c t i v e=1 2⋅γ s o i l,a c t i v e⋅(s t e m h e i g h t+b a s e t h i c k n e s s)2⋅K a H a c t i v e=1 2⋅γ s o i l,a c t i v e⋅(s t e m h e i g h t+b a s e t h i c k n e s s)2⋅K a H a c t i v e=1 2⋅18.85 k N/m 3⋅3.505 2⋅0.271 H a c t i v e=1 2⋅18.85 k N/m 3⋅3.505 2⋅0.271 H a c t i v e=31.377 k N/m H a c t i v e=31.377 k N/m For calculating the horizontal force related to the surcharge presence, an equivalent soil height is calculated first, and then the actual force: h s o i l,e q=s u r c h a r g e v a l u e γ s o i l,a c t i v e=17.237 k N/m 2 18.85 k N/m 3 h s o i l,e q=s u r c h a r g e v a l u e γ s o i l,a c t i v e=17.237 k N/m 2 18.85 k N/m 3 h s o i l,e q=0.914 m h s o i l,e q=0.914 m H s u r c h a r g e=γ s o i l,a c t i v e⋅h s o i l,e q⋅(s t e m h e i g h t+b a s e t h i c k n e s s)⋅K a H s u r c h a r g e=γ s o i l,a c t i v e⋅h s o i l,e q⋅(s t e m h e i g h t+b a s e t h i c k n e s s)⋅K a H s u r c h a r g e=⋅18.85 k N/m 3⋅0.914 m⋅3.505 m⋅0.271 H s u r c h a r g e=⋅18.85 k N/m 3⋅0.914 m⋅3.505 m⋅0.271 H s u r c h a r g e=16.372 k N/m H s u r c h a r g e=16.372 k N/m Overturning moment The overturning moment is calculated as the moment generated by the horizontal loads with respect to the most bottom-left corner of the base. The lever arm distance for each of the horizontal loads will be: One-third of the wall’s height from the bottom of the base for the resultant of the active pressure distribution of the retained soil. It is like that since that pressure follows a triangular distribution with a zero value at the surface level and a maximum value at the bottom of the base level. One-half of the wall’s height from the bottom of the base for the case of the resultant horizontal load from the presence of the surcharge load. It is like that since that pressure follows a rectangular distribution. That said, the overturning moment is calculated as follows: M a c t i v e=H a c t i v e⋅1 3(s t e m h e i g h t+b a s e t h i c k n e s s)M a c t i v e=H a c t i v e⋅1 3(s t e m h e i g h t+b a s e t h i c k n e s s) M a c t i v e=31.377 k N/m⋅1 3 3.505 m M a c t i v e=31.377 k N/m⋅1 3 3.505 m M a c t i v e=36.659 k N m/m M a c t i v e=36.659 k N m/m M s u r c h a r g e,h=H s u r c h a r g e⋅1 2(s t e m h e i g h t+b a s e t h i c k n e s s)M s u r c h a r g e,h=H s u r c h a r g e⋅1 2(s t e m h e i g h t+b a s e t h i c k n e s s) M s u r c h a r g e,h=16.372 k N/m⋅1 2 3.505 m M s u r c h a r g e,h=16.372 k N/m⋅1 2 3.505 m M s u r c h a r g e,h=28.692 k N m/m M s u r c h a r g e,h=28.692 k N m/m Σ M O T M=M a c t i v e+M s u r c h a r g e,h Σ M O T M=M a c t i v e+M s u r c h a r g e,h Σ M O T M=36.659 k N m/m+28.692 k N m/m Σ M O T M=36.659 k N m/m+28.692 k N m/m Σ M O T M=65.351 k N m/m Σ M O T M=65.351 k N m/m Factor of Safety against overturning ACI 318 recommends a factor of safety to be greater than or equal to 2.0 2.0. It is calculated as follows: F S=Σ M R Σ M O T M F S=Σ M R Σ M O T M F S=189.298 k N∗m 65.351 k N∗m=2.897≥2.0 F S=189.298 k N∗m 65.351 k N∗m=2.897≥2.0 PASS! Retaining Wall Calculator In this article, we have discussed overturning moment calculation examples. SkyCiv offers a free Retaining Wall Calculator that will check overturning moment and perform a stability analysis on your retaining walls. The paid version also displays the full calculations, so you can see step-by-step on how to calculate the stability of retaining wall against overturning, sliding and bearing! Retaining Wall Calculator Oscar Sanchez Product Developer BEng (Civil) LinkedIn Still stuck? How can we help?Updated on May 9, 2024 Doc navigation ← Lateral Earth Pressure due to Surcharge LoadsRetaining Wall Sliding Calculation Example → Was this article helpful to you? How can we help? 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12178	https://blog.chatfield.io/simple-method-for-distance-to-ellipse/	Simple Method for Distance to Ellipse Menu Close Home Subscribe Menu Simple Method for Distance to Ellipse 28 August 2017 Analytically finding the smallest distance between a point and an ellipse boils down to solving a quartic equation. The easiest way to see this is by considering that a circle can intersect an ellipse at most four times. In this view, the problem is to find the circle that intersects the ellipse once. If formulated as a quartic, two of the roots will be shared at the intersection, and the other two roots will be complex. Although a closed solution to the quartic does exist, it is unsurprising that iterative methods would be easier to implement and less prone to numeric instabilities. Generic root finders will work, but a robust implementation would be advised. This StackOverflow post shows what can go wrong when using a naive implementation of Newton's Method. Specialised iterative methods also exist, such as described by Maisonobe. The paper's premise is that an oversized circle will intersect the ellipse at least two times. The iteration to minimise the distance between point a on the ellipse and p is as follows: Choose a point a on the ellipse Compute the distance r between a and p On the circle centred at p with radius r, find the other intersection b Set a' to the midpoint of a and b on the ellipse Return to step 2 Visually a' is obviously a very good innovation over a, however computing b and a' is non trivial. In brief, the quartic still has 4 roots, but because the initial guess a is a solution, the quartic can be reduced to a cubic. The cubic is solved to find b. An additional quadratic is then solved to find a'. As noted, this complexity buys a very good estimate for a', and therefore the paper's claim that this algorithm is robust and converges quickly is very believable. What follows is an alternative, much simpler computation for a'. Consider the previous algorithm when the ellipse is a circle. a' then trivially lies on the line between c and p. It can also be seen that a' is optimal. The idea will become to locally approximate the ellipse as a circle, but first we must touch on curvature. As a smooth curve, every point on the ellipse has a perpendicular normal. Additionally, every point has a radius of curvature. Without going into specifics, a tighter curve will result in a smaller radius of curvature, while a shallow curve will result in a larger radius of curvature. The centre of curvature of point e on the ellipse is the centre of the circle with radius equal to the radius of curvature and perpendicular to the ellipse at e. In other words, it is the centre of the circle that locally approximates the ellipse at e. The centre of curvatures can be plotted to form the evolute of the ellipse, which Wolfram and Wikipedia are better suited to explain. It's now time to define the curves mathematically. Although everything can derived in Cartesian coordinates, a parametric definition avoids infinities about the vertices. For the ellipse use the standard definition. x = a cos(t) y = b sin(t) The ellipse evolute also has a parametric form. ev_x = (aa-bb)/a cos^3(t) ev_y = (bb-aa)/b sin^3(t) This is extremely useful for approximating the ellipse as a circle, since we know the centre of curvature for any t. Recall that when the ellipse is a circle, a' lies on the line between the circle centre c and p. Now modify this claim slightly, to state that a' lies on the line between the centre of curvature c and p. This is not as exact as the original a' from the paper since we've approximated the ellipse as a circle, but the beauty is that the closer we get to the optimal a', the better the approximation becomes. Therefore I would expect similar convergence and robustness properties. Intersecting a line with an ellipse is a much simpler problem, however it still requires solving a quadratic. Since we're already estimating, an exact solution isn't required. From the previous figure, define two vectors r and q. Very loosely speaking, r is the radius of curvature and q intersects the ellipse where we would like the radius of curvature to have been. Since we are still using our circle approximation, we can compute the arc length between a and a' as the angle between r and q times the radius of curvature. (r x q) sin(Δc/\|r\|) ≈ ------- \|r\|\|q\| Additionally, since Δc is small, we could further approximate by dropping the sine. Now we would like to know how much to vary t by to achieve the same arc length delta on the ellipse. With a bit of calculus: ``` dx/dt = -a sin(t) dy/dt = b cos(t) dc/dt = sqrt((dx/dt)^2 + (dy/dt)^2) Δc/Δt ≈ sqrt(a^2 sin^2(t) + b^2 cos^2(t)) Δt ≈ Δc / sqrt(a^2 sin^2(t) + b^2 cos^2(t)) Δt ≈ Δc / sqrt(a^2 - a^2 cos^2(t) + b^2 - b^2 sin^2(t)) Δt ≈ Δc / sqrt(a^2 + b^2 - x^2 - y^2) a'x = a cos(t + Δt) a'y = b sin(t + Δt) ``` This proves to be a good enough approximation, but it does suffer at the vertices (pointy ends) of the ellipse. Although in my experiments convergence was not affected, in the code listing I choose to confine t to the first quadrant and then correct the signs of the final result. For clarity, some obvious optimisations have been left unimplemented. ```python def solve(semi_major, semi_minor, p): px = abs(p) py = abs(p) t = math.pi / 4 a = semi_major b = semi_minor for x in range(0, 3): x = a math.cos(t) y = b math.sin(t) ex = (aa - bb) math.cos(t)3 / a ey = (bb - aa) math.sin(t)3 / b rx = x - ex ry = y - ey qx = px - ex qy = py - ey r = math.hypot(ry, rx) q = math.hypot(qy, qx) delta_c = r math.asin((rxqy - ryqx)/(rq)) delta_t = delta_c / math.sqrt(aa + bb - xx - yy) t += delta_t t = min(math.pi/2, max(0, t)) return (math.copysign(x, p), math.copysign(y, p)) ``` One remaining question is why t is initialised to a constant instead of atan2(py, px). This is only a good guess if p is outside of the ellipse. On the inside it is optimally bad. If p is inside the ellipse initialise to atan2(px, py). It turns out that a close to optimally bad guess will require additional iterations to move away from the initial guess in a random fashion. An optimally bad guess will get stuck. I've shown that for rather detailed reasons, the algorithm converges as long as you don't start near the optimally bad guess. You will be fine using atan2 if you correctly case for being inside or outside of the ellipse. The plots show good convergence after three iterations over a wide range of eccentricities. However, at extreme eccentricities the algorithm will eventually fail. At some cutoff the ellipse should be considered a line. The code for the plots is available on GitHub. Carl Chatfield Bringing autonomy to underwater drones. Share this post TwitterFacebookGoogle+Some Methods of Gradient Estimation Are Better Than Others ---------------------------------------------------------- I want to share a plot which took me by surprise. In computer vision, a useful tool for estimating…Carl Chatfield © 2025 Proudly published with Ghost
12179	https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Algebra_Problems?srsltid=AfmBOoqaTTPuVLmZ6QA538Nu_mjUGqKMbwc6ElSYLmPEhUCRR8c59Mo0	Art of Problem Solving Category:Olympiad Algebra Problems - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. 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12180	https://www.mathworks.com/matlabcentral/answers/787269-how-to-substitute-x-by-an-integer-in-a-derivative-function	How to substitute x by an integer in a derivative function - MATLAB Answers - MATLAB Central Skip to content MATLAB Answers MATLAB Help Center Community Learning Get MATLABMATLAB Sign In My Account My Community Profile Link License Sign Out Contact MathWorks Support Visit mathworks.com Search Answers Answers Help Center Answers MathWorks MATLAB Help Center MathWorks MATLAB Answers File Exchange Videos Online Training Blogs Cody MATLAB Drive ThingSpeak Bug Reports Community MATLAB Answers File Exchange Cody AI Chat Playground Discussions Contests Blogs More Communities Treasure Hunt People Community Advisors Virtual Badges About Home Ask Answer Browse MATLAB FAQs More Contributors Recent Activity Flagged Content Manage Spam Help You are now following this question You will see updates in your followed content feed. You may receive emails, depending on your communication preferences. How to substitute x by an integer in a derivative function Follow 4 views (last 30 days) Show older comments Rodrigo Toledoon 29 Mar 2021 Vote 0 Link × Direct link to this question Cancel Copy to Clipboard ⋮ Vote 0 Link × Direct link to this question Cancel Copy to Clipboard Commented:Paul on 30 Mar 2021 Accepted Answer:Andreas Apostolatos Open in MATLAB Online I have a function: Theme Copy syms x f(x) = (tan(x^2 + 3x))^2 a = diff(f(x)) Now i want to calculate the a when x=1. How can i do it? i tried a(2) but didnt work. Thanks 0 Comments Show -2 older comments Hide -2 older comments Sign in to comment. Sign in to answer this question. Accepted Answer Andreas Apostolatoson 29 Mar 2021 Vote 2 Link × Direct link to this answer Cancel Copy to Clipboard ⋮ Vote 2 Link × Direct link to this answer Cancel Copy to Clipboard Open in MATLAB Online Hello, To evaluate symbolic expression 'a' at 'x = 1' you can use function 'subs', namely, Theme Copy a_at_1 = subs(a, 1); I am not sure why you tried typing 'a(2)' since you are trying to evaluate 'a' at 1. More information about function 'subs' you can find in the following documentation page, I hope that this helps. Kind Regards, Andreas 2 Comments Show None Hide None Rodrigo Toledoon 30 Mar 2021 × Direct link to this comment Cancel Copy to Clipboard ⋮ Link× Direct link to this comment Cancel Copy to Clipboard thanks mate. a(2) is a typo. i meant a(1), sorry. Paulon 30 Mar 2021 × Direct link to this comment Cancel Copy to Clipboard ⋮ Link× Direct link to this comment Cancel Copy to Clipboard Open in MATLAB Online You can also define a to be a function of x and then evaluate it at the value of interest Theme Copy syms x f(x) = (tan(x^2 + 3x))^2 a(x) = diff(f(x)) f(x) = tan(x^2 + 3x)^2 a(x) = 2tan(x^2 + 3x)(2x + 3)(tan(x^2 + 3x)^2 + 1) a(1) ans = 10tan(4)(tan(4)^2 + 1) Sign in to comment. More Answers (0) Sign in to answer this question. FEATURED DISCUSSION ###### MATLAB EXPO 2025 Registration is Now Open! November 12 – 13, 2025 Registration is now open for MathWorks annual virtual event MATLAB EXPO 2025... 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12181	https://arup.utah.edu/education/NewbornS2.php/	Continuing Education Short Topics Other Resources MLS Faculty & Student Resource Center Faculty Expertise Collaboration Presentations by Experts How to Start Breadcrumb ARCHIVED: NOT AVAILABLE FOR CREDIT Two Short Talks on Newborn Screening – The Utah Experience Through a decade-long partnership, ARUP and the University of Utah have worked with the Utah Department of Health to implement continued improvements to the state’s newborn screening program. In 2006, Utah implemented the expansion of diseases screened in newborns by use of tandem mass spectrometry technology, and in 2013, Severe Combined Immunodeficiency (SCID) was added to the list. SCID was the first molecular test used as the primary screen in Utah. On the occasion of the 50th anniversary of newborn screening in the U.S., two laboratory experts discuss the history of their collaboration with the Department of Health, their conclusion from the pilot studies, and statistics seen through the newborn screening program in Utah. Originally presented on October 17, 2013, in Salt Lake City, Utah. Lecture Presenters \| \| \| --- \| \| Marzia Pasquali, PhD \| Marzia Pasquali, PhD Medical Director, Biochemical Genetics ARUP Laboratories Medical Director, Newborn Screening ARUP Laboratories Professor of Pathology University of Utah School of Medicine \| Marzia Pasquali, PhD Medical Director, Biochemical Genetics ARUP Laboratories Medical Director, Newborn Screening ARUP Laboratories Professor of Pathology University of Utah School of Medicine Dr. Pasquali is the medical director of the Biochemical Genetics and Newborn Screening laboratories at ARUP. She is also a professor at the University of Utah School of Medicine and a clinical scientist in the area of newborn screening. Born and educated in Italy, Dr. Pasquali earned her degrees of doctor in pharmaceutical chemistry and technology and pharmacy doctor at the University of Parma School of Pharmacy. Dr. Pasquali trained in biochemical genetics at Emory University and later served as the co-director of the Biochemical Genetics Laboratory. Dr. Pasquali is board certified in Clinical Biochemical Genetics. \| \| \| --- \| \| Patricia R. Slev, PhD \| Patricia R. Slev, PhD Medical Director, Serological Hepatitis/Retrovirus Laboratory ARUP Laboratories Assistant Professor of Pathology University of Utah School of Medicine \| Patricia R. Slev, PhD Medical Director, Serological Hepatitis/Retrovirus Laboratory ARUP Laboratories Assistant Professor of Pathology University of Utah School of Medicine Dr. Slev is the medical director of the Serological Hepatitis/Retrovirus Laboratory at ARUP and an assistant professor of pathology at the University of Utah School of Medicine. Dr. Slev earned her PhD in immunology and laboratory medicine from the University of Florida, Gainesville and completed a fellowship in clinical chemistry at the University of Utah. She is board certified by the American Board of Clinical Chemistry. Dr. Slev’s research interests are in immunogenetics and pathogen interactions, particularly HIV and viral hepatitis. Objectives After this presentation, participants will be able to: Understand the aim of state newborn screening programs. Become familiar with recent laboratory technologies used in newborn screening programs. Recognize the relationship between screening, follow up, and confirmatory testing from a laboratory perspective. Sponsored by: University of Utah School of Medicine, and ARUP Laboratories ╳ How do I obtain continuing education credit? To obtain credit, the video lecture must be watched in its entirety, and the quiz must be passed with 80% or higher. You DO NOT have to choose the credit type or enter any user information UNTIL AFTER you have successfully passed the quiz, and only if you want to obtain credit for the video lecture. Here’s how it works: At the conclusion of the video presentation, a dialogue box will appear prompting you to proceed to the quiz. Once the quiz has been passed, you can choose your desired credit type, and enter the information necessary for the credit to be valid and produce a completion certificate. After submitting the user information form, a certificate will be generated based on the credit type you have chosen. 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12182	https://justtothepoint.com/algebra/groupunits/	The Integers modulo n. The Group of Units. Fermat’s Theorem. \| JustToThePoint Reading Maths Calculus Algebra Coding Software SelfHelp Humour Education Social The Integers modulo n. The Group of Units. Fermat’s Theorem. “Life is good and there’s no reason to think it won’t be, right until the moment when a blonde, non experienced nurse is going to insert a catheter in your penis like a Spanish matador,” Dad retorted. [···] “Unless your name is Google, stop pretending you know everything. Unless your name is Facebook, stop sharing it because unless your name is Amazon, we are not buying it,” the class’ bully teased me once again, Apocalypse, Anawim, #justtothepoint. Modular Arithmetic as an Equivalence Relations. Let n be a fixed positive integer. Let’s define a relation on ℤ by a ~ b if and only if n \| (b - a). It is clearly an equivalence relation. It is written as a ≡ b (mod n), “a is congruent to be mod n”. An alternative is a ~ b if and only if b - a = kn for some k ∈ ℤ. Reflexive: a ~ a ↭ a - a = 0 = 0·n, therefore a ≡ a (mod n). Symmetry: a ~ b ↭ ∃k ∈ ℤ: b - a = kn ⇒ ∃-k ∈ ℤ: a -b = -kn Transitivity a ~ b ↭ ∃k ∈ ℤ : b - a = kn, b ~ c ↭ ∃k’ ∈ ℤ: c - b = k’n ⇒∃(k + k’) ∈ ℤ : c - a = (k + k’)n ⇒ a ~ c Examples: 7 ≡ 1 (mod 3), 5 ≡ 2 (mod 3). 12≡5(m o d 7)12 \equiv 5 (mod~ 7)12≡5(m o d 7), but 9≢5(m o d 7)9 \not\equiv 5 (mod~ 7)9≡5(m o d 7) 10 ≡ 0 (mod 5), 12 ≡ 2 (mod 5). 12 ≡ 5 is written in LaTeX as 12 \equiv 5, but 9≢5 9 \not\equiv 5 9≡5 is written as 9 \not\equiv 5. Let [a] be the equivalent class of a. It is called the congruence class or residue class of a mod n. It is the set of integers that are congruent to a modulo n, that is, that differ from a by a multiple of n, i.e., [a] = {b: b ∈ ℤ: b ≡ a (mod n)} = {a + kn: k ∈ ℤ} = {a, a ± n, a ± 2n, a ± 3n, ···}. Notice that there are n-1 possible remainders after division by n ⇒ there are n distinct equivalence classes mod n. Therefore, the integers modulo n are the set of these equivalence classes = {, , , … [n-1]}, usually symbolized by the quotient ℤ/nℤ. The representative is just the remainder of the integer when divided by n. ℤ/nℤ = ℤ n, the integers modulo n = {, , , … [n-1]} ≠ {0, 1, 2, ··· n}. ℤ/nℤ = ℤ n is the set of equivalence classes, not the set of its representatives. Let’s define an addition and multiplication for ℤ/nℤ: [a] + [b] = [a+b] [a] · [b] = [ab] These operations are well-defined, that is, they do not depend on the choices of representatives for the classes involved. Proof: Suppose a 1 ≡ b 1 (mod n) and a 2 ≡ b 2 (mod n) then, ∃s, t ∈ ℤ: a 1 = b 1 + sn, a 2 = b 2 + tn. a 1 + a 2 = b 1 + b 2 + (s + t)n ⇒ a 1 + a 2 ≡ b 1 + b 2 (mod n) ⇒ [a 1 + a 2] = [b 1 + b 2] (the sum of the congruence or residue classes is independent of the representative chosen elements) a 1 a 2 = (b 1 + sn)(b 2 + tn) = b 1 b 2 + (b 1 t + b 2 s + st)n ≡ b 1 b 2 (mod n) ⇒ [a 1 a 2] = [b 1 b 2] ∎ Z r Z\frac{ℤ}{rℤ}r Z Z is an Abelian group with this operation, is the neutral element, and -[n]=[-n]. The Group of Units. The group ℤ n consists of the elements {, , , . . . , [n-1]} with addition mod n as the operation. For convenience’s sake, we identify each equivalence class with its representative. You can also multiply elements of Z n, but you do not obtain a group. For instance, the identity element 0 does not have a multiplicative inverse. ℤ n is a commutative ring with identity 1. However, if you confine your attention to the units in ℤ n, that is, the elements which have multiplicative inverses, u ∈ ℤ n is a unit if ∃v ∈ ℤ n: u·v = 1 and v is called a multiplicative inverse of u, then you indeed get a group under multiplication mod n. It is denoted as U n, and is called the group of units in ℤ n Theorem. An integer “a” has a multiplicative inverse modulo n if and only if a and n are coprime or relative prime. Figure 1.b. Proof. ⇐) Let n be an integer, suppose a is coprime to n ⇒ [By Bézout’s identity] ∃α, β ∈ ℤ: αa + βn = 1 ⇒ αa = 1 - βn ⇒ αa ≡ 1(mod n) ⇒ a has a multiplicative inverse (α). ⇒) Suppose that a has a multiplicative inverse and, for the sake of contradiction, a is not coprime to n ⇒∃d ∈ ℤ, d > 1 such that a = da’, n = db and (a has a multiplicative inverse, αa ≡ 1 (mod n) ) αa = 1 + βn for some α, β ∈ Z ⇒ αda’ = 1 + βnb ⇒ d(αa’ -βb) = 1 ⇒ d = ±1 ⊥ (which is impossible since d > 1)∎ For each n > 1, we define U n to be the set of all positive integers less than n that have a multiplicative inverse modulo n, that is, ∀a ∈ U n, ∃ b ∈ Z n : ab = 1. U n = {a ∈ Z n: a is relatively prime to n, a < n} = {a ∈ Z n: gcd(a, n)=1}. In particular, \|U n\| = φ(n) where φ is the Euler’s totient function. Equivalently, the elements of this group can be thought of as the congruence classes, also known as residues modulo n, that are coprime to n. (ℤ/nℤ)x = {[a] ∈ ℤ/nℤ \| gcd(a, n) = 1} Theorem. ℤ/nℤ ≋ ℤ n. Example. For n = 10, U 10 = {1, 3, 7, 9}. The Cayley table for U 10 is in Figure 1.a. Proposition. U n = {[m]\| gcd(m, n) = 1} is a group under multiplication mod n, i.e., [x][y] = [xy]. It is called the multiplicative group of integers modulo n. is it well defined? [a]·[b] = [ab] have been previously demonstrated that they do not depend on the choices of representatives for the classes involved. Besides, ∀a, b ∈ U n, ∃a-1, b-1. (b-1 a-1)(ab) =[Associative] b-1(a-1 a)b = (b-1 b) = e. Analogously, mutatis mutandis (ab)(b-1 a-1) = e ⇒ ab has a multiplicate inverse, ab ∈ U n. Associativity is inherited from Z n. The identify is 1 with multiplicative inverse to be itself. Every element of U n has a multiplicative inverse by definition. Examples: U 5 = {1, 2, 3, 4} and U 6 = {1, 5}. Their Cayley’s table are shown in Figure 1.c and 1.d respectively. If p is a prime, then all the positive integers smaller than p are relatively prime to p. Then, U(p) = {1, 2, 3, ···, p-1} e.g., U(3) = {1, 2}, U(7) = {1, 2, 3, 4, 5, 6}, and U 11 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. U(14) = {1, 3, 5, 9, 11, 13}. Notice that you multiply elements of U(14) by multiplying as if they were integers, then reducing mod 14, e.g., 11·13 = 143 = 3 (mod 14). U(21) = {1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20}. Definition. The subgroup generated by an element. If k is a divisor of n. U k(n) = {x ∈ U(n): x mod k = 1}. U k(n) is a subgroup of U(n). Examples. U(18) = {1, 5, 7, 11, 13, 17}. 3 is a divisor of 18, U 3(18) = {x ∈ U(18): x mod 3 = 1} = {1, 7, 13} = ⟨7⟩ ≤ U(18). U(105) = {1, 2, 4, 8, 11, 13, 16, 17, 19, 22, 23, 26, 29, 31, 32, 34, 37, 38, 41, 43, 44, 46, 47, 49, 52, 53, 56, 58, 59, 61, 62, 64, 67, 68, 71, 73, 74, 76, 79, 82, 83, 86, 88, 89, 92, 94, 97, 98, 101, 103, 104}. 7 is a divisor of 105. U 7(105) = {x ∈ U(105): x mod 7 = 1} = {1, 8, 22, 29, 43, 64, 71, 92}. The Group of Units Modulo n as an External Direct Product Theorem. Suppose s and t are relatively prime. Then, U(st) ≋ U(s)⊕U(t), i.e., U(st) is isomorphic to the external direct product of U(s) and U(t). Futhermore, U s(st) ≋ U(t) and U t(st) ≋ U(s). Proof. Corollary. Let m = n 1 n 2···n k, ∀i, j, i ≠ j, gcd(n i, n j) = 1. Then, U(m) ≋ U(n 1)⊕U(n 2)⊕···⊕U(n k). Theorem (Gauss, 1801). U(2) = {1} ≋ {0}, U(4) = {1, 3} ≋ ℤ 2. U(2 n) ≋ U(2 2) ⊕ U(2 n-2) ≋ ℤ 2⊕ℤ 2 n-2 ∀n ≥ 3, U(p n) ≋ ℤ p n-p n-1 for p a positive integer and an odd prime. Theorem. Every group U(n) is isomorphic to the external direct product of cyclic groups. Examples: U(55) ≋[Criterion for the direct product to be cyclic] U(5) ⊕ U(11) ≋ ℤ 4⊕ℤ 10 ≋ ℤ 4⊕ℤ 2⊕ℤ 5 U(8) = U(2 3) = {1, 3, 5, 7} has 4 elements and is non-cyclic since each element (except the identity 1) has order 2. U(8) ≋ ℤ 2⊕ℤ 2. If p = 5, n = 2, U(5 2) ≋[U(p n) ≋ ℤ p n-p n-1 for p a positive integer and an odd prime] ℤ 5 2-5 1, i.e., U(25) ≋ ℤ 20. U(16) ≋[U(2 n) ≋ ℤ 2⊕ℤ 2 n-2 ∀n ≥ 3] ℤ 2 ⊕ ℤ 4, U(9) ≈[9 = 3 2, 2 ≱ 3] ℤ 6, U(7) ≋ ℤ 6, U(5) ≋ ℤ 4, U(3) ≋ ℤ 2. U(105) ≋ [105 = 3·5·7] U(3)⊕U(5)⊕U(7) ≋ ℤ 2 ⊕ ℤ 4 ⊕ ℤ 6. \|U(105)\| = 2 × 4 × 6 = 48. U(100) ≋ U(4) ⊕ U(25) ≋ ℤ 2 ⊕ ℤ 20 U(168) ≋ [168 = 3·7·8] U(3)⊕U(7)⊕U(8) ≋ ℤ 2 ⊕ ℤ 6 ⊕ ℤ 2 ⊕ ℤ 2. Euler’s theorem Euler’s theorem. Let a, n ∈ ℤ, n > 0 and gcd(a, n) = 1. Then, a Φ(n)≡1 (mod n). Proof. If gcd(a, n) = 1 ⇒ a ∈ ℤ n = U(n), \|U(n)\| = Φ(n) ⇒ [By Lagrange’s theorem. The order of an element of a finite group divides the order of the group] a Φ(n) = e ↭ a Φ(n) ≡ 1 (mod n)∎ Fermat’s Theorem Fermat’s Theorem. If a and p are integers, p is prime, and p ɫ a, then a p−1 ≡ 1 (mod p). Futhermore, a p≡ a (mod p) for any integer a. Proof. If p is prime, then U p = {1, 2, 3, ···, p-1}, \|U p\| = Φ(p) = p-1. If p is prime and p ɫ a ⇒ a ≡ b (mod p), where b ∈ {1, 2, 3, ···, p-1} -we could exclude the 0 from the list, b≠ 0- ⇒ [By Lagrange’s theorem. The order of an element of a finite group divides the order of the group, a\|G\|=e.] b p-1 ≡ 1 (mod p) ⇒[a ≡ b (mod p)] a p-1 ≡ b p-1 ≡ 1 (mod p) ∎. a p≡ a (mod p) is an immediate consequence by multiplying both sides of a p−1 ≡ 1 (mod p) by a. Besides, if p \| a, then a ≡ 0 (mod p), and a p ≡ 0 p ≡ 0 ≡ 0 (mod p). Examples. Let’s use this recently proven Fermat’s Theorem to reduce power. 77 2401 (mod 97). 97 is prime, 97 ɫ 77 ⇒ [Fermat’s Theorem. p is prime, p ɫ a, then a p−1 ≡ 1 (mod p).] 77 96 ≡ 1(mod 97). 2401/96, quotient = 25, remainder = 1 (77 96)25 ≡ 1 25 (mod 97) ⇒ 77 2400 ≡ 1 (mod 97) ⇒ 77 2401 ≡ 77 (mod 97). 3 100,000 (mod 53). 53 is prime, 53 ɫ 3 ⇒ [Fermat’s Theorem. p is prime, p ɫ a, then a p−1 ≡ 1 (mod p).] 3 52 ≡ 1 (mod 53). 100,000/52, quotient = 1923, remainder = 4 (3 52)1923 ≡ 1 1923 (mod 53) ⇒ 3 99,996 ≡ 1 (mod 53) ⇒ 3 100,000 = 3 4(3 99,996) ≡ 3 4 (mod 53) ⇒ 3 100,000 ≡ 81 (mod 53) ≡ 28 (mod 53). Fermat’s Little Theorem Converse. If m ≥ 2 and ∀a: 1 ≤ a ≤ m-1, a m-1≡ 1 (mod m) ⇒ m is prime. Proof. (Source LibreTexts,Elementary Number Theory (Barrus and Clark), Primality Test.) If m ≥ 2 and ∀a: 1 ≤ a ≤ m-1, a m-1≡ 1 (mod m) ⇒ a has an inverse modulo m, namely a m-2 ⇒ [An integer has an inverse modulo m ↭ (a, m) = 1] ∀a: 1 ≤ a ≤ m-1, (a, m) = 1. Claim: m is prime. For the sake of contradiction, suppose m is not prime ⇒ m = a·b, 1 < a < m, 1 < b < m ⇒ gcd(a, m) = a > 1 ⊥ Corollary. Primarily Test. Given an integer m ≥ 2, for each a between 1 and m -1, test wether a m-1 ≡ 1 (mod m). If the congruence is true for every value of a, then m is prime. Otherwise, if the congruence fails for any value of a between 1 and m-1, then m is not prime. Example: 13 is prime because 1 12 ≡ 1 (mod13); 2 12 = 4096 ≡ 1 (mod13); 3 12 = 531441 ≡ 1 (mod13); [···] and 12 12 ≡ 1 (mod13). Bibliography This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian. NPTEL-NOC IITM, Introduction to Galois Theory. Algebra, Second Edition, by Michael Artin. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson). Field and Galois Theory, by Patrick Morandi. Springer. Michael Penn (Abstract Algebra), and MathMajor. Contemporary Abstract Algebra, Joseph, A. Gallian. Andrew Misseldine: College Algebra and Abstract Algebra. Bitcoin donation JustToThePoint Copyright © 2011 - 2025 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us. This website uses cookies to improve your navigation experience. 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12183	https://math.libretexts.org/Courses/Monroe_Community_College/MTH_165_College_Algebra_MTH_175_Precalculus/05%3A_Trigonometric_Functions_and_Graphs/5.04%3A_The_Other_Trigonometric_Functions	Published Time: 2020-01-24T21:25:51Z 5.4: The Other Trigonometric Functions - Mathematics LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 5: Trigonometric Functions and Graphs MTH 165 College Algebra, MTH 175 Precalculus { "5.4e:Exercises-_Other_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "5.01:_Angles" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.02:_Right_Triangle_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.03:_Sine_and_Cosine_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.04:_The_Other_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.5:_Graphs_of_the_Sine_and_Cosine_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "5.6:_Graphs_of_the_Other_Trigonometric_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "00:_Front_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "00:_Preliminary_Topics_for_College_Algebra" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "01:_Equations_and_Inequalities" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Functions_and_Their_Graphs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Polynomial_and_Rational_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Exponential_and_Logarithmic_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Trigonometric_Functions_and_Graphs" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Analytic_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Further_Applications_of_Trigonometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "zz:_Back_Matter" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sat, 05 Mar 2022 02:31:47 GMT 5.4: The Other Trigonometric Functions 34916 34916 Mary Cameron { } Anonymous Anonymous 2 false false [ "article:topic", "cosecant", "cotangent", "identities", "period", "secant", "tangent", "license:ccby", "showtoc:yes" ] [ "article:topic", "cosecant", "cotangent", "identities", "period", "secant", "tangent", "license:ccby", "showtoc:yes" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. 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MTH 165 College Algebra, MTH 175 Precalculus 5. 5: Trigonometric Functions and Graphs 6. 5.4: The Other Trigonometric Functions Expand/collapse global location MTH 165 College Algebra, MTH 175 Precalculus Front Matter 0: Preliminary Topics for College Algebra 1: Equations and Inequalities 2: Functions and Their Graphs 3: Polynomial and Rational Functions 4: Exponential and Logarithmic Functions 5: Trigonometric Functions and Graphs 6: Analytic Trigonometry 7: Further Applications of Trigonometry Back Matter 5.4: The Other Trigonometric Functions Last updated Mar 5, 2022 Save as PDF 5.3e: Exercises - Sine and Cosine Functions 5.4e: Exercises - Other Trigonometric Functions picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 34916 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Use Definitions to Obtain Trigonometric Function Values 1. Given a point on the terminal side of an angle. 2. Given one trigonometric ratio, find other trigonometric ratios 1. Given a quadrant the angle is in 2. Given a range of angle values 3. Given the sign of another trigonometric ratio 4. Given only one trigonometric ratio Trigonometric Function Values From Identities Reciprocal and Quotient Identities Trigonometric Functions of Special Angles Pythagorean Identities Given a Trigonometric Ratio and a Quadrant Trigonometric Function Values from Reference Angles Trigonometric Function Values from a Calculator Even and Odd Trigonometric Functions Cofunction Identities Key Equations Key Concepts Glossary Contributors and Attributions In the previous section, we defined the sine and cosine functions as ratios of sides of a right triangle formed by the point of intersection (x,y)(x,y) between a circle of radius r r and the terminal side of an angle in standard position. Since a triangle has 3 sides, there are 6 possible combinations of ratios. While the sine and cosine are the two prominent ratios that can be formed, there are four others, and together they define the 6 trigonometric functions. Definition: Sine, Cosine, Tangent, Secant, Cosecant, and Cotangent Functions For the point (x x, y y) on a circle of radius r r at an angle of θ θ, six trigonometric functions are defined as the ratios of the sides of the corresponding triangle. They are given below. The sinefunction: sin(θ)=y r sin⁡(θ)=y r The cosecant function: csc(θ)=r y csc⁡(θ)=r y The cosine function: cos(θ)=x r cos⁡(θ)=x r The secant function: sec(θ)=r x sec⁡(θ)=r x The tangent function: tan(θ)=y x tan⁡(θ)=y x The cotangent function: cot(θ)=x y cot⁡(θ)=x y The signs that x x and y y have depend on the quadrant the terminal side of the angle is in.The sign of r r is always positive and r=x 2+y 2−−−−−−√r=x 2+y 2.Geometrically, notice that the definition of tangent corresponds with the slope of the line segment between the origin (0, 0) and the point (x x, y y). This relationship can be very helpful in thinking about tangent values. Use Definitions to Obtain Trigonometric Function Values If the coordinates of a point on the terminal side of an angle are known, the values of the trigonometric function can be found using the definitions. Given a point on the terminal side of an angle. Example 5.4.1 a 5.4.1 a: Find trigonometric ratios given a point on the terminal side of an angle The point (-1, 7) is on the terminal side of some angle θ θ. Evaluate the six trigonometric functions for this angle. Solution We know a triangle is formed in the 2nd Quadrant, with sides x=−1 x=−1 and y=7 y=7. The Pythagorean Theorem can be used to find the third side: r=x 2+y 2−−−−−−√=(−1)2+7 2−−−−−−−−−√=50−−√=5 2–√r=x 2+y 2=(−1)2+7 2=50=5 2. Now the definitions of the six trigonometric functions can be used to obtain the desired function values. sin θ=y r=7 5 2–√=7 5 2–√⋅2–√2–√=7 2–√10 sin⁡θ=y r=7 5 2=7 5 2⋅2 2=7 2 10 cos θ=x r=−1 5 2–√=−1 5 2–√⋅2–√2–√=−2–√10 cos⁡θ=x r=−1 5 2=−1 5 2⋅2 2=−2 10 tan θ=y x=7−1=−7 tan⁡θ=y x=7−1=−7 csc θ=r y=5 2–√7 csc⁡θ=r y=5 2 7 sec θ=r x=5 2–√−1=−5 2–√sec⁡θ=r x=5 2−1=−5 2 cot θ=x y=−1 7 cot⁡θ=x y=−1 7 Example 5.4.1 b 5.4.1 b Figure 5.4.1 b 5.4.1 b The point (2–√2,−2–√2)(2 2,−2 2) is on the unit circle, as shown in Figure 5.4.1 b 5.4.1 b. Find sin t,cos t,tan t,sec t,csc t,sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t, and cot t cot⁡t. Solution Because this is a point on the unit circle, r=1 r=1, and furthermore, x=2–√2 x=2 2 and y=−2–√2 y=−2 2. Using these values for x x, y y, and r r and the definitions, the values of the six trigonometric ratios are obtained. (Because this is a point on the unit circle, we could have directly stated the values of sin t=y sin⁡t=y and cos t=x cos⁡t=x, but the definitions were used below.) sin t=y r=−2√2 1=−2–√2 sin⁡t=y r=−2 2 1=−2 2 cos t=x r=2√2 1=2–√2 cos⁡t=x r=2 2 1=2 2 tan t=y x=−2√2 2√2=−1 tan⁡t=y x=−2 2 2 2=−1 csc t=r y=1−2√2=−2 2–√⋅2–√2–√=−2–√csc⁡t=r y=1−2 2=−2 2⋅2 2=−2 sec t=r y=1 2√2=2–√sec⁡t=r y=1 2 2=2 cot t=x y=2√2−2√2=−1 cot⁡t=x y=2 2−2 2=−1 Try It 5.4.1 5.4.1 Figure 5.4.1 c 5.4.1 c Suppose the terminal side of θ θ, when plotted in standard position, contains the point Q(3,−4)Q(3,−4). Evaluate the six trigonometric functions for this angle. The point (−3–√2,1 2)(−3 2,1 2) is on the unit circle, as shown in Figure 5.4.1 c 5.4.1 c.Find sin t,cos t,tan t,sec t,csc t,sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t, and cot t cot⁡t. Answers 1. Since x=3 x=3 and y=−4 y=−4, then r=x 2+y 2−−−−−−√=(3)2+(−4)2−−−−−−−−−−√=25−−√=5 r=x 2+y 2=(3)2+(−4)2=25=5. sin(θ)=−4 5 sin⁡(θ)=−4 5, cos(θ)=3 5 cos⁡(θ)=3 5, tan(θ)=−4 3 tan⁡(θ)=−4 3, csc(θ)=−5 4 csc⁡(θ)=−5 4, sec(θ)=5 3 sec⁡(θ)=5 3, cot(θ)=−3 4 cot⁡(θ)=−3 4. We have x=−3–√2 x=−3 2 and y=1 2 y=1 2. Because this is a point on the unit circle, r=1 r=1. sin(θ)=1 2 sin⁡(θ)=1 2, cos(θ)=−3–√2 cos⁡(θ)=−3 2, tan(θ)=−3–√3 tan⁡(θ)=−3 3, csc(θ)=2 csc⁡(θ)=2, sec(θ)=−2 3–√3 sec⁡(θ)=−2 3 3, cot(θ)=−3–√cot⁡(θ)=−3. Given one trigonometric ratio, find other trigonometric ratios When given a trigonometric ratio, it is possible to narrow down the possible quadrants the angle is in to two, based on the definition of the trigonometric ratio and the signs that x x and y y take on in the different quadrants. sin θ=y r,csc θ=r y sin⁡θ=y r,csc⁡θ=r y QII:+QI:+QII:+QI:+cos θ=x r,sec θ=r x cos⁡θ=x r,sec⁡θ=r x QI:+QIV:+QI:+QIV:+tan θ=y x,cot θ=x y tan⁡θ=y x,cot⁡θ=x y QIII:+QI:+QI:+QIII:+ Figure 5.4.2 5.4.2. Quadrants in which trigonometric functions are positive If given a trigonometric ratio, the definitions can be used 'in reverse' to determine the values of two of the three legs (x x, y y, and r r ) of a right triangle inscribed in the circle. Remember r r is ALWAYS positive, so the sign of the ratio depends entirely on the signs of x x and y y. How to: Find trig ratios given one trig ratio. Reference Triangle or Pythagorean Theorem Approach. Given a trigonometric ratio and perhaps some other information. Determine the quadrant the angle is in, and the signs of x x and y y for any point on the terminal side of the angle. Use the definition of the given trigonometric function and the quadrant it is in to determine two of the three sides of a reference triangle and their appropriate signs. Use the Pythagorean Theorem to find the third side, and its appropriate sign. Now all three sides of the reference triangle are known. Use the definitions to determine the other trigonometric ratios. Given a quadrant the angle is in Example 5.4.2 5.4.2 Find Trigonometric Functions given one Trig Ratio and the quadrant the angle is in Given sec(θ)=2 6–√3 sec⁡(θ)=2 6 3 with θ θ in Quadrant IV. Find the value of sin θ sin⁡θ and cot θ cot⁡θ. Solution Step 1. Angle θ θ is in QIV where x>0 x>0 and y<0 y<0. Step 2. sec(θ)=2 6–√3=r x sec⁡(θ)=2 6 3=r x, therefore we can choose r=2 6–√r=2 6 and x=3 x=3 Step 3. Find y y with the Pythagorean Theorem: x 2+y 2=r 2⇒3 2+y 2=(2 6–√)2⇒y 2=4(6)−9=15.x 2+y 2=r 2⇒3 2+y 2=(2 6)2⇒y 2=4(6)−9=15. Since y y is known to be negative, y=−15−−√y=−15. Step 4. Using the definitions, sin θ=y r=−15−−√2 6–√=−15−−√2 6–√⋅6–√6–√=−90−−√2(6)=−3 10−−√12=−10−−√4 sin⁡θ=y r=−15 2 6=−15 2 6⋅6 6=−90 2(6)=−3 10 12=−10 4 and cot θ=x y=3−15−−√=3−15−−√⋅15−−√15−−√=−3 15−−√15=−15−−√5 cot⁡θ=x y=3−15=3−15⋅15 15=−3 15 15=−15 5 Try It 5.4.2 5.4.2 Suppose θ θ is a Quadrant IV angle with cot(θ)=−4 cot⁡(θ)=−4. Find the values of the five remaining circular functions of θ θ. Answers We need to find a point Q(x,y)Q(x,y) which lies on the terminal side of θ θ, when θ θ is plotted in standard position. Since θ θ is a Quadrant IV angle, we also know x>0 x>0 and y<0 y<0. We have that cot(θ)=−4=x y cot⁡(θ)=−4=x y.We may choose any values x x and y y so long as x>0 x>0, y<0 y<0 and x y=−4 x y=−4. (For example, we could choose x=8 x=8 and y=−2 y=−2 or we could choose x=4 x=4 and y=−1 y=−1. Using x=4 x=4 and y=−1 y=−1 we obtain r=x 2+y 2−−−−−−√=(4)2+(−1)2−−−−−−−−−−√=17−−√r=x 2+y 2=(4)2+(−1)2=17. Thus sin(θ)=−1 17−−√=−17−−√17 sin⁡(θ)=−1 17=−17 17, cos(θ)=4 17−−√=4 17−−√17 cos⁡(θ)=4 17=4 17 17, tan(θ)=−1 4 tan⁡(θ)=−1 4, sec(θ)=17−−√4 sec⁡(θ)=17 4, csc(θ)=−17−−√csc⁡(θ)=−17. Given a range of angle values If a range of values an angle can have is given, that information can be translated to the quadrant the angle is in, and therefore the signs that x x and y y can take on. This follows from the definition of an angle in standard position. QII:π 2<θ<π 90∘<θ<180∘QIII:π<θ<3 π 2 180∘<θ<270∘QI:0<θ<π 2 0∘<θ<90∘QIV:3 π 2<θ<2 π 270∘<θ<360∘QII:π 2<θ<π QI:0<θ<π 2 90∘<θ<180∘0∘<θ<90∘QIII:π<θ<3 π 2 QIV:3 π 2<θ<2 π 180∘<θ<270∘270∘<θ<360∘ Figure 5.4.3 5.4.3. Values of Angles in Standard Position in the Different Quadrants Example 5.4.3 5.4.3: Use Definitions to Evaluate Trigonometric Functions If sin(ϕ)=−3 7 sin⁡(ϕ)=−3 7 and 3 π 2<ϕ<2 π 3 π 2<ϕ<2 π, find cos(ϕ)cos⁡(ϕ) and tan(ϕ)tan⁡(ϕ). If tan(θ)=3 tan⁡(θ)=3 and π<θ<3 π 2,π<θ<3 π 2, find sin(θ)sin⁡(θ). Solution Angle ϕ ϕ is in QIV where x>0 x>0 and y<0 y<0. sin(ϕ)=−3 7=y r sin⁡(ϕ)=−3 7=y r, therefore we can choose r=7 r=7 and y=−3 y=−3 Find x x with the Theorem of Pythagoras: x 2+y 2=r 2⇒x 2+(−3)2=7 2⇒x 2=49−9=40 x 2+y 2=r 2⇒x 2+(−3)2=7 2⇒x 2=49−9=40. Thus x=±40−−√x=±40. Because x x is known to be positive, x=2 10−−√x=2 10. Using the definitions, cos(ϕ)=x r=2 10−−√7 cos⁡(ϕ)=x r=2 10 7 and tan(ϕ)=y x=−3 2 10−−√=3 10−−√20 tan⁡(ϕ)=y x=−3 2 10=3 10 20. Angle θ θ is in QIII where x<0 x<0 and y<0 y<0. tan(θ)=3=3 1=y x,tan⁡(θ)=3=3 1=y x, and because θ θ is in QIII, x=−1 x=−1 and y=−3 y=−3 Find r r with the formula r=x 2+y 2−−−−−−√⇒r=(−1)2+(−3)2−−−−−−−−−−−√⇒r=10−−√r=x 2+y 2⇒r=(−1)2+(−3)2⇒r=10. Using the definitions, sin(θ)=y r=−3 10−−√=−3 10−−√10.sin⁡(θ)=y r=−3 10=−3 10 10. Try It 5.4.3 5.4.3 If sec(t)=−17 8 sec⁡(t)=−17 8 and 0<t<π,0<t<π,find the values of the other five trigonometric functions. Answers t t is in QII where x<0 x<0 and y>0 y>0. x=−8 x=−8, y=15 y=15, and r=17 r=17. sin t=15 17,cos t=−8 17,tan t=−15 8,csc t=17 15,cot t=−8 15 sin⁡t=15 17,cos⁡t=−8 17,tan⁡t=−15 8,csc⁡t=17 15,cot⁡t=−8 15. Given the sign of another trigonometric ratio Example 5.4.4 5.4.4; Find other trig ratios given one trig ratio and the sign of another ratio If csc(θ)=−10 7 csc⁡(θ)=−10 7 and tan(θ)>0 tan⁡(θ)>0, determine the exact values of cos(θ)cos⁡(θ) and tan(θ)tan⁡(θ). If cot(t)=−6 cot⁡(t)=−6 and cos(t)<0 cos⁡(t)<0, determine the exact values of sin(t)sin⁡(t) and sec(t)sec⁡(t). 1. Solution Step 1. Angle θ θ is either in QIII or QIV because csc θ<0 csc⁡θ<0. Angle θ θ is in either QI or QIII because tan θ>0 tan⁡θ>0. Therefore θ θ must be in QIII where x<0 x<0 and y<0 y<0. Step 2. csc(θ)=−10 7=r y csc⁡(θ)=−10 7=r y, therefore we can choose r=10 r=10 and y=−7 y=−7 Step 3. Find x x with the Theorem of Pythagoras: x 2+y 2=r 2⇒x 2+(−7)2=10 2 x 2+y 2=r 2⇒x 2+(−7)2=10 2⇒x 2=100−49=51.⇒x 2=100−49=51.Thus x=±51−−√x=±51. Because x x is known to be negative, x=−51−−√x=−51. Step 4. Using the definitions, cos(θ)=x r=−51−−√10 cos⁡(θ)=x r=−51 10 and tan(θ)=y x=−7−51−−√=7 51−−√51 tan⁡(θ)=y x=−7−51=7 51 51 2. Solution Step 1. Angle t t is either in QII or QIV because cot t<0 cot⁡t<0. Angle t t is in either QII or QIII because cos t>0 cos⁡t>0. Therefore t t must be in QII where x<0 x<0 and y>0 y>0. Step 2. cot(t)=−6=−6 1=x y cot⁡(t)=−6=−6 1=x y, and because t t is in QII, x=−6 x=−6 and y=1 y=1 Step 3. Find r r with the formula r=x 2+y 2−−−−−−√⇒r=(−6)2+1 2−−−−−−−−−√⇒r=37−−√r=x 2+y 2⇒r=(−6)2+1 2⇒r=37. Step 4. Using the definitions, sin(t)=y r=1 37−−√=37−−√37 sin⁡(t)=y r=1 37=37 37 and sec(t)=r x=−37−−√6 sec⁡(t)=r x=−37 6 Try It 5.4.4 5.4.4 If sec(θ)=3 sec⁡(θ)=3 and sin(θ)<0 sin⁡(θ)<0, find the values of the other five trigonometric functions. Answers θ θ is in QIV where x>0 x>0 and y<0 y<0. x=1 x=1, y=−2 2–√y=−2 2, and r=3 r=3. sin(θ)=−2 2–√3 sin⁡(θ)=−2 2 3, cos(θ)=−1 3 cos⁡(θ)=−1 3, tan(θ)=−2 2–√tan⁡(θ)=−2 2, csc(θ)=−3 2–√4 csc⁡(θ)=−3 2 4, cot(θ)=−2–√4 cot⁡(θ)=−2 4. Given only one trigonometric ratio Notice in the problems above we were given both a trigonometric ratio and the quadrant the angle was in. When the quadrant is not given, there are two possible solutions, as the example below illustrates. Example 5.4.5 5.4.5 Given cos θ=−4 5 cos⁡θ=−4 5, draw possible angles for θ θ in standard position and determine the value of sin θ sin⁡θ and tan θ tan⁡θ. Solution Figure 5.4.5 5.4.5: cos θ=−4 5 cos⁡θ=−4 5 Given cos θ=x r=−4 5 cos⁡θ=x r=−4 5 we can conclude a point on the terminal side of θ θ has coordinate (−4,y)(−4,y) on a circle of radius r=5 r=5. Using the Pythagorean Theorem to determine y y we find x 2+y 2=r 2⇒(−4)2+y 2=5 2⇒y 2=9⇒y=±3 x 2+y 2=r 2⇒(−4)2+y 2=5 2⇒y 2=9⇒y=±3. Thus there are actually two possible angles for θ θ. If the point (−4,3)(−4,3) is on the terminal side of θ θ, then θ θ is in QII, and so we have x=−4 x=−4, y=3 y=3, and r=5 r=5. Thus, sin θ=y r=3 5 sin θ=y r=3 5 and tan θ=y x=3−4.tan θ=y x=3−4. If the point (−4,−3)(−4,−3) is on the terminal side of θ θ, then θ θ is in QIII, and so we have x=−4 x=−4, y=−3 y=−3, and r=5 r=5. Thus, sin θ=y r=−3 5 sin⁡θ=y r=−3 5 and tan θ=y x=−3−4=3 4.tan⁡θ=y x=−3−4=3 4. Trigonometric Function Values From Identities Whenever a trigonometric ratio is given, and other function values are desired, there are two different approaches that can be used. We have been illustrating the use of the definitions and the Pythagorean Theorem. Another approach which will now be discussed uses identities. Reciprocal and Quotient Identities You may have noticed that the ratios defining the secant, cosecant, and cotangent are the reciprocals of the ratios defining the cosine, sine, and tangent functions, respectively. Additionally, notice that tan(θ)=y x=r sin(θ)r cos(θ)=sin(θ)cos(θ)tan⁡(θ)=y x=r sin⁡(θ)r cos⁡(θ)=sin⁡(θ)cos⁡(θ) Applying this concept to the other trigonometric functions we can state the Reciprocal and Quotient Identities. Identities Reciprocal Identities csc(θ)=1 sin(θ)csc⁡(θ)=1 sin⁡(θ)sec(θ)=1 cos(θ)sec⁡(θ)=1 cos⁡(θ)cot(θ)=1 tan(θ)cot⁡(θ)=1 tan⁡(θ) sin(θ)=1 csc(θ)sin⁡(θ)=1 csc⁡(θ)cos(θ)=1 sec(θ)cos⁡(θ)=1 sec⁡(θ)tan(θ)=1 cot(θ)tan⁡(θ)=1 cot⁡(θ) Quotient Identities tan(θ)=sin(θ)cos(θ)tan⁡(θ)=sin⁡(θ)cos⁡(θ)cot(θ)=cos(θ)sin(θ)cot⁡(θ)=cos⁡(θ)sin⁡(θ) These relationships are called identities. Identities are statements that are true for all values of the input on which they are defined. Identities are usually something that can be derived from definitions and relationships we already know. Trigonometric Functions of Special Angles Example 5.4.6 a 5.4.6 a: Using Identities to Evaluate Trigonometric Functions Given sin(45°)=2–√2 sin⁡(45°)=2 2 and cos(45°)=2–√2 cos⁡(45°)=2 2, use identities to evaluate tan(45°).tan⁡(45°). Given sin(5 π 6)=1 2 sin⁡(5 π 6)=1 2 and cos(5 π 6)=−3–√2,cos⁡(5 π 6)=−3 2, use identities to evaluate sec(5 π 6)sec⁡(5 π 6). Solution Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions. a.Use the quotient identity for tangent.tan(45∘)=sin(45∘)cos(45∘)=2–√/2 2–√/2=1 a.Use the quotient identity for tangent.tan(45∘)=sin(45∘)cos(45∘)=2/2 2/2=1 Notice this result is consistent with our interpretation of the tangent value as the slope of the line passing through the origin at the given angle: a line at 45 degrees would indeed have a slope of 1. b.Use the reciprocal identity for secant.sec(5 π 6)=1 cos(5 π 6)=1−3√2=−2 3–√=−2 3–√3 b.Use the reciprocal identity for secant.sec(5 π 6)=1 cos⁡(5 π 6)=1−3 2=−2 3=−2 3 3 We know the values of sine and cosine for "special" angles 0 0, π 6 π 6, π 4 π 4, π 3 π 3, and π 2 π 2. Therefore, the values of the other four trigonometric functions for these "special" angles can be found by using the reciprocal and quotient identities. Example 5.4.6 b 5.4.6 b: Finding the Trigonometric Functions of an Angle Find sin t,cos t,tan t,sec t,csc t,sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t, and cot t cot⁡t when t=π 6 t=π 6. Solution We have previously used the properties of equilateral triangles to demonstrate that sin(π 6)=1 2 sin⁡(π 6)=1 2 and cos(π 6)=3–√2 cos⁡(π 6)=3 2. We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values. tan(π 6)sec(π 6)=sin(π 6)cos(π 6)=1 2 3√2=1 3–√=3–√3=1 cos(π 6)=1 3√2=2 3–√=2 3–√3 cot(π 6)csc(π 6)=cos(π 6)sin(π 6)=3√2 1 2=3–√=1 sin(π 6)=1 1 2=2 tan⁡(π 6)=sin⁡(π 6)cos⁡(π 6)=1 2 3 2=1 3=3 3 cot⁡(π 6)=cos⁡(π 6)sin⁡(π 6)=3 2 1 2=3 sec⁡(π 6)=1 cos⁡(π 6)=1 3 2=2 3=2 3 3 csc⁡(π 6)=1 sin⁡(π 6)=1 1 2=2 A common mistake is that although tan(θ)=3–√3=sin(θ)cos(θ)=3 tan⁡(θ)=3 3=sin⁡(θ)cos⁡(θ)=3, it does NOT mean we can take sin(θ)=3–√sin⁡(θ)=3 and cos(θ)=3 cos⁡(θ)=3. Instead, from sin(θ)cos(θ)=3–√3 sin⁡(θ)cos⁡(θ)=3 3 we get: 3 sin(θ)=3–√cos(θ)3 sin⁡(θ)=3 cos⁡(θ). Try It 5.4.6 5.4.6 Find sec(60∘)sec⁡(60∘) Find sin t,cos t,tan t,sec t,csc t,sin⁡t,cos⁡t,tan⁡t,sec⁡t,csc⁡t, and cot t cot⁡t when t=π 3 t=π 3. Answer 1.sec(60∘)=1 cos(60∘)1.sec⁡(60∘)=1 cos⁡(60∘). Hence, sec(60∘)=1(1/2)=2 sec⁡(60∘)=1(1/2)=2. 2.sin(π 3)=3–√2,cos(π 3)=1 2,tan(π 3)=3–√,sec(π 3)=2,csc(π 3)=2 3–√3,cot(π 3)=3–√3 2.sin⁡(π 3)=3 2,cos⁡(π 3)=1 2,tan⁡(π 3)=3,sec⁡(π 3)=2,csc⁡(π 3)=2 3 3,cot⁡(π 3)=3 3 Because we know the sine and cosine values for the common first-quadrant angles, we can find the other four function values for those angles by rewriting the definitions of tangent, secant, cosecant, and cotangent in terms of sine and cosine and evaluating the resulting expressions. The results are shown in Table 5.4.1 5.4.1. It is worth taking the time to memorize the function values for at least the sine, cosine and tangent of the common 'special angles' listed in the table below. Table 5.4.6 5.4.6 Values of Trigonometric functions of "Special Angles"\| Angle \| 0 0 \| π 6 or 30°π 6 or 30° \| π 4 or 45°π 4 or 45° \| π 3 or 60°π 3 or 60° \| π 2 or 90°π 2 or 90° \| --- --- --- \| \| Sine \| 0 \| d 1 2 d 1 2 \| 2–√2 2 2 \| 3–√2 3 2 \| 1 \| \| Cosine \| 1 \| 3–√2 3 2 \| 2–√2 2 2 \| 1 2 1 2 \| 0 \| \| Tangent \| 0 \| 3–√3 3 3 \| 1 \| 3–√3 \| Undefined \| \| Cosecant \| Undefined \| 2 \| 2–√2 \| 2 3–√3 2 3 3 \| 1 \| \| Secant \| 1 \| 2 3–√3 2 3 3 \| 2–√2 \| 2 \| Undefined \| \| Cotangent \| Undefined \| 3–√3 \| 1 \| 3–√3 3 3 \| 0 \| Pythagorean Identities The Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and the definitions of sine and cosine. We can build new identities from previously established identities. For example, if we divide both sides of the Pythagorean Identity by cosine squared, we can create a new identity: cos 2(θ)+sin 2(θ)cos 2(θ)=1 cos 2(θ)cos 2⁡(θ)+sin 2⁡(θ)cos 2⁡(θ)=1 cos 2⁡(θ) Splitting the fraction on the left, cos 2(θ)cos 2(θ)+sin 2(θ)cos 2(θ)=1 cos 2(θ)cos 2⁡(θ)cos 2⁡(θ)+sin 2⁡(θ)cos 2⁡(θ)=1 cos 2⁡(θ) Simplifying and using the definitions of tangent and secant 1+tan 2(θ)=sec 2(θ).1+tan 2⁡(θ)=sec 2⁡(θ). Try It 5.4.7 5.4.7 Use a similar approach to establish that cot 2(θ)+1=csc 2(θ)cot 2⁡(θ)+1=csc 2⁡(θ). Answer cos 2(θ)+sin 2(θ)sin 2 θ=1→cos 2(θ)sin 2(θ)+sin 2(θ)sin 2(θ)=1 sin 2(θ)→cot 2(θ)+1=csc 2(θ)cos 2⁡(θ)+sin 2⁡(θ)sin 2⁡θ=1→cos 2⁡(θ)sin 2⁡(θ)+sin 2⁡(θ)sin 2⁡(θ)=1 sin 2⁡(θ)→cot 2⁡(θ)+1=csc 2⁡(θ) Identities: Alternate forms of the Pythagorean Identity sin 2(θ)+cos 2(θ)=1 sin 2⁡(θ)+cos 2⁡(θ)=1 1+tan 2(θ)=sec 2(θ)1+tan 2⁡(θ)=sec 2⁡(θ)cot 2(θ)+1=csc 2(θ)cot 2⁡(θ)+1=csc 2⁡(θ) How to: Given one trigonometric ratio, find other ratios. Identities Approach. Given a trigonometric ratio and other information. Determine the quadrant the angle must be in, and thus the signs of the various trig ratios in that quadrant. Use identities to find the other trig ratios and their appropriate signs. Typically, use an identity to find the value of sine or cosine or their reciprocal functions, and then use other identities to find the rest of the trigonometric ratios. Given a Trigonometric Ratio and a Quadrant Example 5.4.8 a 5.4.8 a If tan(θ)=2 7 tan⁡(θ)=2 7 and θ θ is in the 3 r d r d quadrant, find cos(θ)cos⁡(θ). Solution There are two approaches to this problem, both of which work equally well. Approach 1 (Use the Pythagorean Theorem) Angle θ θ is in QIII where x<0 x<0 and y<0 y<0. tan(θ)=2 7=−2−7=y x,tan⁡(θ)=2 7=−2−7=y x,and because θ θ is in QIII, x=−2 x=−2 and y=−7 y=−7 Find r r with the formula r=x 2+y 2−−−−−−√⇒r=(−2)2+(−7)2−−−−−−−−−−−√⇒r=53−−√r=x 2+y 2⇒r=(−2)2+(−7)2⇒r=53. Using the definitions, cos(θ)=x r=−7 53−−√=−7 53−−√53 cos⁡(θ)=x r=−7 53=−7 53 53. Approach 2 (Use Identities) Use the 1+tan 2(θ)=sec 2(θ)1+tan 2⁡(θ)=sec 2⁡(θ) form of the Pythagorean Identity because we are given a tangent value, 1+tan 2(θ)1+(2 7)2 53 49 sec(θ)=sec 2(θ)=sec 2(θ)=sec 2(θ)=±53 49−−−√=±53−−√7 Use a Pythagorean Identity Substitute the known tangent value 1+tan 2⁡(θ)=sec 2⁡(θ)Use a Pythagorean Identity 1+(2 7)2=sec 2⁡(θ)Substitute the known tangent value 53 49=sec 2⁡(θ)sec⁡(θ)=±53 49=±53 7 Since the angle is in the third quadrant, the cosine value will be negative so the secant value will also be negative. Keeping the negative result, and using definition of secant, sec(θ)cos(θ)cos(θ)=−53−−√7=−1 sec(θ)=−1 53−−√7=−7 53−−√=−7 53−−√53 Use the Reciprocal Identity for Secant sec⁡(θ)=−53 7 cos⁡(θ)=−1 sec⁡(θ)Use the Reciprocal Identity for Secant cos⁡(θ)=−1 53 7=−7 53=−7 53 53 Example 5.4.8 b 5.4.8 b: Using Identities to Relate Trigonometric Functions Figure 5.4.8 b 5.4.8 b If cos(t)=12 13 cos⁡(t)=12 13 and t t is in quadrant IV, as shown in Figure 5.4.8 b 5.4.8 b, find the values of the other five trigonometric functions. Solution We are given the value of cosine so we can find the sine using the Pythagorean Identity, cos 2 t+sin 2 t=1 cos 2⁡t+sin 2⁡t=1. Since the angle t t is in the fourth quadrant,the sine ratio will be negative. (12 13)2+sin 2 t sin 2 t sin 2 t sin t sin t=1=1−(12 13)2=1−144 169=25 169=±25 169−−−−√=±25−−√169−−−√=±5 13=−5 13(12 13)2+sin 2⁡t=1 sin 2⁡t=1−(12 13)2=1−144 169 sin 2⁡t=25 169 sin⁡t=±25 169=±25 169 sin⁡t=±5 13=−5 13 The remaining functions can be calculated using identities relating them to sine and cosine. tan t=sin t cos t=−5 13 12 13=−5 12 sec t=1 cos t=1 12 13=13 12 csc t=1 sin t=1−5 13=−13 5 cot t=1 tan t=1−5 12=−12 5 tan⁡t=sin⁡t cos⁡t=−5 13 12 13=−5 12 sec⁡t=1 cos⁡t=1 12 13=13 12 csc⁡t=1 sin⁡t=1−5 13=−13 5 cot⁡t=1 tan⁡t=1−5 12=−12 5 Try It 5.4.8 5.4.8 Evaluate cos(θ)cos⁡(θ), where csc(θ)=−5–√csc⁡(θ)=−5 and θ θ is a Quadrant IV angle. Answer Given csc(θ)=−5–√csc⁡(θ)=−5, use the Reciprocal Identity sin(θ)=−1 csc(θ)sin⁡(θ)=−1 csc⁡(θ)to obtain sin(θ)=−1 5√=−5√5 sin⁡(θ)=−1 5=−5 5.Use the Pythagorean Identity, cos 2(θ)+sin 2(θ)=1,cos 2⁡(θ)+sin 2⁡(θ)=1,to find cos(θ)cos⁡(θ). Substituting, we get cos 2(θ)+(−5√5)2=1 cos 2⁡(θ)+(−5 5)2=1, which gives cos 2(θ)+5 25.cos 2⁡(θ)+5 25. Therefore,cos 2(θ)=4 5 cos 2⁡(θ)=4 5, or cos(θ)=±2 5√5 cos⁡(θ)=±2 5 5.Since θ θ is a Quadrant IV angle, cos(θ)>0 cos⁡(θ)>0, so cos(θ)=2 5√5 cos⁡(θ)=2 5 5 Trigonometric Function Values from Reference Angles Figure 5.4.9 5.4.9 Quadrants where trigonometric functions are positive We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x- and y-values in the quadrant where the terminal side is located. Figure 5.4.9 5.4.9 shows which functions are positive in which quadrant. To help us remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “All Students Take Classes.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “A,” all of the six trigonometric functions are positive. In quadrant II, “Students,” only sine and its reciprocal function, cosecant, are positive. In quadrant III, “Take,” only tangent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “Classes,” only cosine and its reciprocal function, secant, are positive. Example 5.4.9 5.4.9: Coterminal Angles and Reference Angles Let θ=928∘θ=928∘. Which angle between 0∘0∘ and 360∘360∘ has the the same trigonometric function values as θ θ? What is the reference angle for θ θ? Draw these angles in standard position. Solution (a) Since 928∘=2×360∘+208∘,928∘=2×360∘+208∘,θ θ has the same terminal side as 208∘208∘, and thus the angles 928∘928∘ and 208∘208∘ are coterminal. This is illustrated in Figure 5.4.9 a 5.4.9 a. Therefore, 208∘208∘ has the same trigonometric function values as θ=928∘θ=928∘. (b)928∘928∘ and 208∘208∘ have the same terminal side in QIII, so the reference angle for θ=928∘θ=928∘ is 208∘−180∘=28∘208∘−180∘=28∘. Figure 5.4.9 a 5.4.9 a Generalized Reference Angle Theorem The absolute value of trigonometric functions of an angle θ θ, are the same as the corresponding trigonometric functions of its reference angle α α. More specifically, if α α is the reference angle for an θ θ, then: sin(θ)=±sin(α)sin⁡(θ)=±sin⁡(α) csc(θ)=±csc(α)csc⁡(θ)=±csc⁡(α)cos(θ)=±cos(α)cos⁡(θ)=±cos⁡(α) sec(θ)=±sec(α)sec⁡(θ)=±sec⁡(α)tan(θ)=±tan(α)tan⁡(θ)=±tan⁡(α) cot(θ)=±cot(α)cot⁡(θ)=±cot⁡(α) The choice of the (±±) depends on the quadrant in which the terminal side of θ θ lies. How to: Find Trigonometric Functions using Reference Angles. Measure the angle formed by the terminal side of the given angle and the horizontal axis. This is the reference angle. The measure of a reference angle is ALWAYS a positive number between 0 0 and π 2 π 2 Observe the quadrant where the terminal side of the original angle is located. Based on the quadrant, determine whether the output is positive or negative. Rewrite the function in terms of the reference angle and the appropriate sign. Evaluate the function at the reference angle. Example 5.4.10 5.4.10: Using Reference Angles to Find Trigonometric Functions Use reference angles to find all six trigonometric functions of −5 π 6−5 π 6. Solution The angle between this angle’s terminal side and the x-axis is π 6 π 6, so that is the reference angle. Since −5 π 6−5 π 6 is in the third quadrant, where both x x and y y are negative, then sine, cosine, secant, and cosecant will be negative, while tangent and cotangent will be positive. sin(−5 π 6)=−sin(π 6)=−1 2 sin⁡(−5 π 6)=−sin⁡(π 6)=−1 2 cos(−5 π 6)=−cos(π 6)=−3–√2 cos⁡(−5 π 6)=−cos⁡(π 6)=−3 2 tan(−5 π 6)=+tan(π 6)=3–√3 tan⁡(−5 π 6)=+tan⁡(π 6)=3 3 csc(−5 π 6)=−csc(π 6)=−2 csc⁡(−5 π 6)=−csc⁡(π 6)=−2 sec(−5 π 6)=−sec(π 6)=−2 3–√3 sec⁡(−5 π 6)=−sec⁡(π 6)=−2 3 3 cot(−5 π 6)=+cot(π 6)=3–√cot⁡(−5 π 6)=+cot⁡(π 6)=3 Try It 5.4.10 5.4.10 Use reference angles to find all six trigonometric functions of −7 π 4−7 π 4. Answer Angle −7 π 4−7 π 4 is in QI where all trigonometric functions are positive. sin(−7 π 4)=2√2,cos(−7 π 4)=2√2,tan(−7 π 4)=1,csc(−7 π 4)=2–√,sec(−7 π 4)=2–√,cot(−7 π 4)=1 sin⁡(−7 π 4)=2 2,cos⁡(−7 π 4)=2 2,tan⁡(−7 π 4)=1,csc⁡(−7 π 4)=2,sec⁡(−7 π 4)=2,cot⁡(−7 π 4)=1 Example 5.4.11 5.4.11 Use Identities to determine the exact value of each of the following function values. sec(7 π 4)sec⁡(7 π 4)2. csc(7 π 6)csc⁡(7 π 6) 3.tan(θ)tan⁡(θ), where θ θ is any angle coterminal with 3 π 2 3 π 2. Solution 1.2.sec(7 π 4)csc(7 π 6)=1 cos(7 π 4)=1+cos(π 4)=1 2√2=2 2–√=2–√=1 sin(7 π 6)=1−sin(π 6)=1−1/2=−2 in QIV so x and cos positive in QIII so y and sine negative 1.sec⁡(7 π 4)=1 cos⁡(7 π 4)=1+cos⁡(π 4)=1 2 2=2 2=2 in QIV so x and cos positive 2.csc⁡(7 π 6)=1 sin⁡(7 π 6)=1−sin⁡(π 6)=1−1/2=−2 in QIII so y and sine negative 3.tan(θ)Or=tan(3 π 2)=sin(3 π 2)cos(3 π 2)=−1 0=Undefined=tan(3 π 2)=±tan(π 2)=Undefined±undefined is undefined 3.tan⁡(θ)=tan⁡(3 π 2)=sin⁡(3 π 2)cos⁡(3 π 2)=−1 0=Undefined Or=tan⁡(3 π 2)=±tan⁡(π 2)=Undefined±undefined is undefined Try It 5.4.11 5.4.11 Use Identities to determine the exact value of each of the following function values. csc(−π 4)csc⁡(−π 4) 2. cot(4 π 3)cot⁡(4 π 3) Answer 1.2.csc(−π 4)cot(4 π 3)Or=1 sin(−π 4)=1−sin(π 4)=1−2√2=−2 2–√=−2–√=cos(4 π 3)sin(4 π 3)=−cos(π 3)−sin(π 3)=1/2 3–√/2=1 3–√=3–√3=1 tan(4 π 3)=1+tan(π 3)=1 3–√=3–√3 in QIV so y&sine neg.in QIII so sine &cos neg.1.csc⁡(−π 4)=1 sin⁡(−π 4)=1−sin⁡(π 4)=1−2 2=−2 2=−2 in QIV so y&sine neg.2.cot⁡(4 π 3)=cos⁡(4 π 3)sin⁡(4 π 3)=−cos⁡(π 3)−sin⁡(π 3)=1/2 3/2=1 3=3 3 in QIII so sine &cos neg.Or=1 tan⁡(4 π 3)=1+tan⁡(π 3)=1 3=3 3 Trigonometric Function Values from a Calculator Values for the six trigonometric functions of "special" first-quadrant angles, and multiples of these angles that lie in other quadrants, can be evaluated using the value of the trig function for its reference angle and modifying the sign when appropriate. For other angles, a calculator can be used to evaluate trigonometric functions. If the calculator has degree mode and radian mode, confirm the correct mode is chosen before making a calculation. There are calculator keys for SIN, COS, and TAN, but typically not for CSC, SEC, or COT. To evaluate these last three functions, the function must be evaluated as the reciprocal of a sine, cosine, or tangent function. Example 5.4.12 5.4.12: Evaluating the Cosecant Using Technology Use a calculator to find the cosecant of 5 π 7 5 π 7.Use a calculator to find the secant of 30°30°. 1. Solution The angle is in radians so be sure the calculator is in radian mode. There is not a cosecant button on the calculator, so the reciprocal identity csc(θ)=1 sin(θ)csc⁡(θ)=1 sin⁡(θ) must be used. Enter the following keystrokes: 1÷(S I N(5×π÷7))E N T E R 1÷(S I N(5×π÷7))E N T E R. The result obtained is 1.2790480...1.2790480... 2. Solution The angle is in degrees so be sure the calculator is in degree mode. Enter the following keystrokes: 1÷C O S(30)E N T E R 1÷C O S(30)E N T E R. The result is 1.15470053...1.15470053... If the calculator does not have a degree mode, the angle must also be converted to radians. When the calculator only has a radian mode, enter the following keystrokes: 1÷C O S(30×π÷180)E N T E R.1÷C O S(30×π÷180)E N T E R. Try It 5.4.12 5.4.12 Use a calculator to evaluate the following. Round to four decimal digits. sec(7 π 8)sec⁡(7 π 8) 2. csc(−π 8)csc⁡(−π 8) 3.cot(3)cot⁡(3) Answer sec(7 π 8)=1 cos(7 π 8)≈−1.0824 csc(−π 8)=1 sin(−π 8)≈−2.6131 cot(3)=1 tan(3)≈−7.015 sec⁡(7 π 8)=1 cos⁡(7 π 8)≈−1.0824 csc⁡(−π 8)=1 sin⁡(−π 8)≈−2.6131 cot⁡(3)=1 tan⁡(3)≈−7.015 Even and Odd Trigonometric Functions To be able to use our six trigonometric functions freely with both positive and negative angle inputs, we should examine how each function treats a negative input. As it turns out, there is an important difference among the functions in this regard. Recall in Section 2.2:Attributes of Functions, even and odd functions were defined. This definition is repeated here. Definition: Odd and Even functions An even function is one in which f(−x)=f(x)f(−x)=f(x). An even function has y y-axis symmetry. An odd function is one in which f(−x)=−f(x)f(−x)=−f(x). An odd function has origin axis symmetry. Figure 5.4.13 5.4.13 We can test whether a trigonometric function is even or odd by drawing a unit circle with a positive and a negative angle, as illustrated in Figure 5.4.13 5.4.13. The sine of the positive angle is sin(t)=y r sin⁡(t)=y r. The sine of the negative angle is sin(−t)=−y r=−y r sin⁡(−t)=−y r=−y r. Thus, sin(−t)=−sin(t)sin⁡(−t)=−sin⁡(t) and thesine function is an odd function. The cosine of the positive angle is cos(t)=x r cos⁡(t)=x r. The cosine of the negative angle is cos(−t)=x r cos⁡(−t)=x r. Thus, cos(−t)=cos(t)cos⁡(−t)=cos⁡(t) and thecosine function is an even function. The tangent of the positive angle is tan(t)=y x tan⁡(t)=y x. The tangent of the negative angle is tan(−t)=−y x=−y x tan⁡(−t)=−y x=−y x. Thus, tan(−t)=−tan(t)tan⁡(−t)=−tan⁡(t) and thetangent function is an odd function. Similar arguments can be made for the secant, cosecant and cotangent functions. Even and Odd Trigonometric Functions; Even/Odd Identities Cosine and secant are even functions: cos(−t)sec(−t)=cos t=sec t cos⁡(−t)=cos⁡t sec⁡(−t)=sec⁡t Sine, tangent, cosecant, and cotangent are odd functions: sin(−t)csc(−t)=−sin t=−csc t tan(−t)cot(−t)=−tan t=−cot t sin⁡(−t)=−sin⁡t tan⁡(−t)=−tan⁡t csc⁡(−t)=−csc⁡t cot⁡(−t)=−cot⁡t Example 5.4.13 5.4.13: Using Even and Odd Properties of Trigonometric Functions If sec t=2 sec⁡t=2, what is sec(−t)sec⁡(−t)? Solution Secant is an even function. The secant of an angle is the same as the secant of its opposite. So if the secant of angle t is 2, the secant of −t−t is also 2. Try It 5.4.13 5.4.13 If the cotangent of angle t t is 3–√3, what is the cotangent of −t?−t? Answer−3–√−3 Cofunction Identities Figure 5.4.14 a 5.4.14 a: The sine of π 3 π 3 equals the cosine of π 6 π 6 and vice versa. If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of π 6 π 6 and π 3 π 3, we see that the sine of π 3 π 3, namely 3√2 3 2, is also the cosine of π 6 π 6, while the sine of π 6 π 6, namely 1 2,1 2, is also the cosine of π 3 π 3 (Figure 5.4.14 a 5.4.14 a). sin π 3 sin π 6=cos π 6=3–√s 2 s=3–√2=cos π 3=s 2 s=1 2 sin⁡π 3=cos⁡π 6=3 s 2 s=3 2 sin⁡π 6=cos⁡π 3=s 2 s=1 2 This result should not be surprising because, as we see from Figure 5.4.14 a 5.4.14 a, the side opposite the angle of π 3 π 3 is also the side adjacent to π 6 π 6, so sin(π 3)sin⁡(π 3) and cos(π 6)cos⁡(π 6) are exactly the same ratio of the same two sides, 3–√s 3 s and 2 s.2 s. Similarly, cos(π 3)cos⁡(π 3) and sin(π 6)sin⁡(π 6) are also the same ratio using the same two sides, s s and 2 s 2 s. Figure 5.4.14 b 5.4.14 b: Cofunction identity The interrelationship between the sines and cosines of π 6 π 6 and π 3 π 3 also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, and the right angle is π 2 π 2, the remaining two angles must also add up to π 2 π 2. That means that a right triangle can be formed with any two angles that add to π 2 π 2—in other words, any two complementary angles. So we may state a cofunction identity: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 5.4.14 b 5.4.14 b. Using this identity, we can state without calculating, for instance, that the sine of π 12 π 12 equals the cosine of 5 π 12 5 π 12, and that the sine of 5 π 12 5 π 12 equals the cosine of π 12 π 12. We can also state that if, for a certain angle t,cos t=5 13,t,cos⁡t=5 13, then sin(π 2−t)=5 13 sin⁡(π 2−t)=5 13 as well. Cofunction Identities The cofunction identities in terms of radians are listed below. sin(t)=cos(π 2−t)sin⁡(t)=cos⁡(π 2−t)cos(t)=sin(π 2−t)cos⁡(t)=sin⁡(π 2−t)tan(t)=cot(π 2−t)tan⁡(t)=cot⁡(π 2−t) csc(t)=sec(π 2−t)csc⁡(t)=sec⁡(π 2−t)sec(t)=csc(π 2−t)sec⁡(t)=csc⁡(π 2−t)sec(t)=csc(π 2−t)sec⁡(t)=csc⁡(π 2−t) Example 5.4.14 5.4.14: Using Cofunction Identities If sin(t)=5 12,sin⁡(t)=5 12, find cos(π 2−t)cos⁡(π 2−t). Solution According to the cofunction identities for sine and cosine,sin(t)=cos(π 2−t).sin⁡(t)=cos⁡(π 2−t).So cos(π 2−t)=5 12.cos⁡(π 2−t)=5 12. Try It 5.4.14 5.4.14 If csc(π 6)=2,csc⁡(π 6)=2, find sec(π 3).sec⁡(π 3). Answer2 Key Equations Tangent function tan t=sin t cos t tan⁡t=sin⁡t cos⁡t Secant function sec t=1 cos t sec⁡t=1 cos⁡t Cosecant function csc t=1 sin t csc⁡t=1 sin⁡t Cotangent function cot t=1 tan t=cos t sin t cot⁡t=1 tan⁡t=cos⁡t sin⁡t Reciprocal Identities csc(θ)=1 sin(θ)csc⁡(θ)=1 sin⁡(θ) sin(θ)=1 csc(θ)sin⁡(θ)=1 csc⁡(θ)sec(θ)=1 cos(θ)sec⁡(θ)=1 cos⁡(θ) cos(θ)=1 sec(θ)cos⁡(θ)=1 sec⁡(θ)cot(θ)=1 tan(θ)cot⁡(θ)=1 tan⁡(θ) tan(θ)=1 cot(θ)tan⁡(θ)=1 cot⁡(θ) Quotient Identities tan(θ)=sin(θ)cos(θ)tan⁡(θ)=sin⁡(θ)cos⁡(θ)cot(θ)=cos(θ)sin(θ)cot⁡(θ)=cos⁡(θ)sin⁡(θ) Pythagorean Identities sin 2(θ)+cos 2(θ)=1 sin 2⁡(θ)+cos 2⁡(θ)=1 1+tan 2(θ)=sec 2(θ)1+tan 2⁡(θ)=sec 2⁡(θ)cot 2(θ)+1=csc 2(θ)cot 2⁡(θ)+1=csc 2⁡(θ) Even/Odd Identities sin(−t)=−sin t csc(−t)=−csc t sin⁡(−t)=−sin⁡t csc⁡(−t)=−csc⁡t cos(−t)=cos t sec(−t)=sec t cos⁡(−t)=cos⁡t sec⁡(−t)=sec⁡t tan(−t)=−tan t cot(−t)=−cot t tan⁡(−t)=−tan⁡t cot⁡(−t)=−cot⁡t Cofunction Identities sin t=cos(π 2−t)sin⁡t=cos⁡(π 2−t) cos t=sin(π 2−t)cos⁡t=sin⁡(π 2−t)sec t=csc(π 2−t)sec⁡t=csc⁡(π 2−t) csc t=sec(π 2−t)csc⁡t=sec⁡(π 2−t)tan t=cot(π 2−t)tan⁡t=cot⁡(π 2−t) cot t=tan(π 2−t)cot⁡t=tan⁡(π 2−t) Key Concepts The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. The secant, cotangent, and cosecant are all reciprocals of other functions. The secant is the reciprocal of the cosine function, the cotangent is the reciprocal of the tangent function, and the cosecant is the reciprocal of the sine function. The six trigonometric functions can be found from a point on the unit circle. Trigonometric functions can also be found from an angle. Trigonometric functions of angles outside the first quadrant can be determined using reference angles. The Pythagorean Identity makes it possible to find a cosine from a sine or a sine from a cosine. Identities can be used to evaluate trigonometric functions. Fundamental identities such as the Pythagorean Identity can be manipulated algebraically to produce new identities. The values of trigonometric functions of special angles can be found by mathematical analysis. To evaluate trigonometric functions of other angles, we can use a calculator or computer software. A function is said to be even if f(−x)=f(x)f(−x)=f(x) and odd if f(−x)=−f(x)f(−x)=−f(x). Cosine and secant are even; sine, tangent, cosecant, and cotangent are odd. Even and odd properties can be used to evaluate trigonometric functions. Glossary cosecant the reciprocal of the sine function: on the unit circle, csc t=1 y,y≠0 csc⁡t=1 y,y≠0, or more generally csc t=r y,y≠0 csc⁡t=r y,y≠0 cotangent the reciprocal of the tangent function: on the unit circle, cot t=x y,y≠0 cot⁡t=x y,y≠0 identities statements that are true for all values of the input on which they are defined secant the reciprocal of the cosine function: on the unit circle, sec t=1 x,x≠0 sec⁡t=1 x,x≠0, or more generally sec t=r x,x≠0 sec⁡t=r x,x≠0 tangent the quotient of the sine and cosine: on the unit circle, tan t=y x,x≠0 tan⁡t=y x,x≠0 Contributors and Attributions Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under aCreative Commons Attribution License 4.0license.Download for free at 5.4: The Other Trigonometric Functions is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts. 5.4: The Other Trigonometric Functions by David Lippman & Melonie Rasmussen is licensed CC BY-SA 4.0. Original source: Current page is licensed CC BY 4.0. 5.3: Points on Circles Using Sine and Cosine by David Lippman & Melonie Rasmussen is licensed CC BY-SA 4.0. Original source: 10.3: The Six Circular Functions and Fundamental Identities is licensed CC BY-NC-SA 3.0. Original source: 5.3: The Other Trigonometric Functions by OpenStax is licensed CC BY 4.0. Original source: 1.6: Other Trigonometric Functions by Ted Sundstrom & Steven Schlicker is licensed CC BY-NC-SA 3.0. Original source: 1.4: Trigonometric Functions of Any Angle by Michael Corral is licensed GNU FDL. 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We have already defined the sine an... 13.4: The Other Trigonometric FunctionsTrigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine an... 1.4: The Other Trigonometric FunctionsTrigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine an... Article typeSection or PageLicenseCC BYShow Page TOCyes Tags cosecant cotangent identities period secant tangent © Copyright 2025 Mathematics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 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12184	https://www.reddit.com/r/C_Programming/comments/12lby6e/what_are_the_differences_between_right_shift_left/	What are the differences between right shift, left shift and unsigned right shift operators? (>>, <<, >>>) : r/C_Programming Skip to main contentWhat are the differences between right shift, left shift and unsigned right shift operators? (>>, <<, >>>) : r/C_Programming Open menu Open navigationGo to Reddit Home r/C_Programming A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to C_Programming r/C_Programming•2 yr. ago Content-Value-6912 What are the differences between right shift, left shift and unsigned right shift operators? (>>, <<, >>>) Hey guys, I'm trying to understand bit shift operators in detail. Please provide some insights regarding where these operators are/can be implemented in C. Read more Share Related Answers Section Related Answers Differences between bit shift operators in C Left shift vs right shift operators in C Usage of << operator in C programming Understanding left shift in C++ Definition of left shift operator in C New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community Top Posts Reddit reReddit: Top posts of April 13, 2023 Reddit reReddit: Top posts of April 2023 Reddit reReddit: Top posts of 2023 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
12185	https://mathworld.wolfram.com/SurfaceofRevolution.html	Surface of Revolution -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Geometry Surfaces Surfaces of Revolution Number Theory Constants Transcendental Root Constants Surface of Revolution Download Wolfram Notebook A surface of revolution is a surface generated by rotating a two-dimensional curve about an axis. The resulting surface therefore always has azimuthal symmetry. Examples of surfaces of revolution include the apple surface, cone (excluding the base), conical frustum (excluding the ends), cylinder (excluding the ends), Darwin-de Sitter spheroid, Gabriel's horn, hyperboloid, lemon surface, oblate spheroid, paraboloid, prolate spheroid, pseudosphere, sphere, spheroid, and torus (and its generalization, the toroid). The area element of the surface of revolution obtained by rotating the curve from to about the x-axis is (1) (2) so the surface area is (3) (4) (Apostol 1969, p.286; Kaplan 1992, p.251; Anton 1999, p.380). If the curve is instead specified parametrically by , the surface area obtained by rotating the curve about the x-axis for if in this interval is given by (5) Similarly, the area of the surface of revolution obtained by rotating the curve from to about the y-axis is given by (6) (7) (Anton 1999, p.380). If the curve is instead specified parametrically by , the surface area obtained by rotating the curve about the y-axis for if in this interval is given by (8) The following table gives the lateral surface areas for some common surfaces of revolution where denotes the radius (of a cone, cylinder, sphere, or zone), and the inner and outer radii of a frustum, the height, the ellipticity of a spheroid, and and the equatorial and polar radii (for a spheroid) or the radius of a circular cross-section and rotational radius (for a torus). surface cone conical frustum cylinder oblate spheroid prolate spheroid sphere torus zone The standard parameterization of a surface of revolution is given by (9) (10) (11) For a curve so parameterized, the first fundamental form has (12) (13) (14) Wherever and are nonzero, then the surface is regular and the second fundamental form has (15) (16) (17) Furthermore, the unit normal vector is (18) and the principal curvatures are (19) (20) The Gaussian and mean curvatures are (21) (22) (Gray 1997). Pappus's centroid theorem gives the volume of a solid of rotation as the cross-sectional area times the distance traveled by the centroid as it is rotated. See also Apple Surface, Catenoid, Cone, Conical Frustum, Cylinder, Darwin-de Sitter Spheroid, Eight Surface, Gabriel's Horn, Hyperboloid, Lemon Surface, Meridian, Minimal Surface of Revolution, Oblate Spheroid, Pappus's Centroid Theorem, Paraboloid, Peninsula Surface, Prolate Spheroid, Pseudosphere, Sinclair's Soap Film Problem, Solid of Revolution, Sphere, Spheroid, Surface of Revolution Parallel, Toroid, Torus, UnduloidExplore this topic in the MathWorld classroom Explore with Wolfram\|Alpha More things to try: surface of revolution surfaces of revolution 5aleph0^aleph0 References Anton, H. Calculus: A New Horizon, 6th ed. New York: Wiley, 1999.Apostol, T.M. Calculus, 2nd ed., Vol.2: Multi-Variable Calculus and Linear Algebra, with Applications to Differential Equations and Probability. Waltham, MA: Blaisdell, 1969.Gray, A. "Surfaces of Revolution." Ch.20 in Modern Differential Geometry of Curves and Surfaces with Mathematica, 2nd ed. Boca Raton, FL: CRC Press, pp.457-480, 1997.Hilbert, D. and Cohn-Vossen, S. "The Cylinder, the Cone, the Conic Sections, and Their Surfaces of Revolution." §2 in Geometry and the Imagination. New York: Chelsea, pp.7-11, 1999.Kaplan, W. Advanced Calculus, 3rd ed. Reading, MA: Addison-Wesley, 1992.Kreyszig, E. Differential Geometry. New York: Dover, p.131, 1991. Referenced on Wolfram\|Alpha Surface of Revolution Cite this as: Weisstein, Eric W. "Surface of Revolution." From MathWorld--A Wolfram Resource. Subject classifications Geometry Surfaces Surfaces of Revolution Number Theory Constants Transcendental Root Constants About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
12186	https://www.combinatorics.org/ojs/index.php/eljc/article/download/v26i2p11/pdf/	Extremal H-free planar graphs Yongxin Lan ∗ Yongtang Shi ∗ Center for Combinatorics and LPMC Nankai University Tianjin, China yxlan0@126.com shi@nankai.edu.cn Zi-Xia Song † Department of Mathematics University of Central Florida Orlando, U.S.A. Zixia.Song@ucf.edu Submitted: Oct 21, 2018; Accepted: Mar 28, 2019; Published: May 3, 2019 c© The authors. Released under the CC BY-ND license (International 4.0). Abstract Given a graph H, a graph is H-free if it does not contain H as a subgraph. We continue to study the topic of “extremal” planar graphs initiated by Dowden [J. Graph Theory 83 (2016) 213–230], that is, how many edges can an H-free planar graph on n vertices have? We define ex P (n, H ) to be the maximum number of edges in an H-free planar graph on n vertices. We first obtain several sufficient conditions on H which yield ex P (n, H ) = 3 n − 6 for all n > \|V (H)\|. We discover that the chromatic number of H does not play a role, as in the celebrated Erd˝ os-Stone Theorem. We then completely determine ex P (n, H ) when H is a wheel or a star. Finally, we examine the case when H is a ( t, r )-fan, that is, H is isomorphic to K1 + tK r−1, where t > 2 and r > 3 are integers. However, determining ex P (n, H ), when H is a planar subcubic graph, remains wide open. Mathematics Subject Classifications: 05C10, 05C35 ∗ Supported by National Natural Science Foundation of China, Natural Science Foundation of Tian-jin (No. 17JCQNJC00300), the China-Slovenia bilateral project “Some topics in modern graph the-ory” (No. 12-6), and the Fundamental Research Funds for the Central Universities, Nankai University (63191516). † Corresponding author. the electronic journal of combinatorics 26(2) (2019), #P2.11 11 Introduction All graphs considered in this paper are finite and simple. We use Kt, Ct and Pt to denote the complete graph, cycle, and path on t vertices, respectively. Given a graph G, we will use V (G) to denote the vertex set, E(G) the edge set, \|G\| the number of vertices, e(G)the number of edges, δ(G) the minimum degree, ∆( G) the maximum degree. For a vertex v ∈ V (G), we will use NG(v) to denote the set of vertices in G which are adjacent to v. Let dG(v) = \|NG(v)\| denote the degree of the vertex v in G and NG[v] = NG(v)∪{ v}. A vertex is a k-vertex in G if it has degree k. We use nk (G) to denote the number of k-vertices in G.For any set S ⊂ V (G), the subgraph of G induced on S, denoted G[S], is the graph with vertex set S and edge set {xy ∈ E(G) : x, y ∈ S}. We denote by G \ S the subgraph of G induced on V (G) \ S. If S = {v}, then we simple write G \ v. For any two disjoint sets A and B of V (G), we use eG(A, B ) to denote the number of edges between A and B in G.The join G + H (resp. union G ∪ H) of two vertex-disjoint graphs G and H is the graph having vertex set V (G) ∪ V (H) and edge set E(G) ∪ E(H) ∪ { xy \| x ∈ V (G), y ∈ V (H)} (resp. E(G)∪E(H)). For a positive integer t, we use tG to denote disjoint union of t copies of a graph G. Given two isomorphic graphs G and H, we may (with a slight but common abuse of notation) write G = H. For any positive integer k, let [ k] := {1, 2, . . . , k }.Given a graph H, a graph is H-free if it does not contain H as a subgraph. One of the best known results in extremal graph theory is Tur´ an’s Theorem , which gives the maximum number of edges that a Kt-free graph on n vertices can have. The celebrated Erd˝ os-Stone Theorem then extends this to the case when Kt is replaced by an arbitrary graph H with at least one edge, showing that the maximum number of edges possible is (1+ o(1)) (n 2 ) (χ(H)−2 χ(H)−1 ) , where χ(H) denotes the chromatic number of H. This latter result has been called the “fundamental theorem of extremal graph theory” . Tur´ an-type problems when host graphs are hypergraphs are notoriously difficult. A large quantity of work in this area has been carried out in determining the maximum number of edges in a k-uniform hypergraph on n vertices without containing k-uniform linear paths and cycles (see, for example, [9, 10, 12]). Surveys on Tur´ an-type problems of graphs and hypergraphs can be found in [8, 11]. In this paper, we continue to study the topic of “extremal” planar graphs, that is, how many edges can an H-free planar graph on n vertices have? We define ex P (n, H ) to be the maximum number of edges in an H-free planar graph on n vertices. Dowden initiated the study of ex P (n, H ) and proved the following results, where each bound is tight. Theorem 1 () . Let n be a positive integer. (a) ex P (n, C 4) 6 15( n − 2) /7 for all n > 4.(b) ex P (n, C 5) 6 (12 n − 33) /5 for all n > 11 . Let Θ 4 and Θ 5 be obtained from C4 and C5, respectively, by adding an additional edge joining two non-consecutive vertices. The present authors studied ex P (n, H ) when the electronic journal of combinatorics 26(2) (2019), #P2.11 2 H ∈ { Θ4, Θ5, C 6} and when H is a path on at most 9 vertices. Results from are summarized below. Theorem 2 () . Let n be a positive integer. Then (a) ex P (n, Θ4) 6 12( n − 2) /5 for all n > 4, with equality when n ≡ 12 (mod 20) .(b) ex P (n, Θ5) 6 5( n − 2) /2 for all n > 5, with equality when n ≡ 50 (mod 120) .(c) ex P (n, C 6) 6 18( n − 2) /7 for all n > 6.(d) ex P (n, P 9) 6 max {9n/ 4, (5 n − 8) /2}. It seems quite non-trivial to determine ex P (n, C t) for all t > 7. In this paper, we first investigate planar graphs H satisfying ex P (n, H ) = 3 n − 6 for all n > \|H\|. This partially answers a question of Dowden . As observed in , for all n > 6, the planar triangulation 2 K1 + Cn−2 is K4-free. Hence, ex P (n, H ) = 3 n − 6 for all graphs H which contains K4 as a subgraph and n > max {\| H\|, 6}. In particular, ex P (n, K − 5 ) = 3 n − 6for all n > 6, where K− p denotes the graph obtained from Kp by deleting one edge. Proposition 3 below describes several sufficient conditions on K4-free planar graphs H such that ex P (n, H ) = 3 n − 6 for all n > \|H\|. Proposition 3. Let H be a K4-free planar graph and let n > \|H\| be an integer. Then ex P (n, H ) = 3 n − 6 if one of the following holds. (a) χ(H) = 4 and n > \|H\| + 2 .(b) ∆( H) > 7.(c) ∆( H) = 6 and either n6 (H) + n5 (H) > 2 or n6 (H) + n5 (H) = 1 and n4 (H) > 5.(d) ∆( H) = 5 and either H has at least three 5-vertices or H has exactly two adjacent 5-vertices. (e) ∆( H) = 4 and n4 (H) > 7.(f ) H is 3-regular with \|H\| > 9 or H has at least three vertex-disjoint cycles or H has exactly one vertex u of degree ∆( H) ∈ { 4, 5, 6} such that ∆( H[N (u)]) > 3.(g) δ(H) > 4 or H has exactly one vertex of degree at most 3. Proposition 3 implies that ex P (n, H ) = 3 n − 6 for all H with n > \|H\| + 2 and either χ(H) = 4 or χ(H) = 3 and ∆( H) > 7. Note that Θ 4 = K− 4 , and both K− 4 and K1 + 2 K2 have chromatic number 3. Theorem 2(a) and Theorem 5 (see below) then demonstrate that the chromatic number of H does not play a role, as in the Erd˝ os-Stone Theorem. By Proposition 3, ex P (n, H ) remains unknown for K4-free planar graphs H with ∆( H) = 6, n6 (H) + n5 (H) = 1 and n4 (H) 6 4; or ∆( H) = 5 and n5 (H) 6 2 (and the two 5-vertices are not adjacent when n5 (H) = 2); or ∆( H) = 4 and n4 (H) 6 6; or ∆( H) 6 3. In particular, by Proposition 3( f ), ex P (n, H ) remains unknown for K4-free planar graphs H with exactly one vertex, say u, of degree ∆( H) 6 6 and ∆( H[N (u)]) 6 2. It seems non-trivial to determine ex P (n, H ) for all such H. We next study ex P (n, W k), where Wk := K1 + Ck is a wheel on k + 1 > 5 vertices. Unlike the classic Tur´ an number of Wk (see [4, 5] for more information), the planar Tur´ an number of Wk can be completely determined. We establish this in Theorem 4. the electronic journal of combinatorics 26(2) (2019), #P2.11 3 Theorem 4. Let n, k be integers with n > k + 1 > 5. Then ex P (n, W k) =  3n − 6 if k > 6, or k = 5 and n 6 = 7 , or k = 4 and n > 12 3n − 7 if k = 4 and n ∈ { 5, 6}, or k = 5 and n = 7 3n − 8 if k = 4 and 7 6 n 6 11 . A graph is a ( t, r )-fan if it is isomorphic to K1 + tK r−1, where t > 2 and r > 2 are integers. The classical Tur´ an number of ( t, r )-fan, namely, ex (n, K 1 + tK r−1), has also been studied when n is sufficiently large (see [6, 2] for more information). We next study ex P (n, H ) when H is a K4-free ( t, r )-fan, in particular, when H ∈ { K1 + 2 K2, K 1,t , K 1 +3K2}. Theorem 5 below establishes a sharp upper bound for ex P (n, K 1 + 2 K2), and Theorem 6 completely determines the value of ex P (n, K 1,t ) for all t > 3. However, the upper bound for ex P (n, K 1 + 3 K2) when n > 15 in Theorem 7 is not tight. Theorem 5. Let n > 5 be an integer. Then 2n − 3 6 ex P (n, K 1 + 2 K2) 6 19 n 8 − 4. Furthermore, ex P (n, K 1 + 2 K2) = 19 n 8 − 4 if and only if n is divisible by 8. Theorem 6. Let n, t be integers with n > t + 1 > 4. Then ex P (n, K 1,t ) =  3n − 6 if t > 7, or t = 6 and n ∈ { 7, 8, 9, 10 , 12 } 3n − 7 if t = 6 and n = 11 3n − 8 if t = 6 and n ∈ { 13 , 14 }, or t = 5 and n = 7 ⌊(t−1) n 2 ⌋ if t ∈ { 3, 4}, or t = 5 and n 6 = 7 , or t = 6 and n > 15 . Theorem 7. Let n > 7 be an integer. Then ⌊5n 2 ⌋ 6 ex P (n, K 1 + 3 K2) < 17 n 6 − 4 for all n > 15 and ex P (n, K 1 + 3 K2) =  3n − 6 if n ∈ { 7, 8, 9, 10 , 12 } 3n − 7 if n = 11 3n − 8 if n ∈ { 13 , 14 }. We need to introduce more notation. Given a plane graph G and an integer i > 3, an i-face in G is a face of order i. Let fi denote the number of i-faces in G. Let Tn denote the electronic journal of combinatorics 26(2) (2019), #P2.11 4 a plane triangulation on n > 3 vertices, and let T − n be obtained from Tn with one edge removed. For every integer n > 5, let On denote the unique outerplane graph with 2 n − 3edges, maximum degree 4, and the outer face of order n; let O′ n be a different drawing of On with one unique inner face of order n; and finally, let O∗ n be the planar triangulation obtained from On and O′ n by identifying the outer face of On with the unique n-face of O′ n in such a way that O∗ n is a simple graph. The proof of Proposition 3 relies heavily on the fact that O∗ n , K1 + On−1 and 2 K1 + Cn−2 are planar triangulations. Finally, we shall make use of the following lemma in the proof of Theorem 6 and Theorem 7. Lemma 8 () . There does not exist a 4-regular planar graph on 7 vertices, or a 5-regular planar graph on 14 vertices, or a planar graph on n ∈ { 11 , 13 } vertices with exactly one vertex of degree 4 and n − 1 vertices of degree 5. 2 Proof of Proposition 3 Let H and n be given as in the statement. To prove ( a), assume χ(H) = 4 and n > \|H\|+2. Since the planar triangulation 2 K1 +Cn−2 has no subgraph on \|H\| vertices with chromatic number 4, we see that 2 K1 + Cn−2 is H-free. Hence, ex P (n, H ) = 3 n − 6 when χ(H) = 4 and n > \|H\| + 2. To prove ( b), assume that ∆( H) > 7. Then n > \|H\| > 8, and the planar triangulation O∗ n is H-free because ∆( O∗ n ) = 6. Hence, ex P (n, H ) = 3 n − 6 for all n > \|H\|.To prove ( c), assume ∆( H) = 6. Then n > \|H\| > 7. Assume first n6 (H) + n5 (H) > 2. Let x, y ∈ V (H) be such that dH (x) = 6 and dH (y) > 5. Then the planar triangulation K1 + On−1 is H-free when xy / ∈ E(H), and the planar triangulation 2 K1 + Cn−2 is H-free when xy ∈ E(H). Hence, ex P (n, H ) = 3 n − 6 when n6 (H) + n5 (H) > 2. Next assume that n6 (H) + n5 (H) = 1 and n4 (H) > 5. Then n5 (H) = 0. Note that the planar triangulation 2 K1 + Cn−2 is H-free when n4 (H) > 6. We may further assume that n4 (H) = 5. Let u, v 1, . . . , v 5 ∈ V (H) with dH (u) = 6 and dH (vi) = 4 for all i ∈ . Let M := H[{u, v 1, . . . , v 5}]. We may assume that uv 1 /∈ E(M ) when dM (u) = 4. Then the planar triangulation 2 K1 + Cn−2 is H-free when dM (u) = 5, or dM (u) 6 3, or dM (u) = 4 and dM (v1) 6 3; and the planar triangulation O∗ n is H-free when NM (u) = NM (v1) = {v2, v 3, v 4, v 5}. It follows that ex P (n, H ) = 3 n − 6 for all n > \|H\|. This proves ( c). To prove ( d), assume ∆( H) = 5 and n5 (H) > 2. Then n > \|H\| > 6. Let u, v be two distinct 5-vertices in H. Then either n5 (H) > 3 or n5 (H) = 2 with uv ∈ E(H). Note that the planar triangulation 2 K1 + Cn−2 has exactly two non-adjacent vertices of degree at least 5 when n > 7 and has maximum degree 4 when n = 6. Hence, 2 K1 + Cn−2 is H-free, and so ex P (n, H ) = 3 n − 6 for all n > \|H\|. This proves ( d). To prove ( e) and ( f ). Assume ∆( H) = 4 and n4 (H) > 7, or H is 3-regular with \|H\| > 9, or H has at least three vertex-disjoint cycles, or H has exactly one vertex u of degree ∆( H) ∈ { 4, 5, 6} such that ∆( H[N (u)]) > 3. Then the planar triangulation 2K1 + Cn−2 is H-free. Hence, ex P (n, H ) = 3 n − 6 for all n > \|H\|. the electronic journal of combinatorics 26(2) (2019), #P2.11 5 It remains to prove ( g). Assume δ(H) > 4 or H has exactly one vertex of degree at most 3. Then n > \|H\| > 5. Note that the planar triangulation K1 + On−1 is 3-degenerate and every subgraph of K1 + On−1 has at least two vertices of degree at most 3, because every subgraph of On−1 has at least two vertices of degree at most 2. Hence, K1 + On−1 is H-free, and so ex P (n, H ) = 3 n − 6 for all n > \|H\|.This completes the proof of Proposition 3. 3 Proof of Theorem 4 Let n, k be given as in the statement. Assume k > 7. By Proposition 3( b), ex P (n, W k) = 3n−6 for all n > k+1. Assume next k ∈ { 5, 6}. Since the planar triangulation 2 K1 +Cn−2 is Wk-free when n > k +3 or n = k +1, we see that ex P (n, W k) = 3 n−6 when n > k +3 or n = k +1. We next determine ex P (n, W k) when n = k + 2. For k = 6 and n = 8, the plane triangulation on 8 vertices depicted in Figure 1(a) is W6-free and so ex P (n, W 6) = 3 n − 6when n = 8. For k = 5 and n = 7, note that the plane graph with 7 vertices and 14 edges Figure 1: Wk-free plane graphs with n = k + 2 vertices and 3 n − 12 + k edges, where k ∈ { 5, 6}.given in Figure 1(b) is W5-free. Thus, ex P (7 , W 5) > 3 · 7 − 6 − 1 = 14. On the other hand, all plane triangulations on seven vertices 1 are depicted in Figure 2, each containing a copy of W5. Hence, ex P (n, W 5) = 3 n − 7 when n = 7. It remains to consider the case when k = 4. To show ex P (n, W 4) = 3 n−6 for all n > 12, assume first that n = 5 t + 2 for some integer t > 2. Let Lt be a plane triangulation on n = 5 t + 2 vertices constructed as follows: for each i ∈ [t], let Ci be a cycle with vertices ui, 1, u i, 2, . . . , u i, 5 in order. Let Lt be the plane triangulation obtained from disjoint union of C1, . . . , C t by adding edges ui,j ui+1 ,j and ui,j ui+1 ,j +1 for all i ∈ [t − 1] and j ∈ , where 1 To find all plane triangulations on 7 vertices, let H be a plane triangulation on 7 vertices. Then e(H) = 15 and H must be 3-connected with maximum degree 5 or 6. Let u ∈ V (H) be a vertex of maximum degree. If d(u) = 6, then H has at most two vertices of degree 6 and H − u has neither K4 nor K2,3 minor. Thus H −u must be outer planar and so H is isomorphic to one of the plane triangulations in Figure 2(a,b,e). If d(u) = 5, then the degree sequence of H is either (5 , 5, 5, 4, 4, 4, 3) or (5 , 5, 4, 4, 4, 4, 4). Since H has no K3,3 minor, one can then check that H is isomorphic to one of the plane triangulations in Figure 2(c,d). the electronic journal of combinatorics 26(2) (2019), #P2.11 6Figure 2: All plane triangulations on 7 vertices. Furthermore, each T7 has a copy of W4 and W5, and each T − 7 has a copy of W4.all arithmetic on the index j + 1 here is done modulo 5, and finally adding two new non-adjacent vertices u and v such that u is adjacent to all vertices of C1 and v is adjacent to all vertices of Ct. The graph Lt when t = 3 is depicted in Figure 3. It is worth noting that Lt is K4-free, dLt (u) = dLt (v) = 5, dLt (u1, j ) = dLt (ut, j ) = 5, dLt (ui, j ) = 6 for 2 6 i 6 t−1and j ∈ . Furthermore, the subgraph induced by the neighborhood of each vertex in Lt is isomorphic to either C5 or C6. Hence, Lt is W4-free and so ex P (n, W 4) = 3 n − 6 when n = 5 t + 2 for some integer t > 2. Figure 3: The plane triangulation Lt when t = 3, where C1, C 2 and C3 are in red, blue and green, respectively. Next assume that n = 5 t + 2 + i for some i ∈ , where t > 2 is an integer. Note that the plane triangulation Lt on 5 t + 2 vertices constructed above contains at least four pairwise vertex-disjoint faces. Let F1, . . . , F i be any i pairwise vertex-disjoint faces of Lt,and let Lit be the plane triangulation obtained from Lt by adding one vertex, say xj , of degree 3 to each Fj for all j ∈ [i]. Clearly, Lit is a plane triangulation on n = 5 t + 2 + i the electronic journal of combinatorics 26(2) (2019), #P2.11 7 vertices. By the choice of F1, . . . , F i, we see that x1, . . . , x i are pairwise non-adjacent in Lit, and no two of x1, . . . , x i have common neighbors in Lit. We next show that Lit is W4-free for all i ∈ . Suppose that Lit contains a copy of W4 for some i ∈ . Let H be a W4 in Lit. Then H must contain exactly one, say x1, of x1, . . . , x i, because Lt is W4-free, and no two of x1, . . . , x i are adjacent or have common neighbors in Lit. Let y, z ∈ V (H) be the two neighbors of x1 such that yz / ∈ E(H). By the choice of x1, we see that yz ∈ E(Lt). But then Lt[V (H \ x1)] = K4 and so Lt contains K4 as a subgraph, a contradiction. Therefore, ex P (n, W 4) = 3 n − 6 for all n > 12. Figure 4: All plane triangulations on 5 and 6 vertices. We next show that ex P (n, W 4) = 3 n − 7 when n ∈ { 5, 6}. Note that all plane tri-angulations on n ∈ { 5, 6} vertices are depicted in Figure 4, each containing a copy of W4. Thus, ex P (n, W 4) 6 3n − 7. On the other hand, for all n ∈ { 5, 6}, the planar graph K2 + ( K2 ∪ Kn−4) has 3 n − 7 edges and is W4-free. Hence, ex P (n, W 4) = 3 n − 7 when n ∈ { 5, 6}.Finally, we show that ex P (n, W 4) = 3 n − 8 for all n ∈ { 7, 8, 9, 10 , 11 }. The plane graph J, given in Figure 5, is W4-free with n = 11 vertices and 3 n − 8 edges. Let B be the set of Figure 5: Graph J.all vertices of degree 3 in J. Then \|B\| = 5. For each n ∈ { 7, 8, 9, 10 }, let Jn be the plane graph obtained from J by deleting 11 −n vertices in B. Then Jn is an induced subgraph of the electronic journal of combinatorics 26(2) (2019), #P2.11 8 J. Clearly, Jn is W4-free with n vertices and 3 n − 8 edges. Hence, ex P (n, W 4) > 3n − 8 for all n ∈ { 7, 8, 9, 10 , 11 }. We next show that ex P (n, W 4) 6 3n−8 for all n ∈ { 7, 8, 9, 10 , 11 }.Suppose this is not true. Let G be a W4-free planar graph on n ∈ { 7, 8, 9, 10 , 11 } vertices with e(G) > 3n − 7. We choose such a G with n minimum. Then G = Tn or G = T − n .Since each T7, depicted in Figure 2, contains a copy of W4, and each T − 7 also contains a copy of W4, it follows that n ∈ { 8, 9, 10 , 11 }. Let u ∈ V (G) with dG(u) = δ(G). Then δ(G) 6 4, else e(G) > 5n 2 3n − 6 because n 6 11, a contradiction. Next, if δ(G) 6 3, then e(G \ u) 6 3( n − 1) − 8 by minimality of n and the fact that ex P (n, W 4) 6 3n − 8when n = 7. Thus, e(G) = e(G \ u) + dG(u) 6 3( n − 1) − 8 + 3 = 3 n − 8, a contradiction. This proves that δ(G) = 4. Since NG[u] does not contain a copy of W4 in G, we see that G 6 = Tn. Thus G = T − n . We may assume that G is a plane drawing of T − n such that the outer face is a 3-face. Let x1, y 1 ∈ V (G) be such that G+x1y1 = Tn. Then x1 and y1 must lie on the boundary of the unique 4-face, say F , in G. Let x1, x 2, y 1, y 2 be the vertices on the boundary of F in order. Then dG(v) > 5 for all v ∈ V (G) \ { x1, x 2, y 1, y 2}, because G = T − n and NG[u] does not contain a copy of W4 in G for any u ∈ V (G) with dG(u) = 4. Thus 2(3 n − 7) = 2 e(G) > 4 · 4 + 5 · (n − 4), which implies that n ∈ { 10 , 11 }. Suppose each vertex in {x1, x 2, y 1, y 2} has degree 4 in G. Since G = T − n , there must exist four distinct vertices z1, z 2, z 3, z 4 ∈ V (G) \ { x1, x 2, y 1, y 2} such that G[A] is isomorphic to the graph given in Figure 6(a), where A = {x1, x 2, y 1, y 2, z 1, z 2, z 3, z 4}. But then G contains K3,3 as Figure 6: The graph G[A]. a minor, because n ∈ { 10 , 11 } and dG(v) > 5 for any v ∈ V (G) \ A. Thus, we may assume that dG(x2) > 5. Then 2(3 n − 7) = 2 e(G) > 4 · 3 + 5 · (n − 3), which implies that n = 11, dG(v) = 4 for all v ∈ { x1, y 1, y 2} and ∆( G) = 5. Thus there exist five distinct vertices z1, z 2, z 3, z 4, z 5 ∈ V (G) \ { x1, x 2, y 1, y 2} such that G[A] is isomorphic to the graph given in Figure 6(b), where A = {x1, x 2, y 1, y 2, z 1, z 2, z 3, z 4, z 5}. But then eG (A, V (G) \ A) = 6, contrary to eG (V (G) \ A, A ) > 8 because n = 11 and dG(v) = 5 for any v ∈ V (G) \ A.This completes the proof of Theorem 4. the electronic journal of combinatorics 26(2) (2019), #P2.11 9 4 Proof of Theorem 5 To establish the desired lower bound, note that the planar graph K2 + ( n − 2) K1 is (K1 + 2 K2)-free for all n > 5. Hence, ex P (n, K 1 + 2 K2) > 2n − 3 for all n > 5. In particular, ex P (5 , K 1 + 2 K2) > 7. We next show that every ( K1 + 2 K2)-free planar graph on n > 5 vertices has at most 19 n/ 8 − 4 edges. We proceed the proof by induction on n.Assume first n = 5. Then ex P (5 , K 1 + 2 K2) = 7, because the only plane triangulation on five vertices, given in Figure 4(a), is not ( K1 + 2 K2)-free, and any T − 5 is not ( K1 + 2 K2)-free. Hence, ex P (n, K 1 + 2 K2) = 7 < 19 n/ 8 − 4 when n = 5. So we may assume that n > 6. Let G be a ( K1 + 2 K2)-free plane graph on n > 6 vertices. Assume there exists a vertex u ∈ V (G) with dG(u) 6 2. By the induction hypothesis, e(G \ u) 6 19( n − 1) /8 − 4and so e(G) = e(G \ u) + dG(u) 6 19( n − 1) /8 − 4 + 2 < 19 n/ 8 − 4, as desired. So we may assume that δ(G) > 3. Next, assume G is disconnected. Then each component of G either is isomorphic to K4 or has at least six vertices because δ(G) > 3. Let G1, . . . , G p, G p+1 , . . . , G p+q be all components of G such that \|G1\| = · · · = \|Gp\| = 4 and 6 6 \|Gp+1 \| 6 · · · 6 \|Gp+q\|, where p > 0 and q > 0 are integers with p + q > 2 and \|Gp+1 \| + · · · + \|Gp+q\| = n − 4p. Then e(Gi) = 6 for all i ∈ [p], and e(Gj ) 6 19 \|Gj \| 8 − 4 for all j ∈ { p + 1 , . . . , p + q} by the induction hypothesis. Therefore, e(G) 6 6p + 19( \|Gp+1 \| + · · · + \|Gp+q\|)8 − 4q = 19 n 8 − 7p 2 − 4q 6 19 n 8 − 7( p + q)2 < 19 n 8 − 4, as desired. So we may further assume that G is connected. Then G has no faces of size at most two. Hence, 2e(G) = 3 f3 + ∑ i>4 if i > 3f3 + 4( f − f3) = 4 f − f3, which implies that f 6 e(G)/2 + f3/4. (1) Note that each 3-vertex is incident with at most three distinct 3-faces in G. Furthermore, since G is ( K1 + 2 K2)-free, we see that for all j > 4, each j-vertex is incident with at most two distinct 3-faces in G. Let U ⊆ V (G) denote the set of 3-vertices each incident with exactly three distinct 3-faces in G. Then U must be an independent set in G because G is connected. Furthermore, no two vertices in U have a common neighbor in G, because G is ( K1 + 2 K2)-free. Thus, 4 \|U \| 6 n and so \|U \| 6 n/ 4. It follows that 3f3 6 3\|U \| + 2( n − \| U \|) = 2 n + \|U \| 6 9n/ 4, (2) which implies that f3 6 3n/ 4. This, together with (1), further implies that f 6 e(G)/2 + 3n/ 16. By Euler’s formula, n − 2 = e(G) − f > e(G)/2 − 3n/ 16. Hence, e(G) 6 19 n/ 8 − 4, as desired. the electronic journal of combinatorics 26(2) (2019), #P2.11 10 Figure 7: The construction of Fk.From the proof above, we see that equality in e(G) 6 19 n/ 8 − 4 is achieved for n if and only if equalities hold in both (1) and (2), and in 4 \|U \| 6 n. This implies that e(G) = 19 n/ 8 − 4 for n if and only if G is a connected, ( K1 + 2 K2)-free plane graph on n vertices satisfying: δ(G) > 3; each 3-vertex in G is incident with exactly three distinct 3-faces; each vertex of degree at least 4 in G is incident with exactly two distinct 3-faces; each face is either a 3-face or a 4-face. We next construct such an extremal plane graph for n and K1 + 2 K2. Let n = 8( k + 1) for some integer k > 0. Let F0 be the graph depicted in Figure 7(a), we then construct Fk of order n recursively for all k > 1via the illustration given in Figure 7(b): the entire graph Fk−1 is placed into the center quadrangle of Figure 7(b) (in such a way that the center bold quadrangle of Figure 7(b) is identified with the outer quadrangle of Fk−1). One can check that Fk is ( K1 + 2 K2)-free with n = 8( k + 1) vertices and 19 n/ 8 − 4 edges for all k > 0. This completes the proof of Theorem 5. 5 Proof of Theorem 6 By Proposition 3(b), ex P (n, K 1,t ) = 3 n − 6 for all n > t + 1 > 8. So we may assume that t 6 6. We next show that ex P (n, K 1,6) = 3 n − 6 for all n ∈ { 7, 8, 9, 10 , 12 }. Let Ja, J b, J c be the plane graphs given in Figure 8. Let J′ a be the plane triangulation obtained from Ja by adding a new vertex adjacent to x1, x 2, x 3, J′ b be the plane triangulation obtained from Jb by first deleting the edge x1x3 and then adding a new vertex adjacent to x1, x 2, x 3, x 4, and J′ c be the plane triangulation obtained from Jc by first deleting the edge x1x3 and then adding one new vertex adjacent to x1, x 2, x 3, x 4, x 5. Then the plane triangulations Ja, J′ a , Jb, J′ b and J′ c are K1,6-free because each of them has maximum degree 5. Hence, ex P (n, K 1,6) = 3 n − 6 for all n ∈ { 7, 8, 9, 10 , 12 }. By Lemma 8, no plane triangulation on 11 vertices has maximum degree at most 5. Hence, every plane triangulation on n ∈ { 11 , 13 , 14 } vertices has maximum degree at least 6. This implies that ex P (n, K 1,6) 6 3n − 7 for all n ∈ { 11 , 13 , 14 }. Since Jc given in Figure 8 is a K1,6-free plane graph with n = 11 vertices and 3 n − 7 edges, we have ex P (n, K 1,6) = 3 n − 7 when the electronic journal of combinatorics 26(2) (2019), #P2.11 11 Figure 8: The graphs Ja, Jb, Jc and Jd. n = 11. By Lemma 8, there does not exist any planar graphs on n ∈ { 13 , 14 } vertices with 3 n − 7 edges and maximum degree at most 5. It follows that ex P (n, K 1,6) 6 3n − 8when n ∈ { 13 , 14 }. Let J′′ c be the plane graph obtained from Jc by first deleting the edge x1x3 and then adding two new adjacent vertices y1, y 2 such that y1 is adjacent to x1, x 2, x 3 and y2 is adjacent to x4, x 5. Then J′′ c and the graph Jd given in Figure 8 are K1,6-free plane graph with n ∈ { 13 , 14 } vertices and 3 n − 8 edges. Hence, ex P (n, K 1,6) = 3 n − 8when n ∈ { 13 , 14 }.It is easy to see that ex P (n, K 1,3) = n for all n > 4, because every K1,3-free planar graph has maximum degree at most 2 and the planar graph Cn is K1,3-free with n edges. We next show that ex P (n, K 1,4) = b3n/ 2c for all n > 5. Clearly, ex P (n, K 1,4) 6 b3n/ 2c for all n > 5, because every K1,4-free planar graph has maximum degree at most 3. Next, for all n > 5, the planar graph obtained from Cn by adding a matching of size bn/ 2c is K1,4-free with b3n/ 2c edges. Hence, ex P (n, K 1,4) = b3n/ 2c for all n > 5. To determine ex P (n, K 1,5) for all n > 6, since every K1,5-free planar graph on n > 6 vertices has maximum degree at most 4, we have ex P (n, K 1,5) 6 2n for all n > 6. Let J′′ a be the plane triangulation obtained from Ja by deleting the unique 3-vertex. Since J′′ a is K1,5-free plane graph on n = 6 vertices with 2 n edges, we have ex P (n, K 1,5) = 2 n when n = 6. By Lemma 8, no planar graph with n = 7 vertices and 2 n edges has maximum degree at most 4. Hence, ex P (n, K 1,5) 6 2n − 1 when n = 7. Let J′′′ a be the plane graph obtained from J′′ a by first deleting the edge x1x2 and then adding a new vertex adjacent to x1, x 2 only. Note that J′′′ a is a K1,5-free plane graph on n = 7 vertices with 2 n − 1 edges, we see that ex P (n, K 1,5) = 2 n − 1 when n = 7. Next, for all n > 8, let C be a cycle on 2 bn/ 2c vertices with vertices u1, . . . , u b n 2c , w b n 2c , . . . , w 1 in order. Let H be the plane graph obtained from C by adding the path with vertices w1, u 2, w 2, u 3, . . . , w b n 2c− 1 , u b n 2c in order. When n is even, the planar graph H + u1ub n 2c u1wb n 2c w1wb n 2c is K1,5-free with 2 n edges. When n is odd, let H′ be obtained from H by first deleting the edge u2u3 and then adding a new vertex u adjacent to u2 and u3. Then the planar graph H′ +uu 1 +uu b n 2c +w1wb n 2c +u1wb n 2c is K1,5-free with 2 n edges. Hence, ex P (n, K 1,5) = 2 n for all n > 8. It remains to show that ex P (n, K 1,6) = b5n/ 2c for all n > 15. Clearly, ex P (n, K 1,6) 6 b5n/ 2c for all n > 15, because every K1,6-free planar graph on n > 15 vertices has maximum degree at most 5. We next show that ex P (n, K 1,6) > b5n/ 2c for all n > 15. the electronic journal of combinatorics 26(2) (2019), #P2.11 12 Let n := 4 q + r > 15, where q > 3 and r ∈ { 0, 1, 2, 3}. Let p ∈ { q, q + 1 }. Let C1 and C2 be two vertex-disjoint cycles with vertices x1, x 2, . . . x q in order and y1, y 2, . . . y p in order, respectively. Let C3 be a cycle of length q + p with vertices b1, a 1, b 2, a 2, . . . b q, a q in order when p = q, and b1, a 1, b 2, a 2, . . . b q, a q, b q+1 in order when p = q + 1. Let Rp be the plane graph on 2 q + 2 p vertices obtained from disjoint copies of C1, C2 and C3 by making xi adjacent to {ai, b i, b i+1 }, and yj adjacent to {bj , a j−1, a j } for all 1 6 i 6 q and 1 6 j 6 p, where all arithmetic on the indices i + 1 and j − 1 here are done modulo Figure 9: Almost 5-regular plane graphs on 4 q + r vertices when q = 4 and r ∈ { 0, 3}. p. Then Rp is K1,6-free planar graph with 10 q edges when p = q and 10 q + 3 edges when p = q + 1. The construction of Rp on 4 q vertices when q = 4 is depicted in Figure 9(a). When n = 4 q for q > 4, the planar graph Rp with p = q is K1,6-free with 10 q edges and so ex P (n, K 1,6) = 10 q = 5 n/ 2. When n = 4 q + 1 for q > 4, let R1 be obtained from Rp − y2y3 − y1yp with p = q by adding a new vertex u adjacent to y2 and y3. Then the planar graph R1 + uy 1 + uy p is K1,6-free with 10 q + 2 edges. Hence, ex P (n, K 1,6) = 10 q + 2 = b5n/ 2c when n = 4 q + 1 for q > 4. When n = 4 q + 2 for q > 4, let R2 be obtained from Rp − y2y3 − y1yp − x2x3 − x1xp − b1ap with p = q by adding two new vertices u adjacent to y2 and y3 and v adjacent to x2 and x3. Then the planar graph R2 + uy 1 + uy p + ua q + vx 1 + vx q + vb 1 is K1,6-free with 10 q + 5 edges. Hence, ex P (n, K 1,6) = 10 q + 5 = 5 n/ 2 when n = 4 q + 2 for q > 4. When n = 4 q + 3 for n > 3, the planar graph obtained from Rp with p = q + 1 by adding a new vertex u adjacent to y1, b 1, y p, b p, given in Figure 9(b) when q = 4, is K1,6-free with 10 q + 7 edges. Hence, ex P (n, K 1,6) = 10 q + 7 = b5n/ 2c when n = 4 q + 3 for q > 3. This completes the proof of Theorem 6. the electronic journal of combinatorics 26(2) (2019), #P2.11 13 6 Proof of Theorem 7 Since the plane triangulations Ja, J′ a , Jb and J′ b constructed in the proof of Theorem 6 is ( K1 + 3 K2)-free, we see that ex P (n, K 1 + 3 K2) = 3 n − 6 for all n ∈ { 7, 8, 9, 10 }. To determine ex P (11 , K 1 + 3 K2), note that the plane graph Jc given in Figure 8 with n = 11 vertices and 3 n − 7 edges is ( K1 + 3 K2)-free. Thus ex P (n, K 1 + 3 K2) > 3n − 7 when n = 11. By Lemma 8, no plane triangulation on 11 vertices has maximum degree at most 5. Hence, every plane triangulation on 11 vertices must contain a vertex of degree at least 6 (and so contains a copy of K1 + 3 K2), which implies that ex P (n, K 1 + 3 K2) = 3 n − 7when n = 11. Since every K1,6-free graph is certainly ( K1 + 3 K2)-free, by Theorem 6, ex P (n, K 1+3 K2) = ex P (n, K 1,6) = 3 n−6 when n = 12, ex P (n, K 1+3 K2) > ex P (n, K 1,6) = 3n − 8 when n ∈ { 13 , 14 }, and ex P (n, K 1 + 3 K2) > ex P (n, K 1,6) = b5n/ 2c when n > 15. Since every plane triangulation on n ∈ { 13 , 14 } vertices has maximum degree at least 6, we see that ex P (n, K 1 + 3 K2) 6 3n − 7 when n ∈ { 13 , 14 }. By Lemma 8, every T − n with n ∈ { 13 , 14 } has maximum degree at least 6 and so contains a copy of K1 + 3 K2. It follows that ex P (n, K 1 + 3 K2) = 3 n − 8 < 17 n/ 6 − 4 when n ∈ { 13 , 14 }.We next show that every ( K1 + 3 K2)-free planar graph G on n > 13 vertices has at most 17 n/ 6 − 4 edges. We proceed the proof by induction on n. This is trivially true when n ∈ { 13 , 14 }. So we may assume that n > 15. Assume there exists a vertex u ∈ V (G) with dG(u) 6 2. By the induction hypothesis, e(G \ u) 6 17( n − 1) /6 − 4 and so e(G) = e(G \ u) + dG(u) 6 17( n − 1) /6 − 4 + 2 < 17 n/ 6 − 4, as desired. So we may assume that δ(G) > 3. Assume next that G is disconnected. Let G1, . . . , G p, G p+1 , . . . , G p+q be all components of G such that \|G1\| 6 · · · 6 \|Gp\| 6 12 and 13 6 \|Gp+1 \| 6 · · · 6 \|Gp+q\|,where p > 0 and q > 0 are integers with p + q > 2 and \|G1\| + · · · + \|Gp+q\| = n. Then e(Gi) 6 3\|Gi\| − 6 for all i ∈ [p], and e(Gj ) 6 17 \|Gj \|/6 − 4 for all j ∈ { p + 1 , . . . , p + q} by the induction hypothesis. Therefore, e(G) 6 3( \|G1\| + · · · + \|Gp\|) − 6p + 17( \|Gp+1 \| + · · · + \|Gp+q\|)6 − 4q = 17 n 6 − (6 p + 4 q) + \|G1\| + · · · + \|Gp\| 6 6 17 n 6 − (6 p + 4 q) + 2 p = 17 n 6 − 4( p + q) < 17 n 6 − 4, as desired. So we may further assume that G is connected. Then G has no faces of size at most two. Hence, 2e(G) = 3 f3 + ∑ i>4 if i > 3f3 + 4( f − f3) = 4 f − f3, which implies that f 6 e(G)/2 + f3/4. Note that n3 (G) > 0 and n5 (G) < n ; and for all i ∈ { 3, 4, 5}, each i-vertex is incident with at most i 3-faces. Furthermore, for all j > 6, each j-vertex is incident with at most four 3-faces because G is ( K1 + 3 K2)-free and n > 15. It follows that 3f3 6 3n3 (G)+4 n4 (G)+5 n5 (G)+4( n−n3 (G)−n4 (G)−n5 (G)) = 4 n−n3 (G)+ n5 (G) < 5n, the electronic journal of combinatorics 26(2) (2019), #P2.11 14 which implies that f3 < 5n/ 3. This, together with the fact that f 6 e(G)/2+ f3/4, further implies that f < e (G)/2 + 5 n/ 12. By Euler’s formula, n − 2 = e(G) − f > e (G)/2 − 5n/ 12. Hence, e(G) < 17 n/ 6 − 4. This completes the proof of Theorem 7. 7 Concluding remarks The lower bound in Theorem 7 can be further improved when n is divisible by 24. To see this, let n = 24( k + 1) for some integer k > 0. Let R5 be the 5-regular plane graph on twelve vertices given in Figure 10(a), and let G0 be the plane graph obtained from two disjoint copies of R5 by adding three independent edges between their outer faces, as depicted in Figure 10(b). We construct Gk of order n recursively for all k > 1 via the illustration given in Figure 10(c): the entire graph Gk−1 is placed into the center quadrangle of Figure 10(c) (in such a way that the center bold quadrangle of Figure 10(c) is identified with the outer quadrangle of Gk−1). One can check that Gk is ( K1 +3 K2)-free with n = 24( k + 1) vertices and 67 n/ 24 − 4 edges for all k > 0. Figure 10: The construction of Gk.As mentioned earlier, it seems non-trivial to determine ex P (n, H ) for all K4-free planar graphs H with exactly one vertex, say u, satisfying dH (u) = ∆( H) 6 6 and ∆( H[N (u)]) 6 We conclude this section by giving an upper bound (but not tight) for ex P (n, K 1 + H), where H is a disjoint union of paths. Theorem 9. Let 4 6 t 6 6 be an integer and let H be a graph on t vertices such that H is a disjoint union of paths. Then ex P (n, K 1 + H) 6 13( t−1) n 4t−2 − 12( t−1) 2t−1 for all n > t + 1 . Proof. Let t and H be given as in the statement. Since H is a subgraph of K1 + Pt,it suffices to show that ex P (n, K 1 + Pt) 6 13( t−1) n 4t−2 − 12( t−1) 2t−1 for all n > t + 1. Let G the electronic journal of combinatorics 26(2) (2019), #P2.11 15 be a ( K1 + Pt)-free planar graph on n > t + 1 vertices. We next show that e(G) 6 13( t−1) n 4t−2 − 12( t−1) 2t−1 by induction on n. This is trivially true when n = t + 1 because e(G) 6 3( t+1) −6 6 13( t−1)( t+1) 4t−2 − 12( t−1) 2t−1 for 5 6 t 6 6 and e(G) 6 3( t+1) −7 6 13( t−1)( t+1) 4t−2 − 12( t−1) 2t−1 for t = 4. So we may assume that n > t + 2. We may further assume that δ(G) > 3 and G is connected. Hence, 2e(G) = 3 f3 + ∑ i>4 if i > 3f3 + 4( f − f3) = 4 f − f3, which implies that f 6 e(G)/2 + f3/4. Note that for all 3 6 i 6 t − 1, each i-vertex is incident with at most i many 3-faces, and for all j > t, each j-vertex is incident with at most ( t − 2) b jt−1 c many 3-faces, because G is ( K1 + Pt)-free. It follows that 3f3 6 t−1 ∑ i=3 i · ni(G) + ∑ j>t (t − 2) ⌊ jt − 1 ⌋ · nj (G) 6 t−1 ∑ i=3 (i · ni(G) t − 1 + (t − 2) i · ni(G) t − 1 ) t − 2 t − 1 ∑ j>t j · nj (G)= t−1 ∑ i=3 i · ni(G) t − 1 + t − 2 t − 1 ∑ `>3 · n(G)= t−1 ∑ i=3 i · ni(G) t − 1 + (t − 2 t − 1 · 2e(G) ) = t−1 ∑ i=3 ni(G) − t−1 ∑ i=3 (t − 1 − i) · ni(G) t − 1 + (t − 2 t − 1 · 2e(G) ) < n + t − 2 t − 1 · 2e(G), which implies that f3 < n 3 2( t−2) 3( t−1) · e(G). This, together with the fact that f 6 e(G)/2 + f3/4, further implies that f 6 e(G)2 + n 12 (t−2) 6( t−1) · e(G) = (4 t−5) 6( t−1) · e(G) + n 12 . By Euler’s formula, n − 2 = e(G) − f > (2 t−1) 6( t−1) · e(G) − n 12 . Hence, e(G) 6 13( t−1) n 4t−2 − 12( t−1) 2t−1 .This completes the proof of Theorem 9. Acknowledgments Zi-Xia Song would like to thank Yongtang Shi and the Chern Institute of Mathematics at Nankai University for hospitality and support during her visit in June 2018. References B. Bollob´ as. Modern Graph Theory. Springer, 2013. the electronic journal of combinatorics 26(2) (2019), #P2.11 16 G. Chen, R. J. Gould, F. Pfender and B. Wei. Extremal graphs for intersecting cliques. J. Combin. Theory Ser. B , 89:159–171, 2003. C. Dowden. Extremal C4-free/ C5-free planar graphs. J. Graph Theory , 83:213–230, 2016. T. Dzido. A note on Tur´ an numbers for even wheels. Graphs Combin. , 29:1305–1309, 2013. T. Dzido and A. Jastrz¸ ebski. Tur´ an numbers for odd wheels. Discrete Math. ,341:1150–1154, 2018. P. Erd˝ os, Z. F¨ uredi, R. J. Gould and D. S. Gunderson. Extremal graphs for inter-secting triangles. J. Combin. Theory Ser. B , 64:89–100, 1995. P. Erd˝ os and A. Stone. On the structure of linear graphs. Bull. Amer. Math. Soc. ,52:1087–1091, 1946. Z. F¨ uredi, Tur´ an type problems, “Surveys in Combinatorics”, London Math. Soc. Lecture Note Ser. 166 , pages 253–300. Cambridge Univ. Press, 1991. Z. F¨ uredi and T. Jiang. Hypergraph Tur´ an numbers of linear cycles. J. Combin. Theory Ser. A , 123:252–270, 2014. Z. F¨ uredi, T. Jiang and R. Seiver. Exact solution of the hypergraph Tur´ an problem for k-uniform linear paths. Combinatorica , 34:299–322, 2014. P. Keevash. Hypergraph Tur´ an problems, “Surveys in Combinatorics”. London Math. Soc. Lecture Note Ser. 392 , pages 83–139. Cambridge Univ. Press, 2011. A. Kostochka, D. Mubayi and J. Verstra¨ ete. Tur´ an problems and shadows I: Paths and cycles. J. Combin. Theory Ser. A , 129:57–79, 2015. Y. Lan, Y. Shi and Z-X. Song. Extremal Theta-free planar graphs. arXiv:1711.01614v2 , 2017. E. F. Schmeichel and S. L. Hakimi. On planar graphical degree sequences. SIAM J. Appl. Math. , 32:598–609, 1977. P. Tur´ an. On an extremal problem in graph theory. Mat. Fiz. Lapok. , 48:436–452, 1941.
12187	https://www.chegg.com/homework-help/questions-and-answers/d-question-2-3-pts-explain-equilibrium-constants-dimensionless-really-dimensionless-must-t-q34686538	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: D Question 2 3 pts Explain why equilibrium constants are dimensionless. They are not really dimensionless, but we must treat them as such in order to be able to take In(K) in the expression AG-RT In K They are dimensionless because the pressures or concentrations we put in are all for the substances in their standard states. e Activities (which are Equilibrium constant is supposed to be… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & ServicesChegg Products & Services Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
12188	https://www.math.uwaterloo.ca/~krdavids/MT/PMath451Notes.pdf	Measure Theory Notes for Pure Math 451 Kenneth R. Davidson University of Waterloo CONTENTS Chapter 1. Measures 1 1.1. Introduction 1 1.2. σ-Algebras 3 1.3. Construction of Measures 6 1.4. Premeasures 8 1.5. Lebesgue-Stieltjes measures on R 10 Chapter 2. Functions 16 2.1. Measurable Functions 16 2.2. Simple Functions 18 2.3. Two Theorems about Measurable Functions 19 Chapter 3. Integration 21 3.1. Integrating positive functions 21 3.2. Two limit theorems 23 3.3. Integrating complex valued functions 26 3.4. The space L1(µ) 28 3.5. Comparison with the Riemann Integral 29 3.6. Product Measures 34 3.7. Product Integration 36 3.8. Lebesgue Measure on Rn 43 3.9. Inﬁnite Product Measures 46 Chapter 4. Differentiation and Signed Measures 49 4.1. Differentiation 49 4.2. Signed Measures 57 4.3. Decomposing Measures 60 Chapter 5. Lp spaces 64 5.1. Lp as a Banach space 64 5.2. Duality for Normed Vector Spaces 68 5.3. Duality for Lp 69 i ii Contents Chapter 6. Some Topology 74 6.1. Topological spaces 74 6.2. Continuity 76 6.3. Compactness 78 6.4. Separation Properties 81 6.5. Nets 85 Chapter 7. Functionals on Cc(X) and C0(X) 88 7.1. Radon measures 88 7.2. Positive functionals on Cc(X) 90 7.3. Linear functionals on C0(X) 94 Chapter 8. A Taste of Probability 97 8.1. The language of probability 97 8.2. Independence 98 8.3. The Law of Large Numbers 100 Bibliography 105 Index 106 CHAPTER 1 Measures 1.1. Introduction The idea of Riemann integration is the following: Convergence of the upper and lower sums to the same limit is required for Riemann integrability. The integral FIGURE 1.1. Riemann Integral is approximated by Riemann sums Z b a f(x) dx ≈ n X i=1 Li(ti −ti−1). The function f(x) has to be ‘almost continuous’ in order for this to succeed. It is too restrictive. Moreover there are no good limit theorems. (Again one needs uniform convergence for most results,) Lebesgue’s approach was to chop up the range instead. Then one gets the FIGURE 1.2. Lebesgue integral approximations Z b a f(x) dx ≈ n X i=1 yi {x : yi−1 < f(x) ≤yi} . The notation 1 2 Measures \|A\| is meant to convey the length of a set A. The complication is that, even for continuous functions, the sets {x : yi−1 < f(x) ≤yi} are neither open nor closed, and can be complicated. It is a Borel set, and that will be seen to be good enough. Lebesgue’s idea was to extend the notion of length or measure from open intervals to a function m on a much larger class of sets so that (1) m((a, b)) = b −a (2) m(A + x) = m(A) for sets A and x ∈R (translation invariance) (3) if A is the disjoint union A = ˙ S∞ n=1An, then m(A) = P∞ n=1 m(An). (countable additivity) It turns out that it is not possible to deﬁne such a function on all subsets of R. To see this, we deﬁne an equivalence relation on [0, 1) by x ∼y if x−y ∈Q. Then use the Axiom of Choice to select a set A which contains exactly one element from each equivalence class. Enumerate Q ∩[0, 1) = {0 = r0, r1, r2, . . . } and deﬁne An = A + rn(mod 1) = (A + rn) ∪(A + rn −1) ∩[0, 1) for n ≥0. Now by translation invariance and ﬁnite additivity, m(A) = m(A + rn) = m (A + rn) ∩[0, 1) + m (A + rn) ∩[1, 2) = m( A + rn) ∩[0, 1) + m (A + rn −1) ∩[0, 1) = m(An). Observe that Am ∩An = ∅if m ̸= n. By construction, S n≥1 An = [0, 1). Hence by countable additivity, 1 = m([0, 1)) = ∞ X n=0 m(An) = ∞ X n=0 m(A). There is no value that can be assigned to m(A) to make sense of this. The way we deal with this is to declare that the set A is not measureable. We will only assign a value to m(A) for a class of ‘nice’ sets. 1.1.1. REMARK. Banach showed that it is possible to deﬁne a ﬁnitely additive, translation invariant function on all subsets of R. However in Rn for n ≥3, even this is not possible. The Banach-Tarski paradox shows that if A and B are two bounded subsets of R3 with interior, then there is a ﬁnite decomposition of each: A = ˙ Sn i=1Ai and B = ˙ Sn i=1Bi and rigid motions (a combination of a rotation and a translation) which carry Ai onto Bi. For example let A be the sphere of radius 1 and let B be the sphere of radius 1010. Clearly this does not preserve volume, in spite of what common sense suggests. This decomposition requires the Axiom of Choice, and can’t be accomplished by hand. 1.2 σ-Algebras 3 1.2. σ-Algebras 1.2.1. DEFINITION. If X is a set, an algebra of subsets is a non-empty col-lection of subsets of X, i.e., A ⊂P(X), which contains the empty set, ∅, and is closed under complements and ﬁnite unions. A σ-algebra is an algebra B of sub-sets of X which is closed under countable unions. We say that (X, B) is a measure space. 1.2.2 SIMPLE PROPERTIES OF σ-ALGEBRAS. Let A be an algebra of sub-sets A ⊂P(X); and let B be a σ-algebra of subsets of X. (1) A is non-empty, so suppose that E ∈A. Then Ec, ∅and X = ∅c all belong to A. (2) If E, F ∈A, then E ∪F, E ∩F = (Ec ∪F c)c, E \ F = E ∩F c and E△F = (E \ F) ∪(F \ E) belong to A. (3) If En ∈B, then T n≥1 En = S n≥1 Ec n c ∈B. (4) If En ∈B, then S n≥1 En = ˙ S n≥1Fn where Fn = En \ Sn−1 i=1 Ei are disjoint elements of B. (5) If Bλ are σ-algebras for λ ∈Λ, then B = T λ∈Λ Bλ is a σ-algebra. So if E ⊂P(X) is any collection of subsets, the intersection of all σ-algebras containing E is the unique smallest σ-algebra containing E. This is called the σ-algebra generated by E. 1.2.3. DEFINITION. If X is a topological space, the Borel sets are the elements of the σ-algebra BorX or Bor(X) generated by the collection of open subsets of X. A Gδ set is a countable intersection of open sets. An Fσ set is a countable union of closed sets. 1.2.4. DEFINITION. A measure on (X, B) is a map µ : B →[0, ∞) ∪{∞} =: [0, ∞] such that µ(∅) = 0 and is countably additive: if A = ˙ S n≥1An where An ∩Am = ∅if m ̸= n, then µ(A) = P i≥1 µ(Ai). A measure µ if ﬁnite if µ(X) < ∞and σ-ﬁnite if X = S n≥1 En such that µ(En) < ∞for n ≥1. It is called a probability measure if µ(X) = 1. A measure µ is semi-ﬁnite if for every F ∈B with µ(F) ̸= 0, there is an E ∈B with E ⊂F and 0 < µ(E) < ∞. 1.2.5 SIMPLE PROPERTIES OF MEASURES. Let µ be a measure on (X, B). (1) (monotonicity) If E, F ∈B with E ⊂F, then µ(F) = µ(E) + µ(F \ E) ≥µ(E). 4 Measures (2) (subadditivity) If En ∈B, write S n≥1 En = ˙ S n≥1Fn where Fn ⊂En. Then µ(E) = X n≥1 µ(Fn) ≤ X n≥1 µ(En). (3) (continuity from below) If En ∈B with En ⊂En+1 for n ≥1, then setting E0 = ∅, µ [ n≥1 En = µ ˙ [ n≥1En \ En−1 = X n≥1 µ(En \ En−1) = lim n→∞ n X i=1 µ(Ei \ Ei−1) = lim n→∞µ(En). (4) (continuity from above) If En ⊃En+1 for n ≥0 and µ(E0) < ∞, then µ \ n≥1 En = lim n→∞µ(En). 1.2.6 SIMPLE EXAMPLES OF MEASURES. Let µ be a measure on (X, B). (1) Counting measure on (X, P(X)) is µc(E) = ( \|E\| if E is ﬁnite ∞ if E is inﬁnite. This measure is semi-ﬁnite. It is σ-ﬁnite if and only if X is countable. (2) point mass on (X, P(X)) for x ∈X is δx(E) = ( 1 if x ∈E 0 if x ̸∈E. This is a probability measure. (3) Deﬁne µ on (X, P(X)) by µ(E) = ( 0 if E = ∅ ∞ if E ̸= ∅. This is a valid measure, but it is essentially useless. Note that it is not even semiﬁnite. 1.2.7. DEFINITION. A measure µ on (X, B) is complete if E ∈B, µ(E) = 0 and F ⊂E implies that F ∈B. 1.2.8. THEOREM. If (X, B, µ) is a measure, then B = {E ∪F : E, N ∈B, F ⊂N, µ(N) = 0} is a σ-algebra and ¯ µ(E ∪F) := µ(E) is a complete measure on (X, B) such that ¯ µ\|B = µ.. Moreover this is the smallest σ-algebra containing B on which µ extends to a complete measure. PROOF. Note that (E ∪F)c = (E ∪N)c ∪(N (E ∪F)). So B is closed under complements. Also if Ei ∪Fi ∈B and Fi ⊂Ni where Ni ∈B and µ(Ni) = 0, 1.3 σ-Algebras 5 then [ i≥1 Ei ∪Fi = [ i≥1 Ei ∪ [ i≥1 Fi = E ∪F where F = S i≥1 Fi ⊂S i≥1 Ni =: N, N ∈B and µ(N) = 0 by subadditivity. So B is a σ-algebra. Moreover this is clearly the smallest σ-algebra containing B and all subsets of null sets. Next we show that ¯ µ is well-deﬁned. Suppose that A = E1 ∪F1 = E2 ∪F2 where Ei ∈B, Fi ⊂Ni and µ(Ni) = 0. Let E = E1 ∩E2 ∈B. Then E ⊂Ei ⊂A ⊂(E1 ∪N1) ∩(E2 ∪N2) ⊂(E1 ∩E2) ∪N1 ∪N2. Therefore µ(E) ≤µ(Ei) ≤µ(E) + µ(N1) + µ(N2) = µ(E). So µ(E1) = µ(E2). Thus ¯ µ(A) = µ(Ei) is well-deﬁned. In particular, if A ∈B, then ¯ µ(A) = µ(A). So ¯ µ\|B = µ. To see that ¯ µ is countably additive, suppose that Ai = Ei ∪Fi are disjoint, and Fi ⊂Ni where Ni are null sets. Then A := ˙ S i≥1Ai = ˙ S i≥1Ei ∪ ˙ S i≥1Fi . Moreover F = S i≥1 Fi ⊂S i≥1 Ni =: N, and µ(N) = 0 by subadditivity. So ¯ µ(A) = µ ˙ [ i≥1Ei = X i≥1 µ(Ei) = X i≥1 ¯ µ(Ai). Thus ¯ µ is a measure. If E ∈B, then ¯ µ(E) = µ(E), so ¯ µ\|B = µ; i.e. ¯ µ extends the deﬁnition of µ. To see that ¯ µ is complete, suppose that ¯ µ(M) = 0 and G ⊂M. Then M = E ∪F where E, N ∈B, F ⊂N and µ(N) = 0. Also µ(E) = ¯ µ(M) = 0. Thus G ⊂M ⊂E ∪N, and µ(E ∪N) = 0. Hence G ∈B. Thus ¯ µ is a complete measure. Clearly B is the smallest σ-algebra containing B and all subsets of null sets. Moreover it is clear that the only way to extend the deﬁnition of µ to B and be a measure is to set ¯ µ(F) = 0 when F ⊂N and µ(N) = 0. Thus, this is the unique smallest complete measure extending µ. ■ 1.2.9. DEFINITION. If µ is a measure on (X, B), then (X, B, ¯ µ) is called the completion of µ. 1.2.10. DEFINITION. Let µ be a measure on (X, B). A property about points in X is true µ-almost everywhere if it is true except on a set of measure 0 (a null set). Write a.e.(µ) or just a.e. if the measure is understood. For example, the statement about f : X →R saying that f = 0 a.e.(µ) means that there is a µ-null set N so that f(x) = 0 for x ∈X \ N. It does not say that {x : f(x) ̸= 0} is measurable, only that it is contained in the set N of measure 0. However if µ is a complete measure, then all subsets of null sets are measurable. So in this case, it is true that {x : f(x) ̸= 0} is measurable. 6 Measures 1.3. Construction of Measures We need to develop some machinery so that we can deﬁne more interesting measures such as Lebesgue measure on the real line. Indeed, the construction of a family of measures related to Lebesgue measure will be our main application of this machinery at this time. 1.3.1. DEFINITION. Let X be a non-empty set. An outer measure on X is a map µ∗: P(X) →[0, ∞] such that (1) µ∗(∅) = 0. (2) (monotonicity) if A ⊂B, then µ∗(A) ≤µ∗(B). (3) (sub-additivity) if {Ai : i ≥1} is a countable collection of sets, then µ∗ S i≥1 Ai ≤P i≥1 µ∗(Ai). In other words, µ∗is a monotone, subadditive function on subsets of X. The following easy proposition shows that outer measures can easily be produced. 1.3.2. PROPOSITION. Suppose that {∅, X} ⊂E ⊂P(X) and µ : E →[0, ∞] is a function with µ(∅) = 0. For A ∈P(X), deﬁne µ∗(A) = inf X i≥1 µ(Ei) : Ei ∈E, A ⊂ [ i≥1 Ei . Then µ∗is an outer measure. PROOF. Note that µ∗(∅) = 0 by deﬁnition, and that monotonicity is imme-diate. To establish sub-additivity, suppose that {Ai : i ≥1} ⊂P(X). There is nothing to prove unless P i≥1 µ∗(Ai) < ∞. In this case, let ε > 0. For each i, ﬁnd sets Eij ∈E for j ≥1 so that Ai ⊂S j≥1 Eij and P j≥1 µ(Eij) < µ∗(Ai) + 2−iε. Then A = S i≥1 Ai is covered by S i≥1 S j≥1 Eij and so µ∗(A) ≤ X i≥1 X j≥1 µ(Eij) < X i≥1 µ∗(Ai) + 2−iε = X i≥1 µ∗(Ai) + ε. Since ε > 0 is arbitrary, this establishes the claim. So µ∗is an outer measure. ■ 1.3.3. EXAMPLE. Let X = R and let E denote the collection of all bounded open intervals. Deﬁne ρ((a, b)) = b −a. This determines an outer measure which will be used to construct Lebesgue measure. 1.3.4. DEFINITION. Let µ∗be an outer measure on X. A subset A ⊂X is µ∗-measurable if µ∗(E) = µ∗(E ∩A) + µ∗(E ∩Ac) for all E ⊂X. 1.3 Construction of Measures 7 Note that subadditivity shows that µ∗(E) ≤µ∗(E ∩A) + µ∗(E ∩Ac). So we only need to show that µ∗(E) ≥µ∗(E ∩A) + µ∗(E ∩Ac) when µ∗(E) < ∞. We are selecting those sets A for which this is always additive. The main result about outer measures is the following important result. 1.3.5. CARATH´ EODORY’S THEOREM. Let µ∗be an outer measure on X. Then the collection B of all µ∗-measurable sets is a σ-algebra, and µ = µ∗\|B is a complete measure. PROOF. It is clear from the deﬁnition that if A is measurable, then so is Ac. Suppose that A, B ∈B and let E ⊂X. Then µ∗(E) = µ∗(E ∩A) + µ∗(E ∩Ac) = µ∗(E ∩A) + µ∗(E ∩Ac ∩B) + µ∗(E ∩Ac ∩Bc) ≥µ∗(E ∩(A ∪B)) + µ∗(E ∩(A ∪B)c). The ﬁrst line follows since A is µ∗-measurable. The second line follows since B is µ∗-measurable. The last line follows from subadditivity of µ∗. This is the non-trivial direction, so this is an equality. Thus A ∪B is µ∗-measurable. Combining these two observations shows that B is an algebra of sets, and thus is closed under ﬁnite unions and intersections. Next suppose that A and B are disjoint. Then µ∗(E ∩(A ˙ ∪B)) = µ∗(E ∩(A ˙ ∪B) ∩A) + µ∗(E ∩(A ˙ ∪B) ∩Ac) = µ∗(E ∩A) + µ∗(E ∩B). By induction, we see that if A1, . . . , An are pairwise disjoint µ∗-measurable sets, then for any E ⊂X, µ∗(E ∩˙ [n i=1Ai) = n X i=1 µ∗(E ∩Ai). Now we consider countable unions. Let Ai ∈B for i ≥1. Set Bn = Sn i=1 Ai and B = S i≥1 Ai. Set A′ i = Ai\Bi−1 so that Bn = ˙ Sn i=1A′ i and B = ˙ S i≥1A′ i. We know that Bn, A′ n ∈B. Take any E ⊂X. Using the additivity from the previous paragraph, µ∗(E) = µ∗(E ∩Bn) + µ∗(E ∩Bc n) ≥µ∗(E ∩Bn) + µ∗(E ∩Bc) = n X i=1 µ∗(E ∩A′ i) + µ∗(E ∩Bc). 8 Measures This is true for all n ≥1 so we can take limits and get µ∗(E) ≥ X i≥1 µ∗(E ∩A′ i) + µ∗(E ∩Bc) ≥µ∗(E ∩B) + µ∗(E ∩Bc) ≥µ∗(E) The last two lines use subadditivity. When µ∗(E) < ∞, we obtain equality, and thus B is µ∗-measurable. Therefore B is a σ-algebra. Next suppose that the Ai are pairwise disjoint. So A′ i = Ai in the previous paragraph. Taking E = B, we get that µ(B) ≥ X i≥1 µ(Ai) ≥µ(B). Therefore µ is countably additive on B. Finally suppose that µ∗(A) = 0. Then for E ⊂X, µ∗(E) ≤µ∗(E ∩A) + µ∗(E ∩Ac) ≤µ∗(A) + µ∗(E) = µ∗(E). Hence this is an equality, showing that A is µ∗-measurable and µ(A) = 0. Any subset F ⊂A also has µ∗(F) = 0, and thus F is µ∗-measurable. Therefore (X, B, µ) is a complete measure. ■ 1.4. Premeasures 1.4.1. DEFINITION. A premeasure is a function µ : A →[0, ∞] deﬁned on an algebra A ⊂P(X) of sets such that µ(∅) = 0 and whenever Ai ∈A are pairwise disjoint and A = ˙ S i≥1Ai ∈A, then µ(A) = P i≥1 µ(Ai). In particular, premeasures are additive: µ(A1 ˙ ∪A2) = µ(A1)+µ(A2) if A1, A2 ∈A are disjoint. A premeasure is an improvement on the arbitrary function used in Proposi-tion 1.3.2 to deﬁne an outer measure. In that earlier construction, the outer measure need not reﬂect much about the original function µ. However in the case of a premeasure, the Carath´ eodory construction yields a measure that extends the pre-measure. 1.4.2. THEOREM. If µ is a premeasure on an algebra A ⊂P(X), then apply-ing Carath´ eodory’s Theorem to the outer measure µ∗yields a complete measure (X, B, ¯ µ) such that B ⊃A and ¯ µ\|A = µ. PROOF. By Proposition 1.3.2, µ∗is an outer measure. So an application of Carath´ eodory’s Theorem 1.3.5, there is a complete measure (X, B, ¯ µ) deﬁned on the σ-algebra B of µ∗-measurable sets. 1.4 Premeasures 9 Suppose that A ∈A and Ai ∈A such that A ⊂S i≥1 Ai. Deﬁne Bi = A ∩Ai \ i−1 [ j=1 Aj for i ≥1, which belong to A since it is an algebra. By construction, the Bi are pairwise disjoint and A = ˙ S i≥1Bi. Therefore, since µ is a premeasure, µ(A) = X i≥1 µ(Bi) ≤ X i≥1 µ(Ai). Therefore µ∗(A) = inf X i≥1 µ(Ai) : A ⊂ [ i≥1 Ai = µ(A). Now we show that A is µ∗-measurable. Let E ⊂X with µ∗(E) < ∞, and let ε > 0. We can ﬁnd Ai ∈A so that E ⊂S i≥1 Ai and P i≥1 µ(Ai) < µ∗(E) + ε. Notice that E ∩A ⊂S i≥1 Ai ∩A and E ∩Ac ⊂S i≥1 Ai ∩Ac. Therefore by the additivity of µ, µ∗(E ∩A) + µ∗(E ∩Ac) ≤ X i≥1 µ(Ai ∩A) + X i≥1 µ(Ai ∩Ac) = X i≥1 µ(Ai) < µ∗(E) + ε. Since ε > 0 is arbitrary, µ∗(E ∩A) + µ∗(E ∩Ac) ≤µ∗(E), which is the non-trivial inequality; so this is an equality. Hence A is µ∗-measurable. We conclude that A ⊂B and for A ∈A, ¯ µ(A) = µ∗(A) = µ(A). So ¯ µ\|A = µ. ■ Here is further detail about the outer measure construction which explains when the extension is unique. 1.4.3. PROPOSITION. Let µ be a premeasure on an algebra A ⊂P(X) and let (X, B, ¯ µ) be the measure of Theorem 1.4.2. Let ν be any measure on a σ-algebra C satisfying A ⊂C ⊂B such that ν\|A = µ. Then ν(E) ≤¯ µ(E) for all E ∈C, with equality if ¯ µ(E) < ∞. Moreover, if E is σ-ﬁnite, then ν(E) = ¯ µ(E). So if µ is σ-ﬁnite, then ¯ µ\|C is the unique extension of µ to a measure on C. PROOF. Let E ∈C. If E ⊂S i≥1 Ai for Ai ∈A, then by subadditivity, ν(E) ≤ X i≥1 ν(Ai) = X i≥1 µ(Ai). Now take the inf over all such covers of E to obtain ν(E) ≤¯ µ(E). 10 Measures If ¯ µ(E) < ∞and ε > 0, we can choose the Ai so that P i≥1 µ(Ai) < ¯ µ(E)+ε. Let Bi = Ai \ Si−1 j=1 Aj, which are in A; and A = S i≥1 Ai = ˙ S i≥1Bi. Then ν(A) = X i≥1 ν(Bi) = X i≥1 µ(Bi) = ¯ µ(A). Therefore ν(E) + ν(A \ E) = ν(A) = ¯ µ(A) = ¯ µ(E) + ¯ µ(A \ E) < ¯ µ(E) + ε. So ν(A\E) ≤¯ µ(A\E) < ε. Whence ν(E) ≥¯ µ(A)−ε ≥¯ µ(E)−2ε. Let ε →0 to get ν(E) = ¯ µ(E). Now if E = S n≥1 En where ¯ µ(En) < ∞, we may replace En with Sn i=1 Ei so that En ⊂En+1. Then by continuity from below, ν(E) = lim n→∞ν(Ei) = lim n→∞¯ µ(Ei) = ¯ µ(E). Finally if µ is σ-ﬁnite, then every set E ∈C is sigma-ﬁnite. Thus ¯ µ\|C is the unique extension of µ. ■ 1.5. Lebesgue-Stieltjes measures on R Suppose that µ is a Borel measure on R such that µ(K) < ∞if K is compact. Then we deﬁne a function F : R →R by F(x) = ( µ([0, x]) if x ≥0 −µ((x, 0)) if x < 0. Note that F is monotone increasing (i.e. non-decreasing if you prefer double neg-atives). For x ≥0 and xn+1 ≤xn with xn ↓x, the continuity from above Prop-erty 1.2.5(4) shows that F(x) = µ( \ n≥1 [0, xn]) = lim n→∞µ([0, xn]) = lim n→∞F(xn). This uses the fact that µ([0, x1]) < ∞. Similarly if x < 0 and 0 > xn ↓x, then the continuity from below property shows that F(x) = −µ( [ n≥1 (xn, 0)) = lim n→∞−µ((xn, 0)) = lim n→∞F(xn). Thus F is right continuous. The function F may fail to be left continuous. For example, for the point mass δ0, F(x) = ( 1 if x ≥0 0 if x < 0. If we knew that there was Lebesgue measure on the line, it would determine the function F(x) = x. 1.5 Lebesgue-Stieltjes measures on R 11 The Lebesgue-Stieltjes construction works in the other direction. Let F : R → R be a monotone increasing, right continuous function. We will extend the deﬁni-tion to include F(∞) = limx→∞F(x) and F(−∞) = limx→−∞F(x), and these values may include ∞or −∞respectively. Set A to be the algebra of sets consisting of all ﬁnite unions of half open in-tervals (a, b], where we allow b = ∞and a = −∞; so that (a, ∞), (−∞, b] and (−∞, ∞) belong to A. Since (a, b]c = (−∞, a]∪(b, ∞) is in A, it is easy to check that A is an algebra. Deﬁne µF ˙ [n i=1(ai, bi] = n X i=1 F(bi) −F(ai) 1.5.1. LEMMA. µF is a premeasure. PROOF. First let’s show that µF is well deﬁned. To this end, suppose that I = (a, b] = ˙ Sn i=1(ai, bi]. Then after rearrranging if necessary, we may suppose that a = a1 < b1 = a2 < · · · < bn−1 = an < bn = b. Then if we set an+1 := bn = b, n X i=1 F(bi) −F(ai) = n X i=1 F(ai+1) −F(ai) = F(b) −F(a). Thus µF (I) does not depend on the decomposition into ﬁnitely many pieces. This readily extends to a ﬁnite union of intervals. Now we consider the restricted version of countable additivity. Suppose that I = (a, b] = ˙ S i≥1(ai, bi]. This is more difﬁcult to deal with because these intervals cannot be ordered, end to end, as in the case of a ﬁnite union. One direction is easy: since I = ˙ Sn i=1(ai, bi] ˙ ∪ I \ Sn i=1(ai, bi] and the last set belongs to A, µF (I) = n X i=1 µF ((ai, bi]) + µF I \ n [ i=1 (ai, bi] ≥ n X i=1 µF ((ai, bi]). Taking a limit as n →∞yields that µF (I) ≥P∞ i=1 µF ((ai, bi]). Conversely, ﬁrst suppose that a, b ∈R. Fix ε > 0. By right continuity, there is a δ > 0 so that F(a + δ) < F(a) + ε. Likewise, there are δi > 0 so that F(bi + δi) < F(bi) + 2−iε. The compact interval [a + δ, b] has the open cover {(ai, bi +δi) : i ≥1}; so there is a ﬁnite subcover (ai1, bi1+δi1), ..., (aip, bip+δip). 12 Measures Therefore Pp j=1 F(bij +δij) −F(aij) ≥F(b) −F(a+δ). Consequently, X i≥1 µF ((ai, bi]) = X i≥1 F(bi) −F(ai) ≥ p X j=1 F(bij) −F(aij) ≥ p X j=1 F(bij + δij) −2−ijε −F(aij) ≥F(b) −F(a + δ) −ε ≥F(b) −F(a) −2ε. Now let ε ↓0 to get P i≥1 µF ((ai, bi]) ≥F(b) −F(a); whence we have equality. Now if b = ∞, we still have P i≥1 µF ((ai, bi]) ≥µF ((a, N]) = F(N) −F(a) for N ∈N. Letting N →∞, we get P i≥1 µF ((ai, bi]) ≥F(∞) −F(a). Similarly we can handle a = −∞. Finally if A = ˙ Sm k=1(ak, bk] is written as a disjoint union of half open intervals, we can split the union into m pieces and use the argument for a single interval on each one. Thus we obtain countable additivity (provided the union remains in A). So µF is a premeasure. ■ 1.5.2. THEOREM. If F : R →R is an increasing, right continuous function, then there is a complete measure (R, B, ¯ µF ) that extends µF . The σ-algebra B contains the σ-algebra BorR of Borel sets. The restriction of ¯ µF to BorR is the unique Borel measure µF on R such that µF ((a, b]) = F(b) −F(a) for all a < b. Conversely, if µ is a Borel measure on R such that µ(K) < ∞for compact sets K ⊂R, then there is an increasing, right continuous function F : R →R such that µ = µF . Also given two increasing, right continuous functions F, G, µF = µG if and only if F −G is constant. PROOF. By Lemma 1.5.1, µF is a premeasure. Thus by Theorem 1.4.2, there is a complete measure ¯ µF on a σ-algebra B containing A which extends µF . The σ-algebra B contains the σ-algebra generated by A. So it contains all intervals (a, b) = S n≥1(a, b−1 n]. Hence it contains all open sets (because every open subset of R is the countable union of intervals), and thus all Borel sets. The restriction µF of ¯ µF to BorR is a Borel measure with µF ((a, b]) = F(b) −F(a) for all a < b. Since µF is σ-ﬁnite, Proposition 1.4.3 shows that this Borel measure is unique. Conversely, we showed that every Borel measure which is ﬁnite on bounded intervals determines an increasing, right continuous function F : R →R such that µF ((a, b]) = F(b) −F(a) for all a < b. By the uniqueness of the construction, we see that µ = µF . Finally, if µG = µF , then µG((a, b]) = G(b) −G(a) = F(b) −F(a) for a < b. 1.5 Lebesgue-Stieltjes measures on R 13 It follows that G(x) = F(x) + (G(0) −F(0)); i.e., F −G is constant. ■ 1.5.3. COROLLARY. Lebesgue measure is the complete measure m = ¯ µF for the function F(x) = x. The σ-algebra L of Lebesgue measurable sets contains all Borel sets, and m((a, b)) = b −a for all a < b in R. Lebesgue measure has some special properties. If E ⊂R and s ∈R, let E + s = {x + s : x ∈E} and sE = {sx : x ∈E}. 1.5.4. THEOREM. Lebesgue measure is translation invariant: m(E + s) = m(E) for all Lebesgue measurable sets E ⊂R. Also m(sE) = \|s\|m(E). PROOF. The set of open intervals is invariant under translation, and hence so is BorR. The measure ms(E) = m(E + s) agrees with m on open intervals. By Theorem 1.5.2, the measures are determined by the functions F(x) = ( m((0, x]) x ≥0 −m((x, 0]) x < 0 and G(x) = ( ms((0, x]) x ≥0 −ms((x, 0]) x < 0 . However as we have observed, these functions are equal. So ms = m. Hence m(E + s) = m(E) for all Borel sets E ⊂R. Now Proposition 1.4.3 shows that there is a unique extension of m\|BorR to the σ-algebra L of Lebesgue measurable sets. If s = 0, then m(0E) = 0 = 0m(E), so suppose that s ̸= 0. Let ns(E) = \|s\|−1m(sE). Observe that ns((a, b)) = \|s\|−1\|sb −sa\| = b −a = m((a, b)). Arguing as in the previous paragraph, we see that ns = m. Hence m(sE) = \|s\|ns(E) = \|s\|m(E) for all measurable sets E. ■ 1.5.5. REMARK. A point {a} has Lebesgue-Stieltjes measure µF ({a}) = lim n→∞µF ((a−1 n, a]) = F(a) −lim x→a−F(x) = F(a) −F(a−). Thus µF (a) > 0 if and only if F has a jump discontinuity at a. In this case, we say that a is an atom of µF . The only discontinuities of a monotone function are jump discontinuities. There are at most countably many jump discontinuities. To see this, consider how many jumps of size at least δ there can be in (a, b]; at most (F(b) −F(a))/δ which is ﬁnite. So there are at most n(F(n) −F(−n)) points in (−n, n] with a jump of more than 1 n. The union of this countable collection of ﬁnite sets is countable, and contains all of the jumps. Note, however that the discontinuities can be dense, say at every rational point. Suppose that the discontinuities occur at ai with jump αi > 0 for i ≥1. Recall that δx is the point mass that sets δ(A) = 1 if x ∈A and δx(A) = 0 otherwise. 14 Measures Then µa = P i≥1 αiδai is called an atomic measure. When we subtract it from µF , we get a measure µc = µF −µa which has no atomic part. There are increasing, right continuous functions Fa and Fc so that µa = µFa and µc = µFc. Moreover Fc is continuous and Fa is entirely determined by its jumps. We can show that (up to a constant) Fa(x) = ( P {i:0≤ai≤x} αi if x ≥0 −P {i:x<ai<0} αi if x < 0 . This converges for every x because P {i:0≤ai≤x} αi ≤F(x) −F(0) < ∞when x ≥0, and similarly P {i:x<ai<0} αi ≤F(0) −F(x) when x < 0. We conclude this section with some regularity properties of ¯ µF -measurable sets. 1.5.6. THEOREM. Let ¯ µF be a Lebesgue-Stieltjes measure. For E ⊂R, the following are equivalent. (1) E is ¯ µF -measurable. (2) For all ε > 0, there is an open set U ⊃E such that µ∗ F (U \ E) < ε. (3) For all ε > 0, there is an closed set C ⊂E such that µ∗ F (E \ C) < ε. (4) There is a Gδ set G ⊃E so that µ∗ F (G \ E) = 0. (5) There is an Fσ set F ⊂E so that µ∗ F (E \ F) = 0. In this case, ¯ µF (E) = inf{µF (U) : E ⊂U open} = sup{µF (K) : E ⊃K compact}. PROOF. First assume that E is bounded. Fix ε > 0. Since E is measurable, ¯ µF (E) = µ∗ F (E) = inf µF (A) where E ⊂A = S i≥1(ai, bi]. Now ¯ µF (E) < ∞ so we can choose A so that µF (A) < ¯ µF (E)+ε/2. For each i ≥1, choose ci > bi so that F(ci) < F(bi) + 2−i−1ε. Let U = S i≥1(ai, ci). Since E is measurable, µ∗ F (A) = µ∗ F (E) + µ∗ F (A \ E). Thus µ∗ F (A \ E) < ε/2. Also µF (U \ A) = µF [ i≥1 (bi, ci) ≤ X i≥1 F(ci) −F(bi) < ε/2. So ¯ µ∗ F (U \ E) = µF (U \ A) + µ∗ F (A \ E) < ε. Now if E is unbounded, let En = E ∩(n −1, n] for n ∈Z. Find an open set Un ⊃En with µ∗ F (Un \ En) < 2−2\|n\|−1ε. Then E ⊂U := S n∈Z Un is an open set, and µ∗ F (U \ E) ≤P n∈Z µ∗ F (Un \ En) < ε 2 + 2 P n≥1 2−2n−1ε < ε. So (1) = ⇒(2). (2) = ⇒(4). Find open sets Un ⊃E with µ∗ F (Un\E) < 1 n. Set G = T n≥1 Un. This is a Gδ set containing E such that µ∗ F (G\E) ≤µ∗ F (Un\E) < 1 n for all n ≥1. Hence µ∗ F (G \ E) = 0. 1.5 Lebesgue-Stieltjes measures on R 15 (4) = ⇒(1). Since ¯ µF is complete and µ∗ F (G \ E) = 0, the set G \ E is ¯ µF -measurable. Since µF is a Borel measure, G is also ¯ µF -measurable. Therefore E = G \ (G \ E) is ¯ µF -measurable. (1) = ⇒(3). If E is measurable, so is Ec. By the equivalence of (1) and (2), there is an open set U ⊃Ec such that µ∗ F (U \ Ec) < ε. Let C = U c, which is closed and C ⊂E. Then µ∗ F (E \ C) = µ∗ F (U \ Ec) < ε. (3) = ⇒(5). Select closed sets Cn ⊂E with µ∗ F (E \ Cn) < 1 n. Let F = S n≥1 Cn. This is an Fσ set contained in E such that µ∗ F (E\F) < µ∗ F (E\Cn) < 1 n for all n ≥1. Hence µ∗ F (E \ F) = 0. (5) = ⇒(1). Since ¯ µF is complete and µ∗ F (E \ F) = 0, the set E \ F is ¯ µF -measurable. Since µF is a Borel measure, F is also ¯ µF -measurable. Therefore E = F ∪(E \ F) is ¯ µF -measurable. For the ﬁnal statement, the ﬁrst statement holds by (2), and the second holds by (4) provided that we use closed sets. If E is bounded, the closed sets are compact. So suppose that E is unbounded. Set En = E ∩[−n, n]. Then ¯ µF (E) = sup n≥1 ¯ µF (En) = sup n≥1 sup En⊃K compact µF (K). ■ CHAPTER 2 Functions 2.1. Measurable Functions 2.1.1. DEFINITION. Let (X, A) and (Y, B) be measure spaces. A function f : X →Y is measurable if f−1(B) ∈A for all B ∈B. In particular, f : X →F ∈{R, C} is measurable if f−1(B) ∈A for all B ∈BorF; i.e. for all Borel sets B ⊂F. We are most often interested in scalar valued functions, i.e. with range in R or C. We always use the σ-algebra of Borel sets, not Lebesgue measurable sets, when deﬁning measurable functions. It is not necessary to verify the measurability condition on all Borel sets, only on a generating family. That is, if E ⊂B generates B as a σ-algebra and f−1(E) ∈ A for all E ∈E, then f is measurable. This follows from the easy facts: f−1(Ec) = (f−1(E))c and f−1( [ n≥1 En) = [ n≥1 f−1(En). So the following proposition is immediate. 2.1.2. PROPOSITION. Let (X, A) and (Y, B) be measure spaces. (a) If f : X →Y , then {B ∈B : f−1(B) ∈A} is a σ-algebra. (b) If f : X →R, the following are equivalent: (1) f is measurable. (2) {x ∈X : f(x) < α} is measurable for all α ∈R. (3) {x ∈X : f(x) ≤α} is measurable for all α ∈R. (4) {x ∈X : f(x) > α} is measurable for all α ∈R. (5) {x ∈X : f(x) ≥α} is measurable for all α ∈R. (c) If f : X →C, then f is measurable if and only if Re f and Im f are measurable R-valued functions. 2.1.3. COROLLARY. Let X be a topological space, and consider (X, BorX). Continuous functions f : X →F are (Borel) measurable. 16 2.1 Measurable Functions 17 PROOF. Since f is continuous, if U ⊂F is open, then f−1(U) is open and thus Borel. ■ The following records some basic operations that preserve measurability. 2.1.4. PROPOSITION. Let (X, B) be a measure space. (a) Suppose that f, g : X →F are measurable. Then f + λg for λ ∈F and fg are measurable; and if g ̸= 0, then f/g is measurable. (b) If fn : X →R are measurable, then sup fn, inf fn, lim sup fn and lim inf fn are measurable. Thus, if f = lim fn exists pointwise, then f is measurable. PROOF. Clearly if λ ̸= 0, then g is measurable if and only if λg is measurable. So consider f + g. Note that {x : f(x) + g(x) < α} = [ r∈Q {x : f(x) < r} ∩{x : g(x) < α −r} ∈B. So f + g is measurable. Now consider fg. If α > 0, then {x : f(x)g(x) < α} = {x : f(x) ≥0} ∩{x : g(x) ≤0} ∪ {x : f(x) ≤0} ∩{x : g(x) ≥0} ∪ [ r∈Q+ {x : \|f(x)\| < r} ∩{x : \|g(x)\| < α/r} This lies in B. Also {x : f(x)g(x) < 0} = {x : f(x) > 0} ∩{x : g(x) < 0} ∪ {x : f(x) < 0} ∩{x : g(x) > 0} And ﬁnally, if α < 0, then {x : f(x)g(x) < α} = [ r∈Q+ {x : f(x) > r} ∩{x : g(x) < α/r} ∪ [ r∈Q+ {x : f(x) < α/r} ∩{x : g(x) > r} Thus fg is measurable. Finally, if g ̸= 0, then for α > 0, {x : 1/g(x) < α} = {x : g(x) > α−1} ∪{x : g(x) < 0}; also {x : 1/g(x) < 0} = {x : g(x) < 0}; and if α < 0, {x : 1/g(x) < α} = {x : α−1 < g(x) < 0}. So 1/g is measurable. Thus f/g is measurable by the product result. Now suppose that fn are measurable for n ≥1. Then {x : sup fn(x) > α} = [ n≥1 {x : fn(x) > α} ∈B. 18 Functions So sup fn is measurable. Similarly inf fn is measurable. Therefore lim sup fn = infk≥1 supn≥k fn is measurable, and similarly for lim inf fn. So f = lim fn exists pointwise means that f = lim sup fn = lim inf fn. So f is measurable. ■ When the measure is complete, changing things on a set of measure 0 has no important consequence. 2.1.5. PROPOSITION. Let µ be a complete measure on (X, B). Then if f is measurable and g = f a.e.(µ), then g is measurable. Also if fn are measurable and f is a function such that fn converge to f a.e.(µ), then f is measurable. PROOF. Suppose that N is a null set such that f = g on X \ N. For any Borel set B ⊂R, f−1(B) △g−1(B) ⊂N. As these two sets differ by a subset of a null set, they differ by a null set by completeness. So one is measurable because the other is. Suppose that f(x) = lim fn(x) except on a null set N. Then f = lim sup fn except on N, and hence is measurable by the previous paragraph. ■ 2.2. Simple Functions 2.2.1. DEFINITION. Let (X, B) be a measure space. A simple function is a function ϕ : X →C of the form ϕ(x) = n X i=1 aiχEi where ai ∈C∗, Ei ∈B and Ei ∩Ej = ∅for i ̸= j. A simple function is just a measurable function with ﬁnite range. Indeed, the range of ϕ is contained in {ai : 1 ≤i ≤n} ∪{0}. Conversely, if the range of a measurable function ϕ is contained in {ai : 1 ≤i ≤n}∪{0} with ai ̸= aj ̸= 0 for i ̸= j, then we can set Ei = ϕ−1({ai}). These are disjoint, and ϕ = Pn i=1 aiχEi. This is called the standard form of a simple function. Note that the set of all simple functions is an algebra, meaning that it is a vector space and is closed under products. The subset of simple functions such that µ(Ei) < ∞for all i is a subalgebra. The main result of this section is that every measurable function is a limit of simple functions in a nice way. 2.2.2. PROPOSITION. (a) If f : X →[0, ∞] is measurable, then there is a sequence of simple functions ϕn so that ϕn ≤ϕn+1, limn→∞ϕn(x) = f(x), and convergence is uniform on {x : f(x) ≤R} for any R ∈R. 2.3 Two Theorems about Measurable Functions 19 (b) If f : X →C is measurable, then there is a sequence of simple functions ϕn so that \|ϕn\| ≤\|ϕn+1\|, limn→∞ϕn(x) = f(x), and convergence is uniform on {x : \|f(x)\| ≤R} for any R ∈R. PROOF. (a) For n ≥1, let Ak,n = x : k 2n ≤f(x) < k+1 2n for 0 ≤k < 4n and A4n,n = {x : f(x) ≥2n}. Deﬁne simple functions ϕn = 4n X k=1 k 2n χAk,n. Observe that Ak,n = A2k,n+1 ˙ ∪A2k+1,n+1 for 0 ≤k < 4n and A4n,n = ˙ [4n+1 k=2·4nAk,n+1. Moreover for 0 ≤k < 4n, k 2n χAk,n ≤ 2k 2n+1 χA2k,n+1 + 2k+1 2n+1 χA2k+1,n+1 and 2nχA4n,n ≤ 4n+1 X k=2·4n k 2n+1 χAk,n+1. Thus ϕn ≤ϕn+1 ≤f. Moreover, if f(x) ≤2N, then f(x) −ϕn(x) ≤2−n for all n ≥N; and if f(x) = ∞, then ϕn(x) = 2n. Thus limn→∞ϕn = f point-wise. Moreover convergence is uniform on {x : \|f(x)\| ≤R} since our estimate is uniformly good once 2n ≥R. (b) Let f = g1−g2+ig3−ig4 where g1 = max{Re f, 0}, g2 = max{−Re f, 0}, g3 = max{Im f, 0} and g4 = max{−Im f, 0}. For each 1 ≤i ≤4, apply part (a) to get sequences of simple functions ψi,n increasing to gi. Then the sequence ϕn = ψ1,n −ψ2,n + iψ3,n −iψ4,n works. Details are left to the reader. ■ 2.3. Two Theorems about Measurable Functions Littlewood had three principles of Lebesgue measure: • every measurable set of ﬁnite mesure is almost a ﬁnite union of intervals (use Theorem 1.5.6(2) and throw out the very small intervals). • Every measurable function is almost continuous (Lusin’s Theorem). • A pointwise convergence sequence of measurable functions is almost uni-formly convergent (Egorov’s Theorem). 2.3.1. EGOROV’S THEOREM. Let (X, B, µ) be a ﬁnite measure space, i.e., µ(X) < ∞. Suppose that fn : X →C are measurable functions and fn →f a.e.(µ). Then for ε > 0, there is a set E ∈B with µ(X \ E) < ε so that fn →f uniformly on E. 20 Functions PROOF. Observe that fn →f uniformly on E provided that for each m ≥1, there is an integer Nm so that \|fn(x) −f(x)\| < 1 m for all x ∈E and all n ≥Nm. Deﬁne Am,N = {x : \|fn(x) −f(x)\| < 1 m ∀n ≥N} = \ n≥N {x : \|fn(x) −f(x)\| < 1 m}. Note that Am,1 ⊂Am,2 ⊂. . . and that [ n≥1 Am,n ⊃{x : fn(x) →f(x)} = X \ N where N is a null set. Since X has ﬁnite measure, we can choose an integer Nm so that µ(Am,Nm) > µ(X) −2−mε. Hence µ(Ac m,Nm) < 2−mε. Let E = T m≥1 Am,Nm. Then, µ(Ec) = µ [ m≥1 Ac m,Nm ≤ X m≥1 µ(Ac m,Nm) < X m≥1 2−mε = ε. By the ﬁrst line of the proof, fn converges uniformly to f on E. ■ 2.3.2. LUSIN’S THEOREM. Let f : [a, b] →C be a Lebesgue measurable function, and let ε > 0. Then there is a continuous function g ∈C[a, b] so that m({x : f(x) ̸= g(x)}) < ε. PROOF. If ϕ = Pm i=1 aiχEi is a simple function and δ > 0, we can ﬁnd compact sets Ai ⊂Ei so that m Sm i=1(Ei \ Ai) < δ. Observe that K = S Ai is compact and ϕ\|K is continuous as a function on K (because it is locally constant). Choose simple functions ϕn converging pointwise to f. Find compact sets Kn so that ϕn\|Kn is continuous and m [a, b] \ Kn < 2−n−1ε. Then K0 = T Kn is compact, each ϕn\|K0 is continuous, and m [a, b] \ K0 ≤ X m [a, b] \ Kn < ε/2. By Egorov’s Theorem, there is a measurable set E ⊂K0 with m(K0 \ E) < ε/4 so that so that ϕn converge uniformly to f on E. Then by Theorem 1.5.6, there is a closed set K ⊂E with m(E \ K) < ε/4. Since ϕn\|K are continuous, and converge uniformly to f\|K, we see that f\|K is continuous. Also m [a, b] \ K < m [a, b] \ K0 + m(K0 \ E) + m(E \ K) < ε 2 + ε 4 + ε 4 = ε. It remains to extend f\|K to a continuous function g on [a, b]. This is a (non-trivial) exercise in a basic real analysis course. Just make g piecewise linear on each open component of the complement. Once achieved, {x : f(x) ̸= g(x)} is contained in [a, b] \ K, thus has measure less than ε. ■ CHAPTER 3 Integration The goal of this chapter is to develop a theory of integration with respect to an arbitrary measure. 3.1. Integrating positive functions We begin by considering the integral only of positive functions. Moreover, we insist that the function be measurable. We ﬁx a measure space (X, B, µ). Let L+ = {f : X →[0, ∞] measurable}. Notice that L+ only depends on the measure space (X, B) and not on µ. The simple functions have a very natural deﬁnition of integral. We bootstrap that into a deﬁnition for positive measurable functions. Remember that simple functions only take ﬁnite values. 3.1.1. DEFINITION. If ϕ = Pn i=1 aiχEi is a simple function in L+, set Z ϕ dµ := n X i=1 aiµ(Ei). We use the convention that 0 · ∞= 0 = ∞· 0. For general f ∈L+, deﬁne Z f dµ := sup n Z ϕ dµ : 0 ≤ϕ ≤f, ϕ simple o . Also deﬁne Z A f dµ := Z fχA dµ for any measurable set A ∈B. The convention that 0 · ∞= 0 means that f = 0χX has Z f dµ = 0 even if µ(X) = ∞; since the integral of a simple function should not be changed by adding a zero function to it. The second identity ∞· 0 = 0 is also needed to coincide with the notion that the value of f on a set of measure 0 should not affect the integral. So Z ∞χ{x} dµ = 0 is µ(x) = 0. At this stage, we allow ∞both as a value of f and of the integral. Later, in some cases, we impose further restrictions. 21 22 Integration In various books, you will see the following notations which all mean the same thing: Z f dµ, Z f(x) dµ(x), Z f(x) µ(dx), Z f. Of course, the last one will only make sense if the measure µ is known. We will use this when convenient to simplify the notation. 3.1.2. LEMMA. Let ϕ, ψ be simple functions in L+; and let f, g ∈L+. (a) R ϕ dµ is well deﬁned. (b) If c ≥0, then R cϕ dµ = c R ϕ dµ. (b′) R cf dµ = c R f dµ. (c) R ϕ + ψ dµ = R ϕ dµ + R ψ dµ. (d) If ϕ ≤ψ, then R ϕ dµ ≤ R ψ dµ. (d′) If f ≤g, R f dµ ≤ R g dµ. (e) ν(A) := R A ϕ dµ for A ∈B deﬁnes a measure on (X, B). PROOF. (a) Suppose that ϕ = Pm i=1 aiχEi = Pn j=1 bjχFj. We can disregard zeros, so we may suppose that ai ̸= 0 ̸= bj. Moreover since the Ei and Fj are pairwise disjoint collections, we have that {ai : 1 ≤i ≤m} = {bj : 1 ≤j ≤n}. Let c1, . . . , cp be the distinct non-zero values of ϕ, and set Ak = {x : ϕ(x) = ck} for 1 ≤k ≤p. Then Ak = [ {Ei : ai = ck} = [ {Fj : bj = ck}. Therefore n X i=1 aiµ(Ei) = p X k=1 ck X ai=ck µ(Ei) = p X k=1 ckµ [ ai=ck Ei = p X k=1 ckµ(Ak) = p X k=1 ckµ [ bj=ck Fj = n X j=1 bjµ(Fj). (b) is trivial, and (b′) follows because {ϕ : ϕ ∈L+, simple, ϕ ≤cf} = {cϕ : ϕ ∈L+, simple, ϕ ≤f}. For (c), suppose that ϕ = Pm i=0 aiχEi and ψ = Pm j=0 bjχFj, where we add in the term a0 = b0 = 0 and E0 = Sm i=1 Ei c and F0 = Sn j=1 Fj c in order that X = Sm i=0 Ei = Sn j=0 Fj. Deﬁne Aij = Ei ∩Fj for 0 ≤i ≤m and 0 ≤j ≤n. Then ϕ = m X i=0 n X j=0 aiχAij, ψ = m X i=0 n X j=0 biχAij and ϕ + ψ = m X i=0 n X ij=0 (ai + bj)χAij. 3.2 Two limit theorems 23 Therefore using (a), we get Z ϕ + ψ dµ = m X i=0 n X j=0 (ai + bj)µ(Aij) = m X i=0 ai n X j=0 µ(Aij) + n X j=0 bj m X i=0 µ(Aij) = m X i=0 aiµ n [ j=0 Aij + n X j=0 bjµ m [ i=0 Aij = m X i=0 aiµ(Ei) + n X j=0 bjµ(Fj) = Z ϕ dµ + Z ψ dµ. (d) Write ϕ and ψ as in (c), and note that ϕ = m X i=0 n X j=0 aiχAij ≤ψ = m X i=01 n X j=0 biχAij means that ai ≤bj if Aij ̸= ∅. Therefore by (a), Z ϕ dµ = m X i=0 n X j=0 aiµ(Aij) ≤ m X i=0 n X j=0 bjµ(Aij) = Z ψ dµ. (d′) follows because {ϕ : ϕ ∈L+, simple ϕ ≤f} ⊂{ϕ : ϕ ∈L+, simple ϕ ≤g}. (e) Clearly ν(∅) = 0 and ν takes values in [0, ∞]. Suppose that Aj are pairwise disjoint sets in B with A = ˙ S j≥1Aj. Then ν(A) = Z A ϕ dµ = n X i=1 aiµ(Ei ∩A) = n X i=1 aiµ ˙ [ j≥1Ei ∩Aj = n X i=1 ai X j≥1 µ(Ei ∩Aj) = X j≥1 n X i=1 aiµ(Ei ∩Aj) = X j≥1 Z Aj ϕ dµ = X j≥1 ν(Aj). Thus ν is countably additive, and hence is a measure. ■ 3.2. Two limit theorems This section establishes two fundamental limit theorems, and extends the addi-tivity of the integral to arbitrary functions in L+. 24 Integration 3.2.1. LEMMA. Let ϕ ≥0 be a simple function,. Let An ∈B with An ⊂An+1 and S n≥1 An = X. Then lim n→∞ Z An ϕ dµ = Z ϕ dµ. PROOF. Let ν(A) = Z A ϕ dµ be the measure deﬁned in Lemma 3.1.2(e). Then lim n→∞ Z An ϕ dµ = lim n→∞ν(An) = ν(X) = Z ϕ dµ by the property of continuity from below. ■ 3.2.2. MONOTONE CONVERGENCE THEOREM. Let (X, B, µ) be a mea-sure space. Let fn ∈L+ such that fn ≤fn+1 for n ≥1. Deﬁne f(x) = limn→∞fn(x). Then Z f dµ = lim n→∞ Z fn dµ. PROOF. Since fn(x) is increasing, the limit always exists in [0, ∞]. Also since fn ≤fn+1 ≤f, we have that Z fn dµ ≤ Z fn+1 dµ ≤ Z f dµ. Thus limn→∞ R fn dµ exists in [0, ∞] and is no larger than R f dµ. Conversely, suppose that ϕ ≤f is a non-negative simple function, and let ε > 0. Deﬁne An = {x : fn(x) ≥(1 −ε)ϕ(x)}. Since ϕ(x) < ∞and fn(x) →f(x) ≥ϕ(x), we have An ⊂An+1 and S n≥1 An = X. Therefore by Lemma 3.2.1, (1−ε) Z ϕ dµ = lim n→∞ Z An (1−ε)ϕ dµ ≤lim n→∞ Z An fn dµ ≤lim n→∞ Z fn dµ. Let ε ↓0 to get that R ϕ dµ ≤limn→∞ R fn dµ. Taking the supremum yields R f dµ ≤limn→∞ R fn dµ. So equality holds. ■ 3.2.3. COROLLARY. The integral is countably additive on L+. That is, if fn ∈ L+ for n ≥1, then Z X n≥1 fn dµ = X n≥1 Z fn dµ. PROOF. Apply MCT to the sequence gk = Pk n=1 fn. ■ 3.2.4. COROLLARY. If f ∈L+, ν(A) = R A f dµ is a measure on (X, B). 3.2 Two limit theorems 25 PROOF. Countable additivity is a special case of the previous corollary; since if A = ˙ S n≥1An, then fχA = P n≥1 fχAn. ■ 3.2.5. LEMMA. If f ∈L+, then R f dµ = 0 if and only if f = 0 a.e.(µ). PROOF. If f = 0 a.e.(µ) and 0 ≤ϕ ≤f where ϕ = Pn i=1 aiχEi is a simple function, then ai > 0 implies µ(Ei) = 0. Therefore R ϕ dµ = Pn i=1 aiµ(Ei) = 0. Hence R f dµ = sup0≤ϕ≤f R ϕ dµ = 0. Conversely, suppose that R f dµ = 0. Let E = {x : f(x) > 0} and En = {x : f(x) ≥1 n}. Then ϕn = 1 nχEn ≤f. So 0 = R f dµ ≥ R ϕn dµ = 1 nµ(En). Therefore µ(E) = limn→∞µ(En) = 0; i.e. f = 0 a.e.(µ). ■ 3.2.6. COROLLARY. If f, fn ∈L+ such that fn ≤fn+1 a.e.(µ) for n ≥1 and f(x) = limn→∞fn(x) a.e.(µ), then Z f dµ = lim n→∞ Z fn dµ. PROOF. Deﬁne En = x : fn(x) > fn+1(x) for n ≥1 and E0 = x : lim n→∞fn(x) ̸= f(x) . Then µ(En) = 0 for n ≥0 by hypothesis. Let E = S n≥0 En. Then µ(E) = 0, fnχEc ≤fn+1χEc and fχEc = limn→∞fnχEc. So by the MCT and Lemma 3.2.5, Z f dµ = Z fχEc dµ = lim n→∞ Z fnχEc dµ = lim n→∞ Z fn dµ. ■ 3.2.7. EXAMPLE. Things don’t work out quite as well for limits which are not monotone. For example, let fn = 1 nχ[0,n] ∈L+(m). Then fn →0 uniformly on R. However, lim n→∞ Z fn dm = lim n→∞1 = 1 ̸= 0 = Z 0 dm. The following theorem provides an important inequality. 3.2.8. FATOU’S LEMMA. Let (X, B, µ) be a measure space. Suppose fn ∈L+ for n ≥1. Then Z lim inf fn dµ ≤lim inf Z fn dµ. PROOF. Let gn(x) = infk≥n fk(x) ≤fn(x). Then gn ≤gn+1 for n ≥1, and lim inf fn = limn→∞gn. Therefore by MCT, 26 Integration Z lim inf fn dµ = lim n→∞ Z gn dµ = lim inf Z gn dµ ≤lim inf Z fn dµ. ■ 3.3. Integrating complex valued functions 3.3.1. DEFINITION. Let (X, B, µ) be a measure space. A measurable function f : X →C is integrable if ∥f∥1 = Z \|f\| dµ < ∞. The set of all integrable functions will be (provisionally) called L1(µ). Write f = f1 −f2 + if3 −if4 where f1 = Re f ∨0 = max{Re f, 0}, f2 = (−Re f) ∨0, f3 = Im f ∨0 and f3 = (−Im f) ∨0. If f is integrable, deﬁne Z f dµ = Z f1 dµ − Z f2 dµ + i Z f3 dµ −i Z f4 dµ. Note that since fi ∈L+ and fi ≤\|f\|, we have R fi dµ ≤ R \|f\| dµ < ∞. So R f dµ is deﬁned as a complex number. The reason for the temporary nature of our deﬁnition of L1(µ) is that this is not a normed vector space, due to the fact that if f = 0 a.e.(µ), then ∥f∥1 = 0. So ∥· ∥1 is just a pseudo-norm. That is, it satisﬁes ∥λf∥1 = \|λ\| ∥f∥1 and the triangle inequality, but is not positive deﬁnite. We will rectify this soon by identifying functions which agree a.e. into an equivalence class representing an element of L1(µ). 3.3.2. PROPOSITION. L1(µ) is a vector space and ∥· ∥1 is a pseudo-norm. That is, it is positive homogeneous: ∥λf∥1 = \|λ\| ∥f∥1 for λ ∈C, and the triangle inequality holds, but ∥f∥1 = 0 if and only if f = 0 a.e.(µ). The map R taking f to R f dµ is linear, and Z f dµ ≤ Z \|f\| dµ = ∥f∥1. PROOF. Showing that L1(µ) is a vector space is straightforward. Multipli-cation by a real number or an imaginary number merely scales and shufﬂes the functions fi, so homogeneity for these scalars is easy. In general, you need to write out λ = a + ib and chase some details which are left to the reader. Lemma 3.2.5 shows that ∥f∥1 = 0 if and only if f = 0 a.e.(µ). The triangle inequality is also straightforward: if f, g ∈L1(µ), then ∥f + g∥1 = Z \|f + g\| dµ ≤ Z \|f\| + \|g\| dµ = ∥f∥1 + ∥g∥1. 3.3 Integrating complex valued functions 27 Linearity follows from linearity of the integral in L+, and is left to the reader. Let f ∈L1(µ) and choose θ so that R f = eiθ R f . Then Z f = Z e−iθf = Re Z e−iθf = Z Re e−iθf. Write Re e−iθf = g1 −g2, where g1 = Re e−iθf ∨0 and g2 = −(Re e−iθf) ∨0. Then 0 ≤ Z Re e−iθf = Z g1 −g2 ≤ Z g1 + g2 ≤ Z \|f\|. Combining these two formulae, we get the desired inequality. ■ We now arrive at the most important limit theorem in measure theory. With this result and MCT, you can deal with most situations that arise. 3.3.3. LEBESGUE DOMINATED CONVERGENCE THEOREM. Suppose that fn, g ∈L1(µ), g ≥0, such that lim n→∞fn = f a.e.(µ) and \|fn\| ≤g a.e.(µ) for n ≥1. Then f ∈L1(µ) and Z f dµ = lim n→∞ Z fn dµ. PROOF. First assume that fn are real valued. Then we apply Fatou’s Lemma 3.2.8 to the sequences g ± fn in L+ to get Z g + f dµ ≤lim inf Z g + fn dµ = Z g dµ + lim inf Z fn dµ and Z g −f dµ ≤lim inf Z g −fn dµ = Z g dµ −lim sup Z fn dµ. Since R g dµ < ∞, it can be cancelled off and we obtain that lim sup Z fn dµ ≤ Z f dµ ≤lim inf Z fn dµ. It follows that R f dµ = limn→∞ R fn dµ. In particular, applying this to \|f\| and \|fn\| we obtain that ∥f∥1 = Z \|f\| dµ = lim n→∞ Z \|fn\| dµ ≤ Z g dµ < ∞. So f ∈L1(µ) and thus is integrable. In general, split f as the sum of its real and imaginary parts, use the real result, and then recombine. ■ 28 Integration 3.3.4. EXAMPLE. Consider lim n→∞ Z b a n 1 + n2x2 dx for 0 ≤a < b ≤∞. Observe that fn(x) = n 1+n2x2 = 1 1 n +nx2 . So lim n→∞fn(x) = 0 for all x > 0. The function h(t) = t + 1 t x2 attains its minimum at t = x. Thus if 0 ≤x ≤1, sup n≥1 fn(x) ≤sup 0<t≤1 1 h(t) = 1 h(x) = 1 2x. On the other hand, if x ≥1, then sup n≥1 fn(x) ≤sup 0<t≤1 1 h(t) = 1 h(1) = 1 1 + x2 . So if we set g(x) = ( 1 2x if 0 < x ≤1 1 1+x2 if x ≥1 , then 0 ≤fn ≤g. The problem is that g ̸∈L1(0, ∞) since R 1 0 1 2x dx = +∞. However if 0 < a ≤1, then Z ∞ a g(x) dx = Z 1 a 1 2x dx+ Z ∞ 1 1 1+x2 dx = 1 2 ln x 1 a+ tan−1(x) ∞ 1 = 1 2 ln a−1+ π 2 < ∞. Hence as long as a > 0, the Lebesgue dominated convergence theorem (LDCT) applies. So limn→∞ R b a fn dµ = 0. However if a = 0, Z b 0 fn(x) dx = Z b 0 n 1 + n2x2 dx = tan−1(nx) b 0 = tan−1(nb). So lim n→∞ Z b 0 fn(x) dx = lim n→∞tan−1(nb) = π 2 ̸= Z b 0 0 dx = 0. Thus the domination condition by an integrable function is crucial. 3.4. The space L1(µ) For the purposes of this section, let L1(µ) denote the vector space of all µ-integrable functions. 3.4.1. DEFINITION. Given (X, B, µ), let N = {f ∈L1(µ) : ∥f∥1 = 0} = {f measurable : f = 0 a.e.(µ)}. Put an equivalence relation on L1(µ) by f ∼g if f = g a.e.(µ); i.e. if f −g ∈N. The normed vector space space L1(µ) = L1/N consists of the equivalence classes with the induced norm ∥[f]∥1 = ∥f∥1. By abuse of notation, we will frequently write f ∈L1(µ) when we mean [f]. 3.4.2. THEOREM. L1(µ) is a complete normed vector space (a Banach space). 3.5 Comparison with the Riemann Integral 29 PROOF. It is easy to check that N is a subspace of the vector space L1(µ), so that the quotient L1(µ) = L1/N is a vector space. Notice that by Proposition 3.3.2, f = g a.e.(µ) if and only if f −g ∈N if and only if ∥f −g∥1 = 0. If [f] is an ele-ment of L1(µ) and f ∼f′, then h = f′ −f = 0 a.e. Thus R \|f′\| dµ = R \|f\| dµ. So ∥[f]∥1 is well deﬁned. The construction has quotiented out all non-zero elements of zero norm, and thus the norm on L1(µ) is positive deﬁnite. It is easily seen to be positive homogeneous. The triangle inequality follows immediately from the triangle inequality for L1(µ). Thus L1(µ) is a normed vector space. Now suppose that ([fn])n≥1 is a Cauchy sequence in L1(µ). Choose repre-sentatives fn ∈[fn] so that we can evaluate them at points. For each k ≥1, there is an nk so that if m, n ≥nk, then ∥fm −fn∥1 < 2−k. In particular, ∥fnk+1 −fnk∥1 < 2−k. Let g = \|fn1\| + P k≥1 \|fnk+1 −fnk\|. By the MCT, ∥g∥1 = Z g dµ = Z \|fn1\| dµ + X k≥1 Z \|fnk+1 −fnk\| dµ = ∥fn1∥1 + X k≥1 ∥fnk+1 −fnk∥1 < ∥fn1∥1 + X k≥1 2−k < ∞. Hence g is integrable, and therefore is ﬁnite a.e.(µ), say on X \ N for µ(N) = 0. It follows that the sum f(x) = fn1(x) + X k≥1 fnk+1(x) −fnk(x) = lim k→∞fnk(x) converges absolutely on X \ N. We deﬁne f\|N = 0. Moreover \|f\| ≤g so that f is integrable, and f = limk→∞fnk a.e.(µ). Also \|fnk\| ≤∥fn1∥1 + k−1 X i=1 ∥fni+1 −fni∥≤g. Therefore \|f −fnk\| ≤2g; and so by LDCT, 0 = ∥f −f∥1 = lim k→∞∥f −fnk∥1. That is, fnk converges to f in L1(µ). Finally a standard argument shows that the whole Cauchy sequence (fn) converges to f in norm. ■ 3.5. Comparison with the Riemann Integral We will do a speedy review of part of the Riemann integral theory. Given a bounded real valued function f on a ﬁnite interval [a, b], we consider a partition P = {a = x0 < x1 < · · · < xn = b} of [a, b]. Deﬁne real numbers Mj = sup xj−1≤x≤xj f(x) and mj = inf xj−1≤x≤xj f(x) for 1 ≤j ≤n. 30 Integration Then lP(x) = m1χ{a} + n X j=1 mjχ(xj−1,xj] ≤f ≤M1χ{a} + n X j=1 Mjχ(xj−1,xj] = uP(x). The upper and lower sums are L(f, P) = n X j=1 mj(xj −xj−1) and U(f, P) = n X j=1 Mj(xj −xj−1). If P and Q are two partitions, let P ∨Q denote the partition using the points in P ∪Q. It is easy to show that L(f, P) ≤L(f, P ∨Q) ≤U(f, P ∨Q) ≤U(f, Q). It follows that sup P L(f, P) ≤inf Q U(f, Q). A function f is Riemann integrable if for every ε > 0, there is a partition P so that U(f, P) −L(f, P) = n X j=1 (Mj −mj)(xj −xj−1) < ε. In this case, one deﬁnes Z b a f(x) dx := sup P L(f, P) = inf Q U(f, Q). If one deﬁnes the mesh of a partition as mesh(P) = max1≤j≤n xj −xj−1, then for any ε > 0, there is a δ > 0 so that if mesh(P) < δ, then U(f, P)−L(f, P) < ε. Hence we can take a nested sequence of partitions P1 ⊂Pn ⊂Pn+1 ⊂. . . with mesh Pn →0 and conclude that Z b a f(x) dx = lim n→∞L(f, Pn) = lim n→∞U(f, Pn). 3.5.1. THEOREM. Every Riemann integrable function f on [a, b] is Lebesgue integrable, and Z f dm = Z b a f(x) dx. PROOF. Notice that a Riemann integrable function is approximated from above and below by a special class of simple functions, the piecewise constant functions, lP ≤f ≤uP. Moreover the Lebesgue and Riemann integrals agree on these piecewise constant functions. Take a nested sequence of partitions Pn as above with mesh Pn →0 and observe that lPn ≤lPn+1 ≤f ≤uPn+1 ≤uPn. 3.5 Comparison with the Riemann Integral 31 Deﬁne l(x) = limn→∞lPn(x) and u(x) = limn→∞uPn(x). Note that l ≤f ≤u. These functions are Lebesgue measurable. Moreover, lPn −LP1 ≥0 and increase to l −lP1. Since lP1 is integrable, the MCT shows that Z l dm = Z lP1 dm + lim n→∞ Z lPn −lP1 dm = lim n→∞ Z lPn dm = lim n→∞L(f, Pn) = Z b a f(x) dx. Similarly, Z u dm = Z b a f(x) dx. In particular, Z u−l dm = 0. By Lemma 3.2.5, u −l = 0 a.e.(m). Therefore l = f = u a.e.(m), so that f is measurable and Z f dm = Z u dm = Z b a f(x) dx. ■ 3.5.2. REMARKS. There is one situation where the Riemann integral can do something that the Lebesgue integral can’t. That is the improper Riemann integrals in which a function which is not Riemann integrable and be integrated as a limit. The typical example is Z ∞ 0 sin x x dx. The integrand extends to be continuous at x = 0, so that is not an issue. In the Riemann theory, this integral is evaluated as Z ∞ 0 sin x x dx = lim r→∞ Z r 0 sin x x dx. This can be shown by a number of techniques to equal π 2 . The reason it is not Lebesgue integrable is that this integral exists only as a conditional limit, and Z ∞ 0 sin x x dx = lim r→∞ Z r 0 sin x x dx = +∞. So this function is neither Lebesgue nor Riemann integrable. However if a function f is Lebesgue integrable, then so is \|f\|. Let’s examine this example in more detail. Z ∞ 0 sin x x dx = ∞ X n=0 Z (n+1)π nπ sin x x dx = ∞ X n=0 (−1)nan where for n ≥1, an = Z (n+1)π nπ \| sin x\| x dx ≤1 nπ Z π 0 sin x dx = 2 nπ and an = Z (n+1)π nπ \| sin x\| x dx ≥ 1 (n + 1)π Z π 0 sin x dx = 2 (n + 1)π. 32 Integration It follows that an ≥ 2 (n+1)π ≥an+1 for n ≥1; so an →0 monotonely. Therefore P∞ n=0(−1)nan converges by the alternating series test. It also shows that Z ∞ 0 sin x x dx = ∞ X n=0 Z (n+1)π nπ \| sin x\| x dx = ∞ X n=0 an. Since an ≥ 2 (n+1)π for n ≥1, this series diverges by comparison with the harmonic series. The other place where an improper Riemann integral is requires is the integra-tion of unbounded functions. The Riemann theory works only for bounded func-tions. A function like f(x) = x−a for 0 < a < 1 has an improper Riemann integral on [0, 1]. Since this function is positive and continuous with a ﬁnite integral, it is Lebesgue integrable. It is possible again to construct a function with alternating sign on (0, 1] that blows up near 0 so that the improper Riemann integral exists as Z 1 0 f(x) dx = lim ε→0 Z 1 ε f(x) dx but so that the absolute value is not integrable. An example is Z 1 0 1 x sin 1 x dx, which after the change of variables u = 1 x converts this to the integral Z ∞ 1 sin u u du which we have already analyzed. When you have to integrate speciﬁc functions, you will usually ﬁnd yourself falling back on the many techniques that have been developed for the Riemann in-tegral. Almost all explicit functions that you will need to integrate will be Riemann integrable or at least locally Riemann integrable. The power of the Lebesgue integral lies in the limit theorems MCT and LDCT. These are much stronger than anything available in the Riemann theory. Also we obtain the completeness of L1(µ) and many other related spaces that allow the power of functional analysis to apply. There is a nice characterization of Riemann integrable functions due to Lebesgue. 3.5.3. THEOREM. If f : [a, b] →R is a bounded function, then f is Riemann integrable if and only if f is continuous except on a set of measure 0. PROOF. Deﬁne functions U(x) = inf δ>0 sup \|y−x\|<δ f(y) and L(x) = sup δ>0 inf \|z−x\|<δ f(z). It is an easy exercise to show that f is continuous at x if and only if L(x) = U(x). The quantity ω(f, x) = U(x) −L(x) is called the oscillation of f at x. 3.6 Comparison with the Riemann Integral 33 Suppose that f is Riemann integrable. We use the notation from the proof of Theorem 3.5.1. Choose an increasing family of partitions Pn with mesh(Pn) →0, and deﬁne lPn, l, uPn and u as before. Let A = {x : l(x) ̸= u(x)} ∪ [ n≥1Pn. The set {x : l(x) ̸= u(x)} has measure 0, and S n≥1 Pn is countable. Therefore m(A) = 0. Moreover lim n→∞uPn(x) −lPn(x) = u(x) −l(x) = 0 for x ∈[a, b] \ A. If x ∈[a, b] \ A, choose n so that uPn(x) −lPn(x) < ε. Then there are adjacent points xi−1 < xi in Pn so that xi−1 < x < xi. Let δ = min{x −xi−1, xi −x}. Then since uPn and lPn are constant on (xi−1, xi] and are upper and lower bounds for f, respectively, it follows that U(x) −L(x) ≤ sup \|y−x\|<δ \|z−x\|<δ f(y) −f(z) ≤uPn(x) −lPn(x) < ε. Since ε > 0 is arbitrary, we have U(x) = L(x), and thus f is continuous on [a, b] \ A, which is a.e.(m). Conversely suppose that f is continuous except on a set A of Lebesgue measure 0. Then U(x) = L(x) for x ∈[a, b] \ A. Note that A = S n≥1 An where An = {x ∈[a, b] : U(x) −L(x) = ω(f, x) ≥2−n}. The set An is closed because its complement is open: if ω(f, x) < r, then there is a δ > 0 so that sup \|y−x\|<δ \|z−x\|<δ f(y) −f(z) < r. So if \|x′ −x\| = d < δ, sup \|y−x′\|<δ−d \|z−x′\|<δ−d f(y) −f(z) < r. Thus ω(f, x′) < r as well. Cover An with a countable family of open intervals of total length at most 2−n. Then since An is compact, there is a ﬁnite subcover I1, . . . , Im. Let B = [a, b] \ Sm i=1 Ii. For each x ∈B, since U(x) −L(x) < 2−n, there is an open inteval Jx ∋x so that the oscillation over Jx is less than 2−n. The collection {Jx : x ∈B} is an open cover of the compact set B. Therefore there is a ﬁnite subcover Jx1, . . . , Jxp. Let P be the partition consisting of the endpoints of I1, . . . , Im, Jx1, . . . , Jxp (together with a, b if necessary). Any of the intervals [xi−1, xi] contained in the union Sp j=1 Jxj will have oscillation less than 2−n, meaning uPn(x)−lPn(x)<2−n on (xi−1, xi]. The remaining intervals are contained in Sm i=1 Ii and thus have total length at most 2−n. The only thing we can say is that uPn(x) −lPn(x) ≤2∥f∥∞ on these intervals. Therefore we obtain the estimate U(f, Pn) −L(f, Pn) < 2−k p X j=1 m(Jxj) + 2∥f∥∞ m X i=1 m(Ii) < 2−kb −a + 2∥f∥∞ . As n →∞, this converges to 0. Hence f is Riemann integrable. ■ 34 Integration 3.6. Product Measures 3.6.1. DEFINITION. If Xλ for λ ∈Λ are non-empty sets, then the product space is X = Y λ∈Λ Xλ = {(xλ) : xλ ∈Xλ, λ ∈Λ}={f :Λ →˙ ∪λ∈ΛXλ : f(λ) = xλ ∈Xλ}. The maps πλ : X →Xλ given by πλ(x) = xλ are the coordinate projections. If (Xλ, Bλ) are σ-algebras, then the product σ-algebra (Q λ∈ΛXλ, N λ∈ΛBλ) is the σ-algebra of subsets of X generated by the sets {π−1 λ (A) : A ∈Bλ. λ ∈Λ}. 3.6.2. REMARK. When Λ = {1, . . . , n} is ﬁnite, there is no problem deﬁning the product space. In this case, the product σ-algebra is generated by all of the “cubes” A1 × · · · × An for Ai ∈Bi, 1 ≤i ≤n. When Λ is inﬁnite, the Axiom of Choice is often needed to guarantee that the product space is non-empty. In this case, the product σ-algebra is generated by sets π−1 λ (Aλ) = Aλ × Q µ∈Λ{λ} Xµ. The intersection of ﬁnitely many yields a cube of the form Aλ1 × · · · × Aλn × Q µ∈Λ{λi:1≤i≤n} Xµ. One can also take the intersection of countably many such slices, but if Λ is uncountable, the cubes formed with proper subsets from each Xλ will generally not belong to the product σ-algebra. Even when Λ is countable, these inﬁnite cubes many turn out to be measure 0 and hence negligible. The following proposition shows that in the familiar case of Borel sets on met-ric spaces, and ﬁnite products, we get the desired result. The separability hypothesis is crucial. 3.6.3. PROPOSITION. If (Xi, di) are separable metric spaces for 1 ≤i ≤n, then n O i=1 Bor(Xi) = Bor( n Y i=1 Xi). PROOF. The product space X = Qn i=1 Xi is also a metric space with the metric d((xi), (yi)) = max{di(xi, yi)}. With this (or any equivalent) metric, the coordi-nate projections are continuous. Thus if Ui is open in Xi, the set π−1 i (Ui) is open in X. These sets generate Nn i=1 Bor(Xi), and thus it is contained in Bor(X). Conversely, since each Xi is separable, so is X. Therefore X is second count-able. Indeed if {xj : j ≥1} is a dense subset, then {br(xj) : j ≥1, r ∈Q+} is a countable neighbourhood base for X. Now br(xj) = Qn i=1 br(xj,i), where πi(xj) = xj,i, belongs to Nn i=1 Bor(Xi). As a σ-algebra is closed under countable unions and every open set in X is the union of those (countably many) sets in the 3.6 Product Measures 35 neighbourhood base that it contains, it follows that every open subset of X belongs to Nn i=1 Bor(Xi). Thus Bor(X) is contained in Nn i=1 Bor(Xi). ■ 3.6.4. COROLLARY. Bor(Rn) = n O i=1 Bor(R) and Bor(C) = Bor(R) ⊗Bor(R). Now let (X, B, µ) and (Y, B′, ν) be two measure spaces. Let A be the collec-tion of all ﬁnite unions of disjoint rectangles of the form A × B for A ∈B and B ∈B′. This is an algebra since it is closed under ﬁnite unions and complements. Indeed, (A × B)c = Ac × Y ˙ ∪A × Bc and A1×B1 ∪A2×B2 = A1×B1 ˙ ∪(A2 \ A1)×B2 ˙ ∪(A1 \ A2)×(B2 \ B1). Now deﬁne a set function π ˙ [n i=1Ai × Bi = n X i=1 µ(Ai)ν(Bi). To apply Carath´ eodory’s Theorem, we need the following lemma. 3.6.5. LEMMA. π is a premeasure on A. PROOF. We need to show that if A × B = ˙ S i≥1Ai × Bi, then π(A × B) = µ(A)ν(B) = X i≥1 µ(Ai)ν(Bi) = X i≥1 π(Ai × Bi). Note that χA(x)χB(y) = X i≥1 χAi(x)χBi(y). If we ﬁx y ∈Y , we obtain a sum of non-negative measurable functions on X. By the Monotone Convergence Theorem, µ(A)χB(y) = Z χA(x) dµ(x) χB(y) = X i≥1 Z χAi(x) dµ(x) χBi(y) = X i≥1 µ(Ai)χBi(y). These functions are non-negative measurable functions on Y , so a second applica-tion of the MCT yields µ(A)ν(B) = µ(A) Z χB(y) dν(y) = X i≥1 µ(Ai) Z χBi(y) dν(y) = X i≥1 µ(Ai)ν(Bi). 36 Integration Hence π is a premeasure. ■ Now we can apply Carath´ eodory’s Theorem via Theorem 1.4.2 to obtain the following, called the product measure. 3.6.6. THEOREM. Let (X, B, µ) and (Y, B′, ν) be two measure spaces. There is a complete measure (X ×Y, B ⊗B′, µ×ν) on X ×Y such that B ⊗B′ ⊃B⊗B′ and µ × ν(A × B) = µ(A)ν(B) for all A ∈B and B ∈B′. We need some more reﬁned information about the sets in B ⊗B′. 3.6.7. DEFINITION. Let A be the algebra of ﬁnite unions of rectangle in B⊗B′ as above. Deﬁne Aσ = E = [ i≥1 Ai : Ai ∈A and Aσδ = G = \ j≥1 Ej : Ej ∈Aσ . 3.6.8. LEMMA. If E ∈B ⊗B′ and µ × ν(E) < ∞, then there is a set G ∈Aσδ such that E ⊂G and µ × ν(G \ E) = 0. PROOF. By deﬁnition of the outer measure π∗which deﬁnes µ × ν, we have that µ × ν(E) = inf n X i≥1 µ × ν(Ai) : Ai ∈A and E ⊂ [ i≥1 Ai o . Thus we can choose Ej = S i≥1 Aj,i ⊃E in Aσ so that µ×ν(Ej) < µ×ν(E)+ 1 j . Hence µ × ν(Ej \ E) < 1 j . Deﬁne G = T j≥1 Ej ∈Aσδ. Then E ⊂G and µ × ν(G \ E) = 0. ■ 3.7. Product Integration An important result from the Riemann theory for integration in multiple vari-ables is that integration over a nice region in Rn is equal to an iterated integral in which the integration is done one variable at a time. There is an important analogue of this for general measures. It actually comes in two ﬂavours. 3.7.1. FUBINI’S THEOREM. Let (X, B, µ) and (Y, B′, ν) be complete mea-sures. If f ∈L1(µ × ν), then 3.7 Product Integration 37 (1) (i) fx(y) = f(x, y) ∈L1(ν) for a.e. x(µ). (ii) fy(x) = f(x, y) ∈L1(µ) for a.e. y(ν). (2) (i) Z Y fx(y) dν =: F(x) ∈L1(µ). (ii) Z X fy(x) dµ =: G(y) ∈L1(ν). (3) Z X×Y f dµ×ν = Z X Z Y f(x, y)dν(y) dµ(x)= Z Y Z X f(x, y) dµ(x) dν(y). 3.7.2. TONELLI’S THEOREM. Let (X, B, µ) and (Y, B′, ν) be complete mea-sures, and suppose that µ × ν is σ-ﬁnite. If f ∈L+(µ × ν), then (1) (i) fx(y) = f(x, y) ∈L+(ν) for a.e. x(µ). (ii) fy(x) = f(x, y) ∈L+(µ) for a.e. y(ν). (2) (i) Z Y fx(y) dν =: F(x) ∈L+(µ). (ii) Z X fy(x) dµ =: G(y) ∈L+(ν). (3) Z X×Y f dµ×ν = Z X Z Y f(x, y)dν(y) dµ(x)= Z Y Z X f(x, y) dµ(x) dν(y). 3.7.3. REMARKS. (1) The hypothesis that µ×ν is σ-ﬁnite in Tonelli’s Theorem is critical, as an example will show. However it is not needed for Fubini’s Theorem because when f ∈L1(µ × ν), we have that Z \|f\| dµ × ν = L < ∞. It follows that Cn = {(x, y) ∈X × Y : \|f(x, y)\| ≥1 n} has µ × ν(Cn) ≤nL. Thus the restriction of µ×ν to C = {(x, y) ∈X ×Y : f(x, y) ̸= 0} = S n≥1 Cn is σ-ﬁnite. (2) Also the hypothesis that f ∈L1(µ × ν) or f ∈L+(µ × ν) is also critical as examples will show. The failure to check this condition leads to common misuses of these theorems. (3) Normal practice is to omit the parentheses and if there is no confusion, also the variables, in multiple integrals. So we write ZZ f dν dµ or ZZ f(x, y)dν(y) dµ(x) for Z X Z Y f(x, y)dν(y) dµ(x). If E ⊂X × Y , deﬁne Ex = {y ∈Y : (x, y) ∈E} for x ∈X; and let Ey = {x ∈X : (x, y) ∈E} for y ∈Y . 38 Integration 3.7.4. LEMMA. If E ∈Aσδ and µ × ν(E) < ∞, then g(x) = ν(Ex) is µ-measurable, g ∈L+ ∩L1(µ), and Z g dµ = µ × ν(E). Similarly, h(y) = µ(Ey) is ν-measurable, h ∈L+ ∩L1(ν), and Z h dν = µ × ν(E). PROOF. If E = A×B for A ∈B and B ∈B′, then Ex = ( B if x ∈A ∅ if x ̸∈A. Therefore g(x) = ν(B)χA is µ-measurable and g ≥0. Moreover, Z g dµ = Z ν(B)χA dµ = µ(A)ν(B) = µ × ν(E) < ∞. Thus g ∈L+ ∩L1(µ). Next, if E ∈Aσ, we can write E = S i≥1 Ai × Bi for Ai ∈B and Bi ∈B′. We can rewrite this is a disjoint union because An × Bn \ Sn−1 i=1 Ai × Bi ∈A and thus can be written as a ﬁnite disjoint union of rectangles, which are clearly also disjoint from Sn−1 i=1 Ai × Bi. Proceeding recursively, E can be rewritten as a disjoint union E = ˙ S i≥1Ci × Di for Ci ∈B and Di ∈B′. Let g(x) = ν(Ex) and gi(x) = ν(Ci)χDi. Then g(x) = P i≥1 gi(x). Hence g is measurable, and thus belongs to L+(µ). By the MCT, Z g dµ = X i≥1 Z gi dµ = X i≥1 µ × ν(Ci × Di) = µ × ν(E) < ∞. Moreover, we now have g ∈L+ ∩L1(µ). Finally, suppose that E = T n≥1 En where En ⊃En+1 all lie in Aσ such that µ × ν(E1) < ∞and µ × ν(En) ↓µ × ν(E). Let gn(x) = ν((En)x) and g(x) = ν(Ex). Then 0 ≤g ≤gn+1 ≤gn ≤g1; and g(x) = inf gn(x) = lim gn(x). Hence g is µ-measurable. It is dominated by g1 which is integrable, so g ∈L+ ∩L1(µ). By the LDCT, Z g dµ = lim n→∞ Z gn dµ = lim n→∞µ × ν(En) = µ × ν(E). The last equality follows by continuity from above since µ × ν(E1) < ∞. Similarly we can interchange the role of x and y. ■ 3.7.5. LEMMA. If E ∈B ⊗B′ has µ × ν(E) = 0, then ν(Ex) = 0 a.e.(µ) and µ(Ey) = 0 a.e.(ν). PROOF. By Lemma 3.6.8, there is a set G ∈Aσδ such that E ⊂G and µ × ν(G) = 0. Therefore by the previous lemma, f(x) = ν(Gx) ∈L+ ∩L1(µ) and R f dµ = 0. Therefore f = 0 a.e.(µ). Now Ex ⊂Gx, and since ν is a complete 3.7 Product Integration 39 measure, g(x) = ν(Ex) = 0 a.e.(µ). It could happen that Ex is not measurable on a set of ν-measure 0. A function which differs from a measurable function on a set of measure 0 is also measurable. Similarly we can interchange the role of x and y. ■ 3.7.6. COROLLARY. If E ∈B ⊗B′ and µ × ν(E) < ∞, then Ex is ν-measurable for a.e.(µ) x ∈X, g(x) = ν(Ex) is µ-measurable, g ∈L+ ∩L1(µ), and Z g dµ = µ × ν(E). Similarly, Ey is µ-measurable for a.e.(ν) y ∈Y , h(y) = µ(Ey) is ν-measurable, h ∈L+ ∩L1(ν), and Z h dν = µ × ν(E). PROOF. Find G ∈Aσδ so that E ⊂G and µ × ν(G \ E) = 0. Then by Lemma 3.7.4, f(x) = ν(Gx) ∈L+ ∩L1(µ), and Z f dµ = µ × ν(G). By Lemma 3.7.5, 0 ≤f −g = 0 a.e.(µ). The result follows. Similarly we can interchange the role of x and y. ■ Now we are ready to prove the theorems. PROOF OF FUBINI’S THEOREM. The various integrals and iterated integrals are linear operations, so we can write f ∈L1(µ × ν) as f = f1 −f2 + if3 −if4 where fi ∈L+ ∩L1(µ × ν) by letting f1 = Re f ∨0, etc. So we may suppose that f ∈L+ ∩L1(µ × ν). There are simple functions ϕn with 0 ≤ϕn ≤ϕn+1 ≤f so that f = limn→∞ϕn and each ϕn is supported on a set of ﬁnite measure (such as the set Cn in Remark 3.7.3(1)). By Corollary 3.7.6 and linearity, Fubini’s Theorem is valid for simple functions with ﬁnite measure support. By the MCT, Z f dµ × ν = lim n→∞ Z ϕn dµ × ν = lim n→∞ Z X Z Y ϕn(x, y)dν(y) dµ(x). The functions Fn(x) = Z Y ϕn(x, y)dν(y) are positive, measurable and monotone increasing to F(x) = Z Y f(x, y)dν(y). Hence this is a measurable function, and by MCT, lim n→∞ Z X Fn(x) dµ(x) = Z X F(x) dµ(x). Thus Z X×Y f dµ × ν = Z X Z Y f(x, y)dν(y) dµ(x). Interchanging the role of X and Y yields the other iterated integral. ■ PROOF OF TONELLI’S THEOREM. Let f ∈L+(µ×ν). Since µ×ν is σ-ﬁnite, there are measurable sets Cn ⊂Cn+1 with µ×ν(Cn) < ∞and X×Y = S n≥1 Cn. 40 Integration For n ≥1, let fn = (f ∧n)χCn, where (f ∧n)(x) = min{f(x), n}. Then 0 ≤fn ≤nχCn, and so fn ∈L+ ∩L1(µ × ν). Now by Fubini’s Theorem and MCT, we have Z f dµ × ν = lim n→∞ Z fn dµ × ν = lim n→∞ Z X Z Y fn(x, y)dν(y) dµ(x). The proof is completed as before. The functions Fn(x) = Z Y fn(x, y)dν(y) are positive, measurable and monotone increasing to F(x) = Z Y f(x, y)dν(y). Hence this is a measurable function, and by MCT, lim n→∞ Z X Fn(x) dµ(x) = Z X F(x) dµ(x). Thus Z X×Y f dµ × ν = Z X Z Y f(x, y)dν(y) dµ(x). Interchanging the role of X and Y yields the other iterated integral. ■ There is a straightforward variant of these results when the measures are not complete. The difference is that given measures (X, B, µ) and (Y, B′, ν), we form µ × ν and restrict it to the σ-algebra B ⊗B′. I will also call this µ × ν. 3.7.7. FUBINI-TONELLI THEOREM WITHOUT COMPLETENESS. Let (X, B, µ) and (Y, B′, ν) be measures with product (X × Y, B ⊗B′, µ × ν). If f ∈L1(µ × ν) (or if µ × ν is σ-ﬁnite and f ∈L+(µ × ν)), then (1) (i) fx(y) = f(x, y) ∈L1(ν) (or L+(ν)) for all x ∈X (ii) fy(x) = f(x, y) ∈L1(µ) (or L+(µ)) for all y ∈Y . (2) (i) Z Y fx(y) dν =: F(x) ∈L1(µ) (or L+(µ)). (ii) Z X fy(x) dµ =: G(y) ∈L1(ν) (or L+(ν)). (3) Z X×Y f dµ×ν = Z X Z Y f(x, y)dν(y) dµ(x)= Z Y Z X f(x, y) dµ(x) dν(y). In Assignment 4, Q5, you showed that if E ∈B ⊗B′, then Ex ∈B′ and Ey ∈B for all x ∈X and y ∈Y . This extends to simple functions, so that (1i) and (1ii) hold for simple functions. This extends to limits, and since every measurable function is a limit of simple functions, we obtain (1i) and (1ii). The remainder of the proof is identical to the proof in the complete case. We now present a few counterexamples that show the limits of these theorems. 3.7 Product Integration 41 3.7.8. EXAMPLE. Consider counting measure on (N, P(N), mc). Form the product, which is (N2, P(N2), mc × mc), where mc × mc is just counting measure on N2. Consider f(m, n) =      1 if n = m −1 if n = m + 1 0 otherwise. Think of this as an N by N array 1 −1 0 0 0 . . . 0 1 −1 0 0 . . . 0 0 1 −1 0 . . . 0 0 0 0 1 . . . . . . . . . . . . . . . . . . ... The ﬁrst iterated integral ﬁrst sums the rows, each absolutely summable with total 0: Z X Z Y f(m, n) dmc(n) dmc(m) = X m≥1 X n≥1 f(m, n) = X m≥1 0 = 0. The second iterated integral ﬁrst sums the columns, each absolutely summable. However the ﬁrst column sums to 1, the rest to 0: Z Y Z X f(m, n) dmc(m) dmc(n) = X n≥1 X m≥1 f(m, n) = 1 + X n≥2 0 = 1. These measures are σ-ﬁnite since N2 is countable and points have measure 1. Tonelli’s theorem is not applicable because f is not positive. More importantly, Fu-bini’s theorem does not apply because f is not integrable. If it were, then \|f\| would have a ﬁnite integral, but clearly Z Z \|f\| dmc × mc = X m≥1 X n≥1 \|f(m, n)\| = ∞. 3.7.9. EXAMPLE. Consider Lebesgue measure (X = [0, 1], B, m) and counting measure on (Y = [0, 1], P([0, 1]), mc). We have ([0, 1]2, B ⊗P([0, 1]), m×mc) as the product. Consider D = {(x, x) : x ∈[0, 1]}, and let f = χD. Observe that D = \ k≥1 2k j=1 j −1 2k , j 2k 2 is a measurable set. Consider the iterated integrals. When integrating fx(y) = δx with respect to counting measure, we get 1. Therefore Z X Z Y f(x, y) dmc(y) dm(x) = Z X 1 dm(x) = 1. 42 Integration When integrating fy(x) = δy with respect to Lebesgue measure, we get 0. There-fore Z Y Z X f(x, y) dm(x) dmc(y) = Z Y 0 dmc(y) = 0. What went wrong? Here f ≥0 is measurable, so belongs to L+(m × mc). The problem is that mc on [0, 1] is not σ-ﬁnite because [0, 1] is uncountable. Finally let’s compute Z Z f dm × mc = m × mc(D). This is computed using the outer measure obtained by covering D with a countable number of rectangles, say D ⊂S i≥1 Ai × Bi where Ai are Lebesgue measurable. Note that this union cannot cover D if for each i ≥1, either m(Ai) = 0 or Bi is ﬁnite. Indeed if m(Ai) = 0 for i ∈J1 and Bi is ﬁnite for i ∈J2, then [ i∈J1 Ai × Bi ⊂A × [0, 1] and [ i∈J2 Ai × Bi ⊂[0, 1] × B where A = S i∈J1 Ai has m(A) = 0, and B = S i∈J2 Bi is countable. But their union omits Ac × Bc, which contains (A ∪B)c × (A ∪B)c and so inter-sects D. Therefore there is some i0 so that m(Ai0) > 0 and Bi0 is inﬁnite, and hence mc(Bi0) = ∞. But then the premeasure of the rectangle is π(Ai0 × Bi0) = m(Ai0)mc(Bi0) = ∞. Since D is measurable, we have that m × mc(D) = ∞. So neither iterated integral yields the value of the product integral even though both exist. 3.7.10. EXAMPLE. Let X = Y = [0, 1] with Lebesgue measure. If we assume the Continuum Hypothesis, there is a well-ordering ≺on [0, 1] with the property that {y ∈[0, 1] : y ≺x} is countable for all x. Let E = {(x, y) : y ≺x} ⊂[0, 1]2. Let f = χE. Clearly f ≥0. For each x ∈[0, 1], Ex = {y ∈[0, 1] : y ≺x} is countable. And for each y ∈[0, 1], Ey = {x ∈[0, 1] : y ≺x} is the complement of a countable set. In particular, these sets are Lebesgue measurable. So we can evaluate the iterated integrals Z X Z Y f(x, y) dm(y) dm(x) = Z X m(Ex) dm(x) = 0 and Z Y Z X f(x, y) dm(x) dm(y) = Z Y m(Ey) dmc(y) = 1. What went wrong? The measures are ﬁnite, and hence σ-ﬁnite. If E were mea-surable, then its measure would be ﬁnite and so f would belong to L+ ∩L1(m2) Then both Fubini’s Theorem and Tonelli’s Theorem would apply! It must be the case that E is not measurable. This is in spite of the fact that Ex and Ey are Borel for all x and y. There is a variant of this example which avoids the Continuum Hypothesis, but requires more detailed knowledge of ordinals. Let Ωbe the ﬁrst uncountable ordinal. This is a well-ordered uncountable set with respect to an order ≺with the 3.8 Lebesgue Measure on Rn 43 property that {y ∈Ω: y ≺x} is countable for all x ∈Ω. The open sets in the order topology are generated by {y ∈Ω: y ≺x} and {y ∈Ω: x ≺y} for x ∈Ω. It is not hard to see that these sets are either countable of have countable complement (co-countable). Moreover one can check that Bor(Ω) consists or all countable and co-countable sets. Put a measure µ on (Ω, Bor(Ω)) by µ(A) = ( 0 if A is countable 1 if A is co-countable. . This is a ﬁnite measure. Form the product space (Ω2, Bor(Ω2), µ × µ). Again we set E = {(x, y) ∈Ω2 : y ≺x} and f = χE. We have the same contradiction. It must be the case that E is not a measurable set. 3.8. Lebesgue Measure on Rn Let (R, L, m) be Lebesgue measure on the real line. Then we can form mn = m × · · · × m on (Rn, L ⊗· · · ⊗L) = (Rn, Bor(Rn)). This is the completion of a Borel measure on Rn such that mn(A1 × · · · × An) = n Y i=1 m(Ai) for Ai ∈L. This is the unique complete measure with this property. Let Ln denote the σ-algebra of Lebesgue measurable sets in Rn. Even though m is a complete measure, the σ-algebra L ⊗L is not complete. For example, if E ⊂R is not measurable and B ⊂L has m(B) = 0, then E × B ⊂R × B = [ n∈Z [n, n + 1) × B. Since m2([n, n + 1) × B) = m(B) = 0, we have that m(R × B) = 0. Then since m2 is complete, m2(E × B) = 0 as well. This set does not belong to L ⊗L. Note that when there is no confusion, Lebesgue measure on Rn is commonly written as m instead of mn. We will adopt this practice. The ﬁrst result is to show that certain regularity properties of Lebesgue measure on the line transfer to mn. 3.8.1. PROPOSITION. Let E ∈Ln, and let ε > 0. (1) There is an open set U ⊃E such that m(U \ E) < ε; and m(E) = inf{m(U) : E ⊂U, U open}. (2) There is a closed set C ⊂E so that m(E \ C) < ε; and m(E) = sup{m(K) : K ⊂E, K compact}. 44 Integration (3) There is an Fσ set F and a Gδ set G so that F ⊂E ⊂G and m(E \ F) = 0 = m(G \ E). (4) If m(E) < ∞, then there is a ﬁnite set of pairwise disjoint rectangles R1, . . . , RN whose sides are intervals so that m E△SN i=1 Ri < ε. PROOF. (1) First assume that E is bounded. The construction of mn comes from the outer measure on the algebra A of ﬁnite unions of disjoint rectangles. Thus there is a countable union of (bounded) rectangles Ri covering E such that P i≥1 m(Ri) < m(E) + ε/2. For each i ≥1, we can use Theorem 1.5.6 to enclose the sides of Ri in open sets with slight increase in measure to obtain an open rectangle Ui ⊃Ri with m(Ui) < m(Ri) + 2−i−1ε. Then E ⊂U = S i≥1 Ui and m(U) ≤ X i≥1 m(Ei) < X i≥1 m(Ri) + 2−i−1ε < m(E) + ε. Hence m(U \ E) < ε. Now if E is unbounded, divide Rn into countably many disjoint unit cubes labeled Ci for i ≥1. Then E = S i≥1 E ∩Ci. For each i, ﬁnd an open set Ui ⊃E ∩Ci so that m(Ui \ (E ∩Ci) < 2−iε. Then U = S i≥1 Ui does the job. The other parts of this proposition are left as an exercise. ■ 3.8.2. LEMMA. If f is Lebesgue measurable, there is a Gδ set G with m(G) = 0 so that fχGc is Borel measurable. PROOF. Write f = f1 −f2 + if3 −if4 where fi are positive and measurable. This reduces the problem to the case of a positive function f. Let {rn : n ≥1} be a dense subset of [0, ∞). Then En = f−1([0, rn)) are measurable. Select Fσ sets Bn and null sets Nn so that En = Bn ∪Nn. Then N = S n≥1 Nn is a null set. Let G be a Gδ set so that N ⊂G and m(G) = 0. Then (fχGc)−1([0, rn)) = En ∪G = Bn ∪G is Borel for all rn. The countable union of Borel sets is Borel, and thus fχGc is Borel. ■ As for Lebesgue measure on the line, Lebesgue measure on Rn is translation invariant. It also behaves well under a linear change of variables just as the Riemann integral does. Let GLn denote the group of invertible n × n real matrices. It is a fact from linear algebra that every invertible matrix is the product of a number of elementary matrices [1,§3.2, Corollary 3]. The elementary matrices have three types: multiplying the ith row by a non-zero scalar c Ri(x1, . . . , xi, . . . , xn) = (x1, . . . , cxi, . . . , xn), adding a multiple of the jth row to the ith row for i ̸= j, 1 ≤i, j ≤n Aij(x1, . . . , xi, . . . , xj, . . . , xn) = (x1, . . . , xi + cxj, . . . , xj, . . . , xn), 3.8 Lebesgue Measure on Rn 45 and interchanging two rows i, j for i ̸= j, 1 ≤i, j ≤n Sij(x1, . . . , xi, . . . , xj, . . . , xn) = (x1, . . . , xj, . . . , xi, . . . , xn). 3.8.3. THEOREM. Let m be Lebesgue measure on Rn. (1) m is translation invariant: m(E + x) = m(E) for E ∈Ln and x ∈Rn. (2) If T ∈GLn and f is measurable, then f ◦T is measurable. If f ∈L1(m) or L+(m), then Z f dm = \| det T\| Z f ◦T dm. In particular, if E ∈Ln, then m(T(E)) = \| det T\|m(E). (3) m is invariant under rotation: m(U(E)) = m(E) for E ∈Ln and U unitary. PROOF. (1) Note that translation of a cube preserves the measure. Thus the premeasure is translation invariant. It follows that the outer measure is translation invariant, and hence so is m. (2) If f is Borel measurable, then so is f ◦T because T is continuous and hence Borel by Corollary 2.1.3. If the result is true for invertible matrices S and T, then Z f dm = \| det S\| Z f ◦S dm = \| det S\| \| det T\| Z (f ◦S) ◦T dm = \| det ST\| Z f ◦ST dm. Hence it is true for their product. Therefore it sufﬁces to establish the result for elementary matrices. Using either the Fubini or Tonelli Theorem, the integral is equal to the iterated integral. For multiplying a row by a non-zero constant, the iterated integral reduces to the one variable fact. Adding a multiple of one row to another, integrate ﬁrst with respect to xi and use translation invariance. Interchanging two coordinates is equivalent to interchanging the order of integration, which results in no change, again by the Fubini or Tonelli theorem. For E Borel, apply the result to χE to get m(E) = Z χE dm = \| det T −1\| Z χE ◦T −1 dm = \| det T\|−1 Z χT(E) dm = \| det T\|−1m(T(E)). The result for a Lebesgue measurable function f now follows because there is a Gδ set G so that fχGc is Borel measurable. Now (fχGc) ◦T = f ◦TχT −1(G)c. 46 Integration By the Borel result, m(T −1(G)) = 0. So f ◦T = (fχGc) ◦T a.e.(m). The result follows. (3) If U is unitary, then \| det U\| = 1. In particular every rotation is unitary; and thus m is rotation invariant. ■ 3.9. Inﬁnite Product Measures We brieﬂy mention the subject of inﬁnite products. Suppose that we are given probability measures (Xi, Bi, µi); i.e. µi(Xi) = 1 for i ≥1. We wish to deﬁne a measure µ on (Q i≥1 Xi, N i≥1 Bi) so that µ Ai × · · · × An × Y i>n Xi = n Y i=1 µi(Ai) for all n ≥1 and Ai ∈Bi. Let A denote the algebra of sets generated by the cylinders π−1 i (Ai) for i ≥1 and Ai ∈Bi, where as usual πi is the coordinate projection of X = Q i≥1 Xi onto Xi. One can show as in section 3.6 that elements of A can be written as a ﬁnite disjoint union of rectangles of the form R(A1, . . . , An) := Ai × · · · × An × Y i>n Xi. Deﬁne µ R(A1, . . . , An) = Qn i=1 µi(Ai) and extend this to ﬁnite disjoint unions by additivity. Note that we have already shown that there are measures µ(n) = µ1 × · · · × µn on (Qn i=1 Xi, Nn i=1 Bi) so that µ(n)(Ai × · · · × An) = µ(R) for all rectangles in Nn i=1 Bi × Q i>n Xi. If we let µ(m,n] = µm+1 × · · · × µn for 1 ≤m < n, we have that µ(n) = µ(m) × µ(m,n]. for 1 ≤m < n. 3.9.1. THEOREM. µ is a premeasure on A. PROOF. It sufﬁces to show that if R, Ri for i ≥1 are rectangles and R = ˙ S i≥1Ri, then µ(R) = P i≥1 µ(Ri). There is an integer m0 so that R ∈ m0 O j=1 Bj × Y j>m0 Xj =: B(m0). Then choose integers m0 ≤mn ≤mn+1 so that Ri ∈B(mn) for 1 ≤i ≤n. Deﬁne Fn = R \ ˙ Sn i=1Ri. Observe that Fn ∈A ∩B(mn), F1 ⊃Fn ⊃Fn+1 ⊃. . . and T n≥1 Fn = ∅. Therefore µ(R) = µ(mn)(R) = n X i=1 µ(mn)(Ri) + µ(mn)(Fn) = n X i=1 µ(Ri) + µ(Fn) 3.9 Inﬁnite Product Measures 47 Our result will follow if we show that limn→∞µ(Fn) = 0. Assuming that this is false, we have that µ(Fn) ≥ε > 0 for all n ≥1. We will produce a contradiction. Given F ∈A and xi ∈Xi for 1 ≤i ≤p, deﬁne F(x1, . . . , xp) = {x ∈ Y i>p Xi : (x1, . . . , xp, x) ∈F}. Then deﬁne G1,n = x1 ∈X1 : µ(1,mn](Fn(x1)) > ε/2 . Note that ε ≤µ(Fn) = Z X1 µ(1,mn](Fn(x1)) dµ1(x1) ≤ Z G1,n 1 dµ1(x1) + Z Gc 1,n ε 2 dµ1(x1) ≤µ(1)(G1,n) + ε 2. Therefore µ(1)(G1,n) ≥ε 2. Now G1,n ⊃G1,n+1 and hence µ T n≥1 G1,n ≥ε 2. Choose a point a1 ∈T n≥1 G1,n. Then µ(1,mn](Fn(a1)) ≥ε 2 for all n ≥1. Recursively we construct points ai ∈Xi so that µ(k,mk](F(a1, . . . , ak)) ≥ε 2k for all n ≥1. Suppose that a1, . . . , ak−1 have been deﬁned with this property. Deﬁne Gk,n = xk ∈Xk : µ(k,mn](Fn(a1, . . . , ak−1, xk)) > ε 2k . Arguing as above, ε 2k−1 ≤µ((k−1,mn](Fn(a1, . . . , ak−1)) = Z Xk µ(k,mn dµk(xk) ≤ Z Gk,n 1 dµk(xk) + Z Gc k,n ε 2k dµk(xk) ≤µ(k)(Gk,n) + ε 2k . Therefore µk T n≥1 Gk,n ≥ ε 2k . Pick a point ak ∈T n≥1 Gk,n. This does the job. It follows that for all k ≥1 and n ≥1, there is a point x so that (a1, . . . , ak, x) is in Fn. In particular, there is a point y so that (a1, . . . , amn, y) ∈Fn. But Fn is a union of cylinder sets of level no greater than mn, and hence (a1, . . . , amn, y) ∈Fn for all y ∈Q i>mn Xi. Therefore a = (a1, a2, a3, . . . ) ∈T n≥1 Fn. This is a contradiction since this intersection in empty. It follows that limn→∞µ(Fn) = 0 and hence µ(R) = P i≥1 µ(Ri). ■ As an immediate consequence of Theorem 1.4.2, we obtain the desired mea-sure. 48 Integration 3.9.2. COROLLARY. There is a unique complete measure µ = Y i≥1 µi on ( Y i≥1 Xi, O i≥1 Bi) so that µ Ai × · · · × An × Y i>n Xi = n Y i=1 µi(Ai) for all n ≥1 and Ai ∈Bi. 3.9.3. EXAMPLE. Fix 0 < p < 1. Take Xi = 2 = {0, 1} and let µi be the probability measure on (Xi, P(Xi)) such that µi(0) = p and µi(1) = 1 −p. Then X = Q i≥1 Xi is homeomorphic to the Cantor set C. Let µp denote the inﬁnite product measure Q i≥1 µi. Identify the Cantor ternary set with {x = 0.(2ε1)(2ε2)(2ε3) . . . base 3} for ε = (ε1, ε2, . . . ) ∈X. The homeomorphism of X onto C is given by h(ε) = 0.(2ε1)(2ε2)(2ε3) . . . base 3 for ε = (ε1, ε2, ε3, . . . ) ∈X. We obtain a Borel measure on C by setting νp(A) = µp(h−1(A)). Let Cn denote the subset of [0, 1] after n operations of removing the middle thirds has been accomplished; so that there remain 2n intervals of length 3−n. With this explicit homeomorphism, the rectangle R(ε1, . . . , εn) corresponds to one of the intervals of Cn intersected with C. For example, R(0, 0) = [0, 1 9] ∩C, R(0, 1) = [ 2 9, 1 3] ∩C, R(1, 0) = [2 3, 7 9] ∩C and R(1, 1) = [8 9, 1] ∩C. Consider the analogue of the Cantor function which is deﬁned on [0, 1] by fp(x) = νp(C ∩[0, x]). Then f(0) = 0 and fp(1) = 1. Observe that this is a monotone increasing function. Moreover on the middle third [1 3, 2 3], it takes the value p. Then on the interval [1 9, 2 9], it takes the value p2 and on [7 9, 8 9], it takes the value p + p(1 −p). Indeed, at level n, we have deﬁned fp(x) on [0, 1] \ int(Cn). If [a, b] is a component of Cn, then the middle third is removed to form Cn+1 and fp(x) = fp(a) + p(fp(b) −fp(a)) for a + b−a 3 ≤x ≤a + 2(b−a) 3 . Notice that fp takes all of the values pk(1 −p)n−k for 0 ≤k ≤n and n ≥1. These values are dense in [0, 1]. As for the usual Cantor function, it follows that fp has no jump discontinuities. So fp is continuous. The measure νp is the Lebesgue-Stieltjes measure for the function fp by Theo-rem 1.5.2. So νp is a probability measure supported on C. Since fp take the same value at each endpoint of a removed interval, we see that νp(R \ C) = 0. The continuity of fp means that νp has no atoms. CHAPTER 4 Differentiation and Signed Measures 4.1. Differentiation Question: Is there a measure theoretic analogue of the Fundamental Theorem of Calculus? That is, to what extent is F(x) = c + R [a,x] f dm differentiable on R? What functions arise in this way? We restrict our attention to real valued functions. In fact, splitting a function f as a difference of two positive functions, we may suppose that f ≥0; and so the integral will be monotone increasing. 4.1.1. DEFINITION. The upper and lower derivative from left and right of a real valued function f are deﬁned as Drf(x) = lim sup h→0+ f(x + h) −f(x) h Drf(x) = lim inf h→0+ f(x + h) −f(x) h Dlf(x) = lim sup h→0+ f(x) −f(x −h) h Dlf(x) = lim inf h→0+ f(x) −f(x −h) h Then f is differentiable at x if Drf(x) = Drf(x) = Dlf(x) = Dlf(x) ∈R. This deﬁnition is clearly equivalent to the familiar deﬁnition from calculus: a function f is differentiable at x if and only if there is a ﬁnite limit f′(x) = lim h→0 f(x + h) −f(x) h . We require a technical deﬁnition. A singleton {a} and the empty set are con-sidered to be degenerate intervals. 4.1.2. DEFINITION. If E ⊂R, a collection J of non-degenerate intervals is a Vitali cover if for every x ∈E and ε > 0, there is I ∈J so that x ∈I and m(I) < ε. 49 50 Differentiation and Signed Measures 4.1.3. VITALI COVERING LEMMA. If E ⊂R has ﬁnite outer measure (m∗(E) < ∞), J is a Vitali cover of E, and ε > 0, then there are disjoint intervals I1, . . . , IN ∈J such that m∗E \ SN j=1 Ij < ε. PROOF. Fix an open U ⊃E with m(U) < ∞. Let J ′ = {I ∈J : I ⊂U}. This is also a Vitali cover since if x ∈E and δ = dist(x, Uc), then any interval I ∈J containing x of length less than δ will belong to J ′. Recursively choose disjoint Ik ∈J ′ so that m(Ik) > αk/2 where αk = sup{m(I) : I ∈J ′, I disjoint from I1, . . . , Ik−1}. Observe that m S k≥1 Ik) = P k≥1 m(Ik) ≤m(U) < ∞. Thus αk is a summable sequence. So we may choose N so that P k>N m(Ik) < ε/5. We claim that I1, . . . , IN works. Let X = E \ SN j=1 Ij. If x ∈X, δ := dist(x, SN j=1 Ij) > 0. Hence there is some I ∈J with x ∈I and m(I) < δ. So I is disjoint from SN j=1 Ij, and hence m(I) ≤αN+1. Pick K > N so that αK+1 < m(I) ≤αK. Then by construction, I cannot be disjoint from SK j=1 Ij. Thus there is some k with N < k ≤K so that Ik ∩I ̸= ∅. Now m(Ik) ≥αk/2 ≥αK/2 ≥m(I)/2. Hence dist(x, midpoint of Ik) ≤1 2m(Ik) + m(I) ≤5 2m(Ik). Let Jk be the closed interval with the same midpoint as Ik but with m(Jk) = 5m(Ik). Then x ∈Jk. Therefore X ⊂S k>N Jk. If follows that m∗(X) ≤ X k>N m(Jk) = 5 X k>N m(Ik) < ε. This proves the claim. ■ 4.1.4. THEOREM. Let f : [a, b] →R be monotone increasing. Then f is con-tinuous except on a countable set, and is differentiable except on a set of measure 0. The derivative f′ is integrable, and Z b a f′ dm ≤f(b) −f(a). PROOF. Deﬁne f(x) = f(a) for x < a and f(x) = f(b) for x > b. Since f is monotone, for c ∈[a, b], we have f(c−) = lim x→c−f(x) = sup xc f(x) = lim x→c+ f(x). Thus f is continuous at c unless it is a jump discontinuity, in which case the jump has length j(c) = f(c+)−f(c−). Clearly P j(c) ≤f(b)−f(a) < ∞. In particular the number of points with jump at least δ > 0 is at most δ−1(f(b) −f(a)) < ∞. It follows that the number of discontinuities is countable. Clearly we have Drf(x) ≤Drf(x) and Dlf(x) ≤Dlf(x). We will show that Dlf(x) ≤Dr(x) a.e. and Drf(x) ≤Dlf(x) a.e. For u, v ∈Q with u < v, 4.1 Differentiation 51 let Eu,v = {x : Dr(x) < u < v < Dlf(x)} and E = [ u<v∈Q Eu,v. To show that m∗(E) = 0, it sufﬁces to show that m∗(Eu,v) = 0 for all u < v. Let m∗(Eu,v) = s and ﬁx an ε > 0. Choose an open set U ⊃Eu,v with m(U) < s + ε. Let J = {I = [x, x + h] ⊂U : f(x + h) −f(x) < uh}. By deﬁnition of Drf(x), this contains arbitrarily small intervals [x, x + h] for each x ∈Eu,v. Thus J is a Vitali cover of E. By the Vitali Covering Lemma, we can ﬁnd I1 = [x1, x1 + h1], . . . , IN = [xN, xN + hN] disjoint intervals in J so that m∗(Eu,v \ SN j=1 Ij) < ε. Therefore s −ε < N X j=1 m(Ij) = N X j=1 hj < m(U) < s + ε, and m∗(Eu,v ∩SN j=1 Ij) > s −ε. Let F = Eu,v ∩ N [ j=1 (xj, xj + hj) ⊂ N [ j=1 (xj, xj + hj) =: V. Consider J ′ = {I = [x −k, x] ⊂V : f(x) −f(x −k) > vk}. As for J , one sees that J ′ is a Vitali cover of F. Choose disjoint intervals Ji = [yi −ki, yi] ∈J ′ for 1 ≤i ≤M so that m∗(F \ SM i=1 Ji) < ε. Therefore M X i=1 ki = M X i=1 m(Ji) > m∗(F) −ε > s −2ε. Since the intervals Ji are disjoint and are contained in SN j=1 Ij, we have that v(s −2ε) < M X i=1 kiv < M X i=1 f(yi) −f(yi −ki) ≤ N X j=1 f(xj + hj) −f(xj) < N X j=1 uhj < u(s + ε). Letting ε →0 yields vs ≤us, and thus s = 0. Hence m∗(Eu,v) = 0. Therefore, since m is complete, m(E) = 0. Consequently, Dlf(x) ≤Drf(x) on Ec, that is a.e. Similarly Drf(x) ≤Dlf(x) a.e. Thus f′(x) exists except on a set of measure 0. Note however that the derivative might be +∞. We will show soon that this also happens only on a set of measure 0. Deﬁne gn(x) = n f(x + 1 n) −f(x) on [a, b]. Monotone functions are Borel, and thus are measurable. So gn is measurable and positive. Moreover lim n→∞gn(x) = lim n→∞ f(x + 1 n) −f(x) 1/n = f′(x) 52 Differentiation and Signed Measures whenever f′(x) is deﬁned, which is almost everywhere. Therefore f′ is measurable and f′ ≥0. We can apply Fatou’s Lemma to this sequence to get Z b a f′ dm ≤lim inf n→∞ Z b a gn dm = lim inf n→∞n Z b+ 1 n a+ 1 n f dm −n Z b a f dm = lim inf n→∞n Z b+ 1 n b f(b) dm −n Z a+ 1 n a f dm ≤f(b) −f(a) < ∞. In particular, f′ is integrable. This implies that f′(x) < ∞a.e. ■ 4.1.5. EXAMPLE. Recall the Cantor ternary function. This is deﬁned on [0, 1] by f(0) = 0, f(1) = 1. Then f(x) = 1 2 for x ∈[1 3, 2 3], the (closure of) the middle third removed in the construction of the Cantor set. Then we set f(x) = 1 4 on [1 9, 2 9] and f(x) = 3 4 on [7 9, 8 9]. This repeats, on each middle third removed, f is deﬁned as the midpoint between the values deﬁned at the endpoints of the (larger) interval. Finally, for x ∈C, where C is the Cantor set, we can deﬁne f(x) = sup{f(y) : 0 ≤y ≤x, y ̸∈C}. This evidently yields a monotone increasing function. Moreover the values attained by f include all diadic rationals 2−nk for 0 ≤k ≤2n. So the range of f is dense in [0, 1]. In particular, f cannot have any jump discontinuities. Therefore f is continuous. Next observe that on each of the removed (open) intervals, ( 1 3, 2 3), (1 9, 2 9) and ( 7 9, 8 9), etc., the function f is constant. Therefore it is differentiable with f′(x) = 0. This occurs on [0, 1] \ C, and since m(C) = 0, we see that f′ is deﬁned and ﬁnite almost everywhere. However Z 1 0 f′ dm = 0 < 1 = f(1) −f(0). So it can certainly be the case that you cannot recover f from its derivative. 4.1.6. DEFINITION. A function f : [a, b] →R is bounded variation (belongs to BV[a, b]) if V b a (f) = sup n n X i=1 \|f(xi)−f(xi−1)\| : n ≥1, a=x0 <x1 <. . .<xn =b o < ∞. 4.1.7. THEOREM. A function f ∈BV[a, b] if and only if f = g −h where g, h are monotone increasing. 4.1 Differentiation 53 PROOF. If f = g −h where g, h are monotone increasing, then for a partition a=x0 <x1 <. . .<xn =b, n X i=1 \|f(xi) −f(xi−1)\| ≤ n X i=1 \|g(xi) −g(xi−1)\| + n X i=1 \|h(xi) −h(xi−1)\| = g(b) −g(a) + h(b) −h(a) =: L. So V b a (f) ≤L. Conversely, if f ∈BV , deﬁne g(x) = V x a (f). Clearly this is an increasing function of x since if a ≤x < y ≤b, V y a (f) = V x a (f) + V y x (f). Let h = g −f, so that f = g −h. If a ≤x < y ≤b, then h(y) −h(x) = V y x (f) −(f(y) −f(x)) ≥\|f(y) −f(x)\| −(f(y) −f(x)) ≥0. So h is also monotone increasing. ■ 4.1.8. COROLLARY. If f ∈BV [a, b], then f is continuous except on a count-able set, f′(x) exists a.e.(m) and f′ is m-integrable. 4.1.9. COROLLARY. If f ∈L1[a, b], then F(x) = Z x a f dm is bounded varia-tion. PROOF. Write f = f+ −f−where f+ = f ∨0 and f−= (−f)∨0 are positive integrable functions. Then g(x) = R x a f+ dm and h(x) = R x a f−dm are monotone increasing. Hence f = g −h is BV. ■ 4.1.10. DEFINITION. A function f : [a, b] →R is absolutely continuous (AC) if for all ε > 0, there is a δ > 0 so that whenever (xi, yi) are disjoint intervals in [a, b] and Pn i=1 yi −xi < δ, then Pn i=1 \|f(yi) −f(xi)\| < ε. 4.1.11. EXAMPLE. The Cantor ternary function f is monotone, and thus BV. However it is not absolutely continuous. If xi, yi for 1 ≤i ≤2n are the endpoints of the intervals remaining after n stages of the construction of the Cantor set, then because f is constant on the complements of these intervals, we have n X i=1 \|f(yi) −f(xi)\| = 1 and 2n X i=1 yi −xi = 2 3 n. Since 2 3 n →0, f is not absolutely continuous. 4.1.12. PROPOSITION. Let f ∈L1(µ) and ε > 0. Then there is a δ > 0 so that whenever µ(A) < δ, Z A \|f\| dµ < ε. 54 Differentiation and Signed Measures PROOF. Choose a simple function 0 ≤ϕ ≤\|f\| so that R ϕ dµ > R \|f\| dµ −ε 2. Since ϕ is simple, there is a constant M so that ϕ ≤M. Set δ = ε 2M . If µ(A) < δ, then Z A \|f\| dµ < Z A ϕ dµ + ε 2 ≤Mµ(A) + ε 2 < ε. ■ 4.1.13. COROLLARY. If f ∈L1[a, b], then F(x) = Z x a f dm is absolutely continuous. PROOF. Given ε > 0, let δ be provided by Proposition 4.1.12. Then if (xi, yi) are disjoint intervals in [a, b] and Pn i=1 yi −xi < δ, then n X i=1 \|F(yi) −F(xi)\| = n X i=1 Z yi xi f dm ≤ Z Sn i=1[xi,yi] \|f\| dm < ε. ■ 4.1.14. LEMMA. If f is absolutely continuous on [a, b], then it has bounded variation. PROOF. Take ε = 1, and ﬁnd δ > 0. Split [a, b] = Sp j=1[aj−1, aj] where aj −aj−1 < δ for 1 ≤j ≤p. If [aj−1, aj] contains disjoint intervals (xi, yi) for 1 ≤i ≤n, then Pn i=1 yi −xi ≤aj −aj−1 < δ. Hence Pn i=1 \|f(yi) −f(xi)\| < 1. Taking the supremum yields V aj aj−1(f) ≤1. Thus V b a (f) = Pp j=1 V aj aj−1(f) ≤p. ■ 4.1.15. LEMMA. If f ∈L1[a, b] and F(x) = Z x a f dm is monotone increasing, then f ≥0 a.e. PROOF. Let E = {x : f(x) < 0} and En = {x : f(x) < −1 n} for n ≥1. Then if m(E) > 0, there is some n so that m(En) > 0. Let ε = m(En)/2n; and choose δ > 0 using Proposition 4.1.12. Select an open set U ⊃En so that m(U \ En) < δ. Write U = ˙ S i≥1(xi, yi). Then 0 ≤ X i≥1 F(yi) −F(xi) = Z U f dm = Z En f dm + Z U\En f dm ≤−m(En) n + ε < 0. This contradiction shows that m(E) = 0, or that f ≥0 a.e. ■ The following consequence is immediate. 4.1 Differentiation 55 4.1.16. COROLLARY. Let f ∈L1[a, b] and F(x) = Z x a f dm. If F = 0, then f = 0 a.e. The main result here is the following result of Lebesgue, his substitute for the Fundamental Theorem of Calculus, which answers the ﬁrst part of the question posed at the beginning of this section. 4.1.17. LEBESGUE DIFFERENTIATION THEOREM. Let f ∈L1[a, b] and F(x) = c + Z x a f dm. Then F ′(x) = f(x) a.e. PROOF. Since F is BV by Corollary 4.1.9, Corollary 4.1.8 shows that F ′ exists a.e. and the derivative is integrable. Extend f by setting f(x) = 0 for x > b; so that F(x) = F(b) for x > b. Let gn(x) = n(F(x + 1 n) −F(x)). Then gn converges pointwise a.e. to F ′. Case 1 \|f\| ≤M. Then gn(x) = n Z x+ 1 n x f dm. So \|gn(x)\| ≤n Z x+ 1 n x M dx = M. Thus \|gn\| ≤Mχ[a,b], which is integrable. By the LDCT and the fact that F is continuous, for c ∈[a, b], Z c a F ′ dm = lim n→∞ Z c a gn dm = lim n→∞n Z c a F(x + 1 n) −F(x) dx = lim n→∞n Z c+ 1 n c F(x) dx −n Z a+ 1 n a F(x) dx = F(c) −F(a) = Z c a f dm. Therefore Z c a (F ′ −f) dm = 0 for all c ∈[a, b]. By Corollary 4.1.16, F ′ = f a.e. Case 2 f ≥0. Let fn = f ∧n for n ≥1. Then fn is bounded and case 1 applies. Observe that F(x) = Z x a fn dm + Z x a f −fn dm. Because the second term is monotone increasing, F ′(x) = fn(x) + d dx Z x a f −fn dm ≥fn(x) a.e. 56 Differentiation and Signed Measures Consequently, F ′(x) ≥limn→∞fn(x) = f(x) a.e. For c ∈[a, b], Z c a F ′(x) dm ≤F(c) −F(a) = Z c a f dm ≤ Z c a F ′(x) dm. Therefore Z c a F ′(x) −f(x) dm = 0 for all c ∈[a, b]. Hence F ′ = f a.e. by Corollary 4.1.16. Case 3. Write f = f+ −f−where f± ∈L+ ∩L1(m). Then by case 2, we obtain that F ′ = f+ −f−= f a.e. ■ Next we wish to characterize which functions are integrals. We ﬁrst need an-other lemma. 4.1.18. LEMMA. If f ∈C[a, b] is absolutely continuous and f′ = 0 a.e., then f is constant. PROOF. Fix c ∈(a, b]. Let ε > 0. Obtain δ > 0 from the deﬁnition of absolute continuity. Set E = {x ∈(a, c) : f′(x) = 0}. So m([a, c] \ E) = 0. Deﬁne J = {[x, x+h] : x ∈E, h > 0, [x, x+h] ⊂(a, c) and \|f(x+h)−f(x)\| < εh}. This collection is a Vitali cover of E because for each x, all sufﬁciently small h work. Therefore there are disjoint intervals I1, . . . , In in J so that m [a, c] \ n [ j=1 Ij = m E \ n [ j=1 Ij < δ. Write Ij = (aj, bj) ordered so that a < a1 < b1 < a2 < b2 < · · · < an < bn < c. Then \|f(c) −f(a)\| ≤ n X j=1 \|f(bj) −f(aj)\| + \|f(a1)−f(a)\| + n−1 X j=1 \|f(aj+1)−f(bj)\| + \|f(c)−f(bn)\| < n X j=1 ε\|bj −aj\| + ε < (c −a + 1)ε. The second term in parentheses is at most ε by absolute continuity since the total length of the intervals is less than δ. Now ε > 0 was arbitrary, and therefore f(c) = f(a); whence f is constant. ■ 4.2 Signed Measures 57 4.1.19. THEOREM. Let F : [a, b] →R. The following are equivalent: (1) There is an f ∈L1(a, b) so that F(x) = c + Z x a f dm. (2) F is absolutely continuous. (3) F is differentiable a.e., F ′ ∈L1(a, b) and F(x) = F(a) + Z x a F ′ dm. PROOF. Lebesgue’s Differentiation Theorem shows that (1) implies (3). Clearly (3) implies (1). Lemma 4.1.9 shows that integrals are AC, so (1) implies (2). If (2) holds, then F is BV by Lemma 4.1.14; and thus by Corollary 4.1.8, F ′ exists a.e. and belongs to L1(m). Let G(x) = F(a) + Z x a F ′ dm. Then Lebesgue’s Differentiation Theorem shows that G′ = F ′ a.e. Both F and G are absolutely continuous, and (G −F)′ = 0 a.e. Therefore Lemma 4.1.18 shows that G −F is constant; and G(a) = F(a). So (3) holds. ■ 4.2. Signed Measures 4.2.1. DEFINITION. A signed measure on (X, B) is a map ν : B →R∪{±∞} such that ν(∅) = 0, ν takes at most one of the values ±∞, and ν is countably additive, meaning that if Ei are disjoint sets in B, then ν ˙ S i≥1Ei = P i≥1 ν(Ei). 4.2.2. REMARK. The deﬁnition implies that if ν ˙ S i≥1Ei < ∞, then the series P i≥1 ν(Ei) converges absolutely. One argument would be that the union is independent of order, and thus the sum should be independent of the order as well. But unconditionally convergent sequences of real numbers converge absolutely. An alternative argument, which is better, is that if the sum converges only conditionally, then the sum of the positive terms diverges, as does the sum of the negative terms. But then if J = {i : ν(Ei) ≥0}, we have that ν ˙ S i∈JEi = P i∈J ν(Ei) = +∞ and ν ˙ S i̸∈JEi = P i̸∈J ν(Ei) = −∞. This is not permitted for signed measures because we cannot make sense of ν(E ˙ ∪F) = ν(E) + ν(F) is ν(E) = +∞and ν(F) = −∞. 4.2.3. EXAMPLES. (1) Let f ∈L1 R(µ) and deﬁne ν(E) = R E f dµ. In this case, neither ±∞arises as a value. The absolute convergence over countably many disjoint sets follows from the LDCT. 58 Differentiation and Signed Measures (2) Let f ∈L+(µ) and g ∈L+ ∩L1(µ). Deﬁne ν(E) = R E f −g dµ. If f is not integrable, then ν(X) = R f dµ−∥g∥1 = +∞. If E = ˙ S i≥1Ei and R E f dµ < ∞, then countable additivity follows as in (1). 4.2.4. DEFINITION. A null set for a signed measure ν is a measurable set E such that ν(F) = 0 for all F ⊂E, F ∈B. A positive set (or negative set) for ν is a measurable set E such that ν(F) ≥0 (or ≤0) for all F ⊂E, F ∈B. 4.2.5. HAHN DECOMPOSITION THEOREM. Let ν be a signed measure on (X, B). Then there are sets P, N ∈B so that X = P ˙ ∪N, P is a positive set, and N is a negative set. If X = P ′ ˙ ∪N′ is another decomposition into positive and negative sets, then P△P ′ is a null set. 4.2.6. LEMMA. If 0 < ν(E) < ∞, then there is a positive set A ⊂E with ν(A) > 0. PROOF. If E contains a set of negative measure, choose a set B1 ⊂E in B so that ν(B1) ≤max −1, 1 2 inf{ν(B) : B ⊂E, B ∈B} . Recursively choose Bn ⊂E \ ˙ Sn−1 i=1 Bi with ν(Bn) ≤max −1, 1 2 inf{ν(B) : B ⊂E \ ˙ [n−1 i=1 Bi, B ∈B} . Either this terminates because A = E \ ˙ Sn i=1Bi is a positive set, or there is an inﬁnite sequence. In this case, let A = E \ ˙ S i≥1Bi. Now by countable additivity, ν(E) = ν(A) + X i≥1 ν(Bi). Since \|ν(E)\| < ∞, this series converges absolutely. Therefore ν(A) = ν(E) − X i≥1 ν(Bi) < ∞. Moreover this shows that ν(A) ≥ν(E) > 0. The convergence also shows that ν(Bi) →0. If B ⊂A had ν(B) < 0, then ν(B) < 2ν(Bn) for some n. But this contradicts the fact that inf{ν(B) : B ⊂E \ ˙ [n−1 i=1 Bi, B ∈B} > 2ν(Bn). Hence if B ⊂A, then ν(B) ≥0. Thus A is a positive set. ■ 4.2.7. LEMMA. If An are positive sets, so is A := S n≥1 An. 4.2 Signed Measures 59 PROOF. If B ⊂A and B ∈B, let Bn = B ∩ An \ Sn−1 i=1 Ai for n ≥1. Then the Bn are pairwise disjoint sets in B such that B = ˙ S n≥1Bn. Since each An is positive, ν(Bn) ≥0. Therefore ν(B) = P n≥1 ν(Bn) ≥0. ■ PROOF OF THE HAHN DECOMPOSITION. We may suppose that ν does not take the value +∞(by considering −ν instead if necessary). Let m = sup{ν(A) : A is positive}. Choose positive sets An with ν(An) →m. Set P = S n≥1 An. By Lemma 4.2.7, P is a positive set. Moreover ν(P) = ν(An) + ν(P \ An) ≥ν(An) for all n ≥1, and hence ν(P) = m. Since ν does not take the value +∞, we see that m < ∞. Let N = P c. Claim: N is negative. If it isn’t, there is a subset E ⊂N such that ν(E) > 0. By Lemma 4.2.6, E contains a positive set A with ν(A) > 0. Then P ˙ ∪A is positive with ν(P) + ν(A) > m. This contradicts the deﬁnition of m. Hence N must be negative. Uniqueness. Suppose that X = P ′ ˙ ∪N′ is a second decomposition of X into a positive and negative set. Then A = P \ P ′ = N′ \ N is both positive and negative, and hence is a null set. Similarly B = P ′ \ P = N \ N′ is a null set. Thus P△P ′ = A ∪B is a null set. ■ 4.2.8. DEFINITION. Two signed measures µ, ν on (X, B) are mutually singular (µ ⊥ν) if there is a decomposition X = A ˙ ∪B such that A is ν-null and B is µ-null. 4.2.9. JORDAN DECOMPOSITION THEOREM. If ν is a signed measure on (X, B), there there is a unique pair of mutually singular (positive) measures ν+ and ν−such that ν = ν+ −ν−. PROOF. Let X = P ˙ ∪N be the Hahn decomposition for ν. Deﬁne ν+(A) = ν(A ∩P) and ν−(A) = −ν(A ∩N). Then by construction, ν+ and ν−are positive measures supported on disjoint sets, so they are mutually singular, and ν = ν+−ν−. For uniqueness, suppose that µ+ ⊥µ−and ν = µ+ −µ−. Let X = P ′ ˙ ∪N′ such that P ′ is µ−-null and N′ is µ+-null. Then P ′ is a positive set for ν and N′ is a negative set. By the uniqueness of the Hahn decomposition, P△P ′ is a ν-null set. For any set A ∈B, µ+(A) = ν(A ∩P ′) = ν(A ∩P) = ν+(A). So µ+ = ν+. Similarly µ−= ν−. ■ 4.2.10. DEFINITION. The absolute value of a signed measure with Jordan de-composition ν = ν+ −ν−is \|ν\| = ν+ + ν−. 60 Differentiation and Signed Measures Note that a set A ∈B is ν-null if and only if \|ν\|(A) = 0. 4.3. Decomposing Measures 4.3.1. DEFINITION. A signed measure ν is absolutely continuous with respect to a (positive) measure µ (ν ≪µ) if A ∈B and µ(A) = 0 implies that ν(A) = 0. Since any measurable subset of A is also a µ-null set, A must be ν-null. This is equivalent to saying that \|ν\|(A) = 0. So ν ≪µ if and only if \|ν\| ≪µ. 4.3.2. RADON-NIKODYM THEOREM. Let µ and ν be σ-ﬁnite measures on (X, B). Suppose that ν ≪µ. Then there is an f ∈L+ so that ν(E) = Z E f dµ for E ∈B. Also f is uniquely determined a.e.(µ). PROOF. First we assume that µ(X) < ∞and ν(X) < ∞. For each r ∈Q+, ν −rµ has a Hahn decomposition (Pr, Nr). Also, let P0 = X and N0 = ∅. Deﬁne f(x) = sup{r ≥0 : x ∈Pr}. Then for t ≥0, f−1(t, ∞] = S r>t Pr is measurable. So f ∈L+. If r < s in Q+, then Ps is a positive set for ν −sµ and thus is positive for ν −rµ = (ν −sµ) + (s −r)µ. Since Nr is a negative set for ν −rµ, we have µ(Nr ∩Ps) = 0. Therefore µ Nr ∩S s>r∈Q+ Ps = 0. Hence f\|Nr ≤r a.e.(µ). Thus µ(f−1(r, ∞]) ≤µ(Pr). Since Pr is positive for ν −rµ, rµ(Pr) ≤ν(Pr). Hence µ(Pr) ≤ν(Pr)/r ≤ν(X)/r. This goes to 0 as r →∞, and thus µ(f−1(∞)) ≤limr→∞µ(Pr) = 0. Therefore f < ∞a.e.(µ). Let E ∈B. Fix N. Deﬁne Ek = E ∩P k N ∩N k+1 N ) for k ≥0; and let E∞= E \ S k≥0 Ek = E ∩T r Pr. Then µ(E∞) = 0 and so ν(E∞) = 0 as well. Observe that (ν −k N µ)(Ek) ≥0 ≥(ν −k+1 N µ)(Ek). Thus k N µ(Ek) ≤ν(Ek) ≤k+1 N µ(Ek). Also k N ≤f(x) ≤k+1 N for a.e. x ∈Ek. So k N χEk ≤fχEk ≤k+1 N χEk. Integrating yields k N µ(Ek) ≤ Z Ek f dν ≤k+1 N µ(Ek). 4.3 Decomposing Measures 61 Summing over k ∈N, we obtain that X k≥0 k N µ(Ek) ≤ X k≥0 ν(Ek) = ν(E) ≤ X k≥0 k+1 N µ(Ek) = X k≥0 k N µ(Ek) + µ(E) N . And similarly X k≥0 k N µ(Ek) ≤ X k≥0 Z Ek f dν = Z E f dν ≤ X k≥0 k+1 N µ(Ek) = X k≥0 k N µ(Ek) + µ(E) N . Therefore ν(E) − Z E f dν ≤µ(E) N . However N is arbitrary, and hence ν(E) = Z E f dν. Suppose that f, g ∈L+ such that ν(E) = Z E f dµ = Z E g dµ for E ∈B. Deﬁne A = {x : f(x) > g(x)} and An = {x : f(x) ≥g(x) + 1 n}. If µ(A) > 0, then µ(An) > 0 for some n. But then ν(An) = Z An f dµ ≥ Z An g + 1 n dµ = Z An g dµ + 1 nµ(An) > ν(An). Hence f ≤g a.e.(µ). Similarly g ≤f a.e.(µ), so that f = g a.e.(µ). For the general case, use the fact that µ and ν are σ-ﬁnite to chop X up into a disjoint union of countable many pieces Xn so that µ(Xn) < ∞and ν(Xn) < ∞ for each n ≥1. Then apply the previous argument on each piece, and combine. ■ 4.3.3. EXAMPLE. On ([0, 1], Bor([0, 1])), consider Lebesgue measure m and counting measure mc. Then m ≪mc but there is no function f ∈L+ so that m(E) = Z E f dmc. But mc is not σ-ﬁnite, and this shows the necessity of this condition in the Radon-Nikodym Theorem. Next we deduce the corresponding results for signed measures. 4.3.4. COROLLARY. If ν is a signed measure on (X, B), there is a measurable function f with \|f\| = 1 so that ν(E) = Z E f d\|ν\|. If in addition, \|ν\| and µ are σ-ﬁnite measures on (X, B) and ν ≪µ, there is a measurable function g = g+ −g−with at least one of g± integrable so that ν(E) = Z E g dµ. PROOF. Let X = P ˙ ∪N be the Hahn decomposition for ν. Deﬁne f(x) = 1 on P and f(x) = −1 on N. Then we have that ν = ν+ −ν−and \|ν\| = ν+ + ν− 62 Differentiation and Signed Measures and ν+(E) = ν(E ∩P). So Z E f d\|ν\| = Z E∩P 1 dν+ + Z E∩N −1 dν−= ν+(E) −ν−(E) = ν(E). If \|ν\| ≪µ are σ-ﬁnite, then by the Radon-Nikodym theorem, there is a mea-surable function h ∈L+ so that \|ν\|(E) = Z E h dµ. Then ν(E) = Z E∩P h dµ − Z E∩N h dµ = Z E fh dµ. So let g = fh = hχP −hχN =: g+ −g−. Then in particular, one of ν(P) and ν(N) is ﬁnite, which means that one of g± is integrable. ■ The Radon-Nikodym Theorem is usually combined with the following decom-position result. 4.3.5. LEBESGUE DECOMPOSITION THEOREM. Let ν, µ be two σ-ﬁnite measures on (X, B). Then there is a unique decomposition ν = νa + νs so that νa ≪µ and νs ⊥µ. PROOF. Let λ = µ + ν. This is also a σ-ﬁnite measure on (X, B). Clearly µ ≪λ and ν ≪λ. By the Radon-Nikodym Theorem, there are functions f, g ∈ L+ so that µ(E) = Z E f dλ and ν = Z E g dλ for E ∈B. Let A = f−1(0, ∞] and B = f−1(0). Set νa(E) = ν(E ∩A) and νs(E) = ν(E ∩B). Clearly ν = νa + νs. Also νs ⊥µ since it is supported on B, which is a µ-null set. If µ(E) = 0, then fχE = 0 a.e.(λ); and hence λ(E ∩A) = 0. Therefore νa(E) = ν(E ∩A) = 0. Thus νa ≪µ. Uniqueness. Suppose that ν = ν′ a+ν′ s is another decomposition so that ν′ a ≪µ and ν′ s ⊥µ. Then there is a set A′ ∈B so that ν′ s(A′) = 0 and µ(B′) = 0, where B′ = A′c. Thus µ(B ∪B′) = 0 and νs(A ∩A′) = ν′ s(A ∩A′) = 0. Now if E ⊂B ∪B′, then µ(E) = 0 and so νa(E) = ν′ a(E) = 0; and so νs(E) = ν′ s(E) = ν(E). On the other hand, if E ⊂A ∩A′, then νs(E) = ν′ s(E) = 0. Combining, we deduce that ν′ s = νs. Hence ν′ a = νa as well. ■ We ﬁnish this section with a discussion of measures taking complex values. These measures are not allowed to take inﬁnite values. 4.3.6. DEFINITION. A complex measure on (X, B) is a map ν : B →C such that ν(∅) = 0 that is countably additive: If E = ˙ S i≥1Ei, then ν(E) = P i≥1 ν(Ei) and this series converges absolutely. 4.3 Decomposing Measures 63 4.3.7. REMARK. As in Remark 4.2.2, when E = ˙ S i≥1Ei, countable additivity together with the fact that every set has a ﬁnite measure forces absolute convergence of the series P i≥1 ν(Ei). Note that Re ν and Im ν are ﬁnite signed measures. Thus there are positive ﬁnite measures νi for 1 ≤i ≤4 so that Re ν = ν1 −ν2 and Im ν = ν3 −ν4. Hence ν = ν1 −ν2 + iν3 −iν4. 4.3.8. EXAMPLE. The main example (and essentially the only example) is ob-tained as follows: let µ be a measure and let f ∈L1(µ). Deﬁne ν(E) = R E f dµ. You can readily check that this is a complex measure. We will write dν = f dµ and dν dµ = f in this case. The second statement of the following result is also sometimes called the Radon-Nikodym Theorem. 4.3.9. THEOREM. If ν is a complex measure on (X, B), there is a unique ﬁnite measure \|ν\| and measurable function h with \|h\| = 1 a.e.(\|ν\|) so that dν = h d\|ν\|. Moreover if µ is a σ-ﬁnite measure on (X, B), then ν decomposes as νa +νs where νa = f dµ for f ∈L1(µ) and νs is supported on a µ-null set. PROOF. With the notation as in the discussion preceding the proposition, let µ = ν1 + ν2 + ν3 + ν4. Then Re ν ≪µ and Im ν ≪µ. By the Radon-Nikodym Theorem, there are f, g ∈L1(µ) so that d Re ν = f dµ and d Im ν = g dµ. Hence ν(E) = Z E f + ig dµ. Deﬁne \|ν\|(E) = Z E \|f + ig\| dµ and h = sign(f + ig). Then \|h\| = 1 a.e.(\|ν\|) and ν(E) = Z E h d\|ν\|. Also \|ν\|(X) = ∥f + ig∥1 < ∞; so \|ν\| is ﬁnite. Uniqueness is left as an exercise. Now if µ is a σ-ﬁnite, the Lebesgue deomposition for \|ν\| yields \|ν\| = \|ν\|a+\|ν\|s where d\|ν\|a = f dµ for f ∈L+ ∩L1(µ) and \|ν\|s is supported on a µ-null set A. Then dν = h d\|ν\| = h d\|ν\|a + h d\|ν\|s = hf dµ + h d\|ν\|s. Thus dνa = hf dµ ≪µ and hf ∈L1(µ) and dνs = h d\|ν\|s is supported on the µ-null set A. ■ Finally, we can integrate with respect to complex measures. If dν = hd\|ν\| is a complex measure and f ∈L1(\|ν\|), then we deﬁne Z f dν := Z fh d\|ν\|. It is easy to check that this is linear. It is advisable to convert to integrals with respect to positive measures when taking limits. CHAPTER 5 Lp spaces 5.1. Lp as a Banach space 5.1.1. DEFINITION. Let (X, B, µ) be a measure space. For 1 ≤p < ∞, let “Lp(µ)” = f measurable, complex valued : ∥f∥p p = Z \|f\|p dµ < ∞ . Set N = {f measurable, complex valued : f = 0 a.e.(µ)} and deﬁne Lp(µ) = “Lp(µ)”/N with the norm ∥[f]∥p = ∥f∥p. “L∞(µ)” = {f measurable, complex valued : ∥f∥∞= ess sup \|f\| < ∞} where ess sup \|f\| = sup{t ≥0 : µ({x : \|f(x)\| > t} > 0}. Set L∞(µ) = “L∞(µ)”/N with norm ∥[f]∥∞= ∥f∥∞. By convention, we write elements of Lp(µ) as f, where the equivalence class is understood. Note that “Lp” is a linear space because if f, g ∈Lp(µ), then λf ∈Lp(µ) with ∥λf∥p = \|λ\| ∥f∥p for λ ∈C; and \|f + g\|p ≤(2 max{\|f\|, \|g\|})p ≤2p(\|f\|p + \|g\|p). whence ∥f + g∥p p ≤2(∥f∥p + ∥g∥p) < ∞. Notice that N is a subspace of “Lp(µ)”, so that Lp(µ) is also a vector space. Moreover, if f is measurable, then ∥f∥p p = R \|f\|p dµ = 0 if and only if f = 0 a.e.(µ) if and only if f ∈N. In particular, ∥· ∥p is only a seminorm on “Lp(µ)”. However, for f ∈Lp(µ), we have ∥f∥p = 0 if and only if f = 0. To verify that ∥· ∥p is a norm on Lp(µ), we need to verify the triangle inequality, which means eliminating the annoying 2 in the inequality above. A function belongs to “L∞(µ)” if it agrees with a bounded function a.e.(µ). The triangle inequality is very easy here. It is also very easy for L1(µ) (see sec-tion 3.4). 5.1.2. MINKOWSKI’S INEQUALITY. Let (X, B, µ) be a measure space. For 1 < p < ∞, the triangle inequality is valid for Lp(µ). Equality holds only when f and g lie in a 1-dimensional subspace. 64 5.1 Lp as a Banach space 65 PROOF. Let f, g ∈Lp(µ). We may suppose that neither is 0, as that case is trivial. Deﬁne A = ∥f∥p and B = ∥g∥p, and set f0 = f/A and g0 = g/B; so ∥f0∥p = 1 = ∥g0∥p. Consider ϕ(x) = xp on [0, ∞). Note that ϕ′′(x) = p(p−1 )xp−2 > 0 on (0, ∞), and thus ϕ(x) is a strictly convex function, meaning that for all x1, x2 ∈[0, ∞) and 0 ≤t ≤1, ϕ(tx1 + (1 −t)x2) ≤tϕ(x1) + (1 −t)ϕ(x2), with equality only when x1 = x2 or t = 0 or t = 1. That is every chord between distinct points on the curve y = ϕ(x) lies strictly above the curve. Hence for any x ∈X, A A+B \|f0(x)\| + B A+B \|g0(x)\| p ≤ A A+B \|f0(x)\|p + B A+B \|g0(x)\|p with equality only when \|f0(x)\| = \|g0(x)\|. Also 1 A + B \|f(x) + g(x)\| ≤\|f(x)\| + \|g(x)\| A + B = A A + B \|f0(x)\| + B A + B \|g0(x)\|, with equality only when sign(f(x)) = sign(g(x)). Integrate the pth power: 1 (A+B)p Z \|f(x) + g(x)\|p dµ ≤ A A+B Z \|f0(x)\|p dx + B A+B Z b a \|g0(x)\|p dµ = A A+B ∥f0∥p p + B A+B ∥g0∥p p = 1. Multiplying through by (A + B)p and take the pth root to get ∥f + g∥p ≤(A + B) = ∥f∥p + ∥g∥p. For equality, we require that \|f0(x)\| = \|g0(x)\| a.e. and sign(f(x)) = sign(g(x)) a.e. This implies that f0 = g0 a.e.(µ), so g = Bf/A; i.e., g is a scalar multiple of f. ■ Now that we have established that Lp(µ) is a normed space, we will show that it is complete, and so is a Banach space. This extends the result for L1(µ), Theorem 3.4.2. This is another important piece of evidence that this is the right context for integration. 5.1.3. RIESZ-FISCHER THEOREM. Let (X, B, µ) be a measure space. For 1 ≤p ≤∞, Lp(µ) is complete. PROOF. Suppose that (fn)n≥1 is a Cauchy sequence in Lp(µ). Select a subse-quence (nj)j≥1 so that ∥fnj −fm∥p < 2−j for all m > nj. Deﬁne hk = \|fn1\| + k−1 X j=1 \|fnj+1−fnj\| and h = lim k→∞hk. 66 Lp spaces Note that 0 ≤hk ≤hk+1. By the triangle inequality, ∥hk∥p ≤∥fn1∥p + k−1 X j=1 2−j < ∥fn1∥p + 1. Since hp k are increasing to hp, the MCT shows that ∥h∥p ≤∥fn1∥p + 1 when p < ∞. For p = ∞, ∥h∥∞≤∥fn1∥p + 1 follows directly. So h ∈Lp(µ), and in particular, h(x) < ∞a.e.(µ). Note that fnk = fn1 + Pk−1 j=1(fnj+1−fnj). Let f = fn1 + X j≥1 (fnj+1−fnj) = lim k→∞fnk. This series converges absolutely whenever h(x) < ∞, and so is deﬁned a.e.(µ). Moreover \|f\| ≤h, so f ∈Lp(µ). Also \|f −fnk\|p ≤ P j≥k \|fnj −fnj+1\| p ≤hp when p < ∞. Thus by the LDCT, lim k→∞∥f −fnk∥p p = lim k→∞ Z \|f −fnk\|p dµ = Z lim k→∞\|f −fnk\|p dµ = 0. When p = ∞, ∥f −fnk∥∞≤P j≥k ∥fnj −fnj+1∥∞< 21−k. Therefore fnk converges to f in Lp(µ). Since (fn)n≥1 is Cauchy, in fact fn →f in Lp(µ). So Lp(µ) is complete. ■ 5.1.4. EXAMPLES. (1) Lp(0, 1) = Lp((0, 1), m). Here Lebesgue measure, m, is a probability measure on (0, 1). If 1 ≤p < r < ∞and f ∈Lr(0, 1), then ∥f∥p p = Z \|f\|p dm ≤ Z \|f\|≤1 1 dm + Z \|f\|>1 \|f\|r dm ≤1 + ∥f∥r r < ∞. Thus f ∈Lp(0, 1). So L1(0, 1) ⊃Lp(0, 1) ⊃Lr(0, 1) ⊃L∞(0, 1). These subspaces are not closed in the larger spaces. In the next section, we will improve the norm inequality. We also write Lp(R) for Lp((R, m)). (2) Consider (N, P(N), mc). Then Lp(N, mc) = lp = (ai)i≥1 : ∥(ai)∥p = X i≥1 \|ai\|p1/p < ∞ for p < ∞, and L∞(N, mc) = l∞= (ai)i≥1 : ∥(ai)∥∞= supi≥1 \|ai\| < ∞ . If 1 ≤p < r < ∞and (ai) ∈lp, then ∥(ai)∥r r = X i≥1 \|ai\|r = X i≥1 \|ai\|p \|ai\|r−p ≤ X i≥1 \|ai\|p sup \|ai\|r−p ≤∥(ai)∥p p ∥(ai)∥r−p p = ∥(ai)∥r p. Thus ∥(ai)∥r ≤∥(ai)∥p. In particular, l1 ⊂lp ⊂lr ⊂l∞. 5.1 Lp as a Banach space 67 (3) Let µ be the measure on (N, P(N)) given by µ(A) = ∞is A ̸= ∅and µ(∅) = 0. Then Lp(µ) = {0} if 1 ≤p < ∞and L∞(µ) = l∞. You need sets of non-zero ﬁnite measure to get interesting functions in Lp(µ). 5.1.5. PROPOSITION. Let (X, B, µ) be a measure and let 1 ≤p < ∞. Then the simple functions of ﬁnite support (ϕ = Pn i=1 aiχAi, where µ(Ai) < ∞) are dense in Lp(µ). For p = ∞, the set of all simple functions is dense in L∞(µ). PROOF. Fix f ∈Lp(µ) for p < ∞. Choose simple functions ϕn so that \|ϕn\| ≤\|ϕn+1\| ≤\|f\| and f(x) = lim n→∞ϕn(x). Then \|ϕn\|p ≤\|f\|p is integrable, so ϕn ∈Lp(µ). In particular, each set Ai in the deﬁnition of ϕn for which ai ̸= 0 must have ﬁnite measure. Also since \|f −ϕn\| ≤ \|f\| + \|ϕn\| ≤2\|f\|, we have \|f −ϕn\|p ≤2p\|f\|p. Therefore by the LDCT, lim n→∞∥f −ϕn∥p p = lim n→∞ Z \|f −ϕn\|p dµ = Z lim n→∞\|f −ϕn\|p dµ = 0. So ϕn converge to f in Lp(µ). When p = ∞, suppose that ∥f∥∞= N < ∞. For n ≥1, and (j, k) ∈ [−nN, nN]2, let Aj,k = {x : j n ≤Re f(x) < j+1 n , k n ≤Im f(x) < k+1 n }. These are measurable sets, and ϕn = PnN j,k=−nN j+ik n χAj,k satisﬁes ∥f −ϕn∥∞≤ √ 2 n . So the simple functions are dense. If µ(A) = ∞, then χA is not the limit of simple functions with ﬁnite support. ■ Recall that if X is a topological space, then Cc(X) denotes the space of con-tinuous functions on X with compact support (i.e. {x : f(x) ̸= 0} is compact). 5.1.6. COROLLARY. If 1 ≤p < ∞, then Cc(R) is dense in Lp(R). This is false for p = ∞. PROOF. Let f ∈Lp(R) and let ε > 0. Use Proposition 5.1.5 to ﬁnd a simple function of ﬁnite support ϕ with ∥f −ϕ∥p < ε/2. Write ϕ = Pn i=1 aiχAi, where µ(Ai) < ∞. Let L = sup \|ai\|. By the regularity of Lebesgue measure (Theo-rem 1.5.6), there are compact sets Ki so that Ki ⊂Ai and m(Ai \ Ki) < ε 4nL. Choose disjoint open sets Ui ⊃Ki so that m(Ui \ Ki) < ε 4nL and Ui is compact. For each 1 ≤i ≤n, let hi = dist(x,Uc i ) dist(x,Ki)+dist(x,Uc i ). Note that this is a continuous function which vanishes off of Ui, has hi(x) = 1 on Ki, and takes values in [0, 1]. Then χAi −hi vanishes on Ki and on U c i ∩Ac i; so ∥χAi −hi∥p p ≤m(Ai \ Ki) + m(Ui \ Ai) ≤ ε 4nL + ε 4nL = ε 2nL. 68 Lp spaces Hence h = Pn i=1 aihi belongs to Cc(R) and ∥ϕ −h∥p ≤ n X i=1 \|ai\|∥χAi −hi∥p < nL ε 2nL = ε 2. Thus ∥f −h∥p < ε. Clearly the constant function 1 is not a limit in L∞(R) of functions of compact support. So this fails for p = ∞. ■ 5.2. Duality for Normed Vector Spaces In this section, we recall some basic facts about linear functionals on normed vector spaces. 5.2.1. DEFINITION. Let (V, ∥· ∥) be a normed vector space over F ∈{R, C}. Let L(V, F) denote the vector space of linear maps of V into the scalars, linear functionals, and let V ∗denote the dual space of V consisting of all continuous linear functionals. 5.2.2. PROPOSITION. Let (V, ∥· ∥) be a normed vector space over F, and let ϕ ∈L(V, F). The following are equivalent: (1) ϕ is continuous. (2) ∥ϕ∥:= sup{\|ϕ(v)\| : ∥v∥≤1} < ∞. (3) ϕ is continuous as v = 0. PROOF. (2) ⇒(1). If u ̸= v ∈V , set w = u−v ∥u−v∥and note that \|ϕ(u) −ϕ(v)\| = \|ϕ(u −v)\| = \|ϕ(w)\| ∥u −v∥≤∥ϕ∥∥u −v∥. Hence ϕ is Lipschitz, and in particular is continuous. (1) ⇒(3) is trivial. (3) ⇒(2). Assume that (2) fails. Then there are vectors vn ∈V with ∥vn∥= 1 and \|ϕ(vn)\| > n2. Thus 1 nvn →0 while \|ϕ( 1 nvn)\| > n diverges. So ϕ is discontinuous at 0. The result follows. ■ 5.2.3. THEOREM. Let (V, ∥· ∥) be a normed vector space. Then (V ∗, ∥· ∥) is a Banach space. PROOF. First we show that ∥· ∥is a norm. Clearly ∥ϕ∥= 0 if and only if ϕ(v) = 0 for all v ∈V with ∥v∥≤1. This forces ϕ = 0 by linearity. Also if 5.3 Duality for Lp 69 λ ∈F, then ∥λϕ∥= sup ∥v∥≤1 \|λϕ(v)\| = \|λ\| sup ∥v∥≤1 \|ϕ(v)\| = \|λ\| ∥ϕ∥. For the triangle inequality, take ϕ, ψ ∈V ∗. ∥ϕ + ψ∥= sup ∥v∥≤1 \|ϕ(v) + ψ(v)\| ≤sup ∥v∥≤1 \|ϕ(v)\| + sup ∥v∥≤1 \|ψ(v)\| = ∥ϕ∥+ ∥ψ∥. To establish completeness, let (ϕn)n≥1 be a Cauchy sequence in V ∗. For each v ∈V , \|ϕm(v)−ϕn(v)\| ≤∥ϕm−ϕn∥∥v∥. It follows that (ϕn(v))n≥1 is a Cauchy sequence in F. Thus we may deﬁne ϕ(v) = limn→∞ϕn(v). Then ϕ(λu+µv) = lim n→∞ϕn(λu+µv) = lim n→∞λϕn(u)+µϕn(v) = λϕ(u)+µϕ(v). Therefore ϕ is linear. Now let ε > 0 and select N so that if m, n ≥N, then ∥ϕm −ϕn∥< ε. In particular, if ∥v∥≤1, we have \|ϕm(v) −ϕn(v)\| < ε. Holding m ﬁxed and letting n →∞, we obtain that \|ϕm(v) −ϕ(v)\| ≤ε. Taking the supremum over all v with ∥v∥≤1 yields ∥ϕm −ϕ∥≤ε when m ≥N. In particular, ∥ϕ∥≤∥ϕm∥+ ∥ϕm −ϕ∥< ∞; so ϕ ∈V ∗. Moreover we have shown that limm→∞ϕm = ϕ in (V ∗, ∥· ∥). So V ∗is complete. ■ 5.3. Duality for Lp In this section, we determine the Banach space dual of the spaces Lp(µ) for 1 ≤p < ∞. We need another import inequality. 5.3.1. LEMMA. If a, b ∈(0, ∞) and 0 ≤t ≤1, then atb1−t ≤ta + (1 −t)b with equality only when a = b or t = 0 or 1. PROOF. This is just the AMGM inequality. The function f(x) = ex is strictly convex. So etα+(1−t)β ≤teα + (1 −t)eβ, with equality only when α = β or t = 0 or 1. Take a = eα and b = eβ and the result follows. ■ 5.3.2. H ¨ OLDER’S INEQUALITY. Let (X, B, µ) be a measure space. Let 1 < p < ∞and deﬁne q so that 1 p + 1 q = 1. If f ∈Lp(µ) and g ∈Lq(µ) , then fg ∈L1(µ) and ∥fg∥1 ≤∥f∥p ∥g∥q. Equality holds if and only if \|f\|p and \|g\|q are collinear. PROOF. We may suppose that f, g are non-zero since the inequality is trivial if fg = 0 a.e.(µ). Let f0 = \|f\|/A where A = ∥f∥p and g0 = \|g\|/B where 70 Lp spaces B = ∥g∥q. Apply Lemma 5.3.1 to a = f0(x)p and b = g0(x)q with t = 1 p and 1 −t = 1 q. Then \|f(x)g(x)\| AB = f0(x)g0(x) ≤1 pf0(x)p + 1 q g0(x)q = 1 pAp \|f(x)\|p + 1 qBq \|g(x)\|q. Integrate to get ∥fg∥1 AB ≤ 1 pAp ∥f(x)∥p p + 1 qBq ∥g(x)∥q q = 1 p + 1 q = 1. Hence ∥fg∥1 ≤AB = ∥f∥p ∥g∥q. Equality holds in the ﬁrst inequality only when f0(x)p = g0(x)q. For it to hold for the integral, this identity must hold a.e.(µ). Hence \|g\|q = B A\|f\|p a.e.(µ). ■ For more elementary reasons, if f ∈L1(µ) and g ∈L∞(µ), we have that fg ∈L1(µ) and ∥fg∥1 ≤∥f∥1 ∥g∥∞. 5.3.3. EXAMPLE. We return to Example 5.1.4(1). Suppose that µ is a probabil-ity measure. If 1 ≤p < r < ∞and f ∈Lr(µ), choose s so that p r + 1 s = 1. Note that \|f\|p ∈Lr/p(µ). Then ∥f∥p p = \|f\|p1 1 ≤ \|f\|p r/p∥1∥s = ∥f∥p r. Hence ∥f∥p ≤∥f∥r. More generally if µ(X) < ∞, the same computation yields ∥f∥p ≤∥1∥1/p s ∥f∥r = µ(X)1/sp ∥f∥r = µ(X) 1 p −1 r ∥f∥r. When every point has measure 1, you get the lp spaces, where the inequalities are reversed. See Example 5.1.4(2). For the spaces Lp(R), there is no containment between Lp(R) and Lr(R) when p ̸= r. We now get to the main result of this section. 5.3.4. THEOREM (Riesz). Let (X, B, µ) be a σ-ﬁnite measure. Suppose that 1 ≤p < ∞, and let q satisfy 1 p + 1 q = 1, where q = ∞when p = 1. Then Lp(µ)∗= Lq(µ) via the isometric pairing Lq(µ) ∋g →Φg, where Φg(f) = Z fg dµ. 5.3.5. LEMMA. Suppose that µ is σ-ﬁnite, 1 ≤p < ∞and g ∈Lq(µ). Then ∥Φg∥= ∥g∥q. PROOF. First suppose that 1 < p < ∞. By H¨ older’s inequality \|Φg(f)\| = Z fg dµ ≤∥fg∥1 ≤∥f∥p ∥g∥q. 5.3 Duality for Lp 71 Hence ∥Φg∥= sup∥f∥p≤1 \|Φg(f)\| ≤∥g∥q. On the other hand, if g ̸= 0, let f = \|g\|q−1sign(g) ∥g∥q−1 q . We will use that p(q−1) = pq(1−1 q) = q. Then \|f\|p = \|g\|q ∥g∥q q . So ∥f∥p p = Z \|g\|q ∥g∥q q dµ = 1. Therefore ∥Φg∥≥Φg(f) = Z \|g\|q−1sign(g) ∥g∥q−1 q g dµ = 1 ∥g∥q−1 q Z \|g\|q dµ = ∥g∥q. For p = 1, take g ∈L∞(µ). As above, \|Φg(f)\| ≤ Z \|f\| ∥g∥∞dµ = ∥f∥1 ∥g∥∞. Conversely, given ε > 0, let A = {x : \|g(x)\| > ∥g∥∞−ε}. Then µ(A) > 0. Since µ is σ-ﬁnite, there is a measurable subset E ⊂A with 0 < µ(E) < ∞. Let f = sign(g) µ(E) χE. Then ∥f∥1 = 1 and Φg(f) = 1 µ(E) Z E \|g\| dµ > ∥g∥∞−ε. Thus ∥Φg∥= ∥g∥∞. ■ 5.3.6. LEMMA. Suppose that µ is σ-ﬁnite, 1 ≤p < ∞and g is a measurable function such that Z ϕg dµ ≤M∥ϕ∥p for all ϕ simple, ﬁnite support. Then g ∈Lq(µ) and ∥g∥q ≤M. PROOF. First take 1 < p < ∞, so that q < ∞. Suppose ﬁrst that g is real valued. Choose simple functions ψn so that \|ψn\| ≤\|ψn+1\| ≤\|g\| and ψn →g. Since µ is σ-ﬁnite, we can write X = S n≥1 Xn where Xn ⊂Xn+1 and µ(Xn) < ∞for all n. Then ϕn = ψnχXn are simple functions with ﬁnite support such that \|ϕn\| ≤\|ϕn+1\| ≤\|g\| and ϕn →g. Analogous to the previous lemma, deﬁne fn = \|ϕn\|q−1 sign(g) ∥ϕn∥q−1 q . Note that sign(g) take only the values ±1, 0 and so fn is a simple function of ﬁnite support. As in the previous lemma, ∥fn∥p = 1. Therefore M ≥sup n≥1 Z fng dµ = sup n≥1 Z \|ϕn\|q−1\|g\| ∥ϕn∥q−1 q dµ ≥sup n≥1 Z \|ϕn\|q ∥ϕn∥q−1 q dµ = sup n≥1 ∥ϕn∥q = ∥g∥q. The last equality follows from the MCT. Now for complex valued g, note that Re g and Im g satisfy the hypotheses of the lemma. Thus they both belong to Lq(µ), and hence g ∈Lq(µ). So by Lemma 5.3.5, 72 Lp spaces ∥Φg∥= ∥g∥q. Proposition 5.1.5 shows that simple functions of ﬁnite support are dense in Lp(µ), and hence the optimal constant M must be ∥Φg∥; so ∥g∥q ≤M. For p = 1, we need to show that ∥g∥∞≤M. If this fails, then for some N > M, {x : \|g(x)\| > N} has positive measure. Hence there is a θ ∈R so that A = {x : Re eiθg(x) > M} has positive measure. Let E ⊂A have positive and ﬁnite measure. Deﬁne f = eiθµ(E)−1χE. Then ∥f∥1 = 1 and M ≥Re R fg dµ > M. This is a contradiction, and hence ∥g∥∞≤M. ■ PROOF OF THEOREM 5.3.4. First assume that µ(X) < ∞. Let Φ ∈Lp(µ)∗. Deﬁne ν(E) = Φ(χE) for E ∈B. Note that ν(∅) = Φ(0) = 0. Also if E = ˙ S i≥1Ei is a disjoint union of sets in B, then χE − n X i=1 χEi p p = µ ˙ [ i>nEi →0 as n →∞. Since Φ is continuous, ν(E) = Φ(χE) = lim n→∞Φ n X i=1 χEi = X i≥1 ν(Ei). Therefore ν is countably additive, and thus is a complex measure. Moreover if µ(E) = 0, then χE = 0 a.e.(µ) and hence ν(E) = Φ(0) = 0. Thus ν ≪µ. By the Radon-Nikodym Theorem 4.3.2, there is a measurable g ∈L1(µ) so that ν(E) = Z E g dµ. If ϕ = Pn i=1 aiχEi is a simple function (with ﬁnite support), then Φ(ϕ) = n X i=1 aiν(Ei) = Z ϕ dν = Z ϕg dµ. Thus Z ϕg dµ = \|Φ(ϕ)\| ≤∥Φ∥∥ϕ∥p. By Lemma 5.3.6, g ∈Lq(µ), and so Φg is a continuous functional on Lp(µ) which agrees with Φ on all simple functions of ﬁnite support. Such functions are dense in Lp(µ) by Proposition 5.1.5. So by continuity, Φ = Φg. By Lemma 5.3.5, ∥Φ∥= ∥Φg∥= ∥g∥q. Moreover this shows that g is unique, because distinct Lp(µ) functions yield distinct functionals. Now consider when µ(X) = ∞. We can write X = S n≥1 Xn where Xn ⊂ Xn+1 and µ(Xn) < ∞. We can restrict Φ to Lp(Xn, µ\|Xn) (since this is a closed subspace of Lp(µ)). Call it Φn. The ﬁrst part of the proof shows that thee is a unique gn ∈Lp(Xn, µ\|Xn) so that Φn = Φgn and ∥gn∥q ≤∥Φ∥. Let g = S n≥1 gn be the unique measurable function such that g\|Xn = gn. The various functions must match up a.e.(µ) because of uniqueness. By the MCT, ∥g∥q = lim n→∞∥gn∥q ≤∥Φ∥. 5.3 Duality for Lp 73 So Φ and Φg agree on the dense subspace S n≥1 Lp(Xn, µ\|Xn) and they are con-tinuous functionals; and so Φ = Φg. Again by Lemma 5.3.5, ∥Φ∥= ∥g∥q. ■ 5.3.7. REMARK. In fact, σ-ﬁniteness is not needed to obtain Lp(µ)∗= Lq(µ) provided that 1 < p < ∞. See Folland’s book for the details. However it is critical to establish that L1(µ)∗= L∞(µ). In Example 5.1.4 (3), each point in N has µ({n}) = ∞and so L1(µ) = {0} while L∞(µ) = l∞. However L1(µ)∗= {0}. A more subtle example is to take (R, P(R), mc). Let B be the σ-algebra of countable and co-countable subsets of R. Let µ = mc\|B. Any function in L1(mc) has countable support, and thus is B-measurable. Therefore L1(µ) = L1(mc). However changing the measure has a dramatic effect on the L∞spaces. The space L∞(mc) = l∞(R) is the space of all bounded functions on R. However for a function to be B-measurable, it must be constant on a co-countable set. In fact, L1(µ)∗= L1(mc)∗= L∞(mc) = l∞(R). One way to see this is that if Φ ∈L1(µ)∗, we can deﬁne g(r) = Φ(χ{r}). Then \|g(r)\| ≤∥Φ∥, so that g is a bounded function; and so g ∈L∞(mc). Moreover, if f ∈L1(µ), then there is a countable (or ﬁnite) collection of real numbers {rn : n ≥1} and scalars αn so that f = P n≥1 αnχ{rn}. Moreover ∥f∥1 = P n≥1 \|αn\|; whence this series converges absolutely. Therefore by continuity, Φ(f) = X n≥1 αnΦ(χ{rn}) = X n≥1 αng(n). Moreover this same computation shows that any element of L∞(mc) determines a distinct continuous functional on L1(µ). . CHAPTER 6 Some Topology This is a brief introduction skewed to certain things needed in the next section. This is not intended as a comprehensive introduction to point set topology. A good general reference is Willard’s book . 6.1. Topological spaces 6.1.1. DEFINITION. A topology τ on a set X is a collection of subsets such that (1) ∅, X ∈τ. (2) If {Uλ : λ ∈Λ} ⊂τ, then S λ∈Λ Uλ ∈τ. (3) If U1, . . . , Un ∈τ, then Tn i=1 Ui ∈τ. The elements U ∈τ are called open sets. 6.1.2. EXAMPLES. (1) If (X, d) is a metric space, then U is open if for every x ∈U, there is an r > 0 so that the open ball br(x) ⊂U. (2) If X is any set, the discrete topology has τd = P(X), the collection of all subsets of X. (3) If X is any set, the trivial topology has τ = {∅, X}. (4) If (X, ≤) is a totally ordered set, the intervals (a, b) = {x ∈X : a < x < b}, (−∞, b) = {x ∈X : x < b} and (a, ∞) = {x ∈X : x > a} are open, and the topology consists of arbitrary unions of such intervals. (5) If (X, τ) is a topology and Y ⊂X, the induced topology on Y is τ\|Y = {U ∩Y : U ∈τ}. 6.1.3. DEFINITION. A set F ⊂X is closed if F c is open. If A ⊂X, the closure of A is A = T{F : A ⊂F, F closed}. A point in A is called a limit point of A. If A ⊂X, then a ∈A is an interior point of A if there exists U ∈τ with a ∈U ⊂A. If A ⊂X, the interior of A is Ao or int A = S{U ∈τ : U ⊂A}. 74 6.1 Topological spaces 75 If x ∈X, a neighbourhood of x is a set N such that x ∈No. 6.1.4. PROPOSITION. (1) Finite unions and arbitrary intersections of closed sets are closed. (2) A is the smallest closed set containing A. (3) x ∈A if and only if every U ∈τ with x ∈U has A ∩U ̸= ∅. (4) A = Acoc is the complement of the interior of Ac. PROOF. Since open sets are closed under arbitrary unions and ﬁnite intersec-tions, the collection of closed sets is closed under arbitrary intersections and ﬁnite unions. Hence the intersection of all closed sets F ⊃A is closed, and is thus the smallest closed set containing A. Now x ∈A if and only if x ∈F for every closed F ⊃A if and only if x ̸∈U if U is open and disjoint from A. Finally X \ A = [ {U ∈τ : U ∩A = ∅} = [ {U ∈τ : U ⊂Ac} = Aco. ■ 6.1.5. DEFINITION. If σ and τ are two topologies on X, we say that σ is a weaker topology than τ, and τ is a stronger topology than σ, if σ ⊂τ. 6.1.6. PROPOSITION. If S ⊂P(X), then there is a weakest topology τ con-taining S. It consists of arbitrary unions of sets which are intersections of ﬁnitely many elements of S. PROOF. Clearly if τ ⊃S is a topology, then it contains all intersections of ﬁnitely many elements of S, and arbitrary unions of these sets. The intersection of no sets is X by convention, and ∅is the union of no sets, so they both belong to τ. This collection is clearly closed under arbitrary unions. To check that it is stable under intersection, observe that if Aα,i and Bβ,j are in S, then [ α∈A Aα,1 ∩· · · ∩Aα,nα ∩ [ β∈B Bβ,1 ∩· · · ∩Bβ,mβ = [ α∈A, β∈B Aα,1 ∩· · · ∩Aα,nα ∩Bβ,1 ∩· · · ∩Bβ,mβ. Hence this collection is a topology. By construction, this is the weakest topology containing S. ■ 6.1.7. DEFINITION. Say that S ⊂P(X) is a base for a topology τ if every open set U ∈τ is the union of elements of S. Also S is a subbase for a topology τ if the collection of ﬁnite intersections of elements of S is a base for τ. 76 Some Topology 6.1.8. EXAMPLES. (1) If (X, d) is a metric space, then b1/n(x) : x ∈X, n ≥1 is a base for the topology. (2) (r, s) : r < s ∈Q is a base for the topology of R. (3) Let C[0, 1] denote the space of continuous functions on [0, 1]. For each x ∈ [0, 1], a ∈C and r > 0, let U(x, a, r) = {f ∈C[0, 1] : f(x) ∈br(a)}. Let τ be the topology generated by these sets. This is the topology of pointwise convergence. An open neighbourhood of f must contain a set of the form {g ∈C[0, 1] : \|g(xi) −f(xi)\| < r for 1 ≤i ≤n} for x1, . . . , xn ∈[0, 1] and r > 0. 6.1.9. DEFINITION. A set A is dense in X if X = A. X is separable if it has a countable dense subset. X is ﬁrst countable if for each x ∈X, there is a countable family {Ui} ⊂τ with x ∈Ui which forms a countable base of neighbourhoods of x; i.e., if x ∈V is open, then there is some i so that Ui ⊂V . X is second countable if there is a countable family of open sets which is a base for τ. 6.1.10. EXAMPLES. (1) If (X, d) is a metric space and x ∈X, then b1/n(x) : n ≥1 is a countable base of neighbourhoods of x. If X is separable, and {xi : i ≥1} is dense in X, then b1/n(xi) : i ≥1, n ≥1 is a base for τ. Indeed, suppose that x ∈U is open. Pick r > 0 so that br(x) ⊂U and xi so that d(x, xi) < 1/n < r/2. Then x ∈b1/n(xi) ⊂U. So X is second countable. In particular, compact metric spaces are separable and so second countable. (2) Consider the discrete topology τd on a set X. Since the topology is generated by {x} : x ∈X , X is always ﬁrst countable. However it is second countable if and only if X is countable if and only if X is separable. 6.2. Continuity 6.2.1. DEFINITION. A function f : (X, τ) →(Y, σ) between topological spaces is continuous if for all V ⊂Y open, the set f−1(V ) is open in X. Say that f is a homeomorphism if f is a bijection such that both f and f−1 are continuous. 6.2.2. EXAMPLES. (1) The identity map (X, discrete) id − →(X, τ) id − →(X, trivial) is a continuous bijection, however in both cases f−1 will be discontinuous provided that τ satisﬁes {∅, X} ⊊τ ⊊P(X). 6.2 Continuity 77 (2) A function f : (X, trivial) →R is continuous only if it is constant, while every function f : (X, discrete) →R is continuous. On the other hand, a function f : R →(X, discrete) is continuous only if it is constant, while every function f : R →(X, trivial) is continuous. (3) f : (−1, 1) →R by f(x) = tan πx 2 is a homeomorphism. (4) Let X = {0, 1}. Let τ = ∅, {0}, X . Then {1} is closed, but {0} is not, and {0} = X. If f : X →R is continuous, then f is constant. (5) Let X = [0, 1) ∪{a, b}. Let the open sets in τ be U ⊂[0, 1) which are open in the usual metric on [0, 1) together with sets U ∪(r, 1) ∪{a}, U ∪(r, 1) ∪{b} and U ∪(r, 1) ∪{a, b} for r < 1. Here the points {a} and {b} are closed because the complement is open. However if a ∈U and b ∈V are open sets, then U ∩V ⊃ (r, 1) for some r < 1. That means that you cannot separate a and b from one another by open sets. If f : X →R is continuous, then f(a) = f(b). 6.2.3. DEFINITION. Let Cb(X) and Cb R(X) or Cb(X, R) denote the normed vector space of bounded continuous functions from X into C and R, respectively, with norm ∥f∥∞= supX \|f(x)\|. Similarly, C(X) and CR(X) or C(X, R) denote the vector space of continuous functions from X into C and R, respectively. 6.2.4. DEFINITION. A topological space is Hausdorff if for all x, y ∈X two distinct points, there are open sets U ∋x and V ∋y so that U ∩V = ∅. 6.2.5. PROPOSITION. If Cb(X) separates points of X, i.e., for x ̸= y in X, there is a continuous function f ∈Cb(X) so that f(x) ̸= f(y), then X is Hausdorff. PROOF. If f(x) = α and f(y) = β and r = \|α −β\|/2 > 0, then x ∈U = f−1(br(α)) and y ∈V = f−1(br(β)) and U ∩V = ∅. ■ Consider the Examples 6.2.2 (4) and (5) in light of this proposition. Recall that fn ∈Cb(X) converge uniformly to a function f if ∥f −fn∥∞→0. The following standard result for metric spaces extends easily. 6.2.6. PROPOSITION. The uniform limit f of a sequence fn ∈Cb(X) is con-tinuous. PROOF. Let U be open in C and let x ∈f−1(U). Then there is an r > 0 so that br(f(x)) ⊂U. Choose n so large that ∥f −fn∥∞< r/3. Then x ∈V = 78 Some Topology f−1 n (br/3(fn(x))) is open. If y ∈V , then \|fn(y) −fn(x)\| < r/3, so \|f(y) −f(x)\| ≤\|f(y) −fn(y)\| + \|fn(y) −fn(x)\| + \|fn(x) −f(x)\| < ∥f −fn∥∞+ r 3 + ∥f −fn∥∞< r. Hence f(y) ∈br(f(x)) ⊂U. Thus V ⊂f−1(U). So f is continuous. ■ The norm ∥f∥∞makes Cb(X) into a normed vector space. In view of Propo-sition 6.2.5, the following is most interesting when X is Hausdorff. 6.2.7. THEOREM. For any topological space, Cb(X) is complete. PROOF. Let (fn)n≥1 be a Cauchy sequence in Cb(X). If ε > 0, there is an N so that if N ≤m < n, then ∥fn −fm∥∞< ε. In particular, for x ∈X, the sequence fn(x) n≥1 is Cauchy in C. So we may deﬁne f(x) = limn→∞fn(x) pointwise. However for m ≥N, \|f(x) −fm(x)\| = lim n→∞\|fn(x) −fm(x)\| ≤ε. Hence ∥f −fm∥∞≤ε. So convergence is uniform. By Proposition 6.2.6, f is continuous. Also ∥f∥∞= limn→∞∥fn∥∞< ∞, and so f lies in Cb(X). Therefore Cb(X) is complete. ■ 6.3. Compactness 6.3.1. DEFINITION. An open cover of a set A ⊂X is a collection of open sets {Uλ : λ ∈Λ} such that A ⊂S Λ Uλ. A set A is compact if every open cover has a ﬁnite subcover, i.e., a ﬁnite subset Uλ1, . . . , Uλn such that A ⊂Sn i=1 Uλi. 6.3.2. EXAMPLE. In 6.2.2 (4), the point {0} is compact but not closed. 6.3.3. PROPOSITION. If X is compact and A ⊂X is closed, then A is com-pact. If X is Hausdorff and A ⊂X is compact, then A is closed. Moreover, if x ̸∈A, there are disjoint open sets U ⊃A and V ∋x. PROOF. If U = {Uλ : λ ∈Λ} is an open cover of A, then U ∪{Ac} is an open cover of X. By compactness, it has a ﬁnite subcover Uλ1, . . . , Uλn, Ac. Hence Uλ1, . . . , Uλn covers A; whence A is compact. Suppose that X is Hausdorff and A ⊂X is compact, and let x ∈Ac. For each a ∈A, there are open sets a ∈Ua and x ∈Va so that Ua ∩Va = ∅. Clearly 6.3 Compactness 79 {Ua : a ∈A} is an open cover of A. By compactness, there is a ﬁnite subcover Ua1, . . . , Uan. Let V = Tn i=1 Vai. Then x ∈V is open, and V ∩A ⊂ n [ i=1 V ∩Uai = ∅. Hence x ̸∈A. So A is closed. Moreover A ⊂U = Sn i=1 Uai, and U ∩V = ∅. ■ 6.3.4. DEFINITION. A family {Aλ : λ ∈Λ} of subsets of X has the ﬁnite intersection property (FIP) if whenever λ1, . . . , λn are ﬁnitely many elements of Λ, then Tn i=1 Aλi ̸= ∅. 6.3.5. PROPOSITION. A topological space X is compact if and only if every family F = {Aλ : λ ∈Λ} of closed sets with FIP has non-empty intersection T F := T Λ Aλ ̸= ∅. PROOF. Suppose that X is compact and F has FIP. Deﬁne open sets Uλ = Ac λ. If T F = ∅, then S Λ Uλ = T F c = X. So U = {Uλ : λ ∈Λ} is an open cover of X. By compactness, there is a ﬁnite subcover Uλ1, . . . , Uλn. Hence Tn i=1 Aλi = Sn i=1 Uλi c = ∅, contradicting FIP. Therefore T F ̸= ∅. Conversely, suppose that U = {Uλ : λ ∈Λ} is an open cover of X. Deﬁne closed sets Aλ = U c λ. If there is no ﬁnite subcover, then Tn i=1 Aλi = Sn i=1 Uλi c ̸= ∅; and thus F = {Aλ : λ ∈Λ} has FIP. But then T F ̸= ∅. Therefore S Λ Uλ = T F c ̸= X, contradicting the fact that U is an open cover. Hence F does not have FIP, so there is a ﬁnite set Aλ1, . . . , Aλn such that Tn i=1 Aλi = ∅. Hence Sn i=1 Uλi = Tn i=1 Aλi c = X. Therefore X is compact. ■ 6.3.6. PROPOSITION. If f : (X, τ) →(Y, σ) is continuous and A ⊂X is compact, then f(A) is compact. PROOF. Let {Vλ : λ ∈Λ} is an open cover of f(A) in Y . Deﬁne Uλ = f−1(Vλ). These are open sets by continuity, and they cover A. Thus there is a ﬁnite subcover Uλ1, . . . , Uλn. Then since Vλ ⊃f(Uλ), it follows that Vλ1, . . . , Vλn covers f(A). Hence f(A) is compact. ■ The following important consequence follows directly. 6.3.7. EXTREME VALUE THEOREM. If (X, τ) is compact and f ∈C(X), then \|f\| attains it maximum. In particular, ∥f∥∞< ∞. 6.3.8. DEFINITION. If (Xλ, τλ) are topological spaces for λ ∈Λ, we deﬁne the product space to be X = Q Λ Xλ = {(xλ) : xλ ∈Xλ} with the weakest topology τ which makes the coordinate projections πλ : X →Xλ by πλ(x) = xλ 80 Some Topology continuous. That is, the sets π−1 λ (U) = Q µ∈Λ{λ} Xµ × U are open and form a subbase for the topology. 6.3.9. REMARKS. (1) The product topology τ consist of arbitrary unions of ﬁnite intersections of the subbase. So if λ1, . . . , λn ∈Λ and Ui ∈τλi, then the sets of the form U1 × · · · × Un × Y µ∈Λ{λi,1≤i≤n} Xµ form a base for the topology. (2) If Λ is ﬁnite, this is a familiar construction in the metric space case. Indeed, if (Xi, di) are metric spaces for 1 ≤i ≤n, then D (x1, . . . , xn), (y1, . . . , yn) = max di(xi, yi) : 1 ≤i ≤n is a metric on the product, and the metric topology coincides with the product topology. (3) When Λ is inﬁnite, it often requires the Axiom of Choice to be able to say that X is non-empty. 6.3.10. THEOREM. If Xi are compact for 1 ≤i ≤n, then X = Qn i=1 Xi is compact. PROOF. It sufﬁces to show that X×Y is compact if both X and Y are compact, as the result follows by ﬁnitely many repetitions. Let W = {Wλ : λ ∈Λ} be an open cover of X × Y . For each (x, y) ∈X × Y , there is a λ(x, y) so that (x, y) ∈ Wλ(x,y). Hence this set contains a basic open set Wλ(x,y) ⊃Ux,y × Vx,y ∋(x, y) for open sets Ux,y in X and Vx,y in Y . Fix y ∈Y . The sets {Ux,y : x ∈X} is an open cover of X. Select a ﬁnite subcover Uxy 1 ,y, . . . , Uxy ny ,y. Then Uxy i ,y × Vxy i ,y for 1 ≤i ≤ny covers X × {y}. Let Vy = Tny i=1 Vxy i ,y. This is open, contains y, and Uxy i ,y × Vxy i ,y for 1 ≤i ≤ny covers X × Vy. Now {Vy : y ∈Y } is an open cover of Y . Let Vy1, . . . , Vym be a ﬁnite subcover. Then the ﬁnite collection {Ux yj i , yj × Vx yj i , yj : 1 ≤i ≤nyj, 1 ≤j ≤m} covers X × Y . Therefore {Wλ(x yj i , yj) : 1 ≤i ≤nyj, 1 ≤j ≤m} covers X × Y . ■ 6.3.11. REMARK. An inﬁnite product of compact spaces is also compact. This is known as Tychonoff’s Theorem. It turns out to be equivalent to the Axiom of Choice. We usually prove it in the functional analysis course. 6.4 Separation Properties 81 6.4. Separation Properties There is a whole hierarchy of separation properties to classify how nice a topo-logical space is. 6.4.1. DEFINITION. A topological space is T0 if x ̸= y ∈X, then there is an open set containing one of these points, but not the other. A topological space is T1 if points are closed. A topological space is T2 if it is Hausdorff. A topological space is T3 if it is T1 and regular: given a closed set A and a point x ̸∈A, there are disjoint open sets U ⊃A and V ∋x. A topological space is T3.5 or Tychonoff if it is T1 and completely regular: given a closed set A and a point x ̸∈A, there is a continuous function f : X →[0, 1] so that f(x) = 1 and f\|A = 0. A topological space is T4 if it is T1 and normal: given disjoint closed sets A, B, there are disjoint open sets U ⊃A and V ⊃B. 6.4.2. REMARKS. (1) The T0 property is very weak. Example 6.2.2(4) is T0 but not T1. (2) T1 is also a very weak property. It implies T0 since if x ̸= y, the set {x}c is an open set containing y but not x. To be T1 it is enough to ﬁnd open sets U ∋x with y ̸∈U and V ∋y with x ̸∈V . (Exercise.) Example 6.2.2(5) is T1 but not Hausdorff. Points are closed in Hausdorff spaces, so T2 implies T1. (3) The trivial topology is regular because there are no points which are disjoint from a non-empty closed set. But throwing in the T1 condition makes a T3 space Hausdorff because you can take your closed set to be {y}. (4) Completely regular spaces are regular, because if x ̸∈A and A is closed, let f : X →[0, 1] be a continuous function with f(x) = 1 and f\|A = 0. Then U = {y : f(y) > 1/2} and V = {y : f(x) < 1/2} are disjoint open sets with x ∈U and A ⊂V . Again we need to add the T1 property to exclude examples like the trivial topology. (5) The T1 property ensures that T4 spaces are Hausdorff. A metric space (X, d) is normal. If A and B are disjoint closed sets, let U = {x : d(x, A) < d(x, B)} and V = {x : d(x, A) > d(x, B)}. It is easy to check that these are disjoint open sets. 6.4.3. PROPOSITION. Compact Hausdorff spaces are normal. PROOF. Let A, B be disjoint closed subsets of a compact Hausdorff space X. Then A and B are compact by Proposition 6.3.3. Moreover since X is Hausdorff, that same Proposition shows that for each point x ∈B, there are disjoint open sets 82 Some Topology Ux ⊃A and Vx ∋x. The collection {Vx : x ∈B} is an open cover of B. Let Vx1, . . . , Vxn be a ﬁnite subcover. Set V = Sn i=1 Vxi and U = Tn i=1 Uxi. These are disjoint open sets with A ⊂U and B ⊂V . ■ Now we prove that T4 spaces have lots of continuous functions. 6.4.4. URYSOHN’S LEMMA. Let X be a normal topological space, and let A and B be disjoint closed sets. Then there is a continuous function f : X →[0, 1] such that f\|A = 0 and f\|B = 1. PROOF. Normality implies the following property: if A is closed and W is open and A ⊂W, then there is an open set U such that A ⊂U ⊂U ⊂W. To see this, take B = W c. Use normality to ﬁnd disjoint open sets U ⊃A and V ⊃B. Then U ⊂V c ⊂W. Start with U1 = Bc. Find an open U1/2 so that A ⊂U1/2 ⊂U1/2 ⊂U1. Repeating this procedure recursively, we ﬁnd open sets Uk/2n for 1 ≤k ≤2n and n ≥1 so that A ⊂Uk/2n ⊂Uk/2n ⊂U(k+1)/2n for 1 ≤k < 2n. Let D = {k/2n : 1 ≤k ≤2n, n ≥1}. Deﬁne f(x) = inf{r ∈D : x ∈Ur} if x ∈U1 and f\|B = 1. Clearly 0 ≤f ≤1 and f\|A = 0. Claim: f is continuous. Note that f−1[0, t) = [ r<t, r∈D Ur is open for t ∈[0, 1]. Also for 0 ≤t < 1, since t < r < s for r, s ∈D implies that Ur ⊂Us, f−1[0, t] = \ r>t, r∈D f−1[0, r) = \ r>t, r∈D Ur = \ r>t, r∈D Ur. This is closed, and therefore f−1(t, 1] = T r>t, r∈D Ur c is open. Hence, f−1(s, t) is open for s < t, and so f is continuous. ■ 6.4.5. REMARK. If (X, d) is a metric space and A and B are disjoint closed sets, deﬁne f(x) = d(x, A) d(x, A) + d(x, B). This satisﬁes the conclusion of Urysohn’s Lemma. 6.4.6. COROLLARY. If X is a T4 space, then it is Tychonoff (T3.5). 6.4.7. COROLLARY. If X is a compact Hausdorff space, C(X) separates points. 6.4 Separation Properties 83 Urysohn’s Lemma implies the following signiﬁcant strengthening. 6.4.8. TIETZE’S EXTENSION THEOREM. Let X be a normal topological space, and let A ⊂X be a closed set. If f : A →[a, b] is continuous, there is a continuous function F : X →[a, b] such that F\|A = f. PROOF. After scaling, we may assume that the range is [−1, 1]. Let A1 = f−1([−1, −1 3]) and B1 = f−1([1 3, 1]). By Urysohn’s Lemma, there is a function g1 : X →[−1 3, 1 3] so that g1\|A1 = −1 3 and g1\|B1 = 1 3. Then f1 = f−g1\|A has range in [−2 3, 2 3]. Repeat the process, setting A2 = f−1 1 ([−2 3, −2 9]) and B2 = f−1 1 ([2 9, 2 3]), and ﬁnding g2 : X →[−2 9, 2 9] with g2\|A2 = −2 9 and g2\|B2 = 2 9. Then f2 = f1 −g2\|A has range in [− 2 3 2, 2 3 2]. Recursively we obtain func-tions gn : X →[−2 · 3−n, 2 · 3−n] so that fn = f −Pn i=1 gn\|A has range in [− 2 3 n, 2 3 n]. Let g = P n≥1 gn. Then g\|A = f and ∥g∥∞≤ X n≥1 ∥gn∥∞= X n≥1 2 · 3−n = 1. ■ 6.4.9. DEFINITION. A Hausdorff space X is locally compact (LCH) if every point has a compact neighbourhood; i.e. for x ∈X, there is a compact set K with x ∈int K. 6.4.10. PROPOSITION. Locally compact Hausdorff spaces are regular. PROOF. Let A be a closed set in a LCH space X, and x ̸∈A. Let K be a compact neighbourhood of x. Then A ∩K is compact. So by Proposition 6.3.3, there are disjoint open sets U ∋x and V ⊃A∩K. Then W = U ∩int K ⊂K \V is an open neighbourhood of x and W ⊂K \ V ⊂Ac. Hence W c ⊃A is open and disjoint from W. Therefore X is regular. ■ 6.4.11. DEFINITION. If f : X →C, the support of f is supp(f) = {x : f(x) ̸= 0}. If X is LCH, Cc(X) denotes the space of continuous C-valued functions with com-pact support. Let C0(X) be the closure of Cc(X) in (Cb(X), ∥· ∥∞). There is a weaker version of Urysohn’s Lemma valid for LCH spaces. 6.4.12. PROPOSITION. Locally compact Hausdorff spaces are completely reg-ular. Moreover, if X is LCH, A ⊂X is closed, B ⊂X is compact and A∩B = ∅, then there is f ∈Cc(X) with 0 ≤f ≤1, f\|A = 0 and f\|B = 1. 84 Some Topology PROOF. Arguing as in the previous proof, for each x ∈B, there is a compact neighbourhood Wx ∋x which is disjoint from A. We can take Wx = int Wx. Then {Wx : x ∈B} is an open cover. Let Wx1, . . . , Wxn be a ﬁnite subcover. Then W = Sn i=1 Wxi is open containing B and W = Sn i=1 Wxi is compact and disjoint from A. Let C = W \ W. Then B and C are disjoint closed sets in the compact Hausdorff space W. By Urysohn’s Lemma, there is a continuous function f : W →[0, 1] with f\|B = 1 and f\|C = 0. Extend f to a function on X by setting f\|W c = 0. Now A ⊂W c, and thus f\|A = 0. To see that f is continuous, note that if a ≥0, f−1(a, b) is open in W, and hence open in X. If a < 0 < b, then D = f−1([b, ∞) is closed in W, and thus f−1(a, b) = Dc is open in X. Since supp(f) ⊂W, we have f ∈Cc(X). ■ There is also a weaker version of Tzietze’s Theorem valid for LCH spaces. 6.4.13. COROLLARY. Suppose that X is LCH, A ⊂X is closed, B ⊂X is compact, A ∩B = ∅and g ∈C(B). Then there is an f ∈Cc(X) with f\|A = 0 and f\|B = g, and ∥f∥∞= ∥g∥∞. PROOF. The setup is the same as the previous proof. Deﬁne f\|B = g and f\|C = 0, and extend this to a continuous function on W using Tietze’s Theorem. Then extend f to all of X by setting f\|W c = 0. This is continuous with compact support as in the previous proof. ■ 6.4.14. DEFINITION. If U is open, say f ≺U for a continuous function f if 0 ≤f ≤1 and supp f ⊂U. Suppose that U = {Uλ : λ ∈Λ} is an open cover of K ⊂X. Then a partition of unity for K relative to U is a collection of functions gλ ≺Uλ which is locally ﬁnite, i.e., for each x ∈X, {λ : gλ(x) ̸= 0} is ﬁnite, and P Λ gλ(x) = 1 for all x ∈K. 6.4.15. PROPOSITION. Let X be LCH, and let K ⊂X be compact. Suppose that U = {U1, . . . , Un} is an open cover of K. Then there are gi ≺Ui in Cc(X) which is a partition of unity for K relative to U. PROOF. For each x ∈K, there is some Ui ∋x. Pick a compact neighbour-hood with x ∈Cx ⊂Ui. Now {int Cx : x ∈K} is an open cover with ﬁnite subcover Cx1, . . . , Cxm. Let Ki = S{Cxj : Cxj ⊂Ui}. Then Ki ⊂Ui and K ⊂Sn i=1 Ki =: C is compact . Let W be an open neighbourhood of C with W compact. By Urysohn’s lemma, there are functions hi so that χKi ≤hi ≺Ui ∩W. Let gi = hi/ Pn i=1 hi. ■ 6.5 Nets 85 6.5. Nets Sequences are not sufﬁcient for dealing with convergence in general topolog-ical spaces, including many that arise in normal contexts. The replacement is the notion of a net, which you can think of as a very wide and very long generalized sequence. This optional section is not required in the next chapter. We will need to call on the Axiom of Choice frequently. Recall that this is the assumption that whenever {Aλ : λ ∈Λ} is a collection of non-empty sets, there is a selection (choice function) ϕ : Λ →S Λ Aλ so that ϕ(λ) ∈Aλ for all λ ∈Λ. 6.5.1. DEFINITION. A partial order on a set Λ is a relation ≤satisfying (1) λ ≤λ for λ ∈Λ (reﬂexive) (2) λ ≤µ and µ ≤λ implies that λ = µ (antisymmetric) (3) λ ≤µ and µ ≤ν implies that λ ≤ν (transitive). Then (Λ, ≤) is called a poset. A poset is upward directed if for λ1, λ2 ∈Λ, there is a µ ∈Λ so that λ1 ≤µ and λ2 ≤µ. 6.5.2. DEFINITION. A net in X is an upward directed poset Λ with a function j : Λ →X, say xλ = j(λ). We usually write the net as (xλ)Λ. A net (xλ)Λ converges to x in (X, τ) if for every open set U ∋x, there is λ0 ∈Λ so that xλ ∈U for every λ ≥λ0. We write limΛ xλ = x. A subnet (yγ)Γ of (xλ)Λ is given by a coﬁnal function ϕ : Γ →Λ so that yγ = xϕ(γ), where we say that ϕ is coﬁnal if for all λ ∈Λ, there is a γ0 ∈Γ so that ϕ(γ) ≥λ for all γ ≥γ0. It is convenient if ϕ is monotone, meaning that γ1 ≤γ2 implies that ϕ(γ1) ≤ϕ(γ2). But this is not necessary. We present a detailed example to explain why nets are needed, and how to use them. 6.5.3. EXAMPLE. Let X = N0×N0. Declare that U ⊂X is open if (0, 0) ̸∈U; and that a set U ∋(0, 0) is open if {m : π−1 1 (m) ∩U is coﬁnite in N0} is coﬁnite in N0. It is easy to verify that this deﬁnes a topology. (a) X is Hausdorff because {(m, n)} is open if m + n ≥1 and {(m, n)}c is an open neighbourhood of (0, 0). (b) (0, 0) ∈X \ {(0, 0)} because every open set U ∋(0, 0) intersects X \ {(0, 0)}. (c) However no sequence xk = (mk, nk) in X \ {(0, 0)} converges to (0, 0). There are two cases. If {mk : k ≥1} is bounded, pick m0 so that mk = m0 inﬁn-itely often. The set U = {(m, n) : m ̸= m0} ∪{(0, 0)} is an open neighbourhood of (0, 0), and the sequence is not eventually in U. Otherwise, there is a sequence 86 Some Topology ki →∞so that mki < mki+1 for i ≥1. Then U = X \ {xki : i ≥1} is an open neighbourhood of (0, 0), and the sequence is not eventually in U. (d) There is a net in X \ {(0, 0)} converging to (0, 0). Let Λ = {U ∈τ : (0, 0) ∈U} where U ≤V if U ⊃V . (We say that Λ is ordered by containment.) This is directed because U, V ≤U ∩V . Order X \ {(0, 0)} by (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (0, 3), (1, 2), (2, 1), (3, 0), . . . . Deﬁne xU to be the least element in this list which belongs to U. (This avoids any issues with the Axiom of Choice.) Then (xU) converges to (0, 0) because given an open neighbourhood U ∋(0, 0), we have xV ∈V ⊂U whenever U ≤V . (e) The sequence (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (0, 3), (1, 2), (2, 1), . . . has a subnet converging to (0, 0). Let Λ be the net just constructed. Deﬁne ϕ(U) = xU considered as an element in this sequence. To see that this map is coﬁnal, let (m0, n0) be in this sequence, and set N0 = m0 + n0. Let U0 = X \ {(m, n) : 1 ≤m + n ≤N0}. Then if U0 ≤U, it follows that xU = (m, n) with m + n > N0 and thus ϕ(U) follows (m0, n0) in the sequence. Therefore this is a subnet of the sequence which converges to (0, 0). Now we show that nets replace sequences in some familiar results. 6.5.4. PROPOSITION. Let A ⊂X. Then x ∈A if and only if there is a net (aλ)Λ in A such that limΛ aλ = x. PROOF. Suppose that x ∈A. By Proposition 6.1.4(3), every open neighbour-hood U of x intersects A. Let O(x) be the open neighbourhoods of x ordered by containment. By the Axiom of Choice, we can pick a point aU ∈A ∩U for each U ∈O(x). The net (aU)O(x) converges to x by construction. Conversely, suppose that (aλ)Λ is a net in A such that limΛ aλ = x. Then for any neighbourhood U of x, there is a λ so that aλ ∈U. In particular, A ∩U ̸= ∅. Hence by Proposition 6.1.4(3), x ∈A. ■ 6.5.5. THEOREM. Let f : (X, τ) →(Y, σ). Then f is continuous if and only if whenever (xλ)Λ is a net in X converging to x, it follows that f(x) = limΛ f(xλ). PROOF. Suppose that f is continuous, and let (xλ)Λ be a net in X converging to x. Let V be an open neighbourhood of f(x). Then U = f−1(V ) is an open neighbourhood of x. By convergence, there is a λ0 ∈Λ so that xλ ∈U for all λ ≥λ0. Hence f(xλ) ∈f(U) ⊂V for all λ ≥λ0. That means that f(x) = limΛ f(xλ). 6.5 Nets 87 Conversely, suppose f is not continuous. Thus there is an open set V ⊂Y such that U = f−1(V ) is not open. Then U c ∩U contains a point x. By Proposi-tion 6.5.4, there is a net (xλ)Λ in U c with limit x. Therefore f(xλ) ∈f(U c) ⊂V c. Since V c is closed, any limit point of this net must remain in V c by Proposi-tion 6.5.4. Therefore it cannot converge to f(x) which lies in V . So f(x) ̸= limΛ f(xλ). ■ 6.5.6. THEOREM. A topological space X is compact if and only if every net in X has a convergent subnet. PROOF. Suppose that every net in X has a convergent subnet. Consider a collection F = {Cα : α ∈A} of closed sets with the FIP. Let Λ = {F ⊂A : F is ﬁnite, non-empty} ordered by inclusion, i.e., F ≤G if F ⊂G. This is an upward directed poset: F1, F2 ≤F1 ∪F2. For each F ∈Λ, use the Axiom of Choice to select a point xF ∈T α∈F Cα. This is possible since the ﬁnite intersection is non-empty. Then (xF )Λ is a net in X. Let (yγ)Γ be a subnet with limit x; where ϕ : Γ →Λ and yγ = xϕ(γ). For any α ∈A, there is a γα ∈Γ so that γ ≥γα implies that ϕ(γ) ≥{α}. Hence yγ ∈Cα for all γ ≥γα. Since Cα is closed, the limit point x ∈Cα. This holds for all α ∈A. Therefore x ∈T F. By Proposition 6.3.5, it follows that X is compact. Conversely suppose that X is compact. Let (xλ)Λ be a net in X. For each λ ∈Λ, deﬁne Cλ = {xµ : µ ≥λ}. Then F = {Cλ : λ ∈Λ} is a collection of non-empty closed sets. It has FIP because if λ1, . . . , λn ∈Λ, the upward directed property ensures that there is some λ0 ∈Λ so that λi ≤λ0 for 1 ≤i ≤n. Hence Tn i=1 Cλi ⊃Cλ0 ̸= ∅. By Proposition 6.3.5, there is a point x ∈T Λ Cλ. Now we build a subnet with limit x. Let O(x) be the set of all open neigh-bourhoods of x. Let Γ = Λ × O(x) with order (λ, U) ≤(µ, V ) if λ ≤µ and U ⊃V . Let Sλ,U = {µ ∈Λ : µ ≥λ and xµ ∈U}. This set is non-empty because x ∈Cλ ∩U = {xµ : µ ≥λ} ∩U; and thus by Proposition 6.5.4, xµ ∈U for some µ ≥λ. Use the Axiom of Choice to select µ = ϕ(λ, U) ∈Sλ,U for each (λ, U) ∈Γ. The map ϕ : Γ →Λ is coﬁnal because if λ0 ∈Λ, then every (λ, U) ≥(λ0, X) will have ϕ(λ, U) = µ ≥λ ≥λ0. So yλ,U = xϕ(λ,U) deﬁnes a subnet (yλ,U)Γ of (xλ)Λ. Finally we claim that limΓ yλ,U) = x. Indeed, let U ∈O(x) be any open neighbourhood of x. Fix some λ0 ∈Λ. Whenever (λ, V ) ≥(λ0, U), we have yλ,V ∈V ⊂U. Thus this net converges to x. ■ CHAPTER 7 Functionals on Cc(X) and C0(X) 7.1. Radon measures Let X be a locally compact Hausdorff space. Let Cc(X) denote the space of continuous functions with compact support (i.e. {x : f(x) ̸= 0} is compact). Let C0(X) be the space of all continuous functions vanishing at inﬁnity, meaning that {x : \|f(x)\| ≥ε} is compact for all ε > 0. Both spaces are endowed with the sup norm ∥f∥∞= sup \|f(x)\|. It is easy to see that C0(X) = Cc(X) ∥·∥∞. When X is compact, Cc(X) = C0(X) = C(X). 7.1.1. DEFINITION. A Borel measure µ is outer regular on E ∈Bor(X) if µ(E) = inf{µ(U) : E ⊂U open} and inner regular on E if µ(E) = sup{µ(K) : E ⊃K compact}. We call µ a regular measure if it is both inner and outer regular on all Borel sets. A Borel measure µ is a Radon measure if µ(K) < ∞for all compact sets K, it is outer regular on all Borel sets and inner regular on open sets. 7.1.2. REMARKS. (1) Some books deﬁne a Radon measure to be a regular measure which takes ﬁnite values on all compact sets. This is somewhat less general (so we would require an additional hypothesis such as σ-compactness), and does not simplify the proof of the main result. (2) If µ(E) < ∞, then µ is regular on E if and only if for ε > 0, there is an open set U and a compact set K so that K ⊂E ⊂U and µ(U \ K) < ε. (3) There are compact Hausdorff spaces which support ﬁnite Borel measures which are not regular. See the exercises in . 7.1.3. PROPOSITION. If X is LCH, µ is a Radon measure on X and E ∈ Bor(X) is σ-ﬁnite, then µ is regular on E. 88 7.1 Radon measures 89 PROOF. Let E ∈Bor(X) such that µ(E) < ∞, and let ε > 0. By outer regularity, there is an open U ⊃E with µ(U \ E) < ε/2. Then there is an open set V ⊃U \ E with µ(V ) < ε/2. Since µ is inner regular on U, there is a compact L ⊂U with µ(U \ L) < ε/2. Then K = L \ V is compact and K ⊂U \ V ⊂E. Finally, µ(E \ K) < µ(U \ K) ≤µ(U \ L) + µ(V ) < ε. Thus µ is regular on E. If E is a σ-ﬁnite set, write E = ˙ S i≥1Ei where µ(Ei) < ∞. Find compact sets Kn ⊂En with µ(En \ Kn) < 2−nε. Then Cp = Sp n=1 Kn are compact. Set K = S n≥1 Kn. Then K ⊂E and µ(E \ K) ≤P n≥1 µ(En \ Kn) < ε. By continuity from below, we get limp→∞µ(Cp) = µ(K) > µ(E) −ε. Therefore µ is regular on E. ■ The following immediate corollary covers many cases of interest. 7.1.4. COROLLARY. Every σ-ﬁnite Radon measure is regular. In particular, if X is σ-compact (i.e. a countable union of compact sets) and µ is a Radon measure on X, then µ is regular. 7.1.5. COROLLARY. If X is a separable, locally compact metric space and µ is a Radon measure on X, then µ is regular. PROOF. We show that X is σ-compact. A separable metric space is second countable (see Example 6.1.10(1)). Indeed, if {xi : i ≥1} is a dense subset, then T = {br(xi) : i ≥1, r ∈Q+} is a countable base for the topology. For each xi, local compactness implies that there is some ri > 0 so that bri(xi) is compact. Then T0 = {br(xi) : i ≥1, r ∈Q+, r ≤ri} is also a base for the topology. Hence X = [ T0 br(xi) = [ T0 br(xi) is the union of countably many compact sets. The result now follows from Corol-lary 7.1.4. ■ 7.1.6. PROPOSITION. If X is a separable, locally compact metric space and µ is a ﬁnite Borel measure on X, then µ is regular and hence Radon. PROOF. Let S be the collection of all Borel sets on which µ is regular. The proof of Corollary 7.1.5 shows that X is σ-compact. Write X = S n≥1 Kn where Kn are compact. If C is closed, then the sets Cn = C ∩Sn i=1 Ki are compact, and 90 Functionals on Cc(X) and C0(X) Cn ⊂Cn+1 ⊂S n≥1 Cn = C. By continuity from below, µ(C) = limn→∞µ(Cn). So µ is inner regular on closed sets. As X is a metric space, a closed set C is a Gδ (let Un = {x : d(x, C) < 1 n}). Since µ is ﬁnite, continuity from above shows that if C = T n≥1 Un and Un ⊃ Un+1, then µ(C) = limn→∞µ(Un). Thus µ is regular on C. Next we claim S is closed under complements. Let A ∈S. By Remark 7.1.2(2), for ε > 0, there is a compact set K and open set U with K ⊂A ⊂U and µ(U \ K) < ε. Therefore U c ⊂Ac ⊂Kc, and µ(Kc \ U c) = µ(U \ K) < ε. The set U c is closed but not necessarily compact. Since U c ∈S, there is a compact set L ⊂U c so that µ(U c \ L) < ε −µ(Kc \ U c). Thus µ(Kc \ L) < ε. Therefore µ is regular on Ac. Now we show that S is closed under countable unions. Let Ai ∈S for i ≥1, and let A = S i≥1 Ai. Given ε > 0, ﬁnd compact sets Ki and open sets Ui so that Ki ⊂Ai ⊂Ui and µ(Ui \ Ki) < 2−iε. Set Cn = n [ i=1 Ki, C = [ i≥1 Ki and U = [ i≥1 Ui. Then Cn ⊂A ⊂U, each Cn is compact, and U is open. Therefore lim n→∞µ(U \ Cn) = µ(U \ C) ≤ X i≥1 µ(Ui \ Ki) < X i≥1 2−iε = ε. The ﬁrst step uses continuity from above, which requires the ﬁniteness of µ. Hence S is a σ-algebra. As S contains all closed sets, it contains all Borel sets. So µ is regular and ﬁnite, whence Radon. ■ 7.2. Positive functionals on Cc(X) 7.2.1. DEFINITION. A positive linear ﬁunctional ϕ on Cc(X) or C0(X) is a linear functional such that ϕ(f) ≥0 if f ≥0. 7.2.2. PROPOSITION. Let X be a locally compact Hausdorff space; and let µ be a Borel measure on X which is ﬁnite on compact sets. (1) Φµ(f) = Z f dµ determines a positive linear functional on Cc(X). (2) Φµ is continuous with respect to the sup norm (i.e. \|Φµ(f)\| ≤C∥f∥∞ for f ∈Cc(X)) if and only if M = sup{µ(K) : K is compact} < ∞. Moreover ∥Φµ∥= M. (3) In particular, if µ is Radon, then Φµ is continuous if and only if ∥µ∥:= µ(X) < ∞. 7.2 Positive functionals on Cc(X) 91 PROOF. If f ∈Cc(X) with ∥f∥∞≤1 and supp(f) ⊂K, where K is com-pact, then \|Φµ(f)\| = Z f dµ ≤ Z K \|f\| dµ ≤µ(K). So this is a linear functional which is norm continuous on the subspace of functions supported on K. Clearly it takes positive functions to positive values. By Proposition 6.4.12, given a compact set K, there is a function h ∈Cc(X) so that χK ≤h ≤1. Hence ∥Φµ∥≥Φµ(h) ≥µ(K). Taking the supremum over all compact K shows that ∥Φµ∥≥M = sup{µ(K) : K is compact}. Therefore if Φµ is continuous with respect to the sup norm, then M < ∞. Conversely, if M is ﬁnite and f ∈Cc(X) with ∥f∥∞≤1, then K = supp f is compact. Thus \|Φµ(f)\| ≤ Z \|f\| dµ ≤ Z K dµ = µ(K) ≤M. Hence ∥Φµ∥= M. If µ is Radon, then it is inner regular on X; so µ(X) = M. Thus Φµ is continuous if and only if ∥µ∥:= µ(X) < ∞. ■ 7.2.3. PROPOSITION. Let X be a LCH space. Suppose that ϕ is a positive linear functional on Cc(X). If K ⊂X is compact, then there is a constant CK so that \|ϕ(f)\| ≤CK∥f∥∞for all f ∈Cc(X) with supp(f) ⊂K. A positive linear functional on C0(X) is continuous, i.e. there is a C < ∞so that \|ϕ(f)\| ≤C∥f∥∞for all f ∈C0(X). PROOF. As in the previous proof, there is a function h ∈Cc(X) such that χK ≤h ≤χL for some compact set L. Let f ∈Cc(X) with supp(f) ⊂K with ∥f∥∞≤1. First suppose that f is real valued. Then −h ≤f ≤h, so 0 ≤f +h ≤2h. By positivity, 0 ≤ϕ(f)+ϕ(h) ≤2ϕ(h); whence \|ϕ(f)\| ≤ϕ(h). If f takes arbitrary complex values, let ϕ(f) = eiθ\|ϕ(f)\|. Replace f by e−iθf so that we may suppose that ϕ(f) > 0. Write f = Re f + i Im f. Since real functions are sent to real values, 0 ≤ϕ(f) = ϕ(Re f) + iϕ(Im f) implies that ϕ(f) = ϕ(Re f). Hence \|ϕ(f)\| ≤ϕ(h). So CK = ϕ(h) is the desired constant. Now suppose that ϕ is deﬁned on C0(X). Let M = sup{ϕ(f) : 0 ≤f ≤1, f ∈C0(X)}. If M = ∞, we can choose fn ∈C0(X) with 0 ≤fn ≤1 and ϕ(fn) > 2n. Deﬁne f = P k≥1 2−kfnk. This converges uniformly, and so lies C0(X). By positivity, ϕ(f) ≥sup p≥1 ϕ p X n=1 2−nfn = sup p≥1 p X n=1 2−nϕ(fn) ≥sup p≥1 p = ∞. This is impossible, and thus M < ∞. We now argue as in the ﬁrst paragraph to deduce that \|ϕ(f)\| ≤ϕ(\|f\|) ≤M∥f∥∞for f in C0(X). ■ 92 Functionals on Cc(X) and C0(X) For convenience of notation, given U ⊂X is open, we will write f ≺U if f ∈Cc(X), 0 ≤f ≤1 and supp(f) ⊂U. Also recall that Cc(X, R) is a lattice, and we write f ∨g = max{f, g} and f ∧g = min{f, g}. 7.2.4. RIESZ-MARKOV THEOREM. Let ϕ be a positive linear functional on Cc(X) for a LCH space X. Then there is a unique Radon measure µ on X so that ϕ = Φµ. PROOF. For U ⊂X open, deﬁne ρ(U) = sup{ϕ(f) : f ≺U}. Use this to deﬁne an outer measure µ∗given by µ∗(E) = inf X i≥1 ρ(Ui) : E ⊂ [ i≥1 Ui, Ui open . Let ¯ µ denote the complete measure on the σ-algebra of µ∗-measurable sets. Claim 1: µ∗(U) = ρ(U) for U open. Since U ⊂U, we have µ∗(U) ≤ρ(U). Suppose that U ⊂S i≥1 Ui for Ui open. Take f ≺U. Then K = supp(f) is a compact subset of U, and hence K ⊂U1 ∪· · · ∪Un for some n. By Proposi-tion 6.4.15, there is a partition of unity gi ∈Cc(X) with gi ≺Ui ∩U such that χK ≤Pn i=1 gi ≺U. Then f = Pn i=1 fgi and fgi ≺Ui ∩U. Therefore ϕ(f) = n X i=1 ϕ(fgi) ≤ n X i=1 ρ(Ui). Taking the supremum over f ≺U, we see that ρ(U) ≤P∞ i=1 ρ(Ui). This means that there is no advantage to using more than one open set to cover E. That is, if E ⊂S i≥1 Ui = U for Ui open, then ρ(U) ≤P i≥1 ρ(Ui). Therefore µ∗(E) = inf{ρ(U) : E ⊂U, U open}. Claim 2: an open set U is µ∗-measurable, i.e., µ∗(E) = µ∗(E∩U)+µ∗(E\U) for all sets E ⊂X. Recall that µ∗(E) ≤µ∗(E ∩U) + µ∗(E \ U) is always valid. Thus is is enough to prove the reverse inequality when µ∗(E) < ∞. First let E be open with µ∗(E) < ∞. Let ε > 0. Then for some f ≺E ∩U, µ∗(E ∩U) = ρ(E ∩U) < ϕ(f) + ε Let K = supp(f). Then µ∗(E \ U) ≤µ∗(E \ K) = ρ(E \ K) < ϕ(g) + ε for some g ≺E \ K. Since f, g have disjoint supports, f + g ≤χE. Therefore µ∗(E ∩U) + µ∗(E \ U) < ϕ(f + g) + 2ε ≤ρ(E) + 2ε. Letting ε →0, we get µ∗(E ∩U) + µ∗(E \ U) ≤µ∗(U), and thus equality holds. Now let E be arbitrary with µ∗(E) < ∞. For ε > 0, pick E ⊂V with V open so that ρ(V ) < µ∗(E) + ε. Then µ∗(E ∩U) + µ∗(E \ U) ≤µ∗(V ∩U) + µ∗(V \ U) = µ∗(V ) < µ∗(E) + ε. 7.2 Positive functionals on Cc(X) 93 Let ε →0 to get the non-trivial inequality. Thus U is µ∗-measurable. Therefore ¯ µ is deﬁned on the σ-algebra generated by all open sets, namely Bor(X). Let µ be the Borel measure obtained by restricting ¯ µ to Bor(X). Claim 3: µ is Radon. First µ is outer regular since for E ∈Bor(X), µ(E) = µ∗(E) = inf{µ(U) : E ⊂U, U open}. If K is compact, there is an open set W ⊃K with L = W compact. By Urysohn’s Lemma, there is a function f with χK ≤f ≺W. Thus µ(K) ≤µ(W) ≤CL < ∞, where CL is the constant from Proposition 7.2.3. By outer regularity, we can choose an open W ⊃K so that µ(W) < µ(K) + ε. If f is chosen with χK ≤f ≺W, then ϕ(f) < µ(W) < µ(K) + ε. On the other hand, suppose f ∈Cc(X) such that χK ≤f. Let ε > 0, and set V = {x : f(x) > 1 −ε}. Then K ⊂V , and V is open. Let g ≺V so that ϕ(g) > µ(V ) −ε ≥µ(K) −ε. Observe that (1 −ε)g ≤f; and hence ϕ(f) ≥(1 −ε)(µ(K) −ε). Letting ε →0 yields ϕ(f) ≥µ(K). Combining these two results shows that µ(K) = inf{ϕ(f) : χK ≤f, f ∈Cc(X)}. Now if U is open, we have that µ(U) = sup{ϕ(f) : f ≺U}. Suppose that r < µ(U). Choose g ≺U with ϕ(g) > r. Set K = supp(g), which is a compact subset of U. By Urysohn’s Lemma, there is an h ∈Cc(X) with χK ≤h ≺U. By the previous paragraph, there is an f ∈Cc(X) with χK ≤f and ϕ(f) < µ(K) + ε. So χK ≤h ∧f ≺U. Therefore r < ϕ(g) ≤ϕ(h ∧f) ≤ϕ(f) < µ(K) −ε. Hence µ(K) > r −ε. It follows that µ(U) = sup{µ(K) : K ⊂U, K compact}. Thus µ is inner regular on open sets. Consequently µ is Radon. Claim 4: ϕ = Φµ. By linearity, it sufﬁces to verify this for 0 ≤f ≤1. Given N ∈N, let ti = i N for 0 ≤i ≤N. Deﬁne fi = (f ∨ti−1) ∧ti −ti−1 and Ki = {x : f(x) ≥ti} for 1 ≤i ≤N, and K0 = supp(f). Then 1 N χKi ≤fi ≤1 N χKi−1 for 1 ≤i ≤N and f = PN i=1 fi. Integrating, we obtain 1 N µ(Ki) ≤Φµ(fi) ≤1 N µ(Ki−1). 94 Functionals on Cc(X) and C0(X) We also obtain 1 N µ(Ki) ≤ϕ(fi). If Ki−1 ⊂U for U open, then Nfi ≺U and so ϕ(fi) ≤1 N µ(U). But µ is outer regular, and hence ϕ(fi) ≤1 N µ(Ki−1). Therefore 1 N µ(Ki) ≤ϕ(fi) ≤1 N µ(Ki−1). Now sum both of these inequalities from 1 to N, we obtain 1 N N X i=1 µ(Ki) ≤Φµ(fi), ϕ(f) ≤1 N N−1 X i=0 µ(Ki). Hence Φµ(f) −ϕ(f) ≤µ(K0) −µ(KN) N ≤µ(K0) N . Since N was arbitrary, we obtain Φµ(f) = ϕ(f). Claim 5: uniqueness. Suppose that ν is a Radon measure such that Φν = ϕ. Then if U is open and f ≺U, then ϕ(f) = Φν(f) ≤ν(U). On the other hand, since ν is inner regular on U, if r < ν(U), there is a compact K ⊂U with ν(K) > r. Take some f ∈Cc(X) with χK ≤f ≺U and observe that ϕ(f) = Φν(f) ≥ν(K) > r. Therefore ν(U) = sup{ϕ(f) : f ≺U} = µ(U). Both measures are outer regular, and therefore they agree on all Borel sets. ■ 7.2.5. COROLLARY. Let X be a LCH space. Suppose that ϕ is a positive linear functional on C0(X). Then there is a unique ﬁnite Radon measure on X so that ϕ = Φµ. PROOF. The restriction of ϕ to Cc(X) is positive. Thus by the Riesz-Markov Theorem, there is a unique Radon measure µ on X so that ϕ = Φµ on Cc(X). By Proposition 7.2.3, ϕ is continuous and thus ϕ = Φµ on C0(X), and moreover µ(X) = sup{ϕ(f) : 0 ≤f ≤1, f ∈Cc(X)} ≤C < ∞. So µ is a ﬁnite measure. ■ 7.3. Linear functionals on C0(X) To discuss the continuous linear functionals on C0(X), we need complex Borel measures. Since they are ﬁnite measures, things simplify somewhat. A complex measure µ is called regular if \|µ\| is regular. 7.3 Linear functionals on C0(X) 95 7.3.1. PROPOSITION. Let X be a LCH space. Let µ be a complex Borel mea-sure on X. Then Φµ(f) = Z f dµ deﬁnes a continuous linear functional on C0(X) such that ∥Φµ∥≤∥µ∥:= \|µ\|(X). If µ is regular, then ∥Φµ∥= ∥µ∥. PROOF. If f ∈C0(X), \|Φµ(f)\| = Z f dµ ≤ Z \|f\| d\|µ\| ≤∥f∥∞\|µ\|(X). Thus ∥Φµ∥≤∥µ∥. Suppose that µ is regular. By the Radon-Nikodym Theorem, there is a Borel function h so that \|h\| = 1 and µ = h\|µ\|. Given ε > 0, chop the unit circle T into disjoint arcs I1, . . . , In of length at most ε and midpoints ζj. Set Ej = h−1(Ij). By regularity, ﬁnd compact Kj ⊂Ej so that \|µ\|(Ej \ Kj) < ε/n. Find disjoint open sets Uj ⊃Kj. By Urysohn’s Lemma, there are functions fj ∈Cc(X) with χKj ≤fj ≺Uj. Let f = Pn j=1 ζjfj. Then ∥f∥∞= 1 and ∥µ∥−Φµ(f) = Z ¯ h −f dµ ≤ Z ¯ h −f d\|µ\| ≤ n X j=1 Z Kj ¯ h −ζj d\|µ\| + n X j=1 2\|µ\|(Ej \ Kj) ≤ε 2∥µ∥+ 2ε. Letting ε →0 yields ∥Φµ∥= ∥µ∥. ■ 7.3.2. COROLLARY. If µ, ν are regular complex Borel measures on a LCH space X, then Φµ = Φν if and only if µ = ν. 7.3.3. DEFINITION. If X is a LCH space, let M(X) be the vector space of all complex regular Borel measures on X with norm ∥µ∥= \|µ\|(X). 7.3.4. RIESZ REPRESENTATION THEOREM. Let X be a LCH space. Then C0(X)∗is isometrically isomorphic to M(X) via the pairing µ →Φµ. PROOF. Proposition 7.3.1 shows that the map from M(X) into C0(X)∗is an isometric linear map. On the other hand, Corollary 7.2.5 of the Riesz-Markov Theorem shows that every positive linear functional on C0(X) is Φµ for a ﬁnite positive regular Borel measure µ. The result will follow if the linear span of the positive linear functional on C0(X) is all of C0(X)∗. We call a linear functional self-adjoint if ϕ(f) ∈R for f ∈C0(X, R). Let ϕ ∈C0(X)∗. Deﬁne ˜ ϕ(f) = ϕ( ¯ f). Set Re ϕ = 1 2(ϕ + ˜ ϕ) and Im ϕ = 1 2i(ϕ −˜ ϕ). 96 Functionals on Cc(X) and C0(X) Then if f ∈C0(X, R) is real valued, (Re ϕ)(f) = ϕ(f) + ϕ(f) 2 = Re ϕ(f) ∈R. Similarly Im ϕ takes real values on C0(X, R). So Re ϕ and Im ϕ are self-adjoint. Note that these functionals are not real valued on C0(X). Observe that ϕ = Re ϕ+ i Im ϕ. So CX 0 )∗is spanned by the self-adjoint linear functionals. Claim: if ϕ is self-adjoint, there are positive linear functionals ϕ1, ϕ2 on C0(X) such that ϕ = ϕ1 −ϕ2. This is an analogue of the Jordan decomposition theorem. If f ≥0, deﬁne ϕ1(f) = sup{ϕ(g) : 0 ≤g ≤f}. Then 0 ≤ϕ1(f) ≤sup0≤g≤f ∥ϕ∥∥g∥= ∥ϕ∥∥f∥. Also for t > 0, ϕ1(tf) = tϕ1(f). Additivity: suppose that f1 ≥0 and f2 ≥0 in C0(X). If 0 ≤gi ≤fi, then 0 ≤g1 +g2 ≤f1 +f2. Hence ϕ1(f1 +f2) ≥ϕ(g1)+ϕ(g2). Taking the supremum over all choices of gi yields ϕ1(f1 + f2) ≥ϕ1(f1) + ϕ1(f2). On the other hand, if 0 ≤g ≤f1 + f2, let g1 = g ∧f1 and g2 = g −g1 = ( 0 if g(x) ≤f1(x) g(x) −f1(x) ≤f2(x) if g(x) > f1(x). So 0 ≤g2 ≤f2. Thus ϕ(g) = ϕ(g1) + ϕ(g2) ≤ϕ1(f1) + ϕ1(f2). Taking the supremum over 0 ≤g ≤f1 + f2 yields the reverse inequality : ϕ1(f1 + f2) ≤ ϕ1(f1) + ϕ1(f2) Therefore ϕ1(f1 + f2) = ϕ1(f1) + ϕ1(f2). Now we extend ϕ1 to C0(X). If f ∈C0(X), we write f = Re f + i Im f = f1 −f2 + if3 −if4 where f1 = Re f ∨0, f2 = −Re f ∨0, f3 = Im f ∨0 and f4 = −Im f ∨0. Deﬁne ϕ1(f) = ϕ1(f1) −ϕ1(f2) + iϕ1(f3) −iϕ1(f4). It is left to the reader to verify that ϕ1 is linear. Once that it checked, it is clear that ϕ1 is a continuous positive linear functional. Set ϕ2 = ϕ1 −ϕ. Then if f ≥0, ϕ2(f) = sup{ϕ(g) −ϕ(f) : 0 ≤g ≤f} ≥ϕ(f) −ϕ(f) = 0. So ϕ2 is also a positive linear functional; and ϕ = ϕ1 −ϕ2. We have shown that the linear span of the positive linear functionals is all of C0(X)∗. Therefore by the Riesz-Markov theorem for C0(X), each is given as in-tegration against a ﬁnite Radon measure, hence regular. Thus the whole functional is given by integration against a regular complex Borel measure. ■ CHAPTER 8 A Taste of Probability This chapter is purely for interest. I expect to give one lecture on some of this, but it won’t be tested. This chapter closely follows Folland’s book . 8.1. The language of probability While the basic theorems of probability theory often coincide with the analysts view of measure theory, the vocabulary is completely different. This chart explains the correspondence for some familiar notions. Analyst’s term Probabalist’s term measure space (X, B, µ) sample space (Ω, B, P) and P(Ω) = 1. σ-algebra σ-ﬁeld measurable set event real measurable function random variable X integral R f dµ expectation E(X) µ({x : f(x) < a}) P(X < a) ∥f∥p < ∞ X has a ﬁnite pth moment almost everywhere (a.e.) almost surely (a.s.) Borel probability measure on R distribution charcteristic function χA indicator function 1A In analysis, we say that a sequence fn of measurable functions on (X, B, µ) converges in measure to f if for all ε > 0, lim n→∞µ {x : \|f(x) −fn(x)\| > ε} = 0. In probability theory, this is called convergence in probability. If X is a random variable, E(X) is the expected value or mean. The standard deviation (which is ﬁnite if X ∈L2(P)) is σ(X) = q E (X −E(X))2 = ∥X −E(X)∥2. If ϕ : (Ω, B) →(Ω′, B′) is a measurable map, deﬁne Pϕ on (Ω′, B′) by Pϕ(B) = P(ϕ−1(B)). For example, if X : Ω→R is a random variable, the 97 98 A Taste of Probability distribution function of X is F(t) = P(X ≤t) = PX (−∞, t] . Moreover PX = dF = µF is the Lebesgue-Stieltjes measure of F. A family of random variables {Xα}α∈A are identically distributed if PXα = PXβ for all α, β ∈A. The joint distribution of (X1, . . . , Xn) is P(X1,...,Xn) where we consider (X1, . . . , Xn) : Ω→Rn. Many properties of random variables can be recovered from their distributions. 8.1.1. PROPOSITION. If ϕ : (Ω, B) →(Ω′, B′) is measurable and f : Ω′ →R is measurable with respect to Pϕ, then Z Ω′ f dPϕ = Z Ω f ◦ϕ dP. PROOF. Let f = χA. Then Z Ω′ χA dPϕ = Pϕ(A) = P(ϕ−1(A)) = Z Ω χϕ−1(A) dP = Z Ω χA ◦ϕ dP. Thus the result is true for simple functions. It extends to all positive measurable functions by the MCT. Then it extends to all integrable functions by the LDCT. ■ 8.1.2. EXAMPLES. (1) E(X) = Z X dP = Z R t dPX. (2) σ2(X) = E (X −E(X))2 = Z R t −E(X) 2 dPX. (3) E(X + Y ) = Z R2 s + t dP(X,Y )(s, t). 8.2. Independence 8.2.1. DEFINITION. Events {Ai : i ∈I} are independent if whenever i1, . . . , ik in I are distinct, then P(Ai1 ∩· · · ∩Aik) = k Y s=1 P(Ais). Random variables {Xi : i ∈I} are independent if for every choice of Borel sets Bi, the collection {X−1 i (Bi) : i ∈I} is independent. That is, whenever i1, . . . , ik ∈I 8.2 Independence 99 are distinct, P(Xi1,...,Xik) k Y s=1 Bis = P k \ s=1 X−1 is (Bi) = k Y s=1 P(Ais) = k Y s=1 PXis k Y s=1 Bis . In other words, P(Xi1,...,Xik) = Qk s=1 PXis. 8.2.2. EXAMPLE. It is not sufﬁcient to check independence for pairs of vari-ables. Let X = {1, 2, 3, 4}, P(i) = 1 4, and A1 = {2, 3}, A2 = {1, 3} and A3 = {1, 2}. You can check that P(Ai) = 1 2 and P(Ai ∩Aj) = 1 4 for i ̸= j, but P(A1 ∩A2 ∩A3) = 0. 8.2.3. PROPOSITION. Let X, Y be independent integrable random variables. Then E(XY ) = E(X) E(Y ). PROOF. Compute E(XY ) = Z st dP(X,Y )(s, t) = Z Z st dPX dPY = Z s dPX(s) Z t dPY (t) = E(X) E(Y ). ■ 8.2.4. PROPOSITION. Let {Xi,j : 1 ≤j ≤Ji, 1 ≤i ≤n} be inde-pendent random variables. If fi : RJi →R are Borel for 1 ≤i ≤n, and Yi = fi(Xi,1, . . . , Xi,Ji), then {Y1, . . . , Yn} are independent. PROOF. Let ¯ Xi = (Xi,1, . . . , Xi,Ji). Then Y −1 i (B) = ¯ X−1 i (f−1 i (B)). Thus if Y = (Y1, . . . , Yn) and X = ( ¯ X1, . . . , ¯ Xn), Y −1(B1 × · · · × Bn) = n \ i=1 ¯ X−1 i (f−1 i (Bi)) = X−1 n Y i=1 f−1 i (Bi) . Therefore P Y −1(B1 × · · · × Bn) = PX n Y i=1 f−1 i (Bi) = n Y i=1 Ji Y j=1 PXi,j n Y i=1 f−1 i (Bi) = n Y i=1 Ji Y j=1 PXi,j f−1 i (Bi) 100 A Taste of Probability = n Y i=1 P ¯ Xi(f−1 i (Bi)) = n Y i=1 Pyi(Bi). So {Y1, . . . , Yn} are independent. ■ 8.2.5. COROLLARY. If {Xi} are independent variables in L2(P), then σ2(X1 + · · · + Xn) = n X i=1 σ2(Xi). PROOF. Let Yi = Xi −E(Xi); so E(Yi) = 0. Then {Yi} are independent, so E(YiYj) = E(Yi) E(Yj) = 0 for i ̸= j. Therefore σ2(X1 + · · · + Xn) = E n X i=1 Xi −E(Xi) 2 = E n X i=1 Yi 2 = n X i=1 E(Y 2 i ) + X i̸=j E(Yi) E(Yj) = n X i=1 σ2(Xi). ■ 8.3. The Law of Large Numbers The following easy lemma is surprisingly useful. 8.3.1. LEMMA (Chebychev’s Inequality). If X ≥0 is a random variable and a > 0, then P(X ≥a) ≤E(X)/a. PROOF. E(X) = Z ∞ 0 t dPX(t) ≥ Z ∞ a a dPX(t) = aP(X ≥a). ■ 8.3.2. COROLLARY. Let X be a random variable such that σ(X) < ∞(i.e. X ∈L2(P)). Then for b > 0, P(\|X −E(X)\| ≥b) ≤σ2(X)/b2. PROOF. Let Y = (X −E(X))2; so E(Y ) = σ2(X). Hence by Chebychev’s inequality, P(\|X −E(X)\| ≥b) = P(Y ≥b2) ≤E(Y )/b2 = σ2(X)/b2. ■ 8.3 The Law of Large Numbers 101 8.3.3. WEAK LAW OF LARGE NUMBERS. Let {Xi}i≥1 be independent L2 random variables. Denote E(Xi) = µi and σ2(Xi) = σ2 i for i ≥1. If lim n→∞ 1 n2 n P i=1 σ2 i = 0, then for any ε > 0, lim n→∞P 1 n n X i=1 Xi −µi > ε = 0. That is, lim n→∞ 1 n n P i=1 Xi −µi = 0 in probability. PROOF. Let Sn = 1 n n P i=1 (Xi −µi); so E(Sn) = 0 and by Corollary 8.2.5, σ2(Sn) = 1 n2 Pn i=1 σ2(Xi −µi) = 1 n2 Pn i=1 σ2 i . Hence by Corollary 8.3.2, P(\|Sn\| > ε) ≤σ2(Sn)/ε2 →0. That is, Sn converges to 0 in probability. ■ 8.3.4. BOREL-CANTELLI LEMMA. Let {An}n≥1 be events in a probability space. Deﬁne B = lim sup An := T k≥1 S n≥k An . (a) If P n≥1 P(An) < ∞, then P(B) = 0. (b) If {An} are independent and P n≥1 P(An) = ∞, then P(B) = 1. PROOF. (a) P(B) ≤P S n≥k An ≤P n≥k P(An) →0. (b) 1 −P [ n≥k An = P \ n≥k Ac n = Y n≥k P(Ac n) = Y n≥k 1 −P(An) ≤ Y n≥k e−P(An) = e−P n≥k P(An) = 0. Thus P(B) = 1. ■ 8.3.5. KOLMOGOROV’S INEQUALITY. Let X1, . . . , Xn be independent random variables with E(Xi) = 0 and σ2(Xi) = σ2 i . Then for a > 0, P max 1≤k≤n \|X1 + · · · + Xk\| ≥a ≤1 a2 n X i=1 σ2 i . PROOF. Let Sk = X1 + · · · + Xk. Let A = max{\|S1\|, . . . , \|Sn\| ≥a and Ak = \|Sj\| < a, 1 ≤j < k, \|Sk\| ≥a . So A = ˙ Sn k=1Ak. 102 A Taste of Probability Since Xi are independent and mean 0, n X i=1 σ2 i = σ2(X1 + · · · + Xn) = E(S2 n) ≥E(S2 nχA) = n X k=1 E(S2 nχAk). Also note that SkχAk depends only on X1, . . . , Xk and Sn −Sk depends only on Xk+1, . . . , Xn. Hence these are independent quantities, so that E (SkχAk)(Sn −Sk) = E(SkχAk)E(Xk+1 + · · · + Xn) = 0. Therefore E(S2 nχAk) = E (SkχAk + (Sn −Sk)χAk)2 = E(S2 kχAk) + 2E (SkχAk)(Sn −Sk) + E (Sn −Sk)2χAk) ≥E(S2 kχAk) ≥a2P(Ak). Consequently, a2P(A) = n X k=1 a2P(Ak) ≤ n X k=1 E(S2 nχAk) ≤ n X i=1 σ2 i . Thus P(A) ≤1 a2 Pn i=1 σ2 i . ■ 8.3.6. LEMMA. Let {Xi : i ≥1} be independent random variables with E(Xi) = 0 and σ2(Xi) = σ2 i . If P i≥1 σ2 i < ∞, then P P i≥1 Xi converges = 1. PROOF. Let Sn = X1+· · ·+Xn. So Sn+k −Sn = Xn+1+· · ·+Xn+k. Given δ > 0, choose N so that P i≥N σ2 i < δ. Then if ε > 0 and n ≥N, Kolmogorov’s inequality yields P max 1≤k≤m \|Sn+k −Sn\| ≥ε ≤1 ε2 n+m X i=n+1 σ2 i . Letting m →∞, we get P max k≥1 \|Sn+k −Sn\| ≥ε ≤1 ε2 X i>n σ2 i < δ ε2 . Choose δ = 2−3j and ε = 2−j and obtain an integer nj so that Aj = max k≥1 \|Snj+k −Snj\| ≥2−j has P(Aj) < 2−j. By the Borel-Cantelli Lemma, since P j≥1 P(Aj) < ∞, we have P T k≥1 S j≥k Aj = 0. Hence P S i≥1 T j≥i Ac j = 1. Therefore a.s., x belongs to T j≥i Ac j for some i, meaning that maxk≥1 \|Snj+k −Snj\| < 2−j for j ≥i. This means that Sn(x) is Cauchy and thus converges with probability 1. ■ 8.3 The Law of Large Numbers 103 8.3.7. LEMMA. If (ai)i≥1 is a sequence of real numbers and P i≥1 ai i converges, then lim n→∞ 1 n Pn i=1 ai = 0. PROOF. Let bn = an n and Sn = Pn i=1 bi. Let L = limn→∞Sn. Then L = limn→∞1 n Pn−1 i=0 Si. Compute 1 n n X i=1 ai = 1 n n X i=1 ibi = 1 n n X i=1 n X k=i bk = 1 n n X i=1 Sn −Si−1 = Sn −1 n n−1 X k=0 Sk →0. ■ 8.3.8. THEOREM (Kolmogorov). Let {Xi : i ≥1} be independent random variables with E(Xi) = µi and σ2(Xi) = σ2 i . If P i≥1 1 i2 σ2 i < ∞, then lim n→∞ 1 n n X i=1 Xi −µi = 0 a.s. PROOF. Let Yi = 1 i (Xi −ai). Then E(Yi) = 0, σ2(Yi) = 1 i2 σ2 i and {Yi} are independent. By hypothesis, P i≥1 σ2(Yi) < ∞. Hence by Lemma 8.3.6, P(P i≥1 Yi converges} = 1. Therefore by Lemma 8.3.7, with ai = P(Xi), P lim n→∞ 1 n n X i=1 (Xi −µi) = 0 = 1. ■ 8.3.9. STRONG LAW OF LARGE NUMBERS. Let {Xi}i≥1 be independent, identically distributed random variables. (a) If Xi are in L1 with E(Xi) = µ, then P lim n→∞ 1 n(X1 + · · · + Xn) = µ = 1. (b) If Xi are not in L1, then P lim sup n→∞ 1 n\|X1 + · · · + Xn\| = ∞ = 1. PROOF. (a) By replacing Xn by Xn −µ, we may suppose that E(Xn) = 0. Let λ = PXn, which is independent of n. Then ∥Xn∥1 = E(\|Xn\|) = Z R \|t\| dλ and 104 A Taste of Probability Z R t dλ = E(Xn) = 0. Let Yn = Xnχ{\|Xn\|≤n} and Zn = Xn −Yn. Then X n≥1 P(Zn ̸= 0) = X n≥1 P(\|Xn\| > n) = X n≥1 Z −n −∞ + Z ∞ n dλ = Z ∞ −∞ ⌊\|t\|⌋dλ ≤ Z ∞ −∞ \|t\| dλ = ∥X1∥1 < ∞. By the Borel-Cantelli Lemma, P(Zn ̸= 0 inﬁnitely often) = 0. Therefore P 1 n n X i=1 Zi →0 = 1. Now σ2(Yn) = E(Y 2 n ) = Z n −n t2 dλ. Hence by the MCT, X n≥1 1 n2 σ2(Yn) = X n≥1 Z n −n t2 n2 dλ = Z ∞ −∞ X n≥1 1 n2 χ[−n,n] t2 dλ for n−1 < \|t\| ≤n, we have P n≥1 1 n2 χ−n,n = P k≥n 1 k2 < 2 n and thus ≤ Z ∞ −∞ 2\|t\| dλ < ∞. Therefore by Kolmogorov’s Theorem, P limn→∞1 n Pn i=1 Yi = 0 = 1. Conse-quently, P lim n→∞ 1 n n X i=1 Xi = 0 ≥P lim n→∞ 1 n n X i=1 Yi = 0 · P lim n→∞ 1 n n X i=1 Zi = 0 = 1. (b) In this case, for any C > 0, X n≥1 P(\|Xn\| ≥Cn) = X n≥1 Z χ{\|t\|≥Cn} dλ ≥ Z R \|t\| C −1 dλ = +∞. By the Borel-Cantelli Lemma, P(\|Xn\| ≥Cn inﬁnitely often) = 1. But then max n\|X1 + · · · + Xn−1\| n −1 , \|X1 + · · · + Xn\| n o ≥C 2 inﬁnitely often a.s. Therefore lim supn→∞ 1 n\|X1 + · · · + Xn\| = ∞a.s. ■ BIBLIOGRAPHY S. Friedberg, A. Insel and L. Spence, Linear Algebra. 5th ed., Prentice Hall, 2018, Englewood Cliffs, NJ G.B. Folland, Real Analysis: modern techniques and their applications, 2nd ed., John Wiley & Sons, New York, 1999. S. Willard, General Topology, Addison-Wesley Pub. Co, Reading, 1970. 105 INDEX Fσ set, 3 Gδ set, 3 Lp(µ), 64 M(X), 95 T0, 81 T1, 81 T2, 81 T3, 81 T3.5, 81 µ∗-measurable, 6 σ-algebra, 3 σ-ﬁnite, 3 absolute value, 59 absolutely continuous, 53, 60 algebra of subsets, 3 almost everywhere, 5 almost surely, 97 atom, 13 atomic measure, 14 base, 75 Borel sets, 3 Borel-Cantelli Lemma, 101 bounded variation, 52 Cantor ternary function, 52 Carath´ eodory’s Theorem, 7 closed, 74 closure, 74 coﬁnal, 85 compact, 78 complete, 4 completely regular, 81 completion, 5 complex measure, 62 continuity from above, 4 continuity from below, 4 continuous, 76 converge uniformly, 77 convergence in probability, 97 converges in measure, 97 coordinate projections, 34 countable base of neighbourhoods of x, 76 countably additive, 3 Counting measure, 4 dense, 76 differentiable, 49 discrete topology, 74 distribution function of X, 98 dual of Lp(µ), 70 Egorov’s Theorem, 19 expectation, 97 expected value, 97 Extreme Value Theorem, 79 Fatou’s Lemma, 25 ﬁnite, 3 ﬁnite intersection property, 79 ﬁrst countable, 76 Fubini’s Theorem, 36 Fubini-Tonelli Theorem without Completenes, 40 H¨ older’s Inequality, 69 Hahn Decomposition Theorem, 58 Hausdorff, 77 homeomorphism, 76 identically distributed, 98 independent, 98 indicator function, 97 induced topology, 74 inner regular, 88 integrable, 26 integral, 21 interior, 74 interior point, 74 joint distribution, 98 Jordan DecompositionTheorem, 59 Kolmogorov’s Inequality, 101 Kolmogorov’s Theorem, 103 Lebesgue Decomposition Theorem, 62 Lebesgue Differentiation Theorem, 55 Lebesgue Dominated Convergence Theorem, 27 limit point, 74 linear functionals, 68 locally compact, 83 106 Index 107 lower derivative, 49 Lusin’s Theorem, 20 measurable, 16 measure, 3 Minkowski’s inequality, 64 Monotone Convergence Theorem, 24 monotonicity, 3 mutually singular, 59 neighbourhood, 75 net, 85 normal, 81 null set, 58 open cover, 78 open sets, 74 oscillation, 32 outer measure, 6 outer regular, 88 partial order, 85 partition of unity, 84 point mass, 4 poset, 85 positive linear ﬁunctional, 90 positive set, 58 premeasure, 8 probability measure, 3 product σ-algebra, 34 product measure, 36 product space, 34, 79 Radon measure, 88 Radon-Nikodym Theorem, 60 random variable, 97 regular, 81, 94 regular measure, 88 Riemann integrable, 30 Riesz Representation Theorem, 95 Riesz-Fischer Theorem, 65 Riesz-Markov Theorem, 92 second countable, 76 self-adjoint, 95 semi-ﬁnite, 3 separable, 76 separation properties, 81 signed measure, 57 simple function, 18 standard deviation, 97 Strong Law of Large Numbers, 103 stronger topology, 75 subadditivity, 4 subbase, 75 subnet, 85 support, 83 Tietze’s Extension Theorem, 83 Tonelli’s Theorem, 37 topology, 74 trivial topology, 74 Tychonoff, 81 upper derivative, 49 upward directed, 85 Urysohn’s Lemma, 82 vanishing at inﬁnity, 88 Vitali cover, 49 Vitali Covering Lemma, 49 Weak Law of Large Numbers, 101 weaker topology, 75
12189	https://openstax.org/books/college-physics-2e/pages/29-3-photon-energies-and-the-electromagnetic-spectrum	Skip to Content Go to accessibility page Keyboard shortcuts menu Log in College Physics 2e 29.3 Photon Energies and the Electromagnetic Spectrum College Physics 2e 29.3 Photon Energies and the Electromagnetic Spectrum Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Explain the relationship between the energy of a photon in joules or electron volts and its wavelength or frequency. Calculate the number of photons per second emitted by a monochromatic source of specific wavelength and power. Ionizing Radiation A photon is a quantum of EM radiation. Its energy is given by E=hf and is related to the frequency f and wavelength λ of the radiation by E=hf=hcλ(energy of a photon), 29.12 where E is the energy of a single photon and c is the speed of light. When working with small systems, energy in eV is often useful. Note that Planck’s constant in these units is h=4.14×10–15eV⋅s. 29.13 Since many wavelengths are stated in nanometers (nm), it is also useful to know that hc=1240 eV⋅nm. 29.14 These will make many calculations a little easier. All EM radiation is composed of photons. Figure 29.9 shows various divisions of the EM spectrum plotted against wavelength, frequency, and photon energy. Previously in this book, photon characteristics were alluded to in the discussion of some of the characteristics of UV, x rays, and γ rays, the first of which start with frequencies just above violet in the visible spectrum. It was noted that these types of EM radiation have characteristics much different than visible light. We can now see that such properties arise because photon energy is larger at high frequencies. Figure 29.9 The EM spectrum, showing major categories as a function of photon energy in eV, as well as wavelength and frequency. Certain characteristics of EM radiation are directly attributable to photon energy alone. \| \| \| --- \| \| Rotational energies of molecules \| 10−5 eV \| \| Vibrational energies of molecules \| 0.1 eV \| \| Energy between outer electron shells in atoms \| 1 eV \| \| Binding energy of a weakly bound molecule \| 1 eV \| \| Energy of red light \| 2 eV \| \| Binding energy of a tightly bound molecule \| 10 eV \| \| Energy to ionize atom or molecule \| 10 to 1000 eV \| Table 29.1 Representative Energies for Submicroscopic Effects (Order of Magnitude Only) Photons act as individual quanta and interact with individual electrons, atoms, molecules, and so on. The energy a photon carries is, thus, crucial to the effects it has. Table 29.1 lists representative submicroscopic energies in eV. When we compare photon energies from the EM spectrum in Figure 29.9 with energies in the table, we can see how effects vary with the type of EM radiation. Gamma rays, a form of nuclear and cosmic EM radiation, can have the highest frequencies and, hence, the highest photon energies in the EM spectrum. For example, a γ-ray photon with f= 1021Hz has an energy E=hf=6.63×10–13J=4.14 MeV. This is sufficient energy to ionize thousands of atoms and molecules, since only 10 to 1000 eV are needed per ionization. In fact, γ rays are one type of ionizing radiation, as are x rays and UV, because they produce ionization in materials that absorb them. Because so much ionization can be produced, a single γ-ray photon can cause significant damage to biological tissue, killing cells or damaging their ability to properly reproduce. When cell reproduction is disrupted, the result can be cancer, one of the known effects of exposure to ionizing radiation. Since cancer cells are rapidly reproducing, they are exceptionally sensitive to the disruption produced by ionizing radiation. This means that ionizing radiation has positive uses in cancer treatment as well as risks in producing cancer. Figure 29.10 One of the first x-ray images, taken by Röentgen himself. The hand belongs to Bertha Röentgen, his wife. (credit: Wilhelm Conrad Röntgen, via Wikimedia Commons) High photon energy also enables γ rays to penetrate materials, since a collision with a single atom or molecule is unlikely to absorb all the γ ray’s energy. This can make γ rays useful as a probe, and they are sometimes used in medical imaging. x rays, as you can see in Figure 29.9, overlap with the low-frequency end of the γ ray range. Since x rays have energies of keV and up, individual x-ray photons also can produce large amounts of ionization. At lower photon energies, x rays are not as penetrating as γ rays and are slightly less hazardous. X rays are ideal for medical imaging, their most common use, and a fact that was recognized immediately upon their discovery in 1895 by the German physicist W. C. Roentgen (1845–1923). (See Figure 29.10.) Within one year of their discovery, x rays (for a time called Roentgen rays) were used for medical diagnostics. Roentgen received the 1901 Nobel Prize for the discovery of x rays. Connections: Conservation of Energy Once again, we find that conservation of energy allows us to consider the initial and final forms that energy takes, without having to make detailed calculations of the intermediate steps. Example 29.2 is solved by considering only the initial and final forms of energy. Figure 29.11 X rays are produced when energetic electrons strike the copper anode of this cathode ray tube (CRT). Electrons (shown here as separate particles) interact individually with the material they strike, sometimes producing photons of EM radiation. While γ rays originate in nuclear decay, x rays are produced by the process shown in Figure 29.11. Electrons ejected by thermal agitation from a hot filament in a vacuum tube are accelerated through a high voltage, gaining kinetic energy from the electrical potential energy. When they strike the anode, the electrons convert their kinetic energy to a variety of forms, including thermal energy. But since an accelerated charge radiates EM waves, and since the electrons act individually, photons are also produced. Some of these x-ray photons obtain the kinetic energy of the electron. The accelerated electrons originate at the cathode, so such a tube is called a cathode ray tube (CRT), and various versions of them are found in older TV and computer screens as well as in x-ray machines. Example 29.2 X-ray Photon Energy and X-ray Tube Voltage Find the maximum energy in eV of an x-ray photon produced by electrons accelerated through a potential difference of 50.0 kV in a CRT like the one in Figure 29.11. Strategy Electrons can give all of their kinetic energy to a single photon when they strike the anode of a CRT. (This is something like the photoelectric effect in reverse.) The kinetic energy of the electron comes from electrical potential energy. Thus we can simply equate the maximum photon energy to the electrical potential energy—that is, hf=qV. (We do not have to calculate each step from beginning to end if we know that all of the starting energy qV is converted to the final form hf.) Solution The maximum photon energy is hf=qV, where q is the charge of the electron and V is the accelerating voltage. Thus, hf=(1.60×10–19C)(50.0×103V). 29.15 From the definition of the electron volt, we know 1 eV=1.60×10–19J, where 1 J=1 C⋅V. Gathering factors and converting energy to eV yields hf=(50.0×103)(1.60×10–19C⋅V)(1 eV1.60×10–19C⋅V)=(50.0×103)(1 eV)=50.0 keV. 29.16 Discussion This example produces a result that can be applied to many similar situations. If you accelerate a single elementary charge, like that of an electron, through a potential given in volts, then its energy in eV has the same numerical value. Thus a 50.0-kV potential generates 50.0 keV electrons, which in turn can produce photons with a maximum energy of 50 keV. Similarly, a 100-kV potential in an x-ray tube can generate up to 100-keV x-ray photons. Many x-ray tubes have adjustable voltages so that various energy x rays with differing energies, and therefore differing abilities to penetrate, can be generated. Figure 29.12 X-ray spectrum obtained when energetic electrons strike a material. The smooth part of the spectrum is bremsstrahlung, while the peaks are characteristic of the anode material. Both are atomic processes that produce energetic photons known as x-ray photons. Figure 29.12 shows the spectrum of x rays obtained from an x-ray tube. There are two distinct features to the spectrum. First, the smooth distribution results from electrons being decelerated in the anode material. A curve like this is obtained by detecting many photons, and it is apparent that the maximum energy is unlikely. This decelerating process produces radiation that is called bremsstrahlung (German for braking radiation). The second feature is the existence of sharp peaks in the spectrum; these are called characteristic x rays, since they are characteristic of the anode material. Characteristic x rays come from atomic excitations unique to a given type of anode material. They are akin to lines in atomic spectra, implying the energy levels of atoms are quantized. Phenomena such as discrete atomic spectra and characteristic x rays are explored further in Atomic Physics. Ultraviolet radiation (approximately 4 eV to 300 eV) overlaps with the low end of the energy range of x rays, but UV is typically lower in energy. UV comes from the de-excitation of atoms that may be part of a hot solid or gas. These atoms can be given energy that they later release as UV by numerous processes, including electric discharge, nuclear explosion, thermal agitation, and exposure to x rays. A UV photon has sufficient energy to ionize atoms and molecules, which makes its effects different from those of visible light. UV thus has some of the same biological effects as γ rays and x rays. For example, it can cause skin cancer and is used as a sterilizer. The major difference is that several UV photons are required to disrupt cell reproduction or kill a bacterium, whereas single γ-ray and X-ray photons can do the same damage. But since UV does have the energy to alter molecules, it can do what visible light cannot. One of the beneficial aspects of UV is that it triggers the production of vitamin D in the skin, whereas visible light has insufficient energy per photon to alter the molecules that trigger this production. Infantile jaundice is treated by exposing the baby to UV (with eye protection), called phototherapy, the beneficial effects of which are thought to be related to its ability to help prevent the buildup of potentially toxic bilirubin in the blood. Example 29.3 Photon Energy and Effects for UV Short-wavelength UV is sometimes called vacuum UV, because it is strongly absorbed by air and must be studied in a vacuum. Calculate the photon energy in eV for 100-nm vacuum UV, and estimate the number of molecules it could ionize or break apart. Strategy Using the equation E=hf and appropriate constants, we can find the photon energy and compare it with energy information in Table 29.1. Solution The energy of a photon is given by E=hf=hcλ. 29.17 Using hc=1240 eV⋅nm, we find that E=hcλ=1240 eV⋅nm100 nm=12.4 eV. Discussion According to Table 29.1, this photon energy might be able to ionize an atom or molecule, and it is about what is needed to break up a tightly bound molecule, since they are bound by approximately 10 eV. This photon energy could destroy about a dozen weakly bound molecules. Because of its high photon energy, UV disrupts atoms and molecules it interacts with. One good consequence is that all but the longest-wavelength UV is strongly absorbed and is easily blocked by sunglasses. In fact, most of the Sun’s UV is absorbed by a thin layer of ozone in the upper atmosphere, protecting sensitive organisms on Earth. Damage to our ozone layer by the addition of such chemicals as CFC’s has reduced this protection for us. Visible Light The range of photon energies for visible light from red to violet is 1.63 to 3.26 eV, respectively (left for this chapter’s Problems and Exercises to verify). These energies are on the order of those between outer electron shells in atoms and molecules. This means that these photons can be absorbed by atoms and molecules. A single photon can actually stimulate the retina, for example, by altering a receptor molecule that then triggers a nerve impulse. Photons can be absorbed or emitted only by atoms and molecules that have precisely the correct quantized energy step to do so. For example, if a red photon of frequency f encounters a molecule that has an energy step, ΔE, equal to hf, then the photon can be absorbed. Violet flowers absorb red and reflect violet; this implies there is no energy step between levels in the receptor molecule equal to the violet photon’s energy, but there is an energy step for the red. There are some noticeable differences in the characteristics of light between the two ends of the visible spectrum that are due to photon energies. Red light has insufficient photon energy to expose most black-and-white film, and it is thus used to illuminate darkrooms where such film is developed. Since violet light has a higher photon energy, dyes that absorb violet tend to fade more quickly than those that do not. (See Figure 29.13.) Take a look at some faded color posters in a storefront some time, and you will notice that the blues and violets are the last to fade. This is because other dyes, such as red and green dyes, absorb blue and violet photons, the higher energies of which break up their weakly bound molecules. (Complex molecules such as those in dyes and DNA tend to be weakly bound.) Blue and violet dyes reflect those colors and, therefore, do not absorb these more energetic photons, thus suffering less molecular damage. Figure 29.13 Why do the reds, yellows, and greens fade before the blues and violets when exposed to the Sun, as with this poster? The answer is related to photon energy. (credit: Deb Collins, Flickr) Transparent materials, such as some glasses, do not absorb any visible light, because there is no energy step in the atoms or molecules that could absorb the light. Since individual photons interact with individual atoms, it is nearly impossible to have two photons absorbed simultaneously to reach a large energy step. Because of its lower photon energy, visible light can sometimes pass through many kilometers of a substance, while higher frequencies like UV, x ray, and γ rays are absorbed, because they have sufficient photon energy to ionize the material. Example 29.4 How Many Photons per Second Does a Typical Light Bulb Produce? Assuming that 10.0% 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. Strategy Power is energy per unit time, and so if we can find the energy per photon, we can determine the number of photons per second. This will best be done in joules, since power is given in watts, which are joules per second. Solution The power in visible light production is 10.0% 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 E=hcλ. This produces E=(6.63×10–34J⋅s)(3.00×108m/s)580×10–9m=3.43×10–19J. The number of visible photons per second is thus photon/s=10.0 J/s3.43×10–19J/photon=2.92×1019photon/s. Discussion This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. It is also a verification of the correspondence principle—on the macroscopic scale, quantization becomes essentially continuous or classical. Finally, there are so many photons emitted by a 100-W lightbulb that it can be seen by the unaided eye many kilometers away. Lower-Energy Photons Infrared radiation (IR) has even lower photon energies than visible light and cannot significantly alter atoms and molecules. IR can be absorbed and emitted by atoms and molecules, particularly between closely spaced states. IR is extremely strongly absorbed by water, for example, because water molecules have many states separated by energies on the order of 10–5eV to 10–2eV, well within the IR and microwave energy ranges. This is why in the IR range, skin is almost jet black, with an emissivity near 1—there are many states in water molecules in the skin that can absorb a large range of IR photon energies. Not all molecules have this property. Air, for example, is nearly transparent to many IR frequencies. Microwaves are the highest frequencies that can be produced by electronic circuits, although they are also produced naturally. Thus microwaves are similar to IR but do not extend to as high frequencies. There are states in water and other molecules that have the same frequency and energy as microwaves, typically about 10–5eV. This is one reason why food absorbs microwaves more strongly than many other materials, making microwave ovens an efficient way of putting energy directly into food. Photon energies for both IR and microwaves are so low that huge numbers of photons are involved in any significant energy transfer by IR or microwaves (such as warming yourself with a heat lamp or cooking pizza in the microwave). Visible light, IR, microwaves, and all lower frequencies cannot produce ionization with single photons and do not ordinarily have the hazards of higher frequencies. When visible, IR, or microwave radiation is hazardous, such as the inducement of cataracts by microwaves, the hazard is due to huge numbers of photons acting together (not to an accumulation of photons, such as sterilization by weak UV). The negative effects of visible, IR, or microwave radiation can be thermal effects, which could be produced by any heat source. But one difference is that at very high intensity, strong electric and magnetic fields can be produced by photons acting together. Such electromagnetic fields (EMF) can actually ionize materials. Misconception Alert: High-Voltage Power Lines Although some people think that living near high-voltage power lines is hazardous to one’s health, ongoing studies of the transient field effects produced by these lines show their strengths to be insufficient to cause damage. Demographic studies also fail to show significant correlation of ill effects with high-voltage power lines. The American Physical Society issued a report over 20 years ago on power-line fields, and additional studies since that time show no consistent, significant link between cancer and power-line fields. It is virtually impossible to detect individual photons having frequencies below microwave frequencies, because of their low photon energy. But the photons are there. A continuous EM wave can be modeled as photons. At low frequencies, EM waves are generally treated as time- and position-varying electric and magnetic fields with no discernible quantization. This is another example of the correspondence principle in situations involving huge numbers of photons. PhET Explorations Color Vision Make a whole rainbow by mixing red, green, and blue light. Change the wavelength of a monochromatic beam or filter white light. View the light as a solid beam, or see the individual photons. Access multimedia content Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Paul Peter Urone, Roger Hinrichs Publisher/website: OpenStax Book title: College Physics 2e Publication date: Jul 13, 2022 Location: Houston, Texas Book URL: Section URL: © Jul 9, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
12190	https://www.physicsforums.com/threads/proof-help-if-a-and-b-are-real-numbers-a-2-b-2-2ab.693415/	General Math Differential Equations Topology and Analysis Linear and Abstract Algebra Differential Geometry Set Theory, Logic, Probability, Statistics MATLAB, Maple, Mathematica, LaTeX Log in More options Style variation System Light Dark Contact us Close Menu You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. Forums Mathematics General Math Proof Help: If a and b are Real Numbers, a^2 + b^2 ≥ 2ab Thread starter woundedtiger4 Start date Tags : Proof AI Thread Summary The discussion centers on proving the inequality a^2 + b^2 ≥ 2ab for real numbers a and b. The proof correctly utilizes the expression (a-b)^2 ≥ 0, leading to the conclusion that a^2 + b^2 - 2ab ≥ 0. Participants clarify that the assumption a, b ≥ 0 is unnecessary, as the proof must hold for all real numbers. There is an emphasis on the importance of not limiting the proof to non-negative values, as the goal is to establish the inequality universally. The conversation ultimately reinforces the validity of the proof while addressing misconceptions about the conditions under which it applies. 1 woundedtiger4 : 188 : 0 If a and b are real numbers, then a^2 + b^2 >= 2 a bProof:if a>=0 and b>=0 then a-b>=0 (a-b)^2>=0a^2 + b^2 -2ab >=0a^2 + b^2 >= 2abQ.E.D.at someone has proved more or less similar question, but at (a + b)^2 > 0 a^2 + 2ab + b^2 > 0 a^2 + b^2 ≥ 2ab when 2ab goes from L.H.S. to R.H.S then it should be -2ab , am I correct? Mathematics news on Phys.org ChatGPT appears to improvise when put through ancient Greek math puzzle Systematic fraud uncovered in mathematics publications The science of sacrifice: How altruism and evolution can work in tandem 2 Fredrik Staff Emeritus Science Advisor Homework Helper Insights Author Gold Member : 10,876 : 423 Yes, and that's why you should use ##(a-b)^2## instead of ##(a+b)^2## to prove this result.You should remove the first two lines from the first proof. Do you see why? 3 woundedtiger4 : 188 : 0 Fredrik said: Yes, and that's why you should use ##(a-b)^2## instead of ##(a+b)^2## to prove this result.You should remove the first two lines from the first proof. Do you see why? Because a and b could be equal, right? If I am correct then it should be: Proof: If a-b>=0 then (a-b)^2 >= 0 a^2 + b^2 - 2ab >= 0 implies a^2 + b^2 >= 2ab Q.E.D. But if a and b could be equal to each other then ">=" justifies that so the proof in my original post remains correct, isn't? 4 HallsofIvy Science Advisor Homework Helper : 42,895 : 984 woundedtiger4 said: If a and b are real numbers, then a^2 + b^2 >= 2 a bProof:if a>=0 and b>=0 then a-b>=0 This is not true. Given that a and b are arbitrary real numbers, a- b can be any number, positive, negative, or 0. However, since a square of a real number is never negative, the next step is correct: (a-b)^2>=0a^2 + b^2 -2ab >=0a^2 + b^2 >= 2abQ.E.D.at someone has proved more or less similar question, but at (a + b)^2 > 0 If a and b are not allowed to be equal, yes. otherwise (a+ b)^2>= 0 as you had before. a^2 + 2ab + b^2 > 0a^2 + b^2 ≥ 2abwhen 2ab goes from L.H.S. to R.H.S then it should be -2ab , am I correct? Yes, but since a and b can be any real numbers, either 2ab< 0, in which case, since a^2+ b^2 is not negative, it is a trivial statement, or 2ab> 0 so that the sign doesn't matter. 5 woundedtiger4 : 188 : 0 HallsofIvy said: This is not true. Ahaan i see your point. Proof: If a-b >=0 then (a-b)^2 >=0 a^2 +b^2 -2ab >=0 a^2 + b^2 >=2ab Q.E.D. What do you say? 6 Office_Shredder Staff Emeritus Science Advisor Gold Member : 5,702 : 1,587 woundedtiger4 said: Ahaan i see your point.Proof:If a-b >=0then (a-b)^2 >=0a^2 +b^2 -2ab >=0a^2 + b^2 >=2abQ.E.D.What do you say? You seem to be stuck on this idea of a-b ≥ 0 that is completely unnecessary. Compare your proof to the following: If the sky is blue then (a-b)^2 >=0 a^2 +b^2 -2ab >=0 a^2 + b^2 >=2ab 7 woundedtiger4 : 188 : 0 Office_Shredder said: You seem to be stuck on this idea of a-b ≥ 0 that is completely unnecessary. Compare your proof to the following:If the sky is bluethen (a-b)^2 >=0a^2 +b^2 -2ab >=0a^2 + b^2 >=2ab Ohh got it. It says "a and b are real numbers". Proof: If (a-b)^2 >=0 then a^2 + b^2 >=0 Q.E.D. Right? 8 Fredrik Staff Emeritus Science Advisor Homework Helper Insights Author Gold Member : 10,876 : 423 woundedtiger4 said: Because a and b could be equal, right? No, the problem is that if the proof relies on the assumption that ##a,b\geq 0##, then the result will only be valid for such values of ##a## and ##b## Don't forget that the statement you want to prove is that "For all real numbers ##a## and ##b##, we have ##a^2+b^2\geq 2ab##. It's for all real numbers, not just for the non-negative ones. woundedtiger4 said: Ohh got it. It says "a and b are real numbers".Proof:If (a-b)^2 >=0then a^2 + b^2 >=0Q.E.D.Right? What makes you think that ##(a-b)^2## may be less than 0, or that it needs to be greater than or equal to zero to ensure that ##a^2+b^2\geq 0##? 9 woundedtiger4 : 188 : 0 Fredrik said: No, the problem is that if the proof relies on the assumption that ##a,b\geq 0##, then the result will only be valid for such values of ##a## and ##b## Proof: If a,b>=0 then (a-b)^2 >=0 a^2 +b^2 -2ab >= 0 a^2 + b^2 >= 2ab QED I don't understand that how come (a+b)^2 >=0 can lead us to prove that a^2 + b^2 >= 2ab because (a+b)^2 >= 0 will lead us to the resut a^2 +b^2 >= -2ab , please explain to me in a simple language becaueee I am a beginner. PS. I want to prove this thm by using direct proof's if & then technique. Thanks in advance. 10 micromass Staff Emeritus Science Advisor Homework Helper Insights Author : 22,169 : 3,327 Can you give me some real numbers such that ##(a-b)^2<0##? 11 Fredrik Staff Emeritus Science Advisor Homework Helper Insights Author Gold Member : 10,876 : 423 Fredrik said: No, the problem is that if the proof relies on the assumption that ##a,b\geq 0##, then the result will only be valid for such values of ##a## and ##b## I didn't express myself very clearly here. I meant that if you assume that ##a,b\geq 0## and then find that ##a^2+b^2\geq 2ab##, you will only have proved that ##a^2+b^2\geq 2ab## for all non-negative real numbers ##a## and ##b##. But your goal was to prove that ##a^2+b^2\geq 2ab## for all real numbers ##a## and ##b##. 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12191	https://www.khanacademy.org/science/in-in-class11th-physics/in-in-phy-gravitation/in-in-gravity-newtonian/v/introduction-to-newton-s-law-of-gravitation	Introduction to Newton's law of gravitation (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content NCERT Physics Class 11 Course: NCERT Physics Class 11>Unit 8 Lesson 1: Newton's law of gravitation Introduction to gravity Mass and weight clarification Gravity for astronauts in orbit Would a brick or feather fall faster? Introduction to Newton's law of gravitation Gravitation (part 2) Acceleration due to gravity at the space station Space station speed in orbit Gravitational field strength Impact of mass on orbital speed Gravity and orbits Newton's law of gravitation review The concept of gravitational field Science> NCERT Physics Class 11> Gravitation> Newton's law of gravitation © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Introduction to Newton's law of gravitation Google Classroom Microsoft Teams About About this video Transcript Gravity is a force of mutual attraction between two objects that both have mass or energy. Newton's universal law of gravitation can be used to approximate the strength of gravitation forces between two objects as a function of the objects' masses and the distance between them.Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Adam Staples 14 years ago Posted 14 years ago. Direct link to Adam Staples's post “Why would G not be consta...” more Why would G not be constant everywhere in the universe? Would it be because the density of matter is different in different parts of space? Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Dan Surerus 14 years ago Posted 14 years ago. Direct link to Dan Surerus's post “G is the universal consta...” more G is the universal constant for the gravitational force. It never changes. The units for G are m^3/(kgs^2) g is the local acceleration due to gravity between 2 objects. The unit for g is m/s^2 an acceleration. The 9.8 m/s^2 is the acceleration of an object due to gravity at sea level on earth. You get this value from the Law of Universal Gravitation. Force = ma = G(Mm)/r^2 Here you use the radius of the earth for r, the distance to sea level from the center of the earth, and M is the mass of the earth. Notice that little m cancels out on both sides of the equation. ma=G(Mm)/r^2 a=GM/r^2 If you put in the mass of the earth and the radius to sea level you will get 9.8 m/s^2 for a. This is what we call little g. Notice if you change your radius that the acceleration(g) will fall off as 1/r^2. If you are twice as far out 2r, you will have 4 times less gravitational acceleration. That is why g can change from place to place on earth. If you are on Mt. Everest your radius will be larger than if you were in Death Valley.. for example. If you want to know how G came about, read on the "Cavendish experiment" Hope that clears it up. Comment Button navigates to signup page (9 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... stutidash 8 years ago Posted 8 years ago. Direct link to stutidash's post “What is the difference be...” more What is the difference between g and G.?? Except the difference in values Answer Button navigates to signup page •1 comment Comment on stutidash's post “What is the difference be...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer musicnarayanan488 5 years ago Posted 5 years ago. Direct link to musicnarayanan488's post “g = 9.8 m/s^2 (accelerati...” more g = 9.8 m/s^2 (acceleration on Earth) whereas G = 6.67430(15)×10^-11 m^3⋅kg^–1⋅s^–2. Hope this helps! Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Mishri 2 years ago Posted 2 years ago. Direct link to Mishri's post “If small earth has 1/2 ma...” more If small earth has 1/2 mass of Earth, then what would be its density? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer sebastian.af.2003 2 years ago Posted 2 years ago. Direct link to sebastian.af.2003's post “Formula for density is M/...” more Formula for density is M/V. If small earth has 1/2 the volume of earth and 1/2 the mass of earth, its density would equal that of earth. Even though volume and mass appear to decrease, its density ratio stays the same Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Rhutuvaruni Kharade 13 years ago Posted 13 years ago. Direct link to Rhutuvaruni Kharade's post “what is the mass of earth...” more what is the mass of earth 5.97.10 rest to 24 or 6.10 rest to 24 Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Rhutuvaruni Kharade 13 years ago Posted 13 years ago. Direct link to Rhutuvaruni Kharade's post “but cant we take round fi...” more but cant we take round figure of 6.10 rest to 10 as given in my science text book Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more nadia adam 8 years ago Posted 8 years ago. Direct link to nadia adam's post “hello, why we have to squ...” more hello, why we have to square r ? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Teacher Mackenzie (UK) 8 years ago Posted 8 years ago. Direct link to Teacher Mackenzie (UK)'s post “its an inverse square rel...” more its an inverse square relationship It means that the force gets shared out over a bigger area In the same way that light does from a point source) the further you go away from the source of the force. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... akki030210 2 years ago Posted 2 years ago. Direct link to akki030210's post “what is centripetal accel...” more what is centripetal acceleration Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Hecretary Bird 2 years ago Posted 2 years ago. Direct link to Hecretary Bird's post “Centripetal acceleration ...” more Centripetal acceleration is acceleration that makes an object follow a circular (or really just a curved) path by pointing to the center of the circle/curvature. If you whirl a yo-yo around, centripetal acceleration (caused by a centripetal force) is the thing that keeps the yo-yo moving in a circle. We can calculate centripetal acceleration by using the formula a_c = v^2 / r, where v here is the tangential velocity of the object. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Glorfindel 8 years ago Posted 8 years ago. Direct link to Glorfindel's post “How can the g represent b...” more How can the g represent both the acceleration due to graviation on Earth and the graviational field? What is the difference between the two gs? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Andrew M 8 years ago Posted 8 years ago. Direct link to Andrew M's post “there is no difference! I...” more there is no difference! It's two ways of describing the same thing. we can say g is 9.8 N/kg or 9.8 m/s^2. To see that those are the same, note that a Newton is a kgm/s^2 So if I tell you that the force on 1 kg is 9.8 N, you also know that the acceleration due to gravity will be 9.8 m/s^2. 5 comments Comment on Andrew M's post “there is no difference! I...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more MatthewGrant55 8 years ago Posted 8 years ago. Direct link to MatthewGrant55's post “I know gravity comes from...” more I know gravity comes from mass, but why? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Mark Zwald 8 years ago Posted 8 years ago. Direct link to Mark Zwald's post “The current understanding...” more The current understanding of gravity is mass (and energy) warp spacetime and warped spacetime creates the gravitational force. This is from Einstein's theory of general relativity which is based on a finite speed of light. Newton's theory of gravity incorrectly assumes an infinite speed of light but works well for certain situations of low speed, but not well at all for others. 1 comment Comment on Mark Zwald's post “The current understanding...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more matesunsharon 10 years ago Posted 10 years ago. Direct link to matesunsharon's post “At 5:46, Isn't (10^6) ^2 ...” more At 5:46 , Isn't (10^6) ^2 (ten to the sixth power squared) supposed to be 36? Answer Button navigates to signup page •Comment Button navigates to signup page (0 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Matt B 10 years ago Posted 10 years ago. Direct link to Matt B's post “Baluur is completely righ...” more Baluur is completely right, and I gave him an upvote too. Just a small addition from my side: Count zeros: (10^6)^2 = 1,000,000 1,000,000 = 1,000,000,000,000 = 10^12 Matesunsharon, you were probably confusing this with 10^(6^2) , which is equal to 10^36 and is not the same as (10^6)^2 which is equal to 10^12 Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more tinaxing2020 5 years ago Posted 5 years ago. Direct link to tinaxing2020's post “When you are calculating ...” more When you are calculating the F, or force of gravity, you are supposed to use the formula and use G as 6.67 10^-11? Is that correct? :) Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer obiwan kenobi 5 years ago Posted 5 years ago. Direct link to obiwan kenobi's post “Yes, that is correct. You...” more Yes, that is correct. You simply substitute the information you have into the formula and then use the Gravitational Constant to calculate the force. Hope this helps! Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript We're now going to learn a little bit about gravity. And just so you know, gravity is something that, especially in introductory physics or even reasonably advanced physics, we can learn how to calculate it, we can learn how to realize what are the important variables in it, but it's something that's really not well understood. Even once you learn general relativity, if you do get there, I have to say, you can kind of say, oh, well, it's the warping of space time and all of this, but it's hard to get an intuition of why two objects, just because they have this thing called mass, they are attracted to each other. It's really, at least to me, a little bit mystical. But with that said, let's learn to deal with gravity. And we'll do that learning Newton's Law of Gravity, and this works for most purposes. So Newton's Law of Gravity says that the force between two masses, and that's the gravitational force, is equal to the gravitational constant G times the mass of the first object times the mass of the second object divided by the distance between the two objects squared. So that's simple enough. So let's play around with this, and see if we can get some results that look reasonably familiar to us. So let's use this formula to figure out what the acceleration, the gravitational acceleration, is at the surface of the Earth. So let's draw the Earth, just so we know what we're talking about. So that's my Earth. And let's say we want to figure out the gravitational acceleration on Sal. That's me. And so how do we apply this equation to figure out how much I'm accelerating down towards the center of Earth or the Earth's center of mass? The force is equal to-- so what's this big G thing? The G is the universal gravitational constant. Although, as far as I know, and I'm not an expert on this, I actually think its measurement can change. It's not truly, truly a constant, or I guess when on different scales, it can be a little bit different. But for our purposes, it is a constant, and the constant in most physics classes, is this: 6.67 times 10 to the negative 11th meters cubed per kilogram seconds squared. I know these units are crazy, but all you have to realize is these are just the units needed, that when you multiply it times a mass and a mass divided by a distance squared, you get Newtons, or kilogram meters per second squared. So we won't worry so much about the units right now. Just realize that you're going to have to work with meters in kilograms seconds. So let's just write that number down. I'll change colors to keep it interesting. 6.67 times 10 to the negative 11th, and we want to know the acceleration on Sal, so m1 is the mass of Sal. And I don't feel like revealing my mass in this video, so I'll just leave it as a variable. And then what's the mass 2? It's the mass of Earth. And I wrote that here. I looked it up on Wikipedia. This is the mass of Earth. So I multiply it times the mass of Earth, times 5.97 times 10 to the 24th kilograms-- weighs a little bit, not weighs, is a little bit more massive than Sal-- divided by the distance squared. Now, you might say, well, what's the distance between someone standing on the Earth and the Earth? Well, it's zero because they're touching the Earth. But it's important to realize that the distance between the two objects, especially when we're talking about the universal law of gravitation, is the distance between their center of masses. For all general purposes, my center of mass, maybe it's like three feet above the ground, because I'm not that tall. It's probably a little bit lower than that, actually. Anyway, my center of mass might be three feet above the ground, and where's Earth's center of mass? Well, it's at the center of Earth, so we have to know the radius of Earth, right? So the radius of Earth is-- I also looked it up on Wikipedia-- 6,371 kilometers. How many meters is that? It's 6 million meters, right? And then, you know, the extra meter to get to my center of mass, we can ignore for now, because it would be .001, so we'll ignore that for now. So it's 6-- and soon. I'll write it in scientific notation since everything else is in scientific notation-- 6.371 times 10 to the sixth meters, right? 6,000 kilometers is 6 million meters. So let's write that down. So the distance is going to be 6.37 times 10 to the sixth meters. We have to square that. Remember, it's distance squared. So let's see if we can simplify this a little bit. Let's just multiply those top numbers first. Force is equal to-- let's bring the variable out. Mass of Sal times-- let's do this top part. So we have 6.67 times 5.97 is equal to 39.82. And I just multiplied this times this, so now I have to multiply the 10's. So 10 to the negative 11th times 10 to the negative 24th. We can just add the exponents. They have the same base. So what's 24 minus 11? It's 10 to the 13th, right? And then what does the denominator look like? It's going to be the 6.37 squared times 10 to the sixth squared. So it's going to be-- whatever this is is going to be like 37 or something-- times-- what's 10 to the sixth squared? It's 10 to the 12th, right? 10 to the 12th. So let's figure out what 6.37 squared is. This little calculator I have doesn't have squared, so I have to-- so it's 40.58. And so simplifying it, the force is equal to the mass of Sal times-- let's divide, 39.82 divided by 40.58 is equal to 9.81. That's just this divided by this. And then 10 to the 13th divided by 10 to the 12th. Actually no, this isn't 9.81. Sorry, it's 0.981. 0.981, and then 10 to the 13th divided by 10 to the 12th is just 10, right? 10 to the first, times 10, so what's 0.981 times 10? Well, the force is equal to 9.81 times the mass of Sal. And where does this get us? How can we figure out the acceleration right now? Well, force is just mass times acceleration, right? So that's also going to just be equal to the acceleration of gravity-- that's supposed to be a small g there-- times the mass of Sal, right? So we know the gravitational force is 9.81 times the mass of Sal, and we also know that that's the same thing as the acceleration of gravity times the mass of Sal. We can divide both sides by the mass of Sal, and we have the acceleration of gravity. And if we had used the units the whole way, you would have seen that it is kilograms meters per second squared. And we have just shown that, at least based on the numbers that they've given in Wikipedia, the acceleration of gravity on the surface of the Earth is almost exactly what we've been using in all the projectile motion problems. It's 9.8 meters per second squared. That's exciting. So let's do another quick problem with gravity, because I've got two minutes. Let's say there's another planet called the planet Small Earth. And let's say the radius of Small Earth is equal to 1/2 the radius of Earth and the mass of Small Earth is equal to 1/2 the mass of Earth. So what's the pull of gravity on any object, say same object, on this? How much smaller would it be on this planet? Well, actually let me save that to the next video, because I hate being rushed. So I'll see you Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. 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12192	https://codehs.com/textbook/apcsa_textbook/3.1	Textbook: AP Computer Science A Textbook \| CodeHS Toggle navigation Products Products Explore what CodeHS has to offer for districts, schools, and teachers. Coding LMS Online IDE CodeHS Pro Computer Science Curriculum Certifications Professional Development AI Creator Use Cases Districts Schools Teachers Products Explore what CodeHS has to offer for districts, schools, and teachers Coding LMS Online IDE CodeHS Pro Computer Science Curriculum Certifications Professional Development AI Creator Use Cases Districts Schools Teachers Platform Assignments Create & configure your course assignments Classroom Manage & organize your class with customizable settings Grading Streamline your grading workflow. 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Primitive Types2. Using Objects3. Boolean Expressions and if Statements3.1 Boolean ExpressionsWhat are Booleans?Working with BooleansComparison OperatorsComparison Operators in a ProgramPitfallsOld Enough To VoteGrade RangeEquality of StringsCheck Your UnderstandingExercise: Number Order3.2 if Statements and Control Flow3.3 if-else Statements3.4 else if Statements3.5 Compound Boolean Expressions3.6 Equivalent Boolean Expressions3.7 Comparing Objects4. Iteration5. Writing Classes6. Arrays7. ArrayLists8. 2D Arrays9. Inheritance10. Recursion Chapter 3 Boolean Expressions and if Statements 3.1 Boolean Expressions How do you write code that tells us whether a user is logged in to our program? Booleans are the solution to these questions. What are Booleans? A boolean refers to a value that is true or false. Those are the only values of a boolean expression, and these are useful if you want to check if something is true or false. Let’s meet the fellow behind the name, “booleans,” George Boole. Mr. Boole was an English-born mathematician, philosopher, and logician. He introduced boolean algebra in The Mathematical Analysis of Logic (1847) and Investigation of the Laws of Thought (1854). Many consider him to be one of the founders of Computer Science. George Boole Working with Booleans Let’s start with an example – create a variable, set it equal to plain true or plain false , then print the value. You first want to declare the variable, plain loggedIn , and initialize its value. java boolean loggedIn = false; Java You can similarly set the variable to true instead: java boolean loggedIn = true; Java You can imagine a boolean variable as a box that can hold only the values plain true or plain false . The box below shows the current value of plain loggedIn : Notice that you do not need to have quotations around plain true or plain false . The full example looks like: java boolean loggedIn = false; System.out.println("User logged in?: " + loggedIn); Java Comparison Operators Comparison operators allow you to compare two values against one another. A comparison returns a boolean result of either plain true or plain false . The table below lists each of the common comparison operators and their usages: \| Operator \| Usage \| --- \| \| plain > \| Greater Than \| \| plain < \| Less Than \| \| plain >= \| Greater Than Or Equal To \| \| plain <= \| Less Than Or Equal To \| \| plain == \| Equal To \| \| plain != \| Not Equal To \| A Basic Comparison In the following example, you compare two variables plain x and plain y . You store the result of this comparison in variable plain z . java int x = 10; int y = 8; boolean z = x > y; System.out.println(z); Java What will get printed to the screen? The above comparison, plain x > y is evaluating if 10 is greater than 8. Because 10 is indeed greater than 8, plain z is assigned a value of plain true . Thus, plain true will get printed to the screen. More Practice with Comparisons Let’s get a little more practice. Take a look at the following code segment below. Pay close attention to each comparison and the operator being used. ```java // Declare some integer variables to use for practice comparisons below. int a = 3; int b = 5; int c = 2; int d = 3; // You store the boolean results of each comparison into boolean variables t-z. boolean t = a > 0; // true boolean u = a == d; // true boolean v = d >= b; // false boolean w = b > c; // true boolean x = a != d; // false boolean y = d < = a; // true boolean z = 4 < = c; // false ``` Java In addition to integers, it is possible to compare other data types too. In the example below, the boolean variables are storing the results of a comparison between some of the boolean values computed above. java boolean boolComparison1 = t == u; // true boolean boolComparison2 = t == w; // true boolean boolComparison3 = t != u; // false boolean boolComparison4 = x != y; // true Java Comparison Operators in a Program Suppose you want to write a program that restricts people under a certain height from riding on a roller coaster. For this particular roller coaster, people who are under 4 feet (48 inches) are not allowed on the ride. How would you do this? java int heightInInches = readInt("How tall are you (in inches)? "); boolean isTallEnough = heightInInches >= 48; System.out.println("Can ride on the roller coaster: " + isTallEnough); Java After getting the potential rider’s height in inches, you do a comparison to ensure that they are over 48 inches. The result of this comparison is then printed out. Pitfalls A common mistake is using plain = when you actually want to use plain == . plain = is used for assignment of variables whereas plain == is used for comparing the equality of two values. For example, plain x = 5 stores the value 5 into the variable x. However, plain x == 5 tests to see if the value 5 is equal to the variable plain x and then returns either true or false. They are not the same thing! Old Enough To Vote Grade Range Equality of Strings Check Your Understanding Incorrect Correct No Answer was selected Invalid Answer Which of the following is NOT a relational operator? Correct Answer Incorrect Answer 2. Correct Answer Incorrect Answer Correct Answer Incorrect Answer != Correct Answer Incorrect Answer <= Check Question Exercise: Number Order Ask the user to enter two plain ints . Create three plain boolean variables, as described below: The first should be plain true if the first number is greater than the second number. The second should be plain true if the first number is equal to the second number. The third should be plain true if the first number is less than the second number. Then print the expressions and the result. Here is an example: plaintext Enter two numbers: 10 12 10 > 12: false 10 == 12: false 10 < 12: true Plain text Previous Section 3 Boolean Expressions and if StatementsNext Section 3.2 if Statements and Control Flow
12193	https://askfilo.com/user-question-answers-smart-solutions/the-number-of-two-digit-prime-numbers-whose-both-digits-are-3135313037393738	Question asked by Filo student The number of two digit prime numbers whose both digits are prime is Views: 5,234 students Updated on: Dec 2, 2024 Text SolutionText solutionverified iconVerified Concepts: Prime numbers, Two-digit numbers Explanation: To find the two-digit prime numbers whose both digits are prime, we first identify the prime digits. The prime digits are 2, 3, 5, and 7. Next, we form two-digit numbers using these digits and check which of them are prime. Step by Step Solution: Step 1 Identify the prime digits: 2, 3, 5, 7. Step 2 List all possible two-digit combinations using these digits: 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, 77. Step 3 Check which of these numbers are prime: 22 (not prime), 23 (prime), 25 (not prime), 27 (not prime), 32 (not prime), 33 (not prime), 35 (not prime), 37 (prime), 52 (not prime), 53 (prime), 55 (not prime), 57 (not prime), 72 (not prime), 73 (prime), 75 (not prime), 77 (not prime). Step 4 The prime numbers from the list are: 23, 37, 53, 73. Step 5 Count the prime numbers found: There are 4 two-digit prime numbers whose both digits are prime. Final Answer: 4 Students who ask this question also asked Views: 5,353 Topic: Smart Solutions View solution Views: 5,598 Topic: Smart Solutions View solution Views: 5,912 \| Heights (inches) \| 57 \| 59 \| 62 \| 63 \| 64 \| 65 \| 55 \| 58 \| 57 \| --- --- --- --- --- \| \| Weights (lbs) \| 113 \| 117 \| 126 \| 126 \| 130 \| 129 \| 111 \| 116 \| 112 \| \| \| A \| B \| C \| D \| E \| --- --- --- \| \| A \| 4 \| 4 \| 2 \| -4 \| -6 \| \| B \| 8 \| 6 \| 8 \| -4 \| 0 \| \| Activity \| A \| B \| C \| D \| E \| F \| G \| H \| I \| J \| K \| L \| M \| N \| O \| P \| Q \| R \| S \| T \| --- --- --- --- --- --- --- --- --- --- \| Prerequisite \| \| B \| A \| C \| B \| A,F \| G \| A,F \| G \| J,H \| J \| L \| M,K \| N \| O \| P \| Q,D,I,R \| E \| Q,D,I,R \| R \| Topic: Smart Solutions View solution Views: 5,160 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| The number of two digit prime numbers whose both digits are prime is \| \| Updated On \| Dec 2, 2024 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 11 \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
12194	http://mathcentral.uregina.ca/QQ/database/QQ.09.06/s/natasha1.html	Simplifying an expression in probability - Math Central ← BACKPRINT+ TEXT SIZE –SEARCHHOME Math Central Quandaries & Queries Question from Natasha, a student: well it is p(4) = 5! ------ X (0.5) ^4 X (1- 0.5)^5-4 = 0.15625 4!(5-4)! Now the ^ is to the power of (I believe) Its written in the book as (1-0.5)4 (but the 4 is little and on the top right corner of the first bracket now the second ^5-4 is also little on the top right corner of the 2nd bracket. This is an equation from my text book but for the life of me I cannot figure out how to solve it. If you can give me step by step procedure I'd like that. The answer is also there I just don't know how to get to the answer. Thank you. Hi Natasha. Let me rewrite that using some equation software: Now presuming I have read your question correctly, here are the steps I would use: Begin by simplifying what's in the parentheses. Note that there are implied parentheses on that exponent expression 5-4, so simplify that too. I prefer to use fractions rather than decimals, so I'll convert to fraction form. Next, expand (or simplify) the exponents. (½)4 is four factors of what is in the parentheses. Remember that 5! (read as "five factorial") is just 5 times 4 times 3 all the way down to 2. So expand the factorials so you can see the multiplication they are made from. Then you can cancel out common factors on the tops and bottoms and simplify them. This gives the final fraction of 5/32, which is equal to 0.15625. Hope this helps, Stephen La Rocque. Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
12195	https://mathbitsnotebook.com/Geometry/TrigApps/TAarea.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| Area of Triangle Using Trigonometry MathBitsNotebook.com Topical Outline \| Geometry Outline \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| \| \| \| --- \| \| \| \| \| A Trigonometric Formula for the Area of a Triangle: \| \| \| The general formula for the area of a triangle is well known. While the formula shows the letters b and h, it is actually the "pattern" of the formula that is important. The area of a triangle equals ½ the length of one side times the height (altitude) drawn to that side (or an extension of that side). \| \| \| \| General Formula for Area of Triangle: b = length of a side (base) h = height draw to that side \| \| \| The area of ΔABC can be expressed as: where a represents the side (base) and h represents the height drawn to that side. \| Using trigonometry, let's take another look at this diagram. In the right triangle CDA, we can state that: The height, h, of the triangle can be expressed as b sin C. \| \| \| Substituting this new expression for the height, h, into the general formula for the area of a triangle gives: where a and b can be any two sides and C is the included angle. \| \| \| \| \| --- \| \| \| \| The area of a triangle can be expressed using the lengths of two sides and the sine of the included angle. AreaΔ = ½ ab sin C. \| \| You may see this referred to as the SAS formula for the area of a triangle. \| \| With this new formula, we no longer have to rely on finding the altitude (height) of a triangle in order to find its area. Now, if we know two sides and the included angle of a triangle, we can find the area of the triangle. This is a valuable new formula! And again, it is this new "pattern" of using two sides and the included angle that is important to remember. The area of a triangle equals ½ the product of two sides and the sine of the angle between these two sides. \| \| \| Given the triangle at the right, find its area. Express the answer to the nearest hundredth of a square unit. \| \| \| \| \| --- \| \| \| \| \| When using your graphing calculator, be sure you are in DEGREE mode, or using the degree symbol. \| \| \| Let a = ST, b = RT, and C = ∠RTS. \| \| \| \| \| --- \| Given the parallelogram shown at the right, find its area to the nearest square unit. The diagonal of a parallelogram divides it into two congruent triangles. So the total area of the parallelogram will be TWICE the area of one of the triangles formed by the diagonal. \| \| \| This example shows that by doubling the triangle area formula, we have created a formula for finding the area of a parallelogram, given 2 adjacent sides (a and b) and the included angle, C. Area of Parallelogram \| \| Let a = PS, b - RS, and C =∠PSR. \| \| \| \| \| --- \| Given the parallelogram shown at the right, find its EXACT area. \| \| \| If a question asks for an EXACT answer, do not use your calculator to find the sin 60º since it will be a rounded value. To get an EXACT value for sin 60º, use the 30º-60º-90º special triangle which gives the sin 60º to be . \| Notice that we are using the formula for the area of a parallelogram we discovered in Example 2. \| Let a = AD, b = AB, and C = ∠BAD. \| \| \| \| --- \| \| \| \| \| For help with this formula on your calculator, click here. \| \| Derive (or Justify) this formula: NOTE: The NY Next Generation Standards ask for students to "justify" this formula for finding the area of any triangle. This is accomplished by drawing an auxiliary line from a vertex perpendicular to the opposite side. While the first statement may be interpreted as applying only to acute triangles, the wording of "any triangle" may imply including obtuse triangles. This site will examine both "acute" and "obtuse" triangles in relation to this formula. Case 1: Acute Triangle At the top of this page, we saw how the new formula was developed by expressing the height, h, as bsinC. Remember, AD = h. \| \| \| If this formula truly works (and it does!), we should be able to apply the formula using any angle in the triangle. So, when attempting to "justify" this formula, we should show that it can be "developed" using any (and every) angle in the triangle. • Using acute ∠C, we have sin C = h/b, which gives h = bsin C, giving the formula \| \| • Now, using acute ∠B, we have sin B = h/c, which gives h = csin B, giving another version of the the formula \| \| \| • Finally, using acute ∠A, with a new height value, h = CD, we have sin A = h/b, which gives h = cbsin A, or h = bcsinA. ,, giving the third version of the the formula \| \| \| \| Case 2: Obtuse Triangle Can we still develop this formula if ∠A is an obtuse angle? The answer is "yes", but it will require more work and some more trigonometric information. We will take a brief look at what is involved when ∠A is an obtuse angle, but these concepts will be more fully developed in upcoming courses. Note: to maintain the use of a single letter to represent the angle in our formula, we will be referring to ∠BAC in the diagram below, as simply∠A. No, this is not normal, but it will make reading the problem easier. \| \| \| We will be justifying the formula: When ∠A is an obtuse angle, the altitude drawn from C or B will be outside of the triangle. Draw the altitude from C to the line containing the opposite side. ΔCAE is a right triangle, but unfortunately it does not contain ∠A that we need for our formula. \| \| We know, however, that ∠CAE is supplementary to ∠A, since they form a linear pair. We can state that m∠CAE = 180 - m∠A and from ΔCAE that . If we apply a trigonometric fact that sin∠A = sin(180 - m∠A), we can substitute and get: (After multiplying both sides of the first equation by b.) Now, substitution into the general formula for the area of a triangle will give us our desired formula: . \| \| \| \| \| \| --- --- \| \| \| \| \| --- \| \| sin∠A = sin (180 - m∠A) Remember that the functions of sine, cosine, and tangent are defined only for acute angles in a right triangle. So, how do we find the sine of an obtuse angle? We cannot use the sides of the triangle to find sin∠BAC because the angle does not reside in a right triangle. We can, however, find sin∠BAD which deals with an acute angle in a right triangle. ∠BAD is the supplement of ∠BAC since they form a linear pair. \| \| \| The sine of an obtuse angle is defined to be the sine of the supplement of the angle. Thus, sin∠A = sin (180 - m∠A). On your graphing calculator, sin(50º) = 0.7660444431 and sin(130º) = 0.7660444431show this fact to be true. These angles are supplementary since 50º + 130º = 180º. \| \| \| \| \| --- \| \| WHY does sin∠A = sin (180 - m∠A)? This topic will be explored in more detail in upcoming courses. To understand "why" this relationship is true, we need a coordinate grid. Right triangle DEF is drawn in quadrant I, as shown. If we draw an angle of 130º, and drop a perpendicular to the x-axis from point H where DH = DF, we will create a reflection of ΔDEF over the y-axis. This reflected triangle (ΔDGH) is congruent to ΔDEF and both triangles have the same lengths for their sides opposite the 50º. It should be noted that both opposite sides deal with positive y-values (designating direction above the x-axis). \| 50º and 130º are supplementary. \| \| When dealing with obtuse angles (such as 130º), the corresponding acute angle (50º) is used to determine the sine, cosine or tangent of that obtuse angle. This corresponding acute angle is called a "reference angle". \| \| \| \| NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". \| Topical Outline \| Geometry Outline \| MathBitsNotebook.com \| MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts \| \|
12196	http://www.damtp.cam.ac.uk/user/po242/pdfs/Mech2015/introduction.pdf	Mathematical Tripos Part IA Michaelmas term 2015 Mechanics (non-examinable) Dr P.J.O’Donnell Mechanics (non-examinable) About this course The purpose of this course is to cover — rapidly and lightly — material from A-level (or the equivalent) which is needed for other courses in the Tripos, in particular for the Lent term Dynamics and Relativity course. The lectures will be fairly informal and fairly short. If you have taken M1, M2 and M3 (or the equivalent), you need not attend the course: the material should all be familiar to you, but you are welcome to attend if you would like to refresh your memory. If you have taken M1 and M2 but not M3, nearly all the material will be familiar. However, depending on your examination board, you may not have covered simple harmonic motion and motion in a circle. You might like to attend the lectures that cover those important topics. Organisation of the course The course consists of 12 lectures, including a short introductory lecture in the ﬁrst week. In each lecture, I will cover a single topic according to the lecture plan overleaf. If the topic looks unfamiliar, you should attend the lecture. If you are not sure, you should look at the relevant example sheet or lecture notes on my web site:1 www.damtp.cam.ac.uk/user/po242/mechanics.html I will write notes on the blackboard, which will be a subset of the printed notes, and you can either take your own notes in the lecture or rely on the printed notes. Examples A short, relatively straightforward, examples sheet will be provided for each topic covered. They are designed to test fundamental understanding but will not require you to do a great deal of manipulation or perform STEP-like acrobatics. If you have already covered the topic at school, you should have no diﬃculty with the examples; if you do ﬁnd diﬃculty with them, you should attend the relevant lecture. Resources Full lecture notes and examples sheets with solutions will appear on my web site (see above) in advance of the relevant lecture. Appropriate books • P. J. O’Donnell Essential Dynamics and Relativity. CRC Press, 2014 (£39.99 paperback) • J. Hebborn and J. Littlewood Mechanics 1, 2 & 3 (Edexel). Heinemann, 2000 (£12.99 paperback) • L. Bostock and F. S. Chandler Applied Mathematics, Vols. 1 & Vol. 2. ST(P) Ltd., 1986 (out of print) 1Dr. Siklos created these notes and I will not deviate from them in any substantial manner. His notes can be found at: www.damtp.cam.ac.uk/user/stcs/mechanics.html 1 Lecture plan L0 (Thursday 8th October) Short introductory lecture No maths — just an outline of the lectures. L1 (Tuesday 13th October) Equilibrium of a single particle The vector nature of forces, addition of forces, examples including gravity, tension in a string, normal reaction, friction. Conditions for equilibrium. L2 (Thursday 15th October) Equilibrium of a rigid body Resultant of several forces, resolving forces, couple, moment of a force. Conditions for equilibrium. L3 (Tuesday 20th October) Centre of mass Deﬁnition and examples of centre of mass of a discrete set of particles in one and three dimensions and of a continuous mass distribution in one dimension. L4 (Thursday 22nd October) Kinematics of a single particle Position, velocity, speed, acceleration. Constant acceleration in one dimension and in three dimen-sions. L5 (Tuesday 27th October) Kinematics of a single particle (continued) Projectile motion in two-dimensions. L6 (Thursday 29th October) Newton’s Laws Newton’s laws of motion. Newton’s second law applied to a single particle. Examples of pulleys. L7 (Tuesday 3rd November) Energy Deﬁnition of energy and work. Kinetic energy, potential energy of a particle in a uniform gravita-tional ﬁeld. Conservation of energy. L8 (Thursday 5th November) Momentum and Impulse Deﬁnition of momentum (as a vector), conservation of momentum, collisions, coeﬃcient of restitu-tion. Impulse; example of oblique impact. L9 (Tuesday 10th November) Springs, strings and SHM Deﬁnition of simple harmonic motion. Elastic springs and strings (Hooke’s law). Oscillations of a particle attached to a spring, and of a particle hanging on a string. L10 (Thursday 12th November) Motion in a circle Derivation of the central acceleration of a particle constrained to move on a circle. Simple pendu-lum. L11 (Tuesday 17th November) Motion in a circle (continued) Examples including motion of a particle sliding on a cylinder. 2
12197	https://blog.csdn.net/2301_80173477/article/details/134685066	分数数列与有规律的数列求和-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作分数数列与有规律的数列求和最新推荐文章于 2025-09-27 22:34:35 发布原创于 2023-11-29 11:02:18 发布·210 阅读 · 0 · 0· CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。文章标签： #算法题目描述已知分数数列{ 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, ...... }，求该数列的第n项。输入一个正整数n（n<=40) 输出分数数列的第n项。样例输入 Copy 3 样例输出 Copy 5/3 ```cs includeint main() { int a = 2, b = 1; int n, i, t; scanf("%d", &n); for (i = 1; i < n; i++) { t = a; a = a + b; b = t; } printf("%d/%d",a,b); return 0;} ``` AI写代码 cs 运行题目描述有一分数序列： 2/1 3/2 5/3 8/5 13/8 21/13...... 求出这个数列的前N项之和，保留两位小数。输入格式 N 输出格式数列前N项和样例输入复制 10 样例输出复制 16.48 ```cs includeint main(){ double Sn=0,an=0,n,i,t,a=2.0,b=1.0;//a为分子，b为分母，an为单项和值，Sn为总和 scanf("%lf",&n); Sn+=a/b;//第一项不在规律中，在循环前先加入 for(i=1;i<n;i++){ t=a; a=a+b;//将前一项的分子加分母给后一项的分子 b=t;//将前一项的分子给后一项的分母 an=a/b; Sn+=an; } printf("%0.2lf",Sn); return 0;} ``` AI写代码 cs 运行确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 Xun苟觉关注关注 0点赞踩 0 收藏觉得还不错? 一键收藏 1评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报输入n，求出分数序列2/1,3/2,5/3,8/5,13/8，21/13，···的前n项之和，并输出。n是大于零的整数大肉包的博客 02-28 7542 思路：根据斐波那契数列，找到规律，简单的循环。 #include<iostream> using namespace std; int main() { int f1=1,f2=2,t,n; cout<<"请输入n:"; cin>>n; cout<<f2<<"/"<<f1<&a 1 条评论您还未登录，请先登录后发表或查看评论有一个分数序列：2/1, 3/2, 5/3, 8/5, 13/8, …编写程序求出这个序列的前n项之和。热门推荐新生啸月的博客 10-11 6万+ 问题与代码：有一个分数序列：2/1, 3/2, 5/3, 8/5, 13/8, …编写程序求出这个序列的前n项之和文件名称：有一个分数序列：2/1, 3/2, 5/3, 8/5, 13/8, …编写程序求出这个序列的前n项之和作者：何知令发表时间：2016年10月11日输入：输入一个整数n，1 输出：输出该数列的前n项和代码如下： #include int main() java 分数数列求和 [编程入门]_有规律的数列求和-题解（Java代码） weixin_35516273的博客 03-07 1047 解题思路:注意事项:进行/运算时，定义的数要是浮点数double或float，不然小数部分会直接被约去参考代码:import java.util.Scanner;//有一分数序列： 2/1 3/2 5/3 8/5 13/8 21/13...... 求出这个数列的前N项之和，保留两位小数public class Main {public static void main(String[] args)... C语言有规律的数列求和 m0_72116652的博客 04-15 2748 有一分数序列： 2/1 3/2 5/3 8/5 13/8 21/13… 求出这个数列的前N项之和，保留两位小数。程序数列求和 c语言,[编程入门]有规律的数列求和-题解（C语言代码） weixin_32165903的博客 05-18 3054 第一种：我们发现其实分子和分母都是斐波那契数列(从第三项开始，每一项等于前两项之和)，所以我们可以定义两个数组，分别来存放分子和分母。然后两个数组里的数按位置相除累加。参考代码：#includeint main (){int N;double total = 0;scanf("%d",&N);int a[N],b[N];a = 2,a = 3; //这里需要手动输入0 1位置的... 有规律数列求和 wasser_7的博客 09-17 240 题目描述：有一分数序列： 2/1 3/2 5/3 8/5 13/8 21/13...... 求出这个数列的前N项之和，保留两位小数。 int main() { double sum; int a, b, n,tmp; scanf("%d", &n); a = 2.0; b = 1.0; sum = 0; sum += a / b; for (int j = 2; j <= n; j++) { tmp = a; a = a + b; b = tmp; sum c语言实现分数数列求和 ret_skd的专栏 03-19 3万+ 例：有一个分数序列 2/1+3/2+5/3+8/5+13/8+… 求出这个数列前20项的和。先找规律，后一个数的分子是前一个数的分子与分母的和。由于是分数数列，所以用浮点型定义变量。下边是代码实现： #include int main() { float a = 1.0; float b = 2.0; float sum = 0.0; int i = 1; flo 有规律的数列求和（ACM练习）加油加油！！！ 08-02 312 ACM练习——递归有规律的数列求和 java 2301_80459560的博客 03-28 346 有一分数序列： 2/1 3/2 5/3 8/5 13/8 21/13……求出这个数列的前N项之和，保留两位小数。 2019届高考数学一轮复习第五章数列课堂达标29 数列求和文新人教版201807234162 09-10 数列求和是高中数学中的一个重要知识点，尤其在高考复习阶段，它经常作为考点出现。数列求和涉及到等差数列、等比数列以及更复杂数列的...在高考复习中，掌握这些方法并多做练习，有助于提高解决数列求和问题的能力。 Java50道经典编程题：（二十）分数数列求和 ——寻找规律解决问题 Oceanst的博客 01-28 3438 1.问题重述题目：有一分数序列：2/1，3/2，5/3，8/5，13/8，21/13…求出这个数列的前20项之和。 2.解析观察后发现，前面一项分子和分母相加的和为下一项的分子，前面一项分子，为下一项的分母；再观察发现，这些数字1、1、2、3、5、8、13、21……正是斐波那锲数列，参照我的第一条博客： c 算法-分数数列求和手中无剑，心中有剑 09-22 3169 题目及代码：#include int feibo(int n) { if (n==1) { return 1; } if (n==2) { return 2; } return feibo(n-1)+feibo(n-2); } int main(int argc, const char argv[]) { Java入门级算法-----有规律的数列求和又菜又爱写 10-09 876 本文将为大家带来“有规律的数列求和”算法的一个解法与详细解释，方法不唯一，本篇文章只列举出一种方法，可以先练习，练习之后查看代码，这样更加有助于提高。文章目录一、题目二、解题思路三、代码(含详细注解)四、代码解读五、致读者一、题目题目描述：有一分数序列： 2/1 3/2 5/3 8/5 13/8 21/13...... 求出这个数列的前N项之和，保留两位小数。输入 N 输出数列前N项和运行结果：样例输入：10 样例输出：16.48 以上是本题的题目，需要求 LeetCode 算法“无重复字符的最长子串”哈希表+滑动窗口+贪心最新发布 W2799401288的博客 09-27 905 本文介绍了一种使用滑动窗口算法结合哈希表高效求解"最长无重复字符子串"问题的方法。通过维护左右指针动态调整窗口范围，利用哈希表快速判断字符重复性，确保时间复杂度为O(n)。详细步骤包括算法原理、示例演示、代码实现及边界处理，适用于各类字符串场景。该方案空间复杂度为O(1)，能正确处理空串、全重复字符等特殊情况。 376.摆动序列 codezhu的博客 09-27 452 f[i] 表示：以 i 位置元素为结尾的所有的子序列中，最后一个位置呈现 “上升” 趋势的最长摆动序列的长度。g[i] 表示：以 i 位置元素为结尾的所有的子序列中，最后一个位置呈现 “下降” 趋势的最长摆动序列的长度。根据初始条件和状态转移方程，确定填表顺序，进而逐步填满dp表，最终返回题目要的结果。dp[i] 表示：以 i 位置元素为结尾的所有的子序列中，最长的摆动序列的长度。 LeetCode:67.寻找旋转排序数组中的最小值未知陨落的博客 09-27 156 旋转排序数组符合上面的图，我们用x来表示数组的最后一个元素，如果大于x的话，说明在上半部分的有序区间，如果小于x的话，说明落在了下半部分的有序区间。通过用一个比较值来判断大小。 2024年CSP-X复赛真题题解（T2：消灭怪兽）王老师青少年编程 09-27 554 2024年CSP-X复赛真题题解（T2：消灭怪兽） Python实现手榴弹爆炸算法（Grenade Explosion Method, GEM）（附完整代码） corn1949的博客 09-23 1435 Python实现手榴弹爆炸算法（Grenade Explosion Method, GEM）（附完整代码）关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司 Xun苟觉博客等级码龄2年 43 原创153 点赞 165 收藏 172 粉丝关注私信 TA的精选新爬虫入门学习 1886 阅读新二叉树层次遍历 814 阅读热 C语言 round函数 12211 阅读热删除字符串中的空格 2532 阅读热字符串压缩（C语言） 2108 阅读查看更多 2025年 1篇 2024年 4篇 2023年 38篇大家在看测试用例设计万能公式揭秘从 “小白懵” 到 “玩得溜”：6 个 Linux 基础指令的硬核拆解，原理 + 实操一次吃透！ PyTorch新手入门指南 606 AI赋能的智能文本编辑器 23 单细胞与空间多组学技术进展 370 上一篇：等差数列（两种解法）下一篇：解方程并编程计算目录已知分数数列{ 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, ...... }，求该数列的第n项。题目描述输入格式输出格式样例输入样例输出展开全部收起目录已知分数数列{ 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, ...... }，求该数列的第n项。题目描述输入格式输出格式样例输入样例输出展开全部收起上一篇：等差数列（两种解法）下一篇：解方程并编程计算登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 ×立即登录评论 1 被折叠的条评论为什么被折叠?到【灌水乐园】发言查看更多评论添加红包祝福语请填写红包祝福语或标题红包数量个红包个数最小为10个红包总金额元红包金额最低5元余额支付当前余额 3.43 元前往充值 > 需支付：10.00 元取消确定成就一亿技术人! 领取后你会自动成为博主和红包主的粉丝规则 hope_wisdom 发出的红包实付元使用余额支付点击重新获取扫码支付钱包余额 0 抵扣说明： 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报选择你想要举报的内容（必选）内容涉黄政治相关内容抄袭涉嫌广告内容侵权侮辱谩骂样式问题其他原文链接（必填）请选择具体原因（必选）包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式补充说明（选填）取消确定下载APP 程序员都在用的中文IT技术交流社区公众号专业的中文 IT 技术社区，与千万技术人共成长视频号关注【CSDN】视频号，行业资讯、技术分享精彩不断，直播好礼送不停！客服返回顶部
12198	https://www.quora.com/How-can-I-prove-that-the-partial-sum-of-the-harmonic-series-is-never-an-integer	Something went wrong. Wait a moment and try again. Partial Sums Divergent Series (mathema... Proofs (mathematics) Harmonic Numbers Number Series Mathematics Number Theory Sums and Series (mathemat... 5 How can I prove that the partial sum of the harmonic series is never an integer? Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by Melchior Grutzmann , Ph.D. Mathematics, Pennsylvania State University (2009) and Alexey Godin , Ph.D. Mathematics & Economics, Moscow State University (1998) · Author has 8.4K answers and 20.7M answer views · 5y We want to show that for any integer n≥2, Hn=∑nk=11k is not an integer. Proof: Let L=lcm(1,2,…,n) so that we can write Hn as a fraction with denominator L. For each k=1,…,n, write L=kak for some positive integer ak; this means that 1k=akL. Hence, we have Hn=n∑k=11k=1Ln∑k=1ak. Since n≥2, L is an even number. In order to conclude this proof, we need to show that ∑nk=1ak is odd so that Hn is not an integer. Let 2r be the largest power of 2 less than or equal We want to show that for any integer n≥2, Hn=∑nk=11k is not an integer. Proof: Let L=lcm(1,2,…,n) so that we can write Hn as a fraction with denominator L. For each k=1,…,n, write L=kak for some positive integer ak; this means that 1k=akL. Hence, we have Hn=n∑k=11k=1Ln∑k=1ak. Since n≥2, L is an even number. In order to conclude this proof, we need to show that ∑nk=1ak is odd so that Hn is not an integer. Let 2r be the largest power of 2 less than or equal to n. Then, the only positive integer less than or equal to n that is divisible by 2r is 2r itself. Then, we can write L=2r⋅b for some odd integer b and thus 2r⋅b=kak for each k=1,2,…,n. When k=2r, we see that ak=b is odd. Otherwise, k is not divisible by 2r which implies that ak is even. Therefore, ∑nk=1ak is an odd integer, as desired. Amitabha Tripathi have more than a working knowledge of Z · Author has 4.7K answers and 13.8M answer views · 7y Related How can I prove 1 1 + 1 3 + 1 5 + ⋯ + 1 2 n − 1 is not an integer? I’ll assume n>1, for ∑nk=112k−1∈Z for . Fix , . Let . Note that no multiple of except belongs to since when . Thus, , where . Hence Since is a multiple of 3 when , the I’ll assume , for for . Fix , . Let . Note that no multiple of except belongs to since when . Thus, , where . Hence Since is a multiple of 3 when , the RHS of is not a multiple of . But the LHS of is a multiple of if is an integer. This contradiction proves that cannot be an integer when . Kenneth Wang Harvard College '17, Graduate Student in Physics at Harvard · 10y Related Why does the harmonic series diverge? There are a couple of ways to see this. The first, rather dry, method is to use the integral test. This allows us to replace this discrete series with an integral (a continuous sum) that is either larger or smaller than our discrete series. In this case, if we want to prove divergence, we are considering an image like this: In this case, the integral will be smaller than our desired sum (pictured in green). Our function will be . Integrating this, we obtain: Therefore the integral diverges, and our series There are a couple of ways to see this. The first, rather dry, method is to use the integral test. This allows us to replace this discrete series with an integral (a continuous sum) that is either larger or smaller than our discrete series. In this case, if we want to prove divergence, we are considering an image like this: In this case, the integral will be smaller than our desired sum (pictured in green). Our function will be . Integrating this, we obtain: Therefore the integral diverges, and our series must diverge. This is a consequence of the fact that our function integrates to the natural log which is not bounded. If we were considering a p-series (as it is called) with , it would integrate to something with an exponent in the denominator greater than zero, which would be bounded. Now, for a more elegant, and classic proof. Consider the following series that is smaller than the harmonic series: The smaller series diverges, because you are just adding one half over and over again, so the harmonic series must diverge, since it is larger than this divergent series! Related questions What is the summation of the harmonic series? What is an intuitive way to see why harmonic series sums to logn? What are some interesting applications of the harmonic series? How can we prove convergence of harmonic series? The almost harmonic series is defined as the harmonic series with all numbers including 9 are omitted, i.e. . Does this series converge? Samim Ul Islam Looking for PhD offer in High Energy Physics · Author has 507 answers and 5.3M answer views · 8y Related Why does the harmonic series diverge? You have heard the name of the Harmonic in Physics. If there is a fair multiplication of the vibrations of the vibrating wire, The prosthesis is called Hormonic. However, the wavelength for Harmonic will be 1, 1/2, 1/3, 1/4 times the wavelength of original tone. Similarly, there is an infinite series called Harmonic in Mathematics. The series is like this: Two Harmonic's are seen to match. Harmonic series terms are exactly like multiplier of harmonic wavelength ! It is understood that the series is unlimited. The question is whether the convergent or divergent is the sequence ? Let, {Sn} = u1 + u 2 You have heard the name of the Harmonic in Physics. If there is a fair multiplication of the vibrations of the vibrating wire, The prosthesis is called Hormonic. However, the wavelength for Harmonic will be 1, 1/2, 1/3, 1/4 times the wavelength of original tone. Similarly, there is an infinite series called Harmonic in Mathematics. The series is like this: Two Harmonic's are seen to match. Harmonic series terms are exactly like multiplier of harmonic wavelength ! It is understood that the series is unlimited. The question is whether the convergent or divergent is the sequence ? Let, {Sn} = u1 + u 2 +u 3 + u 4 + . . . . . .+ u n; partial sum of an infinite series If n driven towards infinity, the value of Sn pushed towards a certain value then it is said that the infinite series convergent.If not, then the infinite series is called Divergent. Looks like the series will be convergent, is not it? Gradually N going towards infinite , u n (Nth terms) is going to be 0. Is not it? But from Comparison Test: 1+ 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + . . . . . . . . ∞ 1+ 1/2 + ( 1/4 + 1/4 ) + ( 1/8 + 1/8 + 1/8 + 1/8 ) + (1/16 +1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 ) + (1/32 +. . . . . . +1/32) + . . . . ∞ (Here 1/3 instead of 1/4, 1/5, 1/6, 1/8 instead of 1/7; Thus, 1/16, 1/32, is written) The inequality is true because we have put a small number on it rather than a large number. The second series is written as such 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + . . . . . . . ∞ The sum of which can not be given as a specific number. That means it's a divergent series. Now we know that any series larger than the divergent series must be divergent and convergent So, There must be a smaller series than the series which is convergent.Therefore, Harmonic series is divergent. Our idea was wrong! Integral Test: y = 1/x Rectangles are drawn in the graph.Then sum of all the rectangles will be 1+ 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + . . . . . . . . Now the area below y = 1 / x curve from x = 1 to x = ∞ is Sponsored by JetBrains Enjoy productive Java with IntelliJ IDEA. Discover instant and clever code completion, on-the-fly code analysis, and reliable refactoring tools. Assistant Bot · 1y To prove that the partial sum of the harmonic series is never an integer for , we can use a contradiction argument based on the properties of rational numbers and their denominators. Step-by-Step Proof Definition of the Harmonic Number: The -th harmonic number is defined as: Common Denominator: To express as a single fraction, we find a common denominator, which is the least common multiple (LCM) of the integers from 1 to , denoted as . Let . Then we can re To prove that the partial sum of the harmonic series is never an integer for , we can use a contradiction argument based on the properties of rational numbers and their denominators. Step-by-Step Proof Definition of the Harmonic Number: The -th harmonic number is defined as: Common Denominator: To express as a single fraction, we find a common denominator, which is the least common multiple (LCM) of the integers from 1 to , denoted as . Let . Then we can rewrite as: where is an integer for each . Numerator Analysis: The numerator is the sum of integers, which is itself an integer. Let’s denote this integer as : Properties of : The denominator is the LCM of all integers from 1 to . Since is at least 2, is at least 2, and it is an even number whenever . Parity Argument: To show that cannot be an integer, we can examine the parity (evenness or oddness) of and . For , is even. Now we need to analyze : If is even, then there are even integers among , making divisible by 2. The contribution to from odd integers will not change the overall parity since adding an even number of odd integers results in an even sum. If is odd, then there are even integers, again making even. The contribution from odd integers will still yield an odd sum, but the even denominator will ensure that cannot simplify to an integer. Conclusion: Since is even and can be either odd or even depending on the count of odd/even integers, cannot be an integer for . Thus, we conclude that: This completes the proof that the partial sum of the harmonic series is never an integer for . Amitabha Tripathi four decades of Number Theory and counting · Upvoted by Pramit Das , B.Stat and M.Stat Statistics & Mathematics, Indian Statistical Institute (2020) · Author has 4.7K answers and 13.8M answer views · 7y Related How do I prove that the sum 1+1/2+1/3+…+1/n is never an integer? Let , and let be the unique positive integer such that . Then , where is an odd positive integer. Write . Therefore , where Note that each is an integer, and that is odd if and only if . Since , if and only if . Hence only the term Let , and let be the unique positive integer such that . Then , where is an odd positive integer. Write . Therefore , where Note that each is an integer, and that is odd if and only if . Since , if and only if . Hence only the term in the numerator of the RHS of eqn. is odd. But then the numerator must be odd, and hence can’t be a multiple of the denominator (which is even). Thus the sum on the LHS of eqn. can’t be an integer. Related questions What is the harmonic series? What is the sum of the harmonic series for all positive integers? How do I prove that the sum 1+1/2+1/3+…+1/n is never an integer? Why do alternating harmonic series converge? In mathematics, when does the harmonic series converge? Mark Barton Researches suspensions for gravity wave detectors · Author has 18.2K answers and 24.4M answer views · Updated 1y Related Why is the harmonic series called so? If you're asking about the Harmonic series (mathematics) then it gets its name from a distant connection with the Harmonic series (music). If you've got a "harmonic series” of musical sounds (strictly a “sequence” in mathematical terminology), then by definition the frequencies will be in the proportion 1::2::3::4::… . Therefore the wavelengths as they propagate in air will be in the proportion 1/1::1/2::1/3::1/4::… . So if something possessed you to add up the wavelengths (creating a mathematical “series” from a “sequence”), you'd have the infinite sum that is called the harmonic series in mat If you're asking about the Harmonic series (mathematics) then it gets its name from a distant connection with the Harmonic series (music). If you've got a "harmonic series” of musical sounds (strictly a “sequence” in mathematical terminology), then by definition the frequencies will be in the proportion 1::2::3::4::… . Therefore the wavelengths as they propagate in air will be in the proportion 1/1::1/2::1/3::1/4::… . So if something possessed you to add up the wavelengths (creating a mathematical “series” from a “sequence”), you'd have the infinite sum that is called the harmonic series in mathematics. As a physicist, I can't think of a plausible physical motivation for doing that, but it's an interesting mathematical puzzle. A physical situation where the series does come up somewhat naturally is the so-called Leaning Tower of Lire. In turn the harmonic series in music is so-called because frequencies with whole number ratios come up in multiple ways in the study of harmony in Western music. First, any sound with a periodic waveform can be analyzed as a Fourier series, i.e., a sum of individual sounds with a harmonic series of frequencies. Because such sounds are often important in nature (e.g., many animal cries), our ears have evolved to do this automatically to some extent, which is probably why we respond instinctively to harmonic series in music. Also, any sound source with a long thin vibrating element (e.g., a violin string or the air column in a wind instrument, but not most bells or drums) will also tend to have a harmonic mix of sounds, even if it's not exactly periodic. Sponsored by CDW Corporation Need solutions to break down silos and support constituents? With Gemini for Google Workspace and CDW, you get collaborative tools that integrate with existing tech. Trevor B.A. with MMath in Mathematics, University of Cambridge (Graduated 2023) · Upvoted by Ermal Feleqi , PhD Mathematics, University of Padua (2010) and Shubhankar Datta , Master of Science Mathematics, Jadavpur University (2022) · Author has 1.3K answers and 5.5M answer views · 7y Related The almost harmonic series is defined as the harmonic series with all numbers including 9 are omitted, i.e. . Does this series converge? Denote Then we have In a similar fashion, if we let Then And so on. So the series Denote Then we have In a similar fashion, if we let Then And so on. So the series Hence, the series must converge, and it's value must be smaller than . Alon Amit PhD in Mathematics; Mathcircler. · Upvoted by Michael Narmo , Msci Mathematics, University College London (2015) and Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) · Author has 8.7K answers and 172.8M answer views · 4y Related What is the explicit value of the first n terms of the harmonic series? The explicit value of is . What other answer would satisfy you? This is an entirely explicit expression of a real number which depends on . Perhaps you’d have been happier if the answer has been, say, ? I bet you would. If it had been true that is exactly , you’d have probably been satisfied with that as an expression for the “explicit value”. But… hang on. Isn’t that insane? You’ve replaced an expression for which asks you to add rational numbers by an expression which is shorthand for an infinite se The explicit value of is . What other answer would satisfy you? This is an entirely explicit expression of a real number which depends on . Perhaps you’d have been happier if the answer has been, say, ? I bet you would. If it had been true that is exactly , you’d have probably been satisfied with that as an expression for the “explicit value”. But… hang on. Isn’t that insane? You’ve replaced an expression for which asks you to add rational numbers by an expression which is shorthand for an infinite series, adding up infinitely many rational numbers and seeking a limit. How exactly is the answer improving upping the question? We are so used to compact expressions such as , we tend to regard them as a “final answer”. A “closed-form expression”. The end of the road. But really, there’s nothing more “explicit” about (a transcendental function) or (The product of simple things, compactly denoted) than . The very definition of the harmonic numbers is an explicit, finite, nice expression. It happens to be true that is very close to , and even closer to . That’s a very valuable thing to know, because there are various manipulations that are more easily achieved with than the raw definition of itself. I’m not saying otherwise. But the difference isn’t that is “explicit” while is not. Sponsored by WebMD Advanced RMS treatments Do you have advanced relapsing multiple sclerosis? Here are some treatments you may want to try. Dale Gray Former Associate Professor, Physics at University of South Dakota (1986–1989) · Author has 763 answers and 1.5M answer views · 2y Related What is proof that the harmonic series is divergent? Let , for . Consider the sum of the areas of the rectangles in the picture below. Call this sum . We will call this sum an upper sum. Notice that . Now, for any positive integer, , we can construct an upper sum, , consisting of the sum of areas of such rectangles ending at . We then have . The sequence, , is the sequence of partial sums for the ha Let , for . Consider the sum of the areas of the rectangles in the picture below. Call this sum . We will call this sum an upper sum. Notice that . Now, for any positive integer, , we can construct an upper sum, , consisting of the sum of areas of such rectangles ending at . We then have . The sequence, , is the sequence of partial sums for the harmonic series. We have , as . Therefore, the sequence of partial sums for the harmonic series diverges, which means, by definition, the harmonic series diverges. Terry Moore M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Author has 16.4K answers and 29.1M answer views · May 17 Related When rearranging series like the alternating harmonic series, why are some terms canceled while others are doubled, and how does this affect the overall sum? When rearranging series like the alternating harmonic series, why are some terms canceled while others are doubled, and how does this affect the overall sum? The question involves a false assumption. The terms of the harmonic series are strictly decreasing so no cancellation or duplication is possible when the signs alternate and the series is rearranged. cannot double or cancel unless , but no two terms have the same denominator. I don’t know if groups of terms may cancel, the following is not an example because terms with even numbered denominators should have a negative When rearranging series like the alternating harmonic series, why are some terms canceled while others are doubled, and how does this affect the overall sum? The question involves a false assumption. The terms of the harmonic series are strictly decreasing so no cancellation or duplication is possible when the signs alternate and the series is rearranged. cannot double or cancel unless , but no two terms have the same denominator. I don’t know if groups of terms may cancel, the following is not an example because terms with even numbered denominators should have a negative sign: . Maybe it’s an interesting question whether groups of terms may cancel, or maybe not. Variations in the overall sum due to rearrangement of the order of terms is not because of cancellation or duplication. It’s because ignoring the signs of the terms leads to a divergent series. It then follows that you can get any sum you like by taking terms of one sign until you pass your favourite number then take as many of the other sign as you like until you get to the other side. As the magnitudes strictly decrease the amount by which you overpass the desired sum decreases each time. This has nothing to do with cancellation or diplication. Robby Goetschalckx Computer scientist for 11+ years and passionate about math since childhood. · Upvoted by David Vanderschel , PhD Mathematics & Physics, Rice (1970) and Justin Rising , PhD in statistics · Author has 6.4K answers and 9.2M answer views · 7y Related What is the proof of the harmonic series going to infinite but not constant? Group together terms in pairs. We get: Note that, for each couple of terms , we have that 3. This means that we get a sum that is larger than , which is exactly the same series again. 4. That means that our original expression, if it were to converge to some value , would have that . By contradiction, we have shown that the series Group together terms in pairs. We get: Note that, for each couple of terms , we have that 3. This means that we get a sum that is larger than , which is exactly the same series again. 4. That means that our original expression, if it were to converge to some value , would have that . By contradiction, we have shown that the series does not converge. Sridhar Ramesh Mathematician/Logician/All-Around Great Guy · Upvoted by Alon Amit , Lover of math. Also, Ph.D. and Anton Fahlgren , M.S. Mathematics, Stockholm University (2018) · Author has 954 answers and 6.6M answer views · 7y Related The almost harmonic series is defined as the harmonic series with all numbers including 9 are omitted, i.e. . Does this series converge? For each term in this series whose denominator has a given number of digits (here, I pretend it’s 3 digits: A, B, C), there are precisely 9 more terms of one digit longer starting with those same digits (the terms ), the largest of which is the original term divided by 10. Thus, the total contribution of these new terms is less than of the original term. Let be the total contribution of all the terms with many digits in the denominator. We’ve just shown that . Thus, the tot For each term in this series whose denominator has a given number of digits (here, I pretend it’s 3 digits: A, B, C), there are precisely 9 more terms of one digit longer starting with those same digits (the terms ), the largest of which is the original term divided by 10. Thus, the total contribution of these new terms is less than of the original term. Let be the total contribution of all the terms with many digits in the denominator. We’ve just shown that . Thus, the total value of our series is less than . Since the parenthetical factor converges (as a convergent geometric series; in fact, it converges to precisely ), we find that our series must converge as well. (And by more careful analysis of this sort, we even discover precise convergence rates for this series; in particular, the tail-contribution from all the terms of at least many digits is at most , in just the same way, and this approaches zero geometrically as gets large.) Related questions What is the summation of the harmonic series? What is an intuitive way to see why harmonic series sums to logn? What are some interesting applications of the harmonic series? How can we prove convergence of harmonic series? The almost harmonic series is defined as the harmonic series with all numbers including 9 are omitted, i.e. . Does this series converge? What is the harmonic series? What is the sum of the harmonic series for all positive integers? How do I prove that the sum 1+1/2+1/3+…+1/n is never an integer? Why do alternating harmonic series converge? In mathematics, when does the harmonic series converge? Why is the harmonic series so wonderful? Why is the harmonic series called so? Why does the harmonic series diverge? Is the harmonic series also a geometric series? What is the proof of the harmonic series going to infinite but not constant? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
12199	https://www.youtube.com/watch?v=s_h30SLjjgY	Determining Whether a Two-Parameter Solution Family Satisfies Given Boundary Values Richard Swearingen Outdoors 1210 subscribers 15 likes Description 1632 views Posted: 13 Apr 2021 In this video, we go over how to determine whether a given solution family is able to satisfy given boundary values in a boundary-value problem. 1 comments Transcript: Introduction so in this video we're going to look at a two parameter a solution family to a differential equation and identify whether that two parameter solution Problem Statement family satisfies given boundary values so i have a differential equation here y double prime plus 16y equals zero and it's a boundary value problem because i've been given um some boundary conditions and again it's not an initial value problem because the inputs for x are not the same so if i had y of 0 and y prime of 0 so that i was inputting x equals zero for each of y and y prime i'd have an initial initial value problem but here i have two distinct values being input for x that makes this a boundary value problem and the claim is that this equation right here is a two parameter solution family for this differential equation so what we're claiming is then that this function y y of x is a solution to this differential equation and we call it a two parameter solution because it contains two constants of integration c sub one and c sub two and we would either need to know initial conditions or boundary values to be able to find real number values for c sub 1 and c sub 2. so the first thing that we ought Solution to do is just verify that this is actually a solution for this differential equation instead of taking it on blind faith so i have y equals c sub 1 cosine of 4x plus c sub 2 times the sine of 4x so i'm going to need the second derivative and the and y in addition to y so i'm going to have to take the first derivative which is going to be 4c minus 4c sub 1 times the sine of 4x got to make my pin work right plus here i'm going to have 4 times c sub 2 times the cosine of 4x and i need the second derivative so we're going to take the take the derivative again so we're going to get negative 16 c sub 1 times the cosine of 4x minus 16 c sub 2 times the sine of 4x and then as you recall to verify that we have a solution we plug in the values for y y prime and y double prime and we verify that they evaluate to zero so if we have y double prime plus 16y it's going to equal replace y double prime with this mess right here so we get minus 16 c sub 1 cosine 4x minus 16 c sub 2 sine of 4x then it's going to be plus 16 times y but y is just c sub 1 cosine of 4x plus c sub 2 sine of 4x and of course when you go ahead and distribute the 16 into the parentheses we get 16 c sub 1 cosine 4x minus 16 c sub 2 sine of 4x and we when we distribute the 16 in here we're going to get 16 c sub 1 cos 4x plus 16 c sub 2 times the sine of 4x and we can see that this will cancel with this and this will combine with this we'll get 0 plus 0 is just 0. so indeed this does satisfy the differential equation so i do have a solution family for the differential equation and then to get a particular members of the family you identify specific values for c sub 1 and c sub 2. and so really the question is are the values of c sub 1 and c sub 2 that are going to satisfy these boundary conditions and to do that we just plug in our values so if i look at my solution family here y equals c sub one cos four x plus c sub two sine of four x when i plug in y equal to zero we want y of zero to equal 0. so plug 0 in here and when we plug 0 in for that we're going to get y equals c sub 1 times the cosine of 4 times 0 is 0 plus c sub 2 times the sine of 4 times 0 is 0. so we're going to get c sub 1 times 1 plus c sub 2 times 0 and this is going to simplify to just c sub 1. so we see that when what we see here is that y of 0 is going to equal c sub 1. so if we want y of 0 to equal 0 and when we plug 0 into the solution family get we get y of 0 equals c sub one putting these together we see that zero has to be c sub one so that's step one we we discover that to satisfy this first boundary condition we can only satisfy it if c sub one is equal to zero and c and zero is a real number so that's a legitimate choice so then our our solution actually winds up for this specific boundary condition simplifies to a single parameter solution family c sub 2 times the sine of 4x and then the question is can we can we find c sub 2 so that the second boundary condition is satisfied so when we plug in y of pi over 2 what are we going to get we're going to get c sub 2 times the sine of 4 times pi over 2 simplifies to 2 pi and the sine of 2 pi is 0. so we get c sub 2 times 0 and we want y of pi over 2 for our boundary condition we want y of pi over 2 to equal 0 so that means 0 needs to equal c sub 2 times 0. and this statement is going to be true no matter what we choose for c sub 2. so for all real numbers c sub 2 so so for c sub 2 any real number y equals c sub 2 times the sine of 4x will satisfy both boundary conditions so notice we're not getting a specific value for c sub 2 that we have to use we're really indicating that it it can be whatever you want and life is going any life's going to be good we have a we have y equals c sub 2 times the sine of 4x as a solution that satisfies the boundary conditions Changing the boundary conditions so we could take the exact same differential equation the exact same two parameter solution family and just change uh change the boundary conditions a little bit keep y of zero equals zero as the first boundary condition and we just did this on the previous slide we saw that our solution family to satisfy this boundary condition that we needed c sub 1 to equal 0 which meant that y needed to equal c sub 2 times the sine of 4x and then let's not even put a 1 here let's put a here where a isn't equal to zero so we're just making the value there something different than it was on the previous slide so if y of pi over 2 equals a where a isn't zero when we plug our y of pi over 2 in for our second boundary condition just like we did on the previous slide we're going to get the sine of 2 pi just like we did on the previous slide so we get c sub 2 times 0 is what's going to happen when we plug in the pi over 2 and if we want y of pi over 2 to equal a where a isn't zero we wind up with the equation a equals c sub 2 times 0. and for for our solution family to satisfy this set of boundary conditions then we would need to find c sub 2 such that 0 times c sub 2 equals a where a isn't equal to 0 and the problem is is no such c sub 2 exist so what we find then is that just making a tweak to the boundary condition we can discover that our solution family for the differential equation is unable to satisfy those boundary conditions

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