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8700	https://www.quora.com/What-is-the-degree-of-a-polynomial-with-two-variables	Something went wrong. Wait a moment and try again. Degree of Polynomial Variables and Constants Specific Polynomials Polynomial Functions Math and Algebra Variables (mathematics) Maths Polynomial 5 What is the degree of a polynomial with two variables? Shivaji Mane Bsc (Bed) mathematics from S.R.M.T. Naded (Graduated 2013) · Author has 99 answers and 69.8K answer views · 4y · Polynomials in two variables are algebraic expressions consisting of terms in the form axnym a x n y m . The degree of each term in a polynomial in two variables is the sum of the exponents in each term and the degree of the polynomial is the largest such sum. 1.3K views · View 1 share · Reuven Harmelin Studied Mathematics at טכניון (Graduated 1978) · Author has 2.3K answers and 1.9M answer views · 2y What is the third degree polynomial with two real zeros? If you are looking for a cubic polynomial with real coefficients with two distinct real zeroes, then for any choice of a pair of distinct real numbers a,b the next cubic real polynomial is an answer to your question. If the coefficients of the required cubic polynomial could be complex numbers, then the answer would be of the form where two of the 3 roots a,b are real numbers and c a complex, non-real number. If you are looking for a cubic polynomial with real coefficients with two distinct real zeroes, then for any choice of a pair of distinct real numbers a,b the next cubic real polynomial is an answer to your question. If the coefficients of the required cubic polynomial could be complex numbers, then the answer would be of the form where two of the 3 roots a,b are real numbers and c a complex, non-real number. 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I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions What is a polynomial in two variables? What is the highest degree of a polynomial in two variables? What is a second degree polynomial? How do you add polynomials with different variables but the same degree? What is the general form of polynomial in two variables? Rajnish Kumar Studied at Anna University, Tamil Nadu, India · Author has 1K answers and 280.2K answer views · 5y Related What is the degree of a polynomial? The degree of a polynomial in one variable is the largest exponent in the polynomial. Polynomials in one variable are algebraic expressions that consist of terms in the form 𝑎𝑥^𝑛 where 𝑛 is a non-negative (i.e. positive or zero) integer and 𝑎 is a real number and is called the coefficient of the term. I found a suitable video content explaining the Polynomial in detail Andrew Winkler PhD in Mathematics, Courant Institute of Mathematical Sciences, NYU (Graduated 1987) · Author has 9.6K answers and 10.9M answer views · 2y Related Can a polynomial have a variable degree? The short answer is no. But there is a subtlety about polynomials, and about variables. In algebra there are actually different kinds of variables, and we are generally sloppy about describing their precise meanings, relying on context to make it clear. When I write a polynomial like ax+b, a, x, and b are all variables in some sense or other. So what’s different about x? If the field of coefficients The short answer is no. But there is a subtlety about polynomials, and about variables. In algebra there are actually different kinds of variables, and we are generally sloppy about describing their precise meanings, relying on context to make it clear. When I write a polynomial like ax+b, a, x, and b are all variables in some sense or other. So what’s different about x? If the field of coefficients is F, say, then a and b are variables taking values in F; they are regarded as arbitrary, but nevertheless fixed. x, on the other hand, doesn’t actually refer to an element of F. The expression ax + b can be evaluated for x in F, x in any field containing F, any linear endomorphism or matrix on a vector space over F, any element of an F algebra. So x is “more variable” than a and b are, and so they aren’t on a par with it. (So the idea that a variable x is “a name for an unknown number” is a limiting one, a limitation that crippled everyone until Galois realized that something far deeper is actually going on.) So when I write a_n x^n, say, that’s a polynomial of degree n. n is a variable, but like a and b and a_n, it’s a variable with a very restricted domain, namely the natural numbers, and it is regarded as fixed. Another word for “regarded as fixed” is “bound”. So when we say “let a be a real number”, a is no longer “varying”, it is bound to a specific value, even though that value hasn’t been specified. A “variable” like x is not regarded as bound. When we say “let x be a real number”, we are not suggesting that x is bound to anything, only that it has all the properties of a real number, and in fact we generally really mean something more subtle, as follows. Formally, variables in polynomials are simply elements of any set X. We create an algebra structure by forming the convolution algebra over F of the monoid of all finitely supported functions on X taking natural number values. Such a function assigns a number to each variable, namely its expone... Assistant Bot · 1y The degree of a polynomial with two variables is defined as the highest total degree of any term in the polynomial. For a term in the form axmyn, where a is a coefficient, x and y are the variables, and m and n are non-negative integers, the total degree of that term is given by m+n. For example, in the polynomial: P(x,y)=3x2y+5xy2+7y+2 The first term 3x2y has a total degree of 2+1=3. The second term 5xy2 has a total degree of 1+2=3. The third term 7y has a total degree of 0+1=1. The constant term 2 has a total degree of 0. In this case, the highest total degree is 3, so the degree of the polynomial P(x,y) is 3. Ask follow-up questions on Poe, a free app from Quora By continuing you agree to create a Poe account, and to its Terms of Service & Privacy Policy Tell me more Can you explain how to find the degree of a polynomial with three variables? What happens if a polynomial has terms with different degrees? Are there any special cases for the degree of polynomials? 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Eric Platt Ph.D in Mathematics · Author has 2.7K answers and 13.3M answer views · 8y Related How can I factor six degree polynomial? Unlike polynomials of degree 2, 3, or 4 there isn’t an analog of the quadratic formula for polynomials of degree 5 or higher. So we have to be extra clever. If you know any of the roots you can factor it out of the polynomial through long division. Synthetic division is a nice shorthand for the same method if you are comfortable with it. The more roots that you know the more you can factor out. Hopefully you can eventually get to the point where you can use the quadratic formula (or the cubic or quartic). Now suppose that you don’t know any of the roots, how would we go about this? Let’s look at Unlike polynomials of degree 2, 3, or 4 there isn’t an analog of the quadratic formula for polynomials of degree 5 or higher. So we have to be extra clever. If you know any of the roots you can factor it out of the polynomial through long division. Synthetic division is a nice shorthand for the same method if you are comfortable with it. The more roots that you know the more you can factor out. Hopefully you can eventually get to the point where you can use the quadratic formula (or the cubic or quartic). Now suppose that you don’t know any of the roots, how would we go about this? Let’s look at the problem backwards. What happens if we have a polynomial already factored out and then we distribute everything (anti-factor?). Without loss of generality I am assuming the leading coefficient is one. If it isn’t we can factor it out. (x+a)=x+a (x+a1)(x+a2)=x2+(a1+a2)x+a1a2 (x+a1)(x+a2)(x+a3)=x3+(a1+a2+a3)x2+(a1a2+a1a3+a2a3)x+a1a2a3 (x+a1)(x+a2)...(x+an)=xn+(a1+a2+...+an)xn−1+...+(a1a2...an) So the coefficient that is the constant term on the end is the product of all of the aj. The second coefficient is the sum of all of the aj, and all of the coefficients in between are the combinations of sums of products of the coefficients. If your coefficients are integers then you can factor the last coefficent and look for rational roots. If this doesn’t lead you anywhere than you may need to use a numerical method such as Newtons method to find a root. Armer Spielmann Knows German · 5y Related What is the degree of a polynomial example? Let p(x)=a0⋅x0+a1⋅x1+a2⋅x2+⋯ be a polynominal with coefficients ai we set to be real numbers. The degree of such a polynominal is then defined as the maximum index n such that an≠0 You asked for an example. So here you are p(x)=4x2+5x4+0x7 is a polynominal of degree 4. Because p(x)=4x2+5x4+0x7=0x0+0x1+4x2+0x3+5x4+ 0x5+0x6+0x7+0x8+⋯ where the points at the end notate that the rest of the terms are all zero And the maximum coefficient unequal 0, which is 5 has index 4, so a4=5 Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Michael Holt IT Manager · Author has 101 answers and 79.7K answer views · 1y Related How do you solve higher degree polynomials? Method 1 of 2:Recognizing Factors 1. Factor out common factors from all terms. If every term in the polynomial has a common factor, factor it out to simplify the problem. This is not possible with all polynomials, but it's a good approach to check first. Solve for x in the polynomial .Each term is divisible by 2x, so factor it out:(2x)(x2)+(2x)(6x)+(2x)(8)=0{\displaystyle (2x)(x^{2})+(2x)(6x)+(2x)(8)=0}=(2x)(x2+6x+8){\displaystyle =(2x)(x^{2}+6x+8)}Now solve the quadratic equation using the quadratic formula or factoring:(2x)(x+4)(x+2)=0{\displaystyle (2x)(x+4)(x+2)=0}The solutions are at 2x = 0 Method 1 of 2:Recognizing Factors 1. Factor out common factors from all terms. If every term in the polynomial has a common factor, factor it out to simplify the problem. This is not possible with all polynomials, but it's a good approach to check first. Solve for x in the polynomial .Each term is divisible by 2x, so factor it out:(2x)(x2)+(2x)(6x)+(2x)(8)=0{\displaystyle (2x)(x^{2})+(2x)(6x)+(2x)(8)=0}=(2x)(x2+6x+8){\displaystyle =(2x)(x^{2}+6x+8)}Now solve the quadratic equation using the quadratic formula or factoring:(2x)(x+4)(x+2)=0{\displaystyle (2x)(x+4)(x+2)=0}The solutions are at 2x = 0, x+4=0, and x+2=0.The solutions are . Identify polynomials that act like a quadratic. You likely already know how to solve second degree polynomials, in the form ax2+bx+c{\displaystyle ax^{2}+bx+c}. You can solve some higher-degree polynomials the same way, if they're in the form ax2n+bxn+c{\displaystyle ax^{2n}+bx^{n}+c}. Here are a couple examples: 3x4+4x2−4=0{\displaystyle 3x^{4}+4x^{2}-4=0}Let a=x2{\displaystyle a=x^{2}}:3a2+4a−4=0{\displaystyle 3a^{2}+4a-4=0}Solve the quadratic using any method:(3a−2)(a+2)=0{\displaystyle (3a-2)(a+2)=0} so a = -2 or a = 2/3Substitute x2{\displaystyle x^{2}} for a: x2=−2{\displaystyle x^{2}=-2} or x2=2/3{\displaystyle x^{2}=2/3}. The other equation, x2=−2{\displaystyle x^{2}=-2}, has no real solution. (If using complex numbers, solve as ). x5+7x3−9x=0{\displaystyle x^{5}+7x^{3}-9x=0} does not follow this pattern, but notice you can factor out an x:(x)(x4+7x2−9)=0{\displaystyle (x)(x^{4}+7x^{2}-9)=0}You can now treat x4+7x2−9{\displaystyle x^{4}+7x^{2}-9} as a quadratic, as shown in example 2. Factor sums or differences of cubes. These special cases look difficult to factor, but have properties which make the problem much easier: A polynomial in the form a3+b3{\displaystyle a^{3}+b^{3}} factors to (a+b)(a2−ab+b2){\displaystyle (a+b)(a^{2}-ab+b^{2})}. A polynomial in the form a3−b3{\displaystyle a^{3}-b^{3}} factors to (a−b)(a2+ab+b2){\displaystyle (a-b)(a^{2}+ab+b^{2})}. Note that the quadratic portion of the result is not factorable. Note that x6{\displaystyle x^{6}}, x9{\displaystyle x^{9}}, and x to any power divisible by 3 all fit these patterns. Look for patterns to find other factors. Polynomials that do not look like the examples above may not have any obvious factors. But before you try the methods below, try looking for a two-term factor (such as "x+3"). Grouping terms in different orders and factoring out part of the polynomial may help you find one. This is not always a feasible approach, so don't spend too much time trying if no common factor seems likely. −3x3−x2+6x+2=0{\displaystyle -3x^{3}-x^{2}+6x+2=0}This has no obvious factor, but you can factor the first two terms and see what happens:(−x2)(3x+1)+6x+2=0{\displaystyle (-x^{2})(3x+1)+6x+2=0}Now factor the last two terms (6x+2), aiming for a common factor:(−x2)(3x+1)+(2)(3x+1)=0{\displaystyle (-x^{2})(3x+1)+(2)(3x+1)=0}Now rewrite this using the common factor, 3x+1:(3x+1)(−x2+2)=0{\displaystyle (3x+1)(-x^{2}+2)=0} Method 2 of 2:Rational Roots and Synthetic Division 1. Try to identify one root of the polynomial. Synthetic division is a useful way to factor high-order polynomials, but it only works if you know one of the roots (or "zeroes") already. You may be able to find this by factoring as described above, or the problem may provide one. If so, skip down to the synthetic division instructions. If you do not know a root, continue to the next step to try to find one. The root of a polynomial is the value of x for which y = 0. Knowing a root c also gives you a factor of the polynomial, (x - c). List the factors of the constant term. The "rational roots" test is a way to guess at possible root values. To begin, list all the factors of the constant (the term with no variable). Example: The polynomial 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} has the constant term 9. Its factors are 1, 3, and 9. List the factors of the leading coefficient. This is the coefficient in the first term of the polynomial, when it is arranged from the highest-degree term to the lowest. List all of that number's factors on a separate line. Example (cont.): 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} has a leading coefficient of 2. Its factors are 1 and 2. Find the possible roots. If the polynomial has a rational root (which it may not), it must be equal to ± (a factor of the constant)/(a factor of the leading coefficient). Only a number c in this form can appear in the factor (x-c) of the original polynomial. Example (cont.): Any rational roots of this polynomial are in the form (1, 3, or 9) divided by (1 or 2). Possibilities include ±1/1, ±1/2, ±3/1, ±3/2, ±9/1, or ±9/2. Don't forget the "±": each of these possibilities could be positive or negative. Test roots until you find one that fits. None of these are guaranteed to be roots, so you'll need to test them with the original polynomial. Example: (1/1=1) is a possible root. If it turns out to be an actual root, plugging it into the polynomial should result in zero.2(1)3+(1)2−12(1)+9=2+1−12+9=0{\displaystyle 2(1)^{3}+(1)^{2}-12(1)+9=2+1-12+9=0}, so 1 is confirmed to be a root.This means the polynomial has the factor (x-1). If none of the possibilities work out, the polynomial has no rational roots and cannot be factored. Set up a synthetic division problem. Synthetic division is a way to find all the factors of a polynomial, if you already know one of them. To set it up, write a root of the polynomial. Draw a vertical line to its right, then write the coefficients of your polynomial arranged from highest degree exponent to lowest. (You do not need to write the terms themselves, just the coefficients.) You may need to insert terms with a coefficient of zero. For example, rewrite the polynomial x3+2x{\displaystyle x^{3}+2x} as x3+0x2+2x+0{\displaystyle x^{3}+0x^{2}+2x+0}. Example (cont.): The rational roots test above told us that the polynomial 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} has the root 1.Write the root 1, followed by a vertical line, followed by the coefficients of the polynomial:(1\|21−129){\displaystyle {\begin{pmatrix}1\|&2&1&-12&9\end{pmatrix}}} Carry down the first coefficient. Copy the first coefficient onto the answer line. Leave a blank line in between the two numbers for later calculations. Example (cont.): Carry the 2 down to the answer line:(1\|21−1292){\displaystyle {\begin{pmatrix}1\|&2&1&-12&9\\&2\end{pmatrix}}} Multiply that number by the root. Write the answer directly below the next term, but not on the answer line. Example (cont.): Multiply the 2 by the root, 1, to get 2 again. Write this 2 in the following column, but on the second row instead of the answer line:(1\|21−12922){\displaystyle {\begin{pmatrix}1\|&2&1&-12&9\&&2\&2\end{pmatrix}}} Add the contents of the column together to get the next portion of the answer. The second coefficient column now contains two numbers. Sum them together and write the result on the answer line directly below them. Example (cont.): 1 + 2 = 3(1\|21−129223){\displaystyle {\begin{pmatrix}1\|&2&1&-12&9\&&2\&2&3\end{pmatrix}}} Multiply the result by the root. Just as you did before, multiply the latest number on the answer line by the root. Write your answer underneath the next coefficient. Example (cont.): 1 x 3 = 3:(1\|21−1292323){\displaystyle {\begin{pmatrix}1\|&2&1&-12&9\&&2&3\&2&3\end{pmatrix}}} Find the sum of the next column. As before, add up the two numbers in the column and write the result on the answer line. Example (cont.): -12 + 3 = -9:(1\|21−1292323−9){\displaystyle {\begin{pmatrix}1\|&2&1&-12&9\&&2&3\&2&3&-9\end{pmatrix}}} Repeat this process until you reach the final column. The last number on your answer line will always be zero. If you get any other result, check your work for mistakes. Example (cont.): Multiply -9 by the root 1, write the answer under the final column, then confirm that the sum of the final column is zero:(1\|21−12923−923−90){\displaystyle {\begin{pmatrix}1\|&2&1&-12&9\&&2&3&-9\&2&3&-9&0\end{pmatrix}}} Use the answer line to find another factor. You have now divided the polynomial by the term (x - c), where c is your factor. The answer line tells you the coefficient of each term in your answer. The x portion of each term has an exponent one lower than the original term directly above it. Example (cont.): The answer line is 2 3 -9 0, but you can ignore the final zero.Since the first term of the original polynomial included an x3{\displaystyle x^{3}}, the first term of your answer is one degree lower: x2{\displaystyle x^{2}}. Therefore, the first term is 2x2{\displaystyle 2x^{2}}Repeat this process to get the answer 2x2+3x−9{\displaystyle 2x^{2}+3x-9}.You have now factored 2x3+x2−12x+9{\displaystyle 2x^{3}+x^{2}-12x+9} into (x−1)(2x2+3x−9){\displaystyle (x-1)(2x^{2}+3x-9)}. Repeat if necessary. You may be able to factor your answer into smaller parts using the same synthetic division method. However, you may be able to use a faster method to finish the problem. For example, once you have a quadratic expression, you can factor it using the quadratic formula. Remember, to start the synthetic division method, you'll need to know one root already. Use the rational roots test again to get this. If none of the rational root possibilities check out, the expression cannot be factored. Example (cont.) You've found the factors (x−1)(2x2+3x−9){\displaystyle (x-1)(2x^{2}+3x-9)}, but the second factor can be broken down further. Try the quadratic equation, traditional factoring, or synthetic division.The final answer is (x−1)(x+3)(2x−3){\displaystyle (x-1)(x+3)(2x-3)}, so the roots of the polynomial are . Alexander Farrugia Ph.D. in Mathematics, University of Malta (Graduated 2016) · Author has 3.2K answers and 27.5M answer views · 9y Related How can I find the leading coefficient and degree of a polynomial? The obvious answer would be 'just read them off the polynomial'. But suppose you are not able to do this. Or you do not know what the polynomial is at all. Then here is an algorithm that extracts the leading coefficient and degree of a polynomial p(x): Input polynomial p(x). If p(x)=0, return leading coefficient 0 and degree −∞. Let p0(x):=p(x) and d=0. Let pd+1(x):=pd(x)−pd(0)x. If pd+1≠0, increase d by 1 and go to step 3. Return leading coefficient pd(0) and degree d. Example: p(x)=2x3−5x+1. p0(x)=2x3−5x+1,d=0. p1(x)=2x2−5,d=1. p2(x)=2x,d=2. p3(x)=2,d=3. p4(x)= The obvious answer would be 'just read them off the polynomial'. But suppose you are not able to do this. Or you do not know what the polynomial is at all. Then here is an algorithm that extracts the leading coefficient and degree of a polynomial p(x): Input polynomial p(x). If p(x)=0, return leading coefficient 0 and degree −∞. Let p0(x):=p(x) and d=0. Let pd+1(x):=pd(x)−pd(0)x. If pd+1≠0, increase d by 1 and go to step 3. Return leading coefficient pd(0) and degree d. Example: p(x)=2x3−5x+1. p0(x)=2x3−5x+1,d=0. p1(x)=2x2−5,d=1. p2(x)=2x,d=2. p3(x)=2,d=3. p4(x)=0. Hence, leading coefficient is p3(0)=2 and degree is d=3. Related questions What is a polynomial in two variables? What is the highest degree of a polynomial in two variables? What is a second degree polynomial? How do you add polynomials with different variables but the same degree? What is the general form of polynomial in two variables? What is the degree of a polynomial in more than one variable? What are the first four terms of a polynomial in two variables? What is the degree of sum of the two polynomials each of degree 4? What is the degree of a constant polynomial? What is the relationship between the degree of a polynomial and its coefficients? What is more than one polynomial variable called? Can two matrices with same characteristic polynomial and same minimal polynomial have two different Jordan form? What is the process for finding the order and degree of a characteristic polynomial? What is the rule for determining the number of terms in a polynomial with unknown variable(s)? Is root 2 a constant polynomial? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
8701	https://www.abebooks.com/9780387975283/Model-Assisted-Survey-Sampling-Springer-0387975284/plp	Model Assisted Survey Sampling (Springer Series in Statistics) - Särndal, Carl-Erik; Swensson, Bengt; Wretman, Jan: 9780387975283 - AbeBooks Skip to main content AbeBooks.com Search Sign in My Account Basket Help Menu Find My Account My Purchases Sign Off Advanced Search Browse Collections Rare Books Art & Collectibles Textbooks Sellers Start Selling Help CLOSE Items related to Model Assisted Survey Sampling (Springer Series in... Lowest Price Model Assisted Survey Sampling Swensson, B., Sarndal, Carl-Er... Särndal, Carl-Erik; Swensson, Bengt; Wretman, JanModel Assisted Survey Sampling (Springer Series in Statistics)ISBN 13:9780387975283 Model Assisted Survey Sampling (Springer Series in Statistics) - Hardcover Särndal, Carl-Erik; Swensson, Bengt; Wretman, Jan 3.86 3.86 out of 5 stars 7 ratings byGoodreads Hardcover ISBN 10:0387975284 ISBN 13:9780387975283 Publisher: Springer, 1997 View all copies of this ISBN edition 3 Used From US$ 55.75 3 New From US$ 351.85 All editions of this title Softcover (1) from US$ 237.02 Hardcover (0) First Edition (0) Signed Copy (0) Synopsis About this title Edition details Synopsis Now available in paperback, this book provides a comprehensive account of survey sampling theory and methodology suitable for students and researchers across a variety of disciplines. It shows how statistical modeling is a vital component of the sampling process and in the choice of estimation technique. The first textbook that systematically extends traditional sampling theory with the aid of a modern model assisted outlook. Covers classical topics as well... More Review "I would recommend that this book be in the office of every survey methodologist." (Journal of Official Statistics) "About this title" may belong to another edition of this title. PublisherSpringerPublication date 1997 Language English ISBN 10 0387975284 ISBN 13 9780387975283 Binding Hardcover Edition number 1 Number of pages 695 Rating 3.86 3.86 out of 5 stars 7 ratings byGoodreads Buy Used Condition: Good Former library book; may include... View this item US$ 55.75 Convert currency FREE shipping within U.S.A. Destination, rates & speeds Add to basket Buy New View this item US$ 351.85 Convert currency US$ 4.25 shipping within U.S.A. 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Destination, rates & speeds Quantity: 1 available Add to basket Stock Image Model Assisted Survey Sampling Sarndal, Carl-Erik; Swensson, Bengt; Wretman, Jan; Wretman, Jan Hakan Published by Springer Verlag, 1992 ISBN 10: 0387975284 / ISBN 13:9780387975283 Used / Hardcover First Edition Seller:bmyguest books, Toronto, ON, Canada (4-star seller)Seller rating 4 out of 5 stars;) Hardcover. Condition: Very Good. 1st Edition. The Boards Are In Very Good Condition. With Some Small Damage At The Corners. Clean Inside With No Remarks Or Highlights. 694 Pages With The Index. Comprehensive Account Of Survey Sampling Theory And Methodology Which Will Be Suitable For Students And Researchers Across A Variety Of Disciplines.books are NOT signed. We will state signed at the description section. we confirm they are signed via email or stated in the description box. - Specializing in academic, collectiblle and historically significant, providing the utmost quality and customer service satisfaction. 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8702	https://bogart.openmathbooks.org/ctgd/s3-3-partitions-int.html	CTGD Partitions of Integers Skip to main content Combinatorics Through Guided Discovery Kenneth P. Bogart Contents IndexPrevUpNext Annotations ContentsPrevUpNext Front Matter Colophon Author Biography Preface Preface to PreTeXt edition 1 What is Combinatorics? About These Notes Basic Counting Principles Some Applications of the Basic Principles Supplementary Chapter Problems 2 Applications of Induction and Recursion in Combinatorics and Graph Theory Some Examples of Mathematical Induction Recurrence Relations Graphs and Trees Supplementary Problems 3 Distribution Problems The idea of a distribution Partitions and Stirling Numbers Partitions of Integers Supplementary Problems 4 Generating Functions The Idea of Generating Functions Generating functions for integer partitions Generating Functions and Recurrence Relations Supplementary Problems 5 The Principle of Inclusion and Exclusion The size of a union of sets Application of Inclusion and Exclusion Deletion-Contraction and the Chromatic Polynomial Supplementary Problems 6 Groups acting on sets Permutation Groups Groups Acting on Sets Pólya-Redfield Enumeration Theory Supplementary Problems Back Matter Relations Mathematical Induction Exponential Generating Functions GNU Free Documentation License Index Authored in PreTeXt Section 3.3 Partitions of Integers ¶ We have now completed all our distribution problems except for those in which both the objects and the recipients are identical. For example, we might be putting identical apples into identical paper bags. In this case all that matters is how many bags get one apple (how many recipients get one object), how many get two, how many get three, and so on. Thus for each bag we have a number, and the multiset of numbers of apples in the various bags is what determines our distribution of apples into identical bags. A multiset of positive integers that add to n n is called a partition of n.n. Thus the partitions of 3 are 1+1+1, 1+2 (which is the same as 2+1) and 3. The number of partitions of k k is denoted by P(k);P(k); in computing the partitions of 3 we showed that P(3)=3.P(3)=3. It is traditional to use Greek letters like λ λ (the Greek letter λ λ is pronounced LAMB duh) to stand for partitons; we might write λ=1,1,1,λ=1,1,1,γ=2,1 γ=2,1 and τ=3 τ=3 to stand for the three partitions we just described. We also write λ=1 3 λ=1 3 as a shorthand for λ=1,1,1,λ=1,1,1, and we write λ⊣3 λ⊣3 as a shorthand for ``λ λ is a partition of three." Problem 157 Find all partitions of 4 and find all partitions of 5, thereby computing P(4)P(4) and P(5).P(5). Subsection 3.3.1 The number of partitions of k k into n n parts ¶ A partition of the integer k k into n n parts is a multiset of n n positive integers that add to k.k. We use P(k,n)P(k,n) to denote the number of partitions of k k into n n parts. Thus P(k,n)P(k,n) is the number of ways to distribute k k identical objects to n n identical recipients so that each gets at least one. Problem 158 Find P(6,3)P(6,3) by finding all partitions of 6 into 3 parts. What does this say about the number of ways to put six identical apples into three identical bags so that each bag has at least one apple? Subsection 3.3.2 Representations of partitions ¶ Problem 159 How many solutions are there in the positive integers to the equation x 1+x 2+x 3=7 x 1+x 2+x 3=7 with x 1≥x 2≥x 3?x 1≥x 2≥x 3? Problem 160 Explain the relationship between partitions of k k into n n parts and lists x 1,x 2,…,x n x 1,x 2,…,x n of positive integers with x 1+x 2+⋯+x n=k x 1+x 2+⋯+x n=k and x 1≥x 2≥…≥x n.x 1≥x 2≥…≥x n. Such a representation of a partition is called a decreasing list representation of the partition. Problem 161 Describe the relationship between partitions of k k and lists or vectors (x 1,x 2,…,x n)(x 1,x 2,…,x n) such that x 1+2 x 2+…k x k=k.x 1+2 x 2+…k x k=k. Such a representation of a partition is called a type vector representation of a partition, and it is typical to leave the trailing zeros out of such a representation; for example (2,1)(2,1) stands for the same partition as (2,1,0,0).(2,1,0,0). What is the decreasing list representation for this partition, and what number does it partition? Problem 162 How does the number of partitions of k k relate to the number of partitions of k+1 k+1 whose smallest part is one? Hint How can you start with a partition of (k) and make it into a new partition of (k+1) that is guaranteed to have a part of size one, even if the original partition didn't? When we write a partition as λ=λ 1,λ 2,…,λ n,λ=λ 1,λ 2,…,λ n, it is customary to write the list of λ i λ i s as a decreasing list. When we have a type vector (t 1,t 2,…,t m)(t 1,t 2,…,t m) for a partition, we write either λ=1 t 1 2 t 2⋯m t m λ=1 t 1 2 t 2⋯m t m or λ=m t m(m−1)t m−1⋯2 t 2 1 t 1.λ=m t m(m−1)t m−1⋯2 t 2 1 t 1. Henceforth we will use the second of these. When we write λ=λ i 1 1 λ i 2 2⋯λ i n n,λ=λ 1 i 1 λ 2 i 2⋯λ n i n, we will assume that λ i>λ i+1.λ i>λ i+1. Subsection 3.3.3 Ferrers and Young Diagrams and the conjugate of a partition ¶ The decreasing list representation of partitions leads us to a handy way to visualize partitions. Given a decreasing list (λ 1,λ 2,…λ n),(λ 1,λ 2,…λ n), we draw a figure made up of rows of dots that has λ 1 λ 1 equally spaced dots in the first row, λ 2 λ 2 equally spaced dots in the second row, starting out right below the beginning of the first row and so on. Equivalently, instead of dots, we may use identical squares, drawn so that a square touches each one to its immediate right or immediately below it along an edge. See Figure 3.3.1 for examples. The figure we draw with dots is called the Ferrers diagram of the partition; sometimes the figure with squares is also called a Ferrers diagram; sometimes it is called a Young diagram. At this stage it is irrelevant which name we choose and which kind of figure we draw; in more advanced work the squares are handy because we can put things like numbers or variables into them. From now on we will use squares and call the diagrams Young diagrams. Figure 3.3.1 The Ferrers and Young diagrams of the partition (5,3,3,2) Problem 163 Draw the Young diagram of the partition (4,4,3,1,1). Describe the geometric relationship between the Young diagram of (5,3,3,2) and the Young diagram of (4,4,3,1,1). Hint Draw a line through the top-left corner and bottom-right corner of the topleft box. Problem 164 The partition (λ 1,λ 2,…,λ n)(λ 1,λ 2,…,λ n) is called the conjugate of the partition (γ 1,γ 2,…,γ m)(γ 1,γ 2,…,γ m) if we obtain the Young diagram of one from the Young diagram of the other by flipping one around the line with slope -1 that extends the diagonal of the top left square. See Figure 3.3.2 for an example. Figure 3.3.2 The Ferrers diagram the partition (5,3,3,2) and its conjugate. What is the conjugate of (4,4,3,1,1)? How is the largest part of a partition related to the number of parts of its conjugate? What does this tell you about the number of partitions of a positive integer k k with largest part m?m? Hint The largest part of a partition is the maximum number of boxes in a row of its Young diagram. What does the maximum number of boxes in a column tell us? Problem 165 A partition is called self-conjugate if it is equal to its conjugate. Find a relationship between the number of self-conjugate partitions of k k and the number of partitions of k k into distinct odd parts. Hint Draw all self conjugate partitions of integers less than or equal to 8. Draw all partitions of integers less than or equal to 8 into distinct odd parts (many of these will have just one part). Now try to see how to get from one set of drawings to the other in a consistent way. Problem 166 Explain the relationship between the number of partitions of k k into even parts and the number of partitions of k k into parts of even multiplicity, i.e. parts which are each used an even number of times as in (3,3,3,3,2,2,1,1). Hint Draw the partitions of six into even parts. Draw the partitions of six into parts used an even number of times. Look for a relationship between one set of diagrams and the other set of diagrams. If you have trouble, repeat the process using 8 or even 10 in place of 6. Problem 167 Show that the number of partitions of k k into four parts equals the number of partitions of 3 k 3 k into four parts of size at most k−1 k−1 (or 3 k−4 3 k−4 into four parts of size at most k−2 k−2 or 3 k−4 3 k−4 into four parts of size at most k k). Hint Draw a partition of ten into four parts. Assume each square has area one. Then draw a rectangle of area 40 enclosing your diagram that touches the top of your diagram, the left side of your diagram and the bottom of your diagram. How does this rectangle give you a partition of 30 into four parts? Problem 168 The idea of conjugation of a partition could be defined without the geometric interpretation of a Young diagram, but it would seem far less natural without the geometric interpretation. Another idea that seems much more natural in a geometric context is this. Suppose we have a partition of k k into n n parts with largest part m.m. Then the Young diagram of the partition can fit into a rectangle that is m m or more units wide (horizontally) and n n or more units deep. Suppose we place the Young diagram of our partition in the top left-hand corner of an m′m′ unit wide and n′n′ unit deep rectangle with m′≥m m′≥m and n′≥n,n′≥n, as in Figure 3.3.4. Figure 3.3.4 To complement the partition (5,3,3,2)(5,3,3,2) in a 6 by 5 rectangle: enclose it in the rectangle, rotate, and cut out the original Young diagram. (a) Why can we interpret the part of the rectangle not occupied by our Young diagram, rotated in the plane, as the Young diagram of another partition? This is called the complement of our partition in the rectangle. (b) What integer is being partitioned by the complement? (c) What conditions on m′m′ and n′n′ guarantee that the complement has the same number of parts as the original one? Hint Consider two cases, (m' \gt m) and (m' = m\text{.}) (d) What conditions on m′m′ and n′n′ guarantee that the complement has the same largest part as the original one? Hint Consider two cases, (n' \gt n) and (n' = n\text{.}) (e) Is it possible for the complement to have both the same number of parts and the same largest part as the original one? (f) If we complement a partition in an m′m′ by n′n′ box and then complement that partition in an m′m′ by n′n′ box again, do we get the same partition that we started with? Problem 169 Suppose we take a partition of k k into n n parts with largest part m,m, complement it in the smallest rectangle it will fit into, complement the result in the smallest rectangle it will fit into, and continue the process until we get the partition 1 of one into one part. What can you say about the partition with which we started? Hint 1 Suppose we take two repetitions of this complementation process. What rows and columns do we remove from the diagram? Hint 2 To deal with an odd number of repetitions of the complementation process, think of it as an even number plus 1. Thus ask what kind of partition gives us the partition of one into one part after this complementation process. Problem 170 Show that P(k,n)P(k,n) is at least 1 n!(k−1 n−1).1 n!(k−1 n−1). Hint How many compositions are there of (k) into (n) parts? What is the maximum number of compositions that could correspond to a given partition of (k) into (n) parts? With the binomial coefficients, with Stirling numbers of the second kind, and with the Lah numbers, we were able to find a recurrence by asking what happens to our subset, partition, or broken permutation of a set S S of numbers if we remove the largest element of S.S. Thus it is natural to look for a recurrence to count the number of partitions of k k into n n parts by doing something similar. Unfortunately, since we are counting distributions in which all the objects are identical, there is no way for us to identify a largest element. However if we think geometrically, we can ask what we could remove from a Young diagram to get a Young diagram. Two natural ways to get a partition of a smaller integer from a partition of n n would be to remove the top row of the Young diagram of the partition and to remove the left column of the Young diagram of the partition. These two operations correspond to removing the largest part from the partition and to subtracting 1 from each part of the partition respectively. Even though they are symmetric with respect to conjugation, they aren't symmetric with respect to the number of parts. Thus one might be much more useful than the other for finding a recurrence for the number of partitions of k k into n n parts. Problem 171 In this problem we will study the two operations and see which one seems more useful for getting a recurrence for P(k,n).P(k,n). (a) How many parts does the remaining partition have when we remove the largest part (more precisely, we reduce its multiplicity by one) from a partition of k k into n n parts? What can you say about the number of parts of the remaining partition if we remove one from each part? Hint These two operations do rather different things to the number of parts, and you can describe exactly what only one of the operations does. Think about the Young diagram. (b) If we remove the largest part from a partition, what can we say about the integer that is being partitioned by the remaining parts of the partition? If we remove one from each part of a partition of k k into n n parts, what integer is being partitioned by the remaining parts? (Another way to describe this is that we remove the first column from the Young diagram of the partition.) Hint Think about the Young diagram. In only one of the two cases can you give an exact answer to the question. (c) The last two questions are designed to get you thinking about how we can get a bijection between the set of partitions of k k into n n parts and some other set of partitions that are partitions of a smaller number. These questions describe two different strategies for getting that set of partitions of a smaller number or of smaller numbers. Each strategy leads to a bijection between partitions of k k into n n parts and a set of partitions of a smaller number or numbers. For each strategy, use the answers to the last two questions to find and describe this set of partitions into a smaller number and a bijection between partitions of k k into n n parts and partitions of the smaller integer or integers into appropriate numbers of parts. (In one case the set of partitions and bijection are relatively straightforward to describe and in the other case not so easy.) Hint Here the harder part requires that, after removal, you consider a range of possible numbers being partitioned and that you give an upper bound on the part size. However it lets you describe the number of parts exactly. (d) Find a recurrence (which need not have just two terms on the right hand side) that describes how to compute P(k,n)P(k,n) in terms of the number of partitions of smaller integers into a smaller number of parts. Hint One of the two sets of partitions of smaller numbers from the previous part is more amenable to finding a recurrence than the other. The resulting recurrence does not have just two terms though. (e) What is P(k,1)P(k,1) for a positive integer k?k? (f) What is P(k,k)P(k,k) for a positive integer k?k? (g) Use your recurrence to compute a table with the values of P(k,n)P(k,n) for values of k k between 1 and 7. (h) What would you want to fill into row 0 and column 0 of your table in order to make it consistent with your recurrence. What does this say P(0,0)P(0,0) should be? We usually define a sum with no terms in it to be zero. Is that consistent with the way the recurrence says we should define P(0,0)?P(0,0)? Hint If there is a sum equal to zero, there may very well be a partition of zero. It is remarkable that there is no known formula for P(k,n),P(k,n), nor is there one for P(k).P(k). This section was are devoted to developing methods for computing values of P(n,k)P(n,k) and finding properties of P(n,k)P(n,k) that we can prove even without knowing a formula. Some future sections will attempt to develop other methods. We have seen that the number of partitions of k k into n n parts is equal to the number of ways to distribute k k identical objects to n n recipients so that each receives at least one. If we relax the condition that each recipient receives at least one, then we see that the number of distributions of k k identical objects to n n recipients is ∑n i=1 P(k,i)∑i=1 n P(k,i) because if some recipients receive nothing, it does not matter which recipients these are. This completes rows 7 and 8 of our table of distribution problems. The completed table is shown in Table 3.3.5. There are quite a few theorems that you have proved which are summarized by Table 3.3.5. It would be worthwhile to try to write them all down! The Twentyfold Way: A Table of Distribution Problems k k objects and conditions on how they are received n n recipients and mathematical model for distribution Distinct Identical 1. Distinct no conditions n k n k functions∑n i=1 S(k,i)∑i=1 n S(k,i) set partitions (≤n≤n parts) 2. Distinct Each gets at most one n k––n k _ k k-element permutations 1 if k≤n;k≤n; 0 otherwise 3. Distinct Each gets at least one S(k,n)n!S(k,n)n! onto functions S(k,n)S(k,n) set partitions (n n parts) 4. Distinct Each gets exactly one k!=n!k!=n! permutations 1 if k=n;k=n; 0 otherwise 5. Distinct, order matters(k+n−1)k––(k+n−1)k _ ordered functions∑n i=1 L(k,i)∑i=1 n L(k,i) broken permutations (≤n≤n parts) 6. Distinct, order matters Each gets at least one(k)n––(k−1)k−n–––––(k)n _(k−1)k−n _ ordered onto functions L(k,n)=(k n)(k−1)k−n–––––L(k,n)=(k n)(k−1)k−n _ broken permutations (n n parts) 7. Identical no conditions(n+k−1 k)(n+k−1 k) multisets∑n i=1 P(k,i)∑i=1 n P(k,i) number partitions (≤n≤n parts) 8. Identical Each gets at most one(n k)(n k) subsets 1 if k≤n;k≤n; 0 otherwise 9. Identical Each gets at least one(k−1 n−1)(k−1 n−1) compositions (n n parts)P(k,n)P(k,n) number partitions (n n parts) 10. Identical Each gets exactly one 1 if k=n;k=n; 0 otherwise 1 if k=n;k=n; 0 otherwise Table 3.3.5 The number of ways to distribute k k objects to n n recipients, with restrictions on how the objects are received Subsection 3.3.4 Partitions into distinct parts ¶ Often Q(k,n)Q(k,n) is used to denote the number of partitions of k k into distinct parts, that is, parts that are different from each other. Problem 172 Show that Q(k,n)≤1 n!(k−1 n−1).Q(k,n)≤1 n!(k−1 n−1). Hint How does the number of compositions of (k) into (n) distinct parts compare to the number of compositions of (k) into (n) parts (not necessarily distinct)? What do compositions have to do with partitions? Problem 173 Show that the number of partitions of 7 into 3 parts equals the number of partitions of 10 into three distinct parts. Hint While you could simply display partitions of 7 into three parts and partitions of 10 into three parts, we hope you won't. Perhaps you could write down the partitions of 4 into two parts and the partitions of 5 into two distinct parts and look for a natural bijection between them. So the hope is that you will discover a bijection from the set of partitions of 7 into three parts and the partitions of 10 into three distinct parts. It could help to draw the Young diagrams of partitions of 4 into two parts and the partitions of 5 into two distinct parts. Problem 174 There is a relationship between P(k,n)P(k,n) and Q(m,n)Q(m,n) for some other number m.m. Find the number m m that gives you the nicest possible relationship. Hint In the case (k=4) and (n=2\text{,}) we have (m=5\text{.}) In the case (k = 7) and (n = 3\text{,}) we have (m = 10\text{.}) Problem 175 Find a recurrence that expresses Q(k,n)Q(k,n) as a sum of Q(k−n,m)Q(k−n,m) for appropriate values of m.m. Hint What can you do to a Young diagram for a partition of (k) into (n) distinct parts to get a Young diagram of a partition of (k-n) into some number of distinct parts? Problem 176 Show that the number of partitions of k k into distinct parts equals the number of partitions of k k into odd parts. Hint For any partition of (k) into parts (\lambda_1\text{,}) (\lambda_2\text{,}) etc. we can get a partition of (k) into odd parts by factoring the highest power of two that we can from each (\lambda_i\text{,}) writing (\lambda_i = \gamma_i\cdot 2^k_i\text{.}) Why is (\gamma_i) odd? Now partition (k) into (2^{k_1}) parts of size (\gamma_1\text{,}) (2^{k_2}) parts of size (\gamma_2\text{,}) etc. and you have a partition of (k) into odd parts. Problem 177 Euler showed that if k≠3 j 2+j 2,k≠3 j 2+j 2, then the number of partitions of k k into an even number of distinct parts is the same as the number of partitions of k k into an odd number of distinct parts. Prove this, and in the exceptional case find out how the two numbers relate to each other. Hint Suppose we have a partition of (k) into distinct parts. If the smallest part, say (m\text{,}) is smaller than the number of parts, we may add one to each of the (m) largest parts and delete the smallest part, and we have changed the parity of the number of parts, but we still have distinct parts. On the other hand, suppose the smallest part, again say (m\text{,}) is larger than or equal to the number of parts. Then we can subtract 1 from each part larger than (m\text{,}) and add a part equal to the number of parts larger than (m\text{.}) This changes the parity of the number of parts, but if the second smallest part is (m+1\text{,}) the resulting partition does not have distinct parts. Thus this method does not work. Further, if it did always work, the case (k \ne \frac{3j^2+j}{2}) would be covered also. However you can modify this method by comparing (m) not to the total number of parts, but to the number of rows at the top of the Young diagram that differ by exactly one from the row above. Even in this situation, there are certain slight additional assumptions you need to make, so this hint leaves you a lot of work to do. (It is reasonable to expect problems because of that exceptional case.) However, it should lead you in a useful direction.
8703	https://blog.csdn.net/baidu_41899377/article/details/127346912	小数点后两位四舍五入及不四舍五入的方法_presto保留两位小数但是不是四舍五入-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小数点后两位四舍五入及不四舍五入的方法文仔酱酱酱已于 2022-10-16 13:58:44 修改阅读量1.4k收藏 1 点赞数 CC 4.0 BY-SA版权文章标签：javascript 于 2022-10-16 13:58:36 首次发布版权声明：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。本文链接：本文介绍了几种实现数值四舍五入及保留特定位数的方法，包括JavaScript中使用toFixed()进行四舍五入处理，以及通过Math.floor()结合乘除法、正则表达式等手段达到非四舍五入截断的效果。四舍五入: 第一种方法 javascript var num =2.446242342; num = num.toFixed(2); // 输出结果为 2.45 AI写代码 javascript 运行不四舍五入: 第一种方法 javascript 先把小数边整数： Math.floor(15.7784514000 100) / 100 AI写代码 javascript 运行第二种，当作字符串，使用正则匹配： javascript Number(15.7784514000.toString().match(/^\d+(?:\.\d{0,2})?/)) // 输出结果为 15.77,不能用于整数如 10 必须写为10.0000 AI写代码 javascript 运行第三种正则替换 javascript var price = '2.446242342'; price.replace(/^(\-)(\d+)\.(\d\d).$/,'$1$2.$3') console.log( price.replace(/^(\-)(\d+)\.(\d\d).$/,'$1$2.$3')) //打印2.44 AI写代码 javascript 运行确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用文仔酱酱酱关注关注 0点赞踩 1 收藏觉得还不错? 一键收藏 0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关于toFixed()保留两位小数--四舍五入不精确问题解决小柠檬爱编程 06-15 5329 近期用到小数点保留两位使用toFixed() 发现不太精确例如10.125 保留两位后为10.13 但使用 var num = 10.125; num.toFixed(2) 返回的结果为10.12 经测试发现大于5会进1 等于5并不会进1 解决办法写个公共方法需要时候调用即可示例: 根据自己需要酌情处理:我是写在main.js中,仅供参考: Vue.prototype.fixed = function (num) { var times = Math.pow(10, 2); va 【presto】函数大全 fuyun6363的博客 03-27 2万+ presto 函数大全参与评论您还未登录，请先登录后发表或查看评论 ...为保留两位小数并拼接百分号_ presto 保留两位小数 9-5 先把数字转换为保留两位小数, 然后转换为字符串, 然后拼接上百分号 cast(cast(todayasdecimal(10,2))asvarchar)\|\|'%' AI写代码sql 四舍五入并且去掉小数点后面多余的零_lua去除小数后多余的0 9-24 number =round(numberpowNum)/powNum ;//四舍五入 return[NSStringstringWithFormat:@"%@",@(number)];//去掉小数点后面多余的零 } presto 字符串转浮点数、整数处理 zhuchunyan_aijia的博客 01-10 1万+ 1. 相除保留小数位 Presto 中两个整数相除会结果为零。当我们相除的两个数都为整数时，比如我想用两个count（）结果相除，想保留 n为小数 select count()1.00 / count() from student; 结果就会自动保留两位小数，想要n位就1.后面n个零 2. 字符串转整数 select cast (bj_score as bigint), bj_score from dcs_input_tab_bj_score 3. 字符串转浮点 select cast 小数点后保留两位四舍五入 iteye_5514的博客 11-29 228 public class Test { / @param args /public static void main(String[] args) { // TODO Auto-generated method stub long lng = 222325546; long lng1= 122222222; double f=lng11.0/lng; java.... ceil,floor,round,sprintf,int_floor round ceil sprintf 8-17 C知道前往体验 > ceil,floor,round,sprintf,int 这些函数的区别以及用法 presto 函数大全中文版 9-19 基于 Presto 0.215版本的中文注释函数大全,参考官方文档: db.github.io/docs/0.215/functions.html 6.1. 逻辑操作 6.2. 比较函数和运算符 6.3. 条件表达式 6.4.Lambda表达式 6.5. 转换函数 6.6. 数学函数和运算符 6.7. 位运算函数 6.8. 小数函数和运算符 ... 解决小数后两位数四舍五入的问题! qq_43097135的博客 07-03 1148 今天上午在写项目的时候出现了一个问题，关于小数点后面四舍五入的问题，这个问题常常出现在我们支付的时候，写脚本里面。现在分享给大家！我们先来看一下问题：大家可以看到我优惠卷能优惠0.9元，总计是1元，那么应付的话应该是0.1。出现了这样的问题，不要急，接着看，这是之前的源码，我是在获取到优惠卷的时候，进行了数学函数的转换。通过.tofixed(2)这个函数来解决这个问题的。.tofixed(... js浮点数保留两位小数点示例代码(四舍五入) 10-26 本文将介绍如何在 JavaScript 中保留浮点数的两位小数点，并采用四舍五入的方式实现这一需求。同时，也会提及 JavaScript 内置的toFixed()方法，它是实现四舍五入功能的一个简便方法。首先，要了解在 JavaScript 中... presto 抛出java.lang.ArithmeticException:Rounding necessary异常 9-24 文章浏览阅读300次。这个异常,是 presto 处理数据发生了精度问题,通常是数据源的精度太高 presto 读不出来,极少情况是舍位运算的时候发生了精读丢失。精度太高上一般是浮点数小数点后的位数太长。 _presto caused by: java.lang.arithmeticexception: rounding necessary at j BigDecimal 不会丢失精度的浮点数_ presto 类型转换bigdecimal 保留... 9-28 BigDecimal 不会丢失精度的浮点数本文介绍了BigDecimal类在Java中用于处理不会丢失精度的浮点数计算,包括定义、加减乘除操作示例,并展示了如何通过setScale 方法进行四舍五入。重点讲解了其在处理高精度浮点数运算中的优势。 BigDecimal 含义:不会丢失精度的浮点数... JS 保留小数点(四舍五入、四舍六入)实现思路及实例 01-19 代码如下: <!DOCTYPE HTML PUBLIC “-//W3C//DTD HTML 4.01 Transitional//EN”>floatDecimal.html<meta http-equiv=”keywords” content=... charset python3 小数位的四舍五入(用两种方法解决round 遇5 不进) 09-19 在Python 3中，进行小数位的四舍五入操作可能会遇到一个问题，即使用内置的round()函数时，当遇到需要四舍五入的数字末尾是5时，不一定会按照通常的四舍五入规则进行。这是因为round()函数在处理浮点数时会受到... ...习题和易错知识点(面试)tofixed函数 _保留两位小数面试题 8-31 1.js 不能用于做科研(因为计算有精度)但是科研用作一般互联网产品 console.log(0.1+0.2) AI写代码 javascript 运行 1 结果 2.1怎么保留几位小数(toFixed) vara=3.1451926;varb=a.toFixed(3)//保留 3位小数 console.log(b) AI写代码 javascript 运行 1 2 3 4 5... 前端 JavaScript 深度学习笔记 9-24 5大主流浏览器:1、chrome 2、IE 3、safari 4、firefox 5、opera 对应内核: 1、trident 2、 webkit blink 3、 webkit 4、gecko 5、 presto 1.2 浏览器历史1990年,蒂姆伯纳斯李超文本分享资讯的人,开发了浏览器:world wide web ,后移植到C,最后更名为 libwww/... PHP 保留两位小数并且四舍五入及不四舍五入的方法 10-26 以上就是PHP中保留两位小数以及进行四舍五入和不四舍五入操作的方法。这些技巧在处理数值时非常实用，能帮助我们得到预期的精确结果。无论是财务计算还是数据分析，理解并掌握这些函数的用法都非常重要。 Javascript 小技巧,去掉小数位并且不会四舍五入 weixin_34335458的博客 01-30 474 1: var n3 = 52.3685; 2: document.writeln(n3 >> 0);// 52 3: 可以去掉小数。 .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier,... Web前端笔试115道题(带答案及解析)_web前端开发笔试题 9-28 使用 Presto 内核的浏览器:Opera7及以上版本; 使用Webkit内核的浏览器:Safari、Chrome。 5、html5有哪些新特性?移除了哪些元素?如何处理HTML5新标签的浏览器兼容性问题?如何区分html和html5? 新增的元素有绘画 canvas ,用于媒介回放的 video 和 audio 元素,本地离线存储 localStorage 长期存储数据,浏览器关闭后数据不丢... java 四舍五入保留小数点后两位 jxs_king的博客 11-13 2194 方法一： double myNum2 = 111231.5585478; java.math.BigDecimal b = new java.math.BigDecimal(myNum2); double myNum3 = b.setScale(4, java.math.BigDecimal.ROUND_HALF_UP).doubleValue(); System.out.pr... PHP 保留两位小数但不四舍五入 JJmaiz-麦嘉俊-Concentrate on Web 12-03 749 echo substr(sprintf("%.3f", 212.65834655), 0, -1); 保留小数点后两位，四舍五入（不用printf.2f) 加0.5int取整 wssjdyhk的博客 11-11 420 保留小数点后两位，四舍五入（不用printf.2f) 加0.5int取整 C#, float.ToString()的一个坑 weixin_33935777的博客 05-13 1604 下面代码的输出竟然是2.0： float a=1.95f;Debug.Log(a.ToString("0.0")); 如果想截取一位小数，可以： float a=1.95f; float t_a=Mathf.Floor(a10)0.1f;Debug.Log(t_a.ToString("0.0")); ... 数字正则表达式（整数、小数、负数、保留两位小数等） lixyovov的博客 09-11 1万+ 原文地址: python中小数点后取2位（四舍五入）以及取2位（四舍五不入）的方法总结热门推荐 Alien-Hu 08-14 26万+ 在很多场景的计算中，最终得到的数值例如123.45678，要截取2位小数得到123.45，而不是默认的四舍五入方法得到123.46，如何实现呢？一.小数点后取2位（四舍五入）的方法方法一：round（）函数方法二：’%.2f’ %f 方法方法三：Decimal（）函数二.小数点后取2位（四舍五不入）的方法方法一：一.小数点后取2位（四舍五入... 取小数不想四舍五入怎么办 Carl_wang3333333的博客 02-29 621 【代码】取小数不想四舍五入怎么办。输入5个整形数据_数据处理之5个舍入类函数，进来康康？ weixin_39706127的博客 12-02 206 数据处理中，经常需要对数据进行一定的舍入处理。Excel存在5个常用舍入类函数，下面一一介绍给大家~ROUND功能：按指定的位数对数值进行四舍五入语法：round(数值，执行四舍五入时采用的位数)实例：通过以下实例可以看出，round()函数是一个标准的四舍五入函数。参数2大于0，则四舍五入到指定的小数位，如参数2等于1，则保留 1位小数位，等于2，则保留两个小数位；参数2等于0，则四舍五入到整数；... 前端算法仅保留后两位或者后几位小数不进位的方法 m0_72663698的博客 06-08 363 同样可以保留两位小数，但是不会进位。保留小数点后两位四舍五入最新发布 08-03 明白了，你现在的需求是： > ✅ 将高频淬火工序的工时从“秒”转为“分钟”，并保留小数点后两位，进行四舍五入。 --- ## ✅ 实现方式在 Java 中，要实现保留小数点后两位 + 四舍五入，推荐使用 Math.round() 或 BigDecimal。我们选择 BigDecimal，因为它在精度要求高的场景下更可靠。 --- ## ✅ 优化后的 getMillingTime() 方法（含四舍五入） java import java.math.BigDecimal; import java.math.RoundingMode; public class MillingTimeUtils { public static float getMillingTime(String processType, float diameter, float length) { AbsMillingWorkTime process = MillingProcessFactory.getProcess(processType); if (process == null) { System.err.println("_不_ 支持的工序类型: " + processType); return 0f; } float rawTime = process.getBaseWorkTime(diameter, length); // 只对“高频淬火”做“秒”转“分钟” if ("高频淬火".equals(processType)) { // 转换为分钟，并 _保留_ _两位_ _小数_，_四舍五入_ BigDecimal timeInMinutes = new BigDecimal(rawTime / 60f) .setScale(2, RoundingMode.HALF_UP); return timeInMinutes.floatValue(); } // 其他工序直接返回（默认是分钟） return rawTime; } // 示例调用 public static void main(String[] args) { float diameter = 55f; float length = 150f; float time = getMillingTime("高频淬火", diameter, length); System.out.printf("工时：%.2f 分钟%n", time); // 输出：工时：2.05 分钟（假设原始值是 123f 秒） } } --- ## ✅ 示例：转换逻辑说明 \| 秒数 \| 转换为分钟 \| 保留 2位小数（四舍五入） \| ------------ \| 90 \| 1.5 \| 1.50 \| \| 123 \| 2.05 \| 2.05 \| \| 125 \| 2.0833... \| 2.08 \| \| 127 \| 2.1166... \| 2.12 \| --- ## ✅ 使用说明 - ✅ BigDecimal：更精确，适合金融、工业、工时计算等场景 - ✅ setScale(2, RoundingMode.HALF_UP)：保留 2 位小数，四舍五入 - ✅ RoundingMode.HALF_UP：Java 中标准的四舍五入方式 - ✅ 不影响其他工序单位（只有高频淬火需要处理） --- ## ✅ 可选封装：格式化输出（如 String）如果你想返回 String 格式，方便显示： java public static String getMillingTimeFormatted(String processType, float diameter, float length) { float minutes = getMillingTime(processType, diameter, length); return String.format("%.2f", minutes); } 这样你可以直接用于界面显示或日志输出。 --- ## ✅ 总结 \| 需求 \| 实现方式 \| -----------\| \| 高频淬火转分钟 \| seconds / 60f \| \| 保留 2位小数 \| BigDecimal.setScale(2, RoundingMode.HALF_UP) \| \| 四舍五入 \| RoundingMode.HALF_UP \| \| 不影响其他工序 \| 仅对 "高频淬火" 做特殊处理 \| --- ✅ 最终推荐做法： > 在 getMillingTime() 中判断工序类型，如果是“高频淬火”，就做“秒转分钟”，并用 BigDecimal 保留 2位小数，四舍五入，确保精度准确。 --- 如果你希望我帮你： - ✅ 生成完整工具类（含日志、异常处理） - ✅ 生成带工厂模式的完整结构 - ✅ 或者帮你生成 JSON 配置文件来控制哪些工序需要“秒转分钟” 也可以继续告诉我，我可以为你定制完整解决方案 😊 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文仔酱酱酱博客等级码龄8年 70 原创74 点赞 142 收藏 51 粉丝关注私信 🔥码云GVP开源项目 16k starUniapp + ElementUI 功能强大支持多语言、二开方便广告分类专栏笔记2篇 city.json1篇上一篇：前端知识点下一篇： rich-text标签富文本解析最新评论高德地图开发手绘地图文仔酱酱酱: 高德地图开发手绘地图 S-Xu:求工具 vue 移动端h5引入vant 和echarts的适配文仔酱酱酱:和web端的echart一样引入。npm 下包。main,js引入 vue 移动端h5引入vant 和echarts的适配 m0_46542859:小民问一下，echart的依赖是哪个，echart怎么引入 git 提交 .gitignore 添加忽略文件不生效处理方法 CSDN-Ada助手:CS入门技能树或许可以帮到你：大家在看 UDS诊断核心知识详解旅游城市气候数据可视化分析 Android特许权限详解文献查阅的网站推荐与使用指南英文文献在哪里找：高效查找英文文献的实用方法与资源推荐最新文章使用 echarts-auto-tooltip自动播放再次获取最新数据时会出现轮播异常解决方法实现图片上图标打点，文字标注处理flex布局下。父盒子宽度不够，子盒子默认换行 2025年 4篇 2024年 8篇 2023年 11篇 2022年 18篇 2021年 27篇 2020年 2篇 🔥码云GVP开源项目 16k starUniapp + ElementUI 功能强大支持多语言、二开方便广告上一篇：前端知识点下一篇： rich-text标签富文本解析分类专栏笔记2篇 city.json1篇展开全部收起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 AI助手 AI 搜索智能体 AI 编程 AI 作业助手下载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
8704	https://ocw.mit.edu/courses/8-21-the-physics-of-energy-fall-2009/b172949e5728814eb3f2f1989e38d570_MIT8_21s09_lec10.pdf	MIT OpenCourseWare 8.21 The Physics of Energy Fall 2009 For information about citing these materials or our Terms of Use, visit: 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Phase Change Energy Conversion I 8.21 Lecture 10 September 30, 2009 2 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances • Thermodynamics of heat extraction • Phase change in pure substances • The vapor compression cycle: heat pumps, refrigeration, air conditioners • The Rankine steam cycle and steam turbines • Some implementations of the Rankine cycle • Why this now? • Thermodynamics of heat extraction • Phase change in pure substances • The vapor compression cycle: heat pumps, refrigeration, air conditioners • The Rankine steam cycle and steam turbines • Some implementations of the Rankine cycle } Part I Part II } 3 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Heat extraction devices are everywhere ⋆Household air conditioning — 5.1% of all U. S. energy use (2001) ⋆Household refrigeration — 4.4% ⋆Commercial AC and refrigeration — 7% ⋆Plus industrial cooling and “actively cooled transport”! Household energy use Kitchen energy use Several aims • Explain how simple thermodynamic cycles can move heat from low to high temperatures • Explain why cycles that make a ﬂuid change from vapor to liquid and back dominate practical applications, and how they work • Construct and evaluate the dominant cooling (“vapor compression”) and power (“Rankine”) cycles in use today 4 Image removed due to copyright restrictions. Similar graph can be found on web page EnergyMgt/EnergyPieChart_500.jpg Image courtesy of EPA. 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Heat pumps are gaining popularity for home heating ⋆Use work to move heat from cold environment to warm environment ⋆Same principle as refrigerator, except aim is to warm rather than to cool! Could describe heat extraction devices absent phase change, but ⋆Heat extraction devices almost universally employ phase change thermodynamic cycle ⋆Which are chosen for thermodynamic properties (eg. liquid ⇔vapor near ambient temperatures) eg. Freon ⋆And phase change power generation is ubiquitous 5 Turbine generator image removed due to copyright restrictions. Please see: Geopresentation/images/img038.jpg Heat pump image removed due to copyright restrictions. 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Many projects involve building solar thermal plants, which use cheaper technology than the solar panels often seen on roofs. In such plants, mirrors heat a liquid to create steam that drives an electricity-generating turbine. As in a fossil fuel power plant, that steam must be condensed back to water and cooled for reuse. 6 Image and text removed due to copyright restrictions. Please see: energy-environment/30water.html 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances • Phase change in pure substances • The vapor compression cycle: heat pumps, refrigeration, air conditioners • The Rankine steam cycle and steam turbines • Some implementations of the Rankine cycle • Why this now? • Thermodynamics of heat extraction 7 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Heat engines and heat extraction devices with ﬂowing ﬂuids • Fluid ﬂows in, bringing heat, ﬂows out removing heat, work gets done, but • In a cycle, the machine returns to its original state and does not store energy or entropy Conservation of energy around a cycle: machine does not store energy QH = QL + W Best possible: Entropy is conserved around a cycle if it’s performed reversibly: “machine does not store entropy” QL TL −QH TH = 0 CoP\|refrigeration = QL W ≤ TL TH −TL CoP\|heat pump = QH W ≤ TH TH −TL Engine “ Coefﬁcient of Performance ” (= Efﬁciency) CoP = W QH ≤TH −TL TH 8 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Reminder: Thermodynamics of an ideal engine ⋆Based on ﬂuid executing a cycle Fluid begins and ends in the same state ⋆Fluid must have same energy and entropy at beginning and end of cycle • Absorbs heat QH at TH. • Does work W • Expels heat QL at TL ⋆First Law Change in internal energy of ﬂuid around cycle must vanish QH = QL + W ⋆Second Law The entropy of the universe can never decrease ⋆When a system absorbs heat Q at temperature T , then it gains entropy ∆S ≥Q T . When it loses heat Q at temperature T it loses entropy \|∆S\| ≤Q T ⋆With entropy: “You always get more than you want and get rid of less than you hope.” ⋆And the equality holds only when heat transfer is reversible ⋆Compute ∆Suniverse ∆Suniverse ≥QL TL −QH TH ≥0 9 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Combine 1st and 2nd laws QH = QL + W QL TL −QH TH ≥0 Substitute and rearrange W QH ≤TH −TL TH “Efﬁciency” = “ Coefﬁcient of Performance ” CoP ≤TH −TL TH And maximum CoP is only reached when heat transfer is reversible: ⋆Minimize temperature and pressured gradients 10 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Thermodynamics of an ideal heat extraction device ⋆Same as engine: ﬂuid executes cycle; begins and ends in same state. ⋆Direction of heat and work ﬂows are reversed: • Absorbs heat QL at TL. • Work done on it, W • Expels heat QH at TH ⋆First Law QH = QL + W ⋆Second Law ∆Suniverse = QH TH −QL TL ≥0 Note signs: Here QH/TH is entropy delivered to universe and QL/TL is entropy removed from universe ) ) 11 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Relations among ideal CoP’s CoP\|ideal heat pump = 1 CoP\|ideal engine CoP\|ideal heat pump = CoP\|ideal refrigerator + 1 CoP\|ideal heat pump is always greater than unity and can be very large Coefﬁcient of Performance? It depends what you’re trying to accomplish ⋆Air conditioning/refrigeration: remove heat from low temperature reservoir: CoP\|refrigeration = QL W ≤ TL TH −TL ⋆Heat pump: provide heat to high temperature reservoir: CoP\|heat pump = QH W ≤ TH TH −TL 12 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Example: A refrigerator • TL = 20◦F • TH = 70◦F CoP = 266 294 −266 = 9.6 Carnot cooling cycle ⋆ [1 →2] Isentropic compression: Work is done on the gas. It heats up to TH. ⋆ [2 →3] Isothermal compression: Work is done on the gas. Heat equal to work is expelled as QH ⋆ [3 →4] Isentropic expansion: Gas does work. It cools to TL ⋆ [4 →1] Isothermal expansion: Gas does work. Heat equal to work is absorbed as QL. ⋆If steps are carried out reversibly, then it’s guaranteed to reach thermodynamic limit for CoP 13 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances 2007/11/water_drop_causing_a_ripple.jpg Cooling based on phase change ⋆Goes back to Michael Faraday in ∼1820: Liquid ammonia left to evaporate in air cools the air! ⋆Make cyclic by condensing ammonia elsewhere and expelling heat ⋆Commercialized by Willis Carrier ∼1930. ⋆Must review thermodynamics of phase change • Liquid/vapor • Phases separated by boiling or condensation curve • Tboiling(P ) or Pboiling(T ) • Other important points: Triple Point and Critical Point 14 Snow flake image removed due to copyright restrictions. Clouds image removed due to copyright restrictions. Please see: Water drop image removed due to copyright restrictions. Please see: 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances • The vapor compression cycle: heat pumps, refrigeration, air conditioners • The Rankine steam cycle and steam turbines • Some implementations of the Rankine cycle • Why this now? • Thermodynamics of heat extraction • Phase change in pure substances 15 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Phase Diagram ⋆Water — but other substances are similar ⋆Phase changes • Solid/liquid Enthalpy (latent heat) of melting/solidiﬁcation • Liquid/gas Enthalpy (latent heat) of vaporization/condensation • Solid/gas Enthalpy of sublimation ⋆Enthalpies of phase change • Enthalpy of melting (at 0◦C) ≈334 kJ/kg • Enthalpy of vaporization (at 100◦C) ≈2.26 MJ/kg ⋆Special points • Triple point: T = 273.16K, P = 611.73Pa • Critical point: T = 647K, P = 22.064MPa 16 Image from 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Why use phase change? 1. Large energy storage potential 2.26 MJ to vaporize 1 kg H2O at 100◦C 2. Energy transfer at constant temperature and pressure! Bring liquid to boiling point, then T and P stay ﬁxed until all liquid →vapor! Copious heat transfer under reversible conditions! 3. Flexibility in inducing phase transition. Adjust pressure to select working T , for example. 4. Enhanced heat transfer in boiling 17 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances 3. Choice of operating set points (T and P ) 1. and 2. Large energy storage potential and constant T and P energy transfer. • Take 1 kg water at 1 atm and 100◦and add heat (example: resistive heating element) Temperature and pressure stay the same until 2.26 MJ has been added and all liquid ⇒vapor. Conditions are ≈reversible: Phase change ceases as soon as heat is removed (turn off current) Most added heat goes into internal energy of vapor (small amount in p dV work). • Compare adding heat to 1 kg of water vapor at 100◦. Heat capacity at constant pressure: ∼2kJ/kg K. So to add same amount of heat to water vapor at constant pressure would raise temperature by ∼2.26 MJ/2 kJ ∼1000◦K! 18 Vapor pressure chart removed due to copyright restrictions. Vapor pressure graph removed due to copyright restrictions. Please see: 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances 4. Enhanced heat transfer in boiling ⋆Compare resistively heated wire in asymptotically laminar liquid ﬂow with same wire in boiling pool ⋆Two advantages: (1) vapor bubble spontaneously migrate away from surface, whereas ﬂuid ﬂow is minimal near surface due to viscosity ⋆(2) Each vapor molecule carries full enthalpy of vaporization with it as it leaves the heated surface. 0 10 20 30 Temperature difference (Ts-Tf) in 0C 1.0 2.0 3.0 Heat Flux (105 W/m2) Laminar flow over a cylinder Heat Flux (W/m2) 10 20 30 106 107 105 Temperature difference (Ts-Tf) in 0C Water Pool Boiling Heat Transfer 19 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Lee Carkner, Department of Physics and Astronomy, Augustana College • So far we looked at phase change in the pT plane. • Need to look at it in the pV and ST planes to get full description • Why? Because one point in the pT plane covers whole process of boiling (or melting) Following phase change in pV, ST, and “quality” 20 Image from 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Phase change in the pV -plane Walk along the isotherms ⋆550◦K ⋆600◦K ⋆650◦K A C D CP B Mixed Phase Isotherm • Choose a T • Slowly lower P • V increases a little • Until you reach Pboiling(T ) • Then all liquid turns to vapor With dramatic increase in volume at ﬁxed P • Then P again can decrease 21 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Saturation dome for water ⋆Phase change curve ⋆Mixed phase below the “dome” ⋆“Saturated vapor” on the right part of curve ⋆“Saturated liquid” on the left part of curve ⋆Quality: χ = mv mv + ml Saturated vapor, quality = 1 Saturated liquid, quality = 0 Mixed phase Critical point 22 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Phase change in the TS-plane Walk along the isobars Saturated vapor, quality = 1 Saturated liquid, quality = 0 Critical point Mixed phase “Superheated vapor” “Subcooled liquid” • Choose a P • Slowly add heat (entropy) • System does work against constant P And T increase (Cp) • Until you reach Tboiling(P ) • Then all liquid turns to vapor With dramatic increase in heat (enthalpy) at ﬁxed T • Then T again can increase 23 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Properties of the mixed phase ⋆To a very good approximation, entensive properties of mixed phase are merely The proportional sum of the liquid and vapor properties. ⋆For example, entropy S = mvSv + mlSl mv + ml = χSv + (1 −χ)Sl ⋆Applies to energy, entropy, enthalpy, volume ⋆Why not exact? Interface properties: 50/50 mix of liquid and vapor water has slightly different properties than a fog. ⋆Where to get properties of saturated liquid and vapor? Not from perfect gas law! Thermodynamic tables! 24 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Steam tables: For each pressure ⋆The saturation temperature ≡boiling point ⋆Properties of saturated liquid and saturated vapor at the boiling point ⋆Table of properties at other temperatures Subcooled liquid Superheated vapor 25 8.21 Lecture 10: Phase change energy conversion I Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances 0. 0.2 0.4 0.6 0.8 1.0 500 1000 1500 2000 2500 418 2675 Quality Enthalpy per kilogram 1418 ⋆Once we know χ = 0.44, (and 1 −χ = 0.56), v = 0.44(1.69) + 0.56(1.04 × 10−3) = 0.744 m3/ kg s = 0.44(7.36) + 0.56(1.30) = 3.97 kJ/kg K Example A kilojoule of heat is added to a kilogram of liquid water at 1 atmosphere pressure and 100◦C. What are the properties of the resultant mixture? ⋆Since the system is at constant pressure, the heat is added as enthalpy. ⋆1 atmosphere is close enough to 1 × 105 N/m 2 to use tables on last slide • hliq = 418 kJ/kg • hvap = 2675 kJ/kg • hmixture = 418 + 1000 kJ/kg, • Which corresponds to a quality of χ = 0.44 26 8.21 Lecture 12: Phase change energy conversion II Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances P1 P1 P2 Heating pipe Throttle ⋆Concept: control volume In a time ∆t, apply ﬁrst (and second) laws on a ﬁxed domain ⋆In a time ∆t, mass ∆m1 = ρ1A1 (v1∆t) enters and ∆m2 = ρ2A2 (v2∆t) leaves ⋆And ∆m1 = ∆m2 ⋆Entering mass brings energy ∆U = u1ρ1A1v1∆t And similarly for enthalpy and entropy entering and leaving Digression: Thermodynamics with ﬂowing ﬂuids ⋆Need to deal with devices where materials ﬂow in and out! Pipe heated by resistive coil: ﬂuid ﬂows in, heats, ﬂows out. “Throttle”: ﬂuid pushed through nozzle ⋆Fluid enters: ρ1, p1, T1, u1, h1, s1 Leaves: ρ2, p2, T2, u2, h2, s2 ⋆These are density (ρ), pressure (p), temperature (T ), speciﬁc energy (u), enthalpy (h), and entropy (s) Speciﬁc energy ≡energy per unit mass, ... 27 8.21 Lecture 12: Phase change energy conversion II Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances ρ1, p1, v1 u1, h1, s1 ρ2, p2, v2 u2, h2, s2 A1 A2 ∆Q ∆W ⋆Apply ﬁrst law to control volume ∆Ein = Internal energy\|in + pdV work\|in = u1ρ1v1A1∆t + p1A1v1∆t ∆Eout = Internal energy\|out + pdV work\|out = u1ρ2v2A2∆t + p2A2v2∆t h2 = h1 + dQ dm −dW dm Result: Enthalpy out = Enthalpy in + Heat in −Work out h2 = h1 + dQ dm −dW dm = h1 + q −w u + p ρ = U m + p m/V = 1 m(U + P V ) = h ⋆First law ∆Eout = ∆Ein + ∆Q −∆W u2ρ2v2A2∆t + p2A2v2∆t = u1ρ1v1A1∆t + p1A1v1∆t + ∆Q −∆W Divide out ∆m = ρ2v2A2∆t = ρ1v1A1∆t u2 + p2 ρ2 = u1 + p1 ρ1 + ∆Q ∆m −∆W ∆m 28 8.21 Lecture 12: Phase change energy conversion II Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Some examples! ⋆Heat exchanger (evaporator): Heat in, no work h2 = h1 + q ⋆Heat exchanger (condensor): Heat out, no work h2 = h1 −q ⋆Throttle: No heat, no work h2 = h1 ⋆Pump (adiabatic): Work in, no heat h2 = h1 + w ⋆Turbine (adiabatic): Work out, no heat h2 = h1 −w In the limit of negligible temperature diﬀerence, steps [ versible. The free expansion, step unavoidably gene this cycle is termed an “ideal” vapor-compression cycle b Notice, importantly, that all heat exchanges occur durin por condensing to liquid in step or liquid evaporatin advantages of phase change heat transfer described in Sec 5.2 Qualitative system of mechanical com s.conv.mechanical The goal of this section is to describe the mechanical co vapor compression air conditioner is constructed. Fig. f.c 14 of the vapor compression cycle, including the four key c four cycle coordinates that correspond to the positions o section. Figure 14: The ideal vapor compression cycle as a system o pressor (stage ), condenser (stage ), throttle (stage [34 conv.systemmodel The four key components in the cycle are the compres t 5.2 Qualitative system of mechanical com ss.conv.mechanical The goal of this section is to describe the mechanical co vapor compression air conditioner is constructed. Fig. f. 14 of the vapor compression cycle, including the four key four cycle coordinates that correspond to the positions o section. Figure 14: The ideal vapor compression cycle as a system o pressor (stage ), condenser (stage ), throttle (stage [34 f.conv.systemmodel The four key components in the cycle are the compres orator. Figure 14: The ideal vapor compression cycle as a system of mechanical components: The com pressor (stage ), condenser (stage ), throttle (stage ) and evaporator (stage ). f.conv.systemmodel The four key components in the cycle are the compressor, condenser, throttle, and evap orator. at the low temperature operating point of th 6.4 System of mechanical comp Moving beyond thermodynamic abstraction neering systems form, showing the four key coordinates that correspond to the positions Figure 19: The ideal Rankine cycle as a system nents in the cycle are the pump (stage ), boil (stage ). Rsystemmodel A pump is a machine that is closely relate section on the vapor compression cycle. W d i d f i i li id mixture with quality less than one. Finally, the low quality vapor from point undergoes an isothermal condensation, re-jecting heat to the external environment and returning back to point as nearly all liquid at the low temperature operating point of the cycle. 6.4 System of mechanical components Moving beyond thermodynamic abstraction, Fig. Rsystemmodel 19 represents the Rankine cycle in engi-neering systems form, showing the four key components of the cycle, and the four cycle coordinates that correspond to the positions on the ST and pV diagrams. Figure 19: The ideal Rankine cycle as a system of mechanical components. The four key compo-nents in the cycle are the pump (stage ), boiler (stage ), expander (stage ) and condenser ( t ) 29 8.21 Lecture 12: Phase change energy conversion II Qulaity and mixed phase Thermo for ﬂowing ﬂuids Vapor compression cycle I Introduction: Why now? Thermodynamics of heat extraction Phase change in pure substances Summary • An engine cycle run backwards ⇒ a refrigerator or a heat pump • CoP heat pump = TH ∕ (TH – TL) = 1 ∕ CoPengine CoP AC = TL ∕ (TH – TL) • Phase change takes place at constant temperature and pressure • Phase change working ﬂuid: (1) High heat capacity; (2) heat transfer at constant T; (3) wide range of (T, p) set points; (4) rapid energy transfer. • Phase change in (T, p), (p,V) and (S,T) planes. • Quality • Saturated vapor, saturated liquid, superheated vapor, subcooled liquid • Quality calculations: properties of the mixed phase are additive (enthalpy for example) • When ﬂuid moves through a device follow the enthalpy! χ = mv mv + ml hmixed(χ) = χhvapor + (1 −χ)hliquid hout = hin + ∆Q ∆m −∆W ∆m HEAT ADDED WORK DONE 30
8705	https://www.sohu.com/a/245754184_563322	高考数学问道篇 \| 导数与对称性的联系？在高中数学当中，我们研究导函数，通常是为了研究原函数的单调性，进而解决极值最值或取值范围等问题。对于常见的关于经过原点的奇函数f(x)=-f(-x)，关于坐标原点中心（点）对称；偶函数f(x)=f(-x)，关于y轴对称。那么，他们的导函数有什么特征呢？相信大部分老师都知道，奇函数的导函数是偶函数，偶函数的导函数是奇函数。这也是我们在这里想给大家带来的导函数与其对称性的关系。简单的证明一下：偶函数的时候同样的证明方法得出，偶函数的导数是奇函数。在证明过程中，我们的小陈博士也出现了一些不严谨的地方，是什么呢？各位老师也一起思考一下，在视频当中，我们《高考数学百题大过关》的名师主编张瑞炳老师就犀利的指出了其中的不足，同时他希望，对这类简单而又不好理解的知识应当给学生做了解，这也是培养学生能力的一种方法。让学生来证明这个简单的数学逻辑定理，学生不一定都能证明出来，让会的同学加深印象，让不会的同学重新学习，不也是我们教学所需要的吗?那么我们能从一个好学生身上能学到什么好的品质传达给我们的学生呢？看完视频或许你能找到答案。对文章视频有想法和建议欢迎留言或者联系我们。网站地图
8706	https://www.scribd.com/document/614699629/1001-4170	The Sum of Digits of $N$ and $N 2$: January 2010 \| PDF \| Theorem \| Mathematical Proof Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 291 views 17 pages The Sum of Digits of $N$ and $N 2$: January 2010 This document summarizes research on the sum of digits of n and n^2 in different bases. It generalizes previous work that examined when the sum of digits base 2 of n equals the sum of digits… Full description Uploaded by Gantumur Choijilsuren AI-enhanced title and description Go to previous items Go to next items Download Save Save 1001.4170 For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save 1001.4170 For Later You are on page 1/ 17 Search Fullscreen See discussions, stats, and author profiles for this publication at: The sum of digits of $n$ and $n^2$ Article · January 2010 Source: arXiv CITATIONS 6 READS 719 3 authors:Some of the authors of this publication are also working on these related projects: Self Affine Sets View project Odd Perfect Numbers View project Kevin G. Hare University of Waterloo 92 PUBLICATIONS 649 CITATIONS SEE PROFILE Shanta Laishram Indian Statistical Institute 56 PUBLICATIONS 302 CITATIONS SEE PROFILE Thomas Stoll University of Lorraine 52 PUBLICATIONS 226 CITATIONS SEE PROFILE All content following this page was uploaded by Thomas Stoll on 05 May 2014. The user has requested enhancement of the downloaded file. adDownload to read ad-free arXiv:1001.4170v1 denote the sum of the digits in the q -ary expansion of an integer n . In 2005, Melﬁ exami ned the structu re of n such that s 2 ( n ) = s 2 ( n 2 ). W e extend this stud y to the more general case of generic q and polynomi als p ( n ), an d ob tai n, in par tic ula r, a reﬁ nem en t of Melﬁ’s res ult. W e als o give a mor e detailed analysis of the special case p ( n ) = n 2 , looking at the subsets of n where s q ( n ) = s q ( n 2 ) = k for ﬁxed k . 1. Introduction Let q ≥ 2 and denote by s q ( n ) the sum of digits in the q -ary represen-tation of an integer n . Recently, considerable progress has been made towards understanding the interplay between the sum-of-digits of some algebraically deﬁned sequences, such as primes and polynomialsor, in particular, squares. In the latter, C. Maudu it and J. Riv at proved an asymptotic expansion of the sum of digits of squares in arit hme tic prog res sio ns. The ir proof heav ily relies on good est ima tes of quadratic Gauss sums. F or the case of general polynomial s p ( n ) of degree h > 2 there is still a great lack of knowledge regarding their distribution with respect to digitally deﬁned functionals .Several authors studied the pointwise properties and relationships of s q ( p ( n )), e.g., K. Stolarsky , B. Lindstr¨om , G. Melﬁ , and M. Drmota and J. Rivat. In particular, a conjecture of Stolarskyabout some extremal distribution properties of the ratio s q ( p ( n )) /s q ( n )has been recen tly settled by the authors. Melﬁ proposed to study the set of n ’s such that s 2 ( n 2 ) = s 2 ( n ), and he obtained that(1) #  n <N : s 2 ( n 2 ) = s 2 ( n )  ≫ N 1 / 40 . Using heuristic arguments, Melﬁ conjectured a much stronger result that(2) #  n <N : s 2 ( n 2 ) = s 2 ( n )  ≈ N β log N with β ≈ 0 . 75488 ... , giving an explicit formula for β . The aim of the present paper is to provide a generalization to general p ( n ) and base q K.G. Hare was partially supported by NSERC.Computational support provided by CFI/OIT grant.Th. Stoll was partially supported by an APART grant of the Austrian Academy of Sciences. 1 adDownload to read ad-free 2 KE VI N G. HA RE, SH AN T A LA IS HR AM, AN D TH OM AS ST OL L of Melﬁ’s result as well as to use the method of proof to sharpen Melﬁ’s exponent in (1). More ov er, we prov ide a local anal ogon, i.e., gett ing a lower bound for the number of n ’s such that s q ( n 2 ) = s q ( n ) = k for some ﬁxed k . Theor em 1.1. Let p ( x ) ∈ Z [ x ] have degr e e at le ast 2, and p osit ive le ading co eﬃcien t. Then ther e exist s an expli citly comp utab le γ > 0 ,dependent only on q and p ( x ) , such that (3) #  n < N, q ∤ n : \| s q ( p ( n )) − s q ( n ) \| ≤ q − 1 2  ≫ N γ , where the implied constant depends only on q and p ( x ) . This result is given in Section2.In the general case of q -ary digits and polynomia ls p ( x ), the bound ( q − 1) / 2 in (3) cannot be improved.This is easily seen by recalling the well-known fact(4) s q ( n ) ≡ n mod ( q − 1) . Indeed, if we set p ( x ) = ( q − 1) x 2 + x + a for a ∈ N then we ﬁnd that s q ( p ( n )) − s q ( n ) ≡ p ( n ) − n ≡ a mod ( q − 1)which could be any of 0 , 1 ,...,q − 2 depending only on the choice of a .The method of proof of Theorem (1.1) allows to improve on Melﬁ’s result (1). Theorem 1.2. (5) #  n <N : s 2 ( n 2 ) = s 2 ( n )  ≫ N 1 / 19 . Following on Melﬁ’s paper, we examine the case when p ( n ) = n 2 and q = 2 in mo re deta il. W e cons id er the set of al l n ’s such that s 2 ( n ) = s 2 ( n 2 ), and partition the set into the subsets dependent upon the value of s 2 ( n ). By no ti ci ng th at s 2 ( n ) = s 2 (2 n ) and s 2 ( n 2 ) = s 2 ((2 n ) 2 ) we see that we can restrict our attention to odd n . Theorem 1.3. Let k ≤ 8 . Then { n <N, n odd : s 2 ( n 2 ) = s 2 ( n ) = k } is a ﬁnite set. This was done by explicit computation of all such n which are given in Tables1a nd2. A discussion of how these computations were made is given in Section3.Based on these initial small values of k , one might expect that this is always true. Let(6) n (2) = 11011 11 00 ... 00    r 1101111 be writt en in base 2. The n s 2 ( n ) = s 2 ( n 2 ) = 12 for all r ≥ Th is is in fact a special case of a more general property. adDownload to read ad-free THE SUM OF DIGITS OF n AND n 2 3 Theorem 1.4. Let k ≥ 16 or k ∈ { 12 , 13 } . Then { n <N, n odd : s 2 ( n 2 ) = s 2 ( n ) = k } is an inﬁnite set. The proof of this result is given in Section4.Despite of great eﬀort we are not able to decide the ﬁniteness problem in the remaining cases k ∈ { 9 , 10 , 11 , 14 , 15 } . Ho we v er, we will comm en t on some heuris tic evidence that it seems unlikely that there are inﬁnitely many solutions in the cases k = 9 and k = 10, respectively, in Section5.Somewhat surprisingly, a similar answer can be given if q ≥ 3. Theorem 1.5. Let q ≥ 3 and assume k ≥ 94( q − 1) . Then the equation (7) s q ( n 2 ) = s q ( n ) = k has inﬁnitely many solutions in n with q ∤ n if and only if (8) k ( k − 1) ≡ 0 m o d ( q − 1) . We show this result in Section6.2. Proof of Theorems1.1and1.2 Following Lindstr¨om we say that terms are noninterfering if we can use the following splitting formulæ: Proposition 2.1. F or 1 ≤ b <q k and a,k ≥ 1 , s q ( aq k + b ) = s q ( a ) + s q ( b ) , (9) s q ( aq k − b ) = s q ( a − 1) + ( q − 1) k − s q ( b − 1) . (10) Proof. See .  Proof of Theorem1.1: The proof uses a construction of a sequence with noninterfering terms which has already been used in . However,to obtain the bound N γ in (3) instead of a logarithmic bound, we have to make a delicate reﬁnement. To begin with, deﬁne the polynomial t m ( x ) = mx 4 + mx 3 − x 2 + mx + m where m ∈ Z . Set m = q l − r with 1 ≤ r ≤ ⌊ q αl ⌋ , q ∤ r and 0 <α < 1.Obviously, for α < 1 there exists l 0 ( α ) such that for all l >l 0 ( α ) we have m ≥ Furthermore let k be such that q k m . By consecutively employing (9) and (10) we see that s q ( t m ( q k )) = ( q − 1) k + s q ( m − 1) + 3 s q ( m )= ( q − 1) k + s q ( q l − ( r 1)) + 3 s q ( q l − r )= ( q − 1) k ( q − 1) l − s q ( r ) + 3(( q − 1) l − s q ( r − 1))(11) ≤ ( q − 1) k 4( q − 1) l. adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Number Theory BSC Notes PDF 50% (2) Number Theory BSC Notes PDF 24 pages 2023 Ant Notes No ratings yet 2023 Ant Notes 50 pages The Inter-Universal Teichmüller Theory and New Diophantine Results Over The Rational Numbers. 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8707	https://www.kany.es/files/probability.pdf	EECS 126 Notes Kanyes Thaker Spring 2019 0.1 Introduction This document is an overview of EECS 126, Probability and Random Pro-cesses, at UC Berkeley. These notes are largely based oﬀof Introduction to Probability by Dimitris P. Bertsekas and John N. Tsitsiklis, and lectures by Shyam Parekh. This is not an introductory class to probability, and these notes assume a basic understanding of probability from CS70, STAT134, or similar. These are not a replacement for lectures, labs, or discussions, but should solid enough for review! 1 Contents 0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 Sample Spaces and Probability 5 1.1 Probabilistic Models . . . . . . . . . . . . . . . . . . . . . . . 6 1.2 Conditional Probability . . . . . . . . . . . . . . . . . . . . . 7 1.3 Total Probability Theorem and Bayes’ Rule . . . . . . . . . . 7 1.4 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.5 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2 Discrete Random Variables 10 2.1 Probability Mass Functions . . . . . . . . . . . . . . . . . . . 10 2.1.1 The Bernoulli Random Variable . . . . . . . . . . . . 10 2.1.2 The Binomial Random Variable . . . . . . . . . . . . . 11 2.1.3 The Geometric Random Variable . . . . . . . . . . . . 11 2.1.4 The Poisson Random Variable . . . . . . . . . . . . . 12 2.2 Functions of Random Variables . . . . . . . . . . . . . . . . . 13 2.3 Expectation, Mean, and Variance . . . . . . . . . . . . . . . . 13 2.3.1 Variance, Moments, and the Expected Value Rule . . 14 2.3.2 Properties of Mean and Variance . . . . . . . . . . . . 14 2.3.3 Mean and Variance of Some Common Random Variables 14 2.4 Joint PMFs of Multiple Random Variables . . . . . . . . . . . 15 2.4.1 Functions of Multiple Random Variables . . . . . . . . 15 2.5 Conditioning . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.5.1 Conditioning a Random Variable on an Event . . . . . 16 2.5.2 Conditioning one Random Variable on Another . . . . 16 2.6 Conditional Expectation . . . . . . . . . . . . . . . . . . . . . 17 2.7 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3 General Random Variables 18 3.1 Exponential Random Variables . . . . . . . . . . . . . . . . . 19 3.2 Cumulative Distribution Functions . . . . . . . . . . . . . . . 19 3.3 Normal Random Variables . . . . . . . . . . . . . . . . . . . . 20 3.4 Joint PDFs of Multiple Random Variables . . . . . . . . . . . 21 3.5 Joint CDFs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.6 Conditioning . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.7 Conditional Expectation . . . . . . . . . . . . . . . . . . . . . 23 3.8 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.9 Continuous Bayes . . . . . . . . . . . . . . . . . . . . . . . . . 24 2 4 Further Topics on Random Variables 25 4.1 Derived Distributions . . . . . . . . . . . . . . . . . . . . . . 25 4.2 Convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 4.3 Covariance and Correlation . . . . . . . . . . . . . . . . . . . 26 4.4 Conditional Expectation and Variance, Revisited . . . . . . . 26 4.5 Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 4.6 Order Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . 27 5 Limit Theorems 29 5.1 The Markov and Chebyshev Inequalities . . . . . . . . . . . . 29 5.2 The ChernoﬀBound . . . . . . . . . . . . . . . . . . . . . . . 30 5.3 The Weak Law of Large Numbers . . . . . . . . . . . . . . . . 31 5.4 Convergence in Probability . . . . . . . . . . . . . . . . . . . 31 5.5 The Central Limit Theorem . . . . . . . . . . . . . . . . . . . 32 5.5.1 The De Moivre-Laplace Approximation of the Binomial 32 5.6 The Strong Law of Large Numbers . . . . . . . . . . . . . . . 33 5.6.1 The Borel-Cantelli Lemma . . . . . . . . . . . . . . . 34 5.7 Binary Erasure Channels . . . . . . . . . . . . . . . . . . . . 34 6 Discrete Time Markov Chains 36 6.1 Discrete-Time Markov Chains . . . . . . . . . . . . . . . . . . 36 6.1.1 The Probability of a Path . . . . . . . . . . . . . . . . 37 6.1.2 k-Step Transition Probabilities . . . . . . . . . . . . . 37 6.2 Classiﬁcation of States . . . . . . . . . . . . . . . . . . . . . . 37 6.2.1 Positive Recurrence and Null Recurrence . . . . . . . 39 6.3 Steady-State Behavior . . . . . . . . . . . . . . . . . . . . . . 40 6.4 Reversibility of Markov Chains . . . . . . . . . . . . . . . . . 42 7 The Bernoulli and Poisson Processes 43 7.1 The Bernoulli Process . . . . . . . . . . . . . . . . . . . . . . 43 7.1.1 Interarrival Times . . . . . . . . . . . . . . . . . . . . 43 7.1.2 Splitting and Merging of Bernoulli Processes . . . . . 44 7.2 The Poisson Process . . . . . . . . . . . . . . . . . . . . . . . 44 7.2.1 Poisson Splitting and Merging . . . . . . . . . . . . . 46 7.2.2 Sums of Random Variables . . . . . . . . . . . . . . . 47 7.2.3 The Random Incidence Paradox . . . . . . . . . . . . 47 8 Erd¨ os-R´ enyi Random Graphs 48 8.1 Sharp Threshold for Connectivity . . . . . . . . . . . . . . . . 48 3 9 Bayesian Statistical Inference 49 9.1 Bayesian Inference . . . . . . . . . . . . . . . . . . . . . . . . 49 9.2 Point Estimation, Hypothesis Testing, and the MAP Rule . . 50 9.2.1 Point Estimation . . . . . . . . . . . . . . . . . . . . . 51 9.2.2 Hypothesis Testing . . . . . . . . . . . . . . . . . . . . 52 9.3 Bayesian Least Mean Squares . . . . . . . . . . . . . . . . . . 52 10 Classical Parameter Estimation 53 10.1 Classical Parameter Estimation . . . . . . . . . . . . . . . . . 53 10.1.1 Maximum Likelihood Estimation (MLE) . . . . . . . . 53 10.2 Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . 54 11 Hilbert Spaces, Estimation, and Kalman Filtering 56 11.1 A Brief Review of Linear Algebra . . . . . . . . . . . . . . . . 56 11.2 Inner Product Spaces and Hilbert Spaces . . . . . . . . . . . 56 11.3 Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 11.4 Gram-Schmidt Orthonormalization . . . . . . . . . . . . . . . 57 11.5 Linear Least Squares Estimate (LLSE) . . . . . . . . . . . . . 58 11.6 Minimum Mean Square Estimation (MMSE) . . . . . . . . . 59 11.6.1 Jointly Gaussian Random Variables . . . . . . . . . . 59 11.7 Kalman Filtering . . . . . . . . . . . . . . . . . . . . . . . . . 60 4 1 Sample Spaces and Probability Probability is an attempt to discuss an uncertain situation. It’s not a con-cept that’s uniformly or universally shared or understood, and we study it in order to create a more concrete understanding of uncertainty. Some people deﬁne probability as a frequency of occurrence, where we try to examine the number of successful occurrences in a large number of trials. We can also deﬁne it as an expression of subjective belief – what would you say the probability is that you’ll do your laundry today? Here we aim to construct a more well-deﬁned notion of probability. Sets A set is a collection of objects (the elements of the set), denoted as S = {x1, ..., xn}. If there exists some mapping f : N 7→S, we call the set countably inﬁnite. We can denote a set by the property it satisﬁes, i.e. S = {x\|P(x)}. If the set does not have elements which can be enumerated by the positive integers (i.e. a continuous range) we call the set uncountable. The universal set is denoted as Ω, and typically we are concerned with S ⊆Ω. The complement of S relative to Ωis the set {x ∈Ω\|x ̸∈S}. The union of two sets is the set of elements which belong to either set or both sets (S ∪T ), and the intersection of two sets is the set of elements which belong to both sets (S ∩T ). We have a general notation for the unions and intersections of multiple sets. ∞ [ n=1 Sn = S1 ∪S2 ∪. . . = {x\|∃Sn ∈S : x ∈Sn} ∞ \ n=1 Sn = S1 ∩S2 ∩. . . = {x\|∀Sn ∈S : x ∈Sn}. Two sets are disjoint if their intersection is empty. A collection of sets is a partition of S if S Sn = S and T S = ∅. Two of the most important properties of sets are given by DeMorgan’s Laws: 5 DeMorgan’s Laws: [ c Sn !c = \ n Sc n, \ n Sn !c = [ n Sc n. 1.1 Probabilistic Models Probabilistic Models consist of a sample space Ωand a probability law that assigns a probability P(A) to each event A that encodes our knowledge about the “likelihood” of the elements in A. In the real world, we use ex-periments (3 coin ﬂips, 2 dice rolls, etc.) in order to produce exactly one of the several possible outcomes. Elements in a sample space should be mutually exclusive and also collectively exhaustive – they shouldn’t overlap and together they should encompass every possible outcome. Probability Axioms: 1. (Nonnegativity) P(A) ≥0 for every event A 2. (Additivity) If A and B are two disjoint events, then the probability of their union satisﬁes P(A ∪B) = P(A) ∪P(B). More generally, for a sequence of disjoint events, P(S Ai) = P PAi. 3. (Normalization) The probability of the entire sample space is 1, that is P(Ω) = 1. Properties of Probability Laws: Consider a probability law and events A, B, and C. 1. If A ⊂B, then P(A) ≤P(B). 2. P(A ∪B) = P(A) + P(B) −P(A ∩B) 3. P(A ∪B) ≤P(A) + P(B) 4. P(A ∪B ∪C) = P(A) + P(Ac ∩B) + P(A∪∩Bc ∩C) 6 We can generalize property (c) in the following: The Union Bound P n [ i=1 Ai ! = n X i=1 P(Ai). 1.2 Conditional Probability Conditional probability gives us a way to reason about an experiment based on partial information. More exactly, given an experiment, sample space, and probability law, say that we know the outcome is part of some event B. We want to know the likelihood that the outcome also belongs to a diﬀerent event A. In other words, we want to know the conditional probability of A given B, or P(A \| B). Conditional probabilities can be thought of as a probability law on a new universe B, since the questions we’re asking are entirely focused on B. We can logically reason that the probability of A given B would be the probability of being in the intersection of A and B divided by the total probability of B, or: P(A\|B) = P(A ∩B) P(B) ⇐ ⇒P(A ∩B) = P(B)P(A\|B). Using this, we can visualize the probability of a large event A as the inter-section of the probability of A1, ..., An, where Ai is one of many events that need to happen for A to happen. This is expressed here: Multiplication Rule P n \ i=1 Ai ! = P(A1) n Y i=2 P  Ai\| i−1 \ j=1 Aj   1.3 Total Probability Theorem and Bayes’ Rule Let A1, ..., An be a partition of Ω, i.e. every single point in Ωis in exactly one of A1, ..., An. Then for any event B, P(B) = P(A1 ∩B) + . . . + P(An ∩B) 7 We’ve discussed the conditional probability of P(A\|B), but we would like to relate this to a similar quantity, P(B\|A). We have a nicely stated rule that helps us form this relationship. Bayes’ Rule P(Ai\|B) = P(Ai)P(B\|Ai) P(B) . Bayes’ rule is often used in inference. We often have a large number of causes that may result in a certain eﬀect – we examine the eﬀect and try to determine the likelihood of each probable cause. P(B\|Ai) is the probability that B occurs given the cause is Ai. Conversely, given that eﬀect B has been observed, we want to ﬁnd P(Ai\|B) that the cause is in fact Ai. We refer to P(Ai\|B) as the posterior of Ai, and we refer to P(Ai) as the prior. 1.4 Independence Conditional probability (P(A\|B)) Allows us to capture the information that B provides about event A. We then have to consider the case when B provides no such information. Two events A and B are independent if P(A ∩B) = P(A)P(B). If P(B) > 0, independence is equivalent to the condition P(A\|B) = P(A), meaning that the fact that B occurred doesn’t give us new information on whether or not A happened. If A and B are independent, then A and Bc are 8 also independent. Two events A and B are conditionally independent to another event C with P(C) > 0 if P(A ∩B\|C) = P(A\|C)P(B\|C). If P(B ∩C) > 0, then we can further equate this to P(A\|B ∩C) = P(A\|C). Keep in mind that just because two events A and B are independent, they are not necessarily conditionally independent. We say that several events A1, A2, ..., An are independent if for every sub-set S of {1, 2, ..., n}, P \ i∈S ! = Y i∈S P(Ai). If any pair of events within a set of events are independent of each other, they are pairwise independent. Note that just because a set of events are pairwise independent, they aren’t necessarily independent (and vice versa). Why? The intuition behind independence is that for a group of events, removing any number of those events doesn’t eﬀect the probability of the remaining events. 1.5 Counting Here are some basic counting rules: 1. Permutations of n objects: n! 2. k-Permutations of n objects: n!/(n −k)! 3. Combinations of k of n objects: n k = n!/[k!(n −k)!] 4. Partitions of n objects into r grouops, with the ith group having ni objects: n n1,n2,...,nr = n!/(n1!n2! . . . nr!) 9 2 Discrete Random Variables In some probabilistic models, the outcomes are always numerical – they’re prices, instrument readings, or gathered data. In many others, however, the outcomes aren’t numerical but they might be related to a number. For instance, we might be looking at how many students get a certain GPA. When dealing with data like this, we want to assign numbers to the notion that a certain event has a certain outcome. We introduce for this the con-cept of the random variable.Given an experiment and a set of outcomes, a random variable assigns each outcome to a number (the value of the random variable). A random variable could be “the sum of two rolls of a die” or “the time needed to make a trip.” While these concepts themselves aren’t numerical, they are tied to real numbers, and random variables assign numbers to those concepts. In more formal terms, a random variable is a function f : Ω7→R of the experimental outcome. A random variable is discrete if the set of values it can take on is either ﬁnite or countably inﬁnite. Random variables that can take on an inﬁnite number of values (i.e. “a2 where a is drawn from [−1, 1]”) are not discrete. 2.1 Probability Mass Functions We are most interested in knowing the probabilities of each of the possible values of the random variable. For a random variable X, we assign the probability mass function (PMF) of X, pX. More speciﬁcally, if x is a value that X can take on, the probability mass of x is pX(x), or the probability of the event {X = x}: pX(x) = P[X = x]. Note that we must follow the additivity and normalization axioms of proba-bility laws, meaning that the events X = x are disjoint and form a partition of Ω, and additionally X x pX(x) = 1. Finding the PMF is simple – for each value x ∈X, collect all the possible values that could result in X = x, and sum their probabilities to get pX(x). 2.1.1 The Bernoulli Random Variable Consider an event with only two possible outcomes, such as the ﬂip of a coin. The Bernoulli random variable takes on the values 1 or 0, with probabilities 10 of p and 1 −p, respectively. Formally, X = ( 1 if a head 0 if a tail. Bernoulli PMF pX(x) = ( p if k = 1 1 −p if k = 0. While the Bernoulli random variable is extremely simple to understand, its real power comes from when several Bernoulli random variables are com-bined. 2.1.2 The Binomial Random Variable Instead of ﬂipping a single coin, now we ﬂip n coins, each with probability p of heads, independent of each other. Let X be the number of heads after n tosses. X is a binomial random variable with parameters n and p. The PMF of X is as follows: Binomial PMF pX(k) = P(X = k) = n k pk(1 −p)n−k, k = 0, 1, ..., n. There are n k ways to select which of the n coins could be heads, and the probability that they will is pk(1 −p)n−k. Note that here we follow the normalization property as well: n X k=0 n k pk(1 −p)n−k = 1. 2.1.3 The Geometric Random Variable Let’s ask a diﬀerent question now: Let’s toss the same coin over, where P[heads] = p. Then X =the number of tosses for a head to come up for the ﬁrst time is a geometric random variable. Geometric PMF pX(k) = (1 −p)k−1p, k = 0, 1, ..., 11 since we’re asking for the probability that there are k −1 consecutive tails and then 1 heads on the kth ﬂip. Again we can see that this follows the normalization axiom: ∞ X k=1 (1 −p)k−1p = p ∞ X k=0 (1 −p)k = p · 1 1 −(1 −p) = 1. Moving away from the coin ﬂip for a second, think of the geometric random variable as doing something over and over until we hit a “success”, where “success” is a loose deﬁnition (basically the ﬁrst time something “meaning-ful” happens). 2.1.4 The Poisson Random Variable Let’s deﬁne the PMF of the Poisson random variable before we provide an intuitive explanation for what it means. Poisson PMF pX(k) = e−λ λk k! , k = 0, 1, 2, ..., where λ > 0 characterizes the PMF. Again, this satisﬁes normalization: ∞ X k=0 e−λ λk k! = e−λ 1 + λ + λ2 2! + . . . = e−λeλ = 1. Think of a binomial random variable with a very small p and very large n. Let X be the number of typos in a book with n words. X is binomial (either a word is a typo or it isn’t), but p =the probability a word is misspelled is very small. Instead of bothering with the complicated system of combinations associated with the binomial PMF, we can approximate X using the Poisson PMF. In general, the Poisson PMF is with parameter λ = np is a good approximation for the binomial PMF with parameters n and p. Here n is very large and p is very small, i.e. e−λ λk k! ≈ n! k!(n −k)!pk(1 −p)n−k, k = 0, 1, ..., n. 12 2.2 Functions of Random Variables Given a certain random variable X, we can generate more random variables by transforming X. For example, let X represent the temperature in degrees Celsius, and Y be the temperature in degrees Fahrenheit – then Y = 1.8X + 32. Here Y is a linear function in X, i.e. Y = g(X) = aX + b. We also consider nonlinear functions of the general form Y = g(X). If X is a random variable, then Y is also a random variable, because it is still taking outcomes from a probability space and assigning values to them. This means that for every value pX(x), we can ﬁnd an equivalent pY (y) by summing all x ∈X such that g(x) = y. pY (y) = X {x\|g(x)=y} pX(x). 2.3 Expectation, Mean, and Variance The PMF of X gives us several numbers, all of the probabilities of every possible value of X. However, this set of numbers is usually too descriptive to be useful. Instead, we try to express the PMF ins a single representative number. We do this through the expectation, which is a weighted (by probability) average of all possible values of X. Expectation The expected value, expectation, or mean of a random variable X with PMF pX(x) is E[X] = X x xpX(x). We interpret the mean as a “representative” value of X, somewhere in the middle of the range. If we take the “mass” portion of “probability mass function” a bit more literally, the expectation is the center of gravity of the PMF. If you’ve taken a physics class, it might be interesting to think about the question of where to place the fulcrum under a beam such that the beam is balanced, and relate it back to this concept! 13 2.3.1 Variance, Moments, and the Expected Value Rule Let’s examine some more metrics we can gather about the PMF. The 2nd moment of X is the expected value of X2 – more generally, the nth mo-ment is E[Xn]. The second most important quantity associated with a random variable (aside from E[X]) is the variance, Var(X). Variance Var(X) = E[(X −E[X])2] = E[X2] −E[X]2. If we examine this expression, we see that we’re measuring how far (on average) X deviates from its mean. The standard deviation is easier to interpret (since it’s in the same units as X): σX = p Var(X). Calculating (X −E[X])2 can be tricky sometimes, even though it’s possible to ﬁnd it using our function properties above and the deﬁnition of E[X]. However, we have an easier method to calculate Var(X). Let X be a random variable with PMF pX, and let g(X) be a function of X. Then E[g(X)] = X x g(x)pX(x). This is the expected value rule for functions of random variables. This simpliﬁes everything greatly – we have now that Var(X) = E (X −E[X])2 = X x (x −E[X])2pX(x). 2.3.2 Properties of Mean and Variance If Y = aX + b, then E[Y ] = aE[X] + b and Var(Y ) = a2Var(X). 2.3.3 Mean and Variance of Some Common Random Variables 1. Bernoulli(p): E[X] = p, Var(X) = p(1 −p) 2. Uniform(k): E[X] = a+b 2 , Var(X) = (b−a)(b−a+2) 12 . a and b are the bounds of the uniform distribution. These can be veriﬁed through induction. 14 3. Pois(λ): E[X] = λ, Var(X) = λ 4. Geom(p, k): E[x] = 1 p, Var(X) = 1−p p2 . 2.4 Joint PMFs of Multiple Random Variables In the real world, we often want to examine models involving multiple ran-dom variables. Consider two discrete random variables X and Y . The probabilities of the values that X and Y can take is the joint PMF of X and Y , written as pX,Y . If (x, y) is a pair of possible values of X and Y , then pX,Y = P(X = x, Y = y). Note that P(X = x, Y = y) = P(X = x ∩Y = y). We can calculate the individual PMFs of X and Y , the marginal PMFs, from the joint PMF: pX(x) = X y pX,Y (x, y), pY (y) = X x pX,Y (x, y). This comes from the law of total probability. 2.4.1 Functions of Multiple Random Variables Now that we have this concept of joint PMFs, we can extend it to functions of multiple variables. Consider a function Z = g(X, Y ). Its PMF can be calculated from the pX,Y , just like we did in the single variable case: pZ(z) = X {(x,y)\|g(x,y)=z} pX,Y (x, y). Likewise, the expected value for rule extends here as well: E[g(X, Y )] = X x X y g(x, y)pX,Y (x, y). We can generalize this: Let Y = g(X1, X2, ..., Xn). Then pY (y) = X {(x1,x2,...,xn)\|g(x1,x2,...,xn)} pX1,X2,...,Xn(x1, x2, ..., xn), E[Y ] = X x1 X x2 . . . X xn g(x1, x2, ..., xn)pX1,X2,...,Xn(x1, x2, ..., xn). Here we also explore the a crucial property of expectation. 15 Linearity of Expectation E[a1X2+a2X2+...+anXn+b] = a1E[X1]+a2E[X2]+...+anE[Xn]+b. 2.5 Conditioning Recall the conditional probabilities of events we discussed in the ﬁrst sec-tion. Here we introduce conditional PMFs, the occurrence of an event given the value of another random variable. This is nothing new, just a natural extension of the concepts introduced earlier. 2.5.1 Conditioning a Random Variable on an Event The conditional PMF of a random variable X, conditioned on an event A with P(A) > 0, is deﬁned as pX\|A(x) = P(X = x\|A) P(X = x ∩A) P(A) . Since the events that compose X are disjoint (by deﬁnition), this leads us to the interesting observation that P(A) = X x P(X = x ∩A). 2.5.2 Conditioning one Random Variable on Another If X and Y are variables in the same experiment, having knowledge of Y gives some color as to the value of X. We capture this in the conditional PMF of X given Y , which is determined by specializing A from the previous example to Y = y. pX\|Y (x\|y) = P(X = x\|Y = y), pX\|Y (x\|y) = P(X = x, Y = y) P(Y = y) = pX,Y (x, y) pY (Y = y). 16 2.6 Conditional Expectation The conditional PMF can be thought of as the normal PMF over a new universe determined by the condition. An analogous case follows for the conditional expectation (and conditional variance). The conditional expec-tation of X given an event A, P(A) > 0, is E[X\|A] = X x xpX\|A(x), E[g(X)\|A] = X x g(x)pX\|A(x). The conditional expectation of X given a value y of Y is E[X\|Y = y] = X x xpX\|Y (x\|y). From the total probability theorem, we get an equivalent total expectation theorem, E[X] = X y pY (y)E[X\|Y = y]. 2.7 Independence Independence with random variables is analogous to independence of events. The independence of a random variable from an event is similar to the independence of two events. X is independent from the event A is P(X = x and A) = P(X = x)P(A) = pX(x)P(A), ∀x. This is essentially saying that for all possible values of X = x, x and A are independent. We can extend this concept to two random variables. Two random variables are independent if pX,Y (x, y) = pX(x)pY (y) ∀x, y. Likewise, X and Y are conditionally independent given an event A if P(X = x, Y = y\|A) = P(X = x\|A)P(Y = y\|A). If X and Y are independent random variables, then E[XY ] = E[X]E[Y ] Var(X + Y ) = Var(X) + Var(Y ). 17 3 General Random Variables Lots of random variables don’t necessarily take on a ﬁnite set of discrete values. For continuous random variables, we try to measure events that can occur on an inﬁnite spectrum, such as velocities or amounts of liquid. Continuous Random Variables and PDFs Continuous Random Variables A random variable X is called continuous if there is a nonnegative function fX, the probability density function or PDF of X, P(X ∈B) = Z B fX(x)dx for every subset B of the real line. Here the integral used is the Riemann/Darboux integral from most calculus classes, and is implicitly assumed to be well-deﬁned. To be more speciﬁc, the probability that the value of X falls within a range [a, b] is P(a ≤X ≤b) = Z b a fX(x)dx. To qualify as a PDF, fX must be nonnegative (fX(x) ≥0 ∀x), and must also satisfy the normalization property: Z ∞ −∞ fX(x)dx = P(−∞< X < ∞) = 1. The expected value or mean of a continuous random variable X is E[X] = Z ∞ −∞ xfX(x)dx. This is identical to the discrete case except the PMF is replaced by the PDF and the summation is replaced by integration. Imagine a variable Y = g(X). Y is a random variable, but it isn’t necessarily continuous. It follows the expected value rule, i.e. E[g(X)] = Z ∞ −∞ g(x)fX(x)dx. 18 The nth moment of a continuous r.v. X is E[Xn], and the variance is deﬁned identically to the discrete case, as E[(X −E[X])2]. 3.1 Exponential Random Variables An exponential random variable has PDF Exponential PDF fX(x) = ( λe−λx, x ≥0 0, x < 0 An exponential random variable is a good model for the amount of time until something important takes place (a message arriving, a lightbulb burn-ing out, etc). It is closely related to the geometric random variable, which is an analog in discrete time. E[X] = 1 λ Var(X) = 1 λ2 3.2 Cumulative Distribution Functions Until right now we’ve been dealing with discrete and continuous random variables diﬀerently, with the PMF and PDF, respectively. We want to be able to describe all kinds of random variables with a single concept. Here we introduce the cumulative distribution function or CDF. Cumulative Distribution Function The CDF of X is denoted FX and provides the quantity P(X ≤x). FX(x) = P(X ≤x) = (P k≤x pX(x), X is discrete, R x −∞fX(t)dt X is continuous. FX(x) “accumulates” the probability “up to” x. Some interesting prop-erties: 1. FX is monotonically nondecreasing since the PDF and PMF are strictly nonnegative. 19 2. FX(x) tends to 0 as x − →−∞and 1 as x − →∞. 3. If X is discrete, then F(X)(x) is a piecewise constant function of x. 4. If X is continuous, then FX(x) is a continuous function of X. It is interesting to note that the CDF of the geometric (1 −(1 −p)n) and exponential (1 −e−λx) random variables are related – the CDF of the expo-nential is the limit of the CDF of the geometric. 3.3 Normal Random Variables Normal Random Variable A continuous random variable X is normal or Gaussian if its PDF is of the form fX(x) = 1 √ 2πσexp −(x −µ)2 2σ2 . Here E[X] = µ and Var(X) = σ2. A normal random variable Y with zero mean and unit variance is called standard normal. Its CDF, denoted by Φ, is Φ(y) = P(Y ≤y) = 1 √ 2π Z y −∞ exp −t2 2 dt. Because the normal distribution is symmetric, note that Φ(y) = 1−Φ(−y), which is useful for negative values of y. Let X ∼N(µ, σ). Then we can “standardize” X by deﬁning a variable Y such that Y = X −µ σ . Y now has mean 0 and variance 1. A vital property to keep in mind: the sum of a large number of independent and identically distributed (i.i.d.) (not necessarily normal) random variables has an approximately normal CDF, regardless of the CDF of the individual random variables. 20 3.4 Joint PDFs of Multiple Random Variables We can now extend the notion of PDFs to the case of multiple random vari-ables. Just like in the discrete case, we introduce the idea of joint, marginal, and conditional PDFs. Two continuous random variables are jointly continuous and can be de-scribed in terms of a joint PDF fX,Y if fX,Y satisﬁes P((X, Y ) ∈B) = ZZ (x,y)∈B fX,Y (x, y)dxdy, For every subset B ∈R2. In the case where B is a rectangle of the form {(x, y)\|a ≤x ≤b, c ≤y ≤d} we have P(a ≤X ≤b, c ≤Y ≤d) = Z d c Z b a fX,Y (x, y)dxdy. We can interpret the joint PDF as a “probability per unit area” in the vicinity of a certain point. To ﬁnd the marginal density with respect to X, we simply take fX(x) = Z ∞ −∞ fX,Y (x, y)dy. Likewise the marginal density of Y is fY (y) = Z ∞ −∞ fX,Y (x, y)dx. 3.5 Joint CDFs If X and Y are two random variables associated with the same experiment, we deﬁne their joint CDF: FX,Y (x, y) = P(X ≤x, Y ≤y). Again, we use the CDF because it works for both discrete and continuous random variables. In particular, if X, Y are described by the joint PDF fX,Y , then FX,Y (x, y) = P(X ≤x, Y ≤y) = Z x −∞ Z y −∞ fX,Y (s, t)dtds ⇕ 21 fX,Y (x, y) = ∂2FX,Y ∂x∂y (x, y). Likewise, by the expected value rule, E[g(X, Y )] = Z ∞ −∞ Z ∞ −∞ g(x, y)fX,Y (x, y)dxdy This can be naturally extended into a a greater number of random variables. 3.6 Conditioning Similar to how we can condition discrete random variables on events (or other random variables), we can do the same for the continuous random variables.The conditional PDF of a continuous random variable X on an event A is deﬁned as a nonnegative function fX\|A that satisﬁes P(X ∈B\|A) = Z B fX\|A(x)dx. In the case where X is an element of the conditioning event, we can use a “Bayesian” approach: P(X ∈B\|X ∈A) = P(X ∈B, X ∈A) P(X ∈A) = R A∩B fX(x)dx P(X ∈A) . Let A1, A2, ..., An be a partition of Ω. Then fX(x) = X P(Ai)fX\|Ai(x). The conditional PDF of X given Y = y is fX\|Y (x\|y) = fX,Y (x, y) fY (y) . This deﬁnition is analogous to the discrete case. This can be generalized in a single expression which relates the joint, marginal, and conditional PDFs: fX,Y (x, y) = fY (y)fX\|Y (x\|y), fX(x) = Z ∞ −∞ fY (y)fX\|Y (x\|y)dy. Additionally we can ﬁnd the conditional probability as an integral of the joint PDF: P(X ∈A\|Y = y) = Z A fX\|Y (x\|y)dx. 22 3.7 Conditional Expectation The justiﬁcations behind the below are all in accordance with the discrete case: E[X\|A] = Z ∞ −∞ xfX\|A(x)dx E[X\|Y = y] = Z ∞ −∞ xfX\|Y (x\|y)dx For a function g(X), E[g(X)\|A] = Z ∞ −∞ g(x)fX\|A(x)dx E[g(X)\|Y = y] = Z ∞ −∞ g(x)fX\|Y (x\|y)dx For a partition of the sample space, E[X] = X P(Ai)E[X\|Ai] E[X] = Z ∞ −∞ E[X\|Y = y]fY (y)dy. For functions of several random variables, E[g(X, Y )\|Y = y] Z g(x, y)fX\|Y (x\|y)dx, E[g(X, Y )] = Z E[g(X, Y )\|Y = y]fY (y)dy. 3.8 Independence Two continuous random variables X and Y are independent if their joint PDF is the product of their marginal PDFs: fX,Y (x, y) = fX(x)fY (y), ∀x, y. The remainder of the properties are the same as in the discrete case, i.e. E[XY ] = E[X]E[Y ], Var(X + Y ) = Var(X) + Var(Y ), etc. 23 3.9 Continuous Bayes Many times, we have some unobserved phenomenon X with PDF fX, but an observed noisy measurement Y which we model with the conditional PDF fX\|Y . We can get information about X through Bayes’ Theorem: fX\|Y (x\|y) = fX(x)fY \|X(y\|x) fY (y) 24 4 Further Topics on Random Variables 4.1 Derived Distributions Say we are given Y = g(X). Given the PDF of X, we should be able to calculate the PDF of Y . We do this in two steps: 1. Calculate the CDF FY of Y using the formula FY (y) = P(g(X) ≤y) = Z {x\|g(x)≤y} fX(x)dx 2. Diﬀerentiate to obtain the PDF of Y fY (y) = dFY dy (y). 4.2 Convolutions Consider the function Z = X+Y of two independent, integer-valued random variables with PMFs pX and pY . Then for any integer z pZ(z) = P(X + Y = z) = X {(x,y)\|x+y=z} P(X = x, Y = y) = X x P(X = x, Y = z −x) = X x pX(x)pY (z −x) pZ is called the convolution of the PMFs of X and Y . Continuous Convolution For X and Y continuous, and Z = X + Y , we get fZ(z) = Z ∞ −∞ fX(x)fY (z −x)dx. Notice that this is equivalent to the deﬁnition of signal convolution from EE120. 25 4.3 Covariance and Correlation So far we’ve been able to examine what random variables are and how we can create relationships (functions) using them. Here we explore how to ﬁnd the direction and inherent relationship between two random variables. Covariance The covariance cov(X, Y ) of two random variables X and Y is de-ﬁned by cov(X, Y ) = E (X −E[X])(Y −E[Y ]) . If cov(X, Y ) = 0, the two random variables are uncorrelated. The correlation coeﬃcient ρ(X, Y ) is deﬁned as ρ(X, Y ) = cov(X, Y ) p Var(X)Var(Y ) We can use the covariance to ﬁnd a general formula for the variance of the sum of several (not necessarily independent) random variables. Var X Xi = X Var(Xi) + X i̸=j cov(Xi, Xj) 4.4 Conditional Expectation and Variance, Revisited We can reformulate the law of total expectation into the law of iterated expectations and create a law of total variance that relates conditional and unconditional variance. As long as X is well deﬁned and has a ﬁnite expectation E[X], the law of iterated expectation says that E[E[X\|Y ]] = E[X]. The law of total variance states: Var(X) = E[Var(X\|Y )] + Var(E[X\|Y ]). 4.5 Transforms Here we introduce a transform associated with X, also known as the mo-ment generating function MX(s) of a scalar s is MX(s) = E[esX]. 26 If X is a discrete random variable, the transform is given by M(x) = X x esxpX(x), and if X is continuous it is M(s) = Z ∞ −∞ esxfX(x)dx. In order to get the moment from the transform, we can diﬀerentiate with respect to s. We take d dsM(s) = d ds Z ∞ −∞ esxfX(x)dx = Z ∞ −∞ d dsesxfX(x)dx = Z ∞ −∞ xesxf(X)(x)dx. Clearly we can see that evaluating this at s = 0 yields E[X]. Likewise, evaluating the nth derivative at s = 0 will yield E[Xn] – try this for yourself if you’re not convinced. The transform MX(s) is invertible, meaning we can determine the probability law (CDF, PDF, or PMF) of X. The proof is beyond the scope of this course. The formulas are diﬃcult and cumbersome to use, so we usually invert them through pattern matching to known tables of distribution-transform pairs. Transforms are particularly useful for the sum of random variables. The ad-dition of independent random variables corresponds to the multi-plication of transforms, which can provide for us a convenient alternative to the convolution formula. Let Z = X + Y . Then MZ(s) = E[esZ] = E[es(X+Y )] = E[esXesY ] = E[esX]E[esY ] = MX(s)MY (s). More generally: Z = X Xi = ⇒MZ(s) = Y MXi(s). 4.6 Order Statistics Let n ∈Z+. Let X1, ..., Xn be i.i.d. continuous random variables with PDF f and CDF F. Then for i = 1, ..., n let X(i) be the ith smallest of the rando 27 mvariables. Then X(i) is known as the ith order statistic. The CDF of the order statistic is P(X(i) ≤x) = n X k=i n k F(x)k(1 −F(x))n−k. Diﬀerentiating the CDF gives us the PDF, namely fX(i)(x) = n n −1 i −1 f(x)F(x)i−1(1 −F(x))n−i. 28 5 Limit Theorems Here we examine what happens to sequences of random variables as the size of the sequence becomes large. For a sequence of i.i.d random variables with mean µ and variacne σ2, we deﬁne the sum of the ﬁrst n variables as Sn = n X i=1 Xn. Since they are independent, Var(Sn) = P Var(Xi) = nσ2, meaning the variance of Sn becomes large linearly as n, so we can’t get any meaningful information here. What we can get meaningful information from is the sample mean, Mn = 1 nSn. Then E[Mn] = µ, Var(Mn) = 1 nσ2, which follows what we expect from some of the laws of large numbers you may have seen in CS70. So now we can see that the variance of the sample mean approaches 0 asymptotically, while the variance of Sn approaches in-ﬁnity asymptotically. Let’s ﬁnd a happy medium between the two. We can create the zero-mean random variable Sn −nµ and divide by √ nσ2 to get Zn = Sn −nµ σ√n . Then E[Zn] = 0, Var(Zn) = 1. The central limit theorem asserts that as n − →∞, the standardized sum of n random variables approaches standard normal. 5.1 The Markov and Chebyshev Inequalities The inequalities mentioned here use the mean and variance of random vari-ables to get an idea about the probabilities of certain events. They’re es-pecially useful when we can’t determine the distribution of X but we can calculate its mean and variance easily. The Markov inequality asserts that if a nonnegative random variable has 29 a small mean, the probability that it takes on a large value is small. Markov’s Inequality P(X ≥a) ≤E[X] a , ∀a > 0. Note that this is not a tight bound, but it is a maximal upper bound. The Chebyshev inequality makes a similar assertion. It states that if the variance of a random variable is small, the probability that it takes on a value far from the mean is also small. i.e. if X has mean µ and variance σ2, then the following holds. Chebyshev’s Inequality P(\|X −µ\| ≥c) ≤σ2 c2 , ∀c > 0. Since the Chebyshev inequality uses both the mean and variance, it allows us to form a slightly tighter bound than the Markov inequality. 5.2 The ChernoﬀBound The Markov and Chebyshev bounds are extremely loose bounds, and as such fare poorly when discussing random variables that fall oﬀexponentially the further you get from the mean. Here we introduce the Chernoﬀbound, which lets us calculate tail probabilities for independent random variables that fall oﬀexponentially from the mean. Let X1, ..., Xn be independent Poisson random variables with P[Xi = 1] = pi, and let X = P Xi as standard. Then if µ = E[X], the general Chernoﬀ bound for δ ∈(0, 1] is P[X < (1 −δ)µ] < e−δ (1 −δ)(1−δ) µ . This bound is a bit clunky to work with, so we can relax it a bit to the following: 30 ChernoﬀBound P[X < (1 −δ)µ] < exp −µδ2/2 . 5.3 The Weak Law of Large Numbers The weak law of large numbers asserts that the sample mean of a large number of i.i.d. random variables is very close to the true mean with high probability. We can use the Chebyshev inequality to elaborate. The Weak Law of Large Numbers For a set of i.i.d. random variables with mean µ, ∀ϵ > 0 we have P(\|Mn −µ\| ≥ϵ) = P 1 n X Xi −µ ≥ϵ − →0, n − →∞. 5.4 Convergence in Probability The weak law of large number essentially states that Mn converges to µ. However, we need to deﬁne convergence a bit more tightly – after all, Mn is a variable, not a constant value. We begin with the traditional deﬁnition of convergence used in analysis: for a sequence of real numbers a1, a2, ... and another real number a we say that an converges to a if: ∀ϵ > 0, ∃N ∈N : n > N = ⇒\|an −a\| ≤ϵ. Intuitively, this means that for a big enough n, an will be within ϵ of a for any ϵ. In probability, we say something similar: for a sequence Y1, Y2, ... of random variables, and a real number a, Yn converges to a if: ∀ϵ > 0, lim n− →∞P(\|Yn −a\| ≥ϵ) = 0. We can also rephrase this more generally to deﬁne the probability to be accurate within some conﬁdence level (this is more similar to the ϵ −δ deﬁnition used in real analysis): ∀ϵ > 0, ∀δ > 0, ∃N ∈N : P(\|Yn −a\| ≥ϵ) ≤δ. 31 5.5 The Central Limit Theorem The Central Limit Theorem Let X1, X2, ... be a sequence of i.i.d. random variables with common mean µ and variance σ2, and deﬁne Zn = X1 + . . . + Xn −nµ σ√n . We are taking the sum of random variables, removing the mean, and keeping the variance ﬁxed. Then the CDF of Zn converges to the standard normal Φ(z) = 1 √ 2 Z z −∞ exp −x2 2 dx, in the sense that lim n− →∞P(Zn ≤z) = Φ(z), ∀z. This theorem is extremely general – aside from requiring the random vari-ables to be independent and with ﬁnite mean and variance, the distribution of the random variables does not matter. The sum of a large number of ran-dom variables is approximately normal. We can use this to treat a large sum Sn = X1+. . .+Xn as if it were normal by normalizing it and approximating with the standard normal CDF. 5.5.1 The De Moivre-Laplace Approximation of the Binomial Remember the way we deﬁned the binomial distribution earlier – as the sum of n independent Bernoulli random variables. We can use the central limit theorem now to approximate the probability of an event {k ≤Sn ≤l}, where k and l are given integers: the derivation is omitted. The De Moivre-Laplace Approximation If Sn is a binomial random variable with parameters n and p, with n large, and k, l ∈Z≥0, then P(k ≤Sn ≤l) ≈Φ l + 1 2 −np p np(1 −p) ! −Φ l −1 2 −np p np(1 −p) ! . 32 5.6 The Strong Law of Large Numbers Here we also show that the sample mean converges to the tru mean, with a diﬀerent type of convergence. In general, we can write the following state-ment. The Strong Law of Large Numbers Let X1, X2, ... be a sequence of i.i.d random variables with mean µ. Then the sequence of sample means Mn = 1 n(X1, X2, ...) converges to µ with probability 1 in the sense that P lim n− →∞ X1 + ... + Xn n = µ = 1. What then, is the diﬀerence betweenn the weak and strong laws? The weak law simply states that the probability that the sample mean deviates from the true mean is low as n − →∞; however, that probability (that it does de-viate signiﬁcatly) still exists, but we don’t know how how many deviations there will be. The strong law gives us this assurance, by saying that Mn converges to µ with probability 1, so for any ϵ, the probability that \|Mn −µ\| will exceed ϵ an inﬁnite number of times is 0. Note the diﬀerence in the convergence emphasized by the strong law versus the weak law. We here deﬁne this new ”almost sure” deﬁnition of conver-gence in more detail. Almost Sure Convergence Let Y1, Y2, ... be a sequence of random variables. Let c ∈R. Yn converges to c with probability 1 or almost surely if P( lim n− →∞Yn = c) = 1. Almost sure convergence implies convergence in probability, but the converse is not generally true! 33 5.6.1 The Borel-Cantelli Lemma Suppose that A1, A2, ... is a series of events in some probability space Ω. Then the event that A occurs inﬁnitely many times (denoted A(i.o.)) is: A(i.o.) = ∞ \ k=1 ∞ [ n=k Ak. The Borel-Cantelli Lemma ∞ X n=1 P(An) < ∞= ⇒P(A(i.o.)) = 0. This is to say that if the sum of individual probabilities of each event Ai is ﬁnite, then the probability that inﬁnitely many of the Ai occur is 0. The converse of the lemma, sometimes called the second Borel-Cantelli lemma, is true if each Ai is independent. Then ∞ X n=1 P(An) = ∞= ⇒P(A(i.o.)) = 1. 5.7 Binary Erasure Channels Suppose we want to send a message over a noisy channel. We do this in three essential steps – we compress the message, add redundancy to deal with noise, and then send the message through the channel. In the 1940’s, Claude Shannon proved that we can design our source and channel coding seperately without impacting the optimal rate. A binary erasure channel (BEC) erases the input to the channel with probability p ∈(0, 1). How many bits can the transmitter send over the channel without error? Say we’re encoding a length L message with a length n > L message (to account for possible erasures). The ratio L/n is the rate of the message. Assume that the receiver can’t contact us in the middle of the transmission to tell us which bits were erased. For an input alphabet X and output alphabet Y, we also send an encoding function fn : X L 7→X n and a decoding function gn : Yn 7→X L. In a BEC, the input alphabet is {0, 1} and the output alphabet is {0, 1, e}. Now we account for the noise in the channel. 34 Maximum Probability of Error Let X(n) and Y (n) be n-length bit strings corresponding to the input and output, respectively. Then Pe(n) := max x∈X L P{gn(Y (n)) ̸= x\|X(n) = fn(x)}. We say a rate R is achievable if for each positive integer n, there exists a set of encoding and decoding functions that encode L to n such that Pe(n) − →0 as n − →∞. The capacity of a channel is the largest achievable rate. The capacity of a BEC with erasure probability p is 1 −p. It is im-portant that we use an intelligent encoding (i.e. Huﬀman coding) to make sure that the receiver has to ask as few questions as possible to decode the information, i.e. go over a small codebook. The general result is stated in terms of the mutual information of ran-dom variables, deﬁned as I(X; Y ) := H(X) + H(Y ) −H(X, Y ). Let X be the channel input and let Y be the channel output, and let P be the set of probability distributions on X. The channel capacity is C := max p∈P,X∼p I(X; Y ). Channel Coding Theorem Any rate below C is achievable. Conversely, any sequence of codes with Pe(n) − →0, n − →∞has rate R ≤C. 35 6 Discrete Time Markov Chains Here we consider processes that depend on and (to an extent) can be pre-dicted using what has happened in the past. We can summarize the eﬀect of the past on the future as a state, which changes over time according to various probabilities. For this course, we only consider models with a ﬁnite number of state values, whose probabilities are time-invariant. 6.1 Discrete-Time Markov Chains We ﬁrst start with Discrete-Time Markov Chains (DTMCs), where the state changes at discrete time intervals n. The state of the chain at time n is a random variable Xn, belonging to a ﬁnite set X of possible states (the state space). The chain is described by its transition probabilities P(xn, xn+1), describing the probability of going from state xn to state xn+1. P(xn, xn+1) = P(Xn+1 = xn+1\|Xn = xn), xn, xn+1 ∈X. Note something interesting in the above – the probability of going between two states does not depend on the time at which we arrived at state xn, and does not depend on how we got to state xn in the ﬁrst place. This is an important property of Markov chains. The Markov Property For all times n, for all states xn, xn+1 ∈S, for all possible sequences of states x0, ..., xn+1, P(Xn+1 = xn+1\|Xn = xn, ..., X0 = x0) = P(Xn+1 = xn+1\|Xn = xn) = P(xn, xn+1). All elements of a Markov chain can be encoded in a transition probability matrix (which is row stochastic, i.e. has row sums of 1) and graph, drawn from a a 2-dimensional array with the element in row i, column j corresponding to pij. 1 2 a b c d ⇐ ⇒ a b c d 36 6.1.1 The Probability of a Path Given a chain, we can compute the probability of a sequence of future states, similar to the multiplication rule in probability tree models. In particular we have P(X0 = x0, X1 = x1, ..., Xn = xn) = π0 n Y k=1 pik−1ik. Think of this as following the path from node to node in the transition graph and multiplying the probabilities at each step. 6.1.2 k-Step Transition Probabilities We often want to ﬁnd the probability law of the state in the future. Pk(x, y) := P(Xk = y\|X0 = x) = P k(x, y). Put another way, this is the probability that after k steps we will reach state y starting from state x, which is the (x, y) entry of the kth power of P We calculate this (the k-step transition probabilities) using the following recursion: Chapman-Kolmogorov Equations The k-step transition probabilities can be generated from Pk(x, y)) = P(Xk = y\|X0 = x) = X x1,...xk−1∈X P(Xk = y, Xk−1 = xk−1, ..., X1 = x1\|X0 = x) = X x1,...xk−1∈X P(x, x1)P(x1, x2) . . . P(xk−2, xk−1)P(xk−1, y) = P k(x, y) 6.2 Classiﬁcation of States The probabilities associated with each state give us diﬀerent information about each state. For instance, some states, once visited, are certain to be visited again, while others may never be visited again. Here we draw a rela-tionship between the states of a Markov chain and the long-term frequency with which they are visited. 37 For each x ∈X, let us deﬁne a random variable T + x := min n ∈N : Xn = x, the hitting time of state x. Additionally deﬁne Px and Ex to indicate that the chain begins at state x, i.e. Px(·) := P(·\|X0 = x) Ex(·) := E[·\|X0 = x]. Then for x, y ∈X, let ρx,y = Px(T + y < ∞), the probability we are guar-anteed to see state state y starting from state x. Let ρx = ρx,x. Clearly, if ρx = 1 (we are guaranteed to see state x from state x), we will visit x inﬁnitely; this state is recurrent. Likewise, if ρx < 1, we are guaranteed to stop seeing x from x after a countably inﬁnite number of steps; this state is transient. Classifying a state as transient or recurrent only depends on whether or not arrows exist in the transition probability graph; they do not depend on what the actual probabilities are. Px-a.s. Let Nx denote the total number of visits to state x, that is, Nx := P n∈N 1{Xn = x}. If x is recurrent, then Nx = ∞Px −a.s., so in particular Ex[Nx] = ∞. If x is transient, then Ex[Nx] < ∞; in fact, Ex[Nx] = ρx 1 −ρx < ∞. In particular, Nx < ∞Px −a.s. Let A(x) be the set of all states accessible (through a series of state transi-tions) from x. If x is recurrent, A(x) forms a recurrent class or commu-nicating class, meaning all states in A(x) are accessible from each other and no state outside A(x) is accessible from them. In graph theory, we call this a strongly-connected component. If a Markov chain is irreducible if it has only a single communicating class. 1 2 3 4 Recurrent Transient Recurrent Recurrent 38 At least one recurrent state must be accessible from any transient state. We can decompose any Markov chain into one or more recurrent classes plus possibly a few transient states. Decompositions let us visualize the evolu-tions of states. Once the state enters in a class of recurrent states, it stays within that class, and since all states in the class are accessible from one another, all states in the class will be visited an inﬁnite number of times. If the initial state is transient, then the state trajectory will contain an initial portion of transient states and a ﬁnal portion consisting of recurrent states within the same class. Periodicity Consider a communicating class R. This class is periodic if its states can be grouped in d > 1 disjoint subsets S1, ..., Sd so that all transitions from Sk lead to Sk+1. The class is aperiodic if and only if there exists a time k such that Pk(x, y) > 0, ∀x, y ∈R. 6.2.1 Positive Recurrence and Null Recurrence A sequence of random variables is stationary if for all k, n ∈Z+, and all events Ai, ..., An, then P(X1 ∈A1, ..., Xn ∈An) = P(Xk+1 ∈A1, ..., Xk+n ∈An). What does this mean? All we’re saying here is that the distribution of X1, ..., Xn stays the same as the joint distribution if we shift the time index by k to Xk+1, ..., Xk+n. Many stochastic processes converge to this notion of stationarity. Stationarity Suppose that a Markov chain is irreducible with a stationary distri-bution π. Then, for teach x ∈X, π(x) = 1 Ex[T + x ]. 39 A Markov chain is, essentially, a way to introduce dependency into the idea of i.i.d. random variables. As such, our best shot at understanding them is by analyzing the i.i.d. structure hidden within the chain. If we’re at state x for the 1st time, there is no diﬀerence in the distribution than if we were at state x for the kth time; let Tx(k) be the kth time we hit state x. Then for all k, we can split the Markov chain into XTx(k), ..., XTx(k+1 −1 and treat each of these as i.i.d random variables, giving rise to the above expression. Positive and Null Recurrence State x is is positive recurrent if x is recurrent and Ex[T + x ] < ∞ (the expected time to return to x is ﬁnite). State x is null recurrent if x is recurrent and Ex[T + x ] = ∞(the expected time to return to x is inﬁnite). 6.3 Steady-State Behavior We’ll examine both the long and short-term behavior of Markov chains. We begin by looking at what the k-step transition probabilities Pk(x, y) are when k is very large. If a Markov chain has multiple classes of recurrent states, the limiting value of Pk(x, y) must depend on our initial state. We focus selectively on chains that have only one recurrent class plus transient states. Even when a chain only has a single recurrent class, Pk(x, y) may not converge! Examine the following Markov chain: 1 2 p = 1 p = 1 ⇐ ⇒Pk(x, y) = ( 1, k even 0, k odd This is what happens when the recurrent class is periodic. Ignoring the two cases mentioned previously (multiple classes and periodicity), we can create the following theorem: 40 Steady-State Convergence Theorem Consider a Markov chain with a single aperiodic recurrent class. Then each state x has an associated steady-state probability π(x): 1. ∀x, lim n− →∞Pk(y, x) = π(x), ∀i 2. π(x) are the unique solutions of the following: π(x) = X y∈X π(y)P(y, x) j = 1, ..., m, 1 = X x∈X π(x) The steady-states sum to 1 and form a distribution, called the stationary distribution. This is because for all n, P(Xn = x) = π(x). This means that if some initial state is chosen according to the stationary distribution, the state at any future time will have the same distribution. The equations X y∈X π(y)P(y, x) are the balance equations. Together with the normalization equation 1 = X x∈X π(x). An irreducible positive reccurent Markov chain has a unique stationary dis-tribution; likewise, an irreducible Markov chain is positive recurrent if and only if a stationary distribution exists. What does it mean if the stationary distribution doesn’t exist? If µ is the solution to the balance equations, then a stationary distribution does not exist if it is impossible to normalize µ, i.e. P µ(x) = 0 or P µ(x) = ∞. Otherwise, if P µ(x) = c, then π := c−1µ. The fraction of time we spend in state x converges almost surely to π(x) as the number of steps we take increases inﬁnitely. Here we also have the ﬁrst-step equations: 41 First-Step Equations For each state x: π(x) = 1 + X y∈X P(x, y)π(y) The expected time to return to a state x: Ex[T + x ] = 1 + X y∈X P(x, y)Ey[T + x ] 6.4 Reversibility of Markov Chains When does a Markov chain look the same whether we run it forwards in time or backwards in time? Fix a positive integer N and deﬁne Yn = XN−n; this is the reversed chain. If the original chain is irreducible and the chain is started from stationary distribution π, then the reversed chain is also irreducible with transition probabilities ˆ P(x, y) = π(y)P(x, y)/π(x). The stationary distribution for the reversed chain is also π. The reversed chain looks the same as the original chain if π(x)P(x, y) = π(y)P(y, x). The above is known as the detailed balance equations. The condition for stationarity is π(y) = X x∈X π(x)P(x, y) This is a stronger condition than the global balance equations; the global equations say the probability mass entering a state equals the probability mass exiting the state. The detailed equations express a local condition where the mass along each edge is balanced. 42 7 The Bernoulli and Poisson Processes What is a process? A stochastic process is a model of a probabilistic experiment over time. Each value in the process is a random variable, so really a process is just a sequence of random variables. In processes, we focus on dependencies in the sequence of values, long-term averages involving the values, and the likelihood or frequencies of boundary events. Here we discuss arrival processes, where we care about the times between arrivals. If the arrivals occur in discrete time, we model this with the Bernoulli process, and if the arrivals occur in continuous time we model them with the Poisson process. 7.1 The Bernoulli Process The Bernoulli process can be visualized as a sequence of independent coin tosses with probability p. Take a sequence of random variables, where suc-cess at the ith variable occurs with probability p. Recall the properties of the binomial and geometric distributions from earlier in this course. Addi-tionally recall the memoryless property of the geometric distribution; the future of the process does not depend on how much time has already passed. 7.1.1 Interarrival Times One of the reasons we study processes is to determine the time of the kth success, or arrival, Yk. Because of this, we can express the events of the process as a sequence of geometric random variables T1, ... with param-eter p, standing for the time in between arrivals. Then the sequence is T1, T1 + T2, T1 + T2 + T3... We then have the following set of observations: 43 Properties of the kth Arrival Time The kth arrival time is the sum of the ﬁrst k interarrival times: Yk = T1 + ... + Tk, where Ti are independent geometric random variables with common parameter p. E[Yk] X E[Ti] = k p, Var(Yk) = X Var(T1) = k(1 −p) p2 . The PMF of Yk is then pYk(t) = t −1 k −1 pk(1 −p)t−k, t = k, k + 1, ..., the Pascal PMF of order k. 7.1.2 Splitting and Merging of Bernoulli Processes Consider splitting a process: When there is an arrival, we keep it with probability q or discard it with probability 1 −q. Then the probability of there being an arrival we keep is pq and the probability of there being an arrival we discard is p(1−q). We are essentially sorting our standard arrival process into two processes. We can also merge two processes, recording an arrival in the merged process if there is an arrival in one of the two processes, with probability p + q −pq 1 −(1 −p)(1 −q). 7.2 The Poisson Process The continuous-time variant of the Bernoulli process is the Poisson pro-cess. An arrival process is a Poisson process with rate λ if if the the prob-ability of k arrivals in τ time, P(k, τ) is the same for all intervals of length τ. The number of arrivals during an interval is independent of the num-ber of arrivals outside that interval. Additionally, it satisﬁes the following 44 small-interval probabilities: P(0, τ) = 1 −λτ + o(τ), P(1, τ) = λτ + o1(τ), P(k, τ) = ok(τ), Here lim τ− →0 o(τ) τ = 0, lim τ− →0 ok(τ) τ = 0. Think of o(τ) as the O(τ 2) terms in the Taylor expansion of P(k, τ). The probability of a single arrival is roughly λτ with a negligible term. The probability of 0 arrivals is roughly 1 −λτ. Random Variables Associated with the Poisson Process The number of arrivals Nτ in a Poisson process with parameter λ over an interval τ: pNτ (k) = P(k, τ) = e−λτ λkτ k k! E[Nτ] = λτ Var(Nτ) = λτ. The time until the ﬁrst arrival T: fT (t) = λe−λt, E[T] = 1 λ, Var(T) = 1 λ2 . The Poisson process is also independent; i.e. two non-overlapping time sets can be considered independent processes, and the distribution of interarrival times is memoryless. Just as the ﬁrst arrival after time t for the Bernoulli process is distributed geometrically, the ﬁrst arrival after time t for the Poisson process is distributed exponentially. 45 The kth Arrival Time As with the Bernoulli process, we can model the Poisson process as a sequence of independent exponential random variables with param-eter λ, and record arrivals at times T1, T1 + T2 , .... Then the kth arrival time is the sum of the ﬁrst k interarrival times: Yk = k X i=1 Ti, where Ti are independent exponential random variables with common parameter λ. The mean and variance of Yk are E[Yk] = k X i=1 E[Ti] = k λ, Var(Yk) = k X i=1 Var(Ti) = k λ2 . The PDF of Yk is fYk(y) = λkyk−1e−λy (k −1)! , y ≥0, the Erlang PDF of order k. 7.2.1 Poisson Splitting and Merging Just as with the Bernoulli case, we can split a Poisson process into two Poisson processes (with rate λp) if we split with probability p (keep with probability p and discard with probability 1 −p). Alternatively, we can merge two Poisson processes into a single process with merged rate λ1 + λ2; the probability that an arrival occurs in the ﬁrst process is λ1 λ1+λ2 , and the probability that an arrival occurs in the other is λ2 λ1+λ2 . In fact, the sum of n Poisson processes is a Poisson process with parameter P λi. 46 7.2.2 Sums of Random Variables Sums of Random Numbers of Random Variables Let N and X1, X2, ... be random variables, and let Y = PN i=1 Xi. 1. If Xi is Bernoulli with parameter p and N is binomial with parameters m, q, then Y is binomial with parameters m and pq. 2. If Xi is Bernoulli with parameter p, and N is Poisson with parameter λ, then Y is Poisson with parameter λp. 3. If Xi is geometric with parameter p, and N is geometric with parameter q, then Y is geometric with parameter pq. 4. If Xi is exponential with parameter λ, and N is geometric with parameter q, then Y is exponential with parameter λq. 7.2.3 The Random Incidence Paradox Here we examine an interesting property of the Poisson process. Assume we take some random time t∗during the process. We call this time a “random incidence,” but be aware that this isn’t a random variable, just an arbitrary time. Assume that t∗happens after the process has been running for a long time, so there has been an arrival sometime in the past. Here we consider the length L interval that contains t∗. We could argue that L is distributed exponentially, like a typical interarrival period. However this is not true – L is distributed according to an Erlang distribution with parameter 2. This is because the time from t∗to the next arrival is a Poisson process with parameter λ; likewise, the time from t∗to the previous arrival is also a Poisson process with parameter λ. Then L is distributed according the Erlang PDF of order 2 by our previous assertions. 47 8 Erd¨ os-R´ enyi Random Graphs Given n ∈Z+, and some probability value p ∈[0, 1], the graph G(n, p) is deﬁned as an undirected graph on n vertices such that each of the n 2 edges of the graph is present (independently) with probability p. This means if p = 0, G is an empty graph and if p = 1 then it is a fully connected graph. We typically deﬁne p as a function of n, p(n), and are especially interested in what happens as n − →∞. Clearly, then, G(n, p) is a distribution over the set of graphs on n vertices! The “PDF” of G ∼G(n, p) can be easily calculated: P(G = G0) = pm(1 −p)(n 2)−m. 8.1 Sharp Threshold for Connectivity Sharp Threshold Let p(n) = λln n n for λ > 0. Then if λ < 1, then P(G(n, p(n) is connected) − →0; if λ > 1, then P(G(n, p(n) is connected) − →1. This is called a sharp thresh-old since a slight deviation in λ around 1 can drastically change the behavior of the limit. 48 9 Bayesian Statistical Inference Here we deviate from our pure mathematical subject of probability (based on the principles from the ﬁrst chapter) and move towards statistics. In probability theory, we assume some underlying model and attempt to pre-dict future events using this information. In statistics, we observe some event, and attempt to ﬁgure out the process which could have led to those observations. If we are drawing red and blue balls from a bag, probability gives us information about what ball we’ll draw next given that we know how many of each are in the bag. Statistics lets us infer how many of each are in the bag based on the balls we draw. There are two schools of thought for looking at statistics – the Bayesian and the classical (or frequentist). In the former, unknown quantities are treated as random variables with known distributions; in the latter, un-known values are treated as deterministic quantities that just happen to have unknown values. There are two forms of inference. In model inference, we take our ob-servations and try to construct a model to explain the process behind those observations. In variable inference, we estimate the value of some un-known variable through some noisy observations. We can roughly classify statistical inference into two buckets. In parame-ter estimation, we have a model with some unknown parameter θ, which we try to estimate using either a Bayesian or classical approach. In m-ary hypothesis testing, we take m possible hypotheses and use our data to determine which is true. 9.1 Bayesian Inference The Bayesian method operates by deﬁning a random variable Θ that rep-resents our model, and a probability distribution pΘ(θ) (the prior). We can then make a set of observations x and derive a posterior distribution pθ\|X(θ\|x). There are three key principal Bayesian inference methods we ex-plore here: Maximum a posteriori (MAP), where we select the hypothe-sis/parameter with the highest posterior probability; Least mean squares (LMS), where we select the hypothesis/parameter with the minimum mean squared error between the parameter and its estimate; and Linear least mean squares (LLMS), where we select an estimator as a linear function 49 of the data that minimizes the mean squared error between the parameter and its estimate. We assume a prior pΘ or fΘ for thee unknown Θ, assume a model pX\|Θ or fX\|Θ for the vector of data or observations X, and then use Bayes’ rule (dif-ferent versions depending on whether X and Θ are discrete or continuous) to determine the posterior pΘ\|X or fΘ\|X. 4 Versions of Bayes’ Rule Let φ = ( pX(x) X discrete, fX(x) X continuous. Then, Bayes rule is as follows: φΘ\|X(θ\|x) =                φΘ(θ)φX\|Θ(x\|θ) X θ′ φΘ(θ′)φX\|Θ(x\|θ′) Θ discrete, φΘ(θ)φX\|Θ(x\|θ) Z φΘ(θ′)φX\|Θ(x\|θ′)dθ′ Θ continuous. 9.2 Point Estimation, Hypothesis Testing, and the MAP Rule Now that we’ve introduced the framework for Bayesian inference, we can introduce a method that we can then apply to estimation and hypothesis testing problems. We’re given a value x of an observation, and we select a value of θ (normally denoted ˆ θ) that maximizes the posterior pΘ\|X(θ\|x) (or fΘ\|X(θ\|x) for continuous Θ). This is known as the Maximum a Posteriori rule (or MAP). 50 Maximum a Posteriori (MAP) Given an observation value x, we select the value ˆ θ that maximizes θ over the posterior pΘ\|X(θ\|x) or fΘ\|X(θ\|x) (depending on whether Θ is discrete or continuous). ˆ θ = arg max θ pΘ\|X(θ\|x) ⇐ ⇒ˆ θ = arg max θ fΘ\|X(θ\|x). By Bayes’ rule, the form of the posterior distribution is a fraction, where the numerator is dependent on both x and θ. However, it is important to note that the denominator is the same for all values of θ (it is dependent on θ′, which takes on all values in Θ). Therefore, we can exclude the denominator entirely, and are left with the following (using the deﬁnition of φ from the previous box): ˆ θ = arg max θ φΘ(θ)φX\|Θ(x\|θ). If Θ takes on a ﬁnite number of values, then MAP essentially mini-mizes the probability of selecting the wrong hypothesis. 9.2.1 Point Estimation Point Estimates An estimator is a random variable ˆ Θ that is a function of our ob-servations, i.e. ˆ Θ = g(X). An estimate is a value ˆ θ of an estimator, which is determined by an actual observation x. The MAP estimator selects ˆ θ to be the θ that maximizes the posterior distribution over all θ given x. The conditional expectation estimator (called the LMS estimator) selects the ˆ θ that is E[Θ\|X = x]. This is called the LMS estimator because it has the property that it minimizes the mean squared error over all estimators. 51 9.2.2 Hypothesis Testing In a hypothesis testing problem, θ takes on one of a small set of values. The ith hypothesis, Hi, is the event Θ = θi. Once we’ve observed x of X, we can calculate the probability of each hypothesis given the outcome. We can then select the hypothesis that maximizes this posterior probability (the MAP rule). The probability of a correct decision is then P(Θ = gMAP (x)\|X = x). 9.3 Bayesian Least Mean Squares Here we discuss the second estimator we mentioned above, the Least Mean Squared or LMS estimator. Consider a simpler problem, where we don’t have any observations x. Then the mean squared error E[(Θ −ˆ θ)2] is minimized when ˆ θ = E[Θ]. We use the MSE because if we were to just use Θ −ˆ θ, we have a random variable that cannot be minimized in θ. In this case: E (Θ −E[Θ])2 ≤E[(Θ −ˆ θ)2]. Now we consider our observation x of X; in this case, E[(Θ −ˆ θ)2\|X = x] is minimized at ˆ θ = E[Θ\|X = x]: E[(Θ −E[Θ]\|X = x])2\|X = x] ≤E[(Θ −ˆ θ)2\|X = x]. Finally, for all esimators g(X), the MSE is minimized when g(X) = E[Θ\|X]: E[(Θ −E[Θ\|X])2] ≤E[(Θ −g(X))2]. Properties of the Estimation Error If we have the LMS estimator ˆ Θ = E[Θ\|X], then the estimation error is ˜ Θ = ˆ Θ −θ. 1. ˜ Θ is unbiased, i.e. it has 0 mean both conditionally and un-conditionally (E[˜ Θ] = Eθ[ˆ Θ] −θ = 0, E[˜ Θ\|X] = 0). The value Eθ[˜ Θ] is the bias, denoted bθ(ˆ Θ). 2. ˜ Θ is uncorrelated with ˆ Θ, i.e. cov(ˆ Θ, ˜ Θ) = 0. 3. Var(Θ) = Var(ˆ Θ) + Var(˜ Θ). 52 10 Classical Parameter Estimation In a classical setting (as opposed to the Bayesian one we previously in-troduced), we view the unknown parameter θ as deterministic rather than random. X, the observation, is then random, and instead of dealing with a single probabilistic model we deal with several possible candidate models. 10.1 Classical Parameter Estimation Recall the deﬁnitions of an estimator and estimate from the previous section. The deﬁnitions of estimation error and bias are the same. It is interesting to note that the mean squared error can also be written as E[˜ Θ2] = b2 θ(ˆ Θ) + Var(ˆ Θ). This leads to an interesting problem in machine learning known as the bias-variance tradeoﬀ, where reducing the bias term increases the variance term and vice versa. 10.1.1 Maximum Likelihood Estimation (MLE) In classical parameter estimation, we can describe a random vector of observations X by a join PMF pX(x; θ), dependent on a deterministic pa-rameter θ. Then a maximum likelihood estimate is a value of ˆ θ that maximizes pX(x = (x1, ...xn); θ) over all θ. Thus, ˆ θ = arg max θ pX(x1, ..., xn; θ). Note that for a continuous X, we replace the PDF pX(x; θ) with the PMF fX(x; θ); we call this the likelihood function. If each observation is inde-pendent (which we usually assume), we can simplify this to ˆ θ = arg max θ Y pXi(xi; θ). Since the logarithm is increasing on its domain, we can equivalently maxi-mize the log likelihood, i.e. ˆ θ = arg max θ X log pXi(xi; θ). Note the diﬀerence between the term likelihood and probability. We are not looking for the probability of θ, since we know that θ is a deterministic (but 53 unknown) quantity. Instead, we’re asking “what would θ need to be to max-imize the chance that we saw the observations we did?” Note our maximization of pX(x; θ) here, and how in our previous MAP formulation we maximized pΘ(θ)pX\|Θ(x\|θ). In this way, we can consider MLE the same as MAP with a uniform prior, meaning that the prior for θ is the same for all θ. 10.2 Hypothesis Testing Recall the issue of hypothesis testing we brought up in the previous section. However, now we assume that we have no prior information. We have two hypothesis we are deciding between – H0, the null hypothesis, and H1, the alternative hypothesis. More precisely, we say we have Θ1 and Θ2, and based on x, we want to know which is more likely; θ ∈Θ0 or θ ∈Θ2 (note the similarity to the parameter estimation problem above!). If θ ∈Θ0, then H0 is correct; otherwise if θ ∈Θ1 then H1 is correct. For this class, we are mainly focused on frequentist hypothesis testing (i.e. H0 being true is unknown but not random). We introduced the Bayesian idea above brieﬂy, but it is not a focus of this class. Let’s be more precise about what accepting a hypothesis means. We can split our total set of possible observations X into two categories: A and Ac. We call A the acceptance region of observations, i.e. we accept H0 ⇐ ⇒x ∈A. Acceptance Region Examples Some examples of acceptance regions: • Reject H0 if x > t • Reject H0 if x = t • Reject H0 with probability γ if x > t As always however, we have to take into account error. If we reject the null hypothesis when the null hypothesis is actually correct, we have a Type-I Error or probability of false alarm (PFA). Formally, let PHi(B) denote the probability of an event B occurring when Hi is true. Then the probability of type-I error (the signiﬁcance level) is α(A) = PH0(H1) = PH0(x ̸∈A). 54 Analogously, we have type-II error, i.e. the probability that we accept the null hypothesis when in reality the alternative is true. Formally: β(A) := PH1(H0) = PH1(x ∈A). Neyman-Pearson Hypothesis Testing Deﬁne the likelihood to be: L(x) = PH1(x) PH0(x), L(x) = fH1(x) fH0(x). Then the Neyman-Pearson Test states that, for some threshhold c: 1. Accept H0 if L(x) < c 2. Reject H0 with probability γ if L(x) = c 3. Reject H0 if L(x) > c Note that this is optimal when the following are true: PH0(L(x) > c) + γPH0(L(x) = c) = α0 PH1(L(x) < c) + (1 −γ)PH1(L(x) = c) = β0. If you’ve taken EE127 or an equivalent optimization course, you may notice that this is a simple optimization problem: max c 1 −PH1(L(x) < c) −(1 −γ)PH1(L(x) = c) s.t.PH0(L(x) > c) + γPH0(L(x) = c) = z This is all very complicated and can be diﬃcult to compute. In this class, there is often a relationship L(x) ⇐ ⇒B for some B which is much simpler to understand; for example, we’ll have L(x) be monotonic in x, meaning L(x) > c ⇐ ⇒x > t or x < t, which makes all of the above much easier. 55 11 Hilbert Spaces, Estimation, and Kalman Fil-tering As we conclude this course with estimation, we draw a parallel between spaces of random variables and vector spaces. Note that this portion is fairly heavy in linear algebra on the level of Math110. Before we begin, let’s introduce once particular set (with X random): H{X : X ∈Rs.t. E[X2] < ∞}. 11.1 A Brief Review of Linear Algebra Recall that a vector space V is a collection of objects (vectors) including the zero vector on which the operations of vector addition and scalar multiplication are deﬁned. Vector addition is commutative, associative, and satisﬁes the identity property with the zero vector. Scalar multiplica-tion is distributive, commutative, and associative, and satisﬁes the identity property with the 1 vector. For a set S ⊆V, span(S) is the set of vectors achievable by only using scalar multiplication and vector addition (using vectors in S), i.e. span(S) = ( m X i=1 civi, m ∈N, vi ∈S, ci ∈R ) . In other words, every element in the span of S is a linear combination of the vectors in S. Additionally recognize that span(S) must be a vector space. Whenever we have U ⊆V and U is a vector space, we call U a sub-space. Note that if v ∈S and v is a linear combination of other vectors in S, then span(S) = span(S{v}). We call v a redundant vector, and if S contains no redundant vectors, we call S linearly independent. If S is linearly independent and spanS = V (i.e. S is a generating set for V) then we call S a basis for V. Every vector space has a basis, and all bases for the same vector space have the same cardinality or size, known as the dimension of V and denoted dimV. 11.2 Inner Product Spaces and Hilbert Spaces An inner product ⟨·, ·⟩maps V ×V 7→R≥. The inner product is symmetric (⟨u, v⟩= ⟨v, u⟩), linear (⟨u + cv, w⟩= ⟨u, w⟩+ c⟨v, w⟩), and positive 56 deﬁnite (⟨u, v⟩> 0, u ̸= 0). Then we call V along with the map ⟨·, ·⟩a real inner product space. The inner product gives us two concepts: a norm and angle: ∥· ∥: V 7→R≥, ∥v∥= p ⟨v, v⟩, ⟨u, v⟩= ∥u∥∥v∥cos θ. We are most interested in when θ = 0, i.e. ⟨u, v⟩= 0; we then call u and v orthogonal. H is a Hilbert space if it is a real inner product space that is com-plete. We don’t need to worry about what completeness means for this course, but I’m putting the full deﬁnition here if you’re interested. Addi-tionally, we behave as if H is ﬁnite-dimensional (even though it usually isn’t). Formal introduction (not in scope): A Hilbert space is a vector space that is complete with respect to an inner product deﬁned on that space (a metric space M is complete if every Cauchy sequence in M con-verges in M). It is an instance of a Banach space (deﬁned by any convex set) wherein the norm is deﬁned via the inner product (ellipse). 11.3 Projection We want to estimate a random variable Y ∈H. However, we only know some X that is correlated with Y . We want to ﬁnd a “best guess” function of X that estimates Y . This problem because much easier if we restrict ourselves to aﬃne functions of the form a + bX. Then our “best guess” is the point x ∈U, U = span{1, X} ⊆H that is closest to Y . We ﬁnd this by ﬁnding the shortest orthogonal distance to Y . Formally, the shortest orthogonal distance is P : V 7→U Py := arg min x∈U ∥y −x∥2. Note that Py is in U and y−Py is orthogonal to U (equivalently, y−Py ∈ U⊥). 11.4 Gram-Schmidt Orthonormalization Since Py ∈U and y −Py ∈U⊥, y −Py must be orthogonal to every basis vector for U. Since Py is linear, Py = P civi for basis vectors vi. We can then compute Py by solving: y − n X i=1 civi, vj + = 0. 57 What is the basis vectors were orthornormal? This makes our job much easier, since we would then be able to get each component of P by computing ⟨y, vi⟩. How do we orthonormalize our basis vectors? We can use the following: Gram-Schmidt Process u = v/\|\|v\|\| for j in 1,...,n-1: let w[j+1] = v[j+1] - sum from i-1 to j of (u[i]) set u[j+1] = w[j+1]/\|\|w[j+1]\|\| 11.5 Linear Least Squares Estimate (LLSE) Recall our linear formulation ˆ Y = a+bX above as our estimate for Y . Then, given X, Y ∈H, we seek to minimize (over b, a ∈R): E[(Y −ˆ Y )2] = E[(Y −a −bX)2]. The solution to this problem is the linear least squares estimator or LLSE. Notice this similarity to the classical parameter estimation problem from earlier; we’re trying to solve Y ∗= arg min ˆ Y ∈U ∥Y −ˆ Y ∥2, U = span{1, X}. Some change of basis and algebra magic gives us the following: LLSE For X, Y ∈H, where X is variable, the LLSE of Y given X is L[Y \|X] = E[Y ] + E Y X −E[X] √ VarX X −E[X] √ VarX = E[Y ] + cov(X, Y ) VarX (X −E[X]). The squared error of the LLSE is then E[(Y −L[Y \|X])2] = VarY −cov(X, Y )2 VarX . 58 11.6 Minimum Mean Square Estimation (MMSE) Say instead of restricting X to linear functions, we let ˆ Y be any function of X, say φ(X). Then ﬁnding the best function φ to minimize E[(Y −φ(X))2] is known as the minimum mean squared error estimator (MMSE). One important condition is placed on this estimator; Y −φ(X) must be orthogonal to all other functions of X. The solution to this problem is the conditional expectation of Y given X, or E[Y \|X], such that for all bounded continuous functions φ: E[(Y −E[Y \|X])φ(X)] = 0. The diﬀerence between MMSE and LLSE is that MMSE deals with all func-tions on X, even nonlinear ones (which is really hard to visualize). The op-timum φ is guaranteed to exist (since the objective function is convex – see more in EE127), but ﬁnding it is often hard. For this course, just knowing these basic facts about MMSE will suﬃce. 11.6.1 Jointly Gaussian Random Variables X and Y are jointly Gaussian if their joint PDFs are multivariate Gaus-sian, or equivalently if each linear combination of X and Y is Gaussian. Generally, L[Y \|X] ̸= E[Y \|X]. However, when X and Y happen to be jointly Gaussian, the MMSE is equivalent to the LLSE, i.e. MMSE for Jointly Gaussian Random Variables If X, Y are jointly Gaussian, then E[Y \|X] = L[Y \|X] = E[Y ] + cov(X, Y ) VarX (X −E[X]). Other important properties that can be used to calculate the LLSE/MMSE are shown below: 59 Key Properties Both the LLSE and MMSE are unbiased, meaning E[X −E[X]] = E[E[Y \|X] −E[Y ]] = 0. Additionally, Y −L[Y \|X] and X are uncorrelated and also indepen-dent (since Y −L[Y \|X] and X are jointly Gaussian). 11.7 Kalman Filtering The Kalman ﬁlter is an optimal state estimation algorithm (akin to the Linear-Quadratic Regulator from EECS127). Consider a system with a state X(n) and an output (n) at time n. We can describe the system as follows: X(n + 1) = AX(n) + V(n) Y(n) = CX(n) + W(n) Here, X(0), V, and W are random and zero-mean, where covV = ΣV and covW = ΣW. We want to recursively be able to recover X to the best of our ability based on our noisy observations Y, i.e. ˆ X(n) = L[X(n)\|(Y(i))n i=0]. The Kalman Filter The ﬁlter is deﬁned recursively as follows: ˆ X(n) = A ˆ X(n −1) + Kn[Y(n) −CA ˆ X(n −1)] Kn = SnC⊤[CSnC⊤+ ΣW]−1 Sn = AΣn−1A⊤+ ΣV = cov(X(n) −A ˆ X(n −1)) Σn = (I −KnC)Sn = cov(X(n) −ˆ X(n)) Some key observations: Kn and Σn can be precomputed at time 0 since they do not depend on the observations (note: even though ˆ X depends on the observations, the residual X(n) −ˆ X(n) does not). Additionally if X(0) and the noise variables V and W are Gaussian, then the Kalman ﬁlter simply computes the MMSE. Additionally, even though this entire box is a computational mess, it can be programmed very easily and it is incredibly simple to solve computationally by a computer. 60
8708	https://askfilo.com/user-question-answers-mathematics/43-the-function-is-differentiable-at-when-36353532373739	Question asked by Filo student 43 The function f(x)=xnsin(1/x) is differentiable at x=0, when Views: 5,458 students Updated on: Jan 3, 2024 Video solutions (1) Learn from their 1-to-1 discussion with Filo tutors. Uploaded on: 1/3/2024 Connect instantly with this tutor Connect now Taught by Total classes on Filo by this tutor - 2,604 Teaches : Mathematics, Algebra, Calculus Connect instantly with this tutor Connect now Notes from this class (1 pages) Practice more questions on Limit, Continuity and Differentiability Views: 5,780 Topic: Integrals Book: Integral Calculus (Amit M. Agarwal) View solution Views: 5,177 Topic: Integrals Book: Integral Calculus (Amit M. Agarwal) View solution Views: 5,445 Topic: Relations and Functions Exam: JEE Advanced 1985 View 3 solutions Views: 5,724 Topic: Relations and Functions View solution Students who ask this question also asked Views: 5,739 Topic: Limit, Continuity and Differentiability View solution Views: 5,920 Topic: Limit, Continuity and Differentiability View solution Views: 5,595 Topic: Limit, Continuity and Differentiability View solution Views: 5,776 Topic: Limit, Continuity and Differentiability View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| 43 The function f(x)=xnsin(1/x) is differentiable at x=0, when \| \| Updated On \| Jan 3, 2024 \| \| Topic \| Limit, Continuity and Differentiability \| \| Subject \| Mathematics \| \| Class \| Class 12 \| \| Answer Type \| Video solution: 1 \| \| Upvotes \| 142 \| \| Avg. Video Duration \| 3 min \| 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
8709	https://www.enjoymathematics.com/blog/find-the-number-of-small-painted-cubes/	Find the Number of Small Painted Cubes Related tags: math-for-computer-science Problem Statement A cube is painted with some colour on all faces. Now, we cut it into 1000 small cubes of equal size. How many small cubes are not painted? Approach 1 One basic approach is to count the number of painted small cubes and subtract it from the total number of small cubes. Note: While solving such questions, the most important part is to visualise the cube. If we visualise the large cube, all smaller cubes will have at least one face facing inside. So none of the smaller cubes will have all faces painted. On the other side, the maximum number of faces of the larger cube that intersect at a point = 3 (at the corners). So, the smaller cubes can have a maximum of 3 faces painted. In other words, there are three types of painted small cubes. Cubes painted on 3 sides These smaller cubes are located at the corners of the large cube. The number of smaller cubes with 3 faces painted = The number of corners in the larger cube = 8. Cubes painted on 2 sides To find the number of smaller cubes with only 2 faces painted, we need to consider the cubes where 2 faces of the larger cube meet, i.e. the edges. So these smaller cubes are positioned on the edges of the large cube. Remember, each edge includes the smaller cubes present at the corners as well, which are painted on 3 sides. So we need to remove those 2 cubes from the number of cubes on each edge. So the number of smaller cubes with 2 faces painted at each edge = n — 2 (Here n is the length of each side of the cube). Overall, there will be 12 such edges on the larger cube. So the number of smaller cubes with 2 faces painted = 12(n — 2) = 12 8 = 96. (Here n = 10). Cubes painted on 1 side These smaller cubes are located at the face of the larger cube, excluding the cubes at corners and edges. At each face, the number of such cubes = (n — 2) (n — 2). There are 6 faces of larger cubes, so the total number of smaller cubes with one face painted = 6 (n — 2) (n — 2) = 6 8 8 = 384. From the above analysis, total number of painted cubes = 8 + 12(n — 2) + 6 (n — 2) (n — 2) = 8 + 96 + 384 = 488. The number of cubes not painted = 1000 – 488 = 512. Approach 2 In n x n x n cube, if we remove the outer layer of all 1 x 1 x 1 painted small cubes, then the dimensions of the hidden cube (not painted) will be (n — 2) x (n — 2) x (n — 2). So number of small 1 x 1 x 1 cubes not painted = (n — 2)³ = 8³ = 512. Critical idea to think! Suppose we have a cuboid of dimension abc painted on all sides which is cut into smaller cubes of dimension 111. Then: Number of cubes with 0 sides painted = (a — 2) (b — 2) (c — 2) Number of cubes with 1 sides painted = 2[(a — 2) (b — 2) + (b — 2)(c — 2) + (a — 2)(c — 2) ] Number of cubes with 2 sides painted = 4(a + b + c — 6) Number of cubes with 3 sides painted = 8 Enjoy learning, Enjoy mathematics! Author: Shubham Gautam Share on social media: More blogs to explore Monty Hall Problem There are 3 doors behind which are two goats and a car. You pick door 1 hoping for the car but don’t open it right away. Monty Hall, the game show host who knows what's behind the doors, opens door 3, which has a goat. Here's the game: do you want to pick door No. 2? Is it to your advantage to switch your choice?Visual Proof: The Sum of Important Mathematical Series The summation formulas are used to calculate the sum of the sequence. In this blog, we have discussed the visual proofs: the sum of numbers from 1 to n (arithmetic series), the sum of infinite geometric series, the sum of squares of numbers from 1 to n, etc.Find GCD of Two Numbers: Euclidean Algorithm Given two non-negative integers, m and n, we have to find their greatest common divisor or HCF. It is the largest number, a divisor of both m and n. The Euclidean algorithm is one of the oldest and most widely known methods for computing the GCD of two integers.The Birthday Paradox The birthday paradox is strange and counter-intuitive. It's a "paradox" because our brain find it difficult to handle the compounding power of exponents. Real-world applications for this include a cryptographic attack called the "birthday attack".Find Square Root of a Number Given a natural number n, find the largest integer less than or equal to √n. This can be seen as a search problem where the search space S is the set {1, . . . , n}, and the number desired is the floor(√n), i.e., the largest integer that is less than or equal to √n.Probability Theory For Machine Learning (Part 1) Originated from the “Games of Chance,” probability in itself is a branch of mathematics concerned about how likely it is that a proposition is true. Follow us on: LinkedinMediumFacebook © 2020 Code Algorithms Pvt. Ltd. All rights reserved.
8710	https://www.geeksforgeeks.org/maths/exact-differential-equations/	Exact Differential Equations Last Updated : 23 Jul, 2025 Suggest changes 2 Likes Equation that involves a differential co-efficient is called a Differential equation. Let's take the equation P(x, y)dx + Q(x, y)dy = 0. This equation is called the exact differential equation if, ∂P/∂y = ∂Q/∂y. In this article, we have covered in detail about exact differential equation. Table of Content Exact Differential Equation Definition Testing for Exactness Solving Exact Differential Equations Solving Exact Differntial Equations by Integrating factor Exact Differential Equation Examples Exact Differential Equation Problems Exact Differential Equation Definition A differential equation Mdx + Ndy = 0 is said to be an exact differential equation, if ∂M/∂y = ∂N/∂x Testing for Exactness Rake functions P(x, y) and Q(x, y) having the continuous partial derivatives in a particular domain, then the differential equation is exact iff, ∂P/∂y = ∂Q/∂y Solving Exact Differential Equations We can solve exact differential equation by following the steps added below as: Step 1: Given differential equation can be written as Mdx + Ndy = 0 form considering as equation 1 Step 2: Check ∂M/∂y = ∂N/∂x Step 3: General solution of equation 1 is ∫Mdx +∫Ndy =C (y= constant) and (do not contain x) Solving Exact Differntial Equations by Integrating factor If Mdx + Ndy = 0 is a homogeneous differential equation and Mdx + Ndy is not equal to 0 then this equation can be solve by multiplying the differential equation with integrating factor and the integrating factor for the same is, 1/(Mx+Ny) for equation of the form Mdx + Ndy = 0. Example: Solve exact differential equation x2ydx - (x3+y3)dy = 0 Solution: Comparing with Mdx + Ndy = 0 M = x2y, and N = -(x3+y3) ∂M/∂y = x2and ∂N/∂x = -3x2 So here ∂M/∂y is not equal to ∂N/∂x Here, given equation is not exact differential equation Considering Integrating factor Integrating Factor = 1/Mx + Ny =1/(x2y)x+(-x3-y3)y = 1/(x3y-x3y-y4)=-1/y4 =-1/y4 is an integrating factor Multiplying the equation with Integrating factor =(-x2y/y4)dx+(x3+y3/y4)dy=0 =(-x2/y3)dx+(x3/y4+1/y)dy=0 Comparing with M1dx + N1dy= 0 M1=-x2/y2 and N1 = x3/y4+1/y Therefore ∂M1/∂y = ∂N1/∂x Finding ∫M1dx +∫N1dy =C ∫(-x2/y3)dx+(x3/y4+1/y)dy=C -x3/3y3+logy = C It is the solution for given equation. Exact Differential Equation Examples Examples of exact differential equations are: (2xy + y3)dx + (x2 + 3xy2)dy = 0 (3x2y - y3)dx + (x3 - 3xy2)dy = 0 (2xy + y2)dx + (x2 + 2xy)dy = 0 (x2cosy - ysinx)dx + (xsiny + ycosx)dy = 0 Read More: Linear Differential Equation Application of Differential Equation Exact Differential Equation Problems Problem 1: Solve (hx + by + f)dy + (ax + hy + g)dx = 0 Solution: Given equation is (hx+by+f)dy+(ax+hy+g)dx=0....(1) Step 1: Comparing with Mdx + Ndy = 0 M= ax+hy+g and N = hx+by+f Step 2: Check ∂M/∂y = ∂N/∂x ∂M/∂y = h ∂N/∂x = h Therefore ∂M/∂y = ∂N/∂x Step 3: The general solution of equation 1 is ∫Mdx +∫Ndy =C (y= constant) and (do not contain x) =∫(ax+hy+g)dy+∫(hx+by+f)dx=C =ax2/2+hy∫dx+g∫dx+0+by2/2+f∫dy=C =ax2/2+hyx+gx+by2/2+fy=C (here ∫dx= x, ∫dy = y) Therefore it is the solution for given differential equation. Problem 2: Solve (y2-2xy)dx-(x2-2xy)dy=0 Solution: Given equation is (y2-2xy)dx-(x2-2xy)dy=0....(1) Step 1: Comparing with Mdx + Ndy = 0 M= y2-2xy and N = -x2+2xy Step 2: Check ∂M/∂y = ∂N/∂x ∂M/∂y = 2y-2x ∂N/∂x = -2x+2y Therefore ∂M/∂y = ∂N/∂x Step 3: The general solution of equation 1 is ∫Mdx +∫Ndy =C (y= constant) and (do not contain x) =∫(y2-2xy)dx+∫(-x2+2xy)dy=C =y2∫dx-2y∫xdx+0=C =xy2-x2y=C Therefore it is a solution for a given differential equation. Problem 3: Solve (y(1+1/x)+cosy)dx+(x+logx-siny)dy=0 Solution: Given equation is (y(1+1/x)+cosy)dx+(x+logx-siny)dy=0 Step 1: Comparing with Mdx + Ndy = 0 M= y+y/x+cosy and N = x+logx-xsiny Step 2: Check ∂M/∂y = ∂N/∂x ∂M/∂y = 1+1/x-siny ∂N/∂x = 1+1/x-siny Therefore ∂M/∂y = ∂N/∂x Step 3: The general solution of equation 1 is ∫Mdx +∫Ndy =C (y= constant) and (do not contain x) ∫(y+y/x)+cosy)dx+∫(x+logx-siny)dy=C =xy+ylogx+xcosy+0=C =y(x+logx)+xcosy=C Therefore it is a solution for a given differential equation. Problem 4: Solve (x2-4xy-2y2)dx+(y2-4xy-2x2)dy=0 Solution: Given equation is (x2-4xy-2y2)dx+(y2-4xy-2x2)dy=0 Step 1: Comparing with Mdx + Ndy = 0 M= x2-4xy-2y2 and N = y2-4xy-2x2 Step 2: Check ∂M/∂y = ∂N/∂x ∂M/∂y = -4x-4y ∂N/∂x = -4x-4y Therefore ∂M/∂y = ∂N/∂x Step 3: The general solution of equation 1 is Mdx +∫Ndy =C (y= constant) and (do not contain x) =∫(x2-4xy-2y2)dx+∫(y2-4xy-2x2)dy=C =x3/3-4yx2/2-2y2x+y3/3=C =x3/3-2x2y-2xy2+y3/3=C Therefore it is a solution for given differential equation. Problem 5: Solve (2xy+y-tany)dx+(x2-xtan2y+sec2y)dy=0 Solution: Given equation is (2xy+y-tany)dx+(x2-xtan2y+sec2y)dy=0 Step 1: Comparing with Mdx + Ndy = 0 M= 2xy+y-tany and N = x2-xtan2y Step 2: Check ∂M/∂y = ∂N/∂x ∂M/∂y = 2x+1-sec2y = 2x-tan2y ∂N/∂x = 2x-tan2y Therefore ∂M/∂y = ∂N/∂x Step 3: The general solution of equation 1 is ∫Mdx +∫Ndy =C (y= constant) and (do not contain x) =∫(2xy+y-tany)dx+∫(x2-xtan2y+sec2y)dy=C =2y∫xdx+y∫dx-tany∫dx+0+0+∫sec2ydy=C =2yx2/2+xy-xtany+tany=C =xy(1+x)+tany(1-x)=C Therefore it is a solution for a given differential equation. 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8711	https://www.quantamagazine.org/simple-geometry-shows-when-two-shapes-are-equal-20221121/	Simple Geometry Shows When Two Shapes Are Equal \| Quanta Magazine Loading [MathJax]/extensions/MathZoom.js Physics Mathematics Biology Computer Science Topics Archive Blog Columns Interviews Podcasts Puzzles Multimedia Videos About Quanta An editorially independent publication supported by the Simons Foundation. Follow Quanta Newsletter Get the latest news delivered to your inbox. Email Subscribe Recent newsletters Gift Store Shop Quanta gear Physics Mathematics Biology Computer Science Topics Archive Saved Articles Create a reading list by clicking the Read Later icon next to the articles you wish to save. See all saved articles #### Log out Change password Type search term(s) and press enter What are you looking for? Popular Searches Mathematics Physics Black Holes Evolution Home The Simple Geometry Behind Brownie Bake Offs and Equal Areas 1 Read Later Share Copied! (opens a new tab) 1 Comments Read Later Read Later Quantized Academy The Simple Geometry Behind Brownie Bake Offs and Equal Areas ByPatrick Honner November 21, 2022 Proving that two polygons have the same area can be as easy as cutting them up and rearranging the pieces. 1 Read Later Robert Neubecker for Quanta Magazine Introduction By Patrick Honner Contributing Columnist November 21, 2022 View PDF/Print Mode geometrymathematicsQuantized AcademyQuantized ColumnsAll topics (opens a new tab) Gina the geometry student stayed up too late last night doing her homework while watching The Great British Bake Off, so when she finally went to bed her sleepy mind was still full of cupcakes and compasses. This led to a most unusual dream. Gina found herself the judge of the Great Brownie Bake Off at Imaginary University, a school where students learn lots of geometry but very little arithmetic. Teams of Imaginary U students were tasked with making the biggest brownie they could, and it was up to Gina to determine the winner. Team Alpha was the first to finish, and they proudly presented their rectangular brownie for judging. Gina pulled out a ruler and measured the brownie: It was 16 inches long and 9 inches wide. Team Beta quickly followed with their square brownie, which measured 12 inches on each side. That’s when the trouble began. “Our brownie is much longer than yours,” said Team Alpha’s captain. “Ours is clearly bigger, so we are the winners!” “But the short side of your rectangle is much shorter than the side of our square,” said a representative from Team Beta. “Our square is clearly bigger. We have won!” Gina found it strange to be arguing about this. “The area of the rectangular brownie is 9 times 16, which is 144 square inches,” she said. “The area of the square brownie is 12 times 12, which is also 144 square inches. The brownies are the same size: It’s a tie.” Quantized Academy Patrick Honner, a nationally recognized high school teacher from Brooklyn, New York, introduces basic concepts from the latest mathematical research. See all Quantized Academy Columns Both teams looked puzzled. “I don’t understand what you mean by ‘times,’” said one student, who had never been taught multiplication. “Me neither,” said another. A third said, “I heard about students at Complex College measuring area using numbers once, but what does that even mean?” Imaginary University was a strange place indeed, even as dreams go. What was Gina to do? How could she convince the teams that their brownies were the same size if they didn’t understand how to measure area and multiply numbers? Luckily, Gina had a genius idea. “Give me a knife,” she said. Gina measured 12 inches down the long side of the rectangular brownie and made a cut parallel to the short side. This turned the large rectangle into two smaller ones: one measuring 9-by-12 and the other 9-by-4. With three quick cuts she turned the 9-by-4 piece into three smaller 3-by-4 pieces. A bit of rearranging resulted in audible oohs and aahs from the crowd: Gina had turned the rectangle into an exact replica of the square. Both teams now had to agree that their brownies were the same size. By dissecting one and rearranging it to form the other, Gina showed that the two brownies occupied the same total area. Dissections like this have been used in geometry for thousands of years to show that figures are the same size, and there are many remarkable results about dissections and equivalence. Even today mathematicians still use dissection and rearrangement to fully understand when certain shapes are equivalent, leading to some surprising recent results. You’ve probably seen geometric dissections in math class when developing the area formulas for basic shapes. For example, you might remember that the area of a parallelogram is equal to the length of its base times its height: This is because a parallelogram can be dissected and rearranged into a rectangle. This dissection shows that the area of the parallelogram is equal to the area of a rectangle with the same base and height, which, as anyone who didn’t attend Imaginary University knows, is the product of those two numbers. Speaking of Imaginary U, the Great Brownie Bake Off was just heating up. Team Gamma approached with a large triangular brownie. “Here is the winner,” they boldly announced. “Both our sides are much longer than the others.” Gina measured the sides. “This has the same area too!” she exclaimed. “This is a right triangle, and the legs measure 18 and 16, and so the area is…” Gina paused for a moment, noticing the baffled looks on everyone’s faces. “Oh, never mind. Just give me the knife.” Gina deftly sliced from the midpoint of the hypotenuse to the midpoint of the longer leg, then rotated the newly formed triangle so that it made a perfect rectangle when nestled into the larger piece. “That’s exactly our brownie!” cried Team Alpha. Sure enough, the resulting rectangle was 9 by 16: exactly the same size as theirs. Team Beta had their doubts. “But how does this triangle compare to our square?” their team leader asked. Gina was ready for that. “We already know the rectangle and the square are the same size, so by transitivity, the triangle and the square are the same size.” Transitivity is one of the most important properties of equality: It says that if a= b and b = c, then a= c. Gina continued, “If the area of the first brownie is equal to the area of the second, and the area of the second brownie is equal to the area of the third, the first and the third brownies must have equal areas too.” But Gina was having too much fun with dissections to stop there. “Or we could just make a few more cuts.” First Gina rotated the rectangle that was formerly a triangle. Then she cut it using the exact same pattern she had used on Team Alpha’s rectangle. Then she showed how this new dissection of Team Gamma’s triangle could be turned into Team Beta’s square, exactly as she had done with Team Alpha’s rectangle. In this situation we say that the triangle and the square are “scissors congruent”: You can imagine using scissors to cut up one figure into finitely many pieces that can then be rearranged to form the other. In the case of the triangle and the square, the brownies show exactly how this scissors congruence works. Notice that the pattern works in either direction: It could be used to turn the triangle into the square or the square into the triangle. In other words, scissors congruence is symmetric: If shape A is scissors congruent to shape B, then shape B is also scissors congruent to shape A. In fact, the above argument involving the triangle, the rectangle and the square shows that scissors congruence is also transitive. Since the triangle is scissors congruent to the rectangle and the rectangle is scissors congruent to the square, the triangle is scissors congruent to the square. The proof is in the patterns: Just overlay them on the intermediate shape, as was done with the rectangle above. If you cut the triangle into pieces that make the rectangle, then cut up the rectangle into pieces that make the square, the resulting pieces can be used to form any of the three shapes. The fact that scissors congruence is transitive is at the heart of an amazing result: If two polygons have the same area, then they are scissors congruent. This means that, given any two polygons with the same area, you can always cut one up into a finite number of pieces and rearrange them to make the other. The proof of this remarkable theorem is also remarkably straightforward. First, slice each polygon into triangles. Second, turn each triangle into a rectangle, similar to how Gina rearranged the triangular brownie. Now comes the tricky technical part: Turn each rectangle into a new rectangle that is one unit wide. To do this, start chopping off pieces from the rectangle that are one unit wide. If you can chop the rectangle into an integral number of pieces of width 1, you’re done: Just stack them on top of each other. Otherwise, stop chopping when the last piece is between 1 and 2 units wide, and stack the rest on top of each other. Don’t worry if the rectangle itself is less than 1 unit wide: Just slice it in half and use the two pieces to make a new rectangle that’s twice as long and half as thick. Repeat as necessary until you’ve got a rectangle between 1 and 2 units wide. Now imagine that this final rectangle has height h and width w, with 1 <w< 2. We’re going to cut up that rectangle and rearrange it into a rectangle with width 1 and height h×w. To do this, overlay the h×w rectangle with the desired hw × 1 rectangle like this. Then cut from corner to corner along the dotted line, and cut off the little triangle at the bottom right following the right edge of the hw × 1 rectangle. This cuts the h×w rectangle into three pieces that can be rearranged into an hw × 1 rectangle. (Justifying this final dissection requires some clever arguments involving similar triangles. See the exercises below for the details.) Finally, put this last rectangle on top of the stack, and you’ve successfully turned this polygon — really, any polygon — into a rectangle of width 1. Now if the area of the original polygon was A, then the height of this rectangle must be A, so every polygon with area A is scissors congruent to a rectangle with width 1 and height A. That means that if two polygons have area A, then they are both scissors congruent to the same rectangle, so by transitivity they are scissors congruent to each other. This shows that every polygon with area A is scissors congruent to every other polygon with area A. But even this powerful result wasn’t enough to successfully complete the judging of Imaginary University’s Brownie Bake Off. There was still one entry left, and no one was surprised at what Team Pi showed up with. The moment Gina saw that circle coming she woke up from her dream in a cold sweat. She knew that it was impossible to cut up a circle into finitely many pieces and rearrange them to form a square, or a rectangle, or any polygon. In 1964 the mathematicians Lester Dubins, Morris Hirsch and Jack Karush proved that a circle is not scissors congruent to any polygon. Gina’s dream had turned into a geometric nightmare. Related: An Ancient Geometry Problem Falls to New Mathematical Techniques When Math Gets Impossibly Hard Why Triangles Are Easy and Tetrahedra Are Hard But as they always seem to do, mathematicians turned this obstacle into new mathematics. In 1990 Miklós Laczkovich proved that it’s possible to slice up a circle and rearrange it into a square, as long as you can use infinitely small, infinitely disconnected, infinitely jagged pieces that couldn’t possibly be produced with a pair of scissors. As surprising and exciting as Laczkovich’s result was, it only proved that such a decomposition is theoretically possible. It didn’t explain how to construct the pieces, only that they could exist. Which is where Andras Máthé, Oleg Pikhurko and Jonathan Noel came in: In early 2022 they posted a paper in which they matched Laczkovich’s accomplishment, but with pieces that are possible to visualize. Unfortunately, you won’t be able to use their result to settle any brownie bake offs. Scissors alone can’t produce the 10 200 pieces needed in their decomposition. But it is another step forward in answering a long line of questions that started when Archimedes first invented, or discovered, π. And it keeps us moving toward inventing, or discovering, new mathematics that previous generations couldn’t dream of. Exercises Explain how we know that in the derivation of the area formula for a parallelogram, the triangle we cut off fits perfectly into the space on the other side of the parallelogram. Explain why any triangle can be dissected into a rectangle. For exercises 3 and 4, consider the diagram used to show that an h×w rectangle is scissors congruent to an hw × 1 rectangle, with points labeled. Explain why △XYQ is similar to △ABX. What does this make the length of QY? Explain why △PCX is congruent to △AZQ. Click for Answer 1: There are many ways to show that the two triangles are congruent. One way is to note that the distance between parallel lines is constant, so the two right triangles have a pair of congruent legs. And in a parallelogram, opposite sides are congruent, which makes the two triangles congruent by the hypotenuse-leg triangle congruence theorem. You could also make an argument using the angle-side-angle triangle congruence theorem. Click for Answer 2: One of the great elementary results in triangle geometry is the triangle midsegment theorem: If you connect the midpoints of two sides of a triangle, the resulting line segment is parallel to, and half the length of, the third side. Because the segment is parallel to the third side, angles 1 and 3 are congruent corresponding angles. And angles 1 and 2 are same-side interior angles, so they are supplementary, which means their measures sum to 180 degrees. Since ∠ 1 is congruent to ∠ 3, that means angles 3 and 2 are also supplementary. Thus, when you flip the top triangle around and to the right, the congruent sides will match up perfectly, and angles 2 and 3 will form a straight line. This turns the triangle into a parallelogram, which, as we already know, can be turned into a rectangle. Click for Answer 3: Since BXYZ is a rectangle, both ∠ZBC and ∠ZYX are right angles. And since opposite sides of a rectangle are parallel, this makes ∠YQX congruent to ∠AXB, as they are alternate interior angles. Thus △XYQ is similar to △ABX by angle-angle similarity. In similar triangles sides are in proportion, so X Y A B=Q Y B X. Thus, h h w=Q Y w, and so QY= 1. Notice that, since ∠ADC is a right angle and ∠DAP and ∠YQX are congruent corresponding angles, this makes △DAP congruent to △YQX. This proves that you can slide △YQX into the spot currently occupied by △DAP, as is needed in the scissors congruence argument. Click for Answer 4: Notice that ∠AZQ and ∠PCX are both right angles, and thus congruent. Using properties of parallel lines as in exercise 3, we can also see that ∠AQZ and ∠PXC are congruent corresponding angles. Also in exercise 3, we showed that QY= 1. This makes QZ = w − 1, which is exactly what CX is equal to. Thus, △PCX is congruent to △AZQ by angle-side-angle triangle congruence. This justifies the other part of the argument that an h × w rectangle is scissors congruent to an hw × 1 rectangle. Share this article Copied! (opens a new tab) Newsletter Get Quanta Magazine delivered to your inbox Subscribe now Recent newsletters (opens a new tab) By Patrick Honner Contributing Columnist November 21, 2022 View PDF/Print Mode geometrymathematicsQuantized AcademyQuantized ColumnsAll topics (opens a new tab) Share this article Copied! 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8712	https://math.stackexchange.com/questions/4630037/what-is-the-neatest-formula-for-the-coefficients-of-the-composition-f-circ-g-w	Skip to main content What is the neatest formula for the coefficients of the composition f∘g where f,g are formal polynomials or generating functions? Ask Question Asked Modified 2 years, 2 months ago Viewed 198 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. There are well-known formulas for the coefficients resulting from multiplying two formal polynomials or generating functions (we define formal polynomial to be a generating function such that all coefficients of sufficiently high index are zero). Adding is even easier, as it is component-wise. Question: What is the neatest formula for the coefficients of the composition f∘g where f,g are formal polynomials or generating functions? I tried approaching it by writing (f∘g)(x)=f(g(x)) and substituting the expression for g into the expression for f. Even with summation notation, the expansion just devolved into an mess involving the multinomial theorem that seems difficult to simplify. I am hopeful that combinatorial identities or other techniques such as exchanging the order of the summations will yield a somewhat satisfactory formula. I'll settle for pretty much any non-trivial simplification. abstract-algebra polynomials generating-functions function-and-relation-composition formal-power-series Share CC BY-SA 4.0 Follow this question to receive notifications asked Feb 1, 2023 at 3:47 FavstFavst 3,46411 gold badge1010 silver badges2828 bronze badges 4 1 I don't know of any "nice" formula for the coefficients of the composite of two generating functions. In the case of exponential generating functions, we can relate composition to an operation on species (see the "composition" section of the linked wikipedia article), but in general I don't know of any results. A quick google brings up a paper of Kruchinin Composition of Ordinary Generating Functions, but I haven't read it so I'm not sure how neat a result it is. – Chris Grossack Commented Feb 1, 2023 at 3:52 2 It's called Faa di Bruno's formula and yes, it is terrible (en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula). It's better not to think about these things directly but conceptually in terms of operations on combinatorial species as HallaSurvivor says. – Qiaochu Yuan Commented Feb 1, 2023 at 4:33 What I am doing is to express composition of functions f(x) and g(x) (having a power series) as matrix-multiplication F⋅G after each function has a "Carleman"-matrix F resp G assigned. The evolution of coefficients in the matrix-operation is now schematic, but is of course nothing else than the explicite coefficients by the Faa di Bruno - formula. But working with it seems to be more transparent to me ... but it is too much to explain this (and its limits and so on ... ) in a comment. – Gottfried Helms Commented Feb 21, 2023 at 17:07 Perhaps an interesting introductory discussion might be in go.helms-net.de/math/tetdocs/ContinuousfunctionalIteration.pdf while the main focus is not function-composition but iteration. But well, perhaps useful anyway... – Gottfried Helms Commented Feb 21, 2023 at 17:41 Add a comment \| 1 Answer 1 Reset to default This answer is useful 4 Save this answer. Show activity on this post. for OGF, let [xi]f(x)=fi and [xj]g(x)=gj. Then you can rewrite the composition as [xk]f(g(x))=∑ifi[xk]gi(x)=∑ifi∑j1+⋯+ji=kgj1…gji For EGF, let [xii!]f(x)=fi and [xjj!]g(x)=gj, then [xkk!]f(g(x))=∑ifii![xkk!]gi(x)=∑ifii!∑j1+⋯+ji=k(kj1,…,ji)gj1…gji I think it doesn't really get prettier than that, and it's indeed better to interpret combinatorially. Informally, if f(x) is the genfunc of species A, and g(x) is the genfunc of species B, then f(g(x)) is the genfunc of species obtained by taking instances of species A and replacing their "atoms" by instances of species B. For example, if A is species of "sets" and B is species of "cycles", then f(g(x)) is the genfunc for the species of "sets of cycles", aka "permutations". The meaning of the composition formula for EGFs is then interpreted as "take an instance of A on i atoms, and replace its atoms with instances of B with total of k atoms across them". Then the multinomial coefficient accounts for the number of ways to distribute k new atoms between i instances of species B. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Jun 15, 2023 at 13:48 Oleksandr KulkovOleksandr Kulkov 1,39777 silver badges2121 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 polynomials generating-functions function-and-relation-composition formal-power-series See similar questions with these tags. 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8713	https://www.chemicalaid.com/tools/molarmass.php?formula=Ra&hl=it	Ra Massa Molare ChemicalAid Calcolatrici Bilanciatore di Equazioni Chimiche Calcolatrice di Massa Molare Calcolatore di Formule Empiriche Significant Figures Calculator Calcolatrice di Stechiometria di Reazione Calcolatore di Reagente Limitante Oxidation Number Calculator Bond Polarity Calculator Tutte le Calcolatrici Elementi Tavola Periodica Proprietà Periodiche Elementi Futuri Informazioni Formule di Chimica Prefissi Chimici Equazioni di Chimica Point Group Symmetry Definizioni Soluzioni delle Domande Chemistry Quizzes Barzelletta Words from Elements Aiuto Chat Forum 🌎 IT italiano Ra Massa Molare La massa molare e il peso molecolare di Ra (Radio) è 226. Si prega di inserire una formula molecolare chimica (l’input tiene conto delle maiuscole) per determinare la massa molecolare del composto. Istruzioni) Formula Chimica Molecolare (ad es. H2O) 🛠️ Calcolare la Massa Molare Ra 1 2 3 4 5 6 7 8 9 0 () [Z]+- +1 -1 +2 -2 +3 -3 +4 -4 H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs BaHf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr RaRf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Scopri di più Calcolatrice Calculator Calcolatrici Istruzioni Istruzioni Questo programma determina la massa molecolare di una sostanza. Inserire la formula molecolare della sostanza. Verrà calcolata la massa totale insieme alla composizione elementare e alla massa di ogni elemento del composto. Utilizzare le maiuscole per il primo carattere nell'elemento e minuscole per il secondo carattere. Esempi: Fe, Au, Co, Br, C, O, N, F. È possibile utilizzare le parentesi () o le parentesi []. Trovare la Massa Molare Leggi il nostro articolo su come calcolare la massa molare. Esempi Fe3(PO4)28H20 Zn(ZnO) AlCo3 4CaCO3 COOHCH2OH CH2ClCONH2 ZnO(H2O) ZnO TiO2 Pb206 Pb204 OCN{-} Masse Molari Calcolate di Recente Peso Molecolare di Ra 226 g/mol La massa molare e il peso molecolare di Ra / Radio è 226. Scopri di più Calculator Calcolatrici Calcolatrice Composizione di Radio - Ra \| Elemento \| Simbolo \| Massa Atomica \| # di Atomi \| Percentuale di Massa \| --- --- \| Radio \| Ra \| 226 g/mol \| 1 \| 100% \| Nome Radio Ra Scopri di più Calculator Calcolatrici Calcolatrice Calcolatrici equazione chimica Bilanciatore di Equazioni Chimiche Calcolatrice di Stechiometria di Reazione Calcolatore di Reagente Limitante Ionic Equation Calculator ossidoriduzione composto chimico Calcolatore di Formule Empiriche Calcolatrice di Massa Molare Oxidation Number Calculator solubilità struttura di Lewis Bond Polarity Calculator calcolo Significant Figures Calculator Calcolatrice di Equazioni Chimiche equazione di stato dei gas perfetti Convertitore di Unità copyright © 2008-2025 Chi siamo · Privacy Policy · Termini Di Servizio 🇮🇹🌎 italiano Lingue 🇺🇸English 🇪🇸español 🇫🇷français 🇩🇪Deutsch 🇵🇹português 🇮🇹italiano 🇷🇺русский 🇵🇱polski 🇨🇿čeština 🇻🇳Tiếng Việt 🇮🇩Indonesia 🇰🇷한국어 🇯🇵日本語 🇨🇳中文 🇸🇦العربية 🇹🇷Türkçe
8714	https://brainly.com/question/47913249	[FREE] Prove that the locus of the midpoint of a chord of the parabola y^2 = 4ax which subtends a right angle at - brainly.com Advertisement 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 +88,9k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +27,1k Ace exams faster, with practice that adapts to you Practice Worksheets +7,3k Guided help for every grade, topic or textbook Complete See more / Mathematics Expert-Verified Expert-Verified Prove that the locus of the midpoint of a chord of the parabola y 2=4 a x which subtends a right angle at the vertex is y 2=2 a(x−4 a). 1 See answer Explain with Learning Companion NEW Asked by Danay9401 • 02/15/2024 Advertisement Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 4197801 people 4M 0.0 0 Upload your school material for a more relevant answer The locus of the midpoint of a chord of the parabola y² = 4ax, which subtends a right angle at the vertex, is proved to be y² = 2a(x-4a) by using the properties of tangents and the geometric attributes of a parabola. Explanation To prove that the locus of the mid-point of a chord of the parabola y² = 4ax which subtends a right angle at the vertex is y² = 2a(x-4a), we start by considering the geometry of a parabola and the properties of chords and tangents. Let's begin with parametric coordinates of a parabola. For a parabola y² = 4ax, the parametric coordinates are x = at² and y = 2at. Given that a chord AB subtends a right angle at the vertex, the tangents at points A and B are perpendicular. According to the property of tangents to a parabola, the product of their slopes is -1. Let the coordinates of A and B be (at_1², 2at_1) and (at_2², 2at_2), respectively. The slope of the tangent at A is (1/t_1), and at B is (1/t_2). For these to be perpendicular, t_1 t_2 = -1. The midpoint M of the chord AB will have coordinates ((at_1² + at_2²)/2, (2at_1 + 2at_2)/2). Simplifying, we get M = (a(t_1² + t_2²)/2, a(t_1 + t_2)). Given t_1 t_2 = -1, t_1 + t_2 = 0 (for any two numbers, if their product is -1, their sum is 0 if they are additive inverses). Substituting t_1 + t_2 = 0 into the coordinates of M, we obtain the x-coordinate as (a(t_1² + t_2²)/2) which simplifies to x = a(t_1² + 1) since t_1 t_2 = -1. To find the relation for y², we substitute the y-coordinate's expression into the parabola equation, deriving y² = 2a(x-4a), proving the locus. Answered by joymathew99608 •40.3K answers•4.2M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 4197801 people 4M 0.0 0 Upload your school material for a more relevant answer The locus of the midpoint of a chord of the parabola y 2=4 a x that subtends a right angle at the vertex is y 2=2 a(x−4 a). This is derived using the parametric coordinates of the parabola and the properties of the tangents. By analyzing the relationships of the parameters of the chord, we successfully find the required locus equation. Explanation To prove that the locus of the midpoint of a chord of the parabola y 2=4 a x, which subtends a right angle at the vertex, is y 2=2 a(x−4 a), we can use the properties of the parabola and the characteristics of tangents. Step-by-Step Explanation: Parametric Representation: For the parabola y 2=4 a x, we can use the parametric equations: x=a t 2 and y=2 a t Here, t represents the parameter, where points on the parabola are defined by a parameter t. Chords Subtending a Right Angle at the Vertex: Let us consider two points on the parabola, say points A and B, corresponding to parameters t 1 and t 2. So, the coordinates of A and B are respectively: A(a t 1 2,2 a t 1)B(a t 2 2,2 a t 2) The condition for these points to subtend a right angle at the vertex (0,0) is given by the slopes of tangents at these points being negative reciprocals: t 1×t 2=−1 Finding the Midpoint M: The midpoint M of the chord AB will have coordinates: M(2 a t 1 2+a t 2 2,2 2 a t 1+2 a t 2)=(a 2 t 1 2+t 2 2,a(t 1+t 2)) Using the relationship t 1×t 2=−1, we determine that: t 1+t 2=0 and t 1 2+t 2 2=(t 1+t 2)2−2 t 1t 2=0−2(−1)=2 Substituting to Find x-coordinate: Plugging t 1 2+t 2 2 into the x-component of the midpoint, we get: x=a 2 2=a Finding the Relation for y-coordinate: Substituting back into the expression for y: y=a(t 1+t 2)=a(0)=0 Therefore, to find the locus, we relate y to x. Deriving the Locus Equation: The derived condition, after substituting back into the parabola equation, leads to: y 2=2 a(x−4 a) Hence, the locus of the midpoints of such chords is characterized by the equation y 2=2 a(x−4 a). Conclusion: Thus, we have proved that the locus of the midpoint of a chord that subtends a right angle at the vertex of the parabola y 2=4 a x is indeed y 2=2 a(x−4 a). Examples & Evidence For example, if we take points on the parabola such as (a, 2√a) and (a, -2√a), both subtend a right angle at the vertex (0, 0), confirming the derived locus equation. Additionally, checking other pairs that satisfy t 1t 2=−1 can provide various midpoints that fit within the locus described by y 2=2 a(x−4 a). The derivation is based on well-established geometric properties of parabolas and tangential relationships, which are commonly taught in high school mathematics courses. The final relation derived follows from consistent algebraic manipulations shaped by the conditions set by the parameters of the parabola. Thanks 0 0.0 (0 votes) Advertisement Danay9401 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer The locus of the midpoints of the chords of the circle x² + y² = 4 which subtend a right angle at the origin is Community Answer The locus of point of intersection of two normals drawn to the parabola y²=4ax which are at right angles is a. y² = a(x-3a) b. y² = a(x-a) c. y² = 3a(x-2a) d. y² = 2a(x-2a) 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? New questions in Mathematics Write an equation of the line that passes through (−2,−3) and is parallel to the line defined by 5 x+y=−5. Write the answer in slope-intercept form and in standard form (A x+B y=C) with smallest integer coefficients. What are the x-intercepts of the function f(x)=−2 x 2−3 x+20? Given the function f(x)=−5 x 2−x+20, find f(3). A circular garden with a radius of 8 feet is surrounded by a circular path with a width of 3 feet. What is the approximate area of the path alone? Use 3.14 for π. A. 172.70 f t 2 B. 178.98 f t 2 C. 200.96 f t 2 D. 379.94 f t 2 (x 2−4)2 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy 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
8715	https://blog.khanacademy.org/impact-from-using-khan-academy/	Skip to content Impact from using Khan Academy by Shantanu Sinha We received an email a couple of months ago from a young man named Mark Halberstadt. In true Khan Academy style, he recorded a YouTube video for us to watch. His story is quite inspirational. Mark is a student who had given up on Math and Science and thought he was incapable of ever pursuing a career in a related field. He claims he was always a “C” student growing up and never had a channel to understand topics that interested him in engineering. He found the Khan Academy in 2007 and started watching videos on Trigonometry, Calculus, and even Arithmetic. He decided to go back to school last year to get a Bachelors degree in Electrical Engineering. He finished his first year in college with a 4.0 GPA for the entire year, including perfect scores on his Calculus and Chemistry final exams. He says, “coming from a background where my GPA graduating from high school was in the 2.0 range. That never would have happened – getting a 4.0 GPA would never have happened without the help I got from the Khan Academy.” He goes on to say, “It has helped me immensely. The impact for me in my life…I see it growing exponentially over the next 20 or 30 years." Students have always liked Khan Academy. The YouTube comments Sal received when he posted his first videos in 2006 are what motivated him to keep going (creating over 2400 as of today). However, it is still astonishing to see the impact our resources can have. Of course, while Mark is thanking us, he deserves 100% of the credit. He did all the work and took control of his own learning after the educational system left him behind. We just played a small role in enabling him. If there were only 1 student like Mark Halberstadt in the world, it would have been worth creating Khan Academy. The fact that this could so easily scale to millions is what makes Khan Academy special. Khan Academy seems to work well with supplemental learners like Mark, but how well does it work in schools? We have always believed that a great teacher can take our resources and push learning to new heights, by better focusing on the individual needs of each student. With the student mastering core skills on the computer, the teacher can leverage the classroom time for more engaging and dynamic activities such as project-based learning, peer tutoring, or lively discussion. Last school year, we started piloting our platform in a few schools in Los Altos, California. Our goal was to create better tools by directly observing how teachers and students interacted with our product. Los Altos was a fantastic partner, and our team built out many significant features based on their feedback (e.g., student knowledge map, teacher dashboards, badging infrastructure, new exercises). At the end of the school year, we all knew it was a success. Teachers could see a dramatic change in their students’ excitement and enthusiasm towards Math. Students who traditionally struggled with the material were more confident and engaged. Other students were challenging themselves to levels we never thought possible. Common sense told everyone involved that we were on to something. We did not do a controlled research study. In part, because our organization was only 5 people for most of the school year, and we were just trying to build something worth researching. Things changed fast for us, and the system the students were using at the end of the school year was very different from the system they started using in November. However, we were curious to see how they did on traditional assessments like the end of year CST exam. It is not the ideal exam since it only tests performance on a narrow set of grade-level skills. Many of our students were remediating topics they should have learned years ago, or challenging themselves with much more advanced topics. None of these gains would be captured. However, the CST clearly matters, so it is worth understanding how our students performed. The initial results were quite promising. Our pilot included a couple of 7th grade classrooms with students who typically struggled in Math. We saw a significant improvement in this group. The number of Advanced or Proficient students increased dramatically, from 23% to 41% This was very heartening. Usually, the performance gap widens with students who struggle in Math, particularly when they get to more advanced topics like Pre-algebra. The fact that these students were closing the gap (non-pilot classrooms saw no significant change in their CST performance) was very promising. Our pilot also included a few 5th grade classrooms. Los Altos is a high performing district, and these students typically do very well on the CST. This year was no different, with 96% of the students in pilot classrooms scoring Advanced or Proficient. While these are great results, they are not statistically different from the non-pilot classrooms in the district. It turns out, in both pilot and non-pilot classrooms, the students were doing great on the exam and didn’t have much room for improvement. However, we could see amazing things happening with the 5th graders. A majority of students were attempting early Algebra, and many students were experimenting with Trigonometry and Calculus. These students were excited, engaged, and loved being challenged. Inadvertently, we highlighted a distinct but not often discussed problem with standardized, age-focused education. Students performing at high levels are often not sufficiently challenged. Teachers shouldn’t take kids who already know the material, and make sure they already know the material. Teachers should be pushing and challenging the students to their full abilities. Los Altos didn’t think everything was perfect because their students were scoring well on standardized exams; they saw significant value in creating an environment that was engaging and challenging for all students. Based on these experiences, Los Altos has now decided to expand the implementation district-wide to over 40 more classrooms. We are also working with a number of additional schools that represent different use cases (e.g., charter, independent, low-income, special needs) to understand how students react to our resources in these different settings. This year we will also look into a better evaluation methodology that reflects learning gains across multiple grade levels. We are completely convinced that our resources can have a huge impact on the learning process. But why exactly does Khan Academy work? Some people have a hard time understanding how online videos and practice exercises can make such a big difference. Or they misunderstand what Khan Academy is all about. Putting videos on YouTube is just a small piece of the equation. What Khan Academy enables is a fundamentally different way for students to approach learning. Here’s my take on the many innovations we are bundling together into a coherent experience. This is what is really making the difference. Students are free to learn anytime, anywhere Students can jump to where help is needed most, and spend as much time as necessary to master concepts The content is short, fun, approachable, and easily digestible There is a clear and continuous path to learning complex topics Students feel an increased sense of ownership – they are learning, not "being taught” The focus on core conceptual understanding ensures students build the necessary skills that are applicable in any curriculum used in schools Interactive practice ensures concepts truly sink in Rich data helps teachers monitor progress and provide focused support Teachers are empowered to make their classroom experiences much more fun, engaging, and social, with less lecturing and more project-based learning and peer tutoring We learned a lot this past year, and I suspect we will learn much more this coming school year. The results so far have been promising. However, in our view, we’re just getting started. We got office space and started building a team only 10 months ago. We still have a long way to go to reach our vision for technology-enabled education. Mapping the Future: Sal Khan’s Day in Ohio August 19, 2025 Fresh Updates to Khan Academy’s Physics Courses Bring More Rigor and Flexibility August 18, 2025 Go Back to School AI-Ready with Khan Academy: Download our Checklist August 18, 2025 Experience the best AI-powered tool in education. For Learners For teachers For parents Visit Khan Academy © 2025 Khan Academy
8716	https://www.teacherspayteachers.com/browse?search=adding%20decimals%20on%20a%20number%20line	Adding Decimals on a Number Line \| TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 2nd grade ELA 3rd grade ELA 4th grade ELA 5th grade ELA 6th grade ELA 7th grade ELA 8th grade ELA High school ELA Elementary ELA Reading Writing Phonics Vocabulary Grammar Spelling Poetry ELA test prep Middle school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep High school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep Math Math by grade PreK math Kindergarten math 1st grade math 2nd grade math 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math High school math Elementary math Basic operations Numbers Geometry Measurement Mental math Place value Arithmetic Fractions Decimals Math test prep Middle school math Algebra Basic operations Decimals Fractions Geometry Math test prep High school math Algebra Algebra 2 Geometry Math test prep Statistics Precalculus Calculus Science Science by grade PreK science Kindergarten science 1st grade science 2nd grade science 3rd grade science 4th grade science 5th grade science 6th grade science 7th grade science 8th grade science High school science By topic Astronomy Biology Chemistry Earth sciences Physics Physical science Social studies Social studies by grade PreK social studies Kindergarten social studies 1st grade social studies 2nd grade social studies 3rd grade social studies 4th grade social studies 5th grade social studies 6th grade social studies 7th grade social studies 8th grade social studies High school social studies Social studies by topic Ancient history Economics European history Government Geography Native Americans Middle ages Psychology U.S. History World history Languages Languages American sign language Arabic Chinese French German Italian Japanese Latin Portuguese Spanish Arts Arts Art history Graphic arts Visual arts Other (arts) Performing arts Dance Drama Instrumental music Music Music composition Vocal music Special education Speech therapy Social emotional Social emotional Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Adding Decimals on a Number Line 850+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search Grade Subject Supports Price Format All filters Filters Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 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 More Not grade specific Subject English language arts ELA test prep Grammar Reading Vocabulary Writing Math Algebra Applied math Arithmetic Basic operations Decimals Fractions Geometry Graphing Math test prep Measurement Mental math Money math Numbers Order of operations Place value Other (math) Science Computer science - technology Instructional technology Social studies European history Other (social studies) World languages Spanish More For all subjects Price Free Under $5 $5-10 $10 and up On sale Format Digital Other (digital) Easel Easel Activities Easel Assessments EBook Google Apps Image More Interactive whiteboards SMART Notebook Microsoft Microsoft PowerPoint Microsoft Word PDF Video Resource type Classroom decor Bulletin board ideas Posters Word walls Forms Classroom forms Teacher tools Lessons Classroom management Homeschool curricula Lectures Outlines Rubrics Tools for common core Unit plans Yearlong curriculum Printables Clip art More Hands-on activities Activities Centers Projects Internet activities Bell ringers Cultural activities Escape rooms Games Simulations Songs Instruction Handouts Interactive notebooks Scaffolded notes Student assessment Assessment Critical thinking and problem solving Study guides Study skills Test preparation Student practice Independent work packet Worksheets Flash cards Graphic organizers Homework Task cards Workbooks Standard Theme Seasonal Autumn Back to school End of year Spring Summer Winter Holiday Christmas-Chanukah-Kwanzaa Easter Halloween New Year St. Patrick's Day Thanksgiving Valentine's Day More Audience TPT sellers Homeschool Parents Staff & administrators More Language En español More Programs & methods Programs GATE / Gifted and Talented Montessori More Supports Special education Life skills Specialty Other (specialty) Speech therapy Language More Adding and Subtracting Decimals on a Number Line Created by Parker Rowland With this activity, students will use number lines to model addition and subtraction of decimals (tenths place only). There are six pages with 2 problems each. Each problem provides movable arrows that allow students to demonstrate understanding of the strategy for adding and subtracting decimals. 4 th - 6 th Decimals, Math CCSS 5.NBT.B.7 $2.50 Original Price $2.50 Add to cart Wish List Adding& Subtracting Rational Numbers Activity 7th Grade Math Escape Room Created by The Great Classroom Escape Practice addition and subtraction of rational numbers in an engaging digital escape room! Students will race to match equivalent expressions involving adding and subtracting positive and negative numbers, figure out the signs of sums and differences of decimals or fractions on a number line, solve word problems, find the distance between points using knowledge of absolute value, and much more to save a submarine that has been sabotaged. Based on seventh grade standard 7.NS.A.1, this online break 6 th - 8 th Basic Operations, Math, Other (Math) CCSS 7.NS.A.1 , 7.NS.A.1b , 7.NS.A.1c +1 Also included in:Integer Operations Activities Add, Subtract, Multiply & Divide Negative Numbers $4.99 Original Price $4.99 Rated 4.85 out of 5, based on 67 reviews 4.9(67) Add to cart Wish List Decimals tenths and hundredths comparing, adding, number lines Math Kits 4th Created by Rachel Lynette BIG SAVINGS ON THIS BUNDLE!Understanding decimals is one of the most important foundations of math. This bundle has everything you need for giving students lots of practice working with 4th grade standards including representing decimals with money and models, comparing and ordering decimals, showing them on a number line, and adding and subtracting them. This resource is perfect for fourth grade students, but is also great for reviewing your fifth graders. This product is great for Blended Lea 4 th - 5 th Decimals, Math, Math Test Prep CCSS 5.NBT.A.3 , 5.NBT.A.3b , 5.NBT.B.7 +2 Bundle (4 products) $23.80 Original Price $23.80 $16.80 Price $16.80 Rated 4.86 out of 5, based on 22 reviews 4.9(22) Add to cart Wish List FREE Adding and Subtracting Fractions with Like Denominators Equivalent Game Created by Promoting Success You will download eight free task cards for your 4th grade and 5th grade math students to practice and review: adding and subtracting fractions with like denominators on a number line equivalent fractionsadding mixed numbers with like denominators with regroupingbuilding fractions from unit fractionsword problemequivalent fractions denominators of 10 and 100decimals as fractions on a number linecomparing decimals to hundredths with visual modelsThese fractions task cards align with 4th grade Com 4 th Basic Operations, Fractions, Math Test Prep Also included in:Fractions 4th Grade Back to School Math Beginning of the Year Review Sub Plans FREE Rated 4.92 out of 5, based on 104 reviews 4.9(104) View 2 Files Wish List Rounding Decimals on a Number Line- Journal Notes Created by Love2Teach Math Rounding Decimals on a Number Line journal notes allows students to think about rounding conceptually. These notes are a wonderful addition for student notebooks, as it explains the thought process involved in rounding decimals, as well as provides a visual explanation. This is a great resource when teaching the following rounding concepts: Rounding decimals to the whole number, the nearest tenth, and the nearest hundredth! I strive to produce accurate, high-quality products. If you noti 4 th - 6 th Decimals, Math $1.00 Original Price $1.00 Rated 5 out of 5, based on 37 reviews 5.0(37) Add to cart Wish List Adding and Subtracting Rational Numbers \| Negative Decimals Word Problems Real Created by Level Their Learning In this real life food truck activity, students will add and subtract rational numbers from real world word problems. Specifically, they will add and subtract negative decimals in real world word problems. 12 problems for students to practice adding and subtracting rational numbers (decimals only) in real world word problems! I used this with kids who were mastering adding and subtracting rational numbers (decimals only) and needed more of a challenge and could handle real world word problems. 7 6 th - 8 th Math CCSS 7.NS.A.1 , 7.NS.A.1a , 7.NS.A.1b +3 Also included in:Real Life Applications 7th Grade Math Year Long CCSS GROWING BUNDLE $2.50 Original Price $2.50 Rated 4.25 out of 5, based on 4 reviews 4.3(4) Add to cart Wish List 4th Grade Decimal Addition/Subtraction Spider Craftivity PLUS EDITABLE VERSION Created by Archer's All Stars -- Rachel Archer Need a fun way to keep your kiddos working, but engaged?? This cut and paste activity might be the perfect solutions for you!! There are two already made 4th grade aligned ready to print spiders. One lesson is fourth grade aligned 4.4A addition and subtraction multi-step word problems. The other ready to print spider is fourth grade aligned 4.2H decimals on a number line tenths and hundredths! If you do not want to use those sheets above, I have included an EDITABLE version too! Make it 2 nd - 6 th Math, Reading, Vocabulary Also included in:EDITABLE Seasonal Cut and Paste Craftivity Projects Growing Bundle $4.00 Original Price $4.00 Rated 4.83 out of 5, based on 71 reviews 4.8(71) Add to cart Wish List NO PREP Decimals on a Number Line Lesson - Click and Teach Student Presentation Created by Rachel Lynette Teach Decimals on a Number Line the easy way with this fun and interactive Click and Teach presentation! This presentation is designed to be used with your whole class and requires almost no teacher prep! It can be used as a PowerPoint or as Google Slides. In addition to direct instruction, this lesson includes cooperative learning (Turn and Talk)learning through movement (Brain Breaks)plenty of practice opportunitiesYour students will be actively involved throughout the entire lesson! There a 3 rd - 5 th Decimals, Math, Place Value $4.00 Original Price $4.00 $3.00 Price $3.00 Rated 4.91 out of 5, based on 11 reviews 4.9(11) Add to cart Wish List Fractions and Decimals on a Number Line (5th Grade EnVision Math Power Point) Created by Fab 5th Fun This Power Point is made to go along with the EnVision Math series for 5th Grade Math. (Topic 9 Lesson 10- Fractions &Decimals on a Number Line). I include transitions and animations in my Power Point Presentations to keep students interested. My students LOVE the added effects! :) 5 th Decimals, Fractions, Math $1.50 Original Price $1.50 Rated 4.87 out of 5, based on 29 reviews 4.9(29) Add to cart Wish List Adding and Subtracting Rational Numbers Decimals Only Add and Subtract Decimals Created by Level Their Learning In this bundle, students will add and subtract rational numbers. Decimals only are used in these addition and subtraction of rational numbers problems. 3 games for students to practice adding and subtracting rational numbers with decimals only. Decimals are used in all adding and subtracting rational numbers games. I used with kids on target when teaching adding and subtracting rational numbers- decimals only. Games:Sink My Ship Board Game Trashketball 7.NS.A.1Apply and extend previous understan 6 th - 8 th Algebra CCSS 7.NS.A.1 , 7.NS.A.1a , 7.NS.A.1b +2 Bundle (3 products) $7.50 Original Price $7.50 $5.99 Price $5.99 Add to cart Wish List Adding Decimal Fractions Created by BudV Education First Adding Decimal Fractions Express a fraction with denominator 10 as an equivalent fraction with denominator 100 and add to another fraction with denominator 100 (4.NF.5)Rewrite a fraction with denominator 100 as a decimal number (4.NF.6)Represent a decimal number or decimal fraction on a number line (4.NF.7)Compare two decimals to hundredths by reasoning about their location on the number line (4.NF.7)Record the results of comparisons with the symbols, greater than, less than, and equal to (4.NF. 4 th - 5 th Decimals, Fractions CCSS 4.NF.C.5 , 4.NF.C.6 , 4.NF.C.7 $10.00 Original Price $10.00 Add to cart Wish List Adding and Subtracting Rational Numbers \| Integers Decimals Real Life Problems Created by Level Their Learning In this Real Life activity, students will add and subtract rational numbers in real world word problems. Decimals and integers are both used in these adding and subtracting rational numbers real world problems. 12 problems for students to practice adding and subtracting rational numbers in real world word problems! I used with kids who were mastering adding and subtracting rational numbers and could handle word problems. Real life word problems for middle schoolers really increases engagement! M 6 th - 8 th Math CCSS 7.NS.A.1 , 7.NS.A.1a , 7.NS.A.1b +2 Also included in:Add and Subtract Rational Numbers \| Integers and Decimals Differentiated Bundle $2.00 Original Price $2.00 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Adding and Subtracting Rational Numbers Decimals Only Add and Subtract Decimals Created by Level Their Learning In this Board Game, students will add and subtract rational numbers. Decimals only are used in these addition and subtraction of rational numbers problems. 16 problems for students to practice adding and subtracting rational numbers with decimals only. Decimals are used in all 16 adding and subtracting rational numbers problems. I used with kids on target when teaching adding and subtracting rational numbers- decimals only. After adding and subtracting rational numbers (decimals only), students 6 th - 8 th Algebra CCSS 7.NS.A.1 , 7.NS.A.1a , 7.NS.A.1b +2 Also included in:Adding and Subtracting Rational Numbers Decimals Only Add and Subtract Decimals $3.50 Original Price $3.50 Add to cart Wish List Decimals on a Number Line Anchor Chart Created by Engaging Upper Elementary Engage your upper elementary students with this colorful anchor chart that reviews the process of plotting decimals on a number line. Important vocabulary is highlighted along with examples of each term, and a closer look at how each point on a number line can be expanded by adding another place value to the decimal number. This anchor chart can be printed as a poster or printed individually for student notebooks. This anchor chart is included in the Decimals On a Number Line Activity! If you l 5 th - 6 th Math CCSS MP4 FREE Rated 5 out of 5, based on 5 reviews 5.0(5) Log in to Download Wish List Decimals on a Number Line Winter Created by Drummer Chick Arithmetic Students will integrate art and creativity into their math work as they graph positive decimals on a number line. They will mark benchmark decimals (fourths) and then draw miniature winter pictures at specified points. All 165 of my students happily complied when I asked them to graph with pictures. No one asked "Do I HAVE to draw?" An added benefit is that you will have a fun time grading these, as their cute and creative pictures are sure to delight! (My favorite was "dog wearing a swea 4 th - 7 th Decimals, Graphing, Math Also included in:Decimals on a Number Line Bundle $2.00 Original Price $2.00 Rated 5 out of 5, based on 24 reviews 5.0(24) Add to cart Wish List Ontario 6th Grade Math - Ordering Decimals on a Number Line - Hands on Activity Created by Clare in the classroom Engage your grade 6 math students with this fun, hands-on Ordering Decimals on a Number Line activity! Perfect for building number sense and practicing decimal ordering, students will work with decimals to the thousandths in both standard and word form. Students will receive paper "clothes" labeled with decimal numbers. Their task is to "hang" the clothes on a string clothesline; ordering the decimals in ascending or descending order. This interactive and visual activity reinforces place val 5 th - 7 th Decimals, Math, Numbers Also included in:Ontario 6th Grade Math Activity Bundle – Ordering Decimals & Measurement BINGO $3.00 Original Price $3.00 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Adding and Subtracting Decimals TASK CARDS Estimating with Number Lines Created by Shiny Gold Star Teaching Materials Adding and Subtracting DECIMALS with ESTIMATING These TASK CARDS are a great way to practice and review adding and subtracting decimals. They include the vital step, often overlooked, of estimating sums and differences, focusing on the use of a number line. With 40 cards, 40 problems, teachers can determine how many problems (10-40) and which problems need the most attention. After printing on cardstock and laminating, students can also use a dry-erase marker to help them ease into mental math 4 th - 5 th Math, Mental Math, Other (Math) CCSS 5.NBT.B.7 $3.50 Original Price $3.50 Rated 4 out of 5, based on 1 reviews 4.0(1) Add to cart Wish List Integers and Opposites on a Number Line 6.NS.C.6.A Math Self-Checking Pixel Art Created by Merola Merchandise Grab this digital SELF-CHECKING math activity on INTEGERS ON A NUMBER LINE AND OPPOSITES with INSTANT FEEDBACK, ENGAGEMENT and NO TEACHER PREP required! Students will add, subtract, multiply, and divide decimals to the thousandths. This resource aligns with the the 6th grade math standard 6.NS.C.6a: "Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., -(-3) 5 th - 7 th Basic Operations, Math, Numbers CCSS 6.NS.C.6a Also included in:6th Grade Math Spiral Review Bundle- The Number System Self-Checking Bundle FREE Rated 4.67 out of 5, based on 3 reviews 4.7(3) Log in to Download Wish List Decimal Card & Dice Games Comparing, Fractions to Decimals, Adding& Subtracting Created by Curious Classroom Adventures Use these 4th grade decimal card and dice games to feed your students' need to play and reinforce taught concepts like converting fractions to decimals, comparing and ordering decimals, and adding and subtracting decimals and fractions at the same time! Unleash the fun with 7 math games for decimals. The specific concepts included cover the 4th grade standards for decimals. Game #1 - Fill the Card: students will roll dice to create a hundredths decimal and then use strategy as they model that 3 rd - 5 th Decimals, Fractions, Math Also included in:No Prep 4th Grade Math Games with Playing Cards & Dice Independent Partner Games $7.97 Original Price $7.97 Add to cart Wish List Add& Subtract Decimals Foldable (TEKS 5.3K) Created by Tay Teaching Texas This is a great introduction for adding and subtracting decimals up to the thousandths place value. My students use this in their interactive math journals. This foldable allows students to practice lining up decimal points and place value using place holder zeros in empty spaces. It also provides practice on where a decimal is on a whole number and how to line up place value before adding or subtracting. TEKS 5.3K 5 th - 6 th Decimals, Math, Other (Math) CCSS 5.NBT.B.7 , 6.NS.B.3 $0.50 Original Price $0.50 Rated 4.82 out of 5, based on 11 reviews 4.8(11) Add to cart Wish List 4th Grade Fraction and Decimal Review \| Equivalent Fractions, Comparing, Adding Created by Taylor Crafted Education Covers 4th Grade standards of Equivalent Fractions, Adding and Subtracting Like Fractions & Mixed Numbers, Comparing Decimals, and Fractions on Number Line practice shown on a ruler. ✦Uses: Test Prep, 5th Grade Back to School Math Review, Fraction or Decimal Review, Sub Plans. Answer Key included. JEOPARDY STYLE GAMES⭐️ Lines and Angles Review \| Interactive Game Show⭐️ Fraction Game Show \| Interactive Game Show ⭐️ Decimal and Fraction Review \| Comparing, Multiplying Fractions⭐️ Area and 4 th - 5 th Fractions, Math, Math Test Prep CCSS 3.NF.A.3b , 4.NF.A.1 , 4.NF.B.3a +3 Also included in:17 Fraction Resources: Equivalent, Number Lines, Comparing, Decimals, Adding $3.00 Original Price $3.00 $2.00 Price $2.00 Rated 5 out of 5, based on 2 reviews 5.0(2) Add to cart Wish List Modelling and Ordering Decimals and Fractions on a Number Line - Shade and Order Created by Sciensanity This bundle includes five sheets that will challenge your students to model fractions and decimals, place them on a number line, then write them in order. Students will explore multiple ways to write and conceptualize decimals and fractions so they understand WHY they go in the order that they do. Viewing decimals and fractions in multiple contexts gives students a deeper understanding of what these numbers actually represent. Each sheet explores one set of four decimal numbers and fractions th 3 rd - 7 th Applied Math, Decimals, Math CCSS 5.NBT.A.1 , 5.NBT.A.2 , 5.NBT.A.3 +13 $3.49 Original Price $3.49 Rated 4 out of 5, based on 2 reviews 4.0(2) Add to cart Wish List Adding& Subtracting Fractions Skittles Activity Created by theMATHisMATHIN In this activity, students write fractions to represent the number of Skittles that they have. They will reduce their fractions, convert them to decimals, represent their numbers on a number line, compare their numbers for given colors using <, >, or =, and will add and subtract the fractions. What a great way to practice a variety of skills! Candy is just the icing on the cake! 6 th - 8 th Algebra, Arithmetic, Fractions FREE Rated 4.88 out of 5, based on 16 reviews 4.9(16) Log in to Download Wish List Adding and Subtracting Rational Numbers \| Decimals and Fractions Bundle Created by Level Their Learning In this differentiated bundle, students will add and subtract rational numbers. Decimals and fractions are both used. Half of every activity is adding and subtracting positive and negative decimals, and half is adding and subtracting positive and negative fractions. Easier choices: This or That?, Coloring Page (with exit tickets) Harder choices: Board Game, Snowmen Decorations 7.NS.A.1Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; repre 6 th - 8 th Math CCSS 7.NS.A.1 , 7.NS.A.1a , 7.NS.A.1b +2 Bundle (4 products) $8.00 Original Price $8.00 $6.25 Price $6.25 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List 1 2 3 4 5 Showing 1-24 of 850+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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8717	https://www.sparkl.me/learn/cambridge-igcse/mathematics-international-0607-advanced/manipulating-algebraic-fractions/revision-notes/2847	Topic 2/3 Revision Notes Flashcards Your Flashcards are Ready! 15 Flashcards in this deck. Start Start How would you like to practise? Easy Continue Shuffle Shuffle Manipulating Algebraic Fractions Introduction Key Concepts Simplifying Algebraic Fractions Simplifying algebraic fractions involves reducing the expression to its simplest form by factoring and canceling common factors from the numerator and the denominator. This process is analogous to simplifying numerical fractions but requires a firm grasp of polynomial factorization. Steps to Simplify: Example: Simplify the algebraic fraction $$\frac{6x^2 - 12x}{3x}$$ Solution: [ \frac{6x^2 - 12x}{3x} = \frac{6x(x - 2)}{3x} = \frac{6}{3} \cdot \frac{x(x - 2)}{x} = 2(x - 2) = 2x - 4 ] \end{p> Addition and Subtraction of Algebraic Fractions Adding and subtracting algebraic fractions require finding a common denominator, much like numerical fractions. The process ensures that the expressions have the same denominator, allowing for the straightforward addition or subtraction of the numerators. Steps to Add or Subtract: Example: Add $$\frac{1}{x}$$ and $$\frac{2}{x + 1}$$ Solution: [ \text{LCD} = x(x + 1) ] [ \frac{1}{x} = \frac{x + 1}{x(x + 1)}, \quad \frac{2}{x + 1} = \frac{2x}{x(x + 1)} ] [ \frac{x + 1 + 2x}{x(x + 1)} = \frac{3x + 1}{x(x + 1)} ] \end{p> Multiplication of Algebraic Fractions Multiplying algebraic fractions is more straightforward compared to addition and subtraction. It involves multiplying the numerators together and the denominators together, followed by simplifying the resulting expression. Steps to Multiply: Example: Multiply $$\frac{2x}{x^2 - 1}$$ by $$\frac{x + 1}{3}$$ Solution: [ \frac{2x}{(x - 1)(x + 1)} \times \frac{x + 1}{3} = \frac{2x(x + 1)}{3(x - 1)(x + 1)} = \frac{2x}{3(x - 1)} ] \end{p> Division of Algebraic Fractions Dividing algebraic fractions is performed by multiplying the first fraction by the reciprocal of the second. This method simplifies the division process and allows for easy cancellation of common factors. Steps to Divide: Example: Divide $$\frac{3x}{2(x + 2)}$$ by $$\frac{x}{x - 1}$$ Solution: [ \frac{3x}{2(x + 2)} \div \frac{x}{x - 1} = \frac{3x}{2(x + 2)} \times \frac{x - 1}{x} = \frac{3(x - 1)}{2(x + 2)} ] \end{p> Solving Equations Involving Algebraic Fractions Solving equations that incorporate algebraic fractions requires clearing the denominators to simplify the equation into a polynomial form. This technique facilitates the application of standard methods for solving polynomial equations. Steps to Solve: Example: Solve $$\frac{2}{x} + \frac{3}{x + 1} = 5$$ Solution: [ \text{LCD} = x(x + 1) ] [ 2(x + 1) + 3x = 5x(x + 1) ] [ 2x + 2 + 3x = 5x^2 + 5x ] [ 5x + 2 = 5x^2 + 5x ] [ 5x^2 = 2 \quad \Rightarrow \quad x^2 = \frac{2}{5} \quad \Rightarrow \quad x = \pm \sqrt{\frac{2}{5}} ] \end{p> Advanced Concepts Partial Fraction Decomposition Partial fraction decomposition is a technique used to break down complex algebraic fractions into simpler, more manageable fractions. This method is particularly useful in integration and solving differential equations, where simpler fractions facilitate easier computation. Theoretical Explanation: Any rational function (a fraction where both numerator and denominator are polynomials) can be expressed as a sum of simpler fractions, provided the degree of the numerator is less than the degree of the denominator. Steps for Decomposition: Example: Decompose $$\frac{5x + 3}{x^2 - x - 6}$$ Solution: [ x^2 - x - 6 = (x - 3)(x + 2) ] [ \frac{5x + 3}{(x - 3)(x + 2)} = \frac{A}{x - 3} + \frac{B}{x + 2} ] [ 5x + 3 = A(x + 2) + B(x - 3) ] [ 5x + 3 = (A + B)x + (2A - 3B) ] [ \begin{cases} A + B = 5 \ 2A - 3B = 3 \end{cases} ] Solving the system: [ A = 5 - B ] [ 2(5 - B) - 3B = 3 \Rightarrow 10 - 2B - 3B = 3 \Rightarrow 10 - 5B = 3 \Rightarrow 5B = 7 \Rightarrow B = \frac{7}{5} ] [ A = 5 - \frac{7}{5} = \frac{18}{5} ] [ \frac{5x + 3}{x^2 - x - 6} = \frac{\frac{18}{5}}{x - 3} + \frac{\frac{7}{5}}{x + 2} = \frac{18}{5(x - 3)} + \frac{7}{5(x + 2)} ] \end{p> Complex Fractions Complex fractions contain fractions within their numerators or denominators. Simplifying complex fractions involves reducing them to simpler forms, often by finding a common denominator or by multiplying the numerator and the denominator by a suitable expression. Methods to Simplify: Example: Simplify $$\frac{\frac{2}{x} + \frac{3}{x + 1}}{\frac{1}{x} - \frac{1}{x + 1}}$$ Solution: [ \text{LCD for numerator and denominator} = x(x + 1) ] [ \frac{\frac{2(x + 1) + 3x}{x(x + 1)}}{\frac{(x + 1) - x}{x(x + 1)}} = \frac{2x + 2 + 3x}{1} = 5x + 2 ] \end{p> Application in Real-world Problems Manipulating algebraic fractions extends beyond pure mathematics, finding applications in various fields such as engineering, economics, and the physical sciences. Understanding these concepts enables students to model and solve real-world problems effectively. Interdisciplinary Connections: Example: In electrical engineering, the impedance of a circuit component can be represented as an algebraic fraction. Manipulating these fractions allows engineers to design circuits with desired electrical properties. Comparison Table \| Operation \| Definition \| Steps Involved \| --- \| Addition/Subtraction \| Combining algebraic fractions with a common denominator. \| 1. Find the LCD. 2. Rewrite each fraction with the LCD. 3. Add or subtract the numerators. 4. Simplify the result. \| \| Multiplication \| Multiplying the numerators and denominators separately. \| 1. Multiply the numerators. 2. Multiply the denominators. 3. Simplify by canceling common factors. \| \| Division \| Dividing one algebraic fraction by another by multiplying by the reciprocal. \| 1. Take the reciprocal of the divisor. 2. Multiply the fractions. 3. Simplify the result. \| Summary and Key Takeaways Coming Soon! Examiner Tip Tips Remember the acronym FACT to simplify algebraic fractions: Factorize the numerator and denominator, Arrange common factors, Cancel out common terms, and Transform the simplified expression. Additionally, always check for restrictions on the variable to avoid division by zero. Using mnemonic devices like "FLAM" (Factor, LCM, Add/Subtract, Multiply) can help remember the steps for operations with algebraic fractions. Did You Know Did you know that the concept of partial fraction decomposition, a key technique in manipulating algebraic fractions, dates back to the 17th century with mathematicians like Isaac Newton? Additionally, algebraic fractions are fundamental in calculus, especially in integral calculus for simplifying complex integrands. These techniques are not only academic but are also applied in engineering fields such as electrical circuit design and fluid dynamics. Common Mistakes One common mistake students make is forgetting to factorize completely before simplifying. For example, incorrectly simplifying $$\frac{x^2 - 4}{2x}$$ as $$\frac{x - 4}{2}$$ instead of the correct $$\frac{(x - 2)(x + 2)}{2x}$$ and then simplifying to $$\frac{x + 2}{2}$$. Another frequent error is neglecting to identify the least common denominator when adding or subtracting fractions, leading to incorrect results. 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8718	https://www.egr.msu.edu/~kdeb/papers/c2020002.pdf	1 Generating Well-Spaced Points on a Unit Simplex for Evolutionary Many-Objective Optimization Julian Blank, Kalyanmoy Deb, Yashesh Dhebar, Sunith Bandaru, and Haitham Seada COIN Report Number 2020002 Abstract—Most evolutionary many-objective optimization (EMaO) algorithms start with a description of a number of pre-deﬁned set of reference points on a unit simplex. So far, most studies have used the Das and Dennis’s structured approach for generating well-spaced reference points. Due to the highly structured nature of the procedure, this method cannot produce an arbitrary number of points, which is desired in an EMaO application. Although a layer-wise implementation has been suggested, EMO researchers always felt the need for a more generic approach. Motivated by earlier studies, we introduce a metric for deﬁning well-spaced points on a unit simplex and propose a number of viable methods for generating such a set. We compare the proposed methods on a variety of performance metrics such as hypervolume, deviation in triangularized sim-plices, distance of the closest point pair, and variance of the geometric means to nearest neighbors in up to 15-dimensional spaces. We show that an iterative improvement based on Riesz s-Energy is able to effectively ﬁnd an arbitrary number of well-spaced points even in higher-dimensional spaces. Reference points created using the proposed Riesz s-Energy method for a number of standard combinations of objectives and reference points as well as a source code written in Python are available publicly at Index Terms—Many-objective optimization, Reference points, Das-Dennis points, Diversity preservation, Riesz s-Energy. I. INTRODUCTION R ECENT evolutionary many-objective optimization al-gorithms (EMaO) use a set of reference directions as guides to parallelly direct their search to ﬁnd a single Pareto-optimal solution along each direction. These so-called decomposition-based EMaO methods, such as MOEA/D , NSGA-III , , are gaining popularity due to their success in handling three to 15-objective problems involving convex, non-convex, multi-modal, disjointed, biased density based, and non-uniformly scaled problems. One of the requirements of these algorithms is the initial supply of a set of reference directions, a matter which has not been pursued much in the literature. Most studies use Das and Dennis’s structured method in which ﬁrst a set of points is initialized on an M-dimensional unit simplex (where M is the number of objectives): z ∈[0, 1]M satisfying PM i=1 zi = 1. Thereafter, a reference direction is constructed by a vector originating from the origin and connected to each of these Julian Blank, Kalyanmoy Deb and Yashesh Dhebar are at Michigan State University, East Lansing, USA (email: {blankjul,kdeb,dhebarya}@msu.edu). Sunith Bandaru is with University of Sk¨ ovde, Sweden (email: sunith.bandaru@his.se), and Haitham Seada is with Ford Motor Company, Dearborn, USA (email: seadahai@msu.edu). points. The number of points on the unit simplex is determined by a parameter p, which indicates the number of gaps between two consecutive points along an objective axis. It turns out that the total number of points (n) on the unit simplex is n = C M+p−1 p . (1) For example, if p = 10 is chosen for a three-objective problem (M = 3), then the total number of points on the unit simplex is C 3+10−1 10 = 12 10 or 66. These 66 points are well-spaced with an identical distance to their nearest neighbor on the unit simplex. If an EMaO algorithm works well to ﬁnd a single Pareto-optimal solution for each of these 66 reference lines (obtained by a vector originating from the origin and passing through each point), a well-spaced set of Pareto-optimal solutions will be expected at the end. If more points are desired, p can be increased by one (or, p = 11), and the total number of points must jump to 78. In other words, if exactly 70 points are desired on the unit simplex, there is no way we can use Das-Dennis approach to achieve them. The jump becomes severe with higher objectives, such as for a 10-objective problem, p values from 2 to 6 requires 55, 220, 715, 2,002, and 5,005 points. Besides the inability to construct an arbitrary number of points, there is another issue with the Das-Dennis approach. As M increases, the number of total points on the unit simplex increases rapidly, as shown for M = 10 in the right vertical axis in Figure 1. Although a sub-linear plot in the semi-log scale indicates weaker than exponential behavior, an extremely large number of reference points is generated for a reasonable value of p. Since the population size of an EMaO is almost the same as the number of reference points, this requires an extremely large population size. Another crucial problem for a large value of M is that all points in a reasonably sized reference set lie on the boundary of the unit simplex and very few points lie in the interior of the simplex. Calculations reveal that, when p < M, there is no interior point, and when p = M, there is exactly one interior point, irrespective of the size of M. With p > M points begin to appear in the interior, but the number of such interior points is nI = p−1 p−M , which is only a tiny fraction ρI = nI n = p! (p −1)! (p −M)! (M + p −1)! (2) of all n Das-Dennis points. Figure 1 shows that for a 10-objective problem (M = 10), the proportion of interior Das-Dennis points grows with p, but the proportion (left vertical 2 axis) is small. To provide an example, with p = 15, there are Interior Points Proportion of Total Points 5 25 Gap, p 10 100 1000 10000 1e+05 1e+06 1e+08 Total Points 15 1e+07 0 0.005 0.01 0.015 0.02 0.025 Proportion of Interior Points 10 20 Fig. 1. Proportion of interior points compared to total Das-Dennis’s points for 10-objective problem (M = 10). a total of 1,307,504 reference points, of which only 0.15% (or only 2,002) points are in the interior. The rest of the points lie on boundary of the unit simplex. In this paper, we extend a previous preliminary study in light of existing literature to propose a number of efﬁcient methods to generate an arbitrary number of well-spaced point-set on the unit simplex. Section II describes a number of existing space-ﬁlling methods which can be used to create points on a hypercube, and a few methods to map those points onto the unit simplex. In Section III we describe a way to measure uniform spacing of a point-set on a unit simplex and describe different approaches to create such points. In Section IV the well-spaced properties of the points obtained by three proposed methods from 3 up to 15-dimensional problems are evaluated by using a variety of metrics. Some practical matters are addressed in Section V and conclusions are drawn in Section VI. II. RELATED WORK The task of creating a well-spaced set of points on the unit simplex has two different aspects as the problem deﬁnition already implies: First, the points should be uniformly and as evenly distributed as possible so that no sparse or crowded areas exist. Second, all points must lie on the unit simplex in the ﬁrst quadrant, which means that each point’s objective values have to be positive and their sum must equal to one. Creating points uniformly without restricting them to be on the unit simplex has been studied extensively and is applied during the design and analysis of computer experiments, where the goal is to capture the effect of inputs on one or more outputs in the best possible manner with as few expensive experiments as possible. The literature distinguishes between model-dependent and model-independent methods. Model-dependent methods, such as full and fractional factorial designs , RSM designs , and D-optimal or G-optimal designs are restricted to small dimensional problems. A. Model-independent Methods With no prior information about the input-output relation-ship, these methods assume that important features of the input-output relationship are equally likely to be present in all parts of the input space. In order to capture these fea-tures through an experimental design, the samples are spread evenly throughout the input space. Such designs are called model-independent or space-ﬁlling designs. Three important categories of space-ﬁlling designs are: 1) Orthogonal arrays: An orthogonal array of strength d and index λ for κ factors (κ > d), each with s levels, is an experimental design that, upon projection onto any subset of d dimensions, resembles a full factorial, with each design point replicated λ times. The total number of designs required is therefore, n = λsd. The corresponding orthogonal array is denoted as (n, κ, s, d). (s3, 4, s, 2). Since, the number of points come as in a structured manner and cannot be set arbitrarily. 2) Latin hypercube design and sampling (LHS): A Latin hypercube design is any orthogonal array of strength d = 1 and index λ = 1. Since N = s, this design gives the ﬂexibility to generate an arbitrary number of samples when the factors are continuous. Latin hypercube sampling involves dividing each factor into N equal intervals. Each of the N required designs is obtained by randomly selecting a previously unselected interval in each factor and sampling a single value from it. 3) Number theoretic methods: Number theoretic methods, originally developed for quasi-Monte Carlo integration, are aimed at creating points uniformly in the input space by minimizing different discrepancy metrics, which are measures of uniformity. Various low-discrepancy (also called quasi-random) sequences are available in the liter-ature. Popular among them are Halton set , Hammer-sley set , Sobol set and Faure sequences. When using low-discrepancy sequences, randomization of samples can be achieved by skipping over used sequences completely or partially. Each of these methods require a way to transform the points inside a hypercube to a unit simplex . B. Mapping onto Unit Simplex All ﬁlling methods described above generate points in an M-dimensional hypercube (zk ∈[0, 1]). However, all points need to lie on the unit simplex to be a feasible solution for our problem. Assuming all coordinates are already positive, a naive approach is to divide each point’s (k-th) coordinate by the sum of all coordinates: z(i) k ←z(i) k . M X j=1 z(i) j . (3) A study has shown that this mapping function is not appropriate, because, despite creating a uniformly distributed set of points on the hypercube, the well-spaced nature of points in the unit simplex is not guaranteed. In order to map these points onto the (M −1)-dimensional unit simplex, the authors proposed: 1) Generate points z(i), i = 1, . . . , n in an (M −1)-dimensional unit hypercube. 3 2) Set i ←1. 3) Sort coordinates {z(i) 1 , z(i) 2 , . . . , z(i) M−1} of z(i) in ascend-ing order. 4) Let {y(i) 1 , y(i) 2 , . . . , y(i) M−1} be the sorted coordinates. 5) Deﬁne y(i) 0 = 0 and y(i) M = 1, so that y(i) 0 < y(i) 1 < y(i) 2 < . . . < y(i) M−1 < y(i) M . 6) Then, set z(i) k ←y(i) k −y(i) k−1 for k = 1, . . . , M to form the mapped point coordinate z(i) k . 7) If i < n, then set i ←(i+1) and go to Step 3, else Stop. Note that PM k=1 z(i) k = 1 is satisﬁed for each point i. C. Probabilistic Filling Methods Instead of creating a point-cloud in the hypercube and then using a mapping function to project it on a unit simplex, points can be directly sampled on the desired unit simplex. This can be achieved by choosing a structured probability distribution for each objective, so that the sum of the objective values is exactly one. One such method was proposed in , in which the ﬁrst objective z1 is chosen in [0,1] with a probability 1 −z1M−2 . This can be achieved by ﬁrst choosing a random number u1 ∈[0, 1] and then computing z1 = 1 − M−1 √u1. Thereafter, zj is computed with another random number u2 ∈[0, 1], as zj = 1 −Pk−1 j=1 zj 1 − M−j √u2 . The process is continued until zM−1 and the ﬁnal objective value is computed as zM = 1 −PM−1 i=1 zi. There exists a number of conformal mapping methods in which uniformly distributed points on a hypercube can be mapped onto a unit simplex. Since we are interested in a well-spaced distribution of points on the unit simplex, it may not be easy to use such methods efﬁciently. A suggestion of creating each member of an M-dimensional vector using an exponential probability distribution and then normalizing it is one way to generate a well-spaced distribution on a unit simplex . Other methods have been proposed to meet the uniform distribution property, however, they cannot be used easily to generate a ﬁnite number of well-spaced points on any dimensional unit simplex. D. Das-Dennis’s Approach In 1958, Scheff´ e ﬁrst proposed the “simplex-lattice” approach to generate a well-spaced point-set on a unit simplex in the context of experimental design. Fang and Wang generalized the approach to create well-spaced points on a contracted unit simplex. However, Das and Dennis used the simplex-lattice approach for multi-objective optimization for the ﬁrst time, hence we call the approach in their name. As mentioned in Section I, Das-Dennis approach is a structured approach which requires an integer gap parameter p (≥1) and then creates M+p−1 p structured points. The method is scalable to any number of objectives (M), but has a few drawbacks: • Number of points cannot be set arbitrarily, • Most points lie on the boundary of the unit simplex, which may not be of interest to decision-makers, and • The approach is not easily changeable to incorporate preference information. As an improvisation, a layer-wise construction process is implemented as shown in Figure 2. A relatively small value of f3 1 1 1 0.4 0.7 0.7 0.4 0.4 0.7 Layer 2 (p=2) Layer 1 (p=3) Layer 3 (p=1) f2 f1 Fig. 2. Layer-wise construction of Das-Dennis’s points for three-objective problem (M = 3). p (< M) is used for each layer. The ﬁrst layer covers the entire unit simplex, but the subsequent layers use a shrunk (i.e. down-scaled) unit simplex. For instance, instead of using p = 12 for a three-objective (M = 3) problem totalling 91 points, the use of three layers with layer-wise p = (3, 2, 1) requires (10, 6, 3) or 19 points in total of which only 9 points are on the boundary. Points from Layer 2 and above are guaranteed to lie in the interior in such a construction, but the uniformity of the points gets lost in the process. In addition, the layered approach can end up with different layouts for the same number of points. For example, with 5 objectives, generating 50 reference points can be done either using p = (3, 2) or p = (2, 2, 2, 1). Creation of a well-spaced distribution with arbitrary number of points is not possible by this method. Also, the user needs to deﬁne the shrinkage factor for each layer, which is not an easy task. Hughes proposed a single-objective optimization method in which a large number of random reference vectors are evaluated based on the proximity of associated non-dominated population members to choose a ﬁnal subset of reference vectors. However, the approach is quite different in principle to the one-time identiﬁcation of an arbitrary and controlled number of reference points, which is the main goal of this paper. III. PROPOSED METHODOLOGIES In EMaO methods, the choice of reference points on the respective unit simplex is an important matter. For an arbitrary number of desired reference points on the unit simplex, a perfect equi-spaced distribution of them is not possible, but a well-spaced distribution is still desired. Elsewhere , a well-spaced distribution of n points (set, z) over a unit simplex was deﬁned using a mean-squared error (MSD) measure, which was proposed to compute by solving an optimization problem. First, a large number of uniformly-distributed points (y(j), j = 1, . . . , K) on a unit simplex is created. Then, the desired 4 set of n well-spaced points (z(i), i = 1, . . . , n) is obtained by minimizing the following MSD measure: Minimize 1 K PK j=1 minn i=1 y(j) −z(i) 2 , subject to PM m=1 z(i) m = 1, i = 1, . . . , n, z(i) m ≥0, i = 1, . . . , n, m = 1, . . . , M. (4) The ﬁnal outcome depends on the chosen initial points and optimization becomes difﬁcult due to the min-operator. Other number-theoretic concepts were used to suggest a discrep-ancy measure to theoretically deﬁne an ‘equi-distributed’ or ‘uniformly-scattered’ point-set , but they did not suggest any simple procedure for arriving at a well-spaced point-set for any arbitrary dimension. Here, we deﬁne a closest neighbor distance (CND) among n points on a unit simplex by maximizing the Euclidean distance (Dmin) between the closest pair of points: Maximize Dmin(z) = mini,j∈{1,...,n},i̸=j z(i) −z(j) , subject to PM m=1 z(i) m = 1, i = 1, . . . , n, z(i) m ≥0, i = 1, . . . , n, m = 1, . . . , M. (5) The constraints force each point z(i) to lie on the unit simplex. The advantage of this formulation is that it does not require any initial point-set (such as y). However, the above problem involves n×M variables (set, z), a non-differentiable objective function involving n(n −1)/2 quadratic functions, n linear equality constraints and non-negativity constraints of all n × M variables. A smoother CND optimization problem can be written as follows, by introducing an additional variable ϵ: Maximize ϵ, subject to ϵ ≤ z(i) −z(j) , ∀i, j ∈{1, . . . , n}, i ̸= j, PM m=1 z(i) k = 1, i = 1, . . . , n, z(i) m ≥0, i = 1, . . . , n, m = 1, . . . , M. (6) The objective function is linear, but there are now n(n −1)/2 quadratic constraints, in addition to n linear equality con-straints and non-negativity of variables. There is no provable algorithm for solving such a nonlinear optimization problem. Hence, ﬁnding a well-spaced distribution for an arbitrary set of points (n) on a unit simplex remains as a challenging task. Note that for Das-Dennis commensurate n points given by Equation 1 for a speciﬁc p, by construction, the Das-Dennis approach produces every point exactly the same distance from its neighbor. This is precisely a property of an optimal solution. Since every boundary line of the unit simplex has a length of √ 2, p gaps will make two neighboring points to have a distance Dmin,∗= √ 2/p between them. When Matlab’s fmincon() routine is used to solve the smoother version for a three-dimensional problem, it can ﬁnd the optimal distribution matching the above Dmin,∗value for up to N = 21 points (with p = 5). For larger p, the number of variables to the above problem is too large for fmincon() to ﬁnd the optimal solution. See Supplementary document for more details. Instead of trying to solve Equation 6, in this section, we introduce different approaches towards ﬁnding a well-spaced distribution and directly present beneﬁts and drawbacks of each method. Also, we present some results to visualize their near-uniformity using standard visualization methods. Moreover, numerical results are presented in Section IV. A. Sampling Methods A naive approach is to use any of the space-ﬁlling sampling methods described in Section II-A to create points in a hypercube and map them onto the unit simplex by using the method described in Section II-B. In Figure 3a, we show a mapping of 91 points using the Latin Hypercube Sampling. Clearly, the resulting distribution shows a lack of uniformity. We have also observed similar outcome with other methods discussed in Section II-A. This indicates that using a space-ﬁlling method and mapping is not sufﬁcient to obtain a well-spaced distribution on the unit simplex. However, Figure 3b indicates that if a large number of points is sampled and then mapped to the unit simplex, the simplex can be covered apparently uniformly. We exploit this observation to propose reduction-based methods (see Section III-C). B. Construction-based Methods Instead of ﬁnding all n points at a time, the construction method uses a bottom-up approach in which the procedure starts with a single point z(0) (or a number of well-spaced points using methods of Sections II-B or II-C) on the unit simplex. Thereafter, points are added one by one in stages until a set of n points is obtained with a well-spaced distribution. The ﬁnal outcome of n points will depend on the initial point(s) chosen. The addition of points can be achieved by using a pre-deﬁned procedure or by an optimization procedure of maximizing the uniformity of points at every stage. At a stage, when k points are already found, the following optimization procedure (Pk+1), simpliﬁed from Equation 5, can be solved to obtain the (k +1)-th point z(k+1) on the unit simplex: Maximize mink i=1 z(k+1) −z(i) , subject to PM m=1 z(k+1) m = 1, z(k+1) m ≥0, m = 1, . . . , M. (7) In the above optimization problem for ﬁnding a single new point z(k+1), there are only M variables and one equality constraint. However, the structure of the problem is same as in Equation 5, but fewer variables may allow each problem Pk+1 to get solved with near-optimality within a reasonable computational effort. Instead of starting with a single initial point, the above procedure can be seeded with more than one well-spaced point on the unit simplex. For example, the process can start with M vertices as initial points or with m points from Layer 1 speciﬁcation (with a small p where p < M), as described in Section II-D. This is a reasonable choice of starting points because the extreme points should exist in the ﬁnal point-set. The construction method has two obvious drawbacks: (i) the above optimization method needs to be solved (n −1) times and the computation of the objective function gets more time 5 (a) Latin Hypercube Sampling of 91 Points mapped onto the unit simplex. (b) 10,000 randomly sampled points mapped onto the unit simplex. (c) Illustration of Construction where 10 points are already obtained (blue) and one more additional point need to be added. Three possible locations are marked in red. Fig. 3. Illustrations of Unit Simplices by using different Methods consuming with increasing k, (ii) each insertion of a new point is dealt independently, and (iii) the procedure may not produce a well-spaced distribution if a small number of points (n) is desired. Figure 3c visualizes the problem for 10 points (blue) which are already added and 3 points (red) which could be added with exactly the same distance to other existing points. It is obvious that if only 11 points are desired, the algorithm will not converge to a well-spaced distribution (corresponding to the optimum of Equation 5) because of the greedy selection at each stage. C. Reduction-based Methods Contrary to the construction-based methods, a completely opposite process can also be devised. First, a set of well-spaced boundary points WB of size nB on the unit simplex is generated and then another well-spaced interior point-set WI of size nI = n −nB (where n is the total number of desired points on the unit simplex) is obtained by using k-means algorithm. The ﬁnal point-set is given by WB ∪WI. Our investigations revealed that this two-step procedure of ﬁrst estimating WB and then WI performs signiﬁcantly better than optimizing the combined distribution of points WB ∪WI all at once. In order to obtain WB, we use the concept proposed in which is based on the partition number p used in the Das-Dennis method. We ﬁnd the largest p (say, ¯ p) which produces Das-Dennis points of size smaller than or equal to n, satisfying the following condition: C ¯ p+M−1 M−1 ≤n < C ¯ p+M M−1. (8) Thus, the number of chosen boundary points nB (< n) is nB = CM+¯ p−1 ¯ p −C ¯ p−1 ¯ p−M. (9) For instance, for creating a total of n = 100 points in a three-objective space (M = 3), ¯ p = 12 satisﬁes the above criterion, because C14 2 = 91 ≤100 < C15 2 = 105. Therefore, nB = C14 12 −C11 9 = 36 and nI = n −nB = 100 −36 = 64. The creation of WB boundary points is executed using the Das-Dennis approach, but the creation of WI interior points is not straightforward and we describe it next. First, the procedure starts with a large sample S of structured or random points on the unit simplex and we set WI = ∅and WB having nB Das-Dennis boundary points. Then, the point ¯ s ∈S which is maximally away (in the Euclidean distance sense) from the already selected points in w ∈W = WB ∪WI is identiﬁed using: ¯ s = argmax s∈S min w∈W z(s) −z(w) . (10) The point ¯ s is then added to WI and is also removed from S for further processing. The procedure is continued until nI points are chosen from S. Please note that this initialization of points is performed similarly by Singh et. al to select a posteriori subset of points from an archive obtained by an MOEA. The resulting distribution of points largely depends on the starting sample set S. A large set will produce a good distribution of interior points, but with an increasing size of S this method also becomes computationally more expensive. However, the points can be improved by executing a few iterations of the well-known k-means algorithm . In an implementation, all n points can be adjusted to ﬁnd a well-spaced distribution on the unit simplex, but the execution of regular k-means algorithm results into the loss of boundary points, as shown in Figure 4a. This happens because the randomly sampled point-set S does not contain any points outside of the unit simplex and, therefore, cluster centers have no incentive to stay on the boundary and are pushed towards the center (see Figure 4a). To overcome this effect, we use points in WI only for cluster-assignment but exclude them from the cluster-center-update step of k-means. The k-means algorithm terminates whenever the average movement of all points is below a certain threshold. We have set this threshold to 10−4 in our implementation. Figure 4b shows that the resulting distribution of points is nearly uniform. However, inspecting the point-set from Figure 4a more carefully reveals that some minor irregularities exist, and the alignment of points is not as well-spaced as produced by the Das-Dennis method. Nevertheless, the Reduction procedure allows any arbitrary number of points to be distributed well, but not to the extent desired. The next subsection discusses a method which optimizes the outcome of the Reduction procedure to get a better well-spaced distri-bution (see the Supplementary document for a comparison). 6 (a) Reduction method, when Edges are not ﬁxed. (b) Reduction method with ﬁxed Edges. (c) s-Energy method with s = 9. Fig. 4. Visualization of distribution of 91 points in 2-D unit simplex obtained by different methods. D. Riesz s-Energy Method The motivation behind the use of an energy concept to obtain a well-spaced distribution comes from nature. Multi-body and interacting physical systems eventually settle on a state which corresponds to the minimum overall potential energy. For two bodies, the potential energy is proportional to the inverse of the distance between them. The minimum poten-tial solution corresponds to a diverse distribution of multiple bodies in the three-dimensional physical space. While dealing with a many-dimensional (s) space, we use a generalization of potential energy called Riesz s-Energy which is deﬁned between two points (z(i) and z(j)) as U(z(i), z(j)) = 1 z(i) −z(j) s . (11) In our context, it is not clear how the dimension s should depend on the number of objectives (M), but some trial-and-error studies (shown in the Supplementary document) motivated us to set s = M 2 for all simulations here. For multiple (n) points, the overall s-Energy can be written as follows: UT (z) = 1 2 n X i=1 n X j=1 j̸=i 1 z(i) −z(j) s , z ∈Rn×M. (12) The concept for our energy method is to ﬁnd the z-matrix of size n×M which minimizes UT subject to every z(i) vector to lie on the unit simplex, that is, PM m=1 z(i) m = 1. We employ a gradient-based optimization method (Adam ) here. Due to very large magnitude of UT , we take a logarithm of UT and then compute the partial derivative of FE = log UT with respect to z(i) m , as follows: ∂FE ∂z(i) m = −d UT    n X j=1, j̸=i z(i) m −z(j) m z(i) −z(j) s+2   . (13) To make sure all points stay on the unit simplex, gradients are projected onto the unit simplex. Figure 4c shows the outcome for M = 3 with 91 points. Riesz s-Energy has been used before to successfully dis-tribute points uniformly on a sphere or other multi-dimensional surfaces , . It has also been shown that an inappropri-ate choice of s can lead to non well-spaced results. However, to the best of our knowledge a concrete implementation of op-timizing the objective function and applying the concept to the unit simplex has not been proposed yet. Moreover, it is worth mentioning that Riesz s-Energy has recently been utilized as a performance metric for multi-objective optimization in and as a reduction method for choosing a set of reference points from a large set in . The overall outline of the s-Energy method is as follows: 1) Generate an initial point-set z(i) (i = 1, . . . , n) on the unit simplex using the Reduction method. 2) Until maximum number of iterations is reached or a convergence criteria is met: a) Calculate gradient with respect to z(i): ∇z(i)FE = ∂FE ∂z(i) 1 , ∂FE ∂z(i) 2 , . . . , ∂FE ∂z(i) M for all points i = 1, 2, . . . , n. b) Project the gradient onto the unit simplex: ∇proj z(i) FE = ∇z(i)FE − 1 M PM m=1 ∂FE ∂z(i) m ˆ u, where ˆ u is M-dimensional vector of ones. This guarantees PM m=1 z(i) m = 1, ∀i. c) Update each z(i) m (for n points and for M dimensions) using the gradient-based optimizer Adam. Whenever a point z(i) is outside of the unit simplex, project it back onto the unit simplex to make sure it satisﬁes z(i) m ≥0, ∀i, m. As mentioned before, since the ﬁnal distribution of points depends on the initial conﬁguration, we use Reduction to generate an initial set of points. We deﬁne the termination criterion based on the average movement of all points (below 10−5) and perform a restart of Adam in case no improvement has been made for 50 iterations. Moreover, the maximum number of iterations is set to 3,000. IV. RESULTS In the following, we evaluate the performance of our pro-posed methods for two different scenarios. In the ﬁrst scenario, we choose the desired number of points in a manner admissible by the Das-Dennis method. We present these results mainly to compare other proposed methods of this paper with the Das-Dennis method. In the second scenario, we consider any desired number of reference points. Since the Das-Dennis method cannot be applied for such cases, we only compare the proposed methods of this paper with each other. 7 TABLE I EXPERIMENTAL RESULTS FOR 3-DIMENSIONAL UNIT SIMPLICES WITH VARYING NUMBER OF POINTS. Das-Dennis PROVIDES THE KNOWN OPTIMUM (Dmin,∗= √ 2/p), THE BEST PERFORMING METHOD (EXCEPT Das-Dennis) IS SHOWN IN BOLD. RIESZ S-ENERGY METHOD PRODUCES CLOSEST Dmin,∗ AND NEAR-ZERO IGD, TRI, AND VGM COMPARED TO OTHER METHODS. 21 (p = 5) 36 (p = 7) 91 (p = 12) 153 (p = 16) 325 (p = 24) % % % % % HV Sampling 1.3980450 100.00 1.2063700 100.00 1.0323220 100.00 0.9788900 100.00 0.9311150 100.00 Construction 1.4440990 7.81 1.2372800 26.41 1.0578780 6.28 0.9982060 1.76 0.9419730 10.80 Reduction 1.4479780 0.04 1.2483510 0.05 1.0596600 -0.25 0.9985990 -0.24 0.9432230 0.53 Riesz s-Energy 1.4480000 0.00 1.2483730 0.00 1.0595890 0.01 0.9985500 0.01 0.9432740 0.11 Das-Dennis 1.4480000 0.00 1.2483730 0.00 1.0595910 0.00 0.9985520 0.00 0.9432870 0.00 Dmin Sampling 5.230e-02 100.00 2.578e-02 100.00 9.831e-03 100.00 5.611e-03 100.00 2.683e-03 100.00 Construction 1.540e-01 55.90 1.500e-01 29.55 8.207e-02 33.13 4.721e-02 49.74 3.959e-02 34.37 Reduction 2.680e-01 6.45 1.829e-01 10.87 8.556e-02 29.90 5.923e-02 35.23 3.519e-02 42.20 Riesz s-Energy 2.819e-01 0.43 2.004e-01 0.91 1.142e-01 3.42 8.467e-02 4.49 5.204e-02 12.24 Das-Dennis 2.828e-01 0.00 2.020e-01 0.00 1.179e-01 0.00 8.839e-02 0.00 5.893e-02 0.00 IGD Sampling 1.053e-01 100.00 8.423e-02 100.00 5.301e-02 100.00 4.018e-02 100.00 2.732e-02 100.00 Construction 8.979e-02 85.23 7.287e-02 86.51 4.169e-02 78.64 3.089e-02 76.87 2.139e-02 78.31 Reduction 2.836e-03 2.69 5.174e-03 6.14 1.785e-02 33.67 1.731e-02 43.08 1.505e-02 55.10 Riesz s-Energy 2.673e-04 0.25 5.092e-04 0.60 1.693e-03 3.19 2.178e-03 5.42 2.486e-03 9.10 Das-Dennis 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 Tri Sampling 3.019e-01 100.00 2.134e-01 100.00 1.203e-01 100.00 8.716e-02 100.00 5.459e-02 100.00 Construction 8.948e-02 29.64 8.834e-02 41.40 4.884e-02 40.59 2.126e-02 24.40 1.908e-02 34.96 Reduction 9.390e-03 3.11 1.287e-02 6.03 2.544e-02 21.15 2.278e-02 26.13 1.712e-02 31.37 Riesz s-Energy 4.104e-04 0.14 5.637e-04 0.26 1.260e-03 1.05 1.229e-03 1.41 1.941e-03 3.56 Das-Dennis 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 VGM Sampling 5.853e-03 100.00 2.777e-03 100.00 8.847e-04 100.00 4.578e-04 100.00 1.978e-04 100.00 Construction 2.905e-03 49.63 7.980e-04 28.74 2.931e-04 33.13 1.913e-04 41.79 1.413e-05 7.15 Reduction 1.374e-05 0.23 2.102e-05 0.76 5.280e-05 5.97 3.583e-05 7.83 2.183e-05 11.04 Riesz s-Energy 7.484e-08 0.00 1.501e-07 0.01 5.906e-07 0.07 4.343e-07 0.09 1.076e-06 0.54 Das-Dennis 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 0.000e+00 0.00 A. Scenario 1: Das-Dennis Commensurate Points We compare our proposed methods – Sampling, Construc-tion, Reduction, and s-Energy – with the Das-Dennis method for a three-objective problem with ﬁve different number of reference points (n), as shown in Table I. The obtained points on the unit simplex are compared with ﬁve different performance metrics, which we describe next. • Hypervolume (HV): The hypervolume measure is frequently used for comparing results of multi and many-objective optimization algorithms. It measures the area or volume of the collective dominated region with the help of a reference point r which is pre-speciﬁed by the user. A suitable reference point r = (r, r, . . . , r)T is essential to provide equal importance to all points in the point-set. We use a suggested value : r = 1 + 1/p, (14) where p is the number of gaps used by Das-Dennis method. The larger the HV, the better is the performance of a method, but we also note that an improper choice of r may not cause the Das-Dennis method to produce the largest HV. • Inverted Generational Distance (IGD): The IGD indi-cator is another popular performance metric used in EMaO studies. It measures the average Euclidean distance (in the objective space) for each supplied Pareto-optimal solution to the corresponding closest solution in the provided solution set. Thus, this metric can only be used in cases where the location of the entire Pareto-optimal front is known. Moreover, since the supplied Pareto-set must be a well-spaced set on the unit simplex, we are only able to use IGD for Das-Dennis commensu-rate scenarios. • Distance of Closest Pair of Points (Dmin): For a per-fectly well-spaced point-set, the distance (Dmin) between the closest pair of points (described in Equation 5) is maximum compared to any other point-set. This metric ignores the distribution of all points except a single iso-lated pair of points which has the shortest distance along the unit simplex. However, for Das-Dennis commensurate n points with a speciﬁc p, this is an appropriate metric and has an upper bound on the objective value of √ 2/p. • Triangulation (Tri): This is another metric which can be used to gauge deviation of the obtained distribution from the optimal Das-Dennis point-set. The obtained point-set is ﬁrst triangulated and then the greatest difference between smallest and largest sides of triangles is used as a measure of uniformity of the triangles. Here, we use Delauney-Triangulation . A point-set obtained by Das-Dennis will make every triangle equilateral, similar to the plot in Figure 4c, thereby making the Tri measure equal to zero. Thus, the smaller the Tri measure, the better 8 is the performance of the method. It is also clear that this metric is governed by a few neighboring points and it fails to capture the uniformity measure of the entire point-set. Since triangulation method is computationally expensive for more than three dimensions, this metric will be difﬁcult to extend for many-objective problems. • Variance of Geometric Mean of k-Nearest Neighbors (VGM): To alleviate the singularity associated with the Dmin metric, we propose to measure the variance of geometric means (VGM). For an M-objective problem, we compute Euclidean distances from k-nearest neigh-boring points (k = M −1) for each obtained point. We then compute the geometric-mean of the k distances. The geometric mean, in this case, is a measure of a repre-sentative size of the hypervolume of the local simplex formed by a point touching its neighbors. For a well-spaced set of points (such as Das-Dennis method), the variance of the neighborhood size (geometric mean of the hypervolume) of each point will be zero. For other distributions, a smaller value of the variance may indicate a more well-spaced distribution. Interestingly, this metric can be extended for any dimension. However, a smaller variance does not ensure a better spread of points on the entire unit simplex. Thus, we propose to use this metric (a small value) along with Dmin metric (with a large value) to determine the superiority of a particular point creation method proposed in this paper. In Table I, the performance measures on a three-objective unit simplex are shown for 21, 36, 91, 153, and 325 points. Corresponding values of p to obtain the above set of points using Das-Dennis method are 5, 7, 12, 16, and 24, respectively. We report the median values of each performance metric over 101 runs, followed by its normalized value in [0, 100]% scale, where 0% corresponds to the Das-Dennis and 100% corresponds to the worst performing point creation method. The best performing method (excluding Das-Dennis method, due to its perceived superiority for chosen commensurate points) is marked in bold. It is observed that s-Energy method outperforms the other methods in almost all cases. Also, from the values of relative performance, it can be observed that s-Energy produces a very similar metric value as that of Das-Dennis method. Except the HV values, s-Energy and Das-Dennis methods are visibly better, hence we do not perform any statistical signiﬁcant test to establish the superiority of these two methods. For 91 and 153 points, the Reduction method obtains slightly better HV values. A negative relative performance implies that the Reduction method is able to outperform even the Das-Dennis method. A similar behavior was recently observed in where the authors have shown that in many cases uniformly-distributed objective vectors are not optimal regarding multi-objective performance indicators (including HV). This implies that HV does not necessarily reﬂect the uniformity of a point-set and, therefore, might not be a reliable performance indicator for the purpose of this paper. The minimum distance Dmin between the closest point pair is a fairly strict metric because one point, not being appropri-ately adjusted, can have a big inﬂuence on the resulting value. However, s-Energy is able to successfully obtain distributions with larger values of Dmin, whereas the other approaches do not reﬂect well on this strict metric. Increasing the number of points increases the overall density of the distribution, thereby resulting into smaller Dmin values. A similar observation is made considering the Tri metric. 9 Small values of VGM, Tri, IGD and large values of HV demonstrated by the s-Energy method with the metric values close to that of Das-Dennis method makes it a suitable ref-erence point creation method for three-objective Das-Dennis commensurate points. Next, we investigate how s-Energy method performs for arbitrary number of desired points for three as well as for larger dimensional spaces. B. Scenario 2: Arbitrary Number of Points Here, we consider three to 15-objective cases involving arbitrary number of desired points on the unit simplex, as shown in Table II. For this scenario, we cannot apply the Das-Dennis method, as no integer p value will correspond to an arbitrary number of desired points for a speciﬁc M-dimensional unit simplex. Thus, we restrict our comparison to four point-creation methods – Sampling, Construction, Reduction, and s-Energy. Also, since an equi-spaced point-set may not be available for an arbitrary n (in fact, ﬁnding a well-spaced set of points on the unit simplex is the goal of this study), we cannot use the IGD metric. The weakness of the HV metric demonstrated in Table I prohibits us to use it as well for the many-objective problems. Also, the computational complexity involved with the Tri metric deters us from using it here. Thus, we use Dmin and VGM metrics to compare the four point-creation methods. Their combination (large Dmin and small VGM) will ensure that points are spaced with a large distance from each other and for each point, its k-closest neighbors are almost equi-spaced. Table II shows the results regarding Dmin. The best per-forming method is marked in bold. The ﬁrst column for each method marks the median Dmin value of 101 runs and the second column indicates its relative performance among all four methods – 0% is marked for the best and 100% for the Sampling method. Other methods are assigned a relative percentage value based on their normalized Dmin values. The results clearly indicate that s-Energy is able to outperform other creation methods in all cases by producing the largest median Dmin value. The reduction of the Riesz s-Energy function with iteration of s-Energy method for M = 3 with 100 points is shown in the Supplementary document. Table II further shows that the Reduction method is not able to perform well, particularly for higher dimensional spaces. For instance, for the 15-dimensional case, its relative performance (about 93% to 100%) is almost similar to the Sampling method (100%), while for three and ﬁve-dimensional cases, its relative performance is below 50%, indicating a reasonable performance. This can be attributed to the fact that we have used a ﬁxed number (10,000) of sampled points on the unit simplex to start the Reduction method. However, for higher dimensional spaces, a larger number of points may be required to uniformly ﬁll the unit simplex. This is not computationally practical due to memory and running time limitations. However, the point-set obtained by the Reduction method can serve as an initial starting set for the s-Energy method. Results indicate that s-Energy is able to morph a point-set to form a well-spaced distribution eventually. Even though Construction is able to scale signiﬁcantly better than Reduc-tion, it is not able to match the performance of the s-Energy method. Moreover, it is worth mentioning that with increasing number of dimensions and points, the overall running time of Construction increases signiﬁcantly, since it demands a dedicated optimization-run for obtaining each single point, one at a time. Table III shows the performance of all four methods using the VGM metric. Results indicate that s-Energy is able to obtain signiﬁcantly smaller VGM values compared to other approaches. Also, similar to the trend observed for Dmin, Reduction is not able to perform well with increasing number of dimensions – for eight-objective and above, it performs signiﬁcantly worse than even the Sampling method. Con-struction performs the second best, after s-Energy method. The combined results of both performance metrics, Dmin and VGM, clearly validate the suitability of our proposed s-Energy method to effectively determine well-spaced point-set on the unit simplex on three to 15-objective problems. Further comparative results are presented in the Supplementary document. In addition to the performance comparisons made above, visualization of points can provide further validation of the uniformity of obtained points by the s-Energy method. It is worth noting here that the s-Energy point-sets can have an arbitrary size, which has been the primary motivation of this study. 1) First, we present results from Das-Dennis commensurate case with p = 12 for a three-objective unit simplex, requiring 91 points. Figure 5 shows the distribution of 91 points on the unit simplex in circles obtained using the s-Energy method. Next, we create a total of 92 points afresh, a case which is not possible to achieve by the Das-Dennis method. The points are marked using a cross. It is interesting to observe from the ﬁgure that only a few points towards the lower right corner of the simplex get readjusted to accommodate the extra (92nd) point. The choice of adjustment at the lower right corner is arbitrary and probably is an outcome of the starting distribution of points (see the Supplementary document for a slightly different distribution obtained in another run of the s-Energy method). The major portion of the overall distribution remains nearly similar to the conﬁguration obtained for n = 91 points. Visually, the resulting point-set with 92 points is uniform, even though it is known to the best of our knowledge that no perfectly well-spaced distribution of points with 92 or any arbitrary number of points exists. This study demonstrates the effectiveness of the proposed s-Energy method. 2) Next, we visually compare the distribution of points obtained by s-Energy for more than three dimensions. We present the points using the parallel coordinate plot (PCP) , which is one of the common visualization techniques in many-objective optimization. All axes are plotted vertically and represent different objectives. Each solution is illustrated by a line marking the objective value on each vertical axis. The crisscrossing of lines between two consecutive axes indicates the existence 10 TABLE II COMPARISON OF FOUR METHODS USING THE CLOSEST DISTANCE PAIR OF POINTS METRIC (DMIN) FOR THREE TO 15-DIMENSIONAL UNIT SIMPLICES. Dmin M n Sampling Construction Reduction Riesz s-energy 3 50 1.753e-02 100.00 1.483e-01 0.21 1.247e-01 18.20 1.485e-01 0.00 100 9.340e-03 100.00 8.115e-02 22.69 7.866e-02 25.37 1.022e-01 0.00 250 3.457e-03 100.00 4.132e-02 36.22 4.193e-02 35.18 6.281e-02 0.00 500 1.673e-03 100.00 2.077e-02 52.74 2.562e-02 40.74 4.208e-02 0.00 5 100 4.417e-02 100.00 2.050e-01 29.59 1.736e-01 43.34 2.726e-01 0.00 250 2.628e-02 100.00 1.487e-01 27.26 1.333e-01 36.42 1.946e-01 0.00 500 1.927e-02 100.00 1.187e-01 27.45 1.089e-01 34.56 1.563e-01 0.00 8 200 6.758e-02 100.00 2.632e-01 29.56 1.461e-01 71.73 3.453e-01 0.00 500 5.293e-02 100.00 2.078e-01 30.86 1.243e-01 68.16 2.770e-01 0.00 1000 4.308e-02 100.00 1.803e-01 27.76 1.134e-01 62.97 2.330e-01 0.00 10 300 7.721e-02 100.00 2.795e-01 28.03 1.303e-01 81.12 3.582e-01 0.00 600 6.571e-02 100.00 2.402e-01 33.92 1.097e-01 83.33 3.297e-01 0.00 1000 5.874e-02 100.00 2.172e-01 28.82 1.121e-01 76.04 2.813e-01 0.00 15 300 1.011e-01 100.00 3.628e-01 26.46 1.013e-01 99.94 4.570e-01 0.00 600 8.792e-02 100.00 3.115e-01 29.27 1.007e-01 95.95 4.040e-01 0.00 1000 8.188e-02 100.00 2.827e-01 27.47 1.015e-01 92.92 3.588e-01 0.00 TABLE III COMPARISON OF FOUR METHODS USING VGM METRIC FOR THREE TO 15-DIMENSIONAL UNIT SIMPLICES. VGM M n Sampling Construction Reduction Riesz s-energy 3 50 1.855e-03 100.00 6.130e-05 0.85 1.892e-04 7.92 4.583e-05 0.00 100 7.899e-04 100.00 1.835e-04 22.33 7.904e-05 8.95 9.130e-06 0.00 250 2.692e-04 100.00 6.638e-05 24.40 3.075e-05 11.12 9.174e-07 0.00 500 1.232e-04 100.00 5.475e-05 44.09 1.603e-05 12.46 7.866e-07 0.00 5 100 4.668e-03 100.00 1.121e-03 23.57 2.227e-03 47.40 2.695e-05 0.00 250 1.687e-03 100.00 3.301e-04 18.55 6.422e-04 37.29 2.099e-05 0.00 500 8.133e-04 100.00 1.676e-04 19.02 8.579e-05 8.76 1.594e-05 0.00 8 200 7.512e-03 100.00 2.813e-03 36.98 9.485e-03 126.45 5.498e-05 0.00 500 2.758e-03 100.00 8.453e-04 30.31 3.922e-03 142.41 1.368e-05 0.00 1000 1.352e-03 100.00 4.327e-04 31.61 1.457e-03 107.87 7.838e-06 0.00 10 300 7.924e-03 100.00 3.305e-03 41.02 9.759e-03 123.44 9.272e-05 0.00 600 3.747e-03 100.00 1.516e-03 40.14 9.803e-03 262.45 1.996e-05 0.00 1000 2.223e-03 100.00 8.517e-04 37.95 4.799e-03 216.58 1.298e-05 0.00 15 300 1.714e-02 100.00 4.445e-03 23.36 3.921e-02 233.17 5.751e-04 0.00 600 8.323e-03 100.00 3.120e-03 36.62 2.248e-02 272.45 1.145e-04 0.00 1000 4.887e-03 100.00 2.290e-03 45.76 1.577e-02 327.16 9.879e-05 0.00 of trade-offs between objective values. Figure 6 shows two PCP plots comparing point-sets obtained using Das-Dennis and s-Energy in a ﬁve-dimensional space. Das-Dennis is initialized with a partition number p = 7 which results in 330 points (see Figure 6a). Seven gaps on each axis is clear from the plot. s-Energy also produces an identical distribution, but is not shown here for brevity. Instead, we reduce the number of points to 320, apply s-Energy method (as Das-Dennis method is no more appli-cable), and present the results in Figure 6b. Again, seven gaps on each axis is clearly visible except some minor adjustments in the ﬁrst and fourth objective values. But apparent crisscrossing patterns are very similar between the two plots, despite the right plot having 10 points less. This study demonstrates the efﬁcacy of our proposed s-Energy method to generate a well-spaced point-set even in higher dimensions. To make well-spaced point-sets directly available for the EMO community, we provide pre-generated results for up to 15 dimensions with some standard size of points online1. A source code of the s-Energy method is available in the multi-objective optimization framework pymoo . V. PRACTICAL MATTERS When no preference information is available, creating a well-spaced set of points is a recommended approach. How-ever, with some known preference information, biased distri-butions are desired. We describe here two biased approaches 1 11 n = 91 n = 92 Fig. 5. Distribution of points obtained by s-Energy with 91 (circle) and 92 (cross) points. Notice how an additional point placed on the bottom-right corner adjusts neighboring points to make a good distribution. z1 z2 z3 z4 z5 0.00 1.00 0.00 1.00 0.00 1.00 0.00 1.00 0.00 1.00 (a) Das-Dennis Sampling with p = 7 resulting in 330 points. z1 z2 z3 z4 z5 0.00 1.00 0.00 1.00 0.00 1.00 0.00 1.00 0.00 1.00 (b) s-Energy generating 320 points. Fig. 6. Well-spaced point-set on the unit simplex in ﬁve dimensions. based on our s-Energy method, which can be of more practical value. A. Points in Region of Interest (ROI) Often, in an optimization scenario, some parts of the objec-tive space are of more interest than other parts. This implies that in addition to creating an overall well-spaced distribution, some preferred regions must be represented with more points. For such scenarios, s-Energy method is suitable because it can be reapplied separately on such regions to create a local distribution with arbitrary number of points. For this purpose, we restrict the shape of ROI as a down-scaled version of the original unit simplex. The size of i-th such preferred region can be deﬁned using a scaling factor Si. To make this even more customized, a user can supply the number of points Ni to be added to the i-th ROI, in addition to the initial set of n points to be distributed uniformly on the entire unit simplex, thereby making the total number of points equal to nT = n+PK i=1 Ni, where K is the number of ROIs. Figure 7 illustrates a case with two user-speciﬁed ROIs with two reference points Ri on a three-dimensional unit simplex: (i) R1 = (0.15, 0.70, 0.15) with S1 = 0.2 and N1 = 12, and (ii) R2 = (0.33, 0.33, 0.33) with S2 = 0.35 and N2 = 20. The initial set contains n = 90 points. The s-Energy method is modiﬁed as follows. First, n points are located using the s-Energy method. Let us call this set S. Thereafter, for i-th ROI, we isolate the points from S in its domain (centered around Ri with shrinkage factor Si). Let us say there are ni points from original n points in its domain. If additional Ni points Fig. 7. Distribution of 112 points with two focused sets using the s-Energy method. z1 z2 z3 z4 z5 z6 z7 z8 Fig. 8. Semi-structured approach with 112 points in eight dimensions on a RadViz plot. are desired in i-th ROI, we apply the s-Energy method in the ROI to ﬁnd the location of (ni + Ni) points, but the energy is now computed from all points and by ﬁxing the location of points outside the ROI. The above procedure is repeated for each ROI one at a time to obtain the ﬁnal distribution of nT points. Another case is shown in the Supplementary document. B. Semi-Structured Point-set Another way to ensure a sufﬁcient number of intermediate points is to use a layer-wise Das-Dennis approach described in Subsection II-D. However, instead of a user specifying the shrinkage factor for each layer, the s-Energy method can be used to ﬁnd the optimal shrinkage factors so that Riesz s-Energy of all n points is maximized. Thus, the user has to only supply information about the gap number p of each layer in an ordered array P, wherein the ﬁrst element represents gap number of the outermost layer and the last element of the inner-most layer. If the user desires to have around 100 points in an 8-objective problem, then p must fall into the range between 0 and 3. Thus, P = {2, 2, 1, 1, 1, 1, 1} is one of the feasible settings which will render a total of 112 points (2 × 36 + 5 × 8) distributed across seven layers. It is recommended to choose a larger p value for outer layers than for the inner layers. The scaling factors for layers are then determined by optimizing the energy function. The points obtained by the method are shown in Figure 8 in a RadViz plot. It is visually evident that the resulting point-set is structured, however the points are not well-spaced. For the layer-wise approach, Dmin and VGM metric values are 9.9 × 10−2 and 4.5 × 10−3, respectively, whereas the s-Energy method produces much improved values: 4.1 × 10−1 and 1.6 × 10−4 (a RadViz plot is presented in the Supplementary document). VI. CONCLUSIONS In this paper, we have addressed an important issue re-lated to decomposition-based evolutionary many-objective op-timization (EMaO) algorithms: creation of arbitrary number of well-spaced reference points for any M dimensional objective space. After deﬁning near-uniformity of a set of points through the outcome of an optimization problem, we have discussed fundamentally different approaches for creating such a set for three to 15-dimensional problems: a baseline approach (Sampling), where points are created in a space-ﬁlling manner and mapped onto the unit simplex; a bottom-up approach 12 (Construction) where a point is incrementally added to an existing point-set; a top-down approach (Reduction) where a large number of points is reduced to a desired number; and a point-set reﬁnement approach (s-Energy) that incremen-tally improves a given point-set with respect to the Riesz s-Energy function. We have compared the results with the existing state-of-the-art method Das-Dennis. Results indicate that the s-Energy method is able to efﬁciently ﬁnd a well-spaced distribution on the unit simplex, based on a number of performance metrics suggested in this study. Moreover, we have provided a scale up study to observe the behavior of our methods to higher dimensional spaces for up to 15 objectives. Across all experiments with different number of points and dimensions, s-Energy method has outperformed other methods. Additionally, we have addressed two practical matters, such as obtaining localized well-spaced distribution in the regions of interest and the generation of a semi-structured point-set through a layer-wise approach, each optimized using the s-Energy method. This study ﬁlls a long-standing gap in the study of EMaO algorithms. While so far researchers have concentrated in developing efﬁcient algorithms by using structured reference directions, this study allows the algorithms to choose an arbitrary number of well-spaced reference directions for any dimensions with and without any preference biases. Thus, it can be used to generate a set of arbitrary number of reference points for computing performance indicators, such as IGD or IGD+. 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8719	http://amsi.org.au/ESA_Senior_Years/SeniorTopic3/3f/3f_2content_9.html	Content Properties of the definite integral Definite integrals obey rules similar to those for indefinite integrals. The following theorem is analogous to one for indefinite integrals. Theorem (Linearity of integration) If (f) and (g) are continuous functions on the interval ([a,b]), then [ \int_a^b \big( f(x) \pm g(x) \big) \; dx = \int_a^b f(x) \; dx \, \pm \int_a^b g(x) \; dx. ] If (f) is a continuous function on ([a,b]), and (k) is a real constant, then> [ \int_a^b kf(x) \; dx = k \int_a^b f(x) \; dx. ] The endpoints on a definite integral obey the following theorem. Theorem (Additivity of integration) Let (f \colon [a,b] \to \mathbb{R}) be continuous, where (a,b) are real numbers. Let (c) be a real number between (a) and (b). Then, [ \int_a^b f(x) \; dx = \int_a^c f(x) \; dx \, + \int_c^b f(x) \; dx. ] As with many mathematical statements, it's useful to understand these two theorems both algebraically (in terms of antiderivatives) and geometrically. For instance, we can think of the second theorem (additivity of integration) as saying geometrically that, if we consider the signed area between (y=f(x)) and the (x)-axis from (x=a) to (x=b), then this signed area is equal to the sum of the signed area from (x=a) to (x=c) and that from (x=c) to (x=b). Alternatively, if we let (F(x)) be an antiderivative of (f(x)), we can regard the theorem as just expressing that [ F(b) - F(a) = \big(F(c) - F(a)\big) + \big(F(b) - F(c)\big). ] This piece of algebra and the fundamental theorem of calculus together give a rigorous proof of the theorem. Exercise 13 Find a geometric interpretation of part (b) of the first theorem of this section (linearity of integration). You may assume the graph of (y=f(x)) lies above the (x)-axis. Also find an interpretation in terms of antiderivatives. We have stated the second theorem (additivity of integration) so that (a < c < b). But in fact, this theorem works when (a,b,c) are in any order, as long as (f,g) are defined and continuous over the appropriate intervals. We just have to make sense of integrals which have their terminals 'in the wrong order'. When (a>b), we define, [ \int_a^b f(x) \; dx = - \int_b^a f(x) \; dx. ] We can think of this as saying that, when (x) goes backwards from (a) to (b), we count areas as negative. In our area estimates, (\Delta x = \frac{1}{n} (b-a)) is the negative of the rectangle widths. This fits with our previous definitions, as you can show in the following exercise. Exercise 14 Let (f \colon \mathbb{R} \to \mathbb{R}) be a continuous function, and let (F(x)) be an antiderivative of (f(x)). Using the fundamental theorem of calculus, show that for any real numbers (a,b) (even when (a>b)), [ \int_a^b f(x) \; dx = F(b) - F(a). ] Exercise 15 Find (\displaystyle\int_5^0 (-2x - 3) \; dx). (Careful with signs! Should your answer be positive or negative?) Next page - Content - Area between two curves \| \| \| --- \| \| This publication is funded by theAustralian Government Department of Education,Employment and Workplace Relations \| ContributorsTerm of use \|
8720	https://tasks.illustrativemathematics.org/blueprints/A2	Illustrative Mathematics Loading [MathJax]/extensions/MathMenu.js Engage your students with effective distance learning resources. ACCESS RESOURCES>> Course High School - A2 High School - A2 Algebra 2 In Algebra 2, students extend the algebra and function work done in Algebra 1. Students continue to develop their picture of the complex number system by investigating how non-real solutions arise and how non-real numbers behave. Exponential functions are considered over a domain of real numbers, necessitating work with fractional exponents. The logarithm is defined as the inverse of exponentiation and from this definition students consider the properties of logarithms in addition to using them to solve for unknown exponents. Students extend their previous work with quadratics and polynomials to achieve a more general understanding of polynomials. Work done in geometry with sin, cos, and tan as operations is extended in a study of the unit circle and sin(x), cos(x), and tan(x) as functions. Students build on their work with probability from grade 7 to admit the notions of independence and conditional probability. Finally, students do further work in statistics where they revisit and extend their understanding of variability in data and of ways to describe variability in data. The normal distribution is studied, and students explore the reasoning that allows them to draw conclusions based on data from statistical studies. Units A2.1 A2.2 A2.3 A2.4 A2.5 A2.6 A2.7 A2.1 Extending the Number System View Details Summary • Work with infinite decimal expansions of numbers on the number line. • Reason about operations with rational and irrational numbers (N-RN.B.3). • Extend properties of integer exponents to rational exponents and write expressions with rational exponents as radicals (N-NR.A.1, N-RN.A.2). • Solve equations and real-world problems involving radicals and fractional exponents (A-REI.A.2). • Note extraneous solutions and explain where they come from (A-REI.A.2). • Discover a new type of number that is outside previously known number systems (N-CN.A.1). • Perform operations with complex numbers (N-CN.A.2). • Solve quadratic equations with complex solutions (N-CN.C.7). View Full Details A2.2 Exponential Functions 2 View Details Summary • Create and analyze a simple exponential function arising from a real-world or mathematical context (F-LE.A.2⋆). • Evaluate and interpret exponential functions at non-integer inputs (N-RN.A.1, F-LE.A.2⋆). • Understand functions of the form f(t)=P(1+r/n)n t and solve problems with different compounding intervals (A-SSE.A.1⋆, F-LE.A.2⋆). • Understand informally how the base e is used in functions to model a quantity that compounds continuously (F-BF.A.1a⋆). • Write exponential expressions in different forms (F-LE.A.2⋆, F-BF.A.1⋆). • Explain what the parameters of an exponential function mean in different contexts (F-LE.B.5⋆). • Use the properties of exponents to write expressions in equivalent forms (A-SSE.B.3a⋆). • Build exponential functions to model real world contexts (F-LE.A.1⋆, F-LE.A.2⋆, F-BF.A.1⋆). • Analyze situations that involve geometric sequences and series (A-SSE.A.1⋆). • Derive the formula for the sum of a finite geometric series (A-SSE.B.4⋆). View Full Details A2.3 Logarithms View Details Summary • Understand the definition of a logarithm as the solution to an exponential equation (F-LE.A.4⋆). • Practice evaluating logarithmic expressions and converting between the exponential form of an equation and the logarithmic form (F-LE.A.4⋆). • Solve exponential equations using logarithms (F-LE.A.4⋆). • Understand the natural logarithm as a special case (F-LE.A.4⋆). • Graph exponential and logarithmic functions, both by hand and using technology (F-IF.C.7e⋆, F-BF.5(+)). • (Optional) Understand and explain the properties of logarithms. • (Optional) Use properties of logarithms to solve problems. • (Optional) Solve problems using the properties of logarithms. View Full Details A2.4 Polynomials and Rational Functions View Details Summary • Add, subtract, and multiply polynomials and express them in standard form using the properties of operations (A-APR.A.1). • Prove and make use of polynomial identities (A-APR.C.4). • Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior (A-APR.B.3⋆, F-IF.C.7c⋆). • Use the remainder theorem to find factors of polynomials (A-APR.B.2, A-APR.D6). • Use various strategies including graphing and factoring to solve problems in contexts that can be modeled by polynomials in one variable. • Build a rational function that describes a relationship between two quantities (F-BF.A.1). • Graph rational functions, interpret features of the graph in terms of a context, and use the graphs to solve problems (A-SSE.A.1a, A-REI.D.11⋆, F-IF.B.4, F-IF.C.7d(+)). • Express rational functions in different forms to see different aspects of the situation they model (A-APR.D.6). • Solve simple rational equations and understand why extraneous roots can arise (A-REI.A.2). View Full Details A2.5 Trigonometric Functions View Details Summary • Understand some real-world situations that demonstrate periodic behavior. • Define coordinates on the unit circle as the sine and cosine of an angle (F-TF.A.2). • Understand radian measure and convert between radians and degree (F-TF.A.1). • Graph basic trigonometric functions using radians as the x-axis scale (F-IF.C.7e⋆). • Understand the relationship between parameters in a trigonometric function and the graph (F-IF.C.7e⋆). • Model with trigonometric functions, including fitting them to data (F-TF.B.5⋆). • Prove the Pythagorean Identity sin 2 θ+cos 2 θ=1 (F-TF.C.8). • Use the unit circle to prove trigonometric identities and relate them to symmetries of the graphs of sine and cosine (F-TF.A.4(+)). •Use the Pythagorean Identity to calculate trigonometric ratios (F-TF.C.8). View Full Details A2.6 Probability View Details Summary •Describe events as subsets of a sample space (the set of outcomes) using characteristics of the outcomes or as unions, intersections, or complements of other subsets (“or,” “and,” “not”) (S-CP.A.1). •Use the Addition Rule to compute probabilities of compound events in a uniform probability model, and interpret the result in terms of the model (S-CP.B.7). •In a uniform probability model, understand the probability of A given B as the fraction of B's outcomes that also belong to A (S-CP.B.6). •Understand the conditional probability of event A given event B as P(A and B)/P(B) (S-CP.A.3). •Understand that A and B are independent if P(A and B) = P(A) • P(B) (S-CP.A.2). •Interpret independence of A and B as saying that the probability of A given B is equal to the probability of A, and the probability of B given A is equal to the probability of B, i.e. P(A\|B) = P(A) and P(B\|A) = P(B) (S-CP.A.5). •Recognize independence in everyday situations and explain it in everyday language (S-CP.A.5). •Determine whether events are independent (S-CP.A.2,S-CP.A.4). •Use data presented in two-way frequency tables to approximate conditional probabilities (S-CP.A.4). View Full Details A2.7 Making Inferences View Details Summary •Understand that statistical methods are used to draw conclusions from data. •Understand that the validity of data-based conclusions depends on the quality of the data and how the data were collected. •Critique and evaluate data-based claims that appear in popular media. •Distinguish between observational studies, surveys and experiments. •Explain why random selection is important in the design of observational studies and surveys. •Explain why random assignment is important in the design of statistical experiments. •Calculate and interpret the standard deviation as a measure of variability. •Use the normal distribution as a model for data distributions that are approximately symmetric and bell-shaped. •Use the least squares regression line to model linear relationships in bivariate numerical data. •Understand sampling variability in the context of estimating a population or a population mean. •Use data from a random sample to estimate a population proportion. •Use data from a random sample to estimate a population mean. •Calculate and interpret margin of error in context. •Understand the relationship between sample size and margin of error. •Given data from a statistical experiment, create a randomization distribution. •Use a randomization distribution to determine if there is a significant difference between two experimental conditions. View Full Details Clusters Typeset May 4, 2016 at 18:58:52. Licensed by Illustrative Mathematics under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
8721	https://www.khanacademy.org/math/geometry-tx/x790e3ac3e338c450:transformations-in-the-plane	Transformations in the plane \| Geometry (TX TEKS) \| Math \| 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. Explore Math: Pre-K - 8th grade Math: Get ready courses Math: High school & college Math: Multiple grades Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search DonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. 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Geometry (TX TEKS)13 units · 101 skillsUnit 1 Introduction to logic and Euclidean geometryUnit 2 Coordinate geometryUnit 3 Transformations in the planeUnit 4 Relationships of lines and transversalsUnit 5 Congruence in trianglesUnit 6 Similarity in trianglesUnit 7 Relationships of right triangles, including trigonometryUnit 8 Relationships of circles, including radian measure and equations of circlesUnit 9 Relationships of two- and three-dimensional figuresUnit 10 Measurement of two-dimensional figuresUnit 11 Measurement of three-dimensional figuresUnit 12 ProbabilityUnit 13 Teacher resources Course challenge Test your knowledge of the skills in this course.Start Course challenge Math Geometry (TX TEKS) Unit 3: Transformations in the plane 900 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test Reflective symmetry of 2D shapes Rotational symmetry of 2D shapes Compose basic transformations Mapping shapes Transformations in the plane: Quiz 1 Algebraic rules for transformations Represent dilations algebraically Describe composed transformations Advanced transformations Compose transformations Transformations in the plane: Quiz 2 Transformations in the plane: Unit test About this unit We'll build from simpler translations, rotations, reflections, and dilations to move figures through the plane and understand their properties. Unit guides are here! Power up your classroom with engaging strategies, tools, and activities from Khan Academy’s learning experts. PDF Symmetry Learn Intro to reflective symmetry (Opens a modal) Intro to rotational symmetry (Opens a modal) Finding a quadrilateral from its symmetries (Opens a modal) Finding a quadrilateral from its symmetries (example 2) (Opens a modal) Practice Reflective symmetry of 2D shapesGet 3 of 4 questions to level up! Rotational symmetry of 2D shapesGet 3 of 4 questions to level up! Composing transformations Learn Mapping shapes (Opens a modal) Practice Compose basic transformationsGet 3 of 4 questions to level up! Mapping shapesGet 3 of 4 questions to level up! Quiz 1 Level up on the above skills and collect up to 320 Mastery points Start quiz Algebraic rules for basic transformations Learn Translations: graph to algebraic rule (Opens a modal) Translations: description to algebraic rule (Opens a modal) Rotations: graph to algebraic rule (Opens a modal) Rotations: description to algebraic rule (Opens a modal) Reflections: graph to algebraic rule (Opens a modal) Reflections: description to algebraic rule (Opens a modal) Representing dilations algebraically, k less than 1 (Opens a modal) Representing dilations algebraically, k greater than 1 (Opens a modal) Practice Algebraic rules for transformationsGet 3 of 4 questions to level up! Represent dilations algebraicallyGet 3 of 4 questions to level up! Building advanced transformations Learn Rotating algebraically about a point (Opens a modal) Dilating algebraically about a point (Opens a modal) Reflecting over y=x (Opens a modal) Reflecting over y=-x (Opens a modal) Graphing a non similar pre image algebraically (Opens a modal) Composing transformations (Opens a modal) Practice Describe composed transformationsGet 3 of 4 questions to level up! Advanced transformationsGet 3 of 4 questions to level up! Compose transformationsGet 3 of 4 questions to level up! Quiz 2 Level up on the above skills and collect up to 400 Mastery points Start quiz Unit test Level up on all the skills in this unit and collect up to 900 Mastery points!Start Unit test Texas Geometry is brought to you with support from the ExxonMobil Foundation logo Texas Geometry is brought to you with support from the Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization. Donate or volunteer today! 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8722	https://math.stackexchange.com/questions/4218853/finding-foot-of-perpendicular-of-centre-of-ellipse-on-its-variable-tangent	analytic geometry - Finding foot of perpendicular of centre of ellipse on its variable tangent - 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 Finding foot of perpendicular of centre of ellipse on its variable tangent Ask Question Asked 4 years, 1 month ago Modified4 years, 1 month ago Viewed 275 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Find the locus of the foot of perpendicular from the centre of the ellipse x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1 on the chord joining the points whose eccentric angles differ by π/2.π/2. Ref:Locus of foot of perpendicular of origin from tangent of ellipse, related The points may be given as (a cos α,b sin α)(a cos⁡α,b sin⁡α) and (−a sin α,b cos α)(−a sin⁡α,b cos⁡α), then their midpoint is given as: (a 2(cos α−sin α),b 2(cos α+sin α))=(κ,β)(a 2(cos⁡α−sin⁡α),b 2(cos⁡α+sin⁡α))=(κ,β), the equation of chord given midpoint of ellipse is: κ X a 2+β Y b 2=κ 2 a 2+β 2 b 2 κ X a 2+β Y b 2=κ 2 a 2+β 2 b 2 Let u=cos α−sin α 2 u=cos⁡α−sin⁡α 2 and v=cos α+sin α 2 v=cos⁡α+sin⁡α 2, then our equation becomes: u X a+v Y b=u 2+v 2 u X a+v Y b=u 2+v 2 Foot of perpendicular from origin for above line is given as: a x u=b y v=(a b)2(u 2+v 2)b 2 u 2+a 2 v 2 a x u=b y v=(a b)2(u 2+v 2)b 2 u 2+a 2 v 2 a x=u((a b)2(u 2+v 2)b 2 u 2+a 2 v 2)a x=u((a b)2(u 2+v 2)b 2 u 2+a 2 v 2) and b y=v((a b)2(u 2+v 2)b 2 u 2+a 2 v 2)b y=v((a b)2(u 2+v 2)b 2 u 2+a 2 v 2), following the last line of this answer, I squared and added: (a x)2+(b y)2=(a b)4[u 2+v 2 b 2 u 2+a 2 v 2]2=(a b)4 4[1 b 2 u 2+a 2 v 2]2(a x)2+(b y)2=(a b)4[u 2+v 2 b 2 u 2+a 2 v 2]2=(a b)4 4[1 b 2 u 2+a 2 v 2]2 How do I write bracketed term in purely (x,y)(x,y)? analytic-geometry conic-sections Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Aug 7, 2021 at 13:04 Clemens BartholdyClemens Bartholdy asked Aug 7, 2021 at 10:46 Clemens BartholdyClemens Bartholdy 14.3k 4 4 gold badges 29 29 silver badges 77 77 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. You are missing a (u 2+v 2)(u 2+v 2) when you add (a x)2(a x)2 and (b y)2(b y)2. Just in case, you want to finish continuing from the point you got to, (a x)2+(b y)2=(a b)4(u 2+v 2)[u 2+v 2 b 2 u 2+a 2 v 2]2(a x)2+(b y)2=(a b)4(u 2+v 2)[u 2+v 2 b 2 u 2+a 2 v 2]2 =(a b)4 2[(u/v)2+1 b 2(u/v)2+a 2]2=(a b)4 2[(u/v)2+1 b 2(u/v)2+a 2]2 Now note that u v=a x b y u v=a x b y That leads to 2(x 2+y 2)2=(a x)2+(b y)2 2(x 2+y 2)2=(a x)2+(b y)2 or we could have substituted u=k a x,v=k b y u=k a x,v=k b y in RHS. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 7, 2021 at 14:27 Math LoverMath Lover 52.2k 3 3 gold badges 24 24 silver badges 46 46 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Here's one method to simplify things. First note that : u 2+v 2=1/2 u 2+v 2=1/2 Now, consider the equation you found: a x u=b y v=a 2 b 2(u 2+v 2)b 2 u 2+a 2 v 2 a x u=b y v=a 2 b 2(u 2+v 2)b 2 u 2+a 2 v 2 Then: a x v=b y u⇒a 2 v 2=b 2 u 2 y 2/x 2⇒b 2 u 2+a 2 v 2=b 2 u 2(x 2+y 2 x 2)a x v=b y u⇒a 2 v 2=b 2 u 2 y 2/x 2⇒b 2 u 2+a 2 v 2=b 2 u 2(x 2+y 2 x 2) Plugging this back into one of the equations gives: a x u=a 2 b 2 x 2 2 b 2 u 2(x 2+y 2)⇒u=a x 2(x 2+y 2)a x u=a 2 b 2 x 2 2 b 2 u 2(x 2+y 2)⇒u=a x 2(x 2+y 2) Similarly you will also find v=b y 2(x 2+y 2)v=b y 2(x 2+y 2) Squaring and adding them, along with the first equation at the extreme top gives: 1 2=u 2+v 2=a 2 x 2 4(x 2+y 2)2+b 2 y 2 4(x 2+y 2)2 1 2=u 2+v 2=a 2 x 2 4(x 2+y 2)2+b 2 y 2 4(x 2+y 2)2 ⇒2(x 2+y 2)2=a 2 x 2+b 2 y 2⇒2(x 2+y 2)2=a 2 x 2+b 2 y 2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 7, 2021 at 13:32 Yuzuriha InoriYuzuriha Inori 748 4 4 silver badges 13 13 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 analytic-geometry conic-sections 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 12Foot of perpendicular to line 1How to derive equation of chord of circle given its midpoint? 1Find the locus of the foot of perpendicular from the centre of the ellipse. 0Foot of perpendicular on a chord of a conic 1Why is each chord of a conic having a unique midpoint corresponding to it? (Excluding centre of symmetries) Related 9Equation of the locus of the foot of perpendicular from any focus upon any tangent to the ellipse x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1 0Foot of perpendicular on a chord of a conic 1If P Q P Q subtends right angle at the centre of ellipse then find 1 O P 2+1 O Q 2.1 O P 2+1 O Q 2. 2Find the locus of foot of perpendicular from origin 3Locus of foot of perpendicular of an ellipse using x=a cos θ x=a cos⁡θ&y=b sin θ y=b sin⁡θ 1Ellipse in which two chords are perpendicular to each other 1Locus of a point of intersecting lines 2Variable pairs of chords at right angles are drawn through a point P P (with eccentric angle π/4 π/4) on the ellipse. Hot Network Questions Numbers Interpreted in Smallest Valid Base Xubuntu 24.04 - Libreoffice With line sustain pedal markings, do I release the pedal at the beginning or end of the last note? An odd question Should I let a player go because of their inability to handle setbacks? 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8723	https://www.wyzant.com/resources/answers/757896/proving-3-points-determine-a-unique-plane	WYZANT TUTORING Licia C. Proving 3 points determine a unique plane I'm trying to prove the early "Postulates" (#6-10 or so) of Euclidean geometry. I'm trying to rely on Euclid's 5 and his definitions - only. I don't want to demonstrate using some sort of manipulative. I can prove the uniqueness of the plane determined by three, given, non-colinear lines. I can't figure out how to prove that such a plane exists. Why MUST there exist a plane that contains any set of three non-colinear points? Mark M. 04/30/20 Licia C. 04/30/20 1 Expert Answer Richard P. answered • 05/01/20 Fairfax County Tutor for HS Math and Science One approach is the following. The three non-colinear points (call them A, B and C) form a triangle. For any triangle there is a circumcenter. There are several theorems that involve the circumcenter. The circumcenter is equidistant from the three corner points. The locus of points equidistant from points A and B is a plane containing the circumcenter. Similarly, the locus of points equidistant from the points B and C is a plane containing the circumcenter. And also, the locus of points equidistant from the points C and A is a plane containing the circumcenter. This is related to the well-known theorem that the perpendicular bisectors of the sides of a triangle all concur at the circumcenter. The intersection of there three planes must define a line. That line is perpendicular to a plane. The direction of that line specifies a normal vector to a plane. Obviously that plane contains the circumcenter. Thus that plane must be the plane containing the three points, A, B and C. Licia C. 05/01/20 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 Does a triangle always have a point where each side subtends equal 120° angles? Answers · 2 How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? Answers · 1 Which undefined geometric term is best described as a location on a coordinate plane that is designed by an ordered pair (x,y) Answers · 1 finding GCD using Euclidean Answers · 1 How would I construct a line perpendicular to a point on a line Answers · 2 RECOMMENDED TUTORS Sotirios K. Geoffrey B. Liz Z. 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
8724	https://community.broadcom.com/vmware-cloud-foundation/discussion/difference-between-provision-and-used-space	Difference Between Provision and Used Space \| vSphere Storage Appliance Products Applications Support Company How To Buy Skip to main content (Press Enter). Sign in Skip auxiliary navigation (Press Enter). Register Skip main navigation (Press Enter). Toggle navigation Home Communities All Communities Application Networking and Security Carbon Black Enterprise Software Mainframe Software Symantec Enterprise Tanzu VMware {code} VMware Cloud Foundation Blogs All Blogs Enterprise Software Mainframe Software Symantec Enterprise VMware Events VMware Explore 2025 All Events Enterprise Software Mainframe Software Symantec Enterprise VMware Cloud Foundation Events Water Cooler Betas Flings Education Groups Enterprise Software Mainframe Software Symantec Enterprise VMware Members vSphere Storage Appliance View Only Community Home Threads 24.9K Library 60 Blogs 0 Events 0 Members 5.3K Back to discussions Expand all\|Collapse all Difference Between Provision and Used Space Jump to Best Answer MrVmware9423 Mar 14, 2014 01:54 PM Dear Team, can some one please explain me the difference between provision and used disk space in ... Alim786 Mar 14, 2014 02:51 PM Hi Mr VMware Provisioned is how much storage has been allocated. Used is how much of that allocation ... MrVmware9423 Mar 20, 2014 11:16 AM Thanks Alim, In the below snap I have thin provision datastore size of almost 1TB, Over here "Provisioned ... Alim786 Mar 20, 2014 12:04 PM Best Answer The above Datastore Capacity Summary is interpreted as follows :- Capacity is the Total Size. ... MrVmware9423 Mar 21, 2014 12:22 PM Many Thanks Alim, You have cleared my confusion . Thanks a ton. regards Mr VMware NORRISC Jun 23, 2015 12:43 PM I am at a loss. I specifically gave each VM enough space to run the OS and application in our ... khorem Feb 22, 2018 09:03 PM what if you don't have enough space? and can't do another snapshot? This is what initially happened. ... AnaghB May 01, 2023 04:47 PM Hello , The screenshot that you have uploaded mentions that there are 2 HDD in the Folder. ... Port2Tech May 01, 2023 11:10 AM Good day all, I realize this topic is old but I am new to dealing with the care and feeding of ... NORRISC Jun 23, 2015 01:56 PM Additional Information - I removed all the SNAPSHOTS (2 of them) from this VM as well, and now with ... AnaghB May 01, 2023 04:45 PM Hello , Vmware prohibits to reduce the size of the disk. This can cause to Data loss. The Windows ... AnaghB May 01, 2023 04:39 PM Hello , The assigned space is the size of the disk that you had specified while creation of the ... #### 1.Difference Between Provision and Used Space Recommend) MrVmware9423 Posted Mar 14, 2014 01:54 PM Reply)Reply PrivatelyOptions Dropdown Dear Team, can some one please explain me the difference between provision and used disk space in simple words. regards Mr VMware #### 2.RE: Difference Between Provision and Used Space Recommend) Alim786 Posted Mar 14, 2014 02:51 PM Reply)Reply PrivatelyOptions Dropdown Hi Mr VMware Provisioned is how much storage has been allocated. Used is how much of that allocation is actually being used by the VM. You may think this should be the same. However, it is only relevant if Thin Provisioning has been used. This is where there will be differences. #### 3.RE: Difference Between Provision and Used Space Recommend) MrVmware9423 Posted Mar 20, 2014 11:16 AM Reply)Reply PrivatelyOptions Dropdown Thanks Alim, In the below snap I have thin provision datastore size of almost 1TB, Over here "Provisioned Space is 847.34 and Free Space is 630.21. if u say provsion is allocated then if we plus Provision and Free then it is more than 1TB. hence need your assistance on the same. Regards Mr VMware #### 4.RE: Difference Between Provision and Used Space Best Answer Recommend) Alim786 Posted Mar 20, 2014 12:04 PM Reply)Reply PrivatelyOptions Dropdown The above Datastore Capacity Summary is interpreted as follows :- Capacity is the Total Size. Provisioned Space is the total size allocated to the VMs. Free Space is the space not used up by VMs yet. Therefore, with thin provisioning, you can over provision the amount of space to VMs based upon the Free Space at that time. A VM can be allocated 100GB but be only using 50GB. The remaining unused 50GB is counted as part of the Free Space. This can be seen also by clicking on the "Virtual Machines" tab of the Datastore to see the "Provisioned Space" and "Used Space". I hope this helps. #### 5.RE: Difference Between Provision and Used Space Recommend) MrVmware9423 Posted Mar 21, 2014 12:22 PM Reply)Reply PrivatelyOptions Dropdown Many Thanks Alim, You have cleared my confusion . Thanks a ton. regards Mr VMware #### 6.RE: Difference Between Provision and Used Space Recommend) NORRISC Posted Jun 23, 2015 12:43 PM Reply)Reply PrivatelyOptions Dropdown I am at a loss. I specifically gave each VM enough space to run the OS and application in our test environment - and now we find that we are running out of space because almost all of the VMs have eaten TWICE the amount I gave them. How do I get this space back, when the VMs will not let me change the HDD size - it is greyed out. :smileysad: We are running ESXi 5.5.0 but all the VMs are VM VERSION 8 machines. Thank You, C #### 7.RE: Difference Between Provision and Used Space Recommend) khorem Posted Feb 22, 2018 09:03 PM Reply)Reply PrivatelyOptions Dropdown what if you don't have enough space? and can't do another snapshot? This is what initially happened. I tried to create a snapshot but after a while it failed, probably because of low storage space (I can't add any additional space either) Went to snapshot manager to see if there are any snapshots available... none went to the storage to see if there are any additional vmdk files I saw one with provisioned.... im assuming thats the corrupted snapshot thats causing the issue... Q: if that is the file that needs to be removed, How can i remove it; Is there a manual way of removing it? I can't create another snapshot due to shortage of space, I can not increase the storage size. please advise. Thanks. #### 8.RE: Difference Between Provision and Used Space Recommend) AnaghB Posted May 01, 2023 04:47 PM Reply)Reply PrivatelyOptions Dropdown Hello, The screenshot that you have uploaded mentions that there are 2 HDD in the Folder. VM_Name.vmdk VM_Name_1.vmdk. These are 2 different HDD which might contain similar or different Data. Please check before deleting. #### 9.RE: Difference Between Provision and Used Space Recommend) Port2Tech Posted May 01, 2023 11:10 AM \| view attached Reply)Reply PrivatelyOptions Dropdown Good day all, I realize this topic is old but I am new to dealing with the care and feeding of VMs and the Dell ESXi systems. I have added more storage and created a new datastore. After the datastore was creates, I looked at the details of the store and noticed that 988MB of space was set as provisioned. After reading the posts in this thread, I am starting to understand but I have not, to my knowledge, actually provisioned any storage to the VMs currently running on the system. So I am trying to understand why the report is showing that 988MB have been used. I do have thin provisioning active, as I copied the settings from the existing datastores on the system. Guessing that I missed a setting when creating this new datastore? Thank you in advance! VMWare1.png × 1 / 1 Show captions Showing image 1 of 1 #### 10.RE: Difference Between Provision and Used Space Recommend) NORRISC Posted Jun 23, 2015 01:56 PM Reply)Reply PrivatelyOptions Dropdown Additional Information - I removed all the SNAPSHOTS (2 of them) from this VM as well, and now with the VM shut down, the HDD size option is no longer greyed out...but it will not allow me to change the size. I have also change the size of the disc in the OS itself. It has been changed from 200gb to 120gb - shows in the OS 80GB of unallocated space. Why can I not release that space by to my VM pool for use in other VMs that I need to increase their HDD sizes? Thanks, C #### 11.RE: Difference Between Provision and Used Space Recommend) AnaghB Posted May 01, 2023 04:45 PM Reply)Reply PrivatelyOptions Dropdown Hello , Vmware prohibits to reduce the size of the disk. This can cause to Data loss. The Windows OS formats the disk with NTFS Filesystem which has Random Writes. So if we write the data then the data is scattered on the disk and is not in sequential order. When you reduce the disk from 200GB to 120 GB then the 80GB sectors that are reduces might contain frangments of data which gets deleted causing data loss. From Vmware Best Practice if you want to reduce the size of the disk then please use Vmware Convertor or create a new HDD of 120GB, copy the data from 200GB Drive to 120GB Drive within the Guest OS and delete the 200GB post confirming that all data is intact. #### 12.RE: Difference Between Provision and Used Space Recommend) AnaghB Posted May 01, 2023 04:39 PM Reply)Reply PrivatelyOptions Dropdown Hello, The assigned space is the size of the disk that you had specified while creation of the VM. I see in the screenshot that you had created Thick Provision disk of 200GB. Thick provision means that even if the disk is empty within the guest OS the whole space is allocated to the disk from Datastore so it will show as 200GB used. For the usage of above 200 GB. When we Power ON the VM there is a .swap file created for memory swapping. The size of the file is equal to the Memory/RAM allocated to the VM. So if the Thick disk is 200 GB and you have assigned 24GB of RAM to the VM then the total usage of that VM on Datastore is 224GB. When you Power OFF the VM the .swap file is deleted and the usage comes back to 200GB of HDD. Please let me know if there are any queries around this. We can discuss this at length. × New Best Answer This thread already has a best answer. Would you like to mark this message as the new best answer? No Products Applications Support Company How To Buy Copyright © 2005-2025 Broadcom. All Rights Reserved. The term "Broadcom" refers to Broadcom Inc. and/or its subsidiaries. Hosted by Higher Logic, LLC on the behalf of Broadcom - Privacy Policy \| Cookie Policy \| Supply Chain Transparency Terms of Use Copyright 2024. All rights reserved. Powered by Higher Logic Cookies Broadcom and third-party partners use technology, including cookies to, among other things, analyze site usage, improve your experience and help us advertise. 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8725	http://physics.bu.edu/~redner/211-sp06/class-thermodynamics/process_adiabatic2.html	Adiabatic Process Revisited In an adiabatic process, the First Law gives: dE = dW. Use this result to determine adiabatic paths (adiabats) in the P-V diagram. dE = dW → nCV dT = - P dV = - n RT dV/V. → CV dT/T = - R dV/V but R = (CP-CV). dT/T = - (dV/V) [(CP-CV)/ CV]. → d ln T = - (γ - 1) d ln V or d [ln T V (γ - 1)] = 0. → T V (γ - 1) = constant. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. Use this result to determine adiabatic paths (adiabats) in the P-V diagram. dE = dW → nCV dT = - P dV = - n RT dV/V. → CV dT/T = - R dV/V but R = (CP-CV). dT/T = - (dV/V) [(CP-CV)/ CV]. → d ln T = - (γ - 1) d ln V or d [ln T V (γ - 1)] = 0. → T V (γ - 1) = constant. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. dE = dW → nCV dT = - P dV = - n RT dV/V. → CV dT/T = - R dV/V but R = (CP-CV). dT/T = - (dV/V) [(CP-CV)/ CV]. → d ln T = - (γ - 1) d ln V or d [ln T V (γ - 1)] = 0. → T V (γ - 1) = constant. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. → CV dT/T = - R dV/V but R = (CP-CV). dT/T = - (dV/V) [(CP-CV)/ CV]. → d ln T = - (γ - 1) d ln V or d [ln T V (γ - 1)] = 0. → T V (γ - 1) = constant. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. dT/T = - (dV/V) [(CP-CV)/ CV]. → d ln T = - (γ - 1) d ln V or d [ln T V (γ - 1)] = 0. → T V (γ - 1) = constant. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. → d ln T = - (γ - 1) d ln V or d [ln T V (γ - 1)] = 0. → T V (γ - 1) = constant. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. → T V (γ - 1) = constant. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. Equivalently use T ∝ PV to give P V γ = constant. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const. Note: adiabats are steeper than isotherms! Adiabat: P V γ = const. or T V (γ - 1) = const. Isotherm: P V = const.
8726	https://artofproblemsolving.com/wiki/index.php/Cyclic_sum?srsltid=AfmBOookYXuwxg5rGA0dGTzo5zGocpS8ujXOYpWicYBOp8ZFs8TSBR_Y	Art of Problem Solving Cyclic sum - 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 Cyclic sum Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Cyclic sum A cyclic sum is a summation that cycles through all the values of a function and takes their sum, so to speak. Rigorous definition Consider a function . The cyclic sum is equal to Note that not all permutations of the variables are used; they are just cycled through. Notation A cyclic sum is often specified by having the variables to cycle through underneath the sigma, as follows: Note that a cyclic sum need not cycle through all of the variables. A cyclic sum is also sometimes specified by . This notation implies that all variables are cycled through. See also Summation Symmetric sum PaperMath’s sum Retrieved from " Categories: Algebra Definition 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.
8727	https://bio.libretexts.org/Bookshelves/Botany/Botany_(Ha_Morrow_and_Algiers)/04%3A_Plant_Physiology_and_Regulation/4.06%3A_Development/4.6.04%3A_Germination	4.6.4: Germination - Biology 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 4.6: Development 4: Plant Physiology and Regulation { } { "4.6.01:_Embryogenesis" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.6.02:_Meristems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.6.03:_Mature_Embryos_and_Seed_Structure" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.6.04:_Germination" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.6.05:_Shoot_Development" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.6.06:_Chapter_Summary" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "4.01:_Photosynthesis_and_Respiration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.02:_Environmental_Responses" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.03:_Nutrition_and_Soils" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.04:_Hormones" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.05:_Transport" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "4.06:_Development" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 28 Jul 2025 22:11:52 GMT 4.6.4: Germination 32044 32044 Jennifer Rogers { } Anonymous Anonymous 2 false false [ "article:topic", "germination", "showtoc:no", "license:ccbysa", "source-bio-5796", "source-bio-5796", "program:oeri", "cid:biol155", "authorname:haetal", "licenseversion:40" ] [ "article:topic", "germination", "showtoc:no", "license:ccbysa", "source-bio-5796", "source-bio-5796", "program:oeri", "cid:biol155", "authorname:haetal", "licenseversion:40" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Botany and Horticulture 4. Botany (Ha, Morrow, and Algiers) 5. 4: Plant Physiology and Regulation 6. 4.6: Development 7. 4.6.4: Germination Expand/collapse global location Botany (Ha, Morrow, and Algiers) Front Matter 1: Introduction to Botany 2: Biodiversity (Organismal Groups) 3: Plant Structure 4: Plant Physiology and Regulation 5: Ecology and Conservation Back Matter 4.6.4: Germination Last updated Jul 28, 2025 Save as PDF 4.6.3: Mature Embryos and Seed Structure 4.6.5: Shoot Development 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 32044 Melissa Ha, Maria Morrow, & Kammy Algiers Yuba College, College of the Redwoods, & Ventura College via ASCCC Open Educational Resources Initiative ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Environmental Requirements for Germination 3. The Mechanism of Germination 1. Germination in Eudicots 2. Germination in Monocots Attributions Learning Objectives Identify the environmental factors that stimulate germination. Distinguish between epigeous and hypogeous germination. Compare germination in eudicots versus monocots. Many mature seeds enter a period of inactivity, or extremely low metabolic activity: a process known as dormancy, which may last for months, years or even centuries. Dormancy helps keep seeds viable during unfavorable conditions. Germinationoccurs when the embryo, which is dormant within a mature seed, resumes growth upon a return to favorable conditions. The embryo becomes a young seedling that is no longer confined within the seed coat. In many seeds, the presence of a thick seed coat can inhibit germination through several mechanisms: (1) the embryo may not be able to break through the thick seed coat; (2) the seed coat may contain chemicals inhibitors; and (3) the seed coat prevents the embryo from accessing water and oxygen. Dormancy is also maintained by the relative hormone concentrations in the embryo itself. Environmental Requirements for Germination The requirements for germination depend on the species. Common environmental requirements include light, the proper temperature, presence of oxygen, and presence of water. Seeds of small-seeded species usually require light as a germination cue. This ensures the seeds only germinate at or near the soil surface (where the light is greatest). If they were to germinate too far underneath the surface, the developing seedling would not have enough food reserves to reach the sunlight. (Recall from 14.5 Dormancy that red light induces germination by converting the inactive form of phytochrome (Pr) to the active form (Pfr), which leads to the production of amylase. This enzyme breaks down the limited food reserves in the seed, facilitating germination.) Not only do some species require a specific temperature to germinate, but they may also require a prolonged cold period prior to germination. In this case, cold conditions gradually break down a chemical germination inhibitor. This mechanism prevents seeds from germinating during an unseasonably warm spell in the autumn or winter in temperate climates. Similarly, plants growing in hot climates may have seeds that need a hot period in order to germinate, an adaptation to avoid germination in the hot, dry summers. Water is always needed to allow vigorous metabolism to begin. Additionally, water can leach away inhibitors in the seed coat. This is especially common among desert annuals. Seeds that are dispersed by animals may need to pass through an animal digestive tract to remove inhibitors prior to germination. Similarly, some species require mechanical abrasion of the seed coat, which could be achieved by water dispersal. Other species are fire adapted, requiring fire to break dormancy (Figure 4.6.4.1). Figure 4.6.4.1: Coffeeberry (Frangula californica) at the Regional Park Botanic Garden in Berkeley Hills, California. This species naturally occurs in the chaparral where wildfires trigger germination of its seeds. Image by Daderot (public domain). The Mechanism of Germination The first step in germination and starts with the uptake of water, also known as imbibition. After imbibition, enzymes are activated that start to break down starch into sugars consumed by embryo. The first indication that germination has begun is a swelling in the radicle. Depending on seed size, the time taken for a seedling to emerge may vary. Species with large seeds have enough food reserves to germinate deep below ground, and still extend their epicotyl all the way to the soil surface while the seedlings of small-seeded species emerge more quickly (and can only germinate close to the surface of the soil). During epigeous germination, the hypocotyl elongates, and the cotyledons extend above ground. During hypogeous germination, the epicotyl elongates, and the cotyledon(s) remain belowground (Figure 4.6.4.2). Some species (like beans and onions) have epigeous germination while others (like peas and corn) have hypogeous germination. In many epigeous species, the cotyledons not only transfer their food stores to the developing plant but also turn green and make more food by photosynthesis until they drop off. Figure 4.6.4.2: Epigeous germination in bean (top) and hypogeous germination in pea (bottom). For the bean, the radicle emerges from the seed. Next, the hypocotyl elongates, and the cotyledons are pushed aboveground. The epicotyl is just above the cotyledon. For the pea, the radicle also emerges from the seed. The epicotyl elongates, and the cotyledons remain below the ground. The hypocotyl remains short and lies between the root and cotyledon. Image by Jen Valenzuela (CC-BY). Germination in Eudicots Upon germination in eudicot seeds, the radicle emerges from the seed coat while the seed is still buried in the soil. For epigeous eudicots (like beans), the hypocotyl is shaped like a hook with the plumule pointing downwards. This shape is called the plumule hook, and it persists as long as germination proceeds in the dark. Therefore, as the hypocotyl pushes through the tough and abrasive soil, the plumule is protected from damage. Additionally, the two cotyledons additionally protect the from mechanical damage. Upon exposure to light, the hypocotyl hook straightens out, the young foliage leaves face the sun and expand, and the epicotyl elongates (Figure 4.6.4.3). Figure 4.6.4.3: A bean seed begins to germinate (left) and a germinated bean seedling (right). Beans are epigeous eudicots, meaning that the hypocotyl elongates, pushing the cotyledons above the ground. On the right, the large cotyledons have just emerged from the seed coat. The oval hilum is next to a tiny, round micropyle on the seed coat. The primary (tap) root has emerged, and secondary (lateral) roots begin to branch from the primary root. On the left, the green hypocotyl has elongated, pushing the cotyledons aboveground. The cotyledons are green and still have the shape of the bean seed. The epicotyl is the part of the stem above the cotyledons. Broad, heart-shaped leaves branch from it. Unlike the cotyledons, these are true leaves. At the tip of the stem, between the leaves is the shoot apical meristem. There is a central, thick root called the primary (tap) root. The roots that branch form it are secondary (lateral) roots. Left image by Doronenko (CC-BY-SA). Right image by Melissa Ha (CC-BY). In hypogeous eudicots (like peas), the epicotyl rather than the hypocotyl forms a hook, and the cotyledons and hypocotyl thus remain underground. When the epicotyl emerges from the soil, the young foliage leaves expand. The epicotyl continues to elongate (Figure 4.6.4.4). Figure 4.6.4.4: A germinated pea seedling. Peas a hypogeous eudicots. The hypocotyl never elongates, and the cotyledons remain below ground. The green and white epicotyl has elongated, giving rise to true leaves, and the cotyledons remain belowground. These true leaves are compound (are composed of smaller leaflets). This seedling has been uprooted and washed, but everything below the epicotyl was belowground. The hypocotyl is the short segment of stem between the cotyledons and roots. It never elongated enough to push the cotyledons above the ground. There is a central, thick root called the primary (tap) root. The roots that branch form it are secondary (lateral) roots. Image by Melissa Ha (CC-BY). The radicle continues to grown downwards and ultimately produces the tap root. Lateral roots then branch off to all sides, producing the typical eudicot tap root system. Germination in Monocots As the seed germinates, the radicle emerges and forms the first root. In epigeous monocots (such as onion), the single cotyledon will bend, forming a hook and emerge before the coleoptile (Figure 4.6.4.5). In hypogeous monocots (such as corn), the cotyledon remains belowground, and the coleoptile emerges first. In either case, once the coleoptile has exited the soil and is exposed to light, it stops growing. The first leaf of the plumule then pieces the coleoptile (Figure 4.6.4.6), and additional leaves expand and unfold. At the other end of the embryonic axis, the first root soon dies while adventitious roots (roots that arise directly from the shoot system) emerge from the base of the stem (Figure 4.6.4.7). This gives the monocot a fibrous root system. Figure 4.6.4.5: The scallion (spring onion) is an epigeous monocot. The curved structure emerging from the ground is the single cotyledon. Later in the process, the coleoptile will emerge, and be pierced by the first leaf of the plumule. Image by Dennis Brown (CC-BY-SA) Figure 4.6.4.6: Germination of an oat seed, a hypogeous monocot. Here the first root formed by the radicle is labeled "primary root", but note that this differs from the primary (main) root of a eudicot tap root system. Note that several adventitious roots have also formed and will ultimately produce a fibrous root system. The coleoptile is the first component of the shoot system to emerge in hypogeous monocots, but it is ultimately pieced by the first (primary) leaf of the plumule. Figure 4.6.4.7: As this monocot grass seed germinates, the radicle, emerges first, followed by the coleoptile, and the adventitious roots. The coleorhiza, a protective sheath that surrounded the radicle in the dormant seed, is now at the tip of the radicle. Attributions Curated and authored by Melissa Ha using the following sources: 16.4B Germination of Seeds from Biology by John W. Kimball (CC-BY) 32.2 Pollination and Fertilization from Biology 2e by OpenStax (licensed CC-BY). Access for free at openstax.org. 7.5 Origin of the Seed from _Introduction to Botany_ by Alexey Shipunov (public domain) This page titled 4.6.4: Germination is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Melissa Ha, Maria Morrow, & Kammy Algiers (ASCCC Open Educational Resources Initiative) . Back to top 4.6.3: Mature Embryos and Seed Structure 4.6.5: Shoot Development Was this article helpful? Yes No Recommended articles 40.6.1: GerminationGermination is the resumption of growth of the embryo plant inside the seed. 15.4: GerminationGermination is the resumption of growth of the embryo plant inside the seed. 7.5: Origin of the SeedWhen plants developed the secondary growth, the almost unlimited perspectives opened for enlarging their body. However, these giants faced a new probl... 16.4: Plant Development - FundamentalsThis page discusses plant growth, emphasizing meristems' role in rapid cell division and germination resuming embryo growth. It describes etiolation, ... 16.4B: Germination of SeedsThis page discusses germination, the process where a seed's embryo resumes growth, influenced by temperature, water, oxygen, and dormancy. Many temper... 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8728	https://www.infocop.es/wp-content/uploads/2023/05/MH_Selective-Mutism-for-Families-2.pdf	Mental Health 1 © 2023 National Association of School Psychologists │ 4340 East West Hwy, Bethesda, MD 20814 \| www.nasponline.org │ 301-657-0270 Selective Mutism: An Overview for Families Selective mutism is an anxiety disorder in which children have difficulty speaking in select situations, such as school, despite their ability to speak appropriately in situations where they are comfortable, such as home. The behaviors you may see from a child with selective mutism can be different for each child, but they can include: • trouble responding to questions from adults or peers, • difficulty sharing information independently with adults or peers, • trouble talking to individuals with whom they are unfamiliar or less comfortable, • difficulty speaking at a full volume, • not participating in activities that require movement, such as gesturing or dancing, or • difficulty asking for help or getting permission to go to the bathroom or nurse. Symptoms are often first noticed in school, because families may not observe the communication difficulties if the child is comfortable speaking at home. However, similar difficulties may be present when the child is in daycare, with extended family members, or out in the community. Selective mutism is more than just being shy, and it is not something that is outgrown. Roughly 0.71% of the population may experience symptoms (Bergman et al., 2002), with a higher rate of symptoms in bilingual populations (Toppelberg et al., 2005). Children diagnosed with selective mutism also may be at greater risk for other anxiety or speech concerns (Driessen et al., 2020; Klein et al., 2013). Symptoms can affect the child’s ability to engage in school tasks, how they are perceived by peers, and their ability to advocate for needs. The long-term impact of not addressing symptoms can include an increased risk for additional anxiety, other mental health concerns, school refusal behaviors, and use of alcohol or other substances to manage anxiety symptoms. GENERAL RECOMMENDATIONS The goal is to create a comfortable and safe environment that helps the child feel supported and included. The child will generally benefit from opportunities for one-on-one and small- group interaction in school, positivity from adults, and support for participating in class activities, such as suggesting games to play during free times or setting up activities where the child can engage with others. In general, adults should avoid asking rapid questions, forcing the child to speak, criticizing the child’s communication difficulties, indicating that the child cannot speak, or predicting what the child wants before they have asked for anything. It is also important to discourage peers from speaking for the child. THE SCHOOL TEAM ADDRESSING SELECTIVE MUTISM Addressing selective mutism in school requires a team approach. Although each team member has a different role, all are important in supporting the child. Role of the School Psychologist A school psychologist may have varying roles in supporting a child with selective mutism in the school setting. Their first role may be the identification of concerns. School psychologists may be more familiar with selective mutism than S e l e c t i v e M u t i s m : A n O v e r v i e w f o r F a m i l i e s Mental Health 2 © 2023 National Association of School Psychologists │ 4340 East West Hwy, Bethesda, MD 20814 \| www.nasponline.org │ 301-657-0270 other educators, and they can be critical in recognizing the symptoms, observing behaviors in the classroom, and sharing information with parents. School psychologists may also be involved in determining if a child qualifies for an IEP/504 Plan or planning interventions that could be completed at school. School psychologists may also collaborate with outside providers—such as a pediatrician or therapist—in order to communicate observations of the child’s functioning at school and use consistent language and approaches across settings. The level of collaboration with outside providers may vary based on the outside providers’ expertise in selective mutism and their knowledge of appropriate treatment goals in school settings. A school psychologists can also act as a consistent support for the child throughout the years. In this role, the school psychologist: • is qualified to help plan out speaking goals and exposure practice; • may participate in fade-in interventions; • may train school staff on appropriate practices; • may facilitate communication between parents, teachers, and outside professionals; • may be involved in progress monitoring and adjusting goals as the child has success; and • may be involved in planning and preparing for transitions to new classrooms each year. Role of Teachers Although teachers can at times act as the key stakeholder for a child, this can be more challenging when planning for transitions from year to year. Teachers should be involved in direct intervention, will provide accommodations in the classroom, will help the child navigate through situations with peers, and may be involved in documenting and selecting goals. The teacher will offer a supportive environment to work on increasing speaking and will find opportunities for the child to practice their communication just outside of their comfort zone. Role of the Parent or Caregiver The caregiver has a critical support role, as they come with the most knowledge about the child, the child’s interests, and helpful ways to engage with the child. The caregiver also gives the school insight into what the child looks like when comfortable, what the child’s communication is like when speaking at home, and approaches that have been successful or unsuccessful for the child. The caregiver can be an advocate in the school setting, helping the school focus on the child’s needs. The caregiver also can be involved in early fade-ins and opportunities to practice speaking—for example, having playdates with peers from school or practicing ordering at a restaurant. Caregivers may also be involved in finding an outside provider with specialty in SM in order to receive a formal diagnosis and further support for managing their child’s anxiety and communication difficulties. Access to a knowledgeable provider may vary based on the family’s proximity to selective mutism care experts, though the Selective Mutism Association does have a directory of SM providers across the country ( Finally, the caregiver may also be involved in the reinforcement system, offering desired rewards (such as a special activity or small item) at home for speaking in school. Role of the Speech/Language Pathologist Not every child with selective mutism will work with a speech/language pathologist, but a speech/language pathologist can provide an assessment that is valuable in helping to determine if symptoms appear consistent with selective mutism and to better understand if there are underlying speech or language concerns for the child. Speech/language pathologists bring knowledge for improving speech and communication for children, and many speech/language pathologists have further knowledge about selective mutism and appropriate treatment approaches. Speech/language pathologists may also be involved in direct intervention and progress monitoring in the school setting. S e l e c t i v e M u t i s m : A n O v e r v i e w f o r F a m i l i e s Mental Health 3 © 2023 National Association of School Psychologists │ 4340 East West Hwy, Bethesda, MD 20814 \| www.nasponline.org │ 301-657-0270 REFERENCES Bergman, R. L., Piacentini, J., & McCracken, J. T. (2002). Prevalence and description of selective mutism in a school-based sample. Journal of the American Academy of Child and Adolescent Psychiatry, 41(8), 938–946. Driessen, J., Blom, J. D., Muris, P., Blashfield, R. K., & Molendijk, M. L. (2020). Anxiety in children with selective mutism: A meta-analysis. Child Psychiatry and Human Development, 51(2), 330–341. Klein, E. R., Armstrong, S. L, & Shipon-Blum, E. (2013). Assessing spoken language competence in children with selective mutism: Using parents as test presenters. Communication Disorders Quarterly, 34(3), 1–12. Toppelberg, C. O., Tabors, P., Coggins, A., Lum, K., & Burger, C. (2005). Differential diagnosis of selective mutism in bilingual children. Journal of American Academy of Child Adolescent Psychiatry, 44, 592–595. RESOURCES More information about selective mutism and interventions can be found through the following books and websites. Books Johnson, M., & Wintgens, A. (2016). The selective mutism resource manual (2nd Ed.). Routledge. Kohlmeier, J. (2016). Learning to play the game: My journey through silence. Lulu Publishing Services. Kotrba, A. (2014). Selective mutism: An assessment and intervention guide for therapists, educators & parents. Pesi Publishing & Media. Kotrba, A., & Saffer, S. J. (2018). Overcoming selective mutism: A parent’s field guide. Summit & Krest. Schum, R. L. (2017). Finding voice: Treating selective mutism and social anxiety. Research Press. Websites • Selective Mutism Association, selectivemutism.org • Kurtz Psychology’s Selective Mutism University, selectivemutismlearning.org • Child Mind Institute’s Complete Guide to Selective Mutism, childmind.org/guide/selective-mutism • SMart Center, selectivemutismcenter.org • Anxiety Canada, • The American Speech-Language-Hearing Association (ASHA), Handout author: Brittany Bice-Urbach, PhD Please cite this document as: Bice-Urbach, B. (2023). Selective mutism: An overview for families [handout]. National Association of School Psychologists.
8729	https://leetcode.com/problems/factorial-trailing-zeroes/	Factorial Trailing Zeroes - LeetCode Problem List Problem List Debugging... Submit 00:00:00 RegisterorLog in Premium Description Description Editorial Editorial Solutions Solutions Submissions Submissions Code Code Testcase Testcase Test Result Test Result 172. Factorial Trailing Zeroes Medium Topics Companies Given an integer n, return the number of trailing zeroes inn!. Note that n! = n (n - 1) (n - 2) ... 3 2 1. Example 1: Input: n = 3 Output: 0 Explanation: 3! = 6, no trailing zero. Example 2: Input: n = 5 Output: 1 Explanation: 5! = 120, one trailing zero. Example 3: Input: n = 0 Output: 0 Constraints: 0 <= n <= 104 Follow up: Could you write a solution that works in logarithmic time complexity? Accepted 582,205/1.3M Acceptance Rate 45.1% Topics Math Companies Similar Questions Number of Digit One Hard Preimage Size of Factorial Zeroes Function Hard Abbreviating the Product of a Range Hard Maximum Trailing Zeros in a Cornered Path Medium Discussion (90) Choose a type Comment 💡 Discussion Rules Please don't post any solutions in this discussion. The problem discussion is for asking questions about the problem or for sharing tips - anything except for solutions. If you'd like to share your solution for feedback and ideas, please head to the solutions tab and post it there. Sort by:Best No comments yet. 1 2 3 4 9 Copyright © 2025 LeetCode. All rights reserved. 3.4K 90 8 Online C++ Auto 1 2 3 4 5 6 class Solution{ public: int trailingZeroes(int n){ } }; Saved Ln 1, Col 1 You need to log in / sign up to run or submit Case 1 Case 2 Case 3 n = 3 9 1 2 3 › 3 5 0 Source FindHeaderBarSize FindTabBarSize FindBorderBarSize Layouts Hints Apply Default Upgrade to premium to unlock this layout Subscribe Note-taking Upgrade to premium to unlock this layout Subscribe Debug Upgrade to turn on custom layouts Subscribe 🧘 Focus Mode NEW
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Cookie yt-remote-device-id Duration Never Expires Description YouTube sets this cookie to store the user's video preferences using embedded YouTube videos. Cookie yt-remote-connected-devices Duration Never Expires Description YouTube sets this cookie to store the user's video preferences using embedded YouTube videos. Reject All Save My Preferences Accept All Powered by Skip to main content AOMRC Medical Training Initiative MTI Evidence-based Interventions EBI AOMRC Medical Training Initiative MTI Evidence-based Interventions EBI Home About Interventions Resources Home About Interventions Resources HomeInterventionsHysterectomy for heavy menstrual bleeding Hysterectomy for heavy menstrual bleeding Statutory Guidance How up to date is this information? Published January 2019 \| Last reviewed September 2024 Using this guidance The guidance set out here was reviewed extensively in the Autumn of 2024. There are no plans for any further reviews. Medicine is constantly evolving and over time it is inevitable that the evidence base will change. Please use your own judgement and/or other sources of clinical guidance alongside the information set out here. Please note this guidance is a recommendation and it should be used in the context of the overall care pathway and when all alternative interventions that may be available locally have been undertaken. Summary Hysterectomy is the surgical removal of the uterus. Recommendation Based on NICE guidelines [NG 88] Heavy menstrual bleeding: assessment and management, hysterectomy should not be used as a first-line treatment solely for heavy menstrual bleeding. It is important that healthcare professionals understand what matters most to each individual and support their personal priorities and choices. Hysterectomy should be considered only when: other treatment options have failed, are contradicted; there is a wish for amenorrhoea (no periods); the individual (who has been fully informed) requests it; the individual no longer wishes to retain their uterus and fertility. 1.13.1.1.1 NICE guideline NG88 1.5 Management of HMB 1.5.1 When agreeing treatment options for HMB with women, take into account: the woman’s preferences, any comorbidities, the presence or absence of fibroids (including size, number and location), polyps, endometrial pathology or adenomyosis, other symptoms such as pressure and pain. 1.13.1.1.2 Treatments for women with no identified pathology, fibroids less than 3 cm in diameter, or suspected or diagnosed adenomyosis 1.5.2 Consider an LNG-IUS (levonorgestrel-releasing intrauterine system) as the first treatment for HMB in women with: no identified pathology or fibroids less than 3 cm in diameter, which are not causing distortion of the uterine cavity or suspected or diagnosed adenomyosis. 1.5.3 If a woman with HMB declines an LNG-IUS or it is not suitable, consider the following pharmacological treatments: non-hormonal: tranexamic acid, NSAIDs (non-steroidal anti-inflammatory drugs), hormonal: combined hormonal contraception, cyclical oral progestogens. 1.5.4 Be aware that progestogen-only contraception may suppress menstruation, which could be beneficial to women with HMB. 1.5.5 If treatment is unsuccessful, the woman declines pharmacological treatment, or symptoms are severe, consider referral to specialist care for: investigations to diagnose the cause of HMB, if needed, taking into account any investigations the woman has already had and alternative treatment choices, including: pharmacological options not already tried (see recommendations 1.5.2 and 1.5.3), surgical options: second-generation endometrial ablation, hysterectomy. 1.5.10 For women with submucosal fibroids, consider hysteroscopic removal. 1.13.1.1.3 Treatments for women with fibroids of 3 cm or more in diameter 1.5.7 Consider referring women to specialist care to undertake additional investigations and discuss treatment options for fibroids of 3 cm or more in diameter. 1.5.8 If pharmacological treatment is needed while investigations and definitive treatment are being organised, offer tranexamic acid and/or NSAIDs. 1.5.9 Advise women to continue using NSAIDs and/or tranexamic acid for as long as they are found to be beneficial. 1.5.10 For women with fibroids of 3 cm or more in diameter, take into account the size, location and number of fibroids, and the severity of the symptoms and consider the following treatments: pharmacological: non-hormonal: tranexamic acid, NSAIDs, hormonal: LNG-IUS, combined hormonal contraception, cyclical oral progestogens, uterine artery embolization, surgical: myomectomy, hysterectomy. 1.5.12 Be aware that the effectiveness of pharmacological treatments for HMB may be limited in women with fibroids that are substantially greater than 3 cm in diameter. 1.5.13 Prior to scheduling of uterine artery embolisation or myomectomy, the woman’s uterus and fibroid(s) should be assessed by ultrasound. If further information about fibroid position, size, number and vascularity is needed, MRI should be considered. 1.5.14 Consider second-generation endometrial ablation as a treatment option for women with HMB and fibroids of 3 cm or more in diameter who meet the criteria specified in the manufacturers’ instructions. 1.5.15 If treatment is unsuccessful: consider further investigations to reassess the cause of HMB, taking into account the results of previous investigations and offer alternative treatment with a choice of the options described in recommendation 5.10. 1.5.16 Pretreatment with a gonadotrophin-releasing hormone analogue before hysterectomy and myomectomy should be considered if uterine fibroids are causing an enlarged or distorted uterus. For further information, please see: NICE Guidance [NG 88] Heavy menstrual bleeding: assessment and management NHS Conditions, Heavy periods Rationale for recommendation NICE’s Guideline Development Group considered the evidence (including 2 reviews, four randomised control trials and one cohort study comparing hysterectomy with other treatments) as well as the views of patients and the public and concluded that hysterectomy should not routinely be offered as first line treatment for heavy menstrual bleeding. The Group placed a high value on the need for education and information provision for indivduals with heavy menstrual bleeding. Complications following hysterectomy are usually rare but infection occurs commonly. Less common complications include: intra-operative haemorrhage; damage to other abdominal organs, such as the urinary tract or bowel; urinary dysfunction –frequent passing of urine and incontinence. Rare complications include thrombosis (DVT and clot on the lung) and very rare complications include death. Complications are more likely when hysterectomy is performed in the presence of fibroids (non-cancerous growths in the uterus). There is a risk of possible loss of ovarian function and its consequences, even if their ovaries are retained during hysterectomy. If oophorectomy (removal of the ovaries) is performed at the time of hysterectomy, menopausal-like symptoms occur. Patient information Information for Patients There are two surgical procedures which have in the past been used to investigate and treat heavy periods. The first, dilation and curettage, was used to establish the cause of heavy periods, but today, the medical evidence tells us this procedure in inappropriate and should not be routinely carried out. The second procedure, a hysterectomy which removes a woman’s womb and therefore ends menstruation completely, can be carried out, but only when specific criteria are met and alternative treatments have been tried first. About the condition Heavy periods are common and can have a significant effect on a woman’s everyday life. In about half of women, no underlying reason is found. But, there are several conditions and some treatments that can cause heavy menstrual bleeding, so you should discuss your symptoms with a clinician if you are concerned. It’s important you and your doctor make a shared decision about what’s best for you if your heavy periods are becoming a problem. When deciding what’s best you should both consider the benefits, the risks, the alternatives and what will happen if you do nothing. What are the BENEFITS of the intervention? There are no diagnostic or treatment benefits with dilation and curettage. A hysterectomy for patients with heavy periods should only be considered in certain circumstances. What are the RISKS? Complications following dilation and curettage are rare, but can include uterine perforation, infection, damaging your cervix. A hysterectomy is a significant operation and therefore inevitably carries a small risk of blood loss or complications from the anaesthetic. Other risks include infection, or a prolapse in later years. It may also cause the early onset of your menopause and should only be considered if you definitely don’t want to have children as your periods will be permanently ended. What are the ALTERNATIVES? A doctor will usually use an ultrasound scan or an instrument which takes a small sample of the lining of your womb to see what’s causing your heavy periods. There are a number of alternative treatment options including hormone treatment and a coil that provides contraception and are good at reducing blood loss. What if you do NOTHING? Doing nothing is not likely to be harmful. However, if heavy periods are having a significant impact on your life, you should seek medical advice to identify the underlying cause and discuss treatment options. ### Download this information ### Make the most of your appointment using BRAN 2020 Please accept cookies to access this content Coding WHEN Primary_Spell_Procedure IN ('Q071','Q072','Q073','Q074','Q075','Q076','Q078','Q079','Q081','Q082','Q083','Q088','Q089') AND ( Any_Spell_Diagnosis like '%N92%' OR Any_Spell_Diagnosis like '%N950%') AND Any_Spell_Diagnosis not like '%D25%' AND not ( Any_Spell_Diagnosis like '%C52%' OR Any_Spell_Diagnosis like '%C53%' OR Any_Spell_Diagnosis like '%C54%' OR Any_Spell_Diagnosis like '%C5%' OR Any_Spell_Diagnosis like '%C57%' OR Any_Spell_Diagnosis like '%C58%') -- Only Elective Activity AND APCS.Admission_Method not like ('2%') THEN 'J_hysterec' Exclusions WHERE 1=1 -- Cancer Diagnosis Exclusion AND (Any_Spell_Diagnosis not like '%C[0-9][0-9]%' AND Any_Spell_Diagnosis not like '%D0%' AND Any_Spell_Diagnosis not like '%D3%' AND Any_Spell_Diagnosis not like '%D4%' OR Any_Spell_Diagnosis IS NULL)-- Private Appointment ExclusionAND apcs.Administrative_Category<>'02' References NICE guidance 2018 Heavy menstrual bleeding: assessment and management [Ng88] NHS information: Heavy periods. Hurskainen R, Teperi J, Rissanen P, et al. Clinical outcomes and costs with the levonorgestrel-releasing intrauterine system or hysterectomy for treatment of menorrhagia: randomized trial 5-year follow-up. JAMA: the journal of the American Medical Association 2004;291(12):1456–63. Learman LA, Summitt Jr RL, Varner RE, et al. Hysterectomy versus expanded medical treatment for abnormal uterine bleeding: Clinical outcomes in the medicine or surgery trial. Obstetrics and Gynecology 2004;103(5 I):824–33. Zupi E, Zullo F, Marconi D, et al. Hysteroscopic endometrial resection versus laparoscopic supracervical hysterectomy for menorrhagia: a prospective randomized trial. American Journal of Obstetrics and Gynecology 2003;188(1):7–12. Lethaby A, Hickey M, Garry R. Endometrial destruction techniques for heavy menstrual bleeding. Cochrane Database Syst Rev. 2005 Oct 19;(4):CD001501. Review. Update in: Cochrane Database Syst Rev. 2009;(4):CD001501. PubMed PMID: 16235284. Hehenkamp WJ, Volkers NA, Donderwinkel PF, et al. Uterine artery embolization versus hysterectomy in the treatment of symptomatic uterine fibroids (EMMY trial): peri- and postprocedural results from a randomized controlled trial. American Journal of Obstetrics and Gynecology 2005;193(5):1618–29. Pinto I, Chimeno P, Romo A, et al. Uterine fibroids: uterine artery embolization versus abdominal hysterectomy for treatment – a prospective, randomized, and controlled clinical Radiology 2003;226(2):425–31. Also in this section How up to date is this information? Using this guidance Summary Recommendation Rationale for recommendation Patient information Coding References Share this page Back to top Evidence-based Interventions The Evidence-based Interventions programme is an initiative led by the Academy of Medical Royal Colleges to improve the quality of care. Created by both doctors and patients, it is designed to reduce the number of medical or surgical interventions as well as some other tests and treatments which the evidence tells us are inappropriate for some patients in some circumstances. In some instances, clinicians recommend we carry out more procedures because this will result in an improved quality of life for patients in the long term. Navigation Home About Interventions Resources Contact Academy of Medical Royal Colleges, 10 Dallington Street, London, EC1V 0DB. Email: england.ebinterventions@nhs.net Charity No. 1056565 Company No. 3166361 © 2025 Academy of Medical Royal Colleges. AOMRC Medical Training Initiative MTI Evidence-based Interventions EBI
8731	https://www.cfuttrup.com/dpc/air.htm	AIR.HTM --- Part of Manual for Driver Parameter Calculator --- by Claus Futtrup.Created 9. September 1996, last revised 16. February 2004. Ported to XHTML 1.0 on 2. October 2004. Last modified 25. October 2004. Table of Contents: Properties of Air Air Density Speed of Sound Sample Calculations for rho and c 0 degC 20 degC 40 degC Summary Additional Questions Acknowledgments Litterature Properties of Air As can be seen from the loudspeaker data found by measurements + calculations depends on two factors: rho : air density c : speed of sound in air Since these factors depends on air pressure, air temperature and more, and are so essential to the accuracy of Vas versus Cms, we will treat them here. Notice that this document concludes that humidity level does not have an influence on Vas versus Cms because changes in rho and c are balanced out. Air Density The air density is defined (as any other density) as "mass per volume," ie. rho = m/V. This expression can be isolated from the gas equation, which assumes that the gas behaves like an ideal gas: pV = nRT (1) which should be familiar to the reader. Since n = m/M, where M is the molar mass of the molecules in air we get: mRT pM pV = --- <===> rho = m/V = ---- (2) M RT p is the gas pressure, in Pa (Pascal) M is the molar mass of air, in kg/mol or alternatively g/mol T is the gas temperature, in K (Kelvin) R is the gas constant, R = 8.314510 J/(mol K) or alternatively Nm/(mol K) m is the mass of the gas n is the number of molecules, measured in mol V is the volume of the gas rho is the mass density of the gas You can measure the temperature on a thermometer and convert to Kelvin: Celcius to Kelvin : K = degC + 273.15 Fahrenheit to Kelvin : K = degF5/9 + 255.37 Reamur to Kelvin : K = 1.25degR + 273.15 Rankine to Kelvin : K = R/1.8 = R5/9 Note that Rankine is not specified on a degree scale, which is because it is measuring absolute temperature similarly to Kelvin (which is not a degree scale either). Note that in some books the shifting value is 273.16 and not 273.15. I have concluded that the variations are due to different temperature scales. I have chosen the 273.15 value. The Reamur degree-scale is not supported by Driver Parameter Calculator. Quite frankly I do not think anybody is using it anymore. Give me a hint if this is not the case and I will support it. Calculations at room temperature, as examples: 20 degC = 293.15 K 70 degF = 294.25 K 16 degR = 293.15 K 530 R = 294.44 K Similarly you can measure the pressure on a barometer (and convert to Pascal). Here the following relations are valid: 101325 Pa = 101.325 kPa = 1.01325 hPa = 1.01325 bar = 1013.25 mbar = 1 atm = 14.69595 psi = 760 torr = 760 mm Hg = 0.76 m Hg = 29.92126 in Hg = 407.189 mm H2O 101325 Pa -- the standard pressure, in Pascal. Then comes kilo-Pascal and hecto-Pascal (which is similar to bar). The standard pressure is also 1 atmosphere. You can also convert from pounds per square-inch or from torricelli or height of a bar of mercury = quicksilver (meter Hg), or height of a similar bar of water (millimeter H20). You most likely can find other measures for pressure, like the gravitational ones. To convert from another unit to Pascal you find the appropriate figure on the right side and divide the Pascal figure by that number. You then read the current pressure and multiply with that figure. Things gets a little harder when it comes to the molar mass because you need to know the distribution of gases in the air. For dry air it approximately is (you can calculate M from a periodic table): gas \| M \| % ------- N_2 \| 28.01 \| 78 \| Nitrogen (M = 28.0134, 78.084% in air) O_2 \| 32.00 \| 21 \| Oxygen (M = 31.9988, 20.947% in air) Ar \| 39.95 \| 1 \| Argon (M = 39.948, 0.9300% in air) This makes up about 100% (and M becomes 28.9673 g/mol), but actually there are more gases in the air, they just do not have any considerable influence, the content of these gases should be measured in ppm (parts per million), not percent: gas \| M ---------\| CO_2 \| 44.010 \| typically around 0.0340% = 340 ppm (M = 44.0098) Ne \| 20.179 \| typically around 0.0018% = 18 ppm He \| 4.003 \| typically around 0.00052% = 5.2 ppm Kr \| 83.80 \| typically around 0.000114% = 1.14 ppm H_2 \| 2.016 \| typically around 0.00005% = 0.5 ppm Xe \| 131.29 \| typically around 0.0000086 = 0.086 ppm O_3 \| 47.997 \| typically around 0.000004% = 0.04 ppm (varies) H2O \| 18.020 \| SO_2 \| 64.065 \| CH_4 \| 16.043 \| NO \| 30.006 \| NO_2 \| 46.006 \| NH_3 \| 17.031 \| CO \| 28.010 \| Some of the gases are reactive and are therefore not stable constituents of air. Others shift toward lower/higher values, eg. the CO_2 level is continuously increasing. The last seen estimate for CO_2 was at 370 ppm, which is the highest level in 420.000 years, and at the end of the 1880'ies the level was measured/estimated at 280 ppm. In the year 2050 it is expected to be between 450 and 650 ppm. To find a value for M that everybody can agree about has proven to be quite difficult. We are talking average values measured over time and at different places. My book (2. reference) says 28.970, which does not match the gas distribution from the same book (see tables above). Other sources agrees that M is as low as 28.964. The difference between the highest and the lowest value is only 0.02% anyway. If we include the other stable gasses and take advantage of the higher precision the more accurate calculation of the molar mass becomes : M = 28.96375 for 99.998% of the air - but we want 100% so by simple scaling we totally get M = 28.96375 / 0.99998 = 28.96433. The initial data does not support this many digits, so from now on the last 2 digits will be cut off and M = 28.965 +/- 0.002 is assumed to be a reasonable value for the molar mass of air as an average value. Though the mixture of air may change for different places on earth, I think we will be surprised how little influence it has on the molar mass. Whether your are in asia, in the jungle or on the ocean. Even if you live in a town and the amount of CO_2 could be much higher when compared to, say the Himalaya Mountains, then CO_2 is only found in small concentations. The typical values of gas concentrations versus location does not vary much because of the mixing of air in the troposphere (from ground level up to a height of 12 km). You would have to live close to a significant source of CO_2 gases to see any real and permanent change of the average molar mass. Local variations in the mixture of gases can be large but they are normally quickly distributed and vary from day to day. For meteorological studies only the 3 most significant gases are used, as far as I know. In a room with a lot of welding the inert gas Argon may be found in slightly higher concentrations too. I believe the above figure for M is as precise as it gets for dry air. For many applications it will suffice to set M = 29 g/mol for the air, whether humid or not. Something which will have an influence is water vapour from the humidity. But at low temperatures the influence is low, since cold air cannot carry much humidity before the relative humidity is close to 100% and then the influence is low anyway. With the above in mind, the approximation for rho and c are usually considered sufficient for a good precision. Below we will include humidity in the calculations and add further to the precision of measuring/calculating driver data. To measure the relative humidity you have several opportunities, of which the easiest is to use a hygrometer (you can buy one wherever you buy barometers). Hygrometers will display the relative humidity. Nowadays electronic hygrometers are inexpensive. Alternatively you can use a thermometer, you read the temperature T_dry. Then moisturize a sock and put it around the thermometer. Then ventilate heavily (with a fan) on the system and reread the temperature T_wet (this method will not give you accurate results). Calculate the drop in temperature (T_drop = T_dry - T_wet) and use the following approximative equation (correct within 1% from 2 degC to 36 degC) to calculate the relative humidity: phi = 100 - (445/(T_dry + 23) - 1.5) (T_drop - (T_drop^2)/80) (3) T_dry and T_drop are in degrees Celcius, and phi is the relative humidity in percent. Since the molar mass of dry air weights more than water, the air actually gets lighter, as it becomes more humid. This is easy to understand when the pressure is assumed to be the same (you measure it), and therefore the total amount of molecules will be the same, only each molecules weighs something different---and since the light water molecules substitute heavier dry-air molecules, totally the air becomes lighter. From thermodynamics we have the following equation: (ML - MH)phiPm rho = rho_0 - ------------------ (4) RL T rho_0 is the density of dry air MH = 18.020 is the molar mass for water ML = 28.965 is the molar mass for dry air (see above) RL = R = 8.314510, R is the gas constant 8.314510 J/(mol K) T is the temperature phi is the relative humidity Pm is the partial pressure for water vapour at similar temperature, but at 100% humidity, given in the table below RL only counts for dry air, but it is usually a good figure for humid air if the water vapour is in thermal balance with the dry air. Besides being in the air as steam, H2O can also be in the air in fluid or solid form because the critical temperature for H2O is higher than for the gases of dry air. Pm can be calculated from the ideal gas equation (though air is not an ideal gas, the results should be valid in the temperature range where you will be measuring, say 5 degC to 60 degC = 41 degF to 140 degF), it is tabulated below (the partial pressure is given in Pascal): T \| 0 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 -----------------------------------------------------------\| 0 \| 611\| 657\| 705\| 757\| 813\| 872\| 935\| 1001\| 1072\| 1147\| 10 \| 1227\| 1312\| 1401\| 1497\| 1597\| 1704\| 1817\| 1936\| 2062\| 2196\| 20 \| 2337\| 2485\| 2642\| 2808\| 2982\| 3166\| 3360\| 3564\| 3778\| 4004\| 30 \| 4242\| 4491\| 4753\| 5029\| 5318\| 5622\| 5940\| 6274\| 6624\| 6991\| 40 \| 7370\| 7780\| 8200\| 8640\| 9100\| 9580\|10090\|10610\|11160\|11740\| 50 \|12330\|12960\|13610\|14290\|15000\|15740\|16510\|17310\|18140\|19010\| For eg. car-fi this table may not cover a sufficient range of temperatures. For this application an approximative (curve-fitted) model should be applied instead. Such an advanced model is implemented into DPC, see for further information. From equation (2) and (4) we can make a single equation for rho: p ML - (ML - MH)phiPm rho = --------------------------- (5) R T This equation (5) is the important one for our future calculations. Speed of Sound Speed of sound can be calculated as: c = sqrt(gammap/rho) = sqrt(gammaRT/M) (6) Since R, T and M have been discussed above the latter mentioned equation (6) at a glance looks better, but the first equation (6) can be used as well. Specifying the air density (rho) we just calculated is actually the easiest way to deal with humid air. for each gas in the air, gamma is given as: cp cv + R gamma = ---- = -------- (7) cv cv where cp is the heat capacity at constant pressure, and cv is the heat capacity at constant volume. This factor is basically given by the structure of the molecule (number of atoms + how well it fits the ideal gas model). From kinetic gas theory cv = f/2 R, where f is the number of freedoms. For triatomic gases the freedom of the third atom is given by the molecule structure: For monoatomic gas : cv = 3/2 R For diatomic gas : cv = 5/2 R This gives for ideal monoatomic gas : gamma = (cv + R) / cv = 5 / 3 = 1.667 For an ideal diatomic gas we get : gamma = 7 / 5 = 1.4 Here is gamma for some gasses, at 20 degC (cv is dependent on temperature): gas \| gamma -------\| N_2 \| 1.40 \| O_2 \| 1.40 \| Ar \| 1.65 \| (for an ideal monoatomic gas, gamma would equal 1.667) CO_2 \| 1.29 \| CO \| 1.40 \| He \| 1.63 \| SO_2 \| 1.27 \| CH_4 \| 1.31 \| NO \| 1.39 \| NO_2 \| 1.17 \| NH_3 \| 1.31 \| H_2 \| 1.40 \| H_2O \| 1.30 \| For dry air gamma = 1.40 is a good estimate, the temperature may even differ from 20 degC, the assumption is valid within 0.3% from about -10 degC to about 50 degC (from 14 degF to 122 degF). I do not know who would measure (or play music on hi-fi loudspeakers) at temperatures outside this range. I assume that the influence on humid air is relatively small too (because c is calculated as a square-root), but having measured it as described above you may as well include it here too, simply use rho instead of rho_0. If rho changes, then M changes similarly, according to the ideal gas equation (2) : P M = rho R T. Here P, R and T are held constant. If M is changed because it now consist of a new mixture of gases we get, say the air is now 2% water vapour (H20) and 98% dry air: Mnew = 0.98 28.965 + 0.02 18.020 = 28.7461 g/mol (8) Then rho will change proportionally: rho Mnew ------- = ------ (9) rho_0 M Remixing gamma as well you get that gamma changes from 1.40 to 1.398, which is a change of about 0.14%, or basically no changes at all, and outside the precision range for the tabulated gamma values. My second reference gives cp and cv explicitly for the air, from which I get gamma = 1.401 at 0 degC (32 degF) and standard pressure 101325 Pa. M has changed 0.76%. The changes work in opposite directions and we get that the expression of c changes 0.31%, which in my opinion may be neglected. Note, though that c usually is squared in the Thiele-Small parameter for Cms, whereby the error in the transformation from Vas to Cms diminishes. This means that when calculating new values for air density and sound speed it is sufficient to base it on temperature and pressure because including humidity will not increase the precision of Cms. This can be seen from the first equation for the speed of sound (6), where c is inverse proportional to the square-root of rho. The sound speed c is used in several other equations, though, like in the calculations of the efficiency of the driver, where no is proportional to 1 / c^3. A change of 0.31% in c will give you 1% error in the reference efficiency, no. Sample Calculations for rho and c The following temperatures will be calculated, always assuming standard pressure p = 101325 Pa, for dry air the molar mass is assumed to be ML = 0.028965 kg/mol and for water vapour MH = 0.018020 kg/mol: 0 degC (32 degF) 20 degC (68 degF) 40 degC (104 degF) at 0 and 100% relative humidity (phi = 0 and phi = 1). 0 degC T = 273.15 K measured on a thermometer Pm = 611 Pa from the above table rho_0 = 1.2923 kg/m3 Substituting 0 with 1 in the above equation (5), for 100% humidity, (we have precalculated ML - MH once and for all) we get: p ML - (ML - MH)Pmphi rho = --------------------------- R T 101325 0.028965 - 0.010945 611 1 = -------------------------------------- (10) 8.314510 273.15 = 1.2893 kg/m3 The speed of sound for the desert (dry air) and for the rain forrest (humid air) becomes: c0 = sqrt(1.40101325/1.2923) = 331.32 m/s (dry) c = sqrt(1.40101325/1.2893) = 331.70 m/s (humid) At 0 degC the changes are really insignificant. Furthermore, please note that the total change in rhoc^2 = 0% and that expression is what is actually used in the Thiele/Small equations for suspension compliance of loudspeakers. The total change in c^3 = 0.3%. 20 degC T = 293.15 K measured on a thermometer Pm = 2337 Pa from the above table Insertion in the equations gives: rho_0 = 1.2041 c0 = 343.23 rho = 1.1936 c = 344.74 The difference on rho are now up to 0.9% and should be taken into considerations. The difference on c is .4% in the opposite direction so that if you calculate rhoc^2 the total error will be zero. If you calculate c^3 the total error becomes 1.3%. 40 degC T = 313.15 K measured on a thermometer Pm = 7370 Pa from the above table Insertion in the equations gives: rho_0 = 1.1273 c0 = 354.74 rho = 1.0963 c = 359.71 The difference on rho is now 2.8% and as we can see the humidity becomes more important for higher temperatures. The change in c^3 = 4%. Summary Considering humidity will in other words change the numbers for rho and c slightly, but totally it has no importance when calculating the loudspeaker suspension compliance, Cms, from its volume equivalent, Vas. Driver Parameter Calculator will continue to support humidity because you may find the calculations of c or rho useful individually in other areas where they are not treatet together, like if you want to know the time-distance in a measuring setup from a loudspeaker driver to a microphone. By the way. If you know the relative humidity out-door, and you know the in-door and out-door temperatures, then you can calculate a new relative humidity. We need: Out-door temperature, T_out In-door temperature, T_in The relative humidity outdoor, H_out Say the relative humidity outdoor is 90%, and the temperature is 9 degC (48.2 degF), but your indoor temperature is 20 degC (68 degF), then from the table of Pm we find Pm_out = 1147 and Pm_in = 2337. Pm_out H_in = H_out ------ = 90 1147/2337 = 44% (11) Pm_in This is a linear equation, which works in the same temperature range as given for the table of partial pressures. This way you suddenly get a completely new figure for the relative humidity. The partial pressure is kept unchanged, and since the temperature is new you will get a different air-density. Assuming a constant Gamma is a basis for small errors, perhaps up to 0.3% and assuming that dry air as well as water vapour is an ideal gas at normal temperatures, which is the basis for the table of partial pressures, may provide a basis for similar errors. Additional Questions What is the weather around the globe? (for designers of speakers, that are exported, this might be of interest). What is needed is the average temperature and pressure at particular locations, and how are they distributed (mean-value + variance?). Perhaps the averages could be weighted in relation to the number of people (who have money to buy speakers) living in the areas. The following internet URL may provide some answers: Outdoor data is one issue, but do not forget, that all figures should be modified to in-door figures. Living in hot and humid areas (India?), you must take air-conditioning into considerations as well. In Denmark, airconditioning is not normal and the temperature is normally approx. 20 degC, with humidity around 40% most of the year being a good approximation. Humidity out-door can be much higher. Following the danish organisation BST, an organisation for health on work places, it is recommended to keep the relative humidity at 35-65 % for comfortable levels. High relative humidity can increase the number of sick days, and normally the relative humidity is kept at the low end. On a measuring instrument at Dynaudio, which displays temperature, pressure and relative humidity, the instrument says DRY until you reach 40% relative humidity, then the instrument says COMFORT. The instrument costs more around 100 USD. In a given room situation the relative humidity could be almost 100% (or at least that is how it feels) but the instrument would show max. 45% RH. The reason was simply wrong placement of the instrument. Take care. If the instrument had been able to go above 70% RH then it would probably specify the environment as WET. A value of 40% RH is probably a good approximation. If 30% RH is chosen, which is the standard setup in DPC, then the values for c and rho are very close to the standard values used in various books. Loudspeaker and Headphone Handbook, ed. John Borwick (ISBN 0 240 51371 1, 2. edition, p. 439) suggests that "measurings should therefore be carried out at normal room temperature, say 25 +- 5 degrees C". I guess this is not necessary anymore, but still---a loudspeaker driver will change behaviour depending on temperature, and these changes is inherent in the materials. Therefore you should pick a day with a reasonably normal temperature. The model of air with a table of partial pressures has been exchanged for an approximative (curve-fitted) model, because such a model is slightly more accurate and useful for a wider range of temperatures. Similarly an approximative model for gamma could be made available within Driver Parameter Calculator, but has not been reworked yet. Acknowledgments Since the creation of this document in September 1996 to the beginning of January 2000 I did not know whether this document was right or wrong regarding to humidity. I finally succeeded to have someone look at the content, and several people replied with suggestions for improvement: Timo Christ, an EE student from University of Bremen. Ron Ennenga, a ME student from University of Minnesota. Ray Hopkins from Cloud and Aerosol Sciences Lab, University of Missouri-Rolla. Claus Petersen from the Research & Development department of the Danish Meteorological Institute, DMI, see Thank you for taking your time to proofread this text. Litterature This document is based on the following litterature: 1. Maskinteknisk Termodynamik, haefte 6 Andreas Andersen & Svend Ishoy Rasmussen DIAM nr. 571, Juli 1988 comment: I have "slaughtered" several bugs in this book. 2. DATABOG fysik kemi Erik Strandgaard Andersen, Paul Jespersgaard Ove Gronbaek Ostergaard 9. edition F & K forlaget ISBN 87-87229-57-9 comment: used for the table of Pm and the ingenious method for measuring relative humidity with a wet sock. This method is actually the "professional" method applied in eg. climate chambers for control of humidity in the chamber.
8732	https://pancreasfoundation.org/pancreas-disease/fcs/	Skip to content FCS FCSSokphal Tun2025-09-11T16:16:30-04:00 Skip To a Section What is Familial Chylomicronemia Syndrome? FCS is a rare, genetic disorder of fat metabolism that is characterized by extremely high triglyceride levels, which are 10- to 100- fold or more above normal. FCS is estimated to occur in 1 of every 1-2 million people. FCS can be diagnosed at any age and affects genders, races, and ethnicities equally. Signs and Symptoms of FCS Symptoms of FCS can vary widely but may include: Severe abdominal pain: Often associated with pancreatitis, a potentially life-threatening condition. Fatty deposits in the skin: Known as eruptive xanthomas, these are small, yellowish bumps often found on the buttocks, arms, and legs. Memory and concentration issues: "Brain fog" is common in patients living with FCS. Enlargement of the liver and spleen due to fat buildup. Chronic tiredness and weakness are often reported by those with FCS. Treatment and Management While there is currently no cure for FCS, treatment focuses on managing symptoms, reducing triglyceride levels, and preventing complications. FCS management can include: Dietary management: A very low-fat diet (often below 10-15% of daily calories) is the cornerstone of FCS management. Foods high in fat can increase chylomicrons, so individuals with FCS often avoid foods like oils, nuts, and fatty fish. In patients with FCS, there is no such thing as "Good Fat," as fat is fat. Medications: Fibrates, omega-3 fatty acids (must be included in your daily fat allowance), and niacin are sometimes prescribed, though their effectiveness in FCS is limited. Currently, Tryngolza the only FDA-approved prescription medicine for adults with FCS. For more information, click HERE. Please discuss with an FCS professional before seeking medication. Regular monitoring: Patients need routine blood tests to monitor triglyceride levels and check for signs of pancreatitis or other complications. Yearly eye exams are needed to ensure the vessels in your eye and optic nerve are not being negatively affected. Dietary Management of FCS High triglyceride levels are managed by eating an extremely restrictive, low-fat diet (10-15% of total caloric intake). People living with FCS may eat vegetables, fruits, whole grains, egg whites, legumes, fat-free dairy products, seafood, and lean poultry. People with FCS can obtain essential fatty acids with supplements that include fat-soluble vitamins (vitamins A, D, E, K), minerals, and medium-chain triglycerides, as needed. Eating small, frequent meals that contain fat-free or low-fat protein is suggested. Know Your TGs – FCS-Friendly Recipes Learn Your Lipids – Low-Fat Cookbook Action FCS – Recipes FCS Foundation Recipes Lifestyle Tips for Living with FCS Managing FCS involves both lifestyle and dietary adjustments. Below are some helpful tips: Maintain a strict low-fat diet (as mentioned above) Stay hydrated: Adequate water intake helps reduce blood viscosity. Engage in gentle exercise: While intense exercise may increase triglyceride levels, moderate activities like walking or yoga are beneficial. Manage stress: Stress can exacerbate symptoms, so practices like meditation and deep breathing exercises can be helpful. Support (group or 1:1) Causes of FCS FCS occurs when the enzyme lipoprotein lipase (LPL), which breaks down triglycerides, doesn't work well or is missing. This can be caused by mutations in the LPL gene or in other genes that help LPL function properly. Without lipoprotein lipase, triglycerides (a type of fat) build up in the blood. These triglycerides are mainly transported in dietary lipids called chylomicrons. In people with FCS, chylomicrons stay in the blood instead of breaking down, causing them to accumulate. This buildup can block blood flow to the pancreas, triggering acute pancreatitis. Acute pancreatitis can cause serious damage to the pancreas, be life-threatening, and lead to other symptoms and complications. Diagnosis of FCS The process of diagnosing FCS can include: Blood tests: Measures extremely high triglyceride levels (often over 880 mg/dL) and confirm the presence of chylomicrons. When blood is drawn, the plasma may have a milky appearance due to excessive lipids. Lipid profile: Helps rule out other lipid disorders and narrows down diagnosis. Early diagnosis is crucial to managing symptoms and reducing the risk of complications. Genetic testing:Identifies mutations in genes related to FCS. Key genes associated with FCS: LPL:Code for the enzyme lipoprotein lipase, responsible for breaking down triglycerides. APOA5, GPIHBP1, APOC2, and LMF1:Mutations in these genes can also contribute to FCS. Risks of FCS & Care Team It is crucial for patients managing FCS to have a diverse care team in place because: Repeated episodes of acute pancreatitis can lead to chronic pancreatitis and signs of exocrine or endocrine pancreatic insufficiency, including pancreatic (type 3c) diabetes. Patients with extremely high levels of triglycerides may have a more severe case of pancreatitis, leading to worse outcomes, including but not limited to longer hospital stays, a higher rate of pancreatic necrosis, more frequent, persistent organ failure, and higher mortality rates. This risk may be lowered by treatment with a healthcare team that understands all aspects of the disease, which may include a lipidologist, pancreatologist, gastroenterologist, primary care physician, registered dietician, psychologist, and/or social worker. FCS Resources FCS Foundation:Living with FCS is the site for all FCS patients, caregivers and family members that offers information, resources and support. Action FCS: Action FCS offers resources for patients and caregivers to learn more about FCS. Animated Pancreas Patient: The NPF's Animated Pancreas Patient offers short, easy-to-understand animations with audio narration, expert video explanations, patient interviews, slide shows, and downloadable information on major pancreatic disease topics. This resource offers expert advice to help you discuss key issues with your healthcare provider and make important decisions related to management and treatment of pancreas disease. EatRight.org: The Academy of Nutrition and Dietetics offers information on nutrition and health, from meal planning and prep to choices that can help prevent or manage health conditions and more. Know Your TGs:Ionis has created a free suite of tools designed to ease the diagnosis, management, and understanding of FCS and the lifestyle it requires. Their helpful resources provide insight into FCS, pancreatitis and diet and lifestyle, in addition to downloadable resources, including the Patient Guide to FCS, the FCS CareBook, and the FCS Nutrition Toolkit. Download available resources HERE. Learn Your Lipids: Learn Your Lipids is the foundation of the National Lipid Association. It is a non-profit focused on providing education and resources to help patients and their families manage and overcome lipid-related health problems that may put them at risk for a heart attack or stroke. They aim to translate scientific and medical progress into effective evidence based guidance for optimal patient care. The Foundation's mission is to improve the welfare of patients and families affected by cholesterol and triglyceride problems. Their website provides resources for FCS patients and families. Lower My TGs: Arrowhead Pharmaceuticals has created a patient and caregiver disease state website that serves as a comprehensive platform offering resources, education, and patient stories. Patients can learn more about lowering Triglycerides to reduce acute pancreatitis attacks. Read their "Spotlight on FCS" paper HERE. National Lipid Association: The National Lipid Association website is a resource dedicated to promoting the understanding and management of lipid disorders and cardiovascular health. It serves healthcare professionals, patients, and researchers with: educational resources, patient care information, and research and advocacy. Emerging Therapies and Research Studies Plozasiran: Arrowhead Pharmaceuticals submitted a new drug application (NDA) to the U.S. FDA in November 2024 for plozasiran to lower triglycerides in adult patients with FCS, with plans to submit to additional regulatory authorities in 2025. A U.S. PDUFA (Prescription Drug User Fee Act) date of November 18, 2025 has been announced. To find out more about the PALISADE trial, please click HERE FCS Patient Education Webinars FCS Patient Stories For Providers Journal of Clinical Lipidology – Familial chylomicronemia syndrome: An expert clinical review from the National Lipid Association Pancreas 101 provides free CME accredited modules on adult pancreatitis, pediatric pancreatitis, and pancreatic cancer. These modules were created for nurses, social workers, dieticians, physician assistants to learn more about signs and symptoms of pancreas disease, surgery options, pain management, research opportunities, as well as hearing from patients living with pancreas disease. Lookout for new modules on FCS that will be added in the Fall, before November 7th (FCS Day). HCP Live is a comprehensive clinical news and information portal that provides physicians with up-to-date FCS resources designed to help them provide better care to patients. HCPLive® videos on Understanding and Managing FCS: Episode 1 \| Familial Chylomicronemia Syndrome Overview Episode 2 \| The Correlation Between Acute Pancreatitis and FCS Episode 3 \| The Clinical Presentation of FCS Episode 4 \| FCS Diagnostic Process FCS Webinars and Podcasts Thank You to Our Sponsors The National Pancreas Foundation 3 Bethesda Metro Center, Suite 700 Bethesda, MD 20814 U.S.A. Please send all donations and mail: The National Pancreas Foundation P.O. Box 750046,Atlanta, GA 30374 U.S.A. Explore Find a Center of Excellence Join a Chapter Volunteer Join Our Newsletter Patient Education Patient Resources Animated Pancreas NPF Cookbook INSPIRE Online Patient Community Camp Hope Black/African American Initiative Mental Health Physician Resources Become a Center of Excellence Research Grants Fellows Symposium Awards Sign up for our newsletter © Copyright 2022 - 2025 \| The National Pancreas Foundation Page load link
8733	https://ocw.mit.edu/courses/5-62-physical-chemistry-ii-spring-2008/resources/31_562ln08/	MIT OpenCourseWare 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: .5.62 Spring 2008 Lecture #31 Page 1 Kinetic Theory of Gases: Mean Free Path and Transport The mean free path λ. The mean free path is the average distance a particle traverses before it experiences a collision. In Lecture #31 we determined the average collision frequency for a particle, Z . The mean time between collisions is simply Z -1. If the mean speed of a particle is v then the mean free path is ! = v Z . As a result, for like particles we find the mean free path equal to: ! = 1 2"#d2 . This is an important and interesting result. For a dilute hard sphere gas, the mean free path depends only on density; it is independent of temperature. However, if the particles had an attractive or repulsive potential between them, the mean free path would depend on T. Typical values at 300K for O 2 d = 0.361 nm πd2 = 41Å 2 Z ZTOT / V λ 1 bar 6.2 × 10 9 coll s –1 7.6 × 10 34 coll m –3s–1 7.1 ×10 –8 m (10 –6 in) 10 –6 bar 6.2 × 10 3 coll s –1 7.6 × 10 22 coll m –3s–1 7.1 × 10 –2 m (3 in) Why is Z proportional to ρ and Z TOT to ρ2 ? The effusion picture described i n Lecture #31 assumed that no collisions occur when molecules pass through a hole of area A and thickness d. This means that it is necessary to assume that d  λ. If this is condition is not satisfied, the description of gas escaping from the vessel must include collisions and transport phenomena. The probability of a particle traveling a distance r before experiencing a collision is revised 4/24 /08 3:49 PM 5.62 Spring 2008 Lecture #31 Page 2 p(r) rp(r) = !e"!r 0 < r < # Is this probability distribution normalized? What are ?r and r2 Variation of the mean free path with density and temperature . The expression we have derived for the m.f.p. is valid only for hard spheres at very low density. How does the m.f.p. behave under more general conditions? The sim ple expression predicts that the m.f.p. goes to zero at infinite density. However we should instead expect that the m.f.p. goes to zero at a critical density ρc which corresponds to the volume per particle: !c = 1 vc = 3 4"d3 so a more general prediction of the behavior of the m.f.p. with density might be: ! = 1 2"d2 1 $ 1 c % & ' ( ) . If the particles attract or repel each other, there will also be an effect on the m.f.p. Qualitatively, we expect that if the potential is repulsive between the particles, the frequency of collisions will decrease and the m.f.p. will increase. If the potential is attractive, we expect that the frequency of collisions will increase and the m.f.p. will decrease. Both of these effects will be more pronounced at lo wer temperatures when the forces between the particles results in a larger fractional change in the velocity ( ! v v ) of the particles, because v ! T . The mean free path and collisions . In dilute gases the m.f.p . is an important quantity because it gives the microscopic distance scale over which collisions will influence the transfer of physical quantities such as mass, momentum, and energy between colliding partners. revised 4/24 /08 3:49 PM 5.62 Spring 2008 Lecture #31 Page 3 1 2 2 ! This pair -colli sion picture lies at the heart of the elementary kinetic theory of transport. Elementary kinetic theory describes the main transport processes in a unified way and provides molecular expressions for the transport coefficients. Transport equations . Thes e are equations that describe non -equilibrium phenomena – the flow of a physical quantity in response to perturbation of the local thermodynamic variable. When the perturbation from equilibrium is “small,” the transport equations have the form: Flux Tran sport coefficient Gradient of thermodynamic variable mass Diffusion coefficient. Concentration momentum Shear viscosity Velocity heat Thermal conductivity Temperature The fluxes are all defined as the physical quantity transported per unit area per un it time . We consider the simple physical situation of the gradient in the “z” direction and because the gas is isotropic, the flux in the “z” direction. The flux (denoted by j with a right superscript that specifies the physical quantity transported an d a right subscript that specifies the direction of flow) flows in the direction opposite to the gradient in order to re -establish equilibrium. Thus, transport equation are often referred to as “relaxation equations.” Transport equations. The transport equations describe the movement of physical quantities on a macroscopic scale. For diffusion (transport of mass), jz m (z, t) = !D "#(z, t) "z . Here D is the diffusion coefficient and the superscript “m” on the flux denotes ma jz m (z, t)ss. Since the units of are mass/area -time, and the units of ρ are mass/volume, the units of D must be area/time, usually cm 2/sec. revised 4/24 /08 3:49 PM 5.62 Spring 2008 Lecture #31 Page 4 For heat conduction (transport of energy): jz e (z, t) = !" #T(z, t) z where κ is the therma l conductivity, T is the temperature, and the superscript “e” on the flux denotes energy. The units of κ are energy/ length -sec -degree. For viscosity (transport of momentum). The situation is more complicated. The prototype physical situation is gas (or a fluid) located between two plates. The bottom plate is stationary and the top plate is pulled with a certain force. The force in the x direction F x per unit area of the plate, will cause the x -component of velocity to v ary with z, v x(z). The transport equation is: Fx = !" dv x (z, t) dz . Note that force/unit area has the same units as momentum per unit time per unit area, so Fx is a momentum flux. The units of the viscosity coefficient, η, are mass/ length -se c. Kinetic theory has two tasks . The first is to explain why the transport equations have the particular form of a flux proportional to a spatial gradient of a local thermodynamic quantity. The second task is to obtain a molecular expression for each tran sport coefficient (e.g. D, κ, η). In the simple form of kinetic theory this is accomplished by considering the microscopic process of collisions in the gas. The analysis is based on some important approximations, but it leads to useful results: the form of the transport equations is explained and approximate values for the transport coefficients are obtained. Once the microscopic analysis justifies the form of the transport equations, it becomes possible to obtain a set of conservation equations that are valid on a macroscopic scale. If a flux of a certain quantity Y has the form revised 4/24 /08 3:49 PM 5.62 Spring 2008 Lecture #31 Page 5 ! jy = !L ! "Y then, choosing a fixed volume element, conservation of Y (flux of Y entering the volume element must be equal to the flux of Y leaving the volume element) leads to the result: !Y !t = " ! $ ! jy = ! $ L ! Y = L#2 Y . Typical values of transport coefficients for dilute gases at 272 °K Gas Self diffusion coefficient D cm 2/sec Thermal conductivity κ 10 5 cal cm –1 sec –1 K–1 Shear viscosity η 10 5 g cm –1 sec Argon 0.156 3.94 20.99 CO 2 0.181 3.49 13.66 CH 4 0.206 7.21 10.30 W. Kauzmann, “Kinetic Theory of Gases,” W. A. Benjamin, New York, 1966, page 209.
8734	https://web.stanford.edu/~lindrew/8.04.pdf	8.04: Quantum Physics I (OCW) Lecturer: Professor Barton Zwiebach Notes by: Andrew Lin IAP 2022 Fact 1 These notes were taken for my own preparation for the 8.04 Advanced Standing Exam, and they are transcribed from video lectures found on the Spring 2016 OCW page. As a result, some content may be less complete than in an ordinary course (because recitations aren’t available on OCW). The main reason for these notes’ existence is to provide more continuity into my 8.05 and 8.06 notes from previous semesters. Lecture 1: Overview of Quantum Mechanics Quantum mechanics is a subject that takes time to learn, and we’ll start by giving some perspective (explaining where we’re going, introducing the basic features and ideas, and describing what’s surprising about quantum mechanics). The subject is almost a hundred years old now, and officially we can think of the subject as starting in 1925, which was the year where Schrodinger and Heisenberg wrote down the equations of quantum mechanics. But the ideas really began earlier – the fundamental puzzles that led to quantum mechanics came from the end of the 19th century (Planck) and the beginning of the 20th century (Einstein and others). Fact 2 Quantum mechanics should be thought of as a framework to do physics which has replaced classical mechanics as the “correct” description at the fundamental level. While classical mechanics provides a good approximation at everyday scales, we know that it breaks down at some point, and thus the way things really work are conceptually very different from what our intuition tells us. What this course (and the ones after this) will do is take the principles and the framework of quantum mechanics and apply them to different physical phenomena, like quantum electrodynamics, quantum chromodynamics, quantum optics, and quantum gravity. In today’s lecture and a bit of the next, we’ll cover five introductory ideas: linearity of quantum mechanics, the necessity of complex numbers, the laws of determinism, interesting features of superposition, and the concept of entanglement. The first idea, linearity, is important whenever we’re considering a physical theory. Such theories have dynamical variables that we’d like to measure (because they are connected with values of physical observations) which are tied to certain equations of motion. 1 Example 3 Maxwell’s theory of electromagnetism is a linear theory, meaning that if we have two different solutions to Maxwell’s equations (such as two plane waves traveling in different directions), their sum (two plane waves propagating simultaneously) is also a valid solution. In other words, solutions don’t “affect” each other, and in practice this is useful because (for example) the air around us is constantly filled with electromagnetic waves. But linearity tells us that a single phone cable can transmit many phone calls at once, and it also tells us that if we’re doing an electromagnetism experiment, it’s okay to have other electromagnetic waves around us in the air. Let’s rigorize this discussion a bit more: in Maxwell’s theory, we have the variables (⃗ E, ⃗ B, ρ, ⃗ J) which must satisfy Maxwell’s equations. What linearity tells us is that (α⃗ E, α⃗ B, αρ, α ⃗ J) (for any real number α) will then also be a solution to the equations of motion, and if (⃗ E1, ⃗ B1, ρ1, ⃗ J1) and (⃗ E2, ⃗ B2, ρ2, ⃗ J2) are both solutions, then so is their sum (⃗ E1 + ⃗ E2, ⃗ B1 + ⃗ B2, ρ1 + ρ2, ⃗ J1 + ⃗ J2). We haven’t written out Maxwell’s equations explicitly, but we’ll mention instead that the fundamental concept necessary for linearity is to write an equation schematically as Lu = 0, where u is some unknown and L is a linear operator. More generally, we can have multiple equations of this form, and u can be a vector of unknowns instead, but what’s really important is that L satisfies the two properties of a linear operator: L(au) = aLu, L(u1 + u2) = Lu1 + Lu2. More generally, these two properties together tell us that L(αu1 + βu2) = L(αu1) + L(βu2) = αLu1 + βLu2, so if u1, u2 are solutions, then L(αu1 + βu2) = α · 0 + β · 0 = 0, and thus any linear combination of solutions is also a solution when we have a linear theory. Example 4 The differential equation du dt + 1 τ u = 0 can be written in the form Lu = 0 by taking L to be the linear operator Lu = du dt + 1 τ u (which we will also write in the form L = d dt + 1 τ ). We can indeed check that this operator L above is linear by verifying the two properties of linearity: indeed, L(au) = d dt (au) + 1 τ (au) = a d dt u + a 1 τ u = aLu, and the other property follows in a similar manner. Fact 5 Linear theories are generally much simpler than nonlinear theories (for example, Maxwell’s theory is linear, while Einstein’s theory of general relativity is very nonlinear and very complicated). However, classical mechanics is actually very nonlinear (for example, Newton could solve the two-body problem but not the three-body one), and we can see this through the following example: 2 Example 6 Consider the classical motion of a one-dimensional particle under the influence of a potential V (x); the equation of motion for the dynamical variable is then Newton’s second law, md2x(t) dt2 = −V ′(x(t)). (Note that V ′ will always denote the derivative of V with respect to its argument.) The issue is that the left-hand side has a linear operation (taking two derivatives), but the right-hand side may not be linear because V (x) can take an arbitrary form (for example, if V (x) ∝x3, then V ′(x) ∝x2 is not linear). And this finally leads us to our discussion point for this class: quantum mechanics is a linear theory, in which the dynamical variable is a wavefunction Ψ (which can depend on time t and other variables), describing the dynamics of a quantum system over time. The equation that governs Ψ is then the linear Schrodinger equation iℏ∂Ψ ∂t = ˆ HΨ, where ˆ H is the Hamiltonian (which is a linear operator). In other words, we can write LΨ = iℏ∂Ψ ∂t −ˆ HΨ = 0 (where because derivatives are linear operations and ˆ H is linear, L is also linear). We then have the advantage (over classical mechanics) that whenever we have solutions in quantum mechanics, we can scale and add them together. And as we go through this course, we’ll see that the quantum mechanics solutions are indeed simpler and more elegant than the classical mechanics ones. Fact 7 The ℏon the left-hand side of the Schrodinger equation is the reduced Planck’s constant (pronounced “h-bar”), originating from when Planck tried to fit the blackbody spectrum and needed a constant. And the Hamiltonian ˆ H is for us to invent or discover (it depends on the physical characteristics of our system). The physical interpretation of the wavefunction Ψ was not obvious for those who first invented quantum mechanics; Max Born later found that it had to do with probability, but we’ll discuss this soon. For now, we’ll first turn to our second idea, which is the role of complex numbers in quantum mechanics. Schrodinger’s equation above includes the constant i = √−1 – such complex numbers were originally invented because they were necessary for solving some equations (like x2 + 1 = 0), and it turns out that “no more numbers need to be invented” after that to solve polynomial equations. Definition 8 A complex number is of the form z = a + ib, where a, b ∈R are real numbers and i = √−1. We define the real part of z to be Re(z) = a, the imaginary part of z to be Im(z) = b, the complex conjugate of z to be z∗= a −ib, and the norm of z to be \|z\| = √ a2 + b2. The set of complex numbers is denoted C. We often represent complex numbers on the complex plane, identifying the point a+ib with (a, b) in the xy-plane (so that the horizontal axis represents real part, and the vertical axis represents imaginary part). The norm of z is then given by the ordinary Euclidean distance formula, and we have \|z\|2 = a2 + b2 = zz∗, 3 which will be pretty important for us as we progress through this course. Notice that by trigonometry, if we want to find the complex number of norm 1 at an angle θ (counterclockwise from the x-axis), we have z = cos θ + i sin θ (because the real and imaginary parts are just the horizontal and vertical projections). What’s important and nontrivial is that we also have Euler’s formula cos θ + i sin θ = eiθ, which we’ll see come up as we work more with the equations of motion. We may have seen complex numbers being used in classical mechanics and in electromagnetism, but they have always been used in auxiliary manners (because positions, velocities, electric fields, and other dynamical variables are always real-valued). But in quantum mechanics, Ψ must be complex-valued because the Schrodinger equation has an i in it (after all, if Ψ were real-valued, the left-hand side would be imaginary and the right-hand side would be real, which could only occur if both sides were 0). And yet, we can never measure complex numbers – any measurement we perform in a lab gives us a real number, and now we come to the point where we explain that \|Ψ2\| is the physically relevant quantity which is proportional to the probability of our systems being in a given configuration. Remark 9. Many physicists disliked or didn’t believe this Born interpretation – Schrodinger’s cat was actually a thought experiment designed to explain that this interpretation was ridiculous. And there were even papers at the time by famous physicists, like the EPR paper, which were later proven wrong. But when very good physicists are wrong, there’s still a lot we can learn from them and a lot of interesting physics that results from the discussion. With that, we’ll turn to a discussion of determinism which must come up when we talk about probabilities. Einstein was the one who (reluctantly) came up with the idea that light is quantized as photons (so not only can we observe light as an electromagnetic wave, we can also see it as a particle through the photoelectric effect). But the difference between a Newtonian and quantum particle is that the former has zero mass and carries a precise energy, position, and momentum, while the latter is some indivisible amount of propagating energy or momentum. So a photon is a quantum mechanical particle in the sense that it is a packet that cannot be decomposed further into smaller packets. What Einstein found was that for any photon, we have the relation E = hν, where ν is the frequency of the light (of which the photon is part of) and h (= 2πℏ) is Planck’s constant. Fact 10 Here, we should remember that c = νλ for any light wave, and because h is very small, individual photons have very low energy. But our eyes are very good at detecting light – if we’re in a completely dark room, we are perhaps able to detect as few as five photons. Example 11 Consider what happens when a beam of light hits a polarizer, which is a material with a preferential direction (which we’ll set to be the x-axis). Recall that this means that light linearly polarized along the x-axis will pass through, but light linearly polarized along the y-axis will all be absorbed. In fact, that linearly polarized light in the x-direction that passes through the polarizer looks identical when it enters and when it exits (same frequency and wavelength, and thus same energy). But now suppose we send light into the 4 polarizer which is polarized at some angle: ⃗ Eα = E0 cos(α)ˆ x + E0 sin(α)ˆ y. From our study of electromagnetism, we know that the component along the y-direction is absorbed, so after going through the polarizer, we just have ⃗ Eα = E0 cos αˆ x. But we know that the energy in an electric field is proportional to \|⃗ E\|2, so this tells us that (because the initial electric field has magnitude E0, and the final electric field has magnitude E0 cos α) the fraction of energy that passes through the polarizer is cos2 α. (This checks out for α = 0, π 2 .) But from the point of view of the photon, what this means is that for individual (identical) photons that hit a polarizer, we must see a fraction cos2 α of the photons pass through and a fraction 1 −cos2 α of them get absorbed (since we can’t have “half a photon” pass through, or else that would change the frequency of the light, which we know doesn’t happen). Classical physics doesn’t like the fact that sending identical particles into a physical situation may yield varying outcomes, but that is indeed what we must have in quantum physics: identically prepared experiments may give different results, and thus we lose predictability as long as photons exist. There are a few ways that we could try to get around this debacle: for example, we could hypothesize that photons will or will not pass through the polarizer based on the interatomic structure of the interaction, but this was found to not be true after repeated experiments. Another suggestion was that there are in fact hidden variables that are unknown to us (which distinguish the photons but make them still look identical), and those hidden variables affect whether or not the photon passes through the polarizer. (In other words, quantum theory is not complete.) It sounds philosophical, to the point where it seems like we wouldn’t be able to refute it without measuring these hidden variables, but in fact the Bell inequality proved that even with them, quantum mechanics could not be made deterministic. Fact 12 The Bell experiment will be discussed more in 8.05, but the main idea is that we can design an experiment in which the hidden variables (under classical mechanics) would imply a certain inequality. Then this inequality was shown experimentally not to hold once the technology was good enough to run the experiment! So in summary, determinism is lost: photons must either go through the polarizer or not, and we can only predict the probability with which they go through. To write down the wavefunction of a photon, we can think of the photon as being in one of several states (we will sometimes also use “vectors” or “wavefunctions” for this same term). The motivation is that we can scale or add vectors together, and that’s also what we can do with solutions to our quantum equations. Dirac invented the notation \|x⟩to represent a vector (where x is just some label that describes the particular state): for example, for our polarizer, we have the two possible states \|photon; x⟩, \|photon; y⟩. Then linearity tells us that we can always create the linear combination of states \|photon; α⟩= cos α \|photon; x⟩+ sin α \|photon; y⟩, which is the quantum mechanical description of a photon polarized in the α-direction. Notice that this is similar to the classical ⃗ Eα = E0 cos(α)ˆ x + E0 sin(α)ˆ y in that we have the same cos α and sin α terms, but we no longer have the E0 coefficient because we’re now representing wavefunctions at the individual photon level. And once the photon makes it through the polarizer, its wavefunction is now just \|photon; x⟩. That leads us to our next discussion of superposition. In classical physics, superposition tells us (for example) that we can add electric fields or force vectors, but in quantum physics we get something much stranger. 5 Example 13 Consider a Mach-Zehnder interferometer, which is a device with two beam splitters, two mirrors, and a detector, as shown below. mirror mirror beam splitter beam splitter detector 0 detector 1 incoming light This device works in the following way: when incoming light hits each beam splitter, half of the light is reflected and half is transmitted. Those resulting light beams are reflected off of two mirrors to recombine at a second beam splitter, and after interference each detector will pick up some signal (in fact, we can adjust the system so that any fraction of the light, between 0 and 1 inclusive, goes to detector 0). Such a device was invented in the 1890s, and it is interesting to think about (given our discussion above) because we know that there must be something probabilistic going on at each beam splitter. In particular, if we have a superposition of different states, “some photons can go both up and down.” And in fact, when we say that there is an interference pattern being formed, it can’t be that different photons are interfering with each other (because energy can’t be created or destroyed); instead, each photon interferes with itself, and we can verify that this is what causes constructive or destructive interference by sending one electron into the device at a time. More mathematically, we can say that each photon is in a superposition between the upper and lower beams in our diagram (where “upper beam” and “lower beam” are now our fundamental states). Remembering that we’re often associating the words “state,” “vector,” and “wavefunction” with each other, we’ll now explain how to associate physical properties with these superpositions. Suppose that we have two states \|A⟩and \|B⟩, and suppose that we’re trying to measure some particular property (such as energy, spin, or position), so that measuring on \|A⟩always gives some value a, and measuring on \|B⟩always gives some value b. Then we can look at a quantum mechanical state of the form \|Ψ⟩= α \|A⟩+ β \|B⟩ for complex numbers α, β ∈C, which is a linear combination (superposition) of the states \|A⟩and \|B⟩. If we measure that same property in \|Ψ⟩, what quantum mechanics tells us is that we will always get either the value a or the value b, but never something intermediate! This is very different from the situation we might have had in classical mechanics – instead of taking a weighted average (as we might be tempted to do), what \|Ψ⟩really represents is that the probability of measuring a is proportional to \|α\|2, and the probability of measuring b is proportional to \|β\|2, and those are the only two possibilities. Example 14 Recalling the example state \|photon; α⟩= cos α \|photon; x⟩+ sin α \|photon; y⟩we had above, we can think of the polarizer as measuring the direction of polarization: it will either measure that the photon is polarized in the x-direction (with probability cos2 α) or that it is polarized in the y-direction (with probability sin2 α). And after we measure the state α \|A⟩+ β \|B⟩once, that state will either collapse to the state \|A⟩or the state \|B⟩, and future measurements will always return the same value – this is the measurement postulate of quantum 6 mechanics. (So if we wanted to know the original value of \|Ψ⟩and didn’t know α and β, we would need to prepare many identical copies of our setup and do our experiment many different times to assess the relevant probabilities). Lecture 2: Superposition, Entanglement, and Interaction-Free Measurements Last lecture, we started discussing superposition, and we explained that the superposition of two states does not have properties that are intermediate between the two initial states (instead, we obtain the properties of one of the two states with certain probabilities). This is the point at which we make a certain physical assumption: superimposing a state with itself does not change the physics from just having the original state. In other words, \|A⟩∼2 \|A⟩∼−\|A⟩∼i \|A⟩∼· · · , where we can put any nonzero coefficient in front of the fundamental state \|A⟩and still have essentially the same object. So it makes sense to pick a convenient representative out of all of these, and typically that is the normalized state (which satisfies properties having to do with the norm of the state). We’ll talk about this more later, but the reason we bring this up is to connect it more to our discussion of states of light from last time. If we think about our photon from last lecture with a superposition of x- and y-direction polarization, we can write it most generally as α \|photon; x⟩+ β \|photon; y⟩, where α, β are complex numbers and thus we have four real parameters to dictate the polarization direction. But we already know (and will review in a second) that there should only be two parameters to dictate the polarization of a particle, and our discussion of normalization helps us out here: if the overall constant doesn’t matter, we can multiply the whole state by 1 α to get a state of the (generic) form \|photon; x⟩+ β α \|photon; y⟩, and thus all of the physics of a polarization state is contained in the ratio γ = β α (which is a single complex parameter, corresponding to two real parameters). And indeed, the most general polarization of an electromagnetic wave is elliptical and dictated by two real parameters, the angle of its major axis and its eccentricity (since the overall size of the ellipse doesn’t matter). This act of normalizing will come up again and again as we get farther into the course. We’ll emphasize one more important concept of superposition using spins. Definition 15 Spin is a fundamental property of elementary particles, dictating their internal (intrisic) angular momentum. While no model of an elementary particle has ever been constructed where there is actually something inside spinning, there is indeed some angular momentum (even for a point particle), and spin is a very quantum mechanical property. Since angular momentum is a vector, we must decide the direction in which spin is pointing. Example 16 Suppose we measure spins along the z-direction. Then if our particle is of spin 1/2 (such as protons, neutrons, and electrons), we will either measure the particle in the spin-up or the spin-down state (corresponding to the respective direction of angular momentum). Curiously, what we’re saying is that whenever we measure the spin of a spin 1/2 particle, it is always either spin up (+1/2) or spin down (-1/2) with full magnitude. We’ll denote the spin up and spin down states as \|↑; z⟩and \|↓; z⟩, 7 respectively, and like in our earlier discussions, if those two are valid quantum mechanical states, so are states arising from superpositions such as \|Ψ⟩= \|↑; z⟩+ \|↓; z⟩. (We won’t talk about normalizing these states at the current moment; it’s not too important.) So if we have a bunch of electrons which are in the state \|Ψ⟩above, and we try to measure each of their spins in the z-direction, then we expect about half of the electrons to be spin up and about half to be spin down, due to the relative magnitude of the \|↑; z⟩and \|↓; z⟩terms. But Einstein might ask the question of how we really know that the particles we measured were in the states \|Ψ⟩ – after all, our current setup does not help us distinguish between a system of all electrons in \|Ψ⟩, versus an ensemble of electrons where about half of them were in the state \|↑; z⟩to start with and the other half were in the state \|↓; z⟩. This is what Einstein meant by realism: if we measure a particle to be spin-up, he would claim that this means the particle was spin-up to begin with. And we won’t resolve this paradox until we learn more about spins (as we will later in the course). Instead, we’ll just point out that if we had an ensemble where we had half spin-up and half spin-down particles, we could take that ensemble and measure its spin along the x-direction instead. It turns out that the particles in the \|Ψ⟩state will all come out + along the x-direction, while the particles in the mixed ensemble will come out half-and-half + and -. So there is indeed an experiment that can tell whether these quantum states exist, and in fact they do. This now prepares us to discuss the last of the fundamental introductory ideas for 8.04, which is entanglement. It turns out that two particles can become entangled without having strong interactions between them – they can even be non-interacting. Example 17 Suppose particle 1 can be in one of the states \|u1⟩and \|u2⟩, while particle 2 can be in the states \|v1⟩and \|v2⟩. We wish to describe the states of the combined system of two particles. To describe the overall system’s state, it seems reasonable to specify each particle’s state – for example, if particle 1 is in state \|u1⟩and particle 2 is in state \|v1⟩, then we represent the whole system by the tensor product \|u1⟩⊗\|v1⟩. In such a product, we always list the first particle’s state on the left and the second particle’s state on the right, so more generally we may have something that looks like (α1 \|u1⟩+ α2 \|u2⟩) ⊗(β1 \|v1⟩+ β2 \|v2⟩). Tensor multiplication then looks similar to regular multiplication, but we never move the states across the product: expanding and moving all of the constants out gives us = α1β1 \|u1⟩⊗\|v1⟩+ α1β2 \|u1⟩⊗\|v2⟩+ α2β1 \|u2⟩⊗\|v1⟩+ α2β2 \|u2⟩⊗\|v2⟩. But notice that we can also be in a state that looks like \|u1⟩⊗\|v1⟩+ \|u2⟩⊗\|v2⟩, which is a superposition of the states \|u1⟩⊗\|v1⟩and \|u2⟩⊗\|v2⟩. But this time, we can’t “factor” this state as a state of particle 1, tensored with a state of particle 2 – after all, matching coefficients with the general multiplication we did above would force α1β1 = α2β2 = 1 but α1β2 = α2β1 = 0, which is a contradiction (since α1β1α2β2 would then have to be both 1 and 0). So we have an unfactorizable state, in which we cannot just describe the configuration as independently analyzing the first and second particle. In other words, we have an entangled state – knowing 8 something about the first particle tells us about the second and vice versa – even though no interactions have occurred between them! Example 18 If we consider two spin 1/2 particles, then we can consider the entangled state \|↑; z⟩1 ⊗\|↓; z⟩2 + \|↓; z⟩1 ⊗\|↑; z⟩2 . People often speak of “Alice and Bob” each having one of these two particles (arranged in their entangled state), with the idea being that Alice and Bob are very far away (perhaps one on Earth and the other on the moon). Then Alice and Bob share an entangled pair, and very interesting experiments can be done (in real labs, this can be done with two photons that are hundreds of kilometers apart), coming from the fact that the two particles’ properties are correlated in strange ways. For example, if Bob measures his particle and finds that it is spin-down, then the whole state must collapse into the \|↑; z⟩1 ⊗\|↓; z⟩2 state, meaning that if Alice also simultaneously measures her particle, she must find that it is spin-up (even before light has had time to travel from one of the entangled particles to the other). It turns out that this does not contradict special relativity – we cannot actually send any information through this process – but the collapse is still instantaneous. Einstein would again object to this concept of entanglement, suggesting instead that some entangled pairs are really in the \|↑; z⟩1 ⊗\|↓; z⟩2 state while others are in the \|↓; z⟩1 ⊗\|↑; z⟩2 state. But that’s where Bell’s inequality comes in – if Alice and Bob can measure in three different directions, then the correlations that result are impossible to explain with classical physics. So what’s going on is really subtle and violates classical mechanics in a peculiar way. Example 19 We’ll now return to the Mach-Zehnder interferometer from Example 13 and perform some further analysis. Recalling that an arbitrary superposition of the top and bottom paths gives us a valid state for an incoming photon, we’ll represent that superposition with a column vector " α β # which give the probability amplitudes of being in the top and bottom paths. (These α, β may depend on time and position, and remember that their norm squared give the actual probabilities.) In particular, this is where we must do a normalization: since probabilities need to add to 1, we must have \|α\|2 + \|β\|2 = 1 . For example, " 1 0 # corresponds to being in the upper beam, and " 0 1 # corresponds to being in the lower beam. Then everything indeed works out: we have " α β # = α " 1 0 # + β " 0 1 # , so that our generic state is indeed a superposition of the “upper beam” and “lower beam” states. If we now think about our beam splitter, we know that an incoming photon from the top would produce some reflection amplitude s (to the top path) and some transmission amplitude t (to the bottom path), meaning that the beam splitter takes " 1 0 # to " s t # . (The values of s, t depend on the particular design of the beam splitter.) Clearly, we must have \|s\|2 + \|t\|2 = 1. Similarly, a photon coming from the bottom results in a change from " 0 1 # to " u v # , where \|u\|2 + \|v\|2 = 1. In other 9 words, we need four numbers to characterize this beam splitter – indeed, for an arbitrary state " α β # passing through the beam splitter, by linearity we have that " α β # = α " 1 0 # + β " 0 1 # 7→α " s t # + β " u v # = " αs + βu αt + βt # = " s u t v # " α β # . In other words, we can think of the beam splitter’s action as multiplying by a certain 2 × 2 matrix! This perspective will be very useful for us going forward for a variety of reasons. For our purposes, we’ll assume we have a balanced beam splitter where half of the light goes through and the other half is reflected, so that \|s\|2 = \|t\|2 = \|u\|2 = \|v\|2 = 1 2. But that doesn’t allow us to determine exactly what s, t, u, v are (because of the phase), and let’s first make the guess that it is " 1 √ 2 1 √ 2 1 √ 2 1 √ 2 # . As long as this matrix satisfies probability conservation, there should theoretically exist a beam splitter that can perform that operation. But this doesn’t work – after all, notice that " 1 √ 2 1 √ 2 1 √ 2 1 √ 2 # " 1 √ 2 1 √ 2 # = " 1 1 # , and the norm of our column vector has changed. Instead, it turns out one answer is " 1 √ 2 1 √ 2 1 √ 2 −1 √ 2 # ; indeed, " 1 √ 2 1 √ 2 1 √ 2 −1 √ 2 # " α β # = 1 √ 2 " α + β α −β # , and this final vector indeed has the same norm as the initial one, because 1 √ 2(α + β) 2 + 1 √ 2(α −β) 2 = 1 2(α + β)(α∗+ β∗) + 1 2(α −β)(α∗−β∗), at which point the cross-terms vanish and we’re left with 1 2(2α2 + β2) = α2 + β2, which is 1 by assumption. So this beam splitter matrix will do the job for us – it’s not the only solution, and we’ll consider two beam splitter matrices BS1 = 1 √ 2 " −1 1 1 1 # , BS2 = 1 √ 2 " 1 1 1 −1 # . corresponding to the two beam splitters in the diagram below. We may also suppose that along the bottom path (hitting the bottom mirror), we place a piece of glass which shifts the probability amplitude by a phase: mirror mirror beam spltr. 1 beam spltr. 2 detector 0 detector 1 β α δ In other words, a probability amplitude of β will turn into a probability amplitude of βeiδ, where δ ∈R. Since \|β\| = \|βeiδ\|, this preserves the norm of the probability amplitude. (We won’t use this phase shifter in today’s lecture, but it’s still a useful concept that will come up in the future.) 10 Problem 20 Consider the setup in the diagram above without the phase shifter. What do we find at the detectors if we have an incoming state " α β # ? We’ll assume that the mirrors do not do anything to the column vector – in fact they each multiply a component by −1, so overall that’s just a constant shift which does nothing – meaning that the detector state comes from applying first the beam splitter matrix BS1, then the beam splitter matrix BS2. In other words, we have output = (BS2)(BS1) " α β # = 1 2 " 1 1 1 −1 # " −1 1 1 1 # " α β # = 1 2 " 0 2 −2 0 # " α β # = " β −α # . Thus, if we send in a photon from the bottom (meaning that we start with the column vector " 0 1 # ), the output will be the column vector " 1 0 # . In particular, this means that even after the photon split at beam splitter 1, there was an interesting interference at beam splitter 2 which caused the amplitudes to cancel out along detector 1 and combine along detector 0, such that every photon will be detected at the top detector! (And indeed, this interference pattern was detected in experiments.) Problem 21 Next, consider the setup in the diagram above, but along the bottom path we place a wall of concrete (at the location of the phase shifter) so that light cannot pass through the bottom path. What do we find at the detectors if we have an incoming state " α β # ? We can no longer use our matrix multiplication formula from above; instead, we’ll look at calculation step by step. After passing through the first beam splitter, we are in the state (BS1) " 0 1 # = 1 √ 2 " −1 1 1 1 # " 0 1 # = " 1 √ 2 1 √ 2 # . However, along the bottom path, we lose the probability amplitude of 1 √ 2 due to the concrete wall – in other words, the input into the second beam splitter is " 1 √ 2 0 # (nothing reaches the splitter from below). Thus, the final detected column vector is (BS2) " 1 √ 2 0 # = 1 √ 2 " 1 1 1 −1 # " 1 √ 2 0 # = " 1 2 1 2 # . Interestingly, this means that by blocking some photons, we’ve created a signal at detector 1, which may seem somewhat counterintuitive. (And in fact we’ll see by the end of this lecture that this is even more unintuitive than we might initially think!) Remembering that the squared components correspond to the probabilities, we can summarize the results of Problem 20 and Problem 21 as follows: 11 Outcome (open paths) Probability Photon is detected at detector 0 12 = 1 Photon is detected at detector 1 02 = 0 Outcome (one blocked path) Probability Photon ends at concrete block 1 √ 2 2 = 1 2 Photon is detected at detector 0 1 2 2 = 1 4 Photon is detected at detector 1 1 2 2 = 1 4 We can take these values and put them into the following thought experiment: Problem 22 (Elitzur-Vaidman bombs) Consider a certain kind of bomb which detonates if a photon passes through it, and suppose that we have several of these bombs but that about half of them have gone defective (so that they will not detonate even if a photon passes through). How can we decide whether a bomb is working without breaking the components apart or having it explode? In classical physics, this is clearly impossible – if we do the measurement to see if the bomb explodes, then either the bomb is defective or we have no more bomb left. But with the Mach-Zehnder interfereometer system that we’ve just described, we can indeed find a working bomb without having it explode! Solution. Place the bomb where our phase shifter is placed in our diagram above, such that a photon passing through the bottom path would set off the bomb. Then if the bomb is defective, it’s as if both paths are open (see left table above), and thus photons must be detected at detector 0 (with probability 1). But if the bomb is working, then we’ve essentially put a block of concrete along the bottom path (with the extra effect that the bomb explodes in that situation). Thus (see right table above), half of the time the bomb will explode, but there is a 1 4 chance that the bomb does not explode and the photon is detected at detector 0, and likewise also a 1 4 chance that the bomb does not explode and the photon is detected at detector 1. Thus, whenever we see a photon at detector 1, we must have a working bomb which has still not detonated. Fact 23 It turns out that adjusting the experiment (placing the bomb in a resonant cavity, for example) can give us arbitrarily high probabilities that the bomb will not go off before we find out that it works! And thus quantum mechanics indeed allows us to perform very surprising measurements. Lecture 3: The Photoelectric Effect, Compton Scattering, and de Broglie Wavelength Last lecture, we discussed the Mach-Zehnder interferometer and what happens to it when we pass a photon through the device – essentially, a relatively simple calculation led to a very surprising result. We’ll continue studying photons today, but we’ll take a step back to 100 years ago, looking at how physicists came up with quantum mechanics through photons’ behavior. 12 Fact 24 The photoelectric effect can be traced back to Hertz’s experiment in 1887, in which high-energy beams of light were shined on polished metal plates and electrons (called photoelectrons because they are created in this way) were released, creating a photoelectric current. The primary interesting feature that was observed was that there was always a certain threshold critical frequency ν0, above which a current is measured but below which nothing happens. And this threshold depended on the metal being irradiated, as well as the roughness and crystalline nature of the surface – it was later discovered that there are free electrons roaming around this crystalline structure, and the critical frequency ν0 thus depends on how much energy is required to free an electron from the metal. Additionally, it was found that the intensity of the light governed the magnitude of the photoelectric current, but that this intensity does not affect the energy of the photoelectrons themselves! So the incoming energy beam affects the number of excited photoelectrons but not their energy – in fact, the energy of the photoelectrons has a linear relation with the frequency ν of the incoming light. These properties are difficult to explain if we treat light solely as a wave, and Einstein’s answer in 1905 was (as previously mentioned) that light must come in quantized bundles of energy (which were later named photons by Gilbert Lewis in the 1920s), with energy E = hν. The picture that we should have in mind is the following: E 0 W Essentially, this graphs the potential energy function for our electrons – while they are in the metal, they are stuck to it unless we can provide some potential energy to get them out of the well. Once we do this, the electron will be able to fly freely and not be affected by the metal. Definition 25 For any surface, the work function (denoted W) is the energy needed to release an electron to the vacuum around the surface. If Einstein’s proposition about photons is correct, then that implies that the energy of a photoelectron satisfies 1 2mv 2 ≈Ee−= Eγ −W = hν −W (the energy of the photon is transferred to kinetic energy of the photoelectron, except the work function W). It wasn’t until 1915 that this was verified experimentally by Millikan, and in fact that experiment gave an estimate of h within 1 percent of the accepted value today (which was the best up until that point). But even then, the photoelectric effect experiment wasn’t enough on its own to really convince people that photons existed, because Maxwell’s theory was very successful and accepting photons meant accepting loss of determinism and all of the other features of quantum physics we’ve been discussing. Let’s see an example computation to see what kind of setting we’re dealing with: 13 Problem 26 Consider ultraviolet light of λ = 290nm shined on a metal with work function W = 4.05 eV. What are the speed and energy of the emitted photoelectrons? Solution. The idea is that we should be able to do this problem without needing to remind ourselves what “eV” means, or searching up the value of ℏ, and in general perform back-of-the-envelope estimates and calculations. First of all, the energy of a photon is Eγ = hν = hc λ = 2πℏc λ (using the definition of the reduced Planck’s constant), and it’s useful to remember that ℏc ≈200 MeV · fm (the actual number is 197.33), where 1 fm = 10−15 m (fm stands for “fermi” or “femtometer”). Thus, we can simplify to = 2π · 197 MeV · 10−15m 290 · 10−9 m = 2π · 197 290 eV = 4.28 eV (where the last step is the only one where we really need to do any calculation). Thus, the energy of the photon is Ee = Eγ −W = 4.28 eV −4.05 eV = 0.23 eV, and this is essentially a nonrelativistic electron because the rest mass of an electron is about 511 eV: 0.23 eV = 1 2mv 2 = 1 2mec2 v c 2 = 1 2 · (511000 eV) v c 2 , yielding v ≈284 km/s . We’ll talk a bit more about Planck’s constant directly: because it shows up in the equation Eγ = hν, it must have units [h] = [E] [ν] = [M][L2]/[T 2] 1/[T] = [M][L2] [T] (where we’re working with units characterized by mass M, length L, and time T – that is all we need). We’ll rearrange this expression slightly to match with another physical quantity: = [L] · [M][L] [T] = [r][p] = [⃗ L]; in other words, Planck’s constant has the same units as angular momentum! This is something for us to keep in mind, and that’s why saying that we have a “particle of spin 1 2” actually means that “the particle has intrinsic angular momentum ℏ 2.” (Unfortunately, as we go forward, we’ll need to be careful about the 2π factor between h and ℏ– we’ll use whichever one looks nicer in the present situation.) And the idea is that once we have a new constant of nature, like h or c or G, we can get inspired and create new quantities with that constant. Since [h] = [r][p], we can invent an associated length for each particle depending on its momentum p (which we’ll do shortly), or even more simply do the following: Definition 27 The Compton wavelength of a particle of mass m is λc = h mc . 14 Example 28 The electron’s Compton wavelength is λc(e) = h mec = 2πℏc mec2 = 2π · 197.33 MeV · fm 0.511MeV ≈2426 fm = 2.426 pm. For comparison, this is smaller than the Bohr radius of an electron (about 50 pm), but much larger than the size of a nucleus. To understand the significance of this length that we’ve just defined, we’ll see its use in both an experiment and a thought experiment: Example 29 Suppose we have a particle of mass m. Then its associated rest mass (energy) is mc2, so a photon with the same energy as the particle has frequency ν = mc2 h and thus wavelength c ν = h mc = λc. This actually has some experimental implications in high-energy particle physics: for example, if a photon hits an electron with a Compton wavelength equal to the wavelength of the light, then the comparable energy scales can create new particles. Thus, it is difficult to isolate a particle to length scales shorter than its Compton wavelength without causing damage to it! Around the time that Einstein discovered general relativity, he was also thinking about photons again – throughout his life, he was very suspicious about the quantum theory and spent much of his time thinking about it. In 1916, he proposed that not only do these photons (which weren’t yet called photons at the time) have energy, but they also serve as quanta of momentum. Specifically, there is a relativistic relation E2 −p2c2 = m2c4 between the energy and the momentum of a particle (this comes from the two formulas E = mc2 q 1−v2 c2 and ⃗ p = m⃗ v q 1−v2 c2 , which reduce to E = 1 2mv 2 and ⃗ p = m⃗ v in the low-velocity limit). Thus, if we know the momentum and the energy of a particle, we also know its mass. Since photons have zero mass, we must have mγ = 0 = ⇒Eγ = pγc, meaning that the photon momentum (treating light as a particle) is pγ = Eγ c = hν c = h λγ . This looks similar to the Compton wavelength relation, but it was still not strong enough evidence until Compton scattering was studied: Example 30 Consider x-rays shining on atoms, where the incoming photons are comparably energetic (100 eV to 100 keV, while binding energies of atoms are on the order of 10 eV). If these photons are high enough in energy, then the resulting excited electrons are almost as if they started off being free. In such a situation, we have a violation of the classical Thompson scattering, which was what caused 15 physicists to finally accept the wave-like nature of photons – those scattering phenomena could be calculated just like scattering phenomena of other particles. In the classical case, we think of light as a wave, in which the magnetic field doesn’t do very much (because we have a low-energy electron) and the electric field shakes the electron: the resulting cross-section is dσ dΩ= e2 mc2 2 1 2(1 + cos2 θ) as a function of the angle θ between the incident angle of the wave and the direction that the photon is scattered in. If we haven’t seen this kind of formula before, the units are area per solid angle, which is the same as just area; what this cross-section measures is, for a given solid-angle region dΩ, the surface area dσ from the incoming beam that scatters into that solid-angle. That area then corresponds to an energy (because a certain amount of energy from the electromagnetic wave goes into each area), and thus with all of this we get a formula for the intensity of radiation as a function of the solid angle. But another property of this classical case is that the frequency of the outgoing wave (after hitting the electron) is the same as the frequency of the incoming wave, because the electron is being driven at the same frequency as the light wave. It turns out that at high energies, the result actually looks very different! If we treat a photon as a particle, so that we have a particle-particle collision, then both particles have some energy and momentum and we should use conservation laws. Before the collision, we have a photon of energy E and momentum p and an electron at rest, and after the collision both particles are moving. It’s an exercise to show that the photon can’t be observed by those conservation laws, and because the electron now has kinetic energy, the photon must lose some energy, meaning that λf > λi (longer wavelength means lower energy), and in fact we have λf = λi + h mec (1 −cos θ) where h mec is the Compton wavelength λc(e) for the electron. In other words, if θ = 0, the photon keeps going and basically doesn’t kick the electron, resulting in no change. However, if the photon bounces totally backwards, then θ = π and the wavelength increases by 2λc. Fact 31 This Compton scattering experiment was in fact performed, and let’s describe briefly how it was set up. Molyb-denum X-rays with λ = 0.0709 nm (corresponding to an energy of Eγ = 17.49 keV) hit a piece of carbon foil. Then the relation between intensity and outgoing λ turns out to have two peaks: the smaller one occurs around λi = 0.0709 nm, while the larger peak occurs at λf = 0.0731 nm. Noticing that the shift in wavelength is 0.0022 nm, this is pretty close to the Compton wavelength calculated in Example 28, meaning that the larger peak is indeed corresponding to what we expect (a shift on the order of λc). But the smaller peak needs slightly more explanation, and this is where Louis de Broglie’s work of 1924 comes into consideration. Treating light as both a particle and a wave means that it will have characteristics of both (light exhibits interference, but it also comes in packets), and de Broglie inferred that this must be a more general property – he took the fundamental step of claiming that all matter particles behave as waves, not just light. This is particularly interesting, because in his description, the wave that’s being associated to light is not the electromagnetic wave – it’s the probability amplitude wave (which are the numbers that we were tracking with the interferometer last lecture)! And that wavefunction was not known about yet, so what was being unsaid in the following conjecture was the question of “of what?” (that is, what is the wave made of). 16 Conjecture 32 (de Broglie (1924)) The wave-particle duality is universal for all matter: for each particle of momentum p, we have an associated plane wave of wavelength λp = h p, known as the de Broglie wavelength. At the time, there was no experimental evidence for the claim, but experiments came a few years later – it was seen that electrons could be diffracted in ways so that they would collide in lattices like waves! (We can read about the Davisson-Germer experiment, for example.) Over time, the two-slit diffraction experiment was performed with larger and larger particles – in fact, in the last few years, molecules of weight around 10000 atomic mass units have exhibited interference patterns. And we should remember that these interference patterns come from a particle’s wavefunction interfering with itself! We’ll see next lecture how this leads us to the Schrodinger equation and the wave-like nature of matter. Lecture 4: de Broglie Waves, Propagation, and the Free Particle Last time, we discussed de Broglie’s conjecture of “matter waves,” in which a free particle with momentum p is associated to a plane wave of de Broglie wavelength λp = h p. It turns out that such a plane wave is actually the wavefunction of a free particle, and its wave equation is basically the Schrodinger equation! So this is really a pillar of quantum mechanics that we’re working with, and the challenges when de Broglie’s matter wave was first proposed came in what physical (e.g. directional) properties it encoded – it turns out that this wavefunction will tell us about spin. But we’ll start with the case where these directional properties don’t matter much – in the case where we have an electron at low velocity or in a small magnetic field, we’ll ignore the spin properties and just write the wavefunction Ψ(⃗ x, t) as being complex-valued. We’re then (historically) curious whether Ψ is measurable and what its meaning is, but first let’s consider what it actually means for a wave to have wavelength h p (inversely proportional to momentum). Example 33 For simplicity, consider non-relativistic (Galilean) physics. Suppose that from our perspective, we have a particle of momentum p, and there is a “boosted observer” moving at constant velocity v relative to us. We wish to consider the differences in observed wavefunctions between the two observers. Because p = h λ = h 2π 2π λ , we can rewrite the momentum as p = ℏk , where ℏis the usual reduced Planck’s constant and k is the wavenumber (the number of periods per unit distance). Call our frame S, and call the other frame S′; fix the coordinates so that S′ is moving with velocity v in the x-direction. Then after t seconds, the two frames are vt units apart, and our particle of mass m will have velocities v and v ′ in frames S and S′, respectively, and suppose that it has momenta p and p′ respectively as well. We can now relate all of our coordinates with Galilean transformations (which are accurate enough for low velocities): x′ = x −vt, t′ = t. Taking a time-derivative, we have dx′ dt′ = dx dt −v, so v ′ = v −v, which we expect from our usual concept of relative motion. Therefore, p′ = mv ′ = mv −mv = p −mv, 17 so the “moving observer” will see a different de Broglie wavelength of λ′ = h p′ = h p −mv ̸= h p = λ. But if this were a familiar propagating wave like a sound or water wave, the relation between λ and λ′ would see a Doppler shift for the frequency, but the wavelength would not change! Indeed, consider an ordinary wave under Galilean transformations, and consider its phase kx −ωt (where k is the wavenumber and ω is the angular frequency). Waves can then be expressed as sines, cosines, or exponentials of this phase, and what’s important is that the phase of a wave is a Galilean invariant (if two people look at the same point on a wave, then they will always agree on the value of that phase – for example, both observers will agree that the wave has a maximum or minimum at a particular point). In particular, we can write φ = k x −ω k t = 2π λ (x −V t) = 2πx λ −2πV λ t. where we use the fact that ω k = 2πf 2π/λ = λf = V is the speed of the wave (not the same as the observer’s relative velocity v). Since this quantity is Galilean invariant, observers at S and S′ should see the same phase for the same point at the same time, meaning that φ = 2π λ (x −vt) = 2π λ (x′ + vt −V t) = 2π λ x′ −(V −v)t′, which further simplifies to = 2π λ x′ −2π λ V 1 −v V t′, should be equal to the phase φ′ = 2π λ x′ −2πV λ t′ seen by the moving observer. In particular, the wavenumber and angular frequency satisfy k′ = 2π λ and ω′ = 2π λ V 1 −v V , so that ω′ = ω 1 −v V , k′ = k = ⇒λ′ = λ for an ordinary wave moving in a medium. So the takeaway is that our de Broglie matter wave does not behave like an ordinary sound wave – instead, two observers may observe different values for the wavefunction, meaning Ψ is not directly measurable. (In fact, we’ve already seen some hints of this – for example, we mentioned that we can multiply our wavefunction by constants, including phases, and that doesn’t change the physical meaning.) Additionally, wavefunctions are not Galilean invariant, but (as an exercise left for us) the values of the wavefunction between two observers are still related. Notice that while we’ve talked a lot about the wavelengths of the de Broglie waves, we haven’t discussed their frequencies yet. de Broglie did in fact answer that as well: since p = ℏk, we’ll also set E = ℏω, and thus the frequency of a matter wave is ω = E ℏ, where E is the particle’s energy. This is one of the postulates of quantum mechanics, and we’ll try to explain why it makes sense now. Definition 34 The phase velocity of a wave with phase kx −ωt is vphase = ω k . The phase velocity is the velocity with which the nodes and extrema of the wave will move. If we’re working 18 non-relativistically, our de Broglie waves satisfy vphase = ω k = E p = 1 2mv 2 mv = 1 2v, and something may seem to be weird because the de Broglie wave looks like it’s moving at half the speed of the particle. But that’s not unexpected – plane waves themselves don’t carry any signal, and instead we should be representing particles with a wave packet. So phase velocity is not a very meaningful physical quantity, and instead we should use the following: Definition 35 The group velocity of a wave with phase kx −ωt is vgroup = dω dk k evaluated at the wavenumber k that we’re propagating. This group velocity, for a de Broglie wave, is then vgroup = dω dk k = dE dp = d(p2/2m) dp = p m = v, as we expect from a propagating particle. Fact 36 If we look at the relativistic version of energy and momentum and do all of these calculations again, we’ll again find the the group velocity vgroup lines up with the velocity v of the particle. And in fact, there is some motivation from relativity here – in special relativity, E c , ⃗ p form a four-vector, and similarly ω c ,⃗ k also form a four-vector, so it makes sense to set the two four-vectors proportional to each other through this constant ℏ. (In our equations above, like p = ℏk, we really have vector identities in multiple dimensions and p, k represent the corresponding vector magnitudes.) The above argument, combined with Einstein’s claim of E = hν = ℏω for photons, provides even more justification for why these de Broglie waves indeed have angular frequency E ℏ. We’ll now return more to our discussion of group velocity. There are certain waves where given the wavenumber k, we can write down the corresponding ω = ω(k): for example, for light waves we have ω = kc, but for other waves the relation is more complicated (for example, ω ∝k2 in mechanics, because E ∝p2). Then in general, the group velocity represents the velocity of a wave packet constructed via superposition, taking the form Ψ(x, t) = Z Φ(k)ei(kx−ω(k)t)dk, where we integrate as a continuous sum, where ei(kx−ω(k)t) represents a wave at wavenumber k and angular frequency ω(k), and Φ(k) is the amplitude of the wave at wavenumber k. Specifically, being able to talk about a group velocity means that we consider the case where Φ(k) is peaked at some value k0, and we want to see how the superposition moves in time. We can answer that question quickly or more rigorously: Quick explanation of group velocity. We can use the principle of stationary phase, which is essentially just mathe-matical intuition. If we look at an expression like f (x) sin x, where f is some positive-valued function, we’ll have a function which is positive half the time and negative half the time, so it will contribute very little to the integral if f is slowly varying relative to sin x (because adjacent parts will cancel). 19 The principle of stationary phase then basically says that integrating something like f (x) sin x only gives us a contribution when the wave varies slowly (and thus the “phase is stationary”). So in the integral above, we can just integrate from k0 −δ to k0 +δ, since the only place where the peaked Φ(k) contributes is around k = k0, and then we require that Φ is stationary with respect to k. And that means our wavefunction is only significantly nonzero around k0 if 0 = dφ(k) dk k0 = x −dω(k) dk k0 t = 0, meaning that x = dω dk k0 t and the group velocity is indeed the speed at which the wave packet is propagating. We’ll now actually evaluate the integral and show that the shape of the wave does move with velocity vgroup: More rigorous derivation of group velocity. Start with the above equation for the wavefunction Ψ(x, t) = Z Φ(k)ei(kx−ω(k)t)dk. Evaluating at t = 0, we have Ψ(x, 0) = Z Φ(k)eikxdk, which we’ll come back to later. Turning back to the original integral, we do a Taylor expansion ω(k) = ω(k0) + (k −k0) dω dk k=k0 + O((k −k0)2), since the values of k that matter are those near k0 anyway. (And with this we already see the group velocity popping up, which is a good sign.) Splitting up the exponential, we find Ψ(x, t) = Z dkΦ(k)eikxe−iω(k0)te −ik dω dk \|k0te ik0 dω dk \|k0te(negligible) (we note here that the negligible term is important if we care about the distortion of the wave pattern, which we won’t talk about for a few lectures). This integral may look difficult, but the e−iω(k0)t factor can come out, and so can the e ik0 dω dk \|k=k0t factor, meaning that we’re left with = e−iω(k0)te ik0 dω dk \|k0t Z dkΦ(k)eikxe −ik dω dk \|k0t. But now the first two terms in front of the integral are pure phases, and the integral resembles the wavefunction at t = 0 (no ω(k) factor), so if we take magnitudes on both sides, we arrive at \|Ψ(x, t)\| = Ψ x −dω dk k0 t, 0 . In other words, the norm of the wave at time t looks like the norm of the wave at time 0, but with an additional displacement distance of vgroupt. So a peak that started at x = 0 at t = 0 will move to x = vgroupt at time t, indeed showing that the shape of the wave packet moves at the group velocity. With all of that discussion done, we’re finally ready to write down the equation of a matter wave. If we have a particle of energy E and momentum p, we know that E = ℏω and p = ℏk, and we’re going to make an argument based on superposition and probability to find the shape of the wave. If we want a plane wave in the x+ direction, we might have a wave of the form sin(kx −ωt), cos(kx −ωt), eikx−iωt, or e−ikx+iωt (all of these are periodic functions that depend on the phase kx −ωt). But suppose we want a wavefunction of a particle with equal probability of 20 traveling to the right or to the left. Then using sin requires us to write down a wavefunction like sin(kx −ωt) + sin(kx + ωt), (since having equal coefficients gives us “equal probability” of moving in each direction), but expanding this gives us = 2 sin kx cos ωt, which is not acceptable because this wavefunction is identically zero at ωt = π 2 , 3π 2 , · · · , meaning that the particle has completely vanished. So that kind of wave (and similarly trying to produce a wavefunction out of cosines, yielding 2 cos kx cos ωt) won’t work for us, and instead we’ll want a wavefunction of the form eikx−iωt + e−ikx−iωt = 2 cos kxe−iωt. This wavefunction now never vanishes for any value of t, since the t-dependence is always a phase, so there’s nothing problematic that occurs here! Similarly, we can also use a +iωt in the exponential and write e−ikx+iωt + eikx+iωt = 2 cos kxeiωt, and again we have not run into trouble yet. But we can’t use both of these kinds of matter waves at once – indeed, if both of them were correct, then we could superimpose eikx−iωt and e−ikx+iωt (both representing a particle moving to the right), and if both represent the same state then the superposition should also give us a particle moving to the right. But instead, we get eikx−iωt + e−ikx+iωt = 2 cos(kx −ωt), and we’ve already seen that we can’t have that be a wavefunction that represents a right-moving particle. All of this is to say that we must choose one of the exponentials to be the wavefunction for a de Broglie matter wave, and by convention the following is decided: Proposition 37 (de Broglie) The matter wave (also wavefunction) for a particle with momentum p = ℏk and energy E = ℏω is Ψ(x, t) = eikx−iωt. (In three dimensions, we replace kx with ⃗ k · ⃗ x and p = hk with ⃗ p = h⃗ k.) Next lecture, we’ll determine the wave equation that this matter wave satisfies, and that will lead us to the Schrodinger equation. Lecture 5: The Schrodinger Equation Last time, we discussed de Broglie waves, and we found that we could write down the equations for plane waves corresponding to particles of a particular momentum p and energy E (Proposition 37) – this gives us the wavefunction for a free particle. Our goal is to use this particular wavefunction to understand the equation governing general potentials (and in particular, we should be able to go from the equation to the wavefunction for a free particle, and then hopefully we can generalize). Our first task is to understand what kind of (differential) equation this wavefunction will satisfy. First of all, if we have this wavefunction Ψ(x, t) = eikx−iωt and we want to extract the momentum of the particle, then we’ll want to 21 take an x-derivative to get out a factor of k: ℏ i ∂ ∂x Ψ(x, t) = ℏkeikx−iωt = pΨ(x, t). In other words, acting with the differential operator ℏ i ∂ ∂x = −iℏ∂ ∂x on the wavefunction gives us the momentum (a number) times that same wavefunction – thus, remember that operators take in functions and give us back functions, so this is the closest we’ll get to extracting a pure quantity out of the wavefunction. Definition 38 The momentum operator, denoted ˆ p, is the differential operator ˆ p = ℏ i ∂ ∂x . For our free particle, we then get the equation ˆ pΨ(x, t) = pΨ(x, t) , and when we have an equation of this form (operator on function equals number times function), the function Ψ(x, t) is called an eigenstate of the operator ˆ p with eigenvalue p. (This language is motivated by linear algebra, in which we have eigenvalues and eigenvectors of a matrix M when there are equations of the form Mv = λv.) Not all wavefunctions are eigenstates, just like most vectors are not eigenvectors of a given matrix (the matrix will usually rotate the vector in some way), but the eigenstates will be important for us in describing physical characteristics. In particular, what we’re saying is that Ψ(x, t) is a state of definite momentum: if we measure the momentum of the state, we will always find p (with no uncertainty). And of course, that’s what we want, because this is supposed to be the wavefunction of a free particle of momentum p. We’ll now consider another aspect of this wavefunction: since we can extract the momentum p, it makes sense to also extract the energy E of the particle. Notice that (because Ψ = eikxe−iωt) taking a time-derivative yields iℏ∂ ∂t Ψ = (iℏ)(−iω)Ψ = ℏωΨ = EΨ. The result iℏ∂ ∂t Ψ = EΨ is another eigenvalue equation, telling us how a wavefunction of energy E evolves over time. But this time, we can extract even more physics: remember that for a nonrelativistic particle, we have E = p2 2m, and we want to capture this dependence in some way. Specifically, we’ll try to write down another operator O satisfying OΨ = EΨ, and because E = p2 2m and pΨ = ˆ pΨ, we have EΨ = p 2m(pΨ) = p 2m ℏ i ∂ ∂x Ψ (where we’ve replaced one copy of p with a ˆ p), and because the other p is still a constant we can move it inside the derivatives and other constants to get = 21 2m ℏ i ∂ ∂x (pΨ) = 1 2m ℏ i ∂ ∂x ℏ i ∂ ∂x Ψ, and now setting the first and last expressions equal gives us the equation −ℏ2 2m ∂2 ∂x2 Ψ = EΨ , which is another eigenvalue equation and motivates the following definition: 22 Definition 39 The energy operator for a free particle, denoted ˆ E, is the differential operator ˆ E = −ℏ2 2m ∂ ∂x2 . In particular, Ψ is an energy eigenstate (of energy E), or equivalently Ψ is a state of definite energy, and from the way we defined the energy operator we also have ˆ E = 1 2m ˆ p2. And if we now look at the two eigenvalue equations involving E, we arrive at the equation −ℏ2 2m ∂ ∂x2 Ψ = iℏ∂ ∂t Ψ . This is the free Schrodinger equation, and it carries a lot of information. For example, it tells us the relation between k and ω for a de Broglie matter wave, because trying the solution Ψ = eikx−iωt gives us iℏ(−iω)Ψ = −ℏ2 2m(ik)2Ψ = ⇒ℏω = ℏ2k2 2m ⇐ ⇒E = p2 2m. So the differential equation admits plane waves as solutions but constrains the momentum and energy of our particles, and thus we really are making progress towards describing our general wavefunctions. In particular, notice that this free Schrodinger equation is linear, so superpositions of solutions Ψ are also solutions, and linearity tells us that sums of plane waves are also solutions to the free Schrodinger equation. Fourier theory then tells us that given plane waves, we can construct whatever wave packet we want, and then we know how to evolve the wavefunction over time because we’ve already described how to evolve each plane wave over time! Remark 40. This kind of logic is common in physics: we take little pieces of evidence and put them together, and even if they don’t completely rigorously lead to a given conclusion, we will have good reason to believe the end result (in this case, the Schrodinger equation). Rephrasing the paragraph above, our most general solution Ψ(x, t) to the free Schrodinger equation is (integrating over waves of different wavenumbers) Ψ(x, t) = Z ∞ −∞ Φ(k)eikx−iω(k)tdk . As previously discussed, such a wave moves with a group velocity dω dk k0 = dE dp = p m = v if Φ(k) is localized at k0 (and if it is not localized then the group velocity does not make sense). We can also make a few more observations: • The “full wavefunction” Ψ cannot be purely real, because the left-hand side of the free Schrodinger equation would then be purely imaginary and the right-hand side would be purely real, but Ψ cannot be identically zero. (Later on, we’ll talk about time-independent wavefunctions, and those will be allowed to be purely real.) • Even though we have a wave moving at some speed v, the free Schrodinger equation is not like the usual wave equation ∂2Ψ ∂x2 = 1 v 2 ∂2φ ∂t2 = 0, both because of the number of derivatives and because there is no i in the ordinary wave equation. Taking another look at the energy operator ˆ E = ˆ p2 2m (which looks like our usual “kinetic energy” in classical mechanics), the next natural step is to add in a potential V (x, t), so that our total energy is E = p2 2m + V . It then makes sense to modify our energy operator definition, and there is a different naming convention in this general case: 23 Definition 41 The Hamiltonian operator, denoted ˆ H, is the differential operator ˆ H = ˆ p2 2m + V (x, t). The Hamiltonian basically represents the energy in terms of position x and momentum p, and thus we’ll need to talk about the position operator as well. But first, this allows us to write down the general Schrodinger equation: Theorem 42 (Schrodinger equation, one dimension) The wavefunction Ψ in one dimension satisfies iℏ∂ ∂t Ψ = ˆ HΨ: iℏ∂Ψ ∂t = −ℏ2 2m ∂ ∂x2 + V (x, t) Ψ. One thing that we may be surprised about is that we multiply our potential V by the wavefunction Ψ, but that’s the simplest way that we can keep the differential equation linear – the Hamiltonian must still be a linear operator. In particular, we should think of V not just as a function but also as an operator (which just multiplies a function f (x, t) by V (x, t) to get f (x, t)V (x, t)). Fact 43 Quantum mechanics is about inventing energy operators, solving the resulting Schrodinger equation and finding the valid wavefunctions. In particular, we can consider different potentials V , and the method for solving the resulting Schrodinger equation will vary and give us very different solutions. As 8.04 goes on, we’ll see many of these methods employed, but now that x has showed up we’ll spend some time talking about its role in all of this. Definition 44 The position operator, denoted ˆ x, multiplies functions by x: ˆ xf (x) = xf (x). The reason we need to be careful with how we write this out is that it illuminates a relation between the position (ˆ x) and momentum (ˆ p) operators. One important property of operators (which we may remember from linear algebra in the context of matrices) is that the order in which they are applied or multiplied matters. Looking at operators is how Heisenberg arrived at quantum mechanics, and that’s discussed more in 8.05, but right now we’re interested in looking at whether two operators commute (that is, whether the order in which they are applied matters). Example 45 We determine whether ˆ x and ˆ p commute by computing the difference between ˆ x ˆ pφ and ˆ pˆ xφ (for some function φ(x, t)). This computation is straightforward, but we need to be careful with it: we wish to evaluate the expression ˆ x ˆ pφ −ˆ pˆ xφ, 24 but when we write down an expression like ˆ A ˆ Bφ for operators ˆ A, ˆ B, we really mean ˆ A( ˆ Bφ) (because we’re first applying ˆ B to φ, and then we apply ˆ A to that result). Thus, we are evaluating ˆ x(ˆ pφ) −ˆ p(ˆ xφ) = ˆ x ℏ i ∂ ∂x φ −ˆ p(xφ), and now both terms in parentheses are functions of x and t, so we can apply the definitions of ˆ x and ˆ p again to get = ℏ i x ∂ ∂x φ −ℏ i ∂ ∂x (xφ). We now use the product rule on the second term, and one of the two parts will cancel with the first term: = ℏ i x ∂ ∂x φ −ℏ i φ −ℏ i x ∂ ∂x φ = iℏφ . Equating the positive parts, and using the fact that ( ˆ A + ˆ B)φ = ˆ Aφ + ˆ Bφ, we find that (ˆ x ˆ p −ˆ pˆ x)φ = iℏφ for any function φ, meaning that the operators acting on φ must also be identical. We’ll use the following notation throughout the rest of the course: Definition 46 The commutator of two operators ˆ A and ˆ B, denoted [ ˆ A, ˆ B], is the operator ˆ A ˆ B −ˆ B ˆ A. Proposition 47 We have the equality of operators [ˆ x, ˆ p] = ˆ x ˆ p −ˆ pˆ x = iℏ. This nonzero commutator really encodes the fact that the two operators interact nontrivially, and that will be the basis of the uncertainty principle later on, as well as the matrix formulation of quantum mechanics. (Remember that operators, wavefunctions, and eigenstates correspond to matrices, vectors, and eigenvectors, respectively.) Example 48 A set of famous matrix commutators comes from the Pauli matrices (which we’ll talk about in 8.05, encoding the spin of spin 1/2 particles) σ1 = " 0 1 1 0 # , σ2 = " 0 −i i 0 # , σ3 = " 1 0 0 −1 # . Specifically, these matrices lead us to the spin operator ⃗ S = ℏ 2⃗ σ. We won’t talk much more about spin here in 8.04, but we’ll compute the matrix commutators: we can verify that σ1σ2 = " i 0 0 −i # , σ2σ1 = " −i 0 0 i # , so that we actually have the relation [σ1, σ2] = 2iσ3. 25 And even though the commutator of Pauli matrices gives us another Pauli matrix, rather than a number like in the [ˆ x, ˆ p] = iℏcase, we’re going to see many more complications with the position and momentum operators – it’s much harder to understand fully what that commutator means! In particular, in 8.05, we’ll talk more about linear algebra and explain how we can write operators like ˆ x and ˆ p using matrix representations, but those matrices will need to be infinite-dimensional. (This is a mathematical exercise – there are no two finite-dimensional matrices whose commutator is a multiple of the identity.) So something strange is going on, but we’ll still get more familiar with position and momentum and see where it leads us. We can generalize the one-dimensional Schrodinger equation from Theorem 42 (to avoid focusing on a simple case, and because we actually live in three dimensions): instead of having ˆ p = ℏ i ∂ ∂x , we now have ˆ px = ℏ i ∂ ∂x (for momentum along the x-direction) and similarly ˆ py = ℏ i ∂ ∂y and ˆ pz = ℏ i ∂ ∂z . (This corresponds to the fact that we can write a de Broglie wave as ei⃗ k·⃗ x−iωt, and then we can expand the dot product out across the three components.) To make the notation easier, we will sometimes replace x, y, z with x1, x2, x3, so that we can just write pk = ℏ i ∂ ∂xk (where k ranges from 1 to 3). Thus, our momentum operator now operates on vectors, and we have ˆ ⃗ p = ℏ i ∇. (Indeed, we can check that ˆ ⃗ pei⃗ k·⃗ x−iωt = ℏ⃗ kei⃗ k·⃗ x−iωt, and thus we still have a valid egienvalue equation.) The Schrodinger equation does not become much more complicated, because the Hamiltonian operator is now ˆ H = ˆ ⃗ p 2 2m + V (⃗ x, t), and the squared vector operator really means that we have ⃗ ˆ p 2 = ℏ i ∇· ℏ i ∇= −ℏ2∇2, where ∇2 is the Laplacian operator. This gives us the following general three-dimensional Schrodinger equation: Theorem 49 (Schrodinger equation, three dimensions) The wavefunction Ψ in three dimensions satisfies iℏ∂ ∂t Ψ = ˆ HΨ: iℏ∂Ψ ∂t = −ℏ2 2m∇2 + V (⃗ x, t) Ψ. This three-dimensional equation follows pretty naturally from the one-dimensional equation, but we’re introducing it now so we keep it in mind throughout all of our work solving differential equations in this class. And the commutation relation [ˆ x, ˆ p] = iℏnow still holds for ˆ xk and ˆ pk for each k, but operators like ˆ px and ˆ y do commute (because y is a constant with respect to the x-derivative). We can summarize the nine different commutators between position and momentum operators in the concise statement [ˆ xi, ˆ pj] = iℏδij, where δij is the Kronecker delta symbol which returns 1 if i = j and 0 otherwise. Our final task in this lecture will be to return to the wavefunction and understand its interpretation. We’ve taken the de Broglie matter wave and used it to obtain a differential equation, but we still don’t know what Ψ actually means. Schrodinger initially thought that the spread of the wavefunction (under the Schrodinger equation) represented disintegrating particles, so that more Ψ represents more of the particle. But physicists found that when a particle 26 comes in and hits a Coulomb potential, the wavefunction falls away as 1 r , but that is in fact not what happens – experimentally, the particle chooses a direction to go in and stays localized. So Born’s interpretation of the wavefunction in terms of probability was the one that made sense – Schrodinger, Einstein, and many others hated it, but over time physicists had to agree that it was right. The statement is essentially that Ψ(x) gives us the probability (density) of finding the particle at position x at time t; more rigorously, this means that if we look at an infinitesimally small box d3x around a point x, then the probability of finding the particle inside that cube is dP(⃗ x, t) = \|Ψ(x, t)\|2d3x. In particular, for this to be a valid probabilistic interpretation, the probability of finding the particle anywhere in space must be 100% (we must measure it to be somewhere), and thus 1 = Z R3 \|Ψ(x, t)\|2d3x . But already this interpretation adds a few complications. In particular, because the Schrodinger equation tells us how to evolve our wavefunction forward in time, knowing Ψ(x, t0) for all x allows us to know Ψ(x, t) for all x and all t > t0, thus specifying the wavefunction for all future times. But for the Born interpretation to make sense, we must ensure that as long as the boxed equation holds at some time t0, it will still hold at all future times t > t0 (in other words, the normalization of our wavefunction preserves probability). We’ll do that discussion next time! Lecture 6: Probability Density, Current, and Conservation Last time, we described the Schrodinger equation and started discussing the interpretation of the wavefunction. Specifically, we mentioned that in one dimension, \|Ψ(x, t)\|2dx gives us the probability for our particle to be found in an interval of length dx around a particular position x (at some particular time t), and thus if we integrate across all possible values x, we must have 1 = Z ∞ ∞ \|Ψ(x, t)\|2dx so that the total probability is 1. But for our discussion to be consistent, we need to ensure that this boxed normalization condition will continue to hold at future times if it holds at a particular time t0, and that’s what we’ll verify today. Along the way, we’ll understand various important features of wavefunctions and Hamiltonian operators, and we’ll also understand the concept of probability current. First, we take another look at the boxed normalization equation – notice that if Ψ approaches some nonzero constant at ±∞, then the integral will definitely be infinite. Thus, we must impose some boundary conditions: we require lim x→±∞Ψ(x, t) = 0. Remark 50. While there are technically mathematical examples where Ψ(x, t) does not have a limit as x →∞, and the integral of \|Ψ\|2 still converges, those examples do not pop up in any physical contexts and thus we are (mostly) safe to ignore them. In addition, we’ll require that the function Ψ does not oscillate too quickly: we also ask that ∂Ψ(x, t) ∂x is bounded as x →±∞. (Since the Schrodinger equation has derivatives in it, we should not be surprised that we need some regularity in Ψ of this form.) We should now turn back to a statement we made a few lectures ago, in which we stated that Ψ and cΨ always 27 represent the same state. Multiplying Ψ by a constant c multiplies the integral R ∞ −∞\|Ψ\|2 by a factor of \|c\|2, so it may seem weird that the normalization condition holds for one but not the other if they’re the same state. That can be clarified through the following statement: Definition 51 For a wavefunction Ψ(x, t) with R ∞ −∞\|Ψ(x, t)\|2dx equal to some finite number N, we say that Ψ is normalizable, and we let Ψ′ = Ψ √ N be the associated normalized wavefunction. In particular, we have Z ∞ −∞ \|Ψ′(x, t)\|2dx = Z ∞ −∞ \|Ψ(x, t)\|2 N dx = N N = 1, so Ψ′ is indeed normalized and is equivalent to the original wavefunction Ψ. And whenever we work with probabilities and probabilistic statements, we’re pre-normalizing our wavefunctions Ψ – in principle, we can always work with normalizable wavefunctions and divide by √ N only when needed. But we’ll be flexible and do whatever is more convenient for us, because multiplying by a constant does not change whether a wavefunction is normalizable or not. We’re now ready to do the check of conservation of probability under the Schrodinger equation. We start with a normalized wavefunction at time t0, so that (noting that \|Ψ\|2 = ΨΨ∗) Z Ψ∗(x, t0)Ψ(x, t0)dx = 1. Here, we’ll let Ψ∗(x, t0)Ψ(x, t0) be called the probability density (whose interpretation we’ve already discussed), and we denote it by ρ(x, t). Then we can define N(t) = Z ρ(x, t)dx , and our goal is to show that N(t) = 1 for all t > t0, given that N(t0) = 1. Rephrasing the question, we thus wish to show that the Schrodinger equation guarantees dN dt = 0. To check this, we write out the definition of N in the derivative: by differentiation under the integral sign and the product rule, we have dN dt = Z ∂ ∂t ρ(x, t)dx = Z ∂Ψ∗ ∂t Ψ + Ψ∗∂Ψ ∂t dx. We now have a time-derivative of Ψ, so here is where we can use the Schrodinger equation iℏ∂Ψ ∂t = ˆ HΨ ⇐ ⇒ ∂Ψ ∂t = −i ℏ ˆ HΨ . But we also have a time-derivative of the complex conjugate Ψ∗, so we want to take a complex conjugate of the equation above, which gives us −iℏ ∂Ψ ∂t ∗ = ( ˆ HΨ)∗⇐ ⇒ ∂Ψ∗ ∂t = i ℏ( ˆ HΨ)∗, because the complex conjugate of the partial derivative is also the partial derivative of the complex conjugate. Plugging those boxed expressions back into our calculation above yields dN dt = Z i ℏ ( ˆ HΨ)∗Ψ −Ψ∗( ˆ HΨ) dx, and in order for this to be zero, it’s equivalent to ask for the identity Z ( ˆ HΨ)∗Ψdx = Z Ψ∗( ˆ HΨ)dx 28 to hold, where on one side the conjugate is with the ˆ H term, and on the other it’s not. So conservation of probability has been rephrased into a condition on the Hamiltonian ˆ H, and it turns out a situation where we do have this condition is when ˆ H is a Hermitian operator. More generally, Hermitian operators should satisfy Z ( ˆ HΨ1)∗Ψ2 = Z Ψ∗ 1( ˆ HΨ2) for all wavefunctions Ψ1, Ψ2 – applying this to Ψ1 = Ψ2 = Ψ gives us dN dt = 0 above. But the actual definition is more general: Definition 52 For any (linear) operator T, the Hermitian conjugate of T, denoted T ∗or T †, is the linear operator that satisfies Z Ψ∗ 1TΨ2 = Z (T †Ψ1)∗Ψ2. We say that T is Hermitian if T = T ∗. It’s useful to think of Hermitian conjugates as similar to complex conjugates, so that Hermitian operators are similar to real numbers. We’ll talk more about this later on, especially in 8.05, but for now we’ll get back to the calculation, since we know the actual formula for our Hamiltonian. Plugging in the expression ˆ H = −ℏ2 2m ∂2 ∂x2 + V (x, t) yields dN dt = Z i ℏ ( ˆ HΨ)∗Ψ −Ψ∗( ˆ HΨ) dx = Z i ℏ ( ˆ HΨ)∗Ψ −Ψ∗( ˆ HΨ) dx = Z i ℏ −ℏ2 2m ∂2Ψ∗ ∂x2 Ψ + V (x, t)Ψ∗Ψ + ℏ2 2mΨ∗∂2Ψ ∂x2 −Ψ∗V (x, t)Ψ dx, where we’ve used the fact that V (x, t) is always real (because it’s some energy). Since V (x, t) is just some number, the second and fourth terms in the parentheses cancel out, and we’re left with = −iℏ 2m Z ∂2Ψ∗ ∂x2 Ψ −Ψ∗∂2Ψ ∂x2 dx. This integrand is not zero, but one common way in physics to show that an integral vanishes is to show that the integrand is actually a total derivative (and show that the boundary terms also vanish). Indeed, we can verify directly that ∂2Ψ∗ ∂x2 Ψ −Ψ∗∂2Ψ ∂x2 = ∂ ∂x ∂Ψ∗ ∂x Ψ −Ψ∗∂Ψ ∂x by expanding out the product rule and noting that two of the terms cancel out and the other two give us the desired integrand. Plugging back in and rewriting −iℏ 2m as ℏ 2im, we have dN dt = Z −∂ ∂x ℏ 2im Ψ∗∂Ψ ∂x −Ψ∂Ψ∗ ∂x dx, so the final integral is just the difference of the bracketed term at x = ∞and x = −∞; both of those are zero by our assumptions that Ψ (and therefore Ψ∗) go to 0 at the boundaries, while ∂Ψ ∂x (and therefore ∂Ψ∗ ∂x ) are bounded, so both product terms go to 0. This finishes our proof, and we’ve indeed verified that the wavefunction remains normalizable. But along the way, we’ve found something useful for our discussion: notice that inside the total derivative, we have an expression of the form z −z∗, where z = Ψ∗∂Ψ ∂x . Since z = a+bi = ⇒z −z∗= (a+bi)−(a−bi) = 2bi = 2iIm(z), what we’ve actually found is that (reminding ourselves that ρ(x, t) is the probability density) ∂ρ ∂t = −∂ ∂x ℏ mIm Ψ∗∂Ψ ∂x . 29 Definition 53 The current density J(x, t) for the wavefunction Ψ(x, t) is given by J(x, t) = ℏ mIm Ψ∗∂Ψ ∂x . We can thus rewrite our calculation above in the simpler form ∂ρ ∂t = −∂J ∂x = ⇒ ∂ρ ∂t + ∂J ∂x = 0 , which is a current conservation statement similar to what we might have seen in electromagnetism – basically, we begin with a charge density ρ(x, t) (which is really a probability density), and we arrive at a current J(x, t). We’ll discuss units briefly here – since R \|Ψ\|2 = 1, and we’re integrating in one dimension, Ψ has units of 1 √ [L], and thus Ψ∗∂ ∂x Ψ = 1 [L]2 , [ℏ] = [M][L]2 [T] = ⇒ ℏ m = [L]2 [T] , so that our current has units of [J] = [L]2 [T] · 1 [L]2 = 1 [T]. This is consistent with what we should expect – probability is unitless, and the probability current gives us a measure of probability (flow) over time. Remark 54. This same probability conservation statement can be derived in three dimensions as well, and that’s mostly left as an exercise to us. Basically, the three-dimensional current is now vector-valued, involving a divergence: ⃗ J(x, t) = ℏ mIm (Ψ∗∇Ψ) , so that our current conservation equation becomes ∂ρ ∂t + ∇· ⃗ J = 0. This last equation should look familiar to us from 8.02 – it’s how Maxwell discovered the displacement current. In this case, we now have [ ⃗ J] = 1 [L]2[T ] (so the units of current are probability per unit area per unit time). Returning to our one-dimensional statement dN dt = Z ∞ −∞ ∂ρ ∂t dx = − Z ∞ −∞ ∂J ∂x dx = −(J(∞, t) −J(−∞, t)) , where J indeed vanishes at x →±∞because of our boundary conditions on the wavefunction Ψ, we’ve checked that N is conserved and thus our wavefunction stays normalizable (thanks to the existence of a probability current). And because the integral canceled out nicely enough when we used the wavefunction Ψ, it’s reasonable to suspect that the Hamiltonian ˆ H is actually Hermitian (satisfying the equation in Definition 52) – that’s an exercise that we can verify ourselves. We’ll finish this lecture by discussing the connection of current conservation to electromagnetism in more detail, which should help clarify some of the intuition. The probability density ρ(x, t) is directly analogous to the charge density, and (integrating that density) the probability to find the particle in some volume is analogous to the charge contained within in some volume. Additionally, the probability current density is just like the current density in electromagnetism. The key point is that the existence of a probability current implies local probability conservation, which is 30 stronger than just saying that dN dt = 0. Indeed, the differential relation ∂ρ ∂t + ∂J ∂x = 0 tells us that probability changes somewhere in space are due to some nonzero current at that point, just like a change of charge density is due to some current in electromagnetism. In particular, the probability that a particle can be found between positions a and b will change only if there is a difference in the currents at the endpoints – let’s see this in more detail. If we have some volume V , and we consider the charge in that volume QV (t) = Z V ρ(x, t)d3x, then we can calculate the change in time of the current via dQV (t) dt = Z V ∂ρ ∂t d3x = − Z V (∇· ⃗ J)d3x (where we’ve used current conservation), and now by Gauss’s law we can relate the divergence to a surface integral on the boundary S of V , = − Z S ⃗ J · d ⃗ A . Indeed, the equality of boxed statements here says that charge is never created or destroyed, so current must escape through the surface in order for the charge inside V to change. And what we’re saying here is that the same holds for probability: if we look at quantum mechanics in one dimension, and we wish to look at the probability Pa,b(t) = Z b a ρ(x, t)dx that a particle is found between a and b on the real line, then dPa,b(t) dt = Z b a ∂ρ(x, t) ∂t dx = − Z b a ∂J ∂x dx = −(J(x = b, t) −J(x = a, t)) = J(a, t) −J(b, t) . In other words, the probability that we find the particle in [a, b] only occurs if probability escapes from the edges: indeed, if a current is positive at a, then probability is moving into the box [a, b] and thus increasing Pa,b, and if a current is positive at b, then probability is escaping [a, b] and thus decreasing Pa,b. Lecture 7: Fourier Transforms, Uncertainty, and Time Evolution We’ll begin today by discussing wave packets and uncertainty – this is our first look into the Heisenberg uncertainty relationships, and we’ll start by just analyzing our setup at a fixed time t = 0. Consider a wavefunction of the form Ψ(x, 0) = 1 √ 2π Z Φ(k)eikxdk which is a superposition of plane waves – in fact, any wavefunction can be written as such a superposition, where the eikx part represents a wave carrying momentum of ℏk. And we can find the coefficients Φ(k) if we’re given the initial wavefunction Ψ(x, 0), because this discussion falls under the realm of Fourier analysis: 31 Theorem 55 (Fourier inversion) Suppose Ψ(x, 0) = 1 √ 2π R Φ(k)eikxdk. Then we can recover the coefficients via Fourier inversion: Φ(k) = 1 √ 2π Z Ψ(x, 0)e−ikxdx. In other words, if we’re given an initial wavefunction (like a sine function, or a Gaussian, or a localized wave packet), we can find Φ(k) by integration, and then we can reconstruct that wavefunction as an integral of plane waves weighted by Φ(k). We’ve already talked about a similar idea when discussing group velocity, but for now we’ll focus on how to understand uncertainties of position and momentum through this result. Example 56 As before, consider some Φ(k) which has a single peak at some wavenumber k = k0, so that it has some uncertainty ∆k which characterizes the typical width of our peak. Assume that Φ(k) is real. We won’t make a precise definition of what uncertainty means (that will come later), but for the purpose of intuition, we can imagine taking the full width at half maximum, which is the distance between the two points where Φ(k) achieves half of its peak value. (We don’t need to worry about things like factors of 2 from the left and right until future lectures when we’re more rigorous.) Recalling our discussion of stationary phase, we can then also think about the resulting wavefunction Ψ(x); the contribution only comes from around k = k0 because that’s the only place where Φ(k) is large. Furthermore, we need to have a stationary phase at k0 to have a wave packet, and because Φ(k) is real it doesn’t contribute to the phase, and we just have phase φ = kx at time t = 0 (there is no ωt term). The k-derivative of the phase is then x, meaning that Ψ(x, 0) must be peaked around x = 0. Thus, we can also associate some uncertainty width ∆x to Ψ(x) at time t = 0, and we wish to relate ∆x and ∆k in our subsequent discussion. Fact 57 We’re glossing over a complication here – we claim that Ψ(x, 0) is not real-valued even though Φ(k) is, so we have to talk about the peak of \|Ψ\|(x, 0) instead. Indeed, to check whether Ψ is real, we must compare it to its complex conjugate (Ψ(x, 0))∗= 1 √ 2π Z (Φ(k))∗e−ikxdk (we take the complex conjugate of each term on the right-hand side). We change variables by replacing k with −k, which leaves = 1 √ 2π Z (Φ(−k))∗eikxdk (here dk became −dk, but the order of integration switches as well, and those two factors cancel out). Then the condition for reality of the wavefunction, meaning that this Ψ(x, 0)∗to be equal to the original Ψ(x, 0) = 1 √ 2π R Φ(k)eikxdk, is that 1 √ 2π Z (Φ(k) −Φ∗(−k))eikxdk = 0 . For this to hold, it turns out that we must have Φ(k) −Φ∗(−k) = 0, and the justification for this actually comes from Theorem 55 (because the function Φ(k) −Φ∗(−k) has a zero Fourier transform, Fourier inversion tells us that 32 it must be the zero function). Therefore, we get reality of the wavefunction as long as Φ(k) = Φ(−k)∗. And indeed, this is not a condition that holds for our wavefunction peaked at some wavenumber k0 ̸= 0, so Ψ(x, 0) will not be real in this case. We’re thus ready to return to our discussion of uncertainty and relating ∆x and ∆k – for this, we write an arbitrary wavenumber as k = k0 + ˜ k, so that we have Ψ(x, 0) = 1 √ 2π eik0x Z Φ(k0 + ˜ k)ei˜ kxd˜ k, where ˜ k is now peaked around 0, and thus the relevant range of integration is in an interval of width on the order of ∆k. As we vary ˜ k across this width of ∆k, the phase in the exponent of ei˜ kx changes by x∆k (we can call this the “total phase excursion”). But as long as this phase excursion is small enough – for example, if ∆k · x ≲1 – we’ll have a large contribution because the phase doesn’t get a chance to wrap around the full 2π. And if ∆k · x ≫1, then the phase has washed out the contributions of Φ, and thus the wavefunction will turn out to be close to 0. So our final conclusion is basically the following: Ψ(x, 0) is sizable in an interval x ∈[−x0, x0], where ∆kx0 ≲1. This means that the uncertainty ∆x is also roughly on the order of 1 ∆k , and overall we have found that ∆k∆x ≈1 for this kind of localized wave packet. This kind of relation turns out to not be restricted to quantum mechanics – it’s only when we start interpreting eikx as a state of definite momentum that we get the physical connection. Since p = ℏk for our matter waves, meaning that ∆p = ℏ∆k multiplying both sides of the boxed equation gives us ∆p∆x ≈ℏ. It turns out the more precise result is that ∆p∆x ≥ℏ 2 for any wavefunction, but we’ll need to define uncertainties more rigorously to explain that in full, and we won’t do that in this lecture. Example 58 Consider a Φ(k) wave which is basically a rectangular pulse: Φ(k) =    1 √ ∆k −∆k 2 < k < ∆k 2 , 0 otherwise. Because this function satisfies Φ(k) = Φ(−k)∗(the conjugate doesn’t matter because Φ is real), the resulting wavefunction will be real; indeed, Ψ(x, 0) = 1 √ 2π Z ∆k 2 −∆k 2 1 √ ∆k dk = 1 √ 2π∆k eikx ix ∆k 2 −∆k 2 , and after a bit of simplification we arrive at the final answer Ψ(x, 0) = r ∆k 2π sin ∆kx 2 ∆kx 2 . 33 The typical width of this wavefunction can then be calculated by looking at when it first hits the x-axis, which is at ± 2π ∆k , and thus we find that ∆x ≈2π ∆k , and indeed ∆x∆k ≈2π regardless of the uncertainty in k. We’ll now move to discussing how our wavepacket evolves, changing shape over time and causing technological complications. Recall that we have Ψ(x, t) = 1 √ 2π Z Φ(k)eikxe−iω(k)t, and we’re still dealing with a wavepacket centered at k0, so it still makes sense to expand ω in a Taylor series: this time, we’ll write ω(k) = ω(k0) + (k −k0) dω dk k0 + 1 2(k −k0)2 d2ω dk2 k0 + · · · , including the second-derivative term. Last time, we only needed the first-derivative term to understand the group velocity, and this time we’re going to see how the next term affects the distortion. (Recall that dω dk k0 = dE dp = p m = ℏk m , so taking another derivative should give us d2ω dk2 k0 = ℏ m, and then all future derivatives are zero – the Taylor series actually terminates! Plugging everything back into the integral gives us an additional e−i 1 2 (k−k0)2 ℏ m t term in the phase, and thus the shape of the wavepacket only significantly changes if this exponent is large. Therefore, roughly as long as (k −k0)2 ℏ m\|t\| ≪1, we will see little shape change in our wave packet. Remembering that k −k0 is on the order of ∆k, we’re really saying that (∆k)2 ℏ\|t\| m ≪1 ⇐ ⇒ (∆p)2\|t\| ℏm ≪1 . Also using the relation ∆p∆x ≈ℏfrom above, we can equivalently phrase this as the condition \|t\| ≪m ℏ(∆x)2 . In other words, knowing the uncertainty in our particle’s momentum or position tells us the amount of time we can wait before the wave starts to significantly distort. Fact 59 There’s yet another way to rewrite these boxed conditions in a more intuitive way, which is ∆p\|t\| m ≪ℏ ∆p = ∆x. This is understandable because it is a statement about lengths: the group velocity is not the same for all of the frequencies, because particles not having a definite momentum means there is some dispersion in the group velocity, corresponding to ∆p m . And if we multiply that uncertainty by the time \|t\| that has passed, our shape is only preserved if the distortion is smaller than the uncertainty in the inherent shape of the wavepacket ∆x. Problem 60 If we have a wavepacket of size ∆x = 10−10 m (which is on the order of the size of an atom, so it corresponds to an electron moving around), how long does that packet stay localized? 34 The answer is that the characteristic timescale here is t ∼m ℏ(∆x)2 = mc2 ℏc (∆x)2 c ≈10−16 seconds, and this turns out to be rather important when we think about applications like particle accelerators trying to keep particle bunches together! With that discussion finished, in the last part of this lecture, we’ll finally discuss time-evolution of a free particle wavepacket. If we know the shape of an initial wavefunction Ψ(x, 0), and we wish to calculate Ψ(x, t), we first calculate the coefficients Φ(k) = 1 √ 2π Z Ψ(x, 0)e−ikxdx, so that we can rewrite our arbitrary function as a superposition Ψ(x, 0) = 1 √ 2π Z Φ(k)eikxdk, and then we can evolve each of the plane waves in the way that we already know: Ψ(x, t) = 1 √ 2π Z Φ(k)ei(kx−ω(k)t)dk , where we should remember that we have the relation ℏω(k) = ℏ2k2 2m because E = p2 2m. The key tool that we’ve used here is linearity: each plane wave ei(kx−ω(k)t) solves the Schrodinger equation, so the full integral also does, and we’ve set up the integral in such a way that it has the correct initial condition at t = 0. So this boxed expression is indeed our answer, and if we’d like the solution more explicitly, we can (sometimes) do the k-integral. But the point is that with Fourier inversion, we’ve found a way to propagate any wavefunction without needing to solve the differential equation directly! Example 61 If we consider an initial condition which is a Gaussian of uncertainty on the order of a, Ψa(x, 0) = (2π)−1/4 1 √a exp −x2 4a2 , we should use the procedure above for time-evolution, finding Φ(k) and evolving plane waves. This is an exercise for us – we’ll find that Φ is also a Gaussian, the wavefunction spreads out over time, and that the relevant timescale for which this change in width occurs is τ = 2ma2 ℏ. Lecture 8: Momentum Space and Expectation Values We’ll talk more about momentum space today, understanding its physical relevance and the intimate relation between position and momentum. In particular, we’ll try to get to a point where we can think of momentum space as being complementary to the usual coordinate space that we’re used to, and that will allow us to discuss expectation values of operators and their time-dependence (which is an important part of quantum dynamics and time-evolution). We’ll begin with a time-independent story, so we suppress the t-dependence in our wavefunction Ψ(x) for simplicity (though everything still works even if we have t). Recall that we had the Fourier decomposition Ψ(x) = 1 √ 2π Z Φ(k)eikxdk, 35 giving us a similar inverse transform Φ(k) = 1 √ 2π Z Ψ(x)e−ikxdk. In particular, these two equations tell us that knowing Φ(k) is equivalent to knowing Ψ(x), since we can always get from one to the other. Thus, the two representations of the wavefunction carry the same amount of information, and Φ(k) can be thought of as carrying the weight of the various plane waves in our superposition for Ψ(x). But we can go deeper, and we’ll need a technical tool to help us manipulate integrals and functions in our subsequent discussion. The idea is to try to substitute one of the two equations above into the other, and that will give us the equation for a delta function. We have Ψ(x) = 1 √ 2π Z 1 √ 2π Z Ψ(x′)e−ikx′dx′ eikxdk, where we had to be careful when we substituted in the bracketed term for Φ(k) because x is a dummy variable in that integral and should not be reused. We’ll further simplify, changing the order of integration freely (if we want to be more mathematically rigorous we need to be careful with that, but we’ll skip the technical details for now), to Ψ(x) = Z ∞ −∞ Ψ(x′) 1 2π Z eik(x−x′)dkdx′. The blue term is then a function of the quantity (x −x′), and it’s what we might recognize as the delta function δ(x −x′) (because integrating it against Ψ(x′) picks out the integrand at the point where x′ −x = 0, meaning we are left with Ψ(x)). Thus, the integral is one way to represent our delta function, and we note that δ(x −x′) and δ(x′ −x) are equivalent (for example, we can do a change of coordinates in the blue integral k 7→−k and preserve everything except the exponent’s sign). Definition 62 The Dirac delta function is a “function” given by δ(x −x′) = 1 2π Z ∞ −∞ eik(x−x′)dk. This delta function is very useful whenever we’re dealing with Fourier transforms, but it’s a strange integral – notice that when x = x′, the integrand is always 1 and the integral diverges, while for nonzero x it’s tricky to make the integral converge to any value either. So this expression is singular in a way that forces us to be careful with it, and usually that means that we will work with the delta function inside integrals. Fact 63 If we haven’t worked with delta functions before, it’s useful to verify using the integral representation that δ(ax) = 1 \|a\|δ(x) for any nonzero real number a. We’ll now turn to our discussion about momentum space, starting with the question of how our normalization condition looks when we switch to Φ(k) from Ψ(x). Recall that we start with the normalization condition Z Ψ∗(x)Ψ(x)dx = 1, and we wish to write something similar for Φ(k); we’ll do that by substituting in our Fourier representation of Ψ(x), 36 using different variables of integration k, k′ for the two terms (but the same x): Z Ψ∗(x)Ψ(x)dx = Z 1 √ 2π Z Φ∗(k)e−ikxdk 1 √ 2π Z Φ(k′)eik′xdk dx. From here, we know that we can’t do the integrals over k in general, since those are in the most general form possible, and thus it makes sense to try to do the x-integral. Rewriting the order of the integrals, we have = Z Φ∗(k) Z Φ(k′) 1 2π Z ei(k′−k)xdx dk′dk, and now the inner integral is indeed the delta function δ(k′−k) as in Definition 62, just with different dummy variables, so this simplifies to = Z Φ∗(k) Z Φ(k′)δ(k −k′)dk′dk = Z Φ∗(k)Φ(k)dk . This has actually given us a pretty important mathematical identity: Theorem 64 (Parseval-Plancherel identity) For any function Ψ(x) with Fourier transform Φ(k), we have Z \|Ψ(x)\|2dx = Z \|Φ(k)\|2dk. In other words, because we get a probabilistic interpretation for the wavefunction when R \|Ψ(x)\|2 = 1, there is also a corresponding probabilistic interpretation for Φ(k): we have the same normalization condition R \|Φ(k)\|2 = 1, so it makes sense that we will also have a probability distribution in this “momentum space.” We’ve been using k so far instead of p = ℏk, so now we’re ready to actually introduce everything in terms of momentum. Since dp ℏ= dk, and we can rewrite functions of k (such as Φ) as functions of p (which we’ll respectively denote ˜ Φ), and our Fourier relations can now be written in language involving x and p: Ψ(x) = 1 √ 2π Z ˜ Φ(p)eipx/ℏdp ℏ, ˜ Φ(p) = 1 √ 2π Z Ψ(x)e−ipx/ℏdx. We lose a bit of symmetry in our Fourier inversion – there’s a factor of ℏin one equation but not the other, so now we’ll define a new function Φ(p) via ˜ Φ(p) = Φ(p) √ ℏ(where there’s some abuse of notation – this Φ(p) does not have the same functional form as the Φ(k) above). Checking our constants, we then arrive at the two equations Ψ(x) = 1 √ 2πℏ Z Φ(p)eipx/ℏdp , Φ(p) = 1 √ 2πℏ Z Ψ(x)e−ipx/ℏdx . Parseval’s theorem also deserves another look: the left-hand side stays the same, and when we convert from Φ(k) to ˜ Φ(p) we also gain no constant factors, but then we switch to Φ(p) and pick up a factor of ( √ ℏ)2 = ℏ, canceling out with the dp ℏfactor when switching integration variables. So the factors actually cancel out: Parseval’s identity now tells us that Z \|Ψ(x)\|2dx = Z \|Φ(p)\|2dp , 37 and in particular Ψ(x) is normalized if and only if Φ(p) is. Fact 65 Much like we did for position, we will thus interpret \|Φ(p)\|2dp as the probability to find a particle’s momentum to be in the range [p, p + dp] (and by conservation of probability and Parseval’s theorem, this is indeed valid under time-evolution too). These equations also have three-dimensional versions as well: since we need to do three integrals instead of one, it turns out that we get Ψ(⃗ x) = 1 (2πℏ)3/2 Z Φ(⃗ p)ei⃗ p·⃗ x/ℏd3⃗ p, Φ(⃗ p) = 1 (2πℏ)3/2 Z Ψ(⃗ x)e−i⃗ p·⃗ x/ℏd3⃗ x. We also similarly have a three-dimensional delta function, also defined with an integral: δ3(⃗ x −⃗ x′) = 1 (2π)3 Z ei⃗ k·(⃗ x−⃗ x′)d3⃗ k, and Parseval’s identity still gives us Z \|Ψ(⃗ x)\|2d3⃗ x = Z \|Φ(⃗ p)\|2d3⃗ p. Basically, everything is very analogous in three dimensions and we don’t have to memorize anything new! With that, we’re finally ready to discuss expectation values of operators, which is one of the first steps towards a full interpretation of quantum mechanics. We’ll start with the basic probabilistic definition: Definition 66 Let Q be a random variable which can take values in a finite set {Q1, · · · , Qn} with probabilities p1, · · · , pn, respectively. Then Q has expectation or expected value ⟨Q⟩= n X i=1 piQi. This is basically a weighted average which tells us the average value obtained by repeatedly taking many samples of Q. Expectations can also be computed for continuous random variables, which is what we need in quantum mechanics for our wavefunction: essentially, because Ψ∗(x, t)Ψ(x, t)dx tells us the probability that our particle is in the range [x, x + dx], we can define the expectation value of ˆ x, the position operator, via ⟨ˆ x⟩= Z xΨ∗(x, t)Ψ(x, t)dx (where x is the value of the position operator, and Ψ∗(x, t)Ψ(x, t)dx is the corresponding probability). What we’re really doing here is treating our position operator as a random variable, and experimentally we’re saying the following: suppose we have a system that is represented by the wavefunction Ψ. Then if we build many identical copies of this system and measure the positions of the particles (at the same time), then the average value should be close to ⟨ˆ x⟩. The key point to always keep in mind is that repeated measurements of the state can result in varying answers! Similarly, because we have a probability density for the momentum p (specifically, \|Φ(p)\|2dp is the probability to find the particle’s momentum in [p, p + dp]), we should define ⟨ˆ p⟩= Z p\|Φ(p)\|2dp. 38 But we know how Φ(p) is related to Ψ(x) through the Fourier relations, and it’s natural to see the expectation value for the momentum operator in terms of the same density as the expectation value for the position operator. So we’ll substitute: = Z pΦ∗(p)Φ(p)dp, and now writing Φ∗and Φ as integrals over x′ and x, respectively, we get = Z ∞ −∞ p Z 1 √ 2πℏeipx′/ℏΨ∗(x′)dx′ Z 1 √ 2πℏe−ipx/ℏΨ(x)dx dp. Like before, our strategy is to do the p-integral because the others were just written out generically: cleaning up the terms gives us = 1 2πℏ Z Ψ∗(x′) Z Ψ(x) Z peipx′/ℏe−ipx/ℏdp dxdx′, and now the insight that we should bring in is that “we know the momentum operator is secretly supposed to be d dx .” Notice that we can rewrite the inner integral as Z peipx′/ℏe−ipx/ℏdp = Z −ℏ i ∂ ∂x eipx′/ℏe−ipx/ℏdp, since the x-partial derivative doesn’t act on the term with the x′ exponent. So now we can move this −ℏ i ∂ ∂x term out of the innermost integral, since it doesn’t have to do with p: a bit more cleaning gives us ⟨ˆ p⟩= Z Ψ∗(x′) Z Ψ(x) −ℏ i ∂ ∂x Z 1 2πℏeip(x′−x)/ℏdp dxdx′. We’re almost there – the inner integral is now the delta function δ(x −x′) (if we just do a change of variables u = p ℏ), and then we can do integration by parts on the x-integral to have the ∂ ∂x act on the other term instead. Here, we’re using the important theme that while integration by parts has boundary terms, they will vanish as long as our wavefunction goes to 0 sufficiently fast (which we will always assume). Thus, we end up with = Z Ψ∗(x′) Z ℏ i ∂ ∂x Ψ(x)δ(x −x′)dxdx′, which we can more clearly write out as = Z ℏ i ∂Ψ ∂x Z Ψ∗(x′)δ(x −x′)dx′dx. The integral over x′ will evaluate Ψ∗(x′) at x, so putting everything back together gives us ⟨ˆ p⟩= Z ℏ i Ψ∗(x)∂Ψ ∂x dx, or in the most revealing form possible, we actually have ⟨ˆ p⟩= Z Ψ∗(x)ˆ pΨ(x)dx , where ˆ p = ℏ i ∂ ∂x is our momentum operator. So even though the expectation value is initially defined by a probabilistic interpretation R p\|Φ(p, t)\|2dp, the end result is that we sandwich ˆ p between Ψ∗and Ψ (and only have it act on Ψ). Notice that if we replace the x in the integral for ⟨ˆ x⟩with an ˆ x operator between Ψ∗and Ψ, that also gives us the same answer, and this motivates the following general definition: 39 Definition 67 The expectation value of a quantum mechanical operator ˆ Q in a state Ψ is defined as ⟨ˆ Q⟩Ψ = Z Ψ∗(x, t)( ˆ QΨ(x, t))dx. (If it is clear which state we are referring to, or if we are speaking generically, we will sometimes omit the Ψ subscript for clarity.) Such expectation values are varying in time, and that time-dependence can be studied. Example 68 Consider the kinetic operator ˆ T = ˆ p2 2m – we’ll write down its expectation value in the position-space representation. If we’re acting on the wavefunction Ψ(x), then we should think of ˆ p as ℏ i ∂ ∂x , and thus we have ⟨ˆ T⟩= Z Ψ∗(x, t) −ℏ2 2m ∂2 ∂x2 Ψ(x, t) dx. But we can also write down the expectation integral in momentum space, and then ˆ p acts as p on a wavefunction Φ(p). Thus, we have ⟨ˆ T⟩= Z p2 2m\|Φ(p)\|2dp = Z Φ∗(p) p2 2mΦ(p)dp, where the first form of the integral comes from the probabilistic definition of expectation values, while the second is thinking of it as a “sandwiched” operator. It might seem like the momentum integral is nicer (in particular, it’s manifestly positive because the integrand is positive), but we can actually partially integrate by parts the x-integral to find that we also have ⟨ˆ T⟩= Z ℏ2 2m ∂Ψ ∂x 2 dx, which is also clearly positive. So both integrals are valid to work with! We can now return to the time-dependence of expectation values, which is a pretty fundamental result in quantum mechanics. We aim to calculate d dt ⟨ˆ Q⟩= d dt Z Ψ∗(x, t)( ˆ QΨ(x, t))dx , and by the product rule this simplifies to = Z ∂Ψ∗ ∂t ˆ QΨ + Ψ∗ˆ Q∂Ψ ∂t dx. But now applying the Schrodinger equation (because that’s how we deal with time-derivatives) in the form iℏ∂Ψ ∂t = ˆ HΨ, we have = Z i ℏ( ˆ HΨ)∗ˆ QΨ −i ℏΨ∗ˆ Q ˆ HΨ dx. Multiplying by iℏto cancel out the constants on the right, we now have iℏd dt ⟨ˆ Q⟩= Z Ψ∗ˆ Q ˆ HΨ −( ˆ HΨ)∗ˆ QΨ dx, and by Hermiticity the ˆ H on the second term can be brought to act on the other term, QΨ, instead: = Z Ψ∗Q ˆ HΨ −Ψ∗ˆ HQΨ dx = Z Ψ∗( ˆ Q ˆ H −ˆ H ˆ Q)Ψdx. But now we see a commutator show up – in fact, this is one of the reasons commutators are so important in quantum 40 physics – and we get the final equation iℏd dt ⟨Q⟩= Z Ψ∗[ ˆ Q, ˆ H]Ψdx = ⟨[ ˆ Q, ˆ H]⟩, where in the last step we’ve noticed that the commutator sandwiches Ψ∗and Ψ and thus gives us the usual formula for an expectation value. One way to phrase what’s happening is that we’ve actually encoded all of the dynamics in the observables – instead of having wavefunctions changing in time and therefore having expectations change as a result, we have the expectations evolving in terms of their commutators with the Hamiltonian. In particular, any operator expectation that commutes with the Hamiltonian has time-derivative zero, and thus commutation relations give rise to conservation laws. For example, since ˆ p commutes with ˆ H = ˆ p2 2m for the free particle, a free particle’s expected momentum must be conserved. We’ll see much more of this concept in the future! Lecture 9: Hermitian Operators, Measurement, and Uncertainty We’ll begin today with the discussion of Hermitian operators as they relate to physical observables. Recall that an operator ˆ Q is Hermitian if it satisfies the equation Z Ψ∗ 1 ˆ QΨ2dx = Z ( ˆ QΨ1)∗Ψ2 for any two sufficiently well-behaved wavefunctions Ψ1, Ψ2. We’ll sometimes use a more concise notation for these kinds of expressions: Definition 69 Define the inner product (. , .), which takes in two wavefunctions Ψ1, Ψ2 and outputs (Ψ1, Ψ2) = Z Ψ∗ 1(x)Ψ2(x)dx. This inner product satisfies some important properties: for example, we can directly check that (aΨ1, Ψ2) = a∗(Ψ1, Ψ2), (Ψ1, aΨ2) = a(Ψ1, Ψ2). Notice that we can rewrite the definition of Hermiticity in a nicer form now: we are asking for (Ψ1, ˆ QΨ2) = ( ˆ QΨ1, Ψ2) for any two well-behaved wavefunctions Ψ1, Ψ2. Last time, we also defined the expectation value of the operator ˆ Q in a state Ψ, given by ⟨ˆ Q⟩Ψ = Z Ψ∗ˆ QΨdx = (Ψ, ˆ QΨ). We’ll connect these two ideas together by making a few observations: Proposition 70 For any Hermitian operator ˆ Q, the expectation value ⟨ˆ Q⟩Ψ is real. Proof. Notice that (⟨ˆ Q⟩Ψ)∗= Z Ψ∗ˆ QΨ ∗dx 41 (if we want to take the complex conjugate of an integral, we can just conjugate the integrand). The integrand is the product of Ψ∗and ˆ QΨ, which are each complex-valued (an important note here: we are not thinking of just conjugating ˆ Q on its own, because it is acting on a wavefunction), and thus we can rewrite this as = Z Ψ( ˆ QΨ)∗dx = Z ( ˆ QΨ)∗Ψdx. where we’ve used the fact that (Ψ∗)∗= Ψ. And now we use the Hermiticity of ˆ Q to move the ˆ Q from one term to the other: = Z Ψ∗( ˆ QΨ)dx = ⟨ˆ Q⟩Ψ. Thus, the expectation value of ˆ Q is its own conjugate, meaning that it is real. Proposition 71 The eigenvalues of a Hermitian operator ⟨ˆ Q⟩are always real. (Recall that if we have a wavefunction Ψ1 that satisfies ˆ QΨ1 = q1Ψ1, then q1 is an eigenvalue of ˆ Q and Ψ1 is the associated eigenvector or eigenfunction.) Proof. Let q1 be an eigenvalue of ˆ Q with associated eigenvector Ψ1. There are many ways to show this result, but we’ll apply Proposition 70 to ˆ Q in the state Ψ1. Then we have the real number ⟨ˆ Q⟩Ψ1 = Z Ψ∗ 1 ˆ QΨ1dx = Z Ψ∗ 1q1Ψ1dx (by the eigenvalue condition), and then taking q1 out of the integral gives us = q1 Z Ψ∗ 1Ψ1dx. The integrand of this integral is always a positive real number, so the integral is real. Because our left-hand side ⟨ˆ Q⟩Ψ1 was also real, our eigenvalue q1 is indeed real. We may notice that Ψ1 can be arbitrarily scaled and it will still be a valid eigenvector; thus, we can always scale so that “being in the state Ψ1” means that we have a normalized state. (In fact, we must do this normalization whenever calculating expectation values of our states!) Then the calculation above simplifies to ⟨ˆ Q⟩Ψ1 = q1, so the eigenvalue of an operator in a state Ψ1 is the expectation value of ˆ Q in that state, and it is always real (as desired). The key point now is that (informally) Hermitian operators are rich: they have as many eigenvectors as we need to span the whole space of states. In other words, if we take any Hermitian operator and write down its set of eigenstates, then any wavefunction is a superposition of those eigenstates! (In linear algebra terms, we can think about a finite-dimensional matrix for a Hermitian matrix – what we’re saying is that the eigenvectors form a basis of the vector space on which the matrix acts on.) This is called the spectral theorem, and we’ll talk about it more in 8.05 when we have more discussion of linear algebra, but we’ll still write out our claim more explicitly. 42 Proposition 72 For any Hermitian operator ˆ Q, consider its eigenvalues and eigenfunctions given by ˆ QΨ1 = q1Ψ1, ˆ QΨ2 = q2Ψ2, · · · (the set of which can be finite or infinite). Then the eigenfunctions can be organized to satisfy the orthonormality relation Z Ψ∗ i (x)Ψj(x)dx = δij In particular, setting i = j tells us that each Ψi is a normalized state, so they are “good eigenvectors” in the expectation value sense. But in addition, for any i ̸= j, we have (Ψi, Ψj) = 0, and we can think of the usual three basis vectors of R3 (the x-, y-, and z-direction unit vectors, all of which have unit length and are orthogonal) as an illustrative example of what’s going on more generally! So the inner product we’ve defined should be thought of as a dot product of wavefunctions. Proof of a simple case. We’ll consider the case where the eigenvalues of different eigenvectors are always different (meaning that qi ̸= qj if i ̸= j). In this case, we can always first normalize our eigenfunctions by rescaling. Then we have Z Ψ∗ i ˆ QΨjdx = Z Ψ∗ i qjΨjdx = qj Z Ψ∗ i Ψj , but also by Hermiticity we have Z Ψ∗ i ˆ QΨjdx = Z ( ˆ QΨi)∗Ψjdx = Z (qiΨi)∗Ψj = qi Z Ψ∗ i Ψj , where we know that q∗ i = qi by Proposition 71. Equating the two boxed expressions, because qi ̸= qj, we must have R Ψ∗ i Ψj = 0, as desired. However, the more general case occurs when we have degeneracy, and this is a very important concept that often comes up in quantum mechanics. Degeneracy occurs when several different eigenfunctions have the same eigenvalue (we’ll in fact see a case where this happens later in the lecture), and in such a case it is important that Proposition 72 includes the clause “can be organized” – we need to pick appropriate eigenfunctions in the eigenspaces, forming linear combinations, so that we do indeed still have orthonormality, and we still do have enough eigenstates to span the whole space. Let’s state that last claim more explicitly: Proposition 73 The eigenfunctions of any Hermitian operator ˆ Q form a set of basis functions, meaning that any (reasonable) wavefunction Ψ can be written as a superposition Ψ = α1Ψ1 + α2Ψ2 + · · · = X i αiΨi. While those coefficients αi may seem mysterious, we can in fact use Proposition 72 to calculate them easily: we have (Ψi, Ψ) = Z Ψ∗ i X j αjΨjdx 43 (we should make sure not to reuse indices here, which is why we sum over j), and now we can switch the sum and integral to get = X j αj Z Ψ∗ i Ψjdx. But orthonormality tells us that the only term that survives is i = j: = X j αjδij = αi . Thus, we compute the coefficient αi (and thus how much of the wavefunction Ψ is “along” the state Ψi) by integrating Ψ∗ i Ψ. Additionally, we can compute through expansion that 1 = Z \|Ψ\|2dx = Z Ψ∗Ψdx = Z X i αiΨi !∗X j αjΨjdx, and now moving the sum over i, j out, we get X i X j α∗ i αj Z Ψ∗ i Ψjdx = X i X j α∗ i αjδij = X i α∗ i αi = X i \|αi\|2 . Phrasing this in words, if we have a state which is a superposition of orthonormal basis vectors, then the sums of squares of coefficients gives us the normalization condition – there’s no mixing between Ψi and Ψj. And this expansion can be done for any state and any Hermitian operator! With that, we’ve done all of the work necessary for stating the measurement postulate which we introduced in our beginning lectures: Theorem 74 (Measurement postulate) Suppose we measure a Hermitian operator ˆ Q in a state Ψ. Then the possible results that we may obtain are the eigenvalues q1, q2, · · · , and the probability of measuring qi is pi = \|αi\|2 = \|(Ψi, Ψ)\|2. If the outcome of the measurement is qi, then the state of the system becomes Ψi (this is called the collapse of the wavefunction). For example, if we measure the kinetic energy of a particle, then we’ll get some eigenvalue of the kinetic operator, and after we measure we’ll be in a state of definite kinetic energy (so that subsequent measurements also give us that same kinetic energy). The key is that this result essentially came from Hermitian operators being rich enough for their eigenvectors to span our space of wavefunctions – if we want to measure some quantity X, we should write our state as a superposition of states with definite X, use the spectral theorem, and then compute the coefficient αi that is relevant. This measurement postulate may seem strange to us, because it essentially divides quantum mechanics into two realms: that of the Schrodinger equation (the natural time-evolution) and that of measurement (the collapse). People have wondered why measurement doesn’t come out of the Schrodinger equation if it’s supposed to govern quan-tum mechanics, but nothing is sufficiently clear for us to question this framing for now, and thus we’ll take this measurement postulate as an additional assumption of quantum mechanics. 44 Example 75 Suppose that we have a state Ψ = P i αiΨi. We’ll compute the expectation of ˆ Q on Ψ in terms of the eigenvectors as a consistency check. First of all, we have (as we’ve done before) ⟨ˆ Q⟩Ψ = Z Ψ∗ˆ QΨ = Z X i αiΨi !∗ ˆ Q  X j αjΨj  dx = X i X j α∗ i αj Z Ψ∗ i QΨjdx, and now using the eigenvector equations and orthonormality yields = X i X j α∗ i αj Z Ψ∗ i qjΨjdx = X i X j α∗ i αjqjδij = X i \|αi\|2qi . But we’ve derived this expected result (achieving qi with probability \|αi\|2) with properties of our eigenvectors, rather than with the definition of an expectation value of a random variable (which is what the end result here is telling us)! Since the measurement postulate tells us that pi = \|αi\|2, everything we’ve said so far is consistent. Let’s do a physical example to see this in exaction: Example 76 Consider a free particle on a (topological) circle x ∈[0, L], where we identify the points at x = 0 and L, and consider the wavefunction at some fixed time Ψ(x) = r 2 L 1 √ 3 sin 2πx L + r 2 3 cos 6πx L ! . We’ll calculate the different values and probabilities that we can obtain when we measure the momentum of the particle. Remark 77. We can verify that Ψ(L) = Ψ(0), so this is a valid wavefunction on the circle. Notice that Ψ(x) is purely real at this point in time, and this is in fact allowed, but our discussion that wavefunctions cannot be purely real (from a previous lecture) tells us that it will not be real forever. Since we’re trying to measure the momentum of our particle, we have to look at the momentum eigenstates, which are plane waves of the form eikx. On the real line, we could never normalize these states (because the norm squared of eikx is 1), but on a circle we’re actually okay, because there’s only a finite length we have to integrate along. Specifically, momentum eigenstates take the form Ψm ∝e2πimx/L for some integer m, because these are plane waves that are periodic with period L (meaning that Ψ(x + L) = Ψ(x)), so they have definite momentum and we just have to normalize them. For that, notice that the squared norm of e2πimx/L is always 1, so to make the integral 1 over the length L we want Ψm = 1 √ Le2πimx/L , 45 for any integer m, to be our normalized momentum eigenstates with corresponding momentum pmΨm = ℏ i ∂ ∂x Ψm = 2πℏm L Ψm = ⇒ pm = 2πℏm L . Thus, we’ve found the momentum eigenfunctions, and all that’s left to do is to rewrite our Ψ as a linear combination of the Ψms. The momenta of the different eigenfunctions are all different, so we have orthonormal eigenfunctions and we can just use the trigonometric identities sin x = eix −e−ix 2i , cos x = eix + e−ix 2 from here to find that Ψ = r 2 3 1 2i Ψ1 − r 2 3 1 2i Ψ−1 + 1 √ 3Ψ3 + 1 √ 3Ψ−3. We can now read off the probabilities and measured momenta: since Ψm has momentum 2πℏm L , our state Ψ will have momentum 2πℏ L with probability q 2 3 1 2i 2 = 1 6, and similarly we’ll get momentum −2πℏ L with probability 1 6, momentum 6πℏ L with probability 1 3, and momentum −6πℏ L with probability 1 3, when measuring in the state Ψ. These probabilities indeed add to 1, and thus we’ve now seen an example of decomposition along eigenstates that we’ve talked about abstractly above. Finally, in the same probabilistic framework, we can now properly define what uncertainty really means in a mathematical context. Definition 78 Let Q be a random variable which takes on values Q1, · · · , Qn with probabilities p1, · · · , pn, and suppose its expectation value is Q = ⟨Q⟩= P i piQi (we use Q and ⟨Q⟩interchangeably). Then the variance of Q is (∆Q)2 = X i pi(Qi −Q)2, and the uncertainty or standard deviation of Q is ∆Q = p (∆Q)2. In other words, the variance calculates the expected (weighted) squared deviation from the average – we can indeed see that the right-hand side is always positive because it’s a sum of positive terms, so ∆Q will be real. Furthermore, if ∆Q = 0, then each term on the right-hand side must be zero, meaning that Qi = Q for any potential value Q takes on (and thus Q is with probability one just its average value Q). We can simplify the variance with some algebraic manipulation: by expanding, we find that (∆Q)2 = X i pi(Qi −Q)2 = X i piQ2 i −2 X i piQiQ + X i piQ 2. Now we can separate out the different interpretations, noting that Q is a constant, and we find that this simplifies to = Q2 −2Q X i piQi + Q 2 X i pi = Q2 −2QQ + Q 2 = Q2 −Q 2 . In particular, because the variance of a random variable is always nonnegative, we find that Q2 ≥Q 2 for any real-valued random variable Q. And now this allows us to turn back to the quantum mechanical interpretation: 46 Definition 79 Let ˆ Q be a Hermitian operator. We define the uncertainty of ˆ Q in the state Ψ, denoted (∆ˆ Q)2 Ψ, to be (∆ˆ Q)2 Ψ = ⟨Q2⟩Ψ −⟨ˆ Q⟩2 Ψ. The point here is that ˆ Q and ˆ Q2 are Hermitian operators, so we can calculate the right-hand side because we always know how to compute expectation values. And from here, we claim the following (which we’ll prove next lecture): Lemma 80 We may write the uncertainty of ˆ Q in the state Ψ as a single expectation (∆ˆ Q)2 Ψ = ⟨( ˆ Q −⟨ˆ Q⟩)2⟩Ψ, and we can also write it as an integral (∆ˆ Q)2 Ψ = Z \|(Q −⟨Q⟩)Ψ\|2 dx. (In the above, ⟨Q⟩is the multiplication-by-a-constant operator.) In addition, we’ll show that analogous to the zero-variance observation in the probabilistic context, a state Ψ has zero uncertainty for ˆ Q if and only if it is an eigenstate for ˆ Q. Lecture 10: Stationary States and the Particle on a Circle We’ll begin today by analyzing the uncertainty of a Hermitian operator in more detail. Last time, we defined (∆ˆ Q)2 Ψ for a Hermitian operator ˆ Q measured in a state Ψ, and today we’ll start by proving the two claims in Lemma 80 from last lecture. Proof of Lemma 80. For the first claim, we do a direct computation (remembering that ⟨ˆ Q⟩is thought of as the operator which multiplies any state by the constant ⟨ˆ Q⟩). We expand the square, noticing that ⟨ˆ Q⟩and ˆ Q commute (because the former is a number) ⟨( ˆ Q −⟨ˆ Q⟩)2⟩= ⟨ˆ Q2 −2⟨Q⟩ˆ Q + ⟨ˆ Q⟩2⟩. (Notice that (A + B)2 = A2 + AB + BA + B2 ̸= A2 + 2AB + B2 if A and B don’t commute.) Now the expectation of a sum is the sum of the individual expectations, so we can further simplify to = ⟨ˆ Q2⟩−2⟨ˆ Q⟩⟨ˆ Q⟩+ ⟨ˆ Q⟩2 (where we’ve used the fact that we can take constants out of expectations: ⟨c ˆ Q⟩= c⟨ˆ Q⟩). Combining the last two terms indeed gives us the definition of the uncertainty of ˆ Q, as desired. For the second claim, we start with the expression ⟨( ˆ Q −⟨ˆ Q⟩)2⟩and plug in the definition of an expectation value: we find that (∆ˆ Q)2 = Z Ψ∗(x)( ˆ Q −⟨ˆ Q⟩)2Ψ(x)dx = Z Ψ∗(x)( ˆ Q −⟨ˆ Q⟩)( ˆ Q −⟨ˆ Q⟩)Ψ(x)dx. We now think of the first ( ˆ Q −⟨Q⟩) as an operator acting on the rest of the integrand – because ˆ Q is Hermitian and so is ⟨Q⟩(multiplying by a real constant is equivalent no matter which part of the integrand it’s done on), the whole 47 term ( ˆ Q −⟨Q⟩) is Hermitian, and thus by the definition of Hermiticity we have = Z ( ˆ Q −⟨ˆ Q⟩)Ψ(x) ∗( ˆ Q −⟨ˆ Q⟩)Ψ(x)dx = Z ( ˆ Q −⟨ˆ Q⟩)Ψ(x) 2 dx, as desired. Notice that our second claim in fact writes the uncertainty (∆ˆ Q)2 as an integral of a positive quantity, meaning that ∆ˆ Q is always real. Corollary 81 A state Ψ is an eigenstate of a Hermitian operator ˆ Q (meaning that ˆ QΨ = λΨ) if and only if (∆ˆ Q)2 = 0. In addition, we have λ = ⟨ˆ Q⟩(so that we can write ˆ QΨ = ⟨ˆ Q⟩ΨΨ). Proof. We show the second claim first: to do so, we integrate ⟨ˆ Q⟩= Z Ψ∗ˆ QΨdx = Z Ψ∗λΨdx = λ\|Ψ\|2dx = λ, so the eigenvalue is indeed the expectation of ˆ Q. So now for the first claim, Ψ is an eigenstate of ˆ Q if and only if ( ˆ Q −⟨Q⟩)Ψ = 0, if and only if (by part two of Lemma 80, since the integral is only zero if the nonnegative integrand is always zero) (∆ˆ Q)2 = 0, if and only if ∆ˆ Q = 0, as desired. We now have several different expressions for the uncertainty of an operator ˆ Q, and they are each useful in different situations. For example, if we have a Gaussian wavefunction that is centered at the origin, then ⟨ˆ x⟩= 0, so the easiest way to calculate the uncertainty in position is (∆x)2 = ⟨x2⟩−⟨x⟩2 = ⟨x2⟩, and that’s just a Gaussian integral for us to do. And now that we’ve defined uncertainty, we can actually state the uncertainty principle precisely: we’re saying that ∆x∆p ≥ℏ 2 under the mathematically rigorous definition. We’ll now move on to our next topic, stationary states, which will keep us busy for a few weeks as we develop intuition for solving the Schrodinger equation. Definition 82 A stationary state is a solution to Schrodinger’s equation with factorized space and time dependence Ψ(x, t) = g(t)ψ(x). Note that these solutions aren’t static – they do have time-dependence, but that time-dependence is very simple. From here on, we’ll be careful to use the capital letter Ψ for full wavefunctions with time-dependence and lowercase ψ for the time-independent wavefunctions. The reason these states are called “stationary states” is that time-independent observables will not have time-dependence in expectation value if we have a state Ψ of this form. We’ll see this in a few minutes, but first let’s use a “separation of variables” technique for solving differential equations to understand Ψ(x, t) more generally. Plugging in this stationary state into Schrodinger’s equation, we start with iℏ∂Ψ(x, t) ∂t = ˆ HΨ(x, t) = −ℏ2 2m ∂2 ∂x2 + V (x) Ψ(x, t). 48 Fact 83 We’re considering here systems where the potential V (x) does not depend on time – otherwise, if the “landscape” of our problem is changing, we can’t easily get stationary states. Substituting Ψ(x, t) = g(t)ψ(x), we find that because the left-hand side only acts on the time-coordinate and the Hamiltonian on the right-hand side only acts on the space-coordinate, iℏψ(x)dg(t) dt = g(t) ˆ Hψ(x), where ˆ Hψ(x) is again just a function of x. Rearranging by dividing the whole equation by Ψ, we find that iℏ 1 g(t) dg(t) dt = 1 ψ(x) ˆ Hψ(x). Since the left-hand side is now only a function of time, and the right-hand side is only a function of space, the only way for these to be equal in general is that both sides are constant, and we’ll call that constant E. This E must have units of energy (because those are the units of ˆ H and also of ℏ t ), and it must be real – we’ll see later that if we try to choose E to be complex, we’ll run into problems. But for now we’ll go ahead and solve the two (decoupled) differential equations: we have iℏdg(t) dt = Eg(t) = ⇒g(t) = Ce−iEt/ℏ, and thus the time-dependence of a stationary state is just the phase e−iEt/ℏ. On the other hand, the other equation becomes ˆ Hψ(x) = Eψ(x) = ⇒ −ℏ2 2m d2 dx2 + V (x) ψ(x) = Eψ(x) (we can use a normal derivative now that we have just functions of a single variable). So the rest of the problem is still complicated – we still need to solve the second-order differential equation for ψ(x), and this is called the time-independent Schrodinger equation. What’s interesting is that when we solve this differential equation, we often find that we are constrained to particular values of E (if we think about the analogy with matrices, the equation ˆ Hψ(x) = Eψ(x) is really an eigenfunction equation, and matrices often only have a discrete set of eigenvalues). Putting things together, we find that once we solve the time-independent Schrodinger equation, we arrive at the full wavefunction Ψ(x, t) = Cψ(x)e−iEt/ℏ. If we try to normalize this state, we can just normalize using ψ (meaning that we absorb C into the time-independent solution). Thus, we are asking for 1 = Z Ψ∗(x, t)Ψ(x, t)dx = Z ψ∗(x)eiEt/ℏψ(x)e−iEt/ℏdx. But the phase terms eiEt/ℏand e−iEt/ℏcancel out here, so in fact the condition for normalization is just Z ψ∗(x)ψ(x)dx = 1, and we can just check normalization for ψ. In particular, the time-dependence doesn’t end up making a contribution for normalization, and this is the key reason why we want E to be real – if the energy weren’t real, the complex conjugate of eiEt/ℏwould not just be e−iEt/ℏ, so we’d have a function of t that would also need to be equal to 1 on 49 the left-hand side. Thus, stationary states must have real E if we want to consider them in actual physical contexts (where we do need normalized states to say meaningful things). We’ll now check out properties of the Hamiltonian operator ˆ H, and we’ll indeed see that this is all connected to energies and energy eigenstates. If Ψ(x, t) is a stationary state, then ⟨ˆ H⟩Ψ(x,t) = Z Ψ∗(x, t) ˆ HΨ(x, t)dx = Z ψ∗(x)eiEt/ℏˆ He−iEt/ℏψ(x), and again the time-dependence cancels out because e−iEt/ℏis a constant from the point of view of ˆ H. Thus, this expectation is just = Z ψ∗(x) ˆ Hψ(x) = ⟨ˆ H⟩ψ(x) ; that is, the expectation value of the Hamiltonian on the full stationary state is the expectation value of the Hamiltonian on just the spatial part. And furthermore, because we have the eigenvalue equation ˆ Hψ(x) = Eψ(x), we can further evaluate this to be = Z ψ∗(x)Eψ(x) = E . So the expectation value of the Hamiltonian is in fact equal to the energy E, and the ψ(x)s are the energy eigenstates of ˆ H. (And there is no uncertainty in the measured energy either, because ⟨ˆ H2⟩= ⟨ˆ H⟩2 = E2.) More generally, we can now verify the claim from above: Proposition 84 The expectation value of any time-independent operator ˆ Q in a stationary state Ψ(x, t) is time-independent. Proof. In exactly the same way, ⟨ˆ Q⟩Ψ(x,t) = Z Ψ∗(x, t) ˆ QΨ(x, t)dx = Z ψ∗(x)eiEt/ℏˆ Qψ(x)e−iEt/ℏdx, and now because of commutativity, we can move the e−iEt/ℏterm past the ψ(x) and ˆ Q terms to cancel with the eiEt/ℏand remove the time-dependence: = Z ψ∗(x) ˆ Qψ(x)dx = ⟨ˆ Q⟩ψ(x), which is time-independent because no t shows up in the final expression. Fact 85 Note that in general, the superposition of two stationary states is not a stationary state, because we can’t factor Ψ1(x, t) + Ψ2(x, t) if the time-components eiEt/ℏlook different (that is, if the energies E are different). In particular, if we take an expectation of a time-independent operator in a superposition of stationary states, we now no longer necessarily have time-independence of that expectation value. Quantum mechanics would be very boring if all expectation values were always constant, and indeed this superposition is how time-dependence ends up occurring in general. We’ll now turn to solving the time-independent equation ˆ Hψ(x) = Eψ(x) for some arbitrary potential V (x). Because this is an eigenvalue equation for a Hermitian operator ˆ H, we have the results from our last lecture: the eigenfunctions can be chosen to form an orthonormal set of functions that span the whole space of (spatial) wavefunc-tions. With that, we are able to obtain the spectrum of the theory (which is the set of eigenvalues and eigenstates 50 associated to the particular Hamiltonian ˆ H). And remember that finding the energy eigenstates ψ1, ψ2, · · · and the associated energy eigenvalues is basically the gold standard for what we want to do, because from there we can write any arbitrary state as a superposition of the ψis. The spectrum may be discrete, or there may be a set of continuous allowed energies, but for any problem (for instance, any potential V (x) that we might be given), our goal is always to find out what that spectrum is. We’ll slightly rewrite the second-order differential equation that we’re trying to solve: we are looking for functions ψ(x) that satisfy d2ψ dx2 = 2m ℏ2 (V (x) −E)ψ(x). The potential V (x) can take on many forms – it might be smooth, or it might have discontinuities and kinks, or it may have delta functions or infinite jumps. We’ll accept all of these different cases in our considerations, because potentials can be as strange as we can imagine depending on the problem that we’re trying to solve. (But there are worse cases, such as potentials discontinuous at every point or derivatives of delta functions, and we won’t think about those cases.) The key is that for each of the different possible kinds of behavior for V (x), we want to understand the boundary conditions that are being imposed on our solution ψ(x). What we really care about is properties of ψ′ and ψ, because those are what allow us to stitch together solutions at interfaces and discontinuities of V . Proposition 86 For a potential V (x) as described above, we have the following: • The wavefunction ψ(x) must be continuous at all x. • ψ′(x) is continuous unless V (x) contains a delta function. Proof. For the first point, if ψ(x) had a discontinuity at some point, then ψ′(x) would contain a delta function, and thus ψ′′(x) would contain the derivative of the delta function. But the right-hand side (involving V (x)) can only have a delta function at worst, so this is not allowed. For the second point, we’re equivalently saying that ψ′ is discontinuous only if V (x) has a delta function. Notice that ψ′(x) is discontinuous only if ψ′′(x) has a delta-function in it; since ψ itself is continuous (by our above argument), the only way for 2m ℏ2 (V (x) −E)ψ(x) to have a delta-function in it is if V (x) does. With this, we’re finally ready to move on to solve the Schrodinger equation in some particular cases. Our first example is one that we’ve already briefly touched on: Example 87 (Particle on a circle) Consider a particle on a circle of total circumference L – equivalently, we can think of solving the Schrodinger equation on the real line with the additional condition that points x and x + L are identified with each other. To make life simple, we’ll have zero potential V (x) = 0. In other words, x = 0, L, 2L, 3L, · · · all correspond to the same point on the circle, and it’s equivalent to think of a circle as “infinitely many copies” of whatever happens between 0 and L also occuring between L and 2L, 2L and 3L, and so on. Thus, we have the boundary condition ψ(x + L) = ψ(x). Because we have zero potential, our Hamiltonian is ˆ H = −ℏ2 2m d2 dx2 , and the time-independent Schrodinger equation we 51 must solve is −ℏ2 2m d2ψ dx2 = Eψ ⇐ ⇒ d2ψ dx2 = −2mE ℏ2 ψ . We claim that all solutions to this equation must have E ≥0. To see this, notice that if we “multiply both sides by R L 0 ψ∗dx,” we have −ℏ2 2m Z L 0 ψ∗(x) d dx d dx ψ(x) = E Z L 0 ψ∗(x)ψ(x)dx. Assuming that ψ is a well-normalized solution, the right-hand side simplifies to E, and now typically we will apply integration by parts to the left-hand side. But we’ll do it slowly here: we can verify by the product rule that we can rewrite as −ℏ2 2m Z L 0 d dx ψ∗dψ dx −dψ∗ dx dψ dx dx = E. The first term in the bracket is now a total derivative, so integrating that term from 0 to L (and using that dψ∗ dx is the conjugate of dψ dx ) gives us −ℏ2 2m ψ∗dψ dx L 0 + ℏ2 2m Z L 0 dψ dx 2 dx = E. But because we’re on a circle here, the points L and 0 are the same point, and thus the boundary term vanishes! (Importantly, we cannot make arguments about ψ going to 0 at the boundary here – we are using the geometry of the problem.) Thus, we find that E is the integral of some positive quantity, and thus the energy is always nonnegative, as claimed. We’ll finish solving this differential equation next time, but for now we’ll just write down a few solutions: looking back at the boxed equation above, we’ll make the definition −2mE ℏ2 = −k2, k ∈R. We know that the left-hand is indeed negative, so this is a valid definition of the constant k, and furthermore we can rearrange to find that E = ℏ2k2 2m , which is p2 2m if p = ℏk. So our notation is actually very good here – the constant k that we’ve defined is actually such that ℏk is the momentum of the particle! So all that’s left to solve is d2ψ dx2 = −k2ψ, and this is the usual wave equation, solved by sines and cosines or by exponentials. For convenience, we’ll use solutions of the form ψ = eikx, and next time we’ll see which values of k work and how to normalize these wavefunctions! Lecture 11: The Infinite and Finite Square Well Today’s lecture is dedicated to solving the Schrodinger equation for certain potentials V (x), beginning to give us insight into the kinds of solutions that we will see in various physical settings. We’ll start by finishing the particle on a circle problem, and then we’ll discuss the infinite and finite square wells. Recall from last time that a particle on a circle can be represented by a particle on the real line, where we identify x with x + L for any real number x. Thus, we want wavefunctions that are periodic mod L (that is, ψ(x + L) = ψ(x) and thus the same is true for the derivative ψ′), and last time we found that for zero potential V (x) = 0, solutions are of the form ψ(x) ∼eikx, where k2 = 2mE ℏ2 for some E ≥0. 52 Our next step is to apply the periodicity condition: since ψ(x + L) = ψ(x), we must have eik(x+L) = eikx = ⇒eikL = 1 = ⇒kL = 2πn for some integer n. We’ll index the allowed values of k as kn = 2πn L , and from this we find that the nth momenta and energy levels are pn = ℏkn = 2πℏn L , En = ℏ2k2 n 2m = ℏ2 2m · 4π2n2 L2 = 2π2ℏ2n2 mL2 . To find our energy eigenstates, we now just need to find the normalization constants for ψn(x) ∼eiknx. And here’s where it’s nice that our particle is on a circle rather than on a line: exponentials like this have \|eiknx\|2 = 1, so we would not be able to normalize the integral over all real numbers, but we can normalize the integral over the range [0, L]. Explicitly, the way we do this is to write ψn(x) = Neiknx, and then require 1 = Z L 0 \|ψn(x)\|2dx = Z L 0 N2dx = LN2 = ⇒N = 1 √ L. (Note that we can choose N to be positive real – an overall phase doesn’t change the state.) This gives us our final answer: ψn(x) = 1 √ Leiknx = 1 √ Le2πinx/L . Thus, we’ve found the energy eigenstates for the time-independent Schrodinger equation, and to obtain our stationary states, we must add in the time phase: for each n, we have Ψn(x, t) = ψn(x)e−iEnt/ℏ. One important point to address here is that as we’ve stated so far, n can be any integer (positive, negative, or zero). Indeed, the momentum of these states, 2πℏn L , are different for all values of n, so they must be different states (and we don’t have any additional degeneracy where different values of n are actually the same state). We might also be suspicious about n = 0, but that just corresponds to the wavefunction ψ0 = 1 √ L, which has no x-dependence and thus no energy or momentum – this is all consistent with the Schrodinger equation we are trying to solve. However, notice that states ψn and ψ−n have the same energy En = E−n, so they are in fact degenerate energy eigenstates. The important concept to keep in mind is the following: Fact 88 If we have two energy eigenstates of the same energy, there must be something physical that distinguishes them, and we must figure out what that is to get a full picture of the physics of our system. In our case, we’ve already figured that out – states ψn and ψ−n have different momenta pn and −pn (informally one represents the “particle moving to the right” and the other represents the “particle moving to the left”). And this additional physical quantity is also good for us because it tells us that ψn and ψ−n are orthonormal – even though they are degenerate in energy, they are eigenstates of the Hermitian operator ˆ p of different momenta, so their inner product is indeed zero. Remark 89. If we hadn’t known about the momentum, we would potentially have had to do some additional work with each ψ−n and ψn, finding linear combinations until we get two orthonormal states of energy En. And if we had tried to use sines and cosines instead of exponentials when we solved this problem, that’s what we would have encountered. Putting everything together, we’ve thus found an orthonormal set of states which spans our time-independent 53 wavefunctions. So if we’re given any periodic ψ(x) (any wavefunction on the circle), we can write ψ(x) = X n∈Z anψn(x). Furthermore, notice that ψk + ψ−k ∼cos(kx) and ψk −ψ−k ∼sin(kx), so we can always rewrite our wavefunction ψ(x) as a sum of sines and cosines as well. Either way, this is secretly getting back to Fourier’s theorem and Fourier series: any periodic function can be written as a sum of sines and cosines, or as a sum of appropriate exponentials. We’re now ready to move on to our next problem of the day: Example 90 (Particle in a box) Suppose we have a particle in one dimension constrained to the segment x ∈[0, a], where there are hard walls to the left and right of this segment that the particle cannot cross. In other words, we have V (x) =          ∞ x ≤0, 0 0 < x < a, ∞ x ≥a. To solve this problem, first notice that if the potential is infinite outside the interval [0, a], then the wavefunction must be zero there as well. Informally, this is because being in an area of infinite potential requires infinite energy – if we want to be more rigorous with that, we can solve the problem with a finite potential outside [0, a] and take that potential to ∞, and we will indeed see the finite square well solution later in this lecture. But for now, let’s just assume ψ(x) = 0 for x < 0 and x > a. Furthermore, we can see that it does not really matter whether the potential V is zero or infinite at the endpoints – since we require that ψ is continuous by Proposition 86 we must also have ψ(0) = ψ(a) = 0 by continuity. So it suffices now to solve the free Schrodinger equation in the region [0, a], and this is very similar to what we had before for the particle on a circle: we have ψ′′(x) = −2mE ℏ2 ψ(x) = −k2ψ(x), and this time it’s more convenient to use sines and cosines instead of exponentials: ψ(x) = c1 cos(kx) + c2 sin(kx). This is because we can now impose boundary conditions – since 0 = ψ(0) = c1, the first term goes away, and then since 0 = ψ(a) = c2 sin(ka) and we can’t also have c2 = 0 (that would be a zero wavefunction), we must similarly have ka = nπ = ⇒kn = πn a , but this time we won’t use all integers n – first of all, n = 0 makes the wavefunction vanish, which is not allowed. (It was allowed for the exponential because n = 0 gave us a nonzero constant there, but having the wavefunction be identically zero means we have no probability of finding the particle anywhere.) In addition, taking kn or k−n gives us the same wavefunction up to a sign, because sin(knx) = −sin(−knx). Thus, we instead index our particle-in-a-box states by integers n ≥1 , and all that’s left is to normalize. Writing ψn(x) = N sin nπx a , we have 1 = Z \|ψn(x)\|2dx = N2 Z a 0 sin2 nπx a dx, 54 and we can either do the integral using double-angle identities or note that the average value of sin2 over any half-period is 1 2, so that 1 = N2 · a 2 = ⇒N = q 2 a. This gives us our final equation for the energy eigenstates: ψn(x) = r 2 a sin nπx a , En = ℏ2k2 n 2m = ℏ2π2n2 2ma2 . This time, every energy state (for n ≥1) has different energy, so there are no degeneracies – in fact, the energy levels become more spaced out as n gets larger. We’ll now extract a few more properties of these infinite square well eigenstates, particularly looking at nodes and symmetries. Here are some sketches of the first 4 (smallest) energy eigenstates: 0 a 0 a 0 a 0 a Notice that each time we add 1 to n, moving up to the next energy eigenstate, the argument of the sin at x = a increases by π, meaning that we gain another half-wavelength per energy level. We call ψ1 the ground state, because it has the lowest energy out of any eigenstate, and we can notice that the ground state has no nodes (that is, no points in the interior, not including the endpoints or infinity, where the wavefunction is zero). Then the next excited state has one node, the one after that has two nodes, and so on. This behavior with nodes is actually a general result for potentials with bound states (that is, normalizable states that decay to zero) – the number of nodes will increase with the energy of the eigenstate. We won’t do a very rigorous proof of this result in this class, but we will see some strong evidence of it through many of our examples. Additionally, we can also pay attention to the symmetry of our energy eigenstates. For simplicity in setting up notation, we set up our infinite well between x = 0 and x = a, but it would have been potentially more enlightening to set it up between x = −a 2 and x = a 2 – in that case, we would have a potential V (x) which is symmetric around x = 0. Correspondingly translating the four energy eigenstates above, notice that the first and third energy eigenstates are symmetric around x = 0, while the second and fourth are antisymmetric! This is also a general fact – if we have bound states in a symmetric potential V (satisfying V (−x) = V (x)), then all energy eigenstates are either odd or even. (We’ll prove this in our next lecture.) Finally, there’s one more important property about bound states in a one-dimensional potential: there are no degeneracies for bound states as long as our potential spans the whole real line or has hard walls. (On the other hand, this assumption did not hold for the particle on a circle because of the identification we performed, so that’s why degeneracy is possible in that case.) Basically, the point of covering the infinite square well is for us to start seeing these kinds of general behaviors for wavefunctions satisfying the Schrodinger equation, and we’ll go into more detail about them in the coming lectures. 55 With that, we’ll turn to our final example of the lecture, where for the first time we won’t be able to write down the solutions directly and will sometimes need to use numerical methods. This time, we’ll set it up so that we have a symmetric potential: Example 91 (Finite square well) Suppose we have a particle in one dimension with a dip between x = −a and x = a: more specifically, our particle is placed in the potential V (x) =          0 x ≤−a, −V0 −a < x < a, 0 x ≥a. for some constant V0 > 0. We will look for bound states, which are normalizable solutions that tend to remain in a particular region (definition to come next lecture). In particular, thinking of this potential classically, if our state is at energy E < 0 and is localized at the origin (inside the well), then it requires some additional energy to escape the well and get to an energy of 0. So what we’ll end up finding with the wavefunction is that a bound state (with energy between 0 and −V0) will have some small but nonzero probability of being found outside the well. Remark 92. It turns out that there are never any solutions to the Schrodinger equation if the energy E of our state is lower than any point in the potential (so in this case, there are no states of energy E < −V0). Recall that our Schrodinger equation can be written as ψ′′ = −2m ℏ2 (E −V (x))ψ, and this is the first time where we have a nonzero V (x) that we must consider when solving the equation. The total energy of a particle can be thought of as the potential plus kinetic energy, so the amount that E exceeds −V0 gives us the kinetic energy of the particle inside the square well. But with that analogy, we have “negative kinetic energy” for the particle outside the well, which is impossible classically. As we do this calculation, we’ll see how this all works out with the “leaking probability” outside the well, and we’ll see that if the barrier gets arbitrarily high, then the wavefunction will get closer and closer to the one for the infinite square well. We’ll now set up the problem and work towards finding the energy eigenstates. The difference between our energy E and the well potential energy −V0 is E −(−V0) = V0 + E = V0 −\|E\| (remember that we’re taking E < 0), and because our potential V (x) is piecewise constant, we can solve the Schrodinger equation in each region and patch the solutions together – in each case, we have an equation of the form ψ′′ = αψ for some constant α. Inside the well, we have α < 0, meaning that solutions will be trigonometric functions, while outside the well, we have α > 0, meaning that solutions will be real exponentials. And when we try to stitch those different forms together into a continuous wavefunction, we expect that we’ll get some quantization of allowed energies, because that was indeed the case in the infinite square well as well. We’ll use the result that symmetric potentials will always give us even or odd solutions for ψ(x), and we’ll first look for even solutions (where ψ(x) = ψ(−x)). The key to having a nice solution here is to make good use of unit-free numbers – we’ll make lots of definitions and it won’t look like we’re solving anything for a while, and then 56 suddenly we’ll see the solution pop up. For an even solution in the region −a < x < a, we’re asking for a solution to d2ψ dx2 = −2m ℏ2 (E −(−V0))ψ = −2m ℏ2 (V0 −\|E\|)ψ. We’ll define k2 = 2m ℏ2 (V0 −\|E\|) > 0, so that our differential equation becomes ψ′′ = −k2ψ = ⇒ ψ(x) = cos(kx) for −a < x < a , the key here being that we only have cosine (and not sine) because we assume our solutions are symmetric. We won’t normalize our solutions because they’re pretty messy (it’s not necessary for obtaining energy eigenstates anyway). Instead, we’ll normalize by fixing the constant of cos(kx) to be 1 – this gives us the form of the wavefunction inside the well, and now we turn to the wavefunction outside the well. Then because V (x) = 0 in this region, we have ψ′′ = −2mE ℏ2 ψ = 2m\|E\| ℏ2 ψ = κ2ψ, where we define the positive constant κ2 = 2m\|E\| ℏ2 . Our solutions are then exponentials: ψ(x) ∼e±κx, and thus we must have ψ(x) = Ae−κx for x > a, ψ(x) = Aeκx for x < −a for some positive constant A. Furthermore, notice that we have the constraining relation κ2 + k2 = 2mV0 ℏ2 , and now we can rewrite it to be unitless: multiplying through by a2 gives us k2a2 + κ2a2 = 2mV0a2 ℏ2 . (We shouldn’t lose track of our goal – we want to find the allowed energies E.) We’ll now define the unit-free constants ξ = κa > 0, η = ka > 0, z2 0 = 2mV0a2 ℏ2 , so that we have η2 + ξ2 = z2 0 . It might look like all we’ve done is traded k, κ for unitless constants η, ξ, but it turns out that z0 is an important quantity which encodes the physical parameters of our system (it’s large if the well is deep or wide and small if the well is shallow or narrow) and actually is all we need to calculate the number of bound states in our well. We’ll soon see that z0 being large gives us many bound states, while z0 being small gives us very few of them! Turning back to calculations, we know that our wavefunction ψ and its derivative ψ′ must be continuous at x = a (and also x = −a, but it’s sufficient to check the former because we know our solution is even). We thus require that cos(ka) = Ae−κa, −k sin(ka) = −κAe−κa. Dividing the second equation by the first and then multiplying by a on both sides gives us k tan ka = κ = ⇒ka tan ka = κa = ⇒ η tan η = ξ . Thus, our problem has now reduced to finding either η or ξ, from which all of the other constants in the problem will 57 follow (since we know z2 0 and a from the physical parameters). Notice that ξ2 = κ2a2 = 2m\|E\|a2 ℏ2 = 2mV0a2 ℏ2 \|E\| V0 = z2 0 \|E\| V0 = ⇒ \|E\| V0 = ξ z0 2 ; in other words, we can get a dimensionless measure of the energy (which essentially tells us the proportionality constant between the depth of the well and our energy) in terms of our newly defined constants! And now we’re ready for the final answer: if we graph η, ξ in their first quadrant (since we know they must both be positive), then the condition ξ = η tan η looks as shown below: π 2 π 3π 2 2π η ξ Then the other curve ξ2 + η2 = z2 0 is just the circle of radius z0 centered at the origin, and we want to find the points where these two curves intersect. For example, here’s the situation for z0 = 5 (it looks like an ellipse below because the axes are scaled differently): π 2 π 3π 2 2π η ξ And just looking at this graph, we can count the number of bound states by looking at how many points to intersection we have, which is given by the number of multiples of π smaller than z0 (because each of those blue branches gives us one intersection). In particular, notice that there will always be a bound state for any positive value of z0! Furthermore, because z0 is fixed and we have the boxed relation between ξ and z0 above, the leftmost solution corresponds to the most deeply bound state (and thus the state of lowest energy), and going from left to right gives us the states in increasing order of energy. We’ve thus solved the finite square well for the even solutions case – a similar solution method works for the odd solutions case, but in that case we have ξ = −η cot η: 58 π 2 π 3π 2 η ξ Thus, the odd solutions don’t always exist – the potential V0 must be sufficiently deep for us to have them, and with a bit more inspection we can see that the even and odd solutions will interleave for the finite square well. The main lesson that we should learn here is the power of unit-free constants and how they can tell us when solutions exist! Lecture 12: Qualitative Properties of Eigenstates, Local Picture of the Wavefunction, and the Shooting Method Last lecture, we computed the energy eigenstates for a few special potentials V (x). In general, if we’re faced with an arbitrary potential, we likely want to know the key features of the resulting eigenstates, and we’ll explore those kinds of qualitative considerations today. We’ll start with three key properties of wavefunctions that are inspired by some of the examples from last lecture: Definition 93 A bound state is a normalizable energy eigenstate over the real line (meaning that ψ →0 as x →±∞). The idea is that if a wavefunction goes to 0 at infinity, then it has a bump in the middle at some point, so the potential is “keeping the particle bound.” Proposition 94 For one-dimensional potentials over the real line, there are no nondegenerate bound states. (This is left as an exercise for us – it’s a problem in Griffiths.) The second result is related to the fact that we were able to construct our wavefunctions to be real-valued, even though generally the full wavefunction is complex-valued. Essentially, just like we saw in last lecture, often we can encode all of the complex-valued behavior in just the phase e−iEt/ℏ, because the time-independent Schrodinger equation ˆ Hψ = Eψ has no i in it: Proposition 95 Energy eigenstates of the time-independent Schrodinger equation, where V (x) is real, can be chosen to be real. What this result says is that in general, we can always have the possibility to work with real solutions for our energy eigenstates. 59 Proof. Consider the Schrodinger equation ˆ Hψ = Eψ. If we take the complex conjugate of both sides, we obtain ˆ Hψ∗= Eψ∗(because the potential V (x) is real, we can check that ( ˆ Hψ)∗= ˆ Hψ∗). Thus, ψ∗and ψ are two energy eigenstates of the same energy. If they are linearly independent, then taking ψr = 1 2(ψ∗+ ψ) and ψi = 1 2i (ψ −ψ∗) gives us two new real, linearly independent energy eigenstates of the same energy (which we can work with instead). And if ψ∗and ψ start off linearly dependent, then we just have one unique energy eigenstate, and at least one of those linear combinations is nonzero. Thus we can always form enough new energy eigenstates that are real as long as V (x) is real. Essentially, these linear combinations of ψ and ψ∗, which we denoted ψr and ψi, give us the real and imaginary parts of ψ, and both of those parts turn out to be solutions to the (time-independent) Schrodinger equation. Corollary 96 All bound states in one-dimensional potentials are real up to a phase. Proof. Proposition 95 tells us that all energy levels are nondegenerate. Following the proof of Proposition 95, ψ and ψ∗must then be proportional to avoid degeneracy. Thus the real and imaginary parts are also proportional to each other, so the full wavefunction ψ is just a complex number times its real part (which is a real wavefunction up to a phase). This is a strong statement – we’re essentially forced to work with real solutions for one-dimensional potentials, and the only complex part of the full wavefunction comes in the time-evolution term e−iEt/ℏof our stationary states. Our third result will be the one that we used last lecture to solve for energy eigenstates of the finite square well: Proposition 97 If a one-dimensional potential V (x) is even (that is, V (x) = V (−x)), then the energy eigenstates can be chosen to be either even or odd (under x 7→−x). Proof. We wish to solve the equation ψ′′(x) + 2m ℏ2 (E −V (x))ψ(x) = 0. (Here, note that ψ′′(x) denotes the second derivative of ψ, evaluated at x.) We essentially wish to show that ψ(−x) solves the same equation, and we’ll make sure we’re clear with what’s going on when we make that substitution. Let φ(x) = ψ(−x), so that the chain rule gives dφ(x) dx (x) = ψ′(−x) · −1 = −ψ′(−x), d2φ dx2 (x) = −ψ′′(−x) · −1 = ψ′′(−x). Thus, evaluating the original time-independent Schrodinger equation at -x yields ψ′′(−x) + 2m ℏ2 (E −V (−x))ψ(−x) = 0, and now replacing ψs with φs and noting that V (−x) = V (x) yields φ′′(x) + 2m ℏ2 (E −V (x))φ(x) = 0, which is the same Schrodinger equation but for φ. Thus, both ψ(x) and φ(x) = ψ(−x) are solutions to the Schrodinger 60 equation with the same energy, so we can form the symmetric and antisymmetric parts of the wavefunction ψs(x) = 1 2(ψ(x) + ψ(−x)), ψa(x) = 1 2(ψ(x) −ψ(−x)), which are even and odd, respectively, and by superposition they are also energy eigenstates of the same energy E. Thus we can use these symmetric and antisymmetric wavefunctions instead (and regardless of whether ψ and φ are linearly dependent or not, doing this operation preserves that fact). Corollary 98 All bound states in one-dimensional potentials are either even or odd. Proof. Again, Proposition 95 tells us that all energy eigenstates are nondegenerate, so the only choice we have for our eigenstates is multiplying by a constant (which does not change whether a function is even, odd, or neither). Thus the result above directly applies. More specifically, ψ(x) and ψ(−x) must be proportional, and we can assume by Corollary 96 that ψ(x) (and thus ψ(−x)) must be real. But if ψ(−x) = cψ(x), then ψ(x) = cψ(−x) by replacing x with −x in the equation above. Thus ψ(x) = c2ψ(x), and thus c = ±1, corresponding to even and odd wavefunctions respectively. This explains why our search for solutions of the finite square well last lecture was divided into two cases (symmetric and antisymmetric) – there are no other energy eigenstates that are possible. We’ll be using this result in future lectures as well! Now that we’ve established some general results, we’ll turn to some qualitative insights of our (now established real) energy eigenstates. Whenever we have a problem with a potential V , we have a total energy E = K + V , and in our classical intuition energy is conserved, so both K and V are functions of x, but E is some fixed number. (And as V gets larger, K gets smaller, and vice versa.) Classically, if we know the kinetic energy of a particle, then we know its momentum (because K = p2 2m). But we know that the corresponding de Broglie wavelength of the particle is λ = h p, so given that the particle has some kinetic energy K, we should expect some kind of “wavelength” of λ to pop up in the wavefunction for our energy eigenstate. And all of that discussion is exactly true if our potential V is constant, because that’s exactly how we derived the Schrodinger equation in the first place. Example 99 Suppose we have a linearly growing potential V (x) and some total energy E, so that as the particle moves to the right, its potential energy increases and thus its kinetic energy decreases. Having the kinetic energy K decrease means that the momentum p gets smaller and that the “wavelength λ gets larger” (at least, we can say this if the potential is not growing too quickly, since that means V is approximately constant. Specifically, we can define a K(x), p(x), and λ(x) in terms of the position of our particle x, and that means we predict that the wavefunction will oscillate more and more slowly as x increases. The exact formula for the solution to the Schrodinger equation in a linear potential relies on Airy functions and so on, so everything we’re saying is only approximate but still useful (and if we plot the wavefunctions that’s indeed what we’ll see). Example 100 Consider a potential V (x) as in the diagram below – we’ll try to extract some more qualitative features of the wavefunction, much like we’ve described above. 61 V (x) E K(x) Classically, we know that the kinetic energy of a particle (represented by the black dashed line) cannot become negative, so the particle cannot move past either intersection point where V (x) = E. (Thus, there are regions on the left and on the right which is classically forbidden.) Those intersection points are called turning points, because in a classical setting those are the positions where (if we imagine a ball rolling around in the potential) the particle would stop and start turning back. What we discussed above is basically that if our potential V were constant, then our wavefunction would take on a very simple form with a fixed wavelength λ. But if V (x) varies slowly enough – that is, if the percentage change in V (x) is small compared to the relevant distance λ(x) – then we can speak of a “local wavelength.” This is actually getting to the discussion of the WKB approximation, which we discuss in 8.06, and the requirement is that λ(x)dV dx ≪V (x), where λ(x) = h p(x) = h p 2mK(x) is the local de Broglie wavelength of the wavefunction at a position x. Really, all we’re saying is that if we were to sketch the wavefunction ψ(x) corresponding to the potential V (x) above, we’d generally have a faster oscillation where V (x) is smaller and a slower oscillation where V (x) is larger. On the other hand, we can also discuss the amplitude of the wavefunction at these different positions, and this is connected to the correspondence principle. Basically, if a particle spends more time in some region, then the wavefunction should have larger magnitude there, so in the example above, we should spend more time where V (x) is larger (because classically that corresponds to a smaller K(x) and thus a smaller velocity for the particle). This turns out to be true, but we’ll explain it in a bit more detail. The probability for a particle to be found in a region [x, x + dx] is \|ψ(x)\|2dx, which is essentially proportional to the fraction of time dt T spent in this interval (where T is the total time in a period, if we imagine the particle rocking back and forth classically). Thus, we have \|ψ\|2dx ∼dt T = dx v(x)T = ⇒\|ψ\|2 ∼ 1 p(x) = λ(x) h , so \|ψ\| ∼ p λ(x) – that is, in areas where the wavelength is longer and the potential V (x) is larger, we will have a larger amplitude for the wavefunction ψ(x). Combining this with our previous discussion, as well as the node theorem from last lecture (telling us how many times ψ intersects zero), we should now be able to estimate the wavelength and amplitude of a wavefunction, giving us a general sketch of ψ, without explicitly solving the Schrodinger equation. We’re now ready to get the full local picture of the wavefunction: we can rewrite the time-independent Schrodinger equation as 1 ψ d2ψ dx2 = −2m ℏ2 (E −V (x)). (Note here that ψ is an energy eigenstate of energy E.) We then have three cases to think about when trying to describe what ψ looks like (and remember that we can always treat it as real-valued): 62 • When E −V (x) < 0 (the classically forbidden region), the right-hand side is positive, so ψ and d2ψ dx2 always have the same sign (unless ψ is zero). In other words, ψ is always convex towards the axis – if ψ is positive, then we have a “parabola going up,” and if ψ is negative, then we have a “parabola going down.” But since we’re in a classically forbidden region, the wavefunction should not get arbitrarily big – instead, what’s often actually happening is that our wavefunction is asymptotically tending to zero as x →∞and x →−∞(and thus the probability of finding the particle deeper into the classically forbidden region decays to zero as well). • when E −V (x) > 0 (the classically allowed region), the right-hand side is negative, so ψ and d2ψ dx2 have different signs. We can’t say anything asymptotically for this kind of behavior, but we should essentially imagine a sine function in the classically allowed region (downward parabola for ψ positive and upward parabola for ψ negative); basically, ψ is always concave towards the axis. • When E −V (x) = 0 (one of the turning points), the right-hand side is zero, so either ψ = 0 or d2ψ dx2 = 0. This corresponds to having inflection points where ψ is nonzero (though notice that when ψ is zero, we also automatically have a point where the second derivative is zero). Thus, the key insight here is to remember that inflection points of occur at turning points and nodes. Example 101 We’ll finish this lecture by returning to a generic smooth symmetric potential V (x) and trying to understand its energy eigenstates. Suppose that V (x) looks as shown below, and as always we want to find bound states with some energy E < 0. We’ve labeled some example energies on the diagram below: V (x) E3 E2 E1 E0 Because our wavefunction will always hit the classically forbidden region for large enough \|x\|, we know that ψ will decay asymptotically to zero on both the left and right side. Furthermore, we can choose ψ to be real-valued, and we can multiply it by a sign so that ψ is positive for large positive x (but decaying to 0), but depending on whether ψ is symmetric or antisymmetric, it may be negative or positive for large negative x. We’ll try to get an understanding for why a potential of this form naturally gives us energy quantization by sketching the wavefunctions that would arise if we tried to form eigenstates at energy E0, E1, E2, and E3 by numerically integrating from −∞and ∞inward to 0. Each energy eigenstate will decay to zero at the endpoints, but between the turning points it will be concave towards the axis. For low energies, the turning points are close to the y-axis, and thus there isn’t enough time for ψ to turn all the way back to 0. This means we must have an even wavefunction, but the turning must happen in such a way that the wavefunction does not have a discontinuity in the derivative (because V is continuous). So it’s possible that in this example, the wavefunction formed by E0 will not work as an energy eigenstate because of a kink, but the one formed by E1 will, as shown below. (Turning points are labeled in black.) 63 E0: E1: In other words, as we slowly increase the energy E, we will finally get to a point where the curve flattens at x = 0, and that’s our first energy eigenstate. If we continue to increase our energy, the turning points will move further out, and we will have even more time to curve back towards the x-axis. This means that we again will not have a valid wavefunction until the ψ we obtain hits the origin – at that point, we can obtain an antisymmetric solution as shown with the energy E3 below. E2: E3: So the second energy eigenstate occurs when the curve matches up to form an antisymmetric wavefunction. Further increasing the energy will create more oscillation in the purple (classically allowed) region, in such a way that for some higher energy E, ψ matches up with continuous derivative after two nodes. Repeating this process gives us higher and higher energy eigenstates – this is essentially the intuition for the node theorem and also for why energy is quantized! Example 102 We can connect this argument to one other piece of intuition through the shooting method for solving differential equations. If we have a symmetric potential and we’re looking for even energy eigenstates, we can do the following procedure: pick some energy E0, and require ψ(0) = 1, ψ′(0) = 0. (We don’t need to normalize our eigenstates, so we’ll use this as the rescaling instead, and the reason for zero derivative is, just like above, to ensure that we have no kinks in ψ.) Since we have a second-order differential equation, these boundary conditions allow us to numerically integrate and find the wavefunction ψ for all x. What we’ll find typically is that after a while, ψ will start to blow up to ±∞for large x, which is bad because that means we won’t have a normalizable energy eigenstate. But if we change E slightly, we might find that ψ now blows up in the opposite direction ∓∞for large x. Then what that means is that in between our two values of E, there’s a single point at which the wavefunction will decay, and that’s the allowed energy eigenvalue for our potential V (x)! Fact 103 Repeatedly searching in a smaller and smaller interval gives us the energy E to any arbitrary accuracy that we want, as long as Mathematica can still integrate for us, and the only thing we need to do is clean up our differential equation so that there are no units. 64 Lecture 13: The Delta Function Potential, Node Theorem, and Harmonic Oscillator We’ll start today by solving the Schrodinger equation for a new potential: Example 104 Suppose we have a particle in one dimension placed in the potential V (x) = −αδ(x) for some α > 0 (this can be represented with a thick arrow pointing in the downward y-axis direction). Essentially, we should imagine that we have a potential which is “infinitely negative” at x = 0, or equivalently the limit of a square well that gets deeper and narrower while keeping the area the same (which is in fact one way to analytically calculate the energy levels of this potential). Just like with the other potentials, we would like to calculate whether there are bound states (in this case, this means E < 0) and what their energy eigenvalues are for the corresponding Schrodinger equation. It turns out we can discover a lot without having to do explicit calculations with the differential equation, and we do so by using our intuition and some of the discussions we’ve been having in the past few lectures. Our first approach will be using units: notice that the three constants that are present in the problem are α, m, and ℏ. If there’s only one way to construct a quantity with units of energy using these constants, then all energy eigenvalues must be proportional to that quantity. When three constants have units that are not “linearly dependent,” that is indeed possible – we build objects with units of length, mass, or time, and then from that we can do anything. Because δ(x) has units of 1 [L] (remember that integrating the delta function over the real line gives us a unitless constant), and V has units of energy, we must have [α] = [E] · [L] = ⇒[α] [L] = [E] = [p]2 [m] = [ℏ]2 [m][L]2 , so that ℏ2 mα has units of length and thus our energy scale can be written as [E] = [α] [L] = [m][α]2 [ℏ]2 . This means that because the bound state energy must be in the forbidden region E < 0 throughout x, we therefore predict that Eb = −#mα2 ℏ2 for some positive number #, and there can’t be any other possibilities just from our unit considerations! And we should expect that this number should not be particularly large or small – for the problem to be “natural,” we should expect # on the order of 1. But we’ll be more precise about that in a second. Another way we can think about this potential is to consider the regularized version of the delta function, which is basically a very deep and very narrow finite square well. Recall that because this is a symmetric potential (because there’s nothing asymmetric about the delta function), the ground state should be even and have no nodes – refer to the pictures under Example 101 for illustration. But as the wavefunction in the classically allowed region becomes narrower and narrower, we expect the “curving back” in that region to occur more and more rapidly, so that in the limit we actually get a discontinuity in the derivative of the wavefunction ψ. With this, we can now write down the differential equation for the delta function potential (though we’ll still not solve it): for x ̸= 0, we have no potential, 65 so −ℏ2 2mψ′′ = Eψ = ⇒ψ′′ = κ2ψ for x ̸= 0, where κ2 = −2mE ℏ2 > 0 just like in the finite square well. Such a differential equation has solutions e±κx (as we already know), or equivalently cosh(κx) and sinh(κx) if we prefer. But we can think now about how many bound states this potential will have – if there’s one of them, it will be even and have no nodes, and if there’s a first excited state after that, it must be odd and vanish at the origin (so that it has one node). That occurs only with a function like ψ(x) ∝sinh(κx), but that’s not good because the function is convex away from the axis even in a forbidden region. So there can only be at most one bound state, and now we’re ready to solve the problem algebraically. Because our ground state needs to be even, we must have ψ(x) = Ae−κ\|x\| for x ̸= 0 , so that our function decays as \|x\| →∞, and this seems to be on the right track because it’s what we expect from a ground state of the form in Example 101 with a very narrow middle region. Our wavefunction is indeed continuous, and now we just need to use the fact that the delta function is of intensity α. (It’s good that our energy scale given in Eb has α in the numerator – indeed, as the potential gets stronger, the bound state should get deeper, so we have some reasonable conditions.) This next part is important because we’ll do it again and again in this course: the Schrodinger equation with the potential term is −ℏ2 2m d2ψ dx2 + V (x)ψ(x) = Eψ(x), and now we integrate both sides from −ε to ε (where eventually we’ll be looking at the limit ε →0) to get −ℏ2 2m dψ dx x=ε −dψ dx x=−ε + Z ε −ε −αδ(x)ψ(x)dx = E Z ε −ε ψ(x)dx, because the integral of the second derivative is the first derivative. Now taking a limit, we have −ℏ2 2m lim ε→0 dψ dx x=ε −dψ dx x=−ε −αψ(0) = lim ε→0 E Z ε −ε ψ(x)dx = 0, because ψ is not divergent and we’re shrinking the limits of integration to a single point. Thus, rearranging this equation gives us what we wanted – we get the differences between the derivatives at 0+ and 0−, which allows us to see the discontinuity in ψ′, which we’ll denote ∆0ψ′: −ℏ2 2m∆0ψ′ −αψ(0) = 0 = ⇒ ∆0ψ′ = −2mα ℏ2 ψ(0) . In other words, as long as the wavefunction doesn’t vanish, it can have a discontinuity due to a delta function, and in fact the change in derivative is proportional to the value of ψ and also to the strength of the delta function α. And returning to our wavefunction, notice that the derivative of our boxed ψ(x) above has discontinuity −2mα ℏ2 A = −2mα ℏ2 ψ(0) = ∆0ψ′ = lim ε→0 −κAe−κε −κAe−κε = −2κA, so that the 2As cancel (we should not expect A to show up because it’s just a normalizing factor in a linear equation), and we find κ = mα ℏ2 . And this is good, because we’ve now specified the energy of our bound state E = −ℏ2κ2 2m = −ℏ2 2m m2α2 ℏ4 = −mα2 2ℏ2 , 66 and indeed the number has been determined to be # = 1 2. And with that, we’ve learned that a single delta function will give us a single bound state – adding more delta functions will give us more bound states, as we’ll examine ourselves. Example 105 We’ll now spend a few minutes talking more about the node theorem, which we’ve stated and used for some of our previous arguments. We won’t do a mathematically rigorous justification, but we’ll give the main intuition for why we should believe it to be true. Suppose we have bound energy eigenstates of a one-dimensional potential, labeled ψ1, ψ2, ψ3, · · · with respective energies E1, E2, E3, · · · in increasing order. We are claiming that ψn has (n −1) nodes, and we’ll understand this by thinking about continuity. Fact 106 Note that the wavefunction can never have ψ(x0) = ψ′(x0) = 0, because for a second-order differential equation the initial conditions ψ and ψ′ tell us everything and the general solution must be ψ = 0. So for any state ψ, the derivative at a node must be nonzero. The idea is that if we have some arbitrary potential V (x), we can truncate it to just the range [−a, a] and get the screened potential Va(x) =    V (x) \|x\| < a, ∞ \|x\| > a. In other words, we have an infinite well, but our potential is no longer flat inside the well. And as a →∞, the bound states of our screened potential will become the bound states of our original potential V (x), because all bound states decay to 0 and thus the potential makes “less and less of an impact” as we move the walls outward. And as a increases, the bound states evolve continuously, but for small a we basically have a very narrow infinite well where the potential is essentially flat. Thus, as a →0, we can use the states of the infinite potential, in which we have a ground state with no nodes, a first excited state with one nodes, and so on, satisfying all of the assumptions of the node theorem. Thus, all that we need to show is that as we increase the length of the screen, we cannot change the number of nodes continuously. Indeed, consider the derivative of the nth energy eigenstate of ψa at x = a. Introducing a new node as we increase a means that we go from approaching the boundary from below to above, or vice versa, meaning that ψa’s derivative at its endpoint must switch signs. Thus, there would need to be some intermediate value of a such that ψ′ = 0 at the endpoint. But because we have an infinite wall at that point, we also have ψ = 0, and by Fact 106 this cannot occur for a nonzero wavefunction. Thus, no new nodes can be created. The same argument works for having a node disappear from our wavefunction, or having a node appear in the middle of the region (which would imply a tangent point when the node first emerges). So at least intuitively, we see that the nth energy eigenstate will continue to have (n −1) nodes as a gets larger, and thus in the limit we have the desired node theorem. Our next step in this class is to look at a classical system that has a lot of deep theory, and we’ll be returning to it in the next few lectures as well. Recall that classically, we have a system with a total potential plus kinetic energy E = K + U = p2 2m + 1 2mω2x2, where ω = q k m for the spring constant k (so that the potential term can also be written as 1 2kx2) and x represents the amount that a spring is stretched from its equilibrium position. To invent a quantum version of this system, we 67 must come up with a Hamiltonian for the Schrodinger equation, and we do this in the purest possible way: Example 107 (Simple harmonic oscillator) Consider a one-dimensional system with Hamiltonian ˆ H = ˆ p2 2m + 1 2mω2ˆ x2, where ˆ x and ˆ p are operators satisfying [ˆ x, ˆ p] = iℏ. It will turn out that the quantization of energies in this problem is pretty interesting – even though the classical oscillator can oscillate with any amplitude, this is not the case for the quantum oscilator. This Hamiltonian corresponds to the quadratic potential V (x) = 1 2mx2, and the reason it is so important is that it is a good approximation for any potential around a local minimum (since by Taylor expansion, the first derivative vanishes), meaning that the logic here can be (and in fact is) applied to many different oscillatory systems, like an electron in a magnetic field or the motion of a diatomic molecule. There are two ways we can find the energy eigenstates of this oscillator – notice that because V is unbounded, all energy eigenstates are bound states for this system, which is not the case for something like the delta function potential. Our goal is to solve the equation ˆ Hφn(x) = Enφn(x) (where φn denotes the nth harmonic oscillator eigenstate), and we can do this either by solving the differential equation directly or by inventing certain clever raising and lowering operators which simplifies the problem to an algebraic trick. The latter of these will be covered more in 8.05, but we’ll get a solution to the harmonic oscillator problem with both methods. First, we write down the equation −ℏ2 2m d2φ dx2 + 1 2mω2x2φ(x) = Eφ(x), and we’ll first remove all of the units from the equation with a simple procedure: change the variable x into one with no units, which will simplify numerical simulation and also make the structure more illuminating. We do this by writing x = au for a unitless u and a constant a with units of length; here, we can use m, ω, ℏto define a via [E] = [ℏ]2 [m][a]2 = [m][ω]2[a]2 = ⇒ a2 = ℏ mω . Plugging in x = q ℏ mωu, we then get −ℏ2 2ma2 d2φ du2 + 1 2mω2a2u2φ = Eφ, and now mω2a2 and ℏ2 ma2 have units of energy and must turn out to be something nice when we substitute in the actual value of a2: indeed, we end up with the equation −1 2ℏω d2φ du2 + 1 2ℏωu2φ = Eφ. And finally, the equation looks nicest if we multiply by 2 ℏω to get −d2φ du2 + u2φ = 2E ℏω φ = Eφ, where E = 2E ℏω is the unit-free energy. Since knowing E and E are equivalent, we’ll find that working with E gives us 68 the nicest answer: we’re essentially left with solving the second-order differential equation d2φ du2 = (u2 −E)φ . Thinking back to the “shooting method” from last lecture, this equation will have solutions for all values of E, but most of them will diverge and not be normalizable. To understand this, we’ll consider the system for large x (that is, large u) and see what the solution must look like there. Since E is a constant, the limiting behavior of the differential equation looks like d2φ du2 = u2φ. The way to get this kind of solution (where we gain powers of u after differentiating) is not by using polynomials in u but by trying expressions of the form φ(u) = uke αu2 2 . Indeed, we find that because we’re just trying to get a rough order of approximation, we have φ′ ∼αuφ (the other term is comparably negligible) and thus φ′′ = (αu)2φ + (subleading terms). Thus, setting α = ±1 are likely to give us approximate solutions, because those get us back to the original differential equation. Going further, we thus expect that our large-\|u\| behavior will look like φ(u) ∼Auke−u2/2 + Bukeu2/2, and now we can see where we might run into trouble: we can’t have the second of these two terms, so we must have E chosen so that B = 0 is the desired solution. (And remember that everything here so far is still \|u\| →∞, so we don’t have any exact solutions, only general behavior for large u.) So without any loss of generality, we’ll write φ(u) = h(u)e−u2/2; we can always multiply back by eu2/2 to get to the original function, but this should encode some of the limiting behavior that we expect, and we hope that h(u) will not diverge too quickly (because then φ(u) will be normalizable) – specifically, now that we’ve isolated the divergence, we hope to see that h(u) is actually a polynomial, because then it is easier to see “when our solution ends.” Our next step is then to get the differential equation for h(u); it will turn out that we must solve d2h du2 −2u dh du + (E −1)h = 0 . Next time, we’ll dive more into this and understand how quantization helps us get to the final answer! Lecture 14: Eigenstates of the Harmonic Oscillator Last time, we took our harmonic oscillator Hamiltonian and simplified it to an equation with a unit-free coordinate u and a unit-free energy E, giving us the differential equation d2φ du2 = (u2 −E)φ. We then analyzed the large-\|u\| behavior of φ, and after making the definition φ(u) = h(u)e−u2/2 (to isolate the most serious divergence of φ), we arrived at a new differential equation d2h du2 −2u dh du + (E −1)h = 0 , which we’ll work towards solving today. To start, we’ll attempt a series expansion of the form h(u) = ∞ X k=0 akuk. 69 We can proceed by directly plugging this into the differential equation, but that involves reindexing sums and can be a bit complicated. Instead, we can just consider the terms with ujs: we must start with a uj+2 in the first term, a uj in the second term, and a uj in the third term, so we find that for all j ≥0, we require (j + 2)(j + 1)aj+2uj −2jajuj + (E −1)ajuj = 0 = ⇒(j + 2)(j + 1)aj+2 −2jaj + (E −1)aj = 0. (Specifically, we’re using the fact that a power series of u can only be identically zero if each term in P ajuj is zero.) In other words, the coefficients of our power series must be related as (j + 2)(j + 1)aj+2 = (2j + 1 −E)aj = ⇒aj+2 = 2j + 1 −E (j + 2)(j + 1)aj, giving us a recurrence relation for the ajs. So now we can start with a0 and successively obtain a2, a4, a6, and so on, which gives us an even solution h(u), or we can start with a1 and successively obtain a3, a5, a7, and so on, giving us an odd solution. (Notice that specifying a0 and a1 is just like specifying the value of h(u) and h′(u) at u = 0, so it makes sense that those two values are all we need to get the full solution of the differential equation.) But now we must ask what happens to these coefficients aj, because as written the solution may go on forever. Then we have aj+2 aj = 2j + 1 −E (j + 2)(j + 1) ∼2j j2 = 2 j , so the coefficients do decay. But they aren’t decaying fast enough – notice that for example that eu2 = ∞ X n=0 1 n!u2n. Looking at the even j = 2n, we have terms of the form 1 (j/2)!uj, so the coefficients cj = 1 (j/2)! satisfy cj+2 cj = 1 j+2 2 = 2 j+2 ∼2 j (no issues with fractional factorials because j is even). So the point is that if we don’t truncate the recursion relation at some point, our wavefunction will actually be unbounded, because h(u) ∼eu2, meaning that φ(u) ∼eu2/2 (the “safety factor” isn’t enough to make our normalization work). So the differential equation cannot work with arbitrary energies – the normalization requirement now quantizes the energies for us, because the only way we can make this work is to have aj+2 aj = 0 for some j, meaning that E = 2j + 1 for some j ∈Z≥0 , and for whichever value of j that is picked, we will have aj+2 = 0 (because the numerator in the recurrence is then 0). Even for j = 0, we’ll have a nonzero a0 but not a2, and that gives us a nontrivial wavefunction as well. So putting everything together, our wavefunctions are of the form h(u) = anun + an−2un−2 + an−4un−4 + · · · for some nonnegative integer n (an even function for even n and an odd function for odd n – remember that energy eigenstates must always be either even or odd), and those correspond to energies of E = 2n + 1 and thus an energy of E = ℏω 2 (2n + 1) = ℏω n + 1 2 . So we’ve arrived at a famous fact: the energy levels of a harmonic oscillator are evenly spaced, except there is an offset of 1 2ℏω so that the ground state is already a little above E = 0. 70 Definition 108 The Hermite polynomial Hn(u) is the polynomial satisfying the Hermite differential equation d2 du2 Hn(u) −2u dHn du + 2nHn(u) = 0, with normalization chosen so that Hn(u) has leading term 2nun. (We know that the leading term is of degree n, and because the Hermite differential equation is linear, we can always scale as we wish.) A few small cases that we can keep in mind are H0(u) = 1, H1(u) = 2u, H2(u) = 4u2 −2, H3(u) = 8u3 −12u. It turns out that we have the generating functional e−z2+2zu = ∞ X n=0 zn n! Hn(u) which we can use to compute Hermite polynomials by expanding out the left-hand side and collecting terms by powers of z. And with this formula, we can indeed see that Hn(u) begins with 2nun, and showing that these z-coefficients indeed satisfy the Hermite differential equation turns out to be pretty simple as well. But turning back to our original problem, we can substitute x a back in for u, so that we can see the form of our original wavefunctions: we have φn(x) = NnHn x a e−x2 2a2 , a2 = ℏ mω for some normalization constant Nn, with corresponding energies En = ℏω n + 1 2 . We’ll now turn to an operator-based algebraic approach: if we look back at the simple harmonic oscillator Hamiltonian again, we can write it as a sum of squares ˆ H = ˆ p2 2m + 1 2mω2ˆ x2 = 1 2mω2 ˆ x2 + ˆ p2 m2ω2 . We’ll now try to factorize the Hamiltonian essentially as the product of two factors: we wish to write ˆ H = ˆ V † ˆ V + c, where ˆ V † is the Hermitian conjugate of ˆ V . In particular, adding the constant c only shifts our energies without doing anything important, and then writing as ˆ V † ˆ V ensures that ˆ H is still Hermitian (because ( ˆ A ˆ B)† = ˆ B† ˆ A†, and (ˆ V †)† = ˆ V ). We’ll soon see why this is a useful thing to do, but first we’ll see how to actually factor in this case: if we had a2 −b2, then we could factor as (a −b)(a + b), and now because we have a2 + b2, it makes sense to factor as (a + ib)(a −ib) instead. Since quantum mechanics inherently needs complex numbers, this is a reasonable thing to try: we in fact almost have ˆ x2 + ˆ p2 m2ω2 ? = ˆ x −i ˆ p mω ˆ x + i ˆ p mω , but this is not exactly true because we have operators instead of numbers, so commutativity matters: indeed, we instead have ˆ x −i ˆ p mω ˆ x + i ˆ p mω = ˆ x2 + ˆ p2 m2ω2 + i mω [ˆ x, ˆ p]. 71 Since the commutator here is iℏ, the constant term is −ℏ mω, so we instead have that ˆ x2 + ˆ p2 m2ω2 = ˆ x −i ˆ p mω ˆ x + i ˆ p mω + ℏ mω I, where I denotes the identity operator. But now we can call the two terms in parentheses ˆ V † and ˆ V , respectively – more specifically, we can define ˆ V = ˆ x + i ˆ p mω and notice that ˆ V † is indeed the other term, because ˆ x and ˆ p are each their own Hermitian conjugate and the i turns into a −i. So putting everything together and simplifying, our Hamiltonian is now ˆ H = 1 2mω2 ˆ V † ˆ V + 1 2ℏω. However, ˆ V and ˆ V † still have units here, so we’ll simplify a bit more. We can do this by computing the commutator [ˆ V , ˆ V †] = ˆ x + i ˆ p mω , ˆ x −i ˆ p mω . The terms of the same type commute, so we’re left with = −i mω [ˆ x, ˆ p] + i mω [ˆ p, ˆ x] = ℏ mω + ℏ mω = 2ℏ mω . Thus, we can renormalize to find r mω 2ℏ ˆ V , r mω 2ℏ ˆ V † = I, and because these normalized operators are nice we’ll give them special names (which will make sense soon): Definition 109 The destruction operator (also annihilation operator) ˆ a and creation operator ˆ a† are given by ˆ a = r mω 2ℏ ˆ V , ˆ a† = r mω 2ℏV †, where ˆ V = ˆ x + i ˆ p mω. We can verify that ˆ a and ˆ a† are unit-free, because we have [a, a†] = 1 . (And because ˆ a† is different from ˆ a, this is not a Hermitian operator.) Since we’re defining ˆ a and ˆ a† in terms of ˆ x and ˆ p, we can also get the equations the other way around, which are ˆ x = r ℏ 2mω (ˆ a + ˆ a†), ˆ p = i r mωℏ 2 (ˆ a† −ˆ a). Notice that from these expressions, it’s consistent that ˆ x and ˆ p are Hermitian (taking the conjugates of the right-hand sides just swap the roles of ˆ a and ˆ a†, and the i in ˆ p’s expression becomes a −i, giving us the original expression again). So now we can write our Hamiltonian in terms of ˆ a and ˆ a† instead of ˆ x and ˆ p, since that was the original goal: we know that ˆ V † ˆ V = 2ℏ mω ˆ a†ˆ a = ⇒ ˆ H = ℏω ˆ a†ˆ a + 1 2 . So we’ve now factorized ˆ H using these creation and destruction operators, and now it turns out we’ll be able to solve the harmonic oscillator while barely needing to solve any differential equations (we just need to solve a single first-order differential equation)! Recall that we have the inner product (φ, ψ) = Z φ∗(x)ψ(x)dx, 72 so the expectation value of the Hamiltonian in a state ψ is ⟨H⟩ψ = (ψ, ˆ Hψ) = ψ, ℏω ˆ a†ˆ a + 1 2 ψ = ℏω(ψ, ˆ a†ˆ aψ) + ℏω 2 (ψ, ψ). Because ψ is normalized, the second inner product is 1, and now by the definition of a Hermitian conjugate, we can move the ˆ a† operator to the other side: = ℏω(aψ, aψ) + ℏω 2 . Now (aψ, aψ) ≥0 for any state ψ (because plugging this into the definition of the integral gives us the integral of a nonnegative quantity). Thus, we find that ⟨H⟩ψ ≥0 + ℏω 2 , and this is the advantage of writing the Hamiltonian as ˆ V † ˆ V : this flipping argument tells us that we always get energies bounded from below. So now for any energy eigenstate ψ, ⟨H⟩ψ ≥ℏω 2 , meaning that any energy eigenvalue is at least ℏω 2 . We know (from our previous method) that there is in fact an energy eigenstate of energy exactly ℏω 2 , and the way we arrive at that fact is by noting that we get equality only if (ˆ aψ, ˆ aψ) = 0 = ⇒ˆ aψ = 0. Thus, if there is a ground state ψ0, it must satisfy ˆ aψ0 = 0 = ⇒ r mω 2ℏ ˆ x + i ˆ p mω ψ0 = 0, by plugging in our definition of ˆ a. Removing the constant and converting everything to x-coordinates gives us the differential equation x + ℏ mω d dx ψ0(x) = 0, and we’ve turned our second-order differential equation into a first-order differential equation by exploiting Hermiticity! This differential equation is much easier to solve – we have dψ0 dx = −mω ℏxψ0 = ⇒ ψ0(x) = N0e−mω 2ℏx2 , which is normalized when N0 = mω πℏ 1/4. So our ground state is a perfect Gaussian, and we can indeed check that ˆ Hψ0 = ℏω ˆ a†ˆ a + 1 2 ψ0 = ℏω 2 ψ0, because ˆ a acting on ψ0 already makes that term 0, and we recover the correct ground state energy for the harmonic oscillator. We’ll continue this discussion next time, since we still need to find all of the excited states. But basically, it turns out ˆ a†ˆ a is an important Hermitian operator called the number operator, and then using the fact that ˆ aˆ a† −ˆ a†ˆ a = 1 allows us to “create” new energy eigenstates with the creation operator ˆ a†. We’ll essentially see that ψ0, ˆ a†ψ0, ˆ a†ˆ a†ψ0, · · · give us the full set of excited states for the harmonic oscillator! Lecture 15: Excited Harmonic Oscillator States and Scattering States Last time, we did an algebraic analysis where we factored the harmonic oscillator Hamiltonian in the form ˆ H = ℏω ˆ a†ˆ a + 1 2 , 73 which allowed us to see that any energy eigenstate has energy at least ℏω 2 and also allowed us to find the unique ground state ψ0 using a first-order differential equation. (Recall that ˆ a is a linear combination of ˆ x and ˆ p, such that [ˆ a, ˆ a†] = 1.) And we do expect to see a nondegenerate ground state, because there are no degeneracies in the bound state spectrum of a one-dimensional potential. Definition 110 The number operator ˆ N for the harmonic oscillator is ˆ N = ˆ a†ˆ a. We can check that ˆ N is Hermitian (because the conjugate of a product is the product of the conjugates, but in reverse order) and that ˆ Nφ0 = 0 (because ˆ Nφ0 = ˆ a†(ˆ aφ) = ˆ a†0 = 0). In particular, ˆ N is basically the Hamiltonian up to a linear shift, and it’s unitless because ˆ a and ˆ a† are, so it’ll be a useful thing to work with (as a dimensionless energy). Specifically, eigenstates of ˆ N and ˆ H are the same, with E = ℏω N + 1 2 . Today, we’ll explore how to use this number operator to get excited states of the harmonic oscillator, completing our solution. We’ll first understand how it interacts with the other operators we have here, and this often means computing relevant commutators. We have that [ ˆ N, ˆ a] = [ˆ a†ˆ a, ˆ a] = [ˆ a†, ˆ a]ˆ a = −ˆ a (because the ˆ a commutes with the ˆ a), which is a simple expression, and similarly [ ˆ N, ˆ a†] = [ˆ a†ˆ a, ˆ a†] = ˆ a†[ˆ a, ˆ a†] = ˆ a†. So the commutator of ˆ N with either ˆ a or ˆ a† gives back a number times either of those two operators, and those numbers are why ˆ a, the destruction operator, is also called the lowering operator and why ˆ a† is similarly called the raising operator. We can also additionally compute (if we’re confused about what’s going on here, we should try some small values of k to have a clear picture) [ˆ a, (ˆ a†)k] = k(ˆ a†)k−1. because we essentially have to move an ˆ a across a string of k ˆ a†s, and each time we move once we gain a contribution of (ˆ a†)k−1[ˆ a, ˆ a†] = (ˆ a†)k−1. This should remind us of the commutators [ˆ p, ˆ xn] from earlier in the course as well. Example 111 Taking a look at the k = 2 case, [ˆ a, (ˆ a†)2] = ˆ aˆ a†ˆ a† −ˆ a†ˆ a†ˆ a = ˆ aˆ a†ˆ a† −ˆ a†ˆ aˆ a† + ˆ a†ˆ aˆ a† −ˆ a†ˆ a†ˆ a , and each of these terms simplifies to ˆ a†: = ([ˆ a, ˆ a†]ˆ a†) + (ˆ a†[ˆ a, ˆ a†]) = ˆ a† + ˆ a† = 2ˆ a†. Similarly, we also have a commutator of the form [ˆ a†, (ˆ a)k] = −k ˆ ak−1, so using these results, we find that we can do more with the number operator: we have [ ˆ N, (ˆ a)k] = −k(ˆ a)k 74 by basically taking the equation in the line above and putting an extra ˆ a in the left argument of the commutator (which commutes with the right argument), and similarly [ ˆ N, (ˆ a†)k] = k(ˆ a†)k. So even with a string of ˆ as or ˆ a†s, taking the commutator with ˆ N still gives us back the same operator with a number, and in fact that counts the number of ˆ as or ˆ a†s – this explains the reason for the name number operator. We’ll now see how this all comes together – define the state φ1 = ˆ a†φ0. (Remember that ˆ aφ0 = 0, so we wouldn’t get an interesting quantum state if we tried applying the other operator.) We may ask if this is an energy eigenstate, and the easiest way to check this is to see if it’s a number eigenstate. Indeed, a useful manipulation is to say that ˆ Nφ1 = ˆ Nˆ a†φ0 = [ ˆ N, ˆ a†]φ0, since ˆ N kills φ0, so the ˆ a† ˆ Nφ0 term is zero. And the formulas for commutators we’ve established above means that it’s often easier to work with them than products – since [ ˆ N, ˆ a†] = ˆ a†, we find that ˆ Nφ1 = ˆ a†φ0 = φ1, meaning that φ1 is an eigenstate of ˆ N (and thus ˆ H) of eigenvalue 1 (and thus ℏω 1 + 1 2 = 3 2ℏω). And this is the reason for the name “creation operator” – acting on the vacuum (lowest energy state) with ˆ a† gives us a new eigenstate. And if we want a more concrete expression for the wavefunction, we can write ˆ a† and φ0 in terms of x or p and get an explicit formula. We defined φ0 to be a normalized energy eigenstate, and it makes sense to ask whether φ1 is normalized. This might seem like it is even more difficult than normalizing φ0, but in fact most of the work has already been done: (φ1, φ1) = (ˆ a†φ0, ˆ a†φ0) = (φ0, ˆ aˆ a†φ0) = (φ0, [ˆ a, ˆ a†]φ0) = (φ0, φ0) = 1 by the properties of the Hermitian conjugate and a similar commutator manipulation as above. (And learning how to do these kinds of algebraic tricks just comes with practice.) To get the next state, we’ll try to define the state φ′ 2 = ˆ a†ˆ a†φ0. Like before, we can check if we have an energy eigenstate by noting that ˆ Nφ′ 2 = ˆ N(ˆ a†)2φ0 = [ ˆ N, (ˆ a†)2]φ0 = 2(ˆ a†)2φ0 = 2φ′ 2, so we have an energy eigenstate of energy ℏω 2 + 1 2 = 5 2ℏω. This time when checking normalization, we find that (φ′ 2, φ′ 2) = (ˆ a†ˆ a†φ0, ˆ a†ˆ a†φ0) = (φ0, ˆ aˆ aˆ a†ˆ a†φ0). Now the blue ˆ a would kill φ0, so we can substitute in a commutator with the subsequent ˆ a†s: = (φ0, ˆ a[ˆ a, (ˆ a†)2]φ0) = (φ0, ˆ a · 2ˆ a†φ0). 75 Pulling out the 2 and then using the same commutator trick as before gives us = 2(φ0, [ˆ a, ˆ a†]φ0) = 2, so this time our energy eigenstate needs to be defined as φ2 = 1 √ 2 ˆ a†ˆ a†φ0. We can now generalize these calculations: Proposition 112 The nth excited state of the simple harmonic oscillator is φn = 1 √ n!(ˆ a†)nφ0. This state has number eigenvalue n and thus energy eigenvalue ℏω n + 1 2 . These calculations can be checked by noting (for the number operator) that ˆ Nφn = 1 √ n! ˆ N(ˆ a†)nφ0 = 1 √ n![ ˆ N, (ˆ a†)n]φ0 = 1 √ n!n(a†)nφ0 = nφn, and (for the normalization) that (φn, φn) = 1 n!(φ0, ˆ an(ˆ a†)n) = 1 n!(φ0, ˆ an−1[ˆ a, (ˆ a†)n]φ0) = 1 n!(φ0, ˆ an−1 · n(ˆ a†)n−1φ0) and then inductively pulling out factors of (n −1), (n −2), · · · , 2, 1 to cancel out with the 1 n!. With this structure in place, we can notice that applying ˆ a† to any of the energy eigenstates gets us the next higher energy eigenstate (with some constant factor). On the other hand, applying ˆ a to an energy eigenstate gets us the next lower energy eigenstate, because it essentially cancels out one of the ˆ a†s. To be more precise, notice that ˆ aφn = 1 √ n! ˆ a ˆ a†n φ0 = 1 √ n![ˆ a, ˆ a†n]φ0 = 1 √ n!n(ˆ a†)n−1φ0 = 1 √ n!n p (n −1)!φn−1 = n √nφn−1 = √nφn−1 , and similarly we have the relation ˆ a†φn = 1 √ n!(ˆ a†)n+1φ0 = 1 √ n! p (n + 1)!φn+1 = √ n + 1φn+1 . This gives us all of the tools we might need to do calculations like the following: Example 113 Suppose we want to calculate the expectation values ⟨ˆ x⟩φn and ⟨ˆ p⟩φn in our energy eigenstates. In our conventional language, this is difficult to compute, because we would need to write φn in terms of a Hermite polynomial and then take a complicated integral. But there are very quick alternative calculations to get us the answer: • For the expectation value of ˆ x, we are essentially calculating an expression like R x(φn(x))2dx. But because we have a symmetric potential, all of the φn(x)s are either even or odd, so their squares are always even. Thus the integrand is odd and the expectation is always zero. • For the expectation value of ˆ p, because we have a stationary state, it doesn’t make sense to have a nonzero 76 momentum because “the state doesn’t move.” (And this is actually a more general result: the expectation of the momentum on a bound state with a real wavefunction is always zero, by integration by parts.) But in preparation for more complicated calculations, we can also write out these expectations in terms of raising and lowering operators: ⟨ˆ x⟩φn = 0 = (φn, ˆ xφn) = r ℏ 2mω φn, (ˆ a + ˆ a†)φn . But now (ˆ a+ ˆ a†)φn is a linear combination of φn−1 and φn+1 (specifically, it is √nφn−1 +√n + 1φn+1), and the overlap of this with φn is zero by orthogonality of the energy eigenstates (because they have different energy eigenvalues). Another way to understand this orthogonality in the harmonic oscillator case is that if we wanted to compute something like (φ3, φ2) = (ˆ a†ˆ a†ˆ a†φ0, ˆ a†ˆ a†φ0) = (φ0, ˆ aˆ aˆ aˆ a†ˆ a†φ0), there are too many ˆ as – two of them can cancel out with the ˆ a†s, but the last one will kill φ0, and this will always vanish. So whenever m ̸= n, (φm, φn) will indeed be zero. Example 114 Suppose that we now want to calculate the uncertainty ∆x in φn, meaning that we must calculate (∆x)2 = ⟨ˆ x2⟩φn −(⟨ˆ x⟩φn)2 = ⟨ˆ x2⟩φn. This time, we will need the full power of the creation and annihilation operators (because doing the integral with Hermite polynomials would require a lot of work or creativity): we find that ⟨ˆ x2⟩φn = (φn, (ˆ x)2φn) = ℏ 2mω φn, (ˆ a + ˆ a†)(ˆ a + ˆ a†)φn . Expanding out (what we really care about is the φn coefficient in the second argument), we have = ℏ 2mω φn, (ˆ aˆ a + ˆ a†ˆ a† + ˆ aˆ a† + ˆ a†ˆ a)φn and now only the last two of the four terms give us a nonzero contribution by orthogonality, so we can use the boxed relations (of ˆ a and ˆ a† acting on φn) above to get = ℏ 2mω φn, ( √ n + 1 · √ n + 1 + √n · √n)φn = ℏ 2mω (2n + 1), or we can rewrite in terms of the number operator, using that ˆ aˆ a† = [ˆ a, ˆ a†] + ˆ a†ˆ a = 1 + ˆ N, to find again that the squared uncertainty is ⟨x2⟩φn = ℏ 2mω φn, (2 ˆ N + 1)φn = ℏ 2mω (2n + 1) . And similarly, we can calculate the uncertainty ∆p, giving us an explicit calculation of ∆x∆p for this state – this is left as an exercise to us. With that, we’ll now turn to a new topic: Definition 115 Scattering states are energy eigenstates that cannot be normalized. The motivating examples to keep in mind here are the eipx/ℏinfinite plane wave solutions – even though they individually cannot be normalized, they can be put together to create (normalizable) wave packets, which can properly represent particles. We’ll start with a particularly illustrative example, the step potential: 77 Example 116 (Step potential) Consider a one-dimensional system given by the potential V (x) =    0 x < 0, V0 x ≥0. Recalling that energy eigenstates must have energy at least min V (x), we know that any eigenstate will have positive energy, and thus there are two qualitative cases, namely E < V0 and E > V0. But it turns out we can just solve one of the two cases and “do the other by analytic continuation” – no matter which case we’re in, the solution will be non-decaying for x < 0 (because E > V ), and thus energy eigenstates will never be normalizable. We’ll first do the case where E > V0 (so that the solution is sinusoidal for both x > 0 and x < 0), and we’ll visualize a solution as “coming in from the left:” ψ(x) =    Aeikx + Be−ikx x < 0, Ceikx x ≥0. This is a wave “coming in from the left” because putting in the e−iEt/ℏfactor into Aeikx gives us an ei(kx−ωt) traveling wave. Analogously to the systems studied in 8.03, this wave will then have a “reflected” and a “transmitted” component, corresponding to the Be−ikx and Ceikx parts, respectively. And we know what k and k must be from the Schrodinger equation: since the kinetic energy of the particle is ℏ2k2 2m , we must have k2 = 2mE ℏ2 and k 2 = 2m(E −V0) ℏ2 . Fur-thermore, the wavefunction and its derivative must be continuous at x = 0 (since there are no delta functions), so A + B = C, ikA −ikB = ikC = ⇒A −B = k k C. So there will be one free parameter among A, B, C, which is okay because A is the “input” strength of the incoming wave – what matters is the values B A = k −k k + k and C A = 2k k + k . And we’ll see next time how to turn this into a wave packet for a physical particle! Lecture 16: Reflection and Transmission Last lecture, we began our study of scattering states (energy eigenstates that are not normalizable). Specifically, we constructed the energy eigenstates for the step potential (jumping from 0 to V0 at x = 0), which we can interpret as an incoming plane wave with a reflected and transmitted component (copied from last lecture): ψ(x) =    Aeikx + Be−ikx x < 0, Ceikx x ≥0. The relevant ratios for reflection and transmission in this potential turn out to be B A = k −k k + k , C A = 2k k + k , where k2 = 2mE ℏ2 , k 2 = 2m(E −V0) ℏ2 (since the particle has different kinetic energies in the two parts of the potential and thus different de Broglie wave-lengths). And if we now look at the limit where E →V0, we find that k →0 and thus B →A and C →2A, yielding 78 the solution ψ(x) =    2A cos kx x < 0, 2A x ≥0. (This isn’t normalizable, but neither is the original solution, so we don’t need to worry too much about that.) We can now discuss the conservation of probability principle from earlier in the course. We can imagine a probability current flowing into the step potential from the left, and if we imagine a narrow window around the discontinuity, we must have the same amount of probability current leaving that window as entering: recall that J(x) = ℏ mIm ψ∗∂ψ ∂x , so for our E = V0 state we have, for x < 0 and x > 0 respectively, JL(x) = ℏk m (\|A\|2 −\|B\|2), JR(x) = ℏk m \|C\|2. Indeed, we see that the current doesn’t depend on the value of x and is (after some computation) equal for x < 0 and x > 0 (keep in mind that all quantities in the following line are real numbers): JL = ℏk m 1 − B A 2! \|A\|2 = ℏk m 1 − k −k k + k 2! \|A\|2 = ℏk m 4kk (k + k)2 \|A\|2 = ℏk m 2k k + k 2 \|A\|2 = ℏk m \|C\|2 = JR. (And in fact, because this conservation law comes from Schrodinger’s equation, it makes sense that the relations between A, B, and C are also encoded in that equation.) But because the solution ψ(x) is made up of two different oscillating components, we can also write JL = JA −JB, where JA = ℏk m \|A\|2 and JB = ℏk m \|B\|2 are the probability currents brought by the incoming and reflected waves alone. (We’ll thus also write JR = JC.) This suggests that we should define reflection and transmission coefficients in terms of this current: R = JB JA = B A 2 , T = JC JA = k k C A 2 Essentially, a reflection coefficient of 0.1 would indicate that if we put particles into this potential from the left, then 10 percent of them would be reflected. (We don’t actually have particles yet, but this is the intuition!) And importantly, T is not just C A 2, as we might be tempted to write – the coefficients originate from probabilities, not amplitudes. With these definitions, we can notice that (because JA −JB = JC) R + T = JB JA + JC JA = 1, as we should expect with reflection and transmission coefficients in general. And when we have a wavepacket instead of a pure energy eigenstate, there will be some uncertainty in position and momentum, but the reflection and transmission probabilities will be basically given by R and T if the packet is localized near some particular energy. Example 117 We’ll now continue our study of the step potential, looking at the E < V0 case. The idea is to trust the principle of analytic continuation here (luckily, in this case it’s not quite as complicated as the mathematical words might suggest), which basically means that we can write down the same solution for x < 0 as before (since the energy E is still positive), with k2 = 2mE ℏ2 . And for x > 0, we just replace the eikx with an e−κx (the form of the solution we know to expect). This is achieved by replacing k with iκ everywhere, which means that 79 κ2 = 2m(V0 −E) ℏ2 (we just pick up a negative sign). We could have also replaced k with −iκ, but the reason for our choice here is that we now have ψ(x) =    Aeikx + Be−ikx x < 0, Ce−κx x ≥0, and ψ decays exponentially to zero as x →∞, as it should. And by doing this, we save the time of having to compute B A and C A by matching continuity of ψ and ψ′ again; instead, we can just replace k with iκ. We thus find that B A = k −iκ k + iκ = −i(κ + ik) i(κ −ik) = −κ + ik κ −ik , which is a ratio of two complex numbers of the same magnitude. In other words (imagining a right triangle formed by the points 0, k, and k + iκ in the complex plane), we have B A = −e2iδ(E), where δ(E) = tan−1 k κ = tan−1 r E V0 −E is a phase shift that depends on the energy of the eigenstate. But C A no longer plays the same role that it originally did – because we have a purely real solution for x > 0 this time, the probability current is zero (because ψ decays, the particle cannot have a positive rates of moving to the right). Instead, we have JC = 0 and JA = JB (consistent with \|A\|2 = \|B\|2 as discussed above). With this, we can rewrite the solution as ψ(x) =    Aeikx −Ae2iδ(E)e−ikx x < 0, Ce−κx x > 0 , and we can rewrite the solution for x < 0 as Aeiδ(E) ei(kx−∆(E)) −e−i(kx−∆(E)) = 2iAeiδ(E) sin(kx −δ(E)) , so that \|ψ(x)\|2 = 4\|A\|2 sin2(kx −δ(E)). If we let x0 = δ(E) k be the first positive x-coordinate where this probability density is zero, then we have the following sketch for \|ψ\|2: x \|ψ\|2 x0 Since δ(E) ranges from 0 to π 2 as E ranges from 0 to V0 (and in fact the derivative dδ dE = 1 2 q 1 E(V0−E) is large near E = 0 and E = V0), we see that near E = 0 the probability density is close to zero at x = 0, but near E = V0 the probability density is near its maximum. Example 118 We can now connect this solution form to a physical problem by considering wavepackets – this is very similar to past discussions with the principle of stationary phase. 80 We’ll start by considering E > V0 (so that we do have a transmitted wave), and we’ll set A = 1 first: ψ(x) =    eikx + k−k k+k e−ikx x < 0, 2k k+k eikx x > 0. We’ll now superimpose waves of this form. First, we need to add the time-dependent factor e−iEt/ℏto our whole solution, then we need to pick the coefficient f (k) with which this solution appears, and finally we superimpose all such solutions by integrating over k: Ψ(x, t) =    R ∞ 0 f (k) eikx + k−k k+k e−ikx e−iEt/ℏdk x < 0, R ∞ 0 f (k) 2k k+k eikxe−iEt/ℏdk x > 0. The key point here is that we can choose the bounds of integration (since we’re creating our own superposition of valid solutions to the Schrodinger equation), and we should not integrate from −∞to ∞because only the k > 0 solutions correspond to an incoming packet moving to the right. Furthermore, to ensure that we have a localized packet, f (k) should be sharply peaked around some k = k0. (In the past, the only way to compute this kind of integral numerically was with a supercomputer, but we can quickly do it on a laptop now.) We’ll split up the integral a bit by letting the incident wave, only defined for x < 0, be Ψinc(x, t) = Z ∞ 0 f (k)eikxe−iEt/ℏdk, letting the reflected wave, also only defined for x < 0, be Ψref(x, t) = Z ∞ 0 f (k) k −k k + k e−ikxe−iEt/ℏdk, and letting the transmitted wave, only defined for x > 0, be Ψtrans(x, t) = Z ∞ 0 f (k) 2k k + k eikxe−iEt/ℏdk. In particular, the total wavefunction is Ψ(x, t) =    Ψinc(x, t) + Ψref(x, t) x < 0, Ψtrans(x, t) x > 0, , and now we need to go a bit further by making the stationary phase approximation to understand how the wave packet is actually moving. Because f (k) appears in every integral in Ψ(x, t) and is very sharply peaked at k = k0, we can say that the wavefunction only has contributions from k ≈k0 and get a pretty accurate estimate. We’ll take f (k) to be real, so that there’s no associated phase to it, and recall that the stationary phase approximation requires that the k-derivative of the phase is zero. • For the incident wave, this means that d dk kx −Et ℏ k=k0 = 0 = ⇒x −ℏk0 m t = 0 = ⇒x = ℏk0 m t, since E = ℏ2k2 2m , and this means that the incident wave is propagating to the right at speed ℏk0 m for t < 0 (this is consistent because Ψinc is only defined for negative x anyway), and it “hits the barrier” at x = 0 at time zero – after that, Ψinc will be approximately zero, since we only look at negative x. 81 • For the reflected wave, our condition is similarly that d dk −kx −Et ℏ k=k0 = 0 = ⇒x = −ℏk0 m t. This also makes sense in that the reflected wave propagates to the left at a speed ℏk0 m for t > 0 and only significantly contributes to the wavefunction after the incident wave hits the barrier. • Finally, for the transmitted wave, the calculation is more complicated because we have k instead of k, and taking a derivative of k with respect to k is more interesting. It turns out that we will have x = ℏk m t (and because Ψtrans only exists for x > 0, this is only significant for t > 0), and this means that the transmitted wave propagates to the right, but at a different speed. We will now consider wavepackets for E < V0 – a similar strategy for superposition gives us the incident wave (for x < 0) Ψinc(x, t) = Z ∞ 0 f (k)eikxe−iEt/ℏdk and the corresponding reflected wave (also for x < 0, this time referring to our discussion of phase δ(E) from above) Ψref(x, t) = − Z ∞ 0 f (k)e−ikxe2iδ(E)e−iEt/ℏdk. The transmitted wave is much less interesting in this case, because there really isn’t a transmitted wave at all. Instead, what’s happening physically here is that we send in a wavepacket, look at the reflected shape, and use that to deduce the type of potential that the particle encountered. The stationary phase approximation again gives us x = ℏk0 m t for the incoming wave, but this time the reflected wave is more complicated: d dk −kx + 2δ(E) −Et ℏ k=k0 = 0 = ⇒ x = −ℏk0 m (t −2ℏδ′(E)) . Since the reflected wave is only defined on x < 0, we see that we get a significant contribution when t > 2ℏδ′(E), or approximately when t is positive, like before. But because of this additional 2ℏδ′(E) term, the wave does not bounce perfectly off the barrier at x = 0; there is a time delay of 2ℏδ′(E) (which is large for E near 0 or V0, as previously discussed), and this type of expression is what is used in scattering theory to determine the type of potential we’re scattering off. To finish this lecture, we’ll think a bit about the forbidden region (where x > 0 and E < V0) and what the decaying exponential wavefunction means. In particular, we may ask what a particle looks like if we find it to have energy E < V0, meaning that it has negative kinetic energy. It would be contradictory to say both that the particle is in the forbidden region and that it has energy less than V0, and quantum mechanics evades this problem (this is not a completely precise argument, but it’s enough to get the intuition) by arguing about uncertainty. Specifically, because the wavefunction decays as e−κx, the length scale at which we are likely to find the particle is on the order of 1 κ (recalling that κ2 = 2m(V0−E) ℏ2 ). But to say that the particle is in the forbidden region, we need to measure the particle with uncertainty ∆x < 1 κ, and thus the particle has an uncertainty in momentum p > ℏ ∆x = ℏk. This uncertainty then gives us a “uncertainty” kinetic energy of KE = p2 2m = ℏ2k2 2m = V0 −E, and thus the “negative kinetic energy” is compensated for by the uncertainty in momentum – the total energy of the particle is now E + (V0 −E) = V0, and no contradiction arises – we’ll just find a normal particle in the region x > 0. 82 Lecture 17: Ramsauer-Townsend and General Scattering We’ll continue our discussion of scattering states today – last lecture, we analyzed the behavior of reflection and transmission under a step potential, and today we’ll explore a different effect called resonance transmission (and how it leads to the Ramsauer-Townsend effect) before turning to more general resonance problems. First, we will do a toy problem with the finite square well: Example 119 Consider the one-dimensional system given by the potential V (x) =    −V0 −a < x < a 0 otherwise, and suppose we send an incoming wave into the potential from the left (meaning that we are looking for scattering states of E > 0). We are interested in understanding reflection and transmission coefficients, but we know from last lecture’s dis-cussion that energy eigenstates tell us most of the story – if a wavepacket is localized around some energy, then it’ll behave like an energy eigenstate of that energy. At both x = −a and x = a, the discontinuity in the potential means there may be reflection and transmission, and physically we can imagine that the particle might “bounce back and forth” repeatedly between those two points. Remark 120. In quantum mechanics, reflection and transmission can occur whether the potential goes up or down – this is different from the classical case in which we can only reflect off of a higher barrier. Even though many reflections may occur, we will still have ψ(x) = Aeikx + Be−ikx in the region x < −a, and similarly we will have ψ(x) = Ceik2x + De−ik2x (different wavenumber because we have a different kinetic energy) in the region −a < x < a. In particular (by the same logic as usual), we have k2 = 2mE ℏ2 and k2 2 = 2m(E+V0) ℏ2 . Finally, we will only have ψ(x) = Feikx in the region (no wave moving to the left) because there is nothing to bounce off of past x = a. Putting everything together, ψ(x) =          Aeikx + Be−ikx, x < −a Ceik2x + De−ik2x, −a < x < a Feikx x > a. We’ll reason about these coefficients by thinking about probability current, since we’ve already seen that this is a good way to know how to define reflection and transmission coefficients. With the step potential, we had an extra factor beyond the C A 2 because the two sides of the barrier were at different energies, but in this case the two sides of the square well are at the same energy. Thus, it’s reasonable to believe that we just have R = B A 2 , T = F A 2 in this case, and this will make sense if it turns out that R + T = 1. Indeed, following the current conservation calculations from last time, we know that JL ∼\|A\|2 −\|B\|2 (the current to the left of x = −a) and JR ∼\|F\|2 (the current to the right of x = a), with the same proportionality constant, so \|A\|2−\|B\|2 = \|F\|2 and we do have R+T = 1. But to actually calculate those ratios, we do need to do some work – there are five variables and four boundary conditions (two at each discontinuity), which makes sense because there’s one free variable of overall normalization. 83 We’ll skip those calculations here – it turns out that 1 T = 1 + 1 4 V 2 0 E(E + V0) sin2(2k2a) (notice that the second term on the right-hand side is nonnegative, so T is indeed between 0 and 1). Looking at some relevant limits, if sin2(2k2a) →0 or E →∞, then we actually have T →1, giving us complete transmission (we’ll discuss this more soon), and if E →0, we have T →0, giving us no transmission. To understand this more, we can rewrite this expression in unit-free language: notice that 2k2a = 2 r 2ma2(E + V0) ℏ2 = 2 s 2ma2V0(1 + E V0 ) ℏ2 = 2 q z2 0(1 + e) = 2z0 √ 1 + e, where we define the unit-free energy e = E V0 (relative to the depth of the potential) and z2 0 = 2ma2V0 ℏ2 should look familiar from our past discussions of the finite square well. Similarly simplifying the prefactor in terms of e gives us 1 T = 1 + 1 4e(1 + e) sin2(2z0 √ 1 + e) . As we mentioned above, we are now interested in the energies E for which T = 1, which means that 2z0 √1 + e = nπ for some integer n. (Furthermore, because √1 + e ≥1, we must have n ≥2z0 π .) Rearranging, we find that 4z2 0(1 + en) = n2π2 = ⇒en = −1 + n2π2 4z2 0 = ⇒E = −V0 + n2π2V0 4z2 0 = −V0 + n2π2V0 2ma2V0/ℏ2 = −V0 + n2π2ℏ2 2m(2a)2 . In this form, the answer can actually make some intuitive sense: we have resonant transmission (meaning T = 1) if the distance between the energy E and the bottom of well −V0 is n2π2ℏ2 2m(2a)2 , which is the energy level of an infinite (rather than finite) square well of width 2a. And this is really because from the point of view of the wavefunction, the condition that k2(2a) = nπ (for resonant transmission) is the same as (2a) λ = n 2. In other words, the de Broglie wavelength of the particle inside the well fits a half-integer number of times inside the well of length 2a, and the energy eigenstate passes completely through the well in this case. Example 121 For a numerical example, suppose we have a square well where z0 = 13π 4 , so that we only get valid energies E when n ≥2z0 π = 13 2 . (In other words, we need to start from the seventh lowest energy of the infinite square well to get a positive E.) We can then plug in numerically to find that e7 = E7 V0 ≈0.15976, e8 = E8 V0 ≈0.51479, e9 = E9 V0 ≈0.91716. More generally, here is a plot of the transmission probability as a function of e = E V0 : T e = E V0 1 1 84 Example 122 We’re now ready to talk about a famous experiment performed by Ramsauer and Townsend (two physicists) in 1921, elastic scattering of electrons off of rare gas atoms. Here, “elastic scattering” means that no particles are created or destroyed, and the gas atoms have completely full outer shells (so that they are inert noble gases and have high ionization energies). That means that we can essentially visualize the gas atoms as a spherical cloud around a central nucleus, and from the point of view of an electron (by Gauss’s law) there will be no electric field until we enter the cloud. At that point, the electric field will point inward, so the electron will curve towards the nucleus and eventually exit the cloud at a different angle. So this is kind of like a finite spherical well, analogous to the finite square well we’ve been discussing above! To turn this into a situation where reflection and transmission coefficients make sense, we can say that particles are reflected if they scatter significantly and are transmitted if they pass through. (More mathematically, we could talk about the scattering cross-section.) But what experiments found was that the reflection coefficient R (as a function of e) also exhibited this “bouncy” behavior as in the plot above, starting off very high but reaching a local minimum around 1 eV (which corresponds to the electron moving at around 600 kilometers per second). And this sensitivity of the reflection coefficient to the energy of the particle comes from a very similar reasoning of resonance within the well! But the explanation didn’t come until years later when quantum mechanics was more fully developed, and if we want the exact numbers we’ll have to do a three-dimensional version of the calculation above. In the rest of this lecture (and the next few as well), we’ll now expand our discussion to more general one-dimensional scattering. Example 123 Suppose we have a one-dimensional system where there is an infinite barrier left of x = 0 and a finite R above which the potential is zero (called the range of the potential): V (x) =          0 x > R, V (x) 0 < x < R, ∞ x < 0. The scattering is performed by having an observer “throw” incoming waves in from x = +∞, which reflect off of the barrier. We then wish to measure the outgoing wave to deduce information about the potential, and this is indeed what is done in particle colliders in modern physics when multiple particles interact with each other. Remark 124. The infinite barrier left of x = 0 makes sense if we think about x as taking the place of the radial coordinate r in three dimensions – it’s never possible to be a negative distance away from the origin. And in a future quantum physics course, we might study scattering in three dimensions, and this will be a useful tool in that discussion. We’ll begin with the simplest case where V (x) = 0 for 0 < x < R, and thus we are just scattering off of an infinite wall with no additional potential. Incident and outgoing waves will then look like e−ikx and eikx respectively (both are energy eigenstates and thus valid solutions), and we can combine these together into a solution to the Schrodinger equation of the form φ(x) ∼eikx −e−ikx so that the wavefunction is zero at x = 0. Remembering that sin x = eix−e−ix 2i , we thus get the solution φ(x) = sin(kx) for x > 0, but we can still think of this as being a combination of an incoming (−e−ikx 2i ) and outgoing ( eikx 2i ) plane wave. 85 This gives us an inspiration for what to write in the more general case with a nonzero potential: we’ll have an incoming wave −e−ikx 2i just like before, but that is only valid for x > R (because plane waves are not energy eigenstates when the potential is nonzero). The outgoing wave is then going to be of the form c 1 2i eikx for some constant c (that’s the only other energy eigenstate at energy E besides e−ikx), but remembering our discussion about probability current, we must actually have \|c\| = 1 by conservation of probability (there is no transmission to x < 0, and in a stationary state we cannot have probability accumulating or depleting inside the potential). In summary, the potential can only influence the outgoing plane wave by a phase, and we’ll write the outgoing wave as 1 2i eikxe2iδ for some phase δ. (And remembering that δ depends on the energy E of the plane waves we’re sending in, we see that we can get δ as a function of E or k, which can give us substantial information about V (x).) The phase is only defined up to a multiple of 2π, but for reasons we’ll see next lecture, it turns out to be convenient to fix the phase δ at k = 0 and then make the phase continuous as a function of k (so if it keeps going clockwise, we keep increasing it even past 2π). The total solution to the Schrodinger equation, incoming plus reflected, is thus ψ(x) = 1 2i eikx+2iδ −e−ikx = eiδ 2i ei(kx+δ) −e−i(kx+δ) = eiδ sin(kx + δ) for x > R. (And if δ = 0, we’re back to the original wavefunction for the “no potential” case.) Comparing the probability density \|φ(x)\|2 = sin2(kx) (no potential) with \|ψ(x)\|2 = sin2(kx + δ) (with potential), we see that the same features of the wavefunctions (maximums, nodes, and so on) occur in both cases, but shifted by a constant distance −δ k . (Specifically, kx = a0 occurs at x = a0 k , but kx + δ = a0 occurs at x = a0 k −δ k .) So when δ > 0, the wavefunction is pulled into the potential (in other words, the potential is attractive), and when δ < 0, the wavefunction is pushed outward (the potential is repulsive). And in preparation for next lecture, we’ll make the following definition: Definition 125 The scattered wave ψs(x) is the extra component of the wavefunction added from the no-additional-potential case, defined via ψ(x) = φ(x) + ψs(x). Remembering that ψ and φ were defined to have the same incoming wave −e−ikx 2i , we see that ψs represents how much more of an outgoing wave we get with the potential between 0 and R, and it must be an outgoing wave itself. And because we have the explicit expressions for the two waves, we see that ψs(x) = ψ(x) −φ(x) = 1 2i eikx+2iδ −e−ikx −1 2i eikx −e−ikx = eikx 2i e2iδ−1 = eikxeiδ sin δ . In other words, this additional scattered wave is zero when δ = 0 (representing that no change occurred), and the amplitude of this additional outgoing wave is eiδ sin δ. We’ll make use of this next time! Lecture 18: Phase Shift and Levinson’s Theorem Last lecture, we introduced the general scattering problem for a finite-range potential of length R, in which we work with a potential V (x) which is infinite for x < 0 and zero for x > R. Specifically, we look at energy eigenstates of energy E = ℏ2k2 2m , and we can compare the solution φ(x) = sin kx (a combination of an incoming and outgoing wave −e−ikx 2i + eikx 2i ) when the potential is zero on (0, R) with the solution ψ(x) = eiδ sin(kx + δ) = −e−ikx 2i + e2iδeikx 2i 86 (only valid on x > R) when we do have a nonzero potential. As mentioned at the end of last lecture, we often write ψ(x) = φ(x) + ψs(x), where the scattered wave turns out to be the outgoing wave ψs(x) = eiδ sin δeikx for x > R. Here, As = eiδ sin δ is called the scattering amplitude, and we are often interested in \|A2 s\| = sin2 δ to quantify the effect of the nonzero potential. Today, we’ll connect this to the concept of time delay by calculating what happens when we send in wavepackets into this potential. (Recall that we’ve previously discussed this in the context of the step potential.) If we have an incident wave of the form Ψinc(x, t) = Z ∞ 0 f (k)e−ikxe−iE(k)t/ℏdk in the region x > R (here we put back the time dependence to get the time-dependent wavefunction), then we know the reflected wave that corresponds to it will be Ψref(x, t) = − Z ∞ 0 f (k)eikxe2iδ(E)e−iE(k)t/ℏdk in the region x > R (negative sign and e2iδ factors coming from the expression for the reflected wave). As usual, f (k) will be a real function that peaks sharply at some k = k0. And now we can keep track of how the peak of the wavepacket moves by doing the usual stationary phase approximation at k = k0 – we’ll skip the calculations because we’ve done them a few times already. The incident wave turns out to satisfy x = −ℏk0 m t = −vgt for a constant group velocity vg (and this is only valid for t < 0), while the reflected wave satisfies x = vg(t −2ℏδ′(E)). So the reflected wave moves at the same group velocity as the incident wave, but it only starts to appear after a delay (which can be positive or negative) of ∆t = 2ℏδ′(E) = 2ℏdδ dk dk dE = 2 1 ℏ dE dk k=k0 dδ dk k=k0 = 2 vg dδ dk k=k0 (we have total rather than partial derivatives, so there’s nothing to worry about with these manipulations). This equation can be further manipulated to 1 R dδ dk = ∆t 2R/vg . This is now a unitless relation, and the right-hand side compares the time delay ∆t to the time it takes to traverse the finite-range potential and back (length 2R) at the group velocity vg! So this ratio of the delay to the free transit time gives us a sense of whether the potential has caused a significant delay in how the wavepacket travels. We’ll now do an example where we can calculate everything explicitly (though the formulas may still be rather messy and are only insightful if we plot with a computer): Example 126 Suppose we have a potential well of length a, meaning that our potential is of the form (for some V0 > 0) V (x) =          ∞ x < 0, −V0 0 < x < a, 0 otherwise, and we wish to explicitly calculate energy eigenstates of energy E > 0. Since this falls under the general setup we’ve been using, we know already that ψ(x) = eiδ sin(kx + δ) for x > a (this is the specific combination of eikx and e−ikx that is necessary), where k2 = 2mE ℏ2 as usual. And for 0 < x < a, we must have a combination of eik′x and e−ik′x, where k′2 = 2m(E+V0) ℏ2 , but it is convenient to use trig functions instead 87 – we can only have something proportional to sin(k′x) because the wavefunction must vanish at 0. This gives us the ansatz (not putting an additional normalization in the x > a case just to remove a variable) ψ(x) =    eiδ sin(kx + δ) x > a, A sin(k′x) x < a, with the additional boundary conditions that ψ and ψ′ must be continuous at x = a, meaning that A sin(k′a) = eiδ sin(ka + δ), Ak′ cos(k′a) = keiδ cos(ka + δ). Much like with the finite square well, we can divide the two equations to get k′ cot(k′a) = k cot(ka + δ) = ⇒cot(ka + δ) = k′ k cot(k′a). We can now isolate δ by taking cot−1 on both sides, but there’s a bit of trigonometric manipulation we can do now: specifically, we can use the fact that cot(A + B) = cot A cot B−1 cot A+cot B and then solve for cot δ, which yields cot δ = tan(ka) + k′ k cot(k′a) 1 −k′ k cot(k′a) tan(ka) . Given any value of the energy E, we can then calculate k′ and k and find the phase shift by plugging in those values. But if we want to plot this on a computer, the right variables are the unitless ones such as ka and k′a: define u2 = (ka)2 = 2mEa2 ℏ2 , so that (k′a)2 = 2mEa2 ℏ2 + 2mV0a2 ℏ2 = u2 + z2 0 with the standard unit-free length scale z0 for the finite square well! So now taking the reciprocal of both sides of the boxed equation, and using the fact that k′a ka = q 1 + z2 0 u2 , we have tan δ = 1 −(k′a) (ka) cot(k′a) tan(ka) tan(ka) + k′a ka cot(k′a) = 1 − q 1 + z2 0 u2 cot p z2 0 + u2 tan u tan u + q 1 + z2 0 u2 cot p z2 0 + u2 , and now z0 is a property of the potential well we’re given – for any well of depth V0 and width a, we can plug in the constant z0 and plot tan δ as a function of u. And notice that as u →0, the numerator approaches a constant (because the q 1 + z2 0 u2 diverging and tan u going to zero balance each other out), while the denominator goes to infinity because of the second term. So tan δ →0 as u →0, and we can pick the phase shift to be δ = 0 for u = 0 (rather than 2π, for example). We can now study the behavior of δ versus u = ka for various values of z0: • When z2 0 = 3.4 (corresponding to z0 ≈0.59π), δ(u) starts off at 0, becomes more negative, and then stabilizes at −π for large u. The scattering amplitude is strongest when δ = −π 2 , so sin2 δ peaks once. Finally, the unit-free delay (the derivative dδ du ) starts around −4 and asymptotically goes to zero. The plots of δ and sin2 δ are shown below: 88 u δ −π 2 −π 0.87 u sin2 δ 1 In particular, notice that a negative time delay corresponds to the packet coming back earlier than ordinarily expected. But this makes sense, because the kinetic energy of the packet increases when the potential is lower and thus the group velocity is larger. As a sidenote, if we plot the amplitude A as a function of u, we get a graph similar to that in Example 121. • The behavior remains very similar as we increase z0, until a sudden change happens. In particular, by the time we are at z0 = 5, the graphs look as shown below (with the two circles indicating the points at which δ crosses −π 2 and −3π 2 : u δ −2π 1.11 7.52 u sin2 δ 1 Exploring the limiting behavior of δ for different values of z0, we see that it either goes back to 0 (for small z0), or it approaches one of −π, −2π, −3π, and so on. And it turns out that the fundamental relation relates the overall excursion in the phase δ to the number of bound states of the potential! We can check as an exercise that for this “half-square well,” this potential has all of the odd bound states of the full finite square well, and we found (see calculations in Example 91) that we determine how many of these solutions we have by seeing if z0 is larger than specified multiples of π 2 . It turns out these numbers will indeed increase at the same time, and this leads us to Levinson’s theorem. This is the most subtle derivation we’ll do in this class, and we’ll need to use the following fact: Fact 127 As we vary a potential, the energy eigenstates will vary, but they do not appear or disappear. This may seem confusing given our discussion above, but when we think about the states of (for example) a finite square well, we should remember that there are an infinite number of scattering states above the finitely many bound states! So if we decrease the depth of a finite square well, eventually the highest-energy bound state will change into a scattering state. And if we increase the depth, eventually the scattering states will “lend” the bound states an extra state. This may seem like an argument that doesn’t make much sense because we have uncountably infinitely many states, but one way to make this rigorous is to imagine putting this whole system in a very large box of length L, making even the positive-energy states discrete. Then as we change the depth of our well, we’ll indeed see the states of positive energy drop down and change into states of negative energy, and still nothing is created or destroyed. 89 Theorem 128 (Levinson’s theorem) Suppose we are in the finite-range potential setting in Example 123. Then the number of bound states N of the potential is predicted by the behavior of scattering via N = 1 π (δ(0) −δ(∞)) . Proof. Motivated by the discussion above, in order to avoid a continuum of states, we will also place an infinitely high potential barrier for x > L, where L ≫R. (This is sometimes called a regulator.) Then our states will be quantized, and for large L, they will be close to each other and look a lot like scattering states (but still be bound states in reality). And we’ll take the limit as L →∞– if the answer doesn’t depend on L, then we can claim the answer holds in the limit as well. If we think about the case with no potential V (x), then all energy eigenstates have positive energy and are of the form φ(x) = sin(kx), where φ(L) = 0 (boundary conditions for an infinite square well). Thus, we require kL = nπ for some positive integer n, and now indeed k can only take on integer multiples of π L (meaning that the energy states are discrete). Within a differential dk, we then have π Ldn positive energy states in that range. On the other hand, in the case where we do have a potential, we can’t solve the Schrodinger equation in general, but we do know that we have the universal solution ψ(x) + eiδ sin(kx + δ) in the region R < x < L, and again we must demand ψ(L) = 0. So this time we require kL + δ(L) = n′π for some integer n′, meaning that for any interval dk, the number of positive energy states is given by Ldk + dδ dk dk = πdn′ = ⇒dn′ = L π dk + 1 π dδ dk dk . If we then look over the entire range of k-values (in the positive real axis), we can count the difference in the number of positive-energy states. In particular, if we imagine doing a slow deformation from V = 0 to V = V (x), the states will also slowly vary (using the logic of Fact 127), and we can count how many energy states are gained or lost in that interval: positive-energy states lost in dk when potential is turned on = dn −dn′ = −1 π dδ dk dk, and thus the total number of energy states lost overall is the integral positive-energy states lost = Z ∞ 0 −1 π dδ dk dk = 1 π (δ(0) −δ(∞)) , because the integrand is just a total derivative. But (again by Fact 127) those states cannot disappear – they must have become bound states. Since we have zero bound states at V = 0, the number of bound states for the potential V (x) must be the boxed quantity above. In particular, this argument gets around the fact that we have infinitely many scattering states in both cases – by counting the differences in number of states in infinitesimal regions, we are able to reason about the count of the finitely many bound states. Lecture 19: Resonance and Time Delay Today’s lecture will explore more on resonances in scattering theory – we’ve seen some of this in the resonant transmission of the Ramsauer-Townsend effect, in which particles move past a potential without any reflection for 90 some particular energies E. We’ll start with the same setting as last lecture, with a finite-range potential within 0 < x < R and an infinite barrier for x < 0. Our first question will concern the time delay 2ℏdδ(E) dE , which we know can be positive or negative – specifically, we want to ask if the time delay can be arbitrarily negative (meaning that the packet comes back much earlier than if we had no potential). The answer is no - essentially, it wouldn’t make sense if the particle came back earlier than if there was a hard (infinite) wall at x = R, because there’s nowhere for the particle to be reflected before that point – this has to do with causality. More explicitly, because the wavepacket travels at speed vg, we have time delay∆t = 2ℏdδ(E) dE ≳−2R vg (≳instead of ≥because our argument isn’t completely rigorous here, and it does turn out there’s a bit of a correction we need to make), and this rearranges to (because dE dk = ℏ2k m = ℏvg 2ℏ1 dE dk dδ dk = 2ℏ1 ℏvg dδ dk ≳−2R vg = ⇒ dδ dk ≳−R . On the other hand, we may ask whether we can have an arbitrarily positive time delay, and this time the answer turns out to be yes. We know that if we have a positive potential V0 in the range 0 < x < R, that will delay particles of energy E > V0, but that isn’t very special because it will also advance particles with energy E < V0 (they will reflect off of the barrier). So what we really need to do is to trap the particle in the potential: Example 129 Consider a potential of the form V (x) =                ∞ x < 0, −V0 0 < x < a, V1 a < x < 2a, 0 x > 2a, where V0, V1 > 0. This potential will have states of energy 0 < E < V1 which look like bound states but will instead act as resonances with very large time delay. We’ve done problems like this before, so we’ll skip some of the calculation – qualitatively the solution is exponentially decaying in the region a < x < 2a and sinuosoidally oscillating in the others. The relevant constants are then k2 = 2mE ℏ2 , (k′)2 = 2m(E + V0) ℏ2 , κ2 = 2m(V1 −E) ℏ2 . Conceptually, we imagine that the wavepacket will come into the potential, spend a lot of time “bouncing” in the region 0 < x < a, before finally bouncing out – this will correspond to a very large amplitude for the wavefunction probability density in that region. So in our ansatz ψ(x) =          A sin(k′x) 0 < x < a, (?) a < x < 2a, eiδ sin(kx + δ) x ≥2a, we will try to find what values of k make A very large. Within the middle region a < x < 2a, we can either use eiκx and e−iκx, or we can use cosh(κx) and sinh(κx). But a third choice is easiest for imposing boundary conditions, namely sinh(κ(x −a)) and cosh(κ(x −a)) (we can check that each of these are also linear combinations of the original 91 exponentials and are solutions to the Schrodinger equation in this region). Since cosh(0) = 1 and sinh(0) = 0, our solution must look like ψ(x) =          A sin(k′x) 0 < x < a, A sin(k′a) cosh(κ(x −a)) + B sinh(κ(x −a)) a < x < 2a, eiδ sin(kx + δ) x ≥2a. We now just need to pick A, B, and δ so that ψ and ψ′ are continuous at x = a and x = 2a (we already ensured continuity of ψ at a). It turns out that the relation between all of our variables is tan(2ka + δ) = k κ · sin(k′a) cosh(κa) + k′ κ cos(k′a) sinh(κa) sin(k′a) sinh(κa) + k′ κ cos(k′a) cosh(κa) (the numerator and denominator are just slight rearrangements of each other). Rewriting this in terms of unitless quantities, we will define u = ka, z2 0 = 2mV0a2 ℏ2 , z2 1 = 2mV1a2 ℏ2 , so that the unit-free energy (comparing the energy of our packet to the height V1 of the tall barrier) can be written as e = E V1 = ℏ2k2a2 2mV1a2 = u2 z2 1 . We can also write the other quantities here in terms of u, namely (k′a)2 = u2 + z2 0, (κa)2 = z2 1 −u2. So if we’re given a fixed potential (in particular the values of a, V0, and V1), the right-hand side now becomes a function of u, because z0 and z1 just become numbers. We can then vary u to see the behavior of δ – it’s messy, but we can do it in principle. Example 130 For a numerical example, let’s consider z2 0 = 1 and z2 1 = 5, and we’ll plot δ as a function of u (which must range between 0 and z1 = √ 5 if we are indeed looking at energies 0 < E < V1). The dot on the curve below indicates the point where tan(2ka + δ) = 0, and u∗≈1.8523 marks the point where δ crosses −π 2 for the second time (scaling on the x-axis is 4 times larger than on the y-axis): u δ(u) √ 5 −π −π 2 u∗ The slope of this graph is about −2 for small u, because for low energies the tall barrier will essentially make the packet bounce back at x = a instead of x = 0, and the delay is proportional to the derivative dδ du . But at a certain 92 point around u∗, δ jumps very quickly. If we then plot sin2 δ, the amplitude of the scattered wave \|A2 s\| will peak around the two values where δ = −π 2 , and will do so more sharply at u∗. However, the first point where δ = −π 2 is not very notable – since the derivative of δ is negative (and not very large), this corresponds to a time advance, and we’ve already discussed above that ∆t cannot be very negative. On the other hand, the peak of \|A2 s\| at u∗instead corresponds to a time delay in which the particle gets trapped in the barrier – this is the resonance that we’re looking for. So if we look at the overall delay, which is proportional to the derivative dδ du , it will be negative for small u, sharply peaking to a large positive value around u∗, and then returning to being slightly negative. And for basically the same reasons, if we plot the amplitude \|A\| of the wavefunction inside the well, we will see a huge jump (to around 3) when u ≈u∗and not anywhere else. We can now ask how we could solve for resonances like this more mathematically, and we’ll do this by modeling what a resonance near some point k = α would look like. We claim that the graph of δ (as a function of k) typically looks something like tan δ = β α −k ⇐ ⇒δ = tan−1 β α −k around k = α, for some β > 0. This is reasonable, because we get a large value of sin2 δ around δ = π 2 , and β α−k goes to ±∞around k = α. Specifically, as k →α−, tan δ →∞, and we can characterize how quickly it goes to ∞ by noting that β α−k = 1 when k = α −β: k tan δ = β α−k α α −β 1 Similarly, tan δ reaches −1 at k = α + β, so most of the action in δ is happening within a narrow band of width 2β. So to get sharp behavior (and strong resonance), we must set up the system to have β small. If we then plot δ (remember that we choose the multiple of π in arctan so that δ is continuous), we’ll see that it is exhibiting the correct behavior as well: k δ = tan−1 β α−k π 2 π α 93 We can then calculate some relevant quantities for this modeled resonance: dδ dk k=α = 1 β , \|ψs\|2 = \|As\|2 = sin2 δ = β2 β2 + (α −k)2 The derivative here makes sense because the phase is changing by almost π over a length scale of β, and the function for the amplitude is actually the famous (non-relativistic) Breit-Wigner (also Cauchy or Lorentzian) distribution, usually written in terms of energy rather than momentum. If we let Eα be the energy of a particle at k = α, we’ll make the approximation E −Eα = ℏ2k2 2m −ℏ2α2 2m = ℏ2 2m(k2 −α2) = ℏ2 2m(k −α)(k + α) ≈ℏ2 2m(k −α) · 2α = ℏ2α m (k −α). Substituting this in, we get the formula (now written in the conventional notation) for the squared magnitude of the scattered wave: \|ψs\|2 = 1 4Γ2 (E −Eα)2 + 1 4Γ2 , Γ = 2αβℏ2 m . Here, Γ is the “full width at half maximum” of the distribution, which is distance between the two points on the curve where \|ψs\|2 = 0.5: E \|ψs\|2 1 0.5 Eα −Γ 2 Eα Eα + Γ 2 Looking more at the physical significance of Γ = 2αβℏ2 m , we notice that it has units of energy, so we can get a characteristic time τ = ℏ Γ = m 2αβℏ. If we then compare this time to the time delay we’ve been studying, ∆t = 2ℏdδ dE = 2ℏdk dE dδ dk , and now because (at resonance k = α) we have dδ dk = 1 β and dk dE = 1 dE dk = 1 ℏ2α/m, substituting in gives us ∆t = 2ℏm ℏ2α 1 β = 2m αβℏ2 = 4ℏ Γ = 4τ. In other words, the half-width of the scattering distribution gives us a characteristic energy, which is associated with a characteristic time that is related to the physical time delay of the scattering problem! So the larger the time-delay, the smaller our width Γ, and the more sharply peaked our distribution will be. (And this corresponds to δ changing very rapidly around k = α.) 94 Fact 131 This kind of analysis is used in nuclear and particle physics – the Higgs boson is an unstable particle which decays very quickly, but it stays within the potential well during its lifetime (about 10−22 seconds). So its cross-section is governed by resonances, and that’s a more accurate description than really thinking of it as a particle. We can then observe that the Higgs boson has a central energy around Eα ≈125 GeV and a width of Γ ≈4 MeV (to about 5 percent accuracy). So we have a very narrow resonance, and the corresponding time ∆t indeed turns out to be around 10−22 seconds with those numbers! So the discovery of the Higgs boson had to do with observing that a resonance did occur at this particular energy. In summary, we find that we cannot have long time advances, but that long time delays can occur in situations of resonance (rapid positive changes of δ). We’ll go a little further to put this topic into a more intriguing footing: we can rewrite the scattering amplitude as As = sin δeiδ = sin δ e−iδ = sin δ cos δ −i sin δ = tan δ 1 −i tan δ . If we want this amplitude to be large, we can say that setting tan δ = −i would make As infinite. Since δ is a phase shift, tan δ is always real and it doesn’t seem to make sense to set it equal to −i. However, we can imagine phase shifts as living in the complex plane, and the idea is that a real phase near an imaginary number with large As will also give a large amplitude. To make this more precise, let’s plug in the tan δ that we’ve been modeling: As = β α−k 1 − iβ α−k = β α −k −iβ = β (α −iβ) −k . And because we usually plot the scattering amplitude As as a function of k, it’s nice that we now have an expression in that form, and we can now imagine plotting As in the complex k-plane. We designed our model so that the resonance occurs at k = α, but As really becomes infinitely large when k = α −iβ, rather than when k = α. (In complex analysis, this is referred to as a pole.) So because we have a pole at α −iβ, we’ll have large values of As near that pole. So this is how we mathematically search for resonances: if we have a formula for δ(k), then we can search for solutions to the equation tan δ(k) = −i. The real part of that solution k will then be the resonance α, and the smaller the imaginary part of k, the more resonant our system actually is. This viewpoint is additionally interesting because in the formula E = ℏ2k2 2m , we can now imagine plugging in a pure imaginary k = iκ (for some real number κ > 0). Then we now have a negative energy E = −ℏ2κ2 2m , which can instead represent bound states of our potential! So the complex k-plane has room for bound states, scattering states, and resonances, and in fact bound states will correspond to poles of As as well. We can even go a bit further and invent anti-bound states on the negative imaginary axis, in which instead of matching our solution to a purely decaying exponential in the forbidden region, we match it to a purely increasing exponential, and this turns out to actually have applications in nuclear physics! Lecture 20: Central Potentials and Angular Momentum We’ll now turn our attention fully to three-dimensional quantum mechanics, in particular starting a study of angular momentum. Just like position, momentum, or energy, angular momentum is an observable, and we will explore many useful properties of it soon. 95 Our starting point today is that there is a certain active view of observables that we haven’t explored much yet. Specifically, we’ve learned that the momentum operator ˆ p = ℏ i ∂ ∂x can be thought of as a differential operator (an x-derivative) which tells us how a function varies. In fact, this can be rephrased to saying that the momentum operator actually generates translations and moves functions in space, and the way it generates translations (and the universal trick with Hermitian operators) is to exponentiate it. If we want to exponentiate ˆ p, we need to make it unitless, and we’ll do this by considering the operator e i ˆ pa ℏfor some length a. Having this act on a wavefunction ψ(x) then yields e i ˆ pa ℏψ(x) = ea ∂ ∂x ψ(x) = ∞ X n=0 1 n! a d dx n ψ(x) = ψ(x) + adψ dx + a2 2! d2ψ dx2 + a3 3! d3ψ dx3 + · · · = ψ(x + a). So this exponential operator takes ψ(x) and translates it, and this characterization of momentum as a “generator of translations” is in some ways more fundamental than what we’ve already discussed! And similarly, angular momentum will then end up being a “generator of rotations” instead, but we’ll need a bit more mathematics to discuss it. There’s another story as well – we’ll see as the class progresses that angular momentum represents not just physical rotations but also spin angular momentum (the mysterious property that particles can have angular momentum without actually rotating). But let’s start from the beginning with the setup. As we’ve been mentioning throughout this class, many of the objects we work with in one dimension generalize to three dimensions as well: for example, the momentum operator becomes px = ℏ i ∂ ∂x , py = ℏ i ∂ ∂y , pz = ℏ i ∂ ∂z = ⇒⃗ p = ℏ i ⃗ ∇, , with the commutators [x, px] = [y, py] = [z, pz] = iℏ. The time-independent Schrodinger equation then becomes −ℏ2 2m∇2ψ(⃗ r) + V (⃗ r)ψ(⃗ r) = Eψ(⃗ r) for a three-dimensional vector ⃗ r, where we are trying to find an energy eigenstate ψ(⃗ r) of energy E. In this class, we’ll simplify the problem to the case where we have a central potential (in other words, spherically symmetric and invariant under rotations) which only depends on the magnitude of r: V (⃗ r) = V (r). This is a strong assumption, but it is the setting of many three-dimensional problems that we will face, and the simplification allows angular momentum to play a role (because “angular momentum generates rotations”). And having a central potential simplifies our calculations, because the Laplacian ∇2 in spherical coordinates looks like ∇2ψ = 1 r 2 ∂ ∂r r 2 ∂ψ ∂r ψ + 1 r 2 1 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 sin2 θ ∂2 ∂φ2 ψ. Our goal today will basically be to build up a structure that allows us to ignore the second term with the angular derivatives. It turns out that we can treat the expression −ℏ2 1 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 sin2 θ ∂2 ∂φ2 as the differential version of angular momentum (much like ℏ i ∂ ∂x was the differential version of momentum), And it makes sense that this expression instead represents squared angular momentum for a variety of reasons: to make the units match up (since ⃗ L should have units of ℏ), because angular momentum should be a first-order derivative (⃗ r × ⃗ p involves one derivative) while there is a second-order derivative in the expression above, and because angular momentum is supposed to be a vector. This will end up being the operator ⃗ L2. 96 Fact 132 The main 3-D system we’re going to be discussing in the remainder of this course is the hydrogen atom, which is a system of two particles. We’ll justify (next lecture) that the Schrodinger equation above is relevant in a two-body problem with potential only depending on distance \|⃗ x1 −⃗ x2\|, so we can reduce the two-body problem (in general) to a one-body problem with a spherically symmetric potential. For now, we’ll focus on developing the theory of angular momentum operators. Since we have ⃗ L = ⃗ r × ⃗ p classically, we can make an analogous definition quantum mechanically (which will turn out to be good): Definition 133 The angular momentum operators are defined in the cyclic manner ˆ Lx = ˆ y ˆ pz −ˆ z ˆ py, ˆ Ly = ˆ z ˆ px −ˆ x ˆ pz, ˆ Lz = ˆ x ˆ py −ˆ y ˆ px. (Notice that because position and momentum operators in different directions commute, it doesn’t matter whether we write ˆ y ˆ pz or ˆ pz ˆ y, and so on.) These angular momentum operators are Hermitian, because ˆ L† x = ˆ p† z ˆ y † −ˆ p† y ˆ z†, but all operators on the right-hand side are Hermitian so the daggers go away, and then we can switch the order back to the original order by commutativity. So ˆ L† x = ˆ Lx, and ˆ Lx is Hermitian (similar logic works for the other two operators). In other words, the operators ˆ Lis must be observables, but they’ll turn out to be not so simple in certain ways. Our first step once we define quantum operators is often to compute their commutators: first, let’s compute [ˆ Lx, ˆ Ly] = [ˆ y ˆ pz −ˆ z ˆ py, ˆ z ˆ px −ˆ x ˆ pz] . We start by looking at the first term in the first argument, ˆ y ˆ pz. Because ˆ y commutes with everything in the sec-ond argument, and ˆ pz only has a nonzero commutator with ˆ z, the only nonzero commutator here is [ˆ y ˆ pz, ˆ z ˆ px] = ˆ y[ˆ pz, ˆ z]ˆ px = −iℏˆ y ˆ px. (Here, we’re using properties that simplify expressions of the form [A, BC] and [AB, C] – we can check these steps more carefully as an exercise.) Similarly, the only nonzero commutator coming from −z ˆ py is [−ˆ z ˆ py, −ˆ x ˆ pz] = ˆ py[ˆ z, ˆ pz]ˆ x = iℏˆ py ˆ x, meaning that [ˆ Lx, ˆ Ly] = iℏ(ˆ x ˆ py −ˆ y ˆ px) = iℏˆ Lz . So even though there were a lot of terms in the commutator, the end result simplifies very nicely! It turns out this is not that miraculous – if we have symmetry transformations and take their commutator, we must get a symmetry back. (And for some classical intuition, we know that when we rotate in different directions, the order does matter, so this commutator should be nonzero.) Furthermore, because we defined our angular momentum operators in a cyclic way, we must also have [ˆ Ly, ˆ Lz] = iℏˆ Lx , [ˆ Lz, ˆ Lx] = iℏˆ Ly . These relations give us the quantum algebra of angular momentum, and this algebra in fact appears in many different fields of mathematics and physics – it’s related to the algebra of generators of the special unitary group SU(2), as well as the orthogonal group of rotations O(3). So the structure is universal and deeper than the derivation – if there are other sets of operators which satisfy these commutation relations but don’t come from ˆ xs and ˆ ps, they can still be 97 angular momentum operators, and that is in fact what happens with spin (in which we have [ ˆ Sx, ˆ Sy] = iℏˆ Sz and so on). Fact 134 Mathematicians approach this topic as the study of Lie algebras, and the job there is basically to classify all potential consistent commutation relations – the one we’ve described is the simplest nontrivial one. Since we have observables, it now makes sense to ask whether we can actually “observe” them, in the sense that there is a state with a definite value of ˆ Lx, ˆ Ly, or Lz. Recall that because position and momentum do not commute, we cannot tell simultaneously the values of both at once. And these angular momentum operators don’t commute either, so we basically cannot have simultaneous eigenstates of ˆ Lx, ˆ Ly, and Lz. More rigorously, suppose that we have a simultaneous eigenstate φ0 of ˆ Lx and ˆ Ly, with corresponding eigenvalues λx and λy. Then ˆ Lx ˆ Lyφ0 = ˆ Lx(λyφ0) = λxλyφ0, ˆ Ly ˆ Lxφ0 = ˆ Ly(λxφ0) = λxλyφ0, so subtracting yields [ˆ Lx, ˆ Ly]φ0 = 0 = ⇒ iℏˆ Lzφ0 = 0. This means that φ0 is then also an eigenstate of ˆ Lz with λz = 0. But now that we’re an eigenstate of all three operators, applying the same logic to a different commutator then tells us that λx = λy = 0 as well, so the only way to be a simultaneous eigenstate is if φ0 is killed by all of the operators ˆ Lx, ˆ Ly, and ˆ Lz, which is not very interesting. Thus, no state is a nontrivial eigenstate of two angular momentum operators, and we can only ever “measure” one of the values of ˆ Lx, ˆ Ly, and ˆ Lz at a given time. But it is possible to know more about the state, and this basically comes down to thinking of another operator that commutes with all of ˆ Lx, ˆ Ly, and ˆ Lz (meaning that it must be rotationally invariant). And this is now connected to an object we introduced earlier in the lecture: Definition 135 The operator ⃗ L2 is defined to be ⃗ L2 = ˆ Lx ˆ Lx + ˆ Ly ˆ Ly + ˆ Lz ˆ Lz. It turns out that this definition is equivalent to the −ℏ2 1 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 sin2 θ ∂2 ∂φ2 mentioned above, but we’ll work with the more algebraic definition for now. In particular (just doing a direct computation), [ˆ Lx, ⃗ L2] = [ˆ Lx, ˆ Lx ˆ Lx + ˆ Ly ˆ Ly + ˆ Lz ˆ Lz] = [ˆ Lx, ˆ Ly ˆ Ly] + [ˆ Lx, ˆ Lz ˆ Lz] = [ˆ Lx, ˆ Ly]ˆ Ly + ˆ Ly[ˆ Lx, ˆ Ly] + [ˆ Lx, ˆ Lz]ˆ Lz + ˆ Lz[ˆ Lx, ˆ Lz] = (iℏˆ Lz)ˆ Ly + ˆ Ly(iℏˆ Lz) + (−iℏˆ Ly)ˆ Lz + ˆ Lz −iℏˆ Ly . (Remember that we can move constants around but should not switch the order of the angular momentum operators.) And now the first and last terms cancel, as do the second and third, and thus ˆ Lx and ⃗ L2 commute, meaning that it is possible to be a simultaneous eigenstate of both operators! And similarly, ⃗ L2 then commutes with ˆ Ly and ˆ Lz as well. Fact 136 We’ll see more of this in 8.05, but it turns out (by linear algebra) that for any two commuting Hermitian operators, we can always find simultaneous eigenstates of both operators. (The reason for wanting simultaneous eigenstates is to discover more specific information about our states, and so we will generally aim for the maximal set of commuting operators. 98 Returning to coordinates now, remembering that in spherical coordinates we have x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, and thus we can also invert these equations to find r = p x2 + y 2 + z2, θ = cos−1 z r , φ = tan−1 y x , and the special role that z plays (over x and y) means that the angular momentum operator ˆ Lz looks the nicest in spherical coordinates, because rotations about z don’t change the value of φ. In particular, the multivariable chain rule tells us that ∂ ∂φ = ∂y ∂φ ∂ ∂y + ∂x ∂φ ∂ ∂x + ∂z ∂φ ∂ ∂z . The last term is zero because z doesn’t depend on φ, and we can compute the derivatives of x and y with respect to φ: we have ∂y ∂φ = r sin θ cos φ = x, and similarly ∂x ∂φ = −y, so ∂ ∂φ = x ∂ ∂y −y ∂ ∂x . Remembering that momentum operators are differential operators, we thus see that ˆ Lz = ℏ i x ∂ ∂y −y ∂ ∂x , meaning that ˆ Lz = ℏ i ∂ ∂φ in spherical coordinates! (The units of this expression are correct, because angles have no units and angular momentum should have units of ℏ.) And using this kind of technique (but more work), we can also get expressions for ˆ Lx, ˆ Ly, and ⃗ L2, the latter of which agrees with our expression above. So the angular momentum operator does indeed relate to the angular component of the Laplacian. Turning our attention back now to simultaneous eigenstates, we now wish to construct functions which are eigenstates of both ⃗ L2 and one of ⃗ Lx, ⃗ Ly, and ⃗ Lz. (For such functions, the angular part of the Laplacian is just a number, which will allow us to simplify the process of solving the Schrodinger equation to just solving a differential equation for the radial component.) Because ˆ Lz looks simple in spherical coordinates, we always choose ˆ Lz instead of the others by convention. We’ll denote the eigenstate ψℓ,m(θ, φ) for some numbers ℓ, m related to the eigenvalues (which are currently arbitrary). Since ˆ Lz should have units of ℏ, it makes sense to require that ˆ Lzψℓ,m = ℏmψℓ,m for some real number m. Similarly, since ⃗ L2 should have units of ℏ2, we’ll require that ⃗ L2ψℓ,m = ℏ2λψℓ,m for some real number λ. Additionally, λ must be positive because ⃗ L2 is a positive operator; specifically, ℏ2λ(ψℓ,m, ψℓ,m) = (ψℓ,m, ⃗ L2ψℓ,m) = (ψℓ,m, ˆ Lx ˆ Lxψℓ,m) + (ψℓ,m, ˆ Lx ˆ Lxψℓ,m) + (ψℓ,m, ˆ Lx ˆ Lxψℓ,m) = (ˆ Lxψℓ,m, ˆ Lxψℓ,m) + (ˆ Lyψℓ,m, ˆ Lyψℓ,m) + (ˆ Lzψℓ,m, ˆ Lzψℓ,m). The right-hand side is now nonnegative (because we’re calculating a squared norm), so the left-hand side must be nonnegative as well, meaning λ ≥0. Anticipating this, we will actually write ⃗ L2ψℓ,m = ℏ2ℓ(ℓ+ 1)ψℓ,m 99 for some real number ℓ∈R – this exact choice will make sense very soon as we return to the differential equation, and we’re not losing any generality here because ℓ(ℓ+ 1) can take on any nonnegative real value (and even some negative ones). In fact, because the range of ℓ(ℓ+ 1) over nonnegative ℓis [0, ∞), we can always pick ℓto be nonnegative. We’ll now solve the two boxed differential equations to wokr towards an explicit formula for ψℓ,m. To satisfy the first equation, we must have ℏ i ∂ ∂φψℓ,m = ℏmψℓ,m = ⇒ ∂ ∂φψℓ,m = imψℓ,m = ⇒ψℓ,m(θ, φ) = eimφP m ℓ(θ) for some arbitrary function P m ℓ. So the φ-dependence is not complicated if our state is an eigenstate of ˆ Lz, but we can actually say a bit more – we must have ψℓ,m(θ, φ+2π) = ψℓ,m(θ, φ), because φ is a coordinate that is only defined modulo 2π anyway. Thus we must require e2im = 1, or in other words that m must be an integer. So the eigenvalues of ˆ Lz actually need to be quantized – they can only be integer multiples of ℏ. For the second equation, a bit more work needs to be done. Substituting in the expression for ⃗ L2, we have −ℏ2 1 sin θ ∂ ∂θ sin θ ∂ ∂θ + 1 sin2 θ ∂2 ∂φ2 ψℓ,m = ℏ2ℓ(ℓ+ 1)ψℓ,m. We already know how the ∂2 ∂φ2 acts on ψℓ,m (each derivative contributes an im factor), and then the ℏ2s and eimφs on both sides cancel out. Further multiplying by −sin2 θ, we are instead solving the differential equation sin θ ∂ ∂θ sin θ ∂ ∂θ −m2 P m ℓ= −ℓ(ℓ+ 1) sin2 θPℓ,m. Remember that sin θ ∂ ∂θ sin θ ∂ ∂θ means that we apply these operators from right to left, so applying them on P m ℓis really sin θ d dθ sin θ ∂P m ℓ ∂θ . So moving everything to one side, and remembering that P is just a function of a single variable, we wish to find solutions to sin θ d dθ sin θdP m ℓ dθ + ℓ(ℓ+ 1) sin2 θ −m2 P m ℓ= 0 . This equation has been studied extensively (because it comes up when we are studying the Laplacian), and one useful parameter substitution to use is to have the variable x = cos θ instead, so that d dx = − 1 sin θ d dθ and sin θ d dθ = −(1−x2) d dx (we should check this as an exercise). But with these facts, plus the fact that sin2 θ = 1−x2, our differential equation can be rewritten to only involve polynomials: after some simplification (dividing by (1 −x2)), we arrive at d dx (1 −x2)dP m ℓ dx + ℓ(ℓ+ 1) − m2 1 −x2 P m ℓ(x) = 0, where our (now normalized) simultaneous eigenstates are ψℓ,m = Nℓ,meimφP m ℓ(cos θ) for some numbers Nℓ,m. This differential equation still looks rather complicated, and the way physicists approach it is usually to start with the m = 0 case (where P 0 ℓis usually just denoted Pℓ): d dx (1 −x2)dPℓ dx + ℓ(ℓ+ 1)Pℓ(x) = 0. We can solve this with a series solution: letting Pℓ(x) = P k akxk and substituting in, the xk coefficient gives us the recursion (much like for the harmonic oscillator) (k + 1)(k + 2)ak+2 + [ℓ(ℓ+ 1) −k(k + 1)]] ak = 0 = ⇒ak+2 ak = −ℓ(ℓ+ 1) −k(k + 1) (k + 1)(k + 2) . And just like with the harmonic oscillator, if our series solution doesn’t determinate, we’ll get singular solutions diverging 100 at x = 1 or x = −1, which is bad (since those points are included in the range of the wavefunction). So the series must terminate at some point, and that can only occur if ℓ(ℓ+ 1) −k(k + 1) = 0 for some nonnegative integer k. And now we see exactly why we chose this form for the eigenvalue of ⃗ L2 – since we’re picking ℓto be nonnegative (as mentioned earlier in the lecture), ℓ= k must be a nonnegative integer. So Pℓ(x) is a polynomial of degree ℓ (these polynomials are known as the Legendre polynomials), and we find quantization of the eigenvalues of ⃗ L2! In other words, both the angular momentum along the z-direction and the overall magnitude of the angular momentum are quantized. Next lecture, we’ll explore these equations some more, understanding what happens for nonzero m and what additional constraints exist for our operators. Lecture 21: Spherical Harmonics, Boundary Conditions, and the Hydrogen Atom Last lecture, we started studying angular momentum by constructing a set of operators that provide observables. Angular momentum becomes particularly important when we have a central potential (only depending on the radial variable r), and we’ll soon use this to help solve the hydrogen atom. But first we’ll review what we discovered last time: the commutators of ˆ Lx, ˆ Ly, ˆ Lz form an algebra of angular momentum, explaining that we cannot have simultaneous eigenstates of these operators. However, we can have simultaneous eigenstates of one of them (we pick ˆ Lz) with ⃗ L2, because [⃗ L2, ˆ Lz] = 0, which we denote ψℓ,m. And we calculated last time that the eigenvalues of this state for ˆ Lz and ⃗ L2 are of the form ℏm and ℏ2ℓ(ℓ+ 1), respectively, where m is an integer and ℓis a nonnegative integer. Remark 137. In Fact 136, we mentioned that the goal is often to be simultaneous eigenstates of as many operators as possible. One additional point here is that we have succeeded if we can uniquely characterize states using the corresponding observables! For example, we might have multiple degenerate states at the same energy, but they may have different values of ℓ, allowing us to tell them apart by measurement. And there might be degenerate states of the same E and ℓ, but then they may have different values of m. This is important because we want to always be able to tell apart two states with some physical difference, and that’s a setting we will see soon with the hydrogen atom. We also started working towards more explicit forms of the ψℓ,m states at the end of the lecture – last time, we found that for m = 0, we have ψℓ,0 = Nℓ,meimφPℓ(cos θ), where Pℓare the Legendre polynomials which turn out to be given by Pℓ(x) = 1 2ℓℓ! d dx ℓ (x2 −1)ℓ. We’ll complete this analysis today, beginning by studying nonzero m in the differential equation for P m ℓ. It turns out there is a simple rule for going from the Legendre polynomials to the Legendre functions: P m ℓ(x) = (1 −x2) \|m\| 2 d dx m Pℓ(x) Notice that the exponent of (1 −x2) can be a half-integer, but this is not actually a problem because (1 −x2) will become sin2 θ when we substitute it back into our coordinates. Checking that this P m ℓ solves the differential equation d dx h (1 −x2) dP m ℓ dx i + h ℓ(ℓ+ 1) − m2 1−x2 i P m ℓ(x) = 0 takes some work, and we won’t check it here, but what’s important is to notice that we can only take ℓderivatives before Pℓ(x) (a polynomial of degree ℓ) vanishes completely, and thus we must have \|m\| ≤ℓ⇐ ⇒ −ℓ≤m ≤ℓ. 101 Fact 138 It turns out that there are no additional regular (non-divergent) solutions to the differential equation besides the P m ℓs that we compute here. So this boxed condition is an actual constraint on our quantum numbers for angular momentum – given any magnitude for the angular momentum, there are only certain values for the angular momentum in the z-direction (which makes sense). In other words, the only state with ℓ= 0 will also have m = 0, but there are three states with ℓ= 1 (with m = −1, 0, 1 respectively), five states with ℓ= 2 (with m = −2, −1, 0, 1, 2 respectively), and so on – in general, there are 2ℓ+ 1 states at total angular momentum quantum number ℓ. This leads us back to the final normalized solution that we want: Theorem 139 The spherical harmonics are simultaneous normalized eigenstates of ˆ Lz and ⃗ L2 given by Yℓ,m(θ, φ) = s 2ℓ+ 1 4π (ℓ−m)! (ℓ+ m)!(−1)meimφP m ℓ(cos θ) when 0 ≤m ≤ℓand Yℓ,m(θ, φ) = (−1)m (Yℓ,−m(θ, φ))∗ when −ℓ≤m ≤0. We can find the explicit formulas for these by looking them up – they are complicated in general, but it’s useful to remember the ones for ℓ= 0, 1: Y0,0 = 1 √ 4π , Y1,±1 = ∓ r 3 8π e±iφ sin θ, Y1,0 = r 3 4π cos θ. And here, normalization means that the spherical harmonics are normalized over all θ, φ, so that (integrating over solid angle) Z (Yℓ,m(θ, φ))2dΩ= Z π 0 sin θ Z 2π 0 \|Yℓ,m(θ, φ)\|2dφdθ = 1. But we can recognize further that sin θdθ = −d(cos θ), so making a change of variables means that we are requiring Z 1 −1 Z 2π 0 \|Yℓ,m\|2dφd(cos θ) = 1 . (We will often represent this double integral by dΩto simplify our notation.) And because the different Yℓ,ms are eigenfunctions of operators with (at least one being) different eigenvalues, they must be orthogonal: Z dΩY ∗ ℓ′,m′(θ, φ)Yℓ,m(θ, φ) = δℓ,ℓ′δm,m′ . We’re now ready to return to the full Schrodinger equation in three dimensions for a central potential, which takes the form −ℏ2 2m∇2ψ + V (r)ψ = Eψ. Plugging in the form of the Laplacian in spherical coordinates, including the definition of ⃗ L2, we now have −ℏ2 2m 1 r d2 dr 2 rψ − 1 ℏ2r 2 ⃗ L2ψ + V (r)ψ = Eψ. 102 (Notice that ⃗ L2 and 1 r 2 commute, because the former only involves θ and φ, so we don’t have to worry about the ordering there. And if we want, we can prove that [⃗ L2, ˆ r] = 0 by direct calculation as well, but that’s more work.) So now we can make the most important simplification: try a factorized solution of the form ψ(⃗ r) = RE(r)Yℓ,m(θ, φ). Plugging this solution in, ⃗ L2 will just act as a number on ψ (because we have an angular momentum eigenstate), so there will be a factor of Yℓ,m that we can cancel everywhere, leaving us with −ℏ2 2m 1 r d2 dr 2 (rRE(r)) −ℓ(ℓ+ 1) r 2 RE(r) + V (r)RE(r) = ERE(r). So we now just need to solve a one-variable differential equation for RE(r), and then multiplying by a spherical harmonic will give us a solution to the Schrodinger equation with energy E, angular momentum ℓ, and z-component of angular momentum m. To make that easier, we’ll clean it up by multiplying everything by r: −ℏ2 2m d2 dr 2 (rRE(r)) + ℏ2 2m ℓ(ℓ+ 1) r 2 (rRE(r)) + V (r)(rRE(r)) = E(rRE(r)). Defining U(r) = rRE(r), we find that the differential equation we actually need to solve is −ℏ2 2m d2U(r) dr 2 + V (r) + ℏ2ℓ(ℓ+ 1) 2mr 2 U(r) = EU(r) . So the radial equation turns out to just be exactly like a a one-dimensional Schrodinger equation, except the radial dependence needs to be obtained by dividing through by r to get to RE(r), and we need to solve the equation repeatedly for all values of ℓ. We can basically think of this additional term ℏ2ℓ(ℓ+1) 2mr 2 as a centrifugal barrier, making it harder to reach the origin, so we now have an effective radial potential Veff(r) = V (r) + ℏ2 2mr 2 ℓ(ℓ+ 1) taking the usual place of V (r). And notice that even for a Coulomb potential V (r) ∝1 r , the second term will dominate near r = 0 for any nonzero ℓ, meaning that a particle with any nonzero angular momentum cannot reach the origin (intuitively because it must spin faster and faster). And now that we’re in the one-dimensional setting, many of the properties and theorems we’ve discovered previously in the class (about bound states and eigenstates and so on) suddenly become useful. Our full wavefunctions are thus of the form ψ(⃗ r) = U(r) r ψℓ,m(θ, φ), and we can notice that U(r) depends on ℓbut does not depend on m – the m-dependence in this wavefunction is really only showing up in the spherical harmonic. (And we can write UE,ℓ(r) to represent this.) We’ve talked about normalization for the spherical harmonics, but now we need to ask about normalization for the full wavefunction: we have that 1 = Z \|ψ\|2d3x = Z ∞ 0 Z \|U(r)\|2 r 2 \|Yℓ,m(θ, φ)\|2r 2drdΩ. But the r 2 factors will cancel, and the solid angle integral R \|Y \|2dΩis just one by normalization of spherical harmonics, and therefore everything simplifies to 1 = Z ∞ 0 \|U(r)\|2. So we truly can think of this as a one-dimensional problem – the normalization condition for ψ is just the usual one-dimensional normalization condition for U! This means that we’re successful, and we’ve reduced a spherical potential 103 to a one-dimensional problem. There is an additional difference, though – because r only runs over the positive reals, we might ask if there are boundary conditions that need to be satisfied as r →0. Our argument here will not be completely general, but it will be good enough – for any potential V (r) where the centrifugal barrier ℏ2ℓ(ℓ+1) 2mr 2 dominates as r →0 (such as the Coulomb potential), our differential equation will become −ℏ2 2m d2u dr 2 + ℏ2ℓ(ℓ+ 1) 2mr 2 u ≈0 = ⇒d2u dr 2 ≈ℓ(ℓ+ 1) r 2 u to leading order. This is essentially a Cauchy-Euler equation, and the solutions are proportional to r ℓ+1 or r −ℓ. But the latter can be ruled out – it’s not normalizable for ℓ≥1, and for ℓ= 0 we can’t actually have a valid exact solution to the Schrodinger equation (though we won’t talk about this in full detail). So instead we must have U(r) ∼r ℓ+1 as r →0 , and in particular U(r) always vanishes as r →0 for any ℓ, vanishing faster and faster for higher values of ℓ. So this is consistent with imagining an infinite barrier at r = 0. With all of the theory set up, we’re now ready to attack the main problem we’d like to solve: Example 140 Consider a proton (of position and momentum ⃗ xp and ⃗ pp, respectively) and electron (of position and momentum ⃗ xe and ⃗ pe, respectively), in which we use canonical variables for each particle. In other words, we have [(⃗ xp)i, (⃗ pp)j] = iℏδij (where i and j range over 1, 2, 3 to represent the x, y, or z-coordinates), and similarly [(⃗ xe)i, (⃗ pe)j] = iℏδij. Since we now have a system of two particles, our wavefunction should account for both, meaning that we now have a function Ψ(⃗ xe,⃗ xp). The interpretation of this wavefunction is still that the squared wavefunction is the probability density, but specifically dP = \|Ψ(⃗ xe,⃗ xp)\|2d3⃗ xed3⃗ xp is the probability that we find the electron in a region d3⃗ xe around ⃗ xe and the proton in a region d3⃗ xp around ⃗ xp. (So if we just care about the position of the proton, we’d integrate this density out across all d3⃗ xe.) And what’s important to note here is that there is still only one wavefunction and one Schrodinger equation, even with multiple particles, but this time our Hamiltonian looks a bit more complicated: we have ˆ H = (⃗ pp)2 2mp + (⃗ pe)2 2me + V (\|⃗ xe −⃗ xp\|), because each particle has a kinetic energy and the potential energy between them depends only on their distance. So in our Schrodinger equation Hψ = Eψ, we treat ⃗ pp as taking derivatives of the proton’s position and ⃗ pe as taking derivatives of the electron’s position. But this Schrodinger equation now has a lot of variables, and our goal will be (as we’ve previously mentioned) to simplify the equation so that it reduces to an equation for the distance between the particles. The change of variables we must perform is now motivated by classical mechanics: because the center-of-mass moves at constant velocity in a classical two-body problem, we should also be able to define a new quantum coordinate and momentum for the center-of-mass. So we’ll define ⃗ P = ⃗ pp + ⃗ pe , 104 and now we need to define a corresponding coordinate ⃗ X – motivated by classical mechanics, we try ⃗ X = mp⃗ xp + me⃗ xe me + mp . Now we must check if these are indeed a valid coordinate-momentum pair; in other words, we must check that [ ⃗ Xi, ⃗ Pj] = iℏδij. Indeed, this commutator is mp(⃗ xp)i + me(⃗ xe)i me + mp , (⃗ pp)j + (⃗ pe)j = mp me + mp [(⃗ xe)i, (⃗ pe)j] + me me + mp [(⃗ xe)i, (⃗ pe)j] (since commutators between proton and electron coordinates and momenta are zero), and this is zero unless i = j, in which case it is iℏ mp me+mp + me me+mp = iℏ. So yes, this is a valid pair of quantum mechanical canonical variables (they have the right units and commutator). To get a second one, it makes sense to use the relative coordinate ⃗ x = ⃗ xe −⃗ xp which appears in our expression for the Hamiltonian. But even when we write this down, we have to be careful – remembering that the variables for the proton and electron commuted, we must also have that the relative and center-of-mass variables commute with each other. This does turn out to be the case for ⃗ x with ⃗ X and ⃗ P – all ⃗ xs commute with each other (so there’s no problem there), and when taking commutators of ⃗ x with ⃗ P, the negative sign for the proton commutator means that the iℏs cancel out. So we just need to define a corresponding combination of the momenta ⃗ p = α⃗ pe −β⃗ pp. This will always commute with ⃗ P, but we need to choose the coordinates so that [⃗ xi, ⃗ pj] = iℏδij (this turns out to enforce α+β = 1) and that all components of ⃗ p commute with ⃗ X (this enforces αmℓ−βmp = 0). Solving the system of two equations gives us ⃗ p = mp me + mp ⃗ pe − me me + mp ⃗ pp. So we now have two new pairs of canonical variables, and to make the notation simpler, it’s useful to define the reduced mass and total mass µ = memp me + mp , M = me + mp. (Since we’re in a setting where the proton is much heavier than the electron, we’ll have µ ≈me and M ≈mp in this case.) In terms of these variables, we then have the unit-free constants α = µ me and β = µ mp , so that another way to write our relative momentum is ⃗ p = µ ⃗ pe me −⃗ pp mp . The point of doing all of these calculations is that we want to write the two-body Hamiltonian in terms of our new coordinates. Since ⃗ P and ⃗ p are both linear combinations of ⃗ pe and ⃗ pp, we can solve to find that ⃗ pp = mp M ⃗ P −⃗ p, ⃗ pe = me M ⃗ P + ⃗ p . And now is where the effort pays off: the kinetic energy terms in the Hamiltonian are (⃗ pp)2 2mp + (⃗ pe)2 2me = 1 2mp mp M ⃗ P −⃗ p 2 + 1 2me me M ⃗ P + ⃗ p 2 , 105 and now the cross-terms vanish: the kinetic energy operator in the Hamiltonian indeed simplifies to (⃗ pp)2 2mp + (⃗ pe)2 2me = 1 2M ⃗ P 2 + 1 2µ ⃗ p2 , a center-of-mass contribution and a relative contribution. And next lecture, we’ll see that this allows us to separate the Schrodinger equation into center-of-mass motion and relative motion, reducing to a one-body problem. Lecture 22: Solving the Hydrogen Atom Last lecture, we rewrote the hydrogen atom Hamiltonian operator in terms of two new pairs of canonical variables (center-of-mass and relative position): ˆ H = (⃗ pp)2 2mp + (⃗ pe)2 2me + V (\|⃗ xe −⃗ xp\|) = 1 2M ⃗ P 2 + 1 2µ ⃗ p2 + V (\|⃗ x\|). In other words, this Hamiltonian describes a system in which the center-of-mass moves as a free particle of mass M = mp + me, and in which the relative position moves as a particle of reduced mass µ = memp me+mp under a central potential V only depending on the magnitude of the distance. But to actually check that, we need to solve the Schrodinger equation, and we have to be a bit careful with the two different coordinates – we have ⃗ P = ℏ i ∇⃗ X and ⃗ p = ℏ i ∇⃗ x (gradients must be taken with respect to the corresponding canonical coordinate). So to solve the (time-independent) Schrodinger equation, we’ll write down our wavefunction ψ( ⃗ X,⃗ x) in terms of the new coordinates, and we’ll do separation of variables ψ( ⃗ X,⃗ x) = ψCM( ⃗ X)ψrel(⃗ x). Substituting this into ˆ Hψ = Eψ (grouping the second and third terms of ˆ H together), we have 1 2M ⃗ P 2ψCM( ⃗ X) ψrel(⃗ x) + 1 2µ ⃗ p2ψrel(⃗ x) + V (\|⃗ x\|)ψrel(⃗ x) ψCM( ⃗ X) = EψCM( ⃗ X)ψrel(⃗ x) (since ⃗ P only acts on the center of mass and ⃗ p only acts on the relative motion). If we now divide by the total wavefunction ψCM( ⃗ X)ψrel(⃗ x), we have 1 ψCM(⃗ x) 1 2M ⃗ P 2ψCM( ⃗ X) + 1 ψrel(⃗ x) 1 2µ ⃗ p2ψrel(⃗ x) + V (\|⃗ x\|)ψrel(⃗ x) = E. But now the first term depends only on ⃗ X, while the second term depends only on ⃗ x, so the only way their sum can be a constant is if each term is a constant – let’s denote those constants ECM and Erel, respectively. So we now have two simpler differential equations to solve, namely (now multiplying back by ψCM and ψrel respectively) 1 2M ⃗ P 2ψCM( ⃗ X) = ECMψCM(⃗ x), 1 2µ ⃗ p2ψrel(⃗ x) + V (\|⃗ x\|)ψrel(⃗ x) = Erelψrel(⃗ x), E = ECM + Erel. So we have two time-independent Schrodinger equations for ψCM( ⃗ X) and ψrel(⃗ x), under no potential and the potential V (\|⃗ x\|), respectively. But we know how wavefunctions look for a free particle (momentum plane waves), and we have been building up the theory for dealing with a particle in a central potential! So intuitively, a hydrogen atom moves as a free particle, but then the behavior of the electron relative to the proton is governed by some central interaction. We can now start to study the actual potential that we have in a hydrogen atom: 106 Example 141 (Hydrogen-like atom) Suppose that instead of a single proton, we have a nucleus of Z protons. Then the potential for the relative motion is given by V (r) = −Ze2 r , where r = \|⃗ x\| is the distance between the particles and e is the charge of a proton (remembering that a proton and electron have opposite charge). The characteristic length scale for this system is the Bohr radius, given by setting equal the units from kinetic and potential energy: since ℏ a0 has units of momentum, we can set equal ℏ2 ma2 0 = e2 a0 = ⇒ a0 = ℏ2 me2 . Intuitively, the e appears in the denominator here, because the stronger the Coulomb force, the smaller the atom should be (because the electron is more tightly bound). To get a rough estimate of this quantity, we can write a0 = ℏ2c2 e2mc2 = ℏc e2 ℏc · mc2 , and now we use the fact that the fine structure constant α = e2 ℏc is approximately 1 137, and we plug in the mass of the electron instead of the reduced mass (it makes only a small percentage difference): = 197 MeV · fm 1 137 · 0.5 MeV ≈52.9 pm, or about 0.529 Angstroms. We then have a corresponding energy scale (plugging in a0 for r into the energy) e2 a0 = e4m ℏ2 = e4 ℏ2c2 mc2 = α2mc2 In other words, the bound state energies in this problem are on the order of α2 of the rest energy 511 keV of the electron, which is about 27.2 eV. Half of this quantity is the potentially-familiar 13.6 eV, which is the negative of the true ground state energy of the hydrogen atom. But of course we wouldn’t know that at this stage, since we’re just doing dimensional analysis! And connecting this to some calculations we did at the beginning of 8.04, note that the Compton wavelength of the electron is a factor 2π off of αa0 = λc 2π = ℏ mc ≈400 fm, much smaller than the Bohr radius, and then we get the classical electron radius with another factor of α: α2a0 ≈2.8 fm. (For comparison, the size of a proton is approximately 1 fm.) With this, we now have a general scale (length and energy) for the problem we’re considering, and we’ll solve the Schrodinger equation for bound states of this potential (meaning that E < 0) – scattering states are usually covered in more advanced courses. This is an interesting physical problem, because we’ll get the exact forms for the energy levels of a system with many applications. Remark 142. There are many other effects that change the energy levels of the actual hydrogen atom, such as the fine structure, relativitistic effects, and so on. But those are higher order effects, and we won’t discuss them for now. 107 Plugging in our V (r) into the radial Schrodinger equation, we are trying to solve for solutions to −ℏ2 2m d2 dr 2 + ℏ2ℓ(ℓ+ 1) 2mr 2 −Ze2 r U(r) = EU(r) with E < 0. (Remember that the solution U depends on E and ℓ, so we need to solve for the allowed energy levels for each of ℓ= 0, 1, 2, · · · – once we find the solutions to the differential equation, our full wavefunction will be ψ(r, θ, φ) = U(r) r Yℓ,m(θ, φ).) Our first step will be to get rid of units and replace r by a unit-free variable – we do this by making use of the characteristic length a0, and it turns out the other constants will simplify best if we define r = a0 2Z x : skipping some of the algebra here (but using that ℏ2 m = a0e2 and that d2 dr 2 = 4Z2 a2 0 d2 dx2 ), we end up with 2Z2e2 a0 −d2 dx2 + ℓ(ℓ+ 1) x2 −1 x U(x) = EU(x) = ⇒ −d2 dx2 + ℓ(ℓ+ 1) x2 −1 x U = −κ2U, which is now a unit-free equation in terms of the unit-free energy κ2 = − E 2Z2e2 a0 – our goal is now to know what values of κ give us valid solutions to the differential equation. The equation now looks a lot simpler than before, but it’s still unfortunately not that easy to solve – in particular, if we try to write down a series solution with the equation in its current form, we will get a three-term recursion (which is more complicated than a two-term one). A good strategy for simplifying is to look at the behavior of the solution near zero or near infinity – because the second and third terms on the left-hand side vanish as x →∞, the differential equation is approximately −d2U dx2 = −κ2U, which is solved by U = e±kappax (we’ll want the decaying solution for normalization reasons). This motivates us to again make a transformation ρ = κx = 2κz a0 r, which turns the differential equation into −d2 dρ2 + ℓ(ℓ+ 1) ρ2 −1 κρ U = −U . (Note that it’s important that κ still shows up somewhere in this equation, since the problem we’re trying to solve fundamentally involves fixing the allowed values of the energy.) And now if we look at the solution as ρ →0 (which is the same as taking r or x to 0), we know that u ∼ρℓ+1 from the discussion last lecture. So we know that the solutions to this equation are not polynomial (because they decay exponentially as e−κx = e−ρ) and also that the don’t start in the power series until ρℓ+1. To encode all of that information, we’ll write down the ansatz U(ρ) = ρℓ+1W(ρ)e−ρ for some unknown function W(ρ), with the hope that W(ρ) will be simpler than U(ρ). Plugging into the boxed differential equation (which just involves a lot of algebra), we get something that initially looks like it’s worse than the original: ρd2W dρ2 + 2(ℓ+ 1 −ρ)dW dρ + 1 κ −2(ℓ+ 1) W = 0. But this is actually a better equation for a series solution, because if we plug in W = ρk into the left-hand side, every term ends up being proportional to either ρk−1 or ρk, meaning that we will get a simple one-step recursion relation (given ak, we can find ak+1). So letting W = P∞ k=0 akρk, we can get our recursion relation by setting the total ρk coefficient to zero, and we find ak+1 ak = 2(k + ℓ+ 1) −1 κ (k + 1)(k + 2ℓ+ 2) . Looking at the limiting behavior as ρ →∞, we find that ak+1 ak ∼2 k . But this doesn’t decay fast enough – in particular, 108 even if ak+1 ak = 2 k+1 (which even decays slightly faster), we would have ak = 2ka0 k! , and that would give us X k akρk ≈a0 X 2k k! ek = a0e2ρ. And this is actually nicely consistent with what we found earlier – if the series does not terminate, then our solution is proportional to a polynomial factor times eρe−2ρ, which means the behavior near ρ →∞has a e+ρ factor instead of an e−ρ one, which still solves the differential equation but just diverges. But if we want normalizable bound states, our series must be truncated at some point. For the sake of characterizing our solutions, we’ll be careful with this step: if we want W to be a polynomial of degree N, then aN ̸= 0 but aN+1 aN = 0 (and then aN+1, aN+2, aN+3, · · · will all automatically be zero because we have a one-step recursion problem). So then we must have 1 κ = 2(N + ℓ+ 1), and because κ is related to the energy E, we see that the energy is quantized – it has to be the negative reciprocal of some integer! And ℓcan take on any nonnegative integer value (because that’s how we set up the problem), and so can N (because it’s the degree of our polynomial W(ρ)). But that means that there is degeneracy in this problem – for any given value of N + ℓ+ 1, there are many ways we can choose N and ℓto get the correct sum. So defining the principal quantum number n = N + ℓ+ 1 = 1 2κ, which is always a positive integer, we now have a formula for the nth energy level of the hydrogen atom: rearranging the definition of κ, En = −2Z2e2 a0 κ2 = −Z2e2 2a0 · 1 n2 , and we’ve derived the famous 1 n2 factor for the energy levels of the hydrogen atom, involving the famous constant e2 2a0 = 13.6 eV. For visualization, if we imagine drawing all of the nonnegative integer lattice points in the (N, ℓ)-plane, then diagonally slanting lines correspond to a fixed value of n: N ℓ n = 1 n = 2 n = 3 n = 4 In general, we see that there are n possible solutions where N + ℓ+ 1 = n (with the constraint 0 ≤ℓ, N ≤n −1), which are (0, n−1), (1, n−2), · · · , (n−1, 0), and all of these solutions are at energy level En. But that does not mean that there is just an n-fold degeneracy, because now we have to remember that m can range over the (2ℓ+ 1) values from −ℓto ℓ, giving us different energy eigenstates even if we have the same function U! So for n = 1, we must have ℓ= 0, which only gives us one state. But then for n = 2, we can have ℓ= 1 or ℓ= 0, giving us 3 + 1 = 4 states. And for n = 3, the allowed values are ℓ= 2, 1, 0, corresponding to 5 + 3 + 1 = 9 overall states – this pattern continues, and there are n2 total states at energy level En. And to summarize, each state in the hydrogen atom comes with three quantum numbers, namely 109 • the principal quantum number n, which fixes the energy of the state and can be any positive integer, • the angular momentum quantum number ℓ(this is more significant physically than the degree of the polynomial N = n −ℓ−1, so it makes sense to use this), which is a positive integer between 0 and (n −1), and • the magnetic quantum number m, representing the z-component of angular momentum, which is an integer between −ℓand ℓ. Returning to the variables we’ve defined along the way, now that we know κ = 1 2n, we see that ρ = 2κZ a0 r = Zr na0 , so the full wavefunction solution in three dimensions is, up to constant factors, ψn,ℓ,m = Un,ℓ r Yℓ,m ∼ρℓ+1 ρ Wn,ℓ(ρ)e−ρYℓ,m(θ, φ) = ρℓW(ρ)e−ρYℓ,m(θ, φ) where W(ρ) is a polynomial of degree N = n −ℓ−1 (these are called the Laguerre polynomials). And converting back to the usual radial coordinate, we have ψn,ℓ,m(r, θ, φ) = A r a0 ℓ polynomial in r a0 of degree (n −ℓ−1) e−Zr/(na0)Yℓ,m(θ, φ) for some normalization constant A. For reference, the simplest solution ψ1,0,0 is spherically symmetric and looks like ψ1,0,0 = 1 p πa3 0 e−r/a0 when Z = 1, so we now have an explicit formula for the ground state of the hydrogen atom! Lecture 23: Spectrum, Degeneracy, and Other Properties of the Hydrogen Atom Last lecture, we found the energy levels of the hydrogen (or hydrogen-like) atom Hamiltonian, and we found that we can describe the eigenstates ψn,ℓ,m with three quantum numbers, where the principal quantum number n tells us the energy En = −Z2e2 2a0n2 of the state, the angular momentum quantum number 0 ≤ℓ≤n −1 tells us the total angular momentum of the state, and the magnetic quantum number −ℓ≤m ≤ℓtells us the state’s angular momentum in the z-direction. And the wavefunction itself (in spherical coordinates) consists of a product of a normalization factor, a r a0 ℓ term (behavior as r →0), a Laguerre polynomial in r a0 of degree (n −ℓ−1), an exponential decay e−Zr/(na0) (behavior as r →∞), and a spherical harmonic Yℓ,m. Remark 143. Only very detailed calculations will actually require us to know the specific polynomial terms, so we won’t talk much about it here. And it’s often faster to just do the recursion relation calculation, rather than trying to search up the wavefunction and figure out discrepancies between units and conventions. It turns out that there’s a classic way to draw the energy levels of bound states for a central potential, in which we represent energy on the y-axis and ℓon the x-axis. We’ll use units of E Z2e2 2a0 for our system, so that the energies are at −1 n2 for positive integers n – the diagram below is not drawn exactly to scale because the energy levels would be much more squashed near E = 0: 110 0 E Z2e2 2a0 −1 ℓ= 0 ℓ= 1 ℓ= 2 ℓ= 3 N = 0 N = 1 N = 2 N = 3 N = 0 N = 1 N = 2 N = 0 N = 1 N = 0 n = 1 n = 2 n = 3 n = 4 Basically, each dash represents a different E and ℓfor which we have an energy eigenstate, so each column essentially corresponds to a different differential equation (because the effective potential changes depending on the value of ℓ). And for our system, incrementing n by one means that we have a state at energy En in a new column (specifically ℓ= n −1), and N always ranges from (n −1) to 0 within any row. But there’s more that we can say: for any given column (that is, any particular value of ℓ), it makes sense that the lowest energy state has N = 0, and the next energy state has N = 1, and so on, because the node theorem for a one-dimensional potential (the radial equation for any given ℓ) tells us that the ground state has no nodes, the next state has one node, and so on, and the only term in ψn,ℓ,m that can vanish is this polynomial. (Here, behavior at r = 0 doesn’t count because that’s the boundary of the wavefunction, which doesn’t count as a node.) So the incrementing values of N should not be surprising to us, and in fact we learn that the Laguerre polynomials of degree N must have N distinct real zeros. On the other hand, the degeneracy across values of ℓis extraordinary and very special to this problem! Usually, when we solve the radial equations for ℓ= 0, 1, 2, · · · , we will get different energy levels for each of those problems, but there is no reason a priori that the energy levels should actually line up for the hydrogen atom (and do so in such an organized way). We do expect some degeneracy to always be present – for example, each line above in the ℓ= 2 column really corresponds to a multiplet of five states with m = 2, 1, 0, −1, −2 – but the additional degeneracy between different ℓs is very special and is not explained by any of the theory we’ve discussed in this class so far. Fact 144 This fact led to developments with the Runge-Lenz vector, which is a conserved vector in planetary orbits in Newton’s theory. Precession of elliptical orbits is not allowed under Newton’s theory (though it is under Einstein’s theory of relativity), so it is not allowed for the hydrogen atom, and in fact that is connected to why we have all of these hidden degeneracies. For comparison, if we had instead solved the infinite spherical well problem (which we do in 8.05), in which the potential is zero inside a sphere and infinite outside, we do not see the coincidences in energy levels between different ℓs, even though the potential looks very simple in that case. The point to take away here is that this kind of hydrogen atom behavior can only be explained by an extra symmetry! 111 Example 145 For some intuition, we’ll start with the Z = 1 ground state, ψ1,0,0 = 1 √ πa3 0 e−r/a0, and understand how to obtain the general-Z ground state without much additional work. Remembering that the potential V (r) goes from −e2 r to −Ze2 r when we go from 1 to Z protons, we should substitute e2 with Ze2 everywhere in our wavefunction. And because a0 = ℏ2 me2 , this means that we should replace a0 with a0 Z everywhere as well. (Often we write solutions in a “mixed way” involving both e and a0, so we should make sure to make both changes.) So in this case, the ground state wavefunction for a general hydrogen-like atom is ψ1,0,0(r, θ, φ) = s Z3 πa3 0 e−Zr/a0. (Intuitively, the fact that the “normalization did not care about a0 appearing in the wavefunction” means that if we replace a0 with any other constant, in this case a0 Z , we will still have a normalized wavefunction.) And we’ve discussed this before in slightly different words, but the reason for the Zr a0 factor in the exponential is that the two of the terms in the radial differential equation go to zero as r →∞, leaving just −ℏ2 2m d2U dr 2 = EU. This should have solutions proportional to e±√ 2mE/ℏ2r, and plugging in the value of E1 (or En in general) indeed recovers the correct exponential decay for the ground state. We can now turn our attention to another aspect of the hydrogen atom: Definition 146 A Rydberg atom is an atom where the outermost electron is in a very high principal quantum number n. In such a setting, we can imagine having a nucleus of charge Ze, a lot of electrons around it, and then the last electron sees the charge from both the nucleus and the (Z −1) other electrons. So essentially by Gauss’s law, the outermost electron (if it’s outside the shell of the remaining electrons) sees an overall charge of +1. (It can be checked that the electrons do appear further and further away from the nucleus for higher n, so this argument is basically valid.) So to a good approximation, a Rydberg atom behaves a lot like a hydrogen atom. A good first step is for us to calculate the size of a Rydberg atom (this doesn’t really make sense precisely, but we can ask for the expectation of the radius), and it turns out the answer is not actually a constant factor times na0, even though the exponential factor in the wavefunction is e−r/(na0). Instead, the intuitive explanation will come from the virial theorem: Theorem 147 (Virial theorem) For the hydrogen atom (or any potential proportional to 1 r ), we have ⟨T⟩= −1 2⟨V ⟩ where T is the kinetic energy and V is the potential energy. (We won’t prove this theorem for now, but it involves some algebraic calculations with commutators.) We can picture this as saying that the kinetic energy ⟨T⟩and the bound state energy Eb = ⟨T⟩+ ⟨V ⟩are negatives of each other, since ⟨V ⟩= −2⟨T⟩. That means that ⟨V ⟩= 2Eb, or for the Rydberg atom we’re discussing (for which Z = 1 as discussed), −e2 r = 2Eb = 2 −e2 a0 1 n2 = ⇒ 1 r = 1 n2a0 . 112 So this suggests that the typical radius is actually on the order of n2a0 instead of na0. Everything we’ve discussed above is exact, but note that 1 r = 1 n2a0 does not imply that ⟨r∠= n2a0 (we can’t manipulate expectation values that way). In fact, the expectation of r actually depends on ℓ, and it can be calculated with a bit more effort to be ⟨r⟩= n2a0 1 + 1 2 1 −ℓ(ℓ+ 1) n2 . (In particular, this expectation is 3 2n2a0 for ℓ= 0 and approximately n2a0 for ℓ= (n −1).) And now we can look back at the wavefunction to understand where this n2 factor comes from: our solution is of the form ψn,ℓ,m = Ar ℓWn,ℓ(r)e−r/(na0)Yℓ,m(θ, φ) = fn,ℓ(r)Yℓ,m(Ω), where fn,ℓis a polynomial of degree (n −ℓ−1) + ℓ= n −1 times an exponential factor. So the probability density of finding the radius to be between r and r + dr is p(r)dr = r 2dr Z \|ψ\|2dΩ= r 2dr\|fn,ℓ\|2 Z \|Yℓ,m\|2dΩ (since we need to integrate out \|ψ\|2d3x along the angular coordinates). But by definition of the spherical harmonics, the integral over solid angle is 1, and then cancelling the drs gives us the radial probability density p(r) = r 2\|fn,ℓ\|2. And it is the polynomial factor that is causing us trouble here – since fn,ℓcontains a degree (n −1) polynomial, let’s just keep the leading order term r n−1 and say that fn,ℓ(r) ∼r n−1e−r/(na0), so that (plugging back in) p(r) ∼r 2ne−2r/(na0). So there is a “fight” between the exponential and the polynomial factor in the probability density – since r 2n is increasing and e−r/(na0) is decreasing, their product will basically be sharply peaked, and it will be have a maximum where the derivative p′(r) is zero: 0 = p′(r) = 2n r − 2 na0 r 2ne−r/(na0) = ⇒r = n2a0, giving us the expected n2 dependence for the radius! Example 148 Rydberg atoms can be observed in nature in interstellar gases – during recombination, when protons capture electrons to form atoms while the universe cooled down, they may capture the electron at a high quantum number. In astrophysics, n = 350 has been measured, which means that the size of the Rydberg atom is about 3502a0 ≈ 6.5µm. This size is on the order of the size of a red blood cell (which is about 8 µm) and just a bit smaller than the diameter of a hair (which is about 50 µm). Also, these atoms are relatively stable – while an electron in the n = 5 energy state would jump to n = 1 within about 10−7 seconds, these Rydberg atoms with very high n sometimes last a millisecond or a tenth of a second (because it takes a long time to spiral down the energy levels when the distance between them is on the order of 1 n3 ). These atoms can be created using lasers in the lab, and they can be detected using ionization (because the energy required to ionize these atoms is much smaller than those for ordinary atoms). In other words, Rydberg atoms are semiclassical, and in fact Bohr’s calculations of the energy levels of the hydrogen atom (which were correct) didn’t 113 require quantum mechanical derivations – he just needed to assume quantization on the allowed photons emitted during energy transitions. Example 149 We’ll finish this lecture by discussing the effect of the effective potential in more detail – specifically, we’ll consider a large n (like 100) and see what happens to the orbit of the electron as ℓvaries. Initially, we might think that ℓ= 0 will be the most circular out of the different states (because it’s spherically symmetric), but in fact it will be the most elliptical (and ℓ= 99 is the most circular). The intuition is as follows: the effective potential for the radial equation is Veff = ℏℓ(ℓ+ 1) 2mr 2 −e2 r , which looks roughly as shown below (with 1 r dominating for large r and 1 r 2 dominating for small r): r Veff En r− r+ If we then imagine a particle at energy En rolling around in this potential, we get a more elliptical orbit if the distance between the two endpoints r+ and r−(where Veff(r) = En) is larger. (After all, if we imagine a planetary orbit, the most elliptical orbits get very close and very far away from the sun, and the most circular ones have a constant r throughout the whole orbit.) So when ℓ= 0, because we don’t even have the upward behavior in the Veff graph as r →0, the electron reaches all the way to r = 0 and we have the largest possible orbit. And when ℓgets larger and larger, the ℏℓ(ℓ+1) 2mr 2 factor will push the blue curve higher and higher, making the endpoints closer and closer. So we get the most circular orbit when Veff is almost tangent to the horizontal line at En, and that comes from the top value of ℓ= n −1. For another way of thinking about this, circular orbits do in fact have a lot of angular momentum ⃗ r × ⃗ p (because the two vectors are orthogonal), but elliptical orbits do not because the angle between ⃗ r and ⃗ p is very small when the particle is far from the center. And next lecture, we’ll do some more calculations with r+ and r−to understand how they relate to the elliptical orbit, but the punchline is that r++r− 2 = n2a0 regardless of the value of ℓ! So in our hydrogen atom spectrum that we drew in the beginning of this lecture, we should think of the energy states of the same n as elliptical orbits of different eccentricity. And this connects to Kepler’s discovery that the total period of an orbit depends only on r−+ r+, the major diameter of the ellipse, meaning that ellipses of the same major diameter have the same energy. Lecture 24: Hydrogen Atom and Spin Angular Momentum Last lecture, we drew the spectrum of the hydrogen atom (which, for central potentials in general, involves drawing the energy levels for the various bound states in different columns for ℓ= 0, 1, 2, and so on). For a generic central 114 potential V (r), solving the Schrodinger equation for the various effective potentials Veff(r) = V (r) + ℏℓ(ℓ+1) 2mr 2 involves solving a bunch of different differential equations, and there is no reason that the energy levels will line up. But in the hydrogen atom, a special symmetry happens to make the energy levels very degenerate (even beyond the (2ℓ+ 1)-fold degeneracy coming from the different values of m we can have for each (n, ℓ)). At the end of the discussion, we started analyzing the various orbits of the hydrogen atom for a fixed n (which is where our “semi-classical” arguments are more valid). In such a situation, we can imagine that our electron is making elliptical orbits around the proton, and the r−and r+ from the diagram below Example 149 correspond to the shortest and longest distance to the proton, which together make up the major axis of the ellipse. (Classically, we can imagine that the electron is oscillating between r−and r+. But quantum mechanically – and thus more accurately – we have a wavefunction \|ψ\|2 which is exponentially decaying outside of the region [r−, r+], and the probability density will mimic the time spent at various rs in an elliptical orbit.) Example 150 We claimed last lecture that the “turning points” where Veff(r) = En satisfy r−+r+ 2 = n2a0, and we’ll verify that fact now with an explicit calculation. Plugging in the formulas for the energy levels and the effective potential, we have ℏ2ℓ(ℓ+ 1) 2mr 2 −e2 r = −e2 2a0 1 n2 . This is a quadratic equation in r, but we can do a transformation r = a0x to get the equation ℏ2ℓ(ℓ+ 1) 2ma2 0 1 x2 −e2 a0x = − e2 2a0n2 , but now ℏ2 2ma2 0 = ℏ2 2ma0 · 1 a0 = ℏ2 2ma0 · me2 ℏ2 = e2 2a0 , so all of the e2 2a0 factors cancel out and we’re left with ℓ(ℓ+ 1) x2 −2 x = −1 n2 = ⇒1 x = 1 ± q 1 −ℓ(ℓ+1) n2 ℓ(ℓ+ 1) by the quadratic formula (treating 1 x as the variable). We thus get the two solutions x = ℓ(ℓ+ 1) 1 ± q 1 −ℓ(ℓ+1) n2 = ℓ(ℓ+ 1) 1 ∓ q 1 −ℓ(ℓ+1) n2 1 − 1 −ℓ(ℓ+1) n2 = n2 1 ∓ r 1 −ℓ(ℓ+ 1) n2 ! by rationalizing the denominator (multiplying numerator and denominator by the radical conjugate), and thus r+ and r−are just a0 times those two solutions. And indeed, the ± factor cancels out, and r+ + r−= 2a0n2, recovering the result from last lecture. Furthermore, r+ ≈r−when ℓis large, because the additional ± term is close to zero in that case, and r−= 0 and r+ = 2n2a0 when ℓis zero, giving us the most elliptical orbit. (But because this is still a semi-classical argument, the idea of an “orbit” is not so accurate at the extreme values.) For a numerical example, in a typical Rydberg atom with n = 100 and ℓ= 60, we have r+ ≈18000a0 and r−≈2000a0, but every other state with n = 100 will also have r+ + r−= 20000. Example 151 To finish the course, we’ll turn our attention to motivating spin angular momentum and the mathematical treatment of spin by considering the “simplest quantum system.” 115 We might initially say that the simplest quantum system is the particle in the box, but that system has infinitely many energy eigenstates, each of which is a function on an interval. And even states with only one bound state, like the delta potential, may have infinitely many scattering states, which are still complicated to deal with. So we’ll go back to the Schrodinger equation iℏ∂Ψ ∂t = ˆ HΨ, for which we can solve for energy eigenstates of the form e−iEt/ℏψ and thus only need to solve the equation ˆ Hψ = Eψ. Picking a quantum system then comes down to picking our Hamiltonian ˆ H, which can be any Hermitian operator with units of energy (which requires defining an inner product). We’ve been solving these kinds of equations in this class when ψ represents a particle living in one dimension (as motivated by classical mechanics), but such wavefunctions take on uncountably many values. So the key to simplifying our system is to be less attached to the physics for a while. What we’ll do is assume that we have a particle which can only live at one of two points x1 or x2, rather than on the whole real line. (It’s not interesting if the particle can only be at one point, because then the probability of finding it there is just one.) So instead of having ψ(x) be a function on the real numbers, we can just think of ψ as having the two pieces of information (which we’ll encode in a vector) " ψ(x1) ψ(x2) # = " α β # . The probability of finding the particle at position x1 is then \|α\|2, and the probability of finding it at x2 is \|β\|2. Fact 152 This setting may look familiar from our discussion with interferometers back in the second lecture of this class! But such a vector can also encode information about other systems, such as a particle which can either be on the left or right side of a box with a partition. And it can also encode a situation where a particle is in one of two states, either “spin down” or “spin up” – that’s what we’re moving toward right now. Such two-state systems turn out to be very rich mathematically – to see this, we need to start by constructing the inner product. Instead of the usual integral (φ, ψ) = R φ∗(x)ψ(x)dx, in which we’re basically adding up the contributions of φ∗ψ, it is natural now to have the inner product be ψ1 = " α1 β1 # , ψ2 = " α2 β2 # = ⇒(ψ1, ψ2) = α∗ 1α2 + β∗ 1β2. In other words, we can think of taking the conjugate transpose of the vector for ψ1 and performing matrix multiplication: (ψ1, ψ2) = h α∗ 1 β∗ 1 i " α2 β2 # (And as we might explore in a future class, we can also think of a wavefunction ψ(x) for a one-dimensional particle as an infinite column vector of values · · · , ψ(−ε), ψ(0), ψ(ε), ψ(2ε), · · · .) If we now think of ˆ H as a 2 × 2 matrix acting on these column vectors by matrix multiplication, the condition for being Hermitian is that (HT )∗= H (we can check this by a direct calculation, but we can think of it as coming from the fact that we are transposing and complex conjugating ψ1 in the definition of the inner product.) The power of this setup will now become clear as we try to classify all Hamiltonians for this two-state system. A 2 × 2 matrix equal to its conjugate transpose must have real entries on the diagonal, and the other two entries must be complex conjugates, so the most general Hamiltonian is ˆ H = " a0 + a3 a1 −ia2 a1 + ia2 a0 −a3 # , where a0, a1, a2, a3 are real numbers. (We’ve chosen to represent the two diagonal entries in this way for reasons that 116 will become clear soon, but we can check that we can indeed pick a0 and a3 to get any real entries on the diagonal.) We can then rewrite this matrix as ˆ H = a0 " 1 0 0 1 # + a1 " 0 1 1 0 # + a2 " 0 −i i 0 # + " 1 0 0 −1 # . The four matrices on the right-hand side are the four basic Hermitian 2 × 2 matrices – scaling them by real numbers and adding them together still gives us a Hermitian matrix. In other words, we can say that there is a four-dimensional space of matrices, spanned by these four basic matrices. They are so important that we give them names – the first one is the 2 × 2 identity matrix, and the next three are called the Pauli matrices σ1, σ2, and σ3, respectively. If we want to write a Hamiltonian down for a physical system, though, we need ˆ H to have units of energy. And there really isn’t a reason to use the identity matrix, because that’s like adding a constant to the Hamiltonian, which doesn’t change the calculations or the states. (For example, we saw this in the harmonic oscillator, whose Hamiltonian was ˆ H = ℏω ˆ N + 1 2 , but for which we could just work with ˆ N instead.) Since ℏω has units of energy, the Hamiltonian we’ll write down here is ˆ H = ℏω1 2 σ1 + ℏω2 2 σ2 + ℏω3 2 σ3 (the factors of 2 will be clear soon) for some real numbers ω1, ω2, ω3. Rewriting this slightly as ˆ H = ω1 ℏ 2σ1 + ω2 ℏ 2σ2 + ω3 ℏ 2ω3 , we see that the three terms in parentheses have units of angular momentum, so we may hope that there is something to do with angular momentum in this setup! Specifically, it is reasonable to try defining the operators ˆ Sx = ℏ 2σ1, ˆ Sy = ℏ 2σ2, ˆ Sz = ℏ 2σ3 as potential components of angular momentum. And in fact, the commutators here are easy to compute because we have matrices: [ ˆ Sx, ˆ Sy] = ℏ 2σ1, ℏ 2σ2 = ℏ 2 · ℏ 2 " 0 1 1 0 # " 0 −i i 0 # − " 0 −i i 0 # " 0 1 1 0 #! = ℏ 2 · ℏ 2 " i 0 0 −i # − " −i 0 0 i #! = ℏ 2 · ℏ 2 " 2i 0 0 −2i # = iℏ· ℏ 2 " 1 0 0 −1 # = iℏℏ 2σz = iℏˆ Sz . So we have the same commutation relation as with the angular momentum operators ˆ Lx, ˆ Ly, ˆ Lz from a few lectures ago – not only do the ˆ S operators have the same units as angular momentum, but they also have the correct commutators (we can check that [ ˆ Sy, ˆ Sz] = iℏˆ Sx and that [ ˆ Sz, ˆ Sx] = iℏˆ Sy as well). So angular momentum does not actually require complicated operators involving ⃗ r and ⃗ p – we can construct it with just 2 × 2 matrices. And what we’ve created is spin one-half, the angular momentum coming from having two discrete degrees of freedom – it’s really a mathematical object, but the interpretation of “spin up” and “spin down” came from physicists. This system is the setting for a substantial fraction of the material of 8.05 – it turns out that the physical 117 interpretation takes some getting used to. Looking back at the Hamiltonian ˆ H = ω1 ˆ Sx + ω2 ˆ Sy + ω3 ˆ Sz = ⃗ ω · ˆ S, where ⃗ ω is a vector of real numbers and ˆ S = ( ˆ Sx, ˆ Sy, ˆ Sz), it turns out this Hamiltonian actually represents a spin in a magnetic field, and solving the corresponding Schrodinger equation will reveal that the spin will precess in the magnetic field – this is the origin of NMR (nuclear magnetic resonance), which is used to detect the density of fluids in the body. Looking ahead to 8.05, we’ll conclude by mentioning some properties of the eigenstates of ˆ H. Recall that for our previous angular momentum problem, we measured the eigenvalues for ⃗ L2 and ˆ Lz. Similarly, we’ll measure the eigenvalues for ⃗ S2 and ˆ Sz here, where ˆ Sz represents the spin in the z-direction. Because ˆ Sz is the matrix ℏ 2 " 1 0 0 −1 # , which is a diagonal matrix, its eigenvectors are \|↑⟩= " 1 0 # , \|↓⟩= " 0 1 # (here “up” and “down” arrows represent being in the upper or lower component), where the eigenvalues are ℏ 2 and −ℏ 2, respectively: ˆ Sz \|↑⟩= ℏ 2 " 1 0 0 −1 # " 1 0 # = ℏ 2 " 1 0 # = ℏ 2 \|↑⟩, ˆ Sz \|↓⟩= ℏ 2 " 1 0 0 −1 # " 0 1 # = −ℏ 2 " 0 1 # = −ℏ 2 \|↓⟩. And the reason for the name of “spin one-half” is that we have the factor of 2 in the eigenvalues and operators that we’ve defined – it might look like that choice was arbitrary, but in fact it was necessary to make the angular momentum commutators work out! In other words, two-state systems are forced to be spin one-half, and such systems must have angular momentum measured to be either ℏ 2 or −ℏ 2. (In contrast, a photon is an example of a spin-one system, in which the two directions of circular polarization correspond to a spin of ℏor −ℏ.) But now that we know what spins look like along the z-direction, we may also be curious about what spin states look like along the x- or y-direction. It may look like we’re running out of states, because we only have a two-dimensional space of possible states and there are three directions of spins that we’re trying to represent. And in fact, the z-spin-up and z-spin-down states were already constructed to be orthogonal and form a full basis of the vector space: any state " a b # is a superposition a \|↑⟩+b \|↓⟩of those two states. But that’s not actually a problem: the operator ˆ Sx = ℏ 2 " 0 1 1 0 # does have two different eigenvectors. Specifically, if we consider the (normalized) state 1 √ 2 " 1 1 # = 1 √ 2 (\|↑⟩+ \|↓⟩) , we see that ˆ Sx 1 √ 2 " 1 1 # = ℏ 2 1 √ 2 " 0 1 1 0 # " 1 1 # = ℏ 2 · 1 √ 2 " 1 1 # , and we do have an eigenstate of ˆ Sx of eigenvalue ℏ 2 (which we’ll denote \|↑; x⟩), formed by taking a superposition of \|↑⟩and \|↓⟩! Similarly, the state \|↓; x⟩= 1 √ 2 " 1 −1 # = 1 √ 2(\|↑⟩−\|↓⟩) will be an eigenstate of ˆ Sx with eigenvalue −ℏ 2. Finally, if we turn our attention to the spin states along the y-direction, we will also be able to construct eigenvectors 118 of ˆ Sy. Specifically, the state 1 √ 2 " 1 i # = 1 √ 2(\|↑⟩+ i \|↓⟩) satisfies ˆ Sy 1 √ 2 " 1 i # = ℏ 2 · 1 √ 2 " 0 −i i 0 # " 1 i # = ℏ 2 · 1 √ 2 " 1 i # , meaning that we have an eigenstate \|↑; y⟩of ˆ Sy of eigenvalue ℏ 2, and a similar calculation shows that the state 1 √ 2 " 1 −i # is the eigenstate \|↑; y⟩of eigenvalue −ℏ 2. So complex numbers play an important role here – there would be no way to get an eigenstate for all directions otherwise. Importantly, all of these eigenstates have nothing to do with our usual functions of x, r, θ, φ, and so on – spin is its own thing with finitely many degrees of freedom and column vectors as wavefunctions. But within this system, we have angular momentum, and this angular momentum is what we use to describe the spins of particles. And spin systems will play a central role in 8.05 – they’ll be used to understand superposition, entanglement, Bell’s inequality, and much more. 119
8735	https://www.youtube.com/watch?v=o76Kg0-4AJI	Worked example: Calculating Molecular weight \| Atoms and Molecules \| Chemistry \| Khan Academy Khan Academy India - English 548000 subscribers 17 likes Description 1268 views Posted: 14 May 2023 Let's understand how to calculate the molecular weight of a molecule! Master the concept of Ionic compounds through practice exercises and videos - Check out more videos and exercises on Atoms, Compounds and Ions - Khan Academy is a free learning platform for Class 1-12 students with videos, exercises, and tests for maths, science, and more subjects. Our content is aligned to CBSE syllabus and available in Hindi, English, and many more regional languages. Experience the joy of easy, seamless, accessible learning anywhere, anytime with Khan Academy. Subscribe to our YouTube channel - As a 501(c)(3) nonprofit organization, we would love your help! Donate here: Created by Vibhor Pandey Transcript: in this video Let's understand how to calculate the molecular weight of a molecule so first of what is molecular weight well we can say that molecular weight is the sum of atomic masses of all the atoms in the molecule for example let's say we take calcium chloride now calcium chloride has two types of atoms it has one of calcium and two atoms of chlorine so molecular weight would be the sum of the atopic mass of calcium and two atoms of chlorine now what is the atomic mass of calcium well for that we can have a look at the periodic table so when we look at the periodic table and try to find calcium this is where it is this right here is calcium and if you look closely the atomic mass is 40.08 so atoic mass is usually written below the symbol of the element right so this is 40.08 plus 2 because there are two chlorine atoms two into the atomic mass of one chlorine atom chlorine is over here and the atoic mass is 30 5.45 so this is this is 35.45 now we can calculate this and find the molecular weight of calcium chloride when we do this this comes out to be equal to 110.98 and and this will be atomic mass unit which is represented by small U because all of these numbers that you see below the symbol of the element they are in atomic mass units that is that is small U one atomic mass unit is equal to this is equal to 1.67 into 10 to the power minus 24 grams I think it's 24 yeah this is this is minus 24 grams all right now let's have a look at some more molecules so now we have c2h4 this is ethene now this one has two types of atoms carbon and hydrogen but it has two carbon atoms and four hydrogen atoms and molecular weight is the sum of the atomic masses of all the atoms in that molecule so this would be 2 into the atomic mass of the atomic mass of carbon plus 4 into the atomic mass of hydrogen so we're talking loss of carbon if you look at a periodic table this is carbon is right over here this is 12.0 1 so 2 into 12.01 plus 4 into the atomic mass of hydrogen which is over here and this is 1.008 now when we solve this this comes out to be equal to 28.052 again atomic mass units now let's take one which is slightly more complicated let's take one with three three types of atoms how about C6 h12 o6 all right now this is really glucose it has six carbon atoms 12 hydrogen and six oxygen why don't you try and calculate the molecular weight of glucose okay hopefully you gave this a shot now this will be six into the atomic mass of carbon 12.01 plus 12 into the atomic mass of hydrogen 1.008 plus 6 into the atomic mass of oxygen and oxygen is over here this is oxygen so atomic mass is 16.00 let's just write sixteen now when we solve this this comes out to be equal to 180.156 atomic mass units
8736	https://www.youtube.com/watch?v=bHqJRvq1NR8	GCSE Maths :\|: Graphical Methods :\|: Linear & Non-linear Inequalities :\|: Linear Programming Mathscool 126 subscribers 3 likes Description 110 views Posted: 8 Jan 2025 LINK For GRID: We've previously learnt to solve linear and non-linear equations, to graph functions and to use graphical methods to find approximate solutions to equations that can't be solved directly... Now we progress to solving INEQUALITIES in ONE-VARIABLE We see that Linear Inequalities can easily be solved without a graph, but that Non-linear inequalities are best solved with a combination of algebraic and graphical methods. We then transition into LINEAR PROGRAMMING (INEQUALITIES in TWO-VARIABLES), learning how to form INEQUALITIES from the CONTRAINTS and identify the FEASIBLE REGION on a grid, and to locate the OPTIMAL SOLUTION Transcript: if you got constraint yo I'll solve them check out my book and let's revolve them okay so we are moving into inequalities and linear programming so let's start off by looking at inequalities in one variable so if I give you an equation like this to solve um you'll be able to solve that easily no problem whatsoever you'd add three to both sides so that will give you 4X is = to 24 you divide by four both sides and that give you X is equal to 6 and so there's your solution okay so that's an equation in one variable in X is the only unknown in that equation and so we solved it what if I give you this inequality well largely it's going to be the same steps so you're going to add three to both sides and so that's going to give you 4X is less than or equal to 24 and you're going to divide by four both sides so the steps are the same that's going me X is less than or equal to six the only conceivable difference is that um we would then draw this on a number line so X is less than or equal to 6 is our answer but we would just show it on a number line so here you see I put in X is 6 I filled in the hole in the center of there if it was if it just said um X is less than 6 and didn't have the equal to on there then that would be hollowed out so that would be um the center of that would be white like that to show that look you can't actually touch six but you can go numbers less than six but that wasn't the situation and that wasn't the situation so I filled in that hole I mean I filled it in I could have filled it in with black that would probably been a bit better oops black okay and then that shows you that X can be less than 6 so this is showing you numbers less than 60 5.2 and 4 and 2.8 and all these numbers root to and all these numbers including the negative numbers are all valid um solutions to that inequality so my question is is it true the only difference between solving a linear equation and a linear inequality is just the number line that we use at the end to demonstrate what the valid answers are is that true no of course it's not true no there's something else what is that something else something peculiar about inequalities that doesn't apply to equations um okay let's have a go at this one here so what we do here is um I mean you could use a flow diagram if you wanted x - 3 -4 + 60 = 24 U minus the 60 from both sides - 60 - 60 and then that gives you um -4 bra x - 3 is = 24 - 60 so - 36 uh that's the opposite of the plus 60 is- 60 the opposite of -4 is divide by -4 divide -4 divide -4 and so that gives you x - oh I didn't know I was going to go red there x - 3 is equal to um so the minuses are going to count out it's just going to give me nine and then just add three to both sides so X is = to 12 now here I'm doing the inequality and there is a difference there's a difference you know what the difference is I hope you do know what the difference is- 60 from both sides and so that gives us -4 braet x - 3 is less than - 36 divide by -4 divide by Min -4 this time I did mean to use the rare divide by Min -4 when you divide by a negative number or multiply by a negative number you have to turn the inequality so this is going to give me x - 3 and now I'm going to rotate the inequality so the inequality going to be facing the opposite direction so I turn the inequality the other way and that's going to give me nine and then just add three to both sides add three add three and so that gives me X is greater than 12 and then just do my number line 12 11 oops 11 13 greater than 12 so not including the 12 so I left it Hollow in the middle and greater than 12 is in that direction okay so that's the only difference when you're multiplying or dividing by a negative number then um you need to rotate the inequality okay but but this is just a linear inequality what happens when we get to nonlinear inequalities so this is a quadratic equation this isn't an inequality yet how would we solve this well we know we'd um maybe collect the three over to the other side so we get x^2 - 2x - 3 = 0 and then we'd see if it factorizes that's going to be an X and an X and A minus 3 and a + one yeah that works um so that gives me x - 3 could be zero which mean X is 3 or x + 1 could be zero which mean X is -1 so we get two Poss possible solutions to that equation now we can't um do the same thing with um a nonlinear inequality we can't just follow the same method and just adjust it um I mean if you wanted to solve this you know you might if you didn't know what to do you might do x^2 - 2x - 3 is less than or equal to Z you might factorize it again x - 3 x + 1 is less than equal to Z you might say x - 3 is less than equal to 0 or x + 1 is less than to 0 which will give you X is less than oral to 3 or X is less than oral to minus1 and so this gives you X is less than or equal to 3 fill in the whole and this gives you X is less than equal to minus one so minus one might be there X is less than and since it's either or both of those then um X is is less than equal to 3 would be your combined solution but that's all wrong that's not correct that's not how you solve a nonlinear inequality you m anything that's nonlinear you must graph it so now you've got different ways you could graph it you could just graph it from the beginning so you could just split it like we were doing last week where we were doing graphical Solutions you could just split it from the beginning so you could just say uh y = x^2 - 2x is less than or equal to or equal to y = 3 then you could sketch this one here so let's just put some axes up and x^2 - 2x you could factorize that that could become x braet x - 2 which means it's got a root at Zer and a root at two and you could sketch that quadratic and then you could sketch y = 3 y = 3 would be like this three and then you could find where they meet now actually where they meet will be the same as the answers we got here because where they meet will be where they're equal to each other and where they're equal to each other is at three and minus1 so we know they meet at minus1 and at three but we want to know when is this one when is this one that's this one here less than when is it below and this is the region where it's below this bit here is where it's below this is where less than it says less than or equal to so we can include the bit where it's equal to it but then this bit up here is no good we don't want this bit up here because then the quadratic is not less than it's not below the line the the black quadratic here is above the line and here is above we only wanted a bit where it's below the line and so you can see the green bit that we want on the x-axis would be the bit from -1 to 3 and so we would combine that into a single inequality and we'd say x is between Min -1 and three and it can equal them both because we filled in those holes so it can equal them and so there's our solution so that's how we solve quadratic inequalities with a graph now in reality in this case you um you already knew the solutions where they meet so in reality you're going to have to make it equal to three and then find the solutions to that and then use those Solutions in your sketch so you're going to have to solve an equation and then use those Solutions in your sketch usually if you're presented with an inequality like this you might actually go about it like this in the first place so everything that I've done there is completely valid but in reality what you might do is if you got an inequality like this you might say oh I haven't already solved this equation we had already solved this equation but in most situations you wouldn't have already solved that equation so let's get rid of that equation and you're just given this inequality right from the start so what you do you X2 - 2x - 3 - 3 is less than or equal to Z then you'd factorize it x - 3 x + 1 is less than equal to0 then you'd sketch this so y = x - 3 x + 1 is less than or equal to or equal to y = 0 and then this we know has a root at three and a root at minus1 and so that looks like that and we're trying to find when that graph is below zer0 y equal 0 is just this line here it's just the x-axis and so when is it below is this bit here and then we'll fill in those because we want it to equal to and so we get exactly the same answer which is X is between -1 and 3 inclusive okay so that's how we would probably solve that we wouldn't we wouldn't have the Y = x^2 - 2x and the y = 3 we' have just do one quadratic do it like that okay so nonlinear inequalities must be solved with a lior sketch it's not that hard you'd be surprised um I normally teach the a level mass and my a level students mess this up all the time so I give them a nonlinear inequality they don't draw a graph then they get it wrong easiest way to trip them up is give them a nonlinear inequality and um they'll do it without a sketch okay so we must use a sketch so this is your question one that you need to do and you've got three um inequalities to solve and you notice they're all nonlinear inequalities okay and um so the first one is this one here you've got to solve it the second one is this one here you've got to solve it and the third one is this one here you've got to solve it um and I've very generously given you a quick sketch of y x cubed a quick sketch of Y X2 and a quick sketch of Y = 1/x so you've been helped out a lot even then I guarantee you none of you will get all three right so I'm willing to um extend a little prize to anyone who gets all three right W without without your teacher helping you or without your your mate who's really good at math helping you I'll test it I'll give you one more to do in the class and if you've got all three right I'll give you one more which is very much like these and if you also get that one right by yourself then I'll be convinced otherwise I'm not so sure I'm going to be convinced then you get a little prize okay that's not going to be anything fantastic so don't expect like uh Tesla or anything like that okay um oh hellbender wonder what that is oh that's what it is okay um so what if you're given an equation like this well you wouldn't try and solve that you might rearrange it and then you might end up sketching it on a grid so to sketch it on the grid you've got a couple of different ways you can do it one way you could do it is to um make it why the subject if you like um so if I make y the subject I would start off by writing it 3 yal 24 - 2x just minusing the 2x from both sides and then I divide by 3 so y = 8 - 23 x and that would tell me the Y intercept is 8 and the gradient is - 2 up 1 2 3 across okay and then uh so that's at 6 then another minus 2 up um and then one another 1 2 3 across and then another min-2 up so we're already down to four now another minus two up is at two and then another three across and then another minus three up and then so we went across three and then six and then 9 and then 12 so 12 must be that route you don't need to have that much detail on there if you don't want to you can just just that point and that Point's enough to sketch the line the alternative way of doing that is to um just use the two Roots method in the first place the the Y intercept and the root method first of all so to find the Y intercept you basically just cross out this term so you cross out this term and you get 3 y = 24 and divide both sides by three is eight so eight is the Y intercept and then to get the root you just do the same thing but you cross out the other one so you cross out this one and so you get 2x = 24 and so X is 12 and then you just draw a straight line through those two points okay oops slightly missed and so there's our line so if we had an equation like that we would draw it as a line on a grid so that's what that's the line we're drawing it's drawn really nicely there on a proper grid um what I'm interested in is um is and and what we're going to be looking at a lot of the time with linear program is we're going to look at integer points so um you might be interested in the integer values of X and Y that lie on that line and so if I look at this point here which lies on the line it's not integer values cuz that's between one and two but this point here is integer value so three corresponds to six so that point there is when X is3 and Y 6 both of them are integers at that point there and also at that point there and again here and then again and then again and so those are the integer values that one that one that one that one that one there the integer points I've kind of restricted myself to um just I haven't gone further down here and and further across there what do we do if we're given an inequality like this so well now it's less than or equal so it's 3 y + 24 is less than or equal to 24 and we could do sort of the same thing as we did before which is we could write it as 3 y is less than or = to 24 - 2x and then divide by 3 so Y is less than oral to 8 - 23 x and 8 - 2/3 x we know that that's this line here we know that that's the same line as we had before in other words if you just change that to an equal sign you get back to that which is that line there so that's the line that we did before again we didn't need to make y the subject we could have just pretended it was an equal sign knocked out the uh y term to get the root knocked out the X term to get the Y intercept we could have just done it like that um so that's but we want to know when Y is not equal to 8 - 2/3 but less than or equal to and and this on this line Y is equal to 8 - 2/3 X above this line Y would be greater than um 8 - 2/38 and below it would be less than 8 - 2/38 so really what we're interested in is all of these points here all of these points and I'm I'm going to stick again to integer points um all of these points here all of these points here anything below the line is what I'm interested in and there's loads of them so I can't I can't fill them all in can I um so we might only be interested in integer values so what I've done here instead is I've I've crossed out the region that I don't want I've shaded out the region that I don't want I've I've sort of you know put lines all over this region here so I don't want this region here so I've crossed it out so what we often do with inequalities is we cross out the region we shade out the region we don't want sometimes they'll tell you to do the opposite and they'll tell you to shade the region that you do want so that will be nice and confusing for you I'll try and be consistent about it and always shade out the region we don't want that's the convention shade out the region you don't want okay so this inequality here would be represented by all of these points now of course it doesn't say anything about integer values so I've said I'm only interested in integer values if you're not just interested in integer values then any point here and here and here and here and here and here they'll all be and here and here they' all be valid but I'm just kind of sticking to integer points uh for the time being and so all of these are kind of the points now you notice I haven't included points like this um because I've also sort of restricted myself to values of X that are bigger than or equal to zero and vales of Y that are bigger than or equal to zero and and the reason I've done that you you'll see in a second when we get there but actually if it doesn't say so then you don't need to restrict yourself but when we do the questions it will often say that actually X has got to be greater than zero and Y is got to be greater than zero when we come to actually doing linear programming okay so right now we're up to the point where we can start learning about linear programming so the father of linear programming is a guy called George danig and he developed this method along with a couple of other very famous mathematicians canr and V Newman they develop these methods for optimizing resource allocation so that might be something that well I think they were doing it you know during World War II for um so optimizing you know they they've got troops that they've got to send to different areas and they've got to supply those troops and they're just kind of trying to work out what the best way of doing that um um he himself discovered something called the Simplex algorithm and he won um the national medal of Science in America for that um because you don't win a Nobel Prize for for um for maths you you you can want win a Fields medal um and then there are other National medals that you can get for math you can't win a Nobel pricee for making some great discovery in MTH which is a bit strange but that's just the way it is okay there's an interesting story about danic while he was a PhD student at Berkeley he arrived a little bit late for a lecture and as he arrived he saw two problems already on the board and he copied them down and just assumed that they were set for homework and handed them in a couple of weeks later complaining that they were a bit hard in fact they weren't homework questions at all the lecturer had put up these two unsolved problems in statistics just to say look you know mathematicians are working on this and we don't have a solution to this at the moment and so there he was as a PhD student assuming they were homework and went off and solved them and handed them in um and uh you know it's kind of a famous story and it's sort of a a a bit of a parable because they say well look you know um if you're not told that this is hard you might not be put off of it whereas if you're told this is unsolved you might be put off of it and thought well well okay well I'm never going to get there so yeah it's an interesting anyway so we're going to we're going to just start an entry into linear programming so it's when you have multiple linear equations and you plot them together on the same grid each time removing The Unwanted region so you're only left with one small region left which is the only region that you're willing to accept it's called the feasible region so we're going to plot each inequality as if it were an equation but we'll use a dash line if it's less than or greater than but we use a solid line if it's less than or equal to or greater than or equal to okay so we're going to shade out the region we don't want so if it's Y is greater than or equal to or Y is greater than we we want the bit where Y is greater so we'll remove the bit where Y is less in other words below the line if it's if the inequality is y is less than or equal to or Y is less than we'll remove the bit above the line for X is greater than or X is greater than or equal to we want the bit where X is greater so we'll remove the bit on the left which is where X would be left and finally for X is less than or equal to or X is less than we'll remove the bit on the right of the line the left the the region that's left unshaded is called the feasible region and it contains all the points that satisfy all the inequalities and you can utilize points that lie on the solid lines but you cannot utilize points that lie on the dash line okay so let's have a go and see how this works so it says on the grid below show each region that satisfies these three show the region that satisfies all of these three in linear inequalities okay so let's start with the first one and what I'm going to do is pretend it's an equal sign so I'm going to pretend it's X um + 2 y = 6 and then I'm going to do what I normally do which is to find the Y intercept by Crossing out the X so that give me the Y intercept is three so the wiep is three whoops I need a 10 so that give me the wi St is three and then I'm going to find the root by um Crossing out the Y term and that will give me X is 6 okay so I'm going to draw a line connecting those two points and because it's got an equal to it I'm going to draw a solid line I'll draw a solid line I'll try to draw in a sort of pinkish color so there's my solid line whoops slightly missed okay just move it into place so there's my solid line there it's still kind of missing a little bit okay uh I want uh so I'm looking at the side where Y is and Y is on the side where it's going to be less than equal to I mean strictly speaking you should make y the subject so let's do that just to be absolutely certain we don't fluff it um so 2 Y is less than or equal to 6 - X and Y is less than oral to 3 - half X so as I said Y is on the less than or equal to side so we want the bit where Y is below this so I'm going to shade out the bit where Y is above so I'm going to cross out this bit here so I'll still be on the pinkish color and I'll cross out this bit here this bit I don't want so that's the region I don't want so I'll cross it out okay we've got another one here so I'll do this one next Y is greater than Z so I'm going to sketch y equals z and yal 0 is just this line here I'll do it a bit thicker and I'll do it in blue y equal 0 I a little bit thicker so I can see it y equal 0 is that line there uh ah but mine's got to be Dash because it doesn't contain any equals on it so let's make this a dash line and let's change that to a dash line okay so there's our dash line okay and we want the bit where go away okay um we want the bit where um Y is greater than so Y is greater than would be above there so we're going to shade out this bit here let just do that and make that thin and make it blue I'm going to shade out this bit here okay um and then now we've got this one here which is the only one that's left so um I'm just going to highlight that one uh which tells me that X's between n and a inclusive I'm going to split that into two separate inequalities I'm going to take this thing here and uh so I'm going to say x is between n and eight and it's got an equal to on that one and I'm going to take that side of it there so not and I'm just going to swap that round so the x is on the left so X is on the left remember the fatter side of the inequality got to be there so X is greater than zero now X is um X's zero would be this line here let's just draw the line remember we want it dashed I'm going to do a green line and I want it Dash X is equal to zero would be that line there uh and I want the bit where X is greater X is greater would be on the right hand side so I'm going to shade out the bit on the left hand side so just do it in green with a thin pan and shade out that bit there and then my second one was this bit here so part of the same thing was this bit here which was X is less than or equal to 8 so I'll start with the line x equal 8 this time I want it a solid line because it's got an equal to sign on that so I do the x = 8 so x = 8 would be there and um I wonder if that a typo I wonder if that should have been 12 I wonder if that should have been 12 if it should have been 12 then I would need to move that line up because then that would be um it would be six so I'm just going to move that line up cuz I think that should have been 12 I'm going to move that line up to six okay which means I just need to rub out this bit here cuz that bit shouldn't have been that bit shouldn't have [Music] been shaded in so I think that might be I just changed it to a 12 here I just changed that to a 12 because I think I think um I'm just looking at the answers and they don't look great so I just changeed that to a 12 okay so X is um less than or equal to 8 was the bit that I was doing X is less than or equal to 8 x is 8 is that line there I want the bit when X is less which which is on this side so I'm going to cross out the bit that I don't want which is on this side here okay and so this is the region that I'm left with okay so it says um what is the greatest value of x it says when 2 y + 1al x what is the greatest value of x if x and y can only be integers X and Y can be any values okay so let's start off with this what does this mean here if 2 y + 1 = X so at the moment I'm only thinking about answers that lie in this region but answers that lie in this region well if you only restrict yourself to integer points then it's effectively these points here like that point there and that point there and that point there and that point and that point and that point and that point you can't touch the dotted line so none of those points count but these are all the integer points that would work you can touch that line so you can touch that line there you can use that point you can use that point um can't touch that line there can't touch you can touch that line you can touch that line okay so those would be all of the integer points that are valid all of these points here that I put with the black dots on them okay so those would be all the integer points that would be valid um but um but they haven't so that's valid according to all these inequalities but now they said no but hold on you've got to also stick to this equation now this isn't an inequality this is an equation so this equation again if we wanted to find out if we wanted to sketch this line what we could do is um we could do the same thing that we always do which is to find the Y intercept to find the Y intercept you cross out the X term so you cross out the X term so that's just going to become zero and so Y is going to become a half so um minus a half by the way minus a half is going to be Y what color should we do this in red minus a half would be Y and then if I want to find the um if I want to find the root I'm just going to cross out that bit there and so that's going to give me X is um one so X is one would be there and so it's a line that goes through those two points I'm going to draw a red line that goes to those two points let's not do it too thick okay so it's sort of like that red line there I'm not 100% sure how accurate my line has ended up so the just going try and move it a little bit that way and move it a little bit that way I think that might be a little bit better okay if you want to check it of course you could you know you could just test out a point so you could say okay well let's work out what the coordinate should be when say x is 7 so if x is 7 um then 7 - 1 is 6 and then divided by 2 is three so X is 7 should be Y is 3 yeah it's sort of missing a little bit isn't it just move it a little bit okay that's a bit better okay all right so you can just put in a a point that's further across you know like here and see if that value matches up okay so now they're saying hold on this rule has got to be obeyed and this is an equation so when an equation's got to be obeyed then that forces you to live only along this line so you can only live on this line okay so if you can only live on this line and you're forced to only be in this region as well because we have those inequalities as well then that means you can be here and if you can if you're only allowed into then that means you could just be here and here they're the only two places which your integers you're within the unshaded region and you're stuck on the line yeah the line is like a railway track and you can only go along that rail you can't you're you're you're Thomas the Tank Engine you can only you can't Thomas Tank Engine can't walk over there yeah so okay so if he's got to be an integer points integer points maybe they're stations I don't know why integer points be stations but they're they're stations so he can only be here and he can only be here so those are the only two points where he can be so uh what is the greatest value of x well this point is the go of the greatest value of x so the answer would be fine the greatest value of x which meets all these inequalities which meets all these inequalities and sticks to that line is going to be X is f what if X and Y can take any values okay so now we're not restricted to integer values so we would go to that point there now the best way to find that point there is to not try and use the graph to find it because it's not an integer so the best way to find that point there is to just do a little bit of algebra that's where this line here which is 2 y + 1 = x meets this line here which is x + 2 y = 12 which is x + 2 y = 12 so I can do a little bit of simultaneous equation I can um leave that as x + 2 y = 12 I can just rearrange this to x - 2 y just minusing the 2 y from both sides equals 1 then I can add the two equations together and so it's going to give me 2x those two are going to cancel out = 13 so x = 13 / 2 13 over 2 which is 6 and A2 so this point here is at 6 and A2 6 and A2 along the x-axis and then if I just put in 6 and A2 back into either of these equations I'll just put it into this one 2 y + 1 = 13 / 2 2 y = 11 / 2 y = 11/ 4 so this point here is 3 and 3/4 is is 11 over 4 which would be um 2 and 3/4 yeah 2 and 3/4 okay so 11 over 4 is 2 and 34 so that would be if it can take any values what's the greatest value of x it will be 6 and a half would be the greatest value of x so using integer values five using any values 6 and a half okay um now of course if you weren't stuck to this line if they hadn't put that and they just asked us what the greatest value of x then the greatest value of x would be those two points there because you know that would be X is a but they didn't they forced us to lie live on that line okay so um this is more of a like a full linear programming question so this one here they'd already given us the equations all we had to do was draw them on the grid now they haven't even given us the equations they've just given us a situation and we've got to analyze this situation a in an exam there are two papers and each is marked out of 100 to pass a student needs to gain 30 or more in paper 1 gain 40 or more in paper two Gain 150 or more when the mark on paper 2 is doubled and added to the mark on paper one form as many inequalities as you can okay so let's make your score in paper one let's call that X so we're going to Define X is your score in paper one so x equal score for paper one and we're going to define y as your score Y is your score for paper 2 okay now to pass you've got to gain 30 or more in paper one so that's easy X has got to be 30 or more so the first inequality I'm going to write is X whoops X whoops X has got to be 30 or more greater than or equal to 30 gain 40 or more in paper two so y has got to be 40 or more G 150 or more when the score in paper two so the score in paper two is y is doubled so you're going to double that you're going to times it by two and you're going to add the score on paper one you're going to add the score on paper one add the score on paper one which is X and it's got to be 150 or more so that has got a total to 150 or more okay uh form as many inequalities as you can now I'm going to add a couple more inequalities that were not given in these instructions but were hidden in here whoops I didn't want to do that marked out of 100 you can't get more than 100 you can't get 110 so X has got to be less than or equal to 100 and Y has got to be less than or equal to 100 okay so uh okay so uh let's add these inequalities to our grid okay so we're going to add the line xal 30 first of all which is just going to be a green line so we'll add a line we'll make it a green line and it will go through 30 so it will go through that let slightly Miss whoops didn't want to move that let's try locking that in place so I can't accidentally move that lock in place okay and let's move that to 30 does that move to 30 yeah oh wow that's a good shot okay uh then y I want X to be greater than 30 so I'm going to cross out the bit where X is less than 30 so I'm going to cross out this bit here so I use a green pen thin green pen and just cross out a bit where it's less than 30 Y is greater than or equal to 40 so I'm going to draw the line y equal 40 so that's going to be a line a blue line at 40 so that's my line for Ys 40 I want y to be greater than 40 greater than 40 be on this side so I'm going to cross out this side here so I'm just going to use a thin blue pen and cross out all of this I also want this line here so I'm going to say that this has got a a root at 150 I can't really draw that root of 150 um and it's got a y intercept at 75 so I can draw the Y intercept at 75 um I'm just going to guess the roote at 150 so Y intercept is at 75 let's just draw that 75 and then the root is at 150 which is sort of like that much on something I'm going to say sort of roughly there ideally I'd have uh a nicer Grid or I could just plot a point I could you know maybe it' be better better if I said right if uh X is 100 y would be 25 and so you could do 100 comma 25 and you could plot that point but I'm equally going to struggle with plotting that point um so I'm just going to leave as as it is okay so this line is going to be let's do this line in pinky color and let's not do it too thick and so there's that line there slightly missed again okay and if you rearrange that you'd get Y is greater than or equal to 150 -/ x so Y is on the greater than side so that means we don't want the region where Y is less so we don't want this region here so we don't want that region there whoops some of that kind of crossed over the boundary let's just rub out a bit cross over the boundary okay uh and we need these two lines here which should be obvious that's just a green line at 100 let's get it green and do it at 100 oh didn't come out green let's instruct it again to be green and do it at 100 and we want X to be less than 100 so we want the left bit so we're going to cross out the right bit so we're going to cross out this bit here and then Y is less than 100 so that's going to be a blue line a line in blue at 100 which is going to be there and um we're going to shade out this bit up here which we don't want blue okay okay so so the feasible region that we're left with the feasible region is the points that are still the unshaded bit it's called the Fe feasible region identify the feasible region on the grid what is the minimum total marks for a path now I don't have a grid here I mean I've got a I've got some axes but you can't see the squared paper or the graph paper lines so I'm going to switch to this so I've put it onto JRA and I've put it onto a grid and you can see now that there is graph paper behind there and it's asking me so let's just go back to here it's asking me um let's let's rub out all of this so we can see the question so um uh okay I'm just going to get I'm just going to move that out the way what is the minimum total marks that for a part I think it means for a part pass what's the minimum total marks for a pass so you're passing now what I've done here oh hold on I've done on next what I've done here is um I've asked geogebra to highlight all the points so all the actual integer points have got these little crosses on them okay and so all of these points are passing points so you're you're you're scoring a pass in all of these points and um what I'm doing is I'm looking at the coordinates of the point and I'm thinking okay well here you've passed but if I go to that corner there you've passed because you got 100 score in paper one and you got 100 score in paper to so you weren't aiming for a minimum pass there you're a superb student who gets absolutely everything right so minimum marks would be more like down here and usually when you're trying to find an answer within this region here for the minimum or maximum of something the the thing is going to line somewhere near one of these Corners so this shape that we're left with has got corners and the corners for the minimum marker are sort of these two here so it's these two points here that we're looking at and so what I'm going to do is I'm going to say well it's going to be one of these two so this corner here that little cross point there I can zoom it in I'll try zooming it in but it's the J is going a bit slow today okay maybe I shouldn't have tried that okay um let's try with the mouse instead okay let's try zooming it in it's not really liking me zooming it in when when I set it up um it should have been possible to zoom it but it might have just got a bit locked at some stage it's not letting me Zoom it okay let's forget about zooming it okay um so when I put these on the uh homework I'll put the J applet in and hopefully you'll be able to zoom it in there so I'm looking at that point there and I can see that that's when X is 30 and y is 60 so your total score in that case would be 30 marks in paper one and 60 marks in paper two so you'd have a total score of um 90 marks so could it be 90 that's the answer let's have a look at the other Corner the other corner is here and here you got 70 in paper 1 and 40 in paper 2 so here you got 110 so here you got 110 well well that's more marks so that's the lowest possible Mark you can get and actually pass so to to pass with the lowest possible mark you've got to aim for uh 30 in paper 1 and 60 in paper 2 now if paper 2 is harder than paper one maybe that's a tricky thing to achieve because you don't have to do so well in paper one but you have to do really well in paper two whereas if paper 2 is the harder one uh well so if I keep with my uh suggestion that paper two might be the harder one then maybe the actually easier thing to do is score really well in paper one and then you can get away with not scoring so well in paper too so maybe this is actually easier even though you have to score more more marks [Music] overall okay Mindy scored 40 marks in paper one so we know that for her X was equal to 40 so we know that for her X was equal to 40 so definitely for her X was equal to 40 so what I'm going to do is I'm just going to type in here x is 40 so I'm going to type in there xal 40 and so now we have to live along this line where X is 40 so if so now it's not the whole region with all of these points that we can use it's just the points that lie on that line that we can we X is 40 first you can't get more than 40 or less than 40 she already got her score in paper one so for her the minimum Mark she can get in paper 2 is whatever that point there is obviously as you go up she's getting small marks in PTY so that minimum Point looks like 55 to me me and again I can't zoom it in so I'm going to say that that is exactly 55 so for her um m is the point and it's uh it's this point here they want us to mark on this point here um whoops let's just put that back again and then put that at the front they want us to mark up this point here so that point is M so if I could write on top of there I'd say that this is the point M okay Mor is confident he scored 50% more marks in paper 2 than paper 1 find the minimum Mark Mor must managing paper one in okay I've done a few alliterations here just for fun Mark Point M to represent the minimum Mark Mindy must make in paper one um he scored 50% more marks in paper two than paper one so that means that whatever he got in paper one so let's just give it a number let's say he got uh 50 marks in paper one then that would then he would have got 75 marks in paper two because it's 50% more in other words Y is equal to 1.5 X 1.5 1 X is 100% of X 1.5 X is 150% of X so that's 50% more so it's y = 1.5x so we know that he's got 50% more marks in paper 2 than paper 1 in other words he lives on the line yal 1.5x so let's add in that line so let's change that to Y = 1.5x so I'll just change that to y = 1.5x y = 32x and so that's this line here now so he's got to live on this line here and these points further up here is him scoring more and more and more in both of the papers and these points towards the bottom I'm soring left now where that meets is not one of my crossed points so it's not an integer Point okay so it's um it's he's got to live on this line we want to go as low down on this line as possible because that will give us the lowest marks in both X and Y paper 1 and paper 2 that point there is not an integer point cuz it's not got a little pink cross next to it that point there is an integer point so it's that point that I need which if I'm trying to read it looks like um well as far as I can see that's 56 along the Y AIS and um 38 on the x axis so 56 along the Y AIS y = 56 and X is 38 x is 38 so um his the minimum Mark he needs in paper 1 is 38 then when he does does better in paper two and gets one and a half times the marks 56 then he's going to have um pass the exam with the minimum possible he can do okay so that's a proper linear programming question where we had to form all the inequalities we had to plot them all this was a particularly nasty one because the the grid was huge there were a lot of points to consider so you wouldn't really necessarily get that in an actual question okay so this is now um your questions two and your question three that you've got to submit so this is combines two and three um I'm moving from my penthouse flat to a house with a huge 9 m x 16 M Garden I'm Uber excited I replant the garden in the perfect garden maximum 2/3 of the area is planted so 1/3 is going to be decked or well more than one3 is one3 or more is going to be dead so um so the area of this is going to be 9 16 so that's going to be 144 M squar and we're only going to plant a maximum of 2/3 of that so 2/3 of 144 would be um uh 96 yeah 96 so 96 me squared is the area that you're going to plant maximum so the area is going to be uh uh 96 or less less than or equal to 96 I'm going to use a mixture of shrubs X is how many shrubs and trees why is the number of trees they'll grow as my garden matures I'm told to allow 6 M squar for each shrub and 4 m squar per tree oh I get it so if you had two shrubs that would be 12 M squar or if you had four shrubs that would be 24 M squar or if you had X shrubs that would be 6X M squar and then for the trees you can do the same thing but it would be Y and then that total has got to be less than or equal to 96 so there's your first inequality ideally one should aim for more shrubs than trees okay so that's another inequality more shrubs than trees how you going to write that as an inequality um if you made equal to each other you'd say x equals y that would be equal numbers of shrubs and tree but it says no more shrubs and trees so you want this to be more than that so you maybe you change it to an inequality like that okay just helping out a little bit I'd like a minimum of eight shrubs that's an inequality but less than 12 shrubs that's an inequality I'm a tree hugger so oh look you can see men are hugging the tree there uh so I need at least four because you sometimes kill them when you hug the Trees Too Much form as many inequalities as you can and identify the feasible region on the grid remember the grid grid will be linked below State all the integer points that satisfy my constraints in other words the feasible Solutions the integer points that work are called the feasible Solutions okay so that one you've got to do entirely by yourself use this guide here and have a go at it tricky question but see how far you get with it okay and lastly it's our classroom practice so this is what we're going to do in the classroom so there's going to be some inequalities that want you to solve that's a linear inequality and then all of these are nonlinear inequalities and then we've got a grid where the the the feasible region is shown here but they haven't told you what inequalities they use to define that feasible region so you got to work backwards same here and then this is like a proper question a school with 10 teachers so I suppose we're going to make um X be the number of kids and Y be the number of teachers and they're going on a school trip and then we need need to form these so I'll be asking you come on help me with some or rather men will be asking you help me with form some inequalities here and if you've looked at it beforehand then you'll be able to help with that okay so reminder um Mena will put up the homework in the lesson and it will be listed in the students portal don't forget you need to hand in your corrections to the previous homework at the weekend and be be sure to prepare for this weekend's test okay that's it from me many will see you at the weekend bye
8737	https://proofwiki.org/wiki/Sine_of_Supplementary_Angle	Sine of Supplementary Angle - ProofWiki Sine of Supplementary Angle From ProofWiki Jump to navigationJump to search [x] Contents 1 Theorem 2 Proof 3 Also see 4 Sources Theorem sin(π−θ)=sin θ sin⁡(π−θ)=sin⁡θ where sin sin denotes sine. That is, the sine of an angle equals its supplement. Proof sin(π−θ)sin⁡(π−θ)==sin π cos θ−cos π sin θ sin⁡π cos⁡θ−cos⁡π sin⁡θSine of Difference ==0×cos θ−(−1)×sin θ 0×cos⁡θ−(−1)×sin⁡θSine of Straight Angle and Cosine of Straight Angle ==sin θ sin⁡θ ■◼ Also see Cosine of Supplementary Angle Tangent of Supplementary Angle Cotangent of Supplementary Angle Secant of Supplementary Angle Cosecant of Supplementary Angle Sources 1953:L. Harwood Clarke: A Note Book in Pure Mathematics... (previous)... (next): V V. Trigonometry: Angles larger than 90∘90∘: Examples 1968:Murray R. Spiegel: Mathematical Handbook of Formulas and Tables... (previous)... (next): §5§5: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I 2014:Christopher Claphamand James Nicholson: The Concise Oxford Dictionary of Mathematics(5th ed.)... (previous)... (next): Appendix 12 12: Trigonometric formulae: Symmetry 2021:Richard Earland James Nicholson: The Concise Oxford Dictionary of Mathematics(6th ed.)... (previous)... (next): Appendix 14 14: Trigonometric formulae: Symmetry Retrieved from " Categories: Proven Results Sine Function Supplementary Angles Navigation menu Personal tools Log in Request account Namespaces Page Discussion [x] English Views Read View source View history [x] More Search Navigation Main Page Community discussion Community portal Recent changes Random proof Help FAQ P r∞f W i k i P r∞f W i k i L A T E X L A T E X commands ProofWiki.org Proof Index Definition Index Symbol Index Axiom Index Mathematicians Books Sandbox All Categories Glossary Jokes To Do Proofread Articles Wanted Proofs More Wanted Proofs Help Needed Research Required Stub Articles Tidy Articles Improvements Invited Refactoring Missing Links Maintenance Tools What links here Related changes Special pages Printable version Permanent link Page information This page was last modified on 25 April 2023, at 17:56 and is 1,942 bytes Content is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Privacy policy About ProofWiki Disclaimers
8738	https://www.sohu.com/a/345971718_100182824	原创这6道题全懂了，求对数函数的定义域和值域再不作难了高考数学复习，这6道题全懂了，求对数函数的定义域和值域再不作难了。求对数函数的定义域，相对来说比较简单，主要考虑的是真数必须大于0。求对数函数的值域要难不少，对数函数的最大特点是：要么是增函数，要么是减函数，也就是说，对数函数是单调函数，求值域的一般步骤是先确定真数的取值范围，然后根据单调性或者图像求出函数值的取值范围，即值域。第1题首先，x是真数，所有x必须大于0，见①；其次1/3为底的对数也是真数，所以它也必须大于0，见②；然后解这两个不等式，并求交集。其中不等式②的解法一定要熟悉，下面列出了两种解法，解法一，就是把0用真数为1的同底对数来表示，然后根据1/3为底的对数单调递减来求x的范围，这种解法是常规解法；我更愿意使用解法二，即最后一行给出的推荐解法。第2题首先x＋1是真数，故应令其大于0，见①；对数在根号内，又是分母，所以对数必须大于0，解对数不等式即可求出x的范围。从本题的计算过程可以看出，如果对数计算熟练的话，第一步即①是可以省去的。第3题求对数函数的值域，一般分两步。第一步：求出真数的取值范围，如下①；第二步：根据对数函数的图像或者单调性求出值域，现在是把整个真数部分u看成自变量来求值域，容易得出当真数u∈[1,＋∞)时，函数值f(u)∈(－∞,0]，这就是要求的值域。第4题解：和上题一样分两步。第一步：求真数x＋1的取值范围为(0,＋∞)，这里解释一下，有学生可能会有疑问，x＋1不是可以取任意实数吗？本来确实如此，但它正好位于对数的真数部分，所以它只能取大于0的实数；第二步：根据对数的图像或者单调性求值域，容易得到值域为(－∞, ＋∞)。更快的解法：f(x)的图像是由1/3为底，x为真数的对数函数图像沿x轴平移得到的，平移前后值域是不会变化的，所以值域为(－∞, ＋∞)。第5题因为x－2可以取大于0的一切实数，所以本来①式可以取任意实数，但它处于对数的真数部分，所有和x－2一样，取值范围应为(0,＋∞)，得出了真数的取值范围，根据图像即可求出值域。第6题请认真体会本题和上题的不同之处。从这几道题可以看出，求对数的值域，最主要的工作是确定出真数的取值范围，理解了这一点，求对数的值域问题再也难不住你。高中、高考、基础、提高、真题讲解，专题解析；孙老师数学，全力辅助你成为数学解题高手。加油！孙老师微信公众号：slsh2018；名称“爱做数学题”。返回搜狐，查看更多
8739	https://www.quora.com/What-exactly-is-a-parabola-What-is-the-purpose	Something went wrong. Wait a moment and try again. Conic Sections Parabola Precalculus PLANE GEOMETRY Mathematical Concepts Parabolas (geometry) Concept of Geometry Maths Geometry Circle and Conics 5 What exactly is a parabola? What is the purpose? · A parabola is a U-shaped curve that is a specific type of conic section. It can be defined mathematically as the set of all points (x, y) in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix. The standard equation of a parabola that opens upwards or downwards is given by: y=ax2+bx+c where a, b, and c are constants, and a≠0. If the parabola opens sideways, the equation takes the form: x=ay2+by+c Purpose and Applications of Parabolas Parabolas have several important purposes and applications in various fields: Physics: Parabolas d A parabola is a U-shaped curve that is a specific type of conic section. It can be defined mathematically as the set of all points (x, y) in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix. The standard equation of a parabola that opens upwards or downwards is given by: y=ax2+bx+c where a, b, and c are constants, and a≠0. If the parabola opens sideways, the equation takes the form: x=ay2+by+c Purpose and Applications of Parabolas Parabolas have several important purposes and applications in various fields: Physics: Parabolas describe the trajectory of projectiles under the influence of gravity (ignoring air resistance). This is known as projectile motion. Engineering: Parabolic shapes are used in the design of satellite dishes and reflectors because they can focus signals or light to a single point (the focus). Mathematics: Parabolas are studied in algebra and calculus to understand their properties, such as vertex, axis of symmetry, and intersections with other functions. Architecture: Parabolic arches are used in construction for their strength and aesthetic appeal, distributing weight evenly. Optics: Parabolic mirrors are used in telescopes and headlights to focus light. Economics: Parabolic functions can model certain types of cost and revenue relationships in business scenarios. In summary, parabolas are not just mathematical abstractions; they have practical applications across many disciplines, making them a crucial concept in both theoretical and applied contexts. Paul Graig Ellis Sometimes looks at things from strange angles · Author has 782 answers and 939.4K answer views · Updated 4y Consider a simple pendulum, almost any weight on the end of a string or rope: a child on a garden swing, or the weight of a pendulum clock. Extending the first seven answers to date, the parabola provides a simple but fundamental model in wide areas of theoretical physics - in particular, the changing potential energy of a pendulum as it swings, known more generally as a “harmonic oscillator”, which is a useful general model in quantum physics as well as classical physics. (The more realistic “anharmonic” oscillator - such as the potential energy of a vibrating chemical bond or even a corpse sw Consider a simple pendulum, almost any weight on the end of a string or rope: a child on a garden swing, or the weight of a pendulum clock. Extending the first seven answers to date, the parabola provides a simple but fundamental model in wide areas of theoretical physics - in particular, the changing potential energy of a pendulum as it swings, known more generally as a “harmonic oscillator”, which is a useful general model in quantum physics as well as classical physics. (The more realistic “anharmonic” oscillator - such as the potential energy of a vibrating chemical bond or even a corpse swinging on a noose, perhaps - is considered further below.) The potential energy of a harmonic oscillator, V, as a function of the displacement from the horizontal equilibrium, X, as a result of the swinging motion (k is just a simple constant that depends on the exact physical setup), is given by: which has the shape of a parabola, as drawn in the figure below A simple computation shows that the oscillator moves between positive and negative turning points ±X(max) where the total energy E equals the potential energy 1/2.k.[X(max)]^2 while the kinetic energy is momentarily zero. In contrast, when the oscillator moves past X=0 the kinetic energy reaches its maximum value while the potential energy equals zero. Potential energy function and first few energy levels for harmonic oscillator (quantum mechanical case). Note, however, that for a more realistic model, say the vibrational oscillations of a diatomic molecule such as oxygen, we use an “anharmonic oscillator” which is approximately parabolic at its lowest point, but gradually changes as the elasticity of the restoring force (k, which was constant for the harmonic case) changes, until the bond holding the structure together finally ‘snaps’ as the system breaks apart (here, a molecule, which ‘dissociates’ by losing an atom) . A harmonic oscillator described above obeys Hooke's Law and is an idealized expression that assumes that a system displaced from equilibrium responds with a restoring force whose magnitude is proportional to the displacement. In nature, idealized situations break down and fail to describe linear equations of motion. Anharmonic oscillation is described as the restoring force is no longer proportional to the displacement. Two forms of nonlinearity are used to describe real-world situations: elastic anharmonicity damping anharmonicity Anharmonic oscillators can be approximated to a harmonic oscillator and the anharmonicity can be calculated using perturbation theory. [CLICK on the diagram if the model isn’t oscillating] Figure (used with permission from Wikipedia). shows the ground state potential well and is calculated using the energy levels of a harmonic oscillator with the first anharmonic correction. Do is the dissociation energy, which is different from the well depth De. The vibrational energy levels of this plot are calculated using the harmonic oscillator model: FROM: Chapter 5: Harmonic Oscillator 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? Please be reliable and detailed. · Yes, there is a way. 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Dating Safety Check – Review your date’s background to help keep you safe. Theodore Schaft Jr. Former Retired Math Teacher · 4y A parabola is defined as a plane figure consisting of all points equidistant from a point and a line. There are other definitions such as the intersection of a cone and a plane where the plane is parallel to a side of the cone, namely it only intersects one side of the con. Objects thrown near earth follow a near parabolic path. More interesting is a parabloid, the three dimensional version. It has the characteristic that rays coming directly in go through a focal point and rays emanating from the focus leave parallel to each other. Flashlights and high beams are parabloids. Your dish receiver A parabola is defined as a plane figure consisting of all points equidistant from a point and a line. There are other definitions such as the intersection of a cone and a plane where the plane is parallel to a side of the cone, namely it only intersects one side of the con. Objects thrown near earth follow a near parabolic path. More interesting is a parabloid, the three dimensional version. It has the characteristic that rays coming directly in go through a focal point and rays emanating from the focus leave parallel to each other. Flashlights and high beams are parabloids. Your dish receiver is a parabloid and the rays coming from the satellites collect at the focal point. Related questions What is so special about the focus of a parabola? And what are its applications in real life? What are some applications of parabola in real life? What is a parabola? What are some real-world applications of parabolas? What is the application of a parabola? Tejaswini Naik Lives in India · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Daniel McLaury , Ph.D. Student in Mathematics at University of Illinois at Chicago · Author has 82 answers and 225.4K answer views · Updated 11y Originally Answered: What is a parabola? · A parabola is the set of all points in the plane equidistant from a given line (the conic section directrix ) and a given point not on the line (the focus ). The focal parameter (i.e., the distance between the directrix and focus) is therefore given by , where is the distance from the vertex to the directrix or focus. The surface of revolution obtained by rotating a parabola about its axis of symmetry is called a paraboloid . The parabola was studied by Menaechmus in an attempt to achieve cube duplication . Menaechmus solved the problem by finding the intersection of the two parabolas and A parabola is the set of all points in the plane equidistant from a given line (the conic section directrix) and a given point not on the line (the focus). The focal parameter (i.e., the distance between the directrix and focus) is therefore given by , where is the distance from the vertex to the directrix or focus. The surface of revolution obtained by rotating a parabola about its axis of symmetry is called aparaboloid. The parabola was studied by Menaechmus in an attempt to achieve cube duplication. Menaechmus solved the problem by finding the intersection of the two parabolas and . Euclid wrote about the parabola, and it was given its present name by Apollonius. Pascal considered the parabola as a projection of a circle, and Galileo showed that projectiles falling under uniform gravity follow parabolic paths. Gregory and Newton considered the catacaustic properties of a parabola that bring parallel rays of light to a focus (MacTutor Archive), as illustrated above. For a parabola opening to the right with vertex at (0, 0), the equation in Cartesian coordinates is (1) (2) (3) (4) The quantity is known as the latus rectum. If the vertex is at instead of (0, 0), the equation of the parabola is (5) If the parabola instead opens upwards, its equation is (6) Three points uniquely determine one parabola with directrix parallel to the -axis and one with directrix parallel to the -axis. If these parabolas pass through the three points , , and , they are given by equations (7) and (8) In polar coordinates, the equation of a parabola with parameter and center (0, 0) is given by (9) (left figure). The equivalence with the Cartesian form can be seen by setting up a coordinate system and plugging in and to obtain (10) Expanding and collecting terms, (11) so solving for gives (◇). A set of confocal parabolas is shown in the figure on the right. In pedal coordinates with the pedal point at the focus, the equation is (12) The parabola can be written parametrically as (13) (14) or (15) (16) A segment of a parabola is a Lissajous curve. A parabola may be generated as the envelope of two concurrent line segments by connecting opposite points on the two lines (Wells 1991). In the above figure, the lines , , and are tangent to the parabola at points , , and , respectively. Then (Wells 1991). Moreover, the circumcircle of passes through the focus (Honsberger 1995, p. 47). In addition, the foot of the perpendicular to a tangent to a parabola from the focus always lies on the tangent at the vertex (Honsberger 1995, p. 48). Given an arbitrary point located "outside" a parabola, the tangent or tangents to the parabola through can be constructed by drawing the circle having as a diameter, where is the focus. Then locate the points and at which the circle cuts the vertical tangent through . The points and (which can collapse to a single point in the degenerate case) are then the points of tangency of the lines and and the parabola (Wells 1991). The curvature, arc length, and tangential angle are (17) (18) (19) The tangent vector of the parabola is (20) (21) The plots below show the normal and tangent vectors to a parabola. Source: Weisstein, Eric W. "Parabola." From MathWorld--A Wolfram Web Resource. Mohammad Zita Fuady Former Student at SMAN 48 Jakarta (2017–2020) · 7y Originally Answered: What is meant by parabola? · In mathematics [ ], a parabola is a plane curve [ ]which is mirror-symmetrical [ ] and is approximately U-shaped [ ]. It fits any of several superficially different mathematical [ In mathematics [ ], a parabola is a plane curve [ ]which is mirror-symmetrical [ ] and is approximately U-shaped [ ]. It fits any of several superficially different mathematical [ ]descriptions, which can all be proved to define exactly the same curves. Part of a parabola (blue), with various features (other colours). The complete parabola has no endpoints. In this orientation, it extends infinitely to the left, right, and upward. One description of a parabola involves a point and a line [ ] (the directrix [ ]). The focus does not lie on the directrix. The parabola is the locus of points [ ] in that plane that are equidistant [ ] from both the directrix and the focus. Another description of a parabola is as a conic section [ ], created from the intersection of a right circular conical surface [ ] and a plane [ ] which is parallel Promoted by Grammarly Grammarly Great Writing, Simplified · Updated 2y How can I effectively edit my own writing? So, you think you’ve drafted a tweet, an email, a short story, or even a novel. These are different forms of communication, but the process of bringing them to fruition has a necessary, sometimes overlooked step: editing! Unless you’re a professional writer, it’s unlikely that you have an editor who can review your writing regularly. Here are some tips to help you review your own work. Give your writing some space. Have you ever felt a mix of pure relief and joy when you’ve finished a draft of something? Don’t downplay that feeling and the ability to walk away from your work before you start ed So, you think you’ve drafted a tweet, an email, a short story, or even a novel. These are different forms of communication, but the process of bringing them to fruition has a necessary, sometimes overlooked step: editing! Unless you’re a professional writer, it’s unlikely that you have an editor who can review your writing regularly. Here are some tips to help you review your own work. Give your writing some space. Have you ever felt a mix of pure relief and joy when you’ve finished a draft of something? Don’t downplay that feeling and the ability to walk away from your work before you start editing it. You may need minutes, hours, or days, but once you sit back down with what you originally had on the page, you’ll have the thrill of looking at it with fresh eyes. You’ll notice errors you may not have seen the first time. You’ll come to new realizations about its overall tone and structure. If it’s a text or email, maybe you only need a few minutes away from it. If it’s a story or essay, perhaps you’ll need longer. Regardless of what type of work it is, it will help your writing tremendously. Don’t use overachieving synonyms. Looking at your work for the second, third, or fourth time around may inspire you to spice up your language with longer, more uncommon words. There’s nothing wrong with having a thesaurus nearby, but try to limit the repetition of long, pretentious-feeling words so your work flows well and doesn’t feel too bogged down. At the end of the day, you want it to feel true to you and the message you’re conveying. Remember who the reader is. Don’t forget your own voice as the writer—but don’t forget who your reader is. Many writers get too close to their work; editing is a chance to try to get out of your own head. Who is your ideal reader? What do you want them to take away from the writing? It’s a unique time to step in their shoes, to make sure your communication is as effective as you’d like it to be. Kill your darlings. Don’t be scared to remove chunks of your work, even if it feels precious to you. If it’s a passage that’s really tough to part with, try saving it somewhere else, so you can return to it later in your piece or for another work. Use Grammarly. Last but not least, Grammarly has countless resources for editing your work. Our writing assistant helps you find areas of your writing that are unclear or too wordy, as well as help you find mistakes you might not have caught. Editing may feel tedious, but it’s just as important as writing itself. For an extra pair of editing eyes on everything you write, download the free Grammarly for Windows and Mac today. Joseph Heavner mathematics student at the University of Maryland · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 943 answers and 5.5M answer views · Updated 11y Originally Answered: What is a parabola? · A conic section, meaning a curve that is achieved when a plane intersects a cone. When the cutting plane is parallel to exactly one generating line of the cone, then the conic is unbounded and is called a parabola. Parabolas are U-shaped and go by the standard quadratic equation: y=ax2+bx+c Also, a parabola has a focus (a point) and a directrix (a line), and any point on the parabola is equidistant from the focus and directrix. A parabola is found in quadratic equations, meaning ones of the general form ax2+bx+c. A parabola is also just one of four conic sections, the other three being the ci A conic section, meaning a curve that is achieved when a plane intersects a cone. When the cutting plane is parallel to exactly one generating line of the cone, then the conic is unbounded and is called a parabola. Parabolas are U-shaped and go by the standard quadratic equation: y=ax2+bx+c Also, a parabola has a focus (a point) and a directrix (a line), and any point on the parabola is equidistant from the focus and directrix. A parabola is found in quadratic equations, meaning ones of the general form ax2+bx+c. A parabola is also just one of four conic sections, the other three being the circle, the ellipse and the hyperbola. Conic sections, like the parabola, are introduced in basic mathematics, often in a Algebra I or Algebra II class, and are important to all mathematics. For more information comment, or Google this because sites like Wikipedia have tons of information on mathematical concepts like this. Related questions Is a catenary a parabola? What is the purpose of a parabola that is located under a bridge? What are some examples of a parabola? What is the difference btween parabola and parabola of safety? What are the points of a parabola that never move? Jay Jacob Wind Director (2001–present) · Author has 123 answers and 57.2K answer views · 4y One way to view a parabola is to graph this function: y = x^2 One way to view a parabola is to graph this function: y = x^2 Your response is private Was this worth your time? This helps us sort answers on the page. 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It’s quick and easy toopen an account with SoFi Checking and Savings (member FDIC) and watch your money grow faster than ever. Read Disclaimer Mark Regan Chemistry, Trig, and Calc in Engineering (college major) & Mathematics, Bakersfield College · Author has 1.5K answers and 2.9M answer views · 4y Parabolas are used in the design of headlights, radar antenna, and in the study of objects in free fall like ballistic missiles, golf balls, and rocks. The reason they are used in headlights and radio antenna, is that a parabolic shaped dish, with the receiver located at the focus of the parabola, will direct every signal that hits the dish right back to the focus. They are used in the study of objects in free fall because on object thrown into the area with both horizontal and vertical velocity will generally follow the path of a parabola. The definition of a parabola is the set points equidis Parabolas are used in the design of headlights, radar antenna, and in the study of objects in free fall like ballistic missiles, golf balls, and rocks. The reason they are used in headlights and radio antenna, is that a parabolic shaped dish, with the receiver located at the focus of the parabola, will direct every signal that hits the dish right back to the focus. They are used in the study of objects in free fall because on object thrown into the area with both horizontal and vertical velocity will generally follow the path of a parabola. The definition of a parabola is the set points equidistant from a point and a line. It is convenient to place the point just above the Origin on the coordinate plane and the line, which is called the directrix, parallel to the x-axis the same distance below the Origin to get an idea of how a parabola behaves. This parabola will have a vertex at the Origin and open up to the top. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Pavel Juranek Technician, Programmer, Analyst, Consultant · Upvoted by Richard Goldstone , PhD Mathematics, The Graduate Center, CUNY (1995) · Author has 3.7K answers and 1.2M answer views · 3y Originally Answered: What is a parabola, and how does it work in a real life application? · 2py=x^2 … parabola py = x ……… its derivative a) every parabolic mirror return rays in one point, or inverse, of course. it has application for directional arrays for communication between two points in big distances refraction optic in for telescopes, cameras, … directional light sources (see your car) b) have a bridge with a carrying beam. Then every segment of the beam is loaded with a force that is proportional to the horizontal distance (weight of the bridge deck section), then the parabola solves this problem (beam shape), because of the difference in the derivative values (related to 2py=x^2 … parabola py = x ……… its derivative a) every parabolic mirror return rays in one point, or inverse, of course. it has application for directional arrays for communication between two points in big distances refraction optic in for telescopes, cameras, … directional light sources (see your car) b) have a bridge with a carrying beam. Then every segment of the beam is loaded with a force that is proportional to the horizontal distance (weight of the bridge deck section), then the parabola solves this problem (beam shape), because of the difference in the derivative values (related to resulting force) is constant on each section of the beam. Sponsored by OrderlyMeds Is Your GLP-1 Personalized? Find GLP-1 plans tailored to your unique body needs. Software Developer at Tata Consultancy Services (company) · Author has 86 answers and 692.9K answer views · 8y Parabola is one of the interesting chapters in Geometry…. Definition : The locus of a point moves in a plane such that the ratio of its distance from a fixed point to its perpendicular distance from a fixed straight line is equals to one…. I.e SP/PM = 1 Where , S is focus , P is any point on the parabola.. M is perpendicular distance from any point… Angelos Tsirimokos M.A. in Mathematics, Harvard University (Graduated 1976) · Author has 1.9K answers and 2.7M answer views · 7y They are second-degree curves of a particular kind. They can be defined in several ways: as the intersection of a circular cone with a plane parallel to a generator, as the locus (=set) of points equidistant from a given point and a given line, as the graph of a quadratic polynomial… The trajectory of a stone or other projectile thrown in the air (not vertically, but at an angle to the horizon) or of water ejected from a hose under pressure is a parabola, or rather, it would be a parabola if air resistance could be neglected. The words ‘parabola’, ‘parable’ and ‘parole’ all come from one and the They are second-degree curves of a particular kind. They can be defined in several ways: as the intersection of a circular cone with a plane parallel to a generator, as the locus (=set) of points equidistant from a given point and a given line, as the graph of a quadratic polynomial… The trajectory of a stone or other projectile thrown in the air (not vertically, but at an angle to the horizon) or of water ejected from a hose under pressure is a parabola, or rather, it would be a parabola if air resistance could be neglected. The words ‘parabola’, ‘parable’ and ‘parole’ all come from one and the same Greek word, παραβολή, originally meaning ‘comparison’. Mike Roberts Author has 788 answers and 525.6K answer views · 4y You can look for a formal definition, but my memory is that a parabola is a collection of points that are equidistant from a point (called the focus) and a straight line. Practically, I think it’s useful in optics because a lot of the light that hits a parabolic mirror tend to collect at the focus - think of the weapon on the Death Star. Prashant Rastogi It is a shape of a curve or a path followed by a particle, Its equations are :- Y^2=4ax, and Y^2=-4ax, where ‘a’ is the half of latus rectum as the fornula of latus rectum is 2a and latus rectum is the perpendicular drawn from the focus which interests parabola at two points Related questions What is so special about the focus of a parabola? And what are its applications in real life? What are some applications of parabola in real life? What is a parabola? What are some real-world applications of parabolas? What is the application of a parabola? Is a catenary a parabola? What is the purpose of a parabola that is located under a bridge? What are some examples of a parabola? What is the difference btween parabola and parabola of safety? What are the points of a parabola that never move? What is the length of a parabola? What does the P stand for in a parabola? [What is π ? What is a parabola, and how does it work? What is the purpose of determining the vertex, focus, and directrix in a parabola? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
8740	https://math.stackexchange.com/questions/1140096/repeated-operations-on-a-b	elementary number theory - Repeated operations on $(a,b)$ - 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 Repeated operations on (a,b)(a,b) Ask Question Asked 10 years, 7 months ago Modified9 years, 10 months ago Viewed 139 times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. For each pair of integers (a,b)(a,b), define the sequence S(a,b)S(a,b) as: If a n−1<b n−1 a n−1<b n−1 then (a n,b n)=(2 a n−1,b n−1+1)(a n,b n)=(2 a n−1,b n−1+1); otherwise if b n−1<a n−1 b n−1<a n−1 then (a n,b n)=(a n−1+1,2 b n−1)(a n,b n)=(a n−1+1,2 b n−1). If a n=b n a n=b n then we are done. In other words, at each step we double the lesser and add one to the greater of (a,b)(a,b). We stop when a n=b n a n=b n, i.e. we reach a pair (c,c)(c,c). My question is: for any a,b a,b are we guaranteed to terminate at a pair (c,c)(c,c)? elementary-number-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Dec 1, 2015 at 9:12 user26857 53.4k 14 14 gold badges 76 76 silver badges 166 166 bronze badges asked Feb 9, 2015 at 2:58 james hjames h 882 6 6 silver badges 18 18 bronze badges 3 Hmm, I see now that I possibly misunderstood the focus of your question. The word "for any" is ambiguous to me. Are you searching (at least) one solution "where we reach a pair (c,c)" ? Or are you asking "for each a,b, are we guaranteed...? "?Gottfried Helms –Gottfried Helms 2015-02-09 20:41:23 +00:00 Commented Feb 9, 2015 at 20:41 Well I was wondering if every such pair (a,b)(a,b) has a solution but it doesn't work except for certain values.james h –james h 2015-02-09 20:45:37 +00:00 Commented Feb 9, 2015 at 20:45 Ok, I see. Thanks!Gottfried Helms –Gottfried Helms 2015-02-09 21:29:01 +00:00 Commented Feb 9, 2015 at 21:29 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. If you start with a pair of the form (a,2 r a−r)(a,2 r a−r), then you get (2 a,2 r a−(r−1))(2 a,2 r a−(r−1)), ie a pair of the form (a′,2 r−1 a′−(r−1)). So repeating the operation you will get a pair of the form (c,c). I claim this is the only way. Otherwise, we would have to get a pair of the form (a,2 r a−r) from something other than (a/2,2 r a−(r+1)). The only possibility for this is (a−1,2 r a−r 2). But for r≥2, a>0 we have 2 r a−r 2>a; when r=1 2 r a−r is not divisible by 2; and when r=0 it's just (a−1,a/2), so not actually different to the pair we know works. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 9, 2015 at 11:09 ChristopherChristopher 7,217 1 1 gold badge 24 24 silver badges 32 32 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I doubt that the termination at (c,c) occurs for every pair. I just started with (4,3) and got the following matrix showing the trajectory of the first couple of iterations as result. Here each row is one iterate of the initial pair and the difference between the elements of the pair: a b a−b 4 3 1 5 6−1 10 7 3 11 14−3 22 15 7 23 30−7 46 31 15 47 62−15 94 63 31 95 126−31 190 127 63 191 254−63⋮⋮⋮ The differences (in the third column) follow an obvious pattern and without analyzing this in detail I'm confident that this initial pair of numbers has an infinite trajectory. [update] If we look at the numbers along the trajectory of iterations, then it seems that the following is important. Assume a 0<b 0 . Then let's write all iterations while a j<b j as one step and we want a 1=a 0⋅2 k>b 1=b 0+k>b 0+k−1>a 0⋅2 k−1>a 0 We see, that b 1 lies in the interval of a 1..a 1/2. The next iteration, to build b 2=2 j⋅b 1>a 1+j needs then only a j smaller than k and along further iterations the required exponent at 2 converges to 1. So in the long run the transformation reduces to a k+1=2⋅a k,b k+1=b k+1 and vice versa a k+2=a k+1+1,b k+2=2⋅b k+1 . I think, this is the crucial observation (and should be made more rigorous: there's a smallest k from where the above rule is valid if not already a=b) Then proceeding on this path, writing two steps in one this is a k+2=2 1⋅a k+1 b k+2=2 1⋅b k+2 a k+4=2 2⋅a k+3 b k+4=2 2⋅b k+6 a k+6=2 3⋅a k+7 b k+6=2 3⋅b k+14 Generalizing this gives a k+2 r+1=2 r⋅(a k+1)b k+2 r+2=2 r⋅(b k+2) and this shows, that the differences d k=(b k+2)−(a k+1) expand to d k+2 r=2 r⋅d k (if not a k−b k=1). And if the differences change, then the trajectories of (a k,b k) cannot be constant. If a k−b k=1 at some k then all the further differences are 1 as well. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 9, 2015 at 15:45 answered Feb 9, 2015 at 10:50 Gottfried HelmsGottfried Helms 35.9k 3 3 gold badges 73 73 silver badges 152 152 bronze badges 5 All numbers of the form (x,x+1) have a solution using 10 steps. Assume a,b, and L doubles a and adds 1 to b, while R doubles b and adds 1 to a, then the operations are R R L L L L R L R R and the solution reaches the diagonal at 32 x+68. See a complete solution at reddit.com/r/mathriddles/comments/2v6eaj/doubling_and_adding_1Foo Barrigno –Foo Barrigno 2015-02-09 22:28:55 +00:00 Commented Feb 9, 2015 at 22:28 @Foo : ??? How can you begin on a pair (a,b)=(x,x+1) with the transformation R : because x+1>x the OP's scheme requires that we begin with L on this. Thus after the first step we have (2 x,x+2). Then we have to apply R giving (2 x+1,2 x+2). Then we have again L and so on alternating L R L R L R L R...Gottfried Helms –Gottfried Helms 2015-02-09 22:38:29 +00:00 Commented Feb 9, 2015 at 22:38 That's why there's a separate solution for k=1. I didn't write the solution presented there down as a proof because I'm skeptical that there isn't a shorter method, but there is certainly a solution for k=1 as I listed above.Foo Barrigno –Foo Barrigno 2015-02-09 22:41:31 +00:00 Commented Feb 9, 2015 at 22:41 @Foo : in the Reddit-forum there is less restriction: it is not required that the 2 x-operation is always on the smaller number of the pair. So that is a different problem...Gottfried Helms –Gottfried Helms 2015-02-09 22:43:15 +00:00 Commented Feb 9, 2015 at 22:43 I see, I assumed this was the same question and missed the fact that this one was restricted to the smaller number. Makes sense and I agree, it does seem that not every pair is solvable in this manner.Foo Barrigno –Foo Barrigno 2015-02-09 22:47:05 +00:00 Commented Feb 9, 2015 at 22:47 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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8741	https://www.bamo.org/archives/examfiles/bamo2023examsol.pdf	24th Bay Area Mathematical Olympiad Problems and Solutions Mar 1, 2023 The problems from BAMO-8 are A–E, and the problems from BAMO-12 are 1–5. A A tangent line to a circle is a line that intersects the circle in exactly one point. A common tangent line to two circles is a line that is tangent to both circles. As an example, in the ﬁgure to the right, line a is a common tangent to both circles, but line b is only tangent to the larger circle. a b Given two distinct circles in the plane, let n be the number of common tangent lines that can be drawn to these two circles. What are all possible values of n? Your answer should include drawings with explanations. Solution: The possible values of n are 0, 1, 2, 3, and 4. These cases are illustrated below. n = 0 Two nonintersecting circles, one contained inside the other n = 1 Two circles internally tangent to each other n = 2 Two circles intersecting at two points n = 3 Two circles externally tangent to each other n = 4 Two nonintersecting circles, neither contained inside the other B Ara and Bea play a game where they take turns putting numbers from 1 to 5 into the cells of the X-shaped diagram on the right. Each number must be played exactly once, and a cell cannot have more than one number placed in it. Ara’s goal is for the two diagonals of the X diagram to have the same sum when the game is over; Bea’s goal is for these two sums to be unequal. (a) Show that Ara can always win if he goes ﬁrst. (b) Show that Bea can always win if she goes ﬁrst. Solution: (a) Suppose Ara goes ﬁrst. He can begin by placing a 5 in the center of the board. This is the only space shared by both diagonals, so when the game is over, the two diagonal sums are guaranteed to add up to 1+2+3+4+5+5 = 20. For her ﬁrst move, Bea must play some number n in one of the four corner spaces, where n = 1, 2, 3, or 4. Ara can then respond by playing 5−n in the opposite corner, completing a diagonal. (We know this move is still available, because 5−n is either 1, 2, 3, or 4, and cannot be equal to n.) After these three moves, the sum of the completed diagonal is 5 + n + (5 −n) = 10. Since the two diagonal sums will add up to 20 at the end of the game, the other diagonal sum will also be 10. Thus Ara will win. (b) Suppose Bea goes ﬁrst. She can begin by placing a 2 in the center of the board. By the same reasoning as in part (a), the two diagonal sums are now guaranteed to add up to 1+2+3+4+ 5+2 = 17 at the end of the game. For the diagonal sums to be equal, each diagonal sum would have to be 17 2 . But the diagonal sums are integers, so this cannot happen. Therefore, no matter what the remaining moves are, Bea will win. C/1 Mr. Murgatroyd decides to throw his class a pizza party, but he’s going to make them hunt for it ﬁrst. He chooses eleven locations in the school, which we’ll call 1,2,...,11. His plan is to tell students to start at location 1, and at each location n from 1 to 10, they will ﬁnd a message directing them to go to location n+1; at location 11, there’s pizza! Mr. Murgatroyd sends his teaching assistant to post the ten messages in locations 1 to 10. Unfor-tunately, the assistant jumbles up the message cards at random before posting them. If the students begin at location 1 as planned and follow the directions at each location, show that they will still get to the pizza. Solution: If the students never visit the same room twice, then their hunt lasts a ﬁnite number of steps. In that case, they must reach the pizza (since the hunt always continues if they have not yet reached the pizza). Therefore, the only way for the students to not reach the pizza is for them to visit the same room twice, which gets them stuck in a loop. Such a loop must consist of n rooms containing n messages that collectively point to that set of n rooms. But this means that all the messages pointing into the loop are in rooms that are part of the loop, so there’s no way to enter the loop from outside. Room 1 can’t be part of a loop, since no message points to room 1. Thus, the students do not begin in a loop. Since they cannot enter a loop, they eventually get to the pizza. (In fact, they get to the pizza at least as quickly as Mr. Murgatroyd intended, since the worst case is that they have to visit every room once!) D/2 Given a positive integer N (written in base 10), deﬁne its integer substrings to be integers that are equal to strings of one or more consecutive digits from N, including N itself. For example, the integer substrings of 3208 are 3, 2, 0, 8, 32, 20, 320, 208, and 3208. (The substring 08 is omitted from this list because it is the same integer as the substring 8, which is already listed.) What is the greatest integer N such that no integer substring of N is a multiple of 9? (Note: 0 is a multiple of 9.) Solution: The answer is 88,888,888. In our solution, we’ll make use of the well-known fact that an integer is divisible by 9 if and only if the sum of its digits (in base 10) is divisible by 9. It was permissible to use this fact without proof on the contest, but for the sake of completeness, a proof can be found in the appendix following this solution. No integer substring of 88,888,888 is divisible by 9, since 9 does not divide 8k for any k = 1,...,8. We now show that every N > 88,888,888 has an integer substring divisible by 9. Suppose N > 88,888,888. If N has 8 digits, then one of those digits must be 9, which constitutes an integer substring by itself, so we are done. Thus, we assume from now on that N has 9 or more digits. We claim that for any such N, there is some integer substring divisible by 9. In fact, we will describe an algorithm to ﬁnd such a substring, using N = 328,346,785 as an illustrative example. For 0 ≤k ≤ 9, let sk be the sum of the ﬁrst k digits of N, where we deﬁne s0 to be 0. We can think of sk as a “running total” of the digits; here is our example number with s0,...,s9 written below it: 3 2 8 3 4 6 7 8 5 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 0 3 5 13 16 20 26 33 41 46 Next we consider the remainders left by s0,...,s9 when they are divided by 9. We have ten remainders with only nine possible values. By the pigeonhole principle, some two remainders must be equal. For instance, in our example, two of the remainders are equal to 5: 3 2 8 3 4 6 7 8 5 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 0 3 5 13 16 20 26 33 41 46 ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 0 3 5 4 7 2 8 6 5 1 Suppose it is sj and sk that leave the same remainder (where j < k). Then sk −sj is divisible by 9. But sk −sj is the sum of the digits of the integer substring consisting of the (j+1)th through kth digits of N. In the example above, for instance, this substring is 834678: 3 2 8 3 4 6 7 8 5 ↓ ↓ 5 41 and we have 8+3+4+6+7+8 = 41−5 = 36, a multiple of 9. Since we have an integer substring whose digits add up to a multiple of 9, that substring is itself a multiple of 9, and we are ﬁnished. Appendix. Here is the proof of the “well-known fact” mentioned at the beginning. To keep the notation simple, we state the proof for 4-digit integers; it should be clear how to generalize to integers with any number of digits. Let n = abcd, where the underline means we are writing the digits of n. By the nature of base 10 representation, we have n = 1000a+100b+10c+d = (999+1)a+(99+1)b+(9+1)c+d = (999a+99b+9c)+(a+b+c+d). The ﬁrst bracketed quantity is a multiple of 9. Thus, n and a+b+c+d differ by a multiple of 9. In particular, if n is a multiple of 9, then so is a+b+c+d (the sum of its digits), and vice versa. This completes our proof. E/3 In the following ﬁgure—not drawn to scale!—E is the midpoint of BC, triangle FEC has area 7, and quadrilateral DBEG has area 27. Triangles ADG and GEF have the same area, x. Find x. A B C D E F G Solution: The answer is x = 8. Use the notation [·] to denote the area of a polygon. Draw GB; notice that triangles GBE and GEC have equal bases and altitudes, so [GBE] = [GEC] = x+7. Since [ABE] = 27+x, we have [GDB] = 20−x. Likewise, if we draw AC, we see that [ABE] = [AEC] = 27 + x, so [AGC] = 20, which implies that [CAD] = 20+x. A B C D E F G Now triangles GAD and GDB have the same altitude (from G to AB), so their bases are proportional to their respective areas. In other words, AD DB = [GAD] [GDB] = x 20−x. But AD and DB are also the bases of triangles CAD and CBD, which have the same altitude (from C to AB). Hence AD DB = [CAD] [CDB] = 20+x 34+x. Equating these two fractions leads to the quadratic equation 34x + x2 = 400 −x2; the only positive solution is x = 8. 4 Zaineb makes a large necklace from beads labeled 290,291,292,...,2023. She uses each bead ex-actly once, arranging the beads in the necklace any order she likes. Prove that no matter how the beads are arranged, there must be three beads in a row whose labels are the side lengths of a triangle. Solution: More generally, we will prove that if there are 6n beads labeled n+1,n+2,...,7n, there must be three beads in a row whose labels are the side lengths of a triangle. (When n = 289, this coincides with the problem statement.) Aiming for a contradiction, assume there are no three beads in a row whose labels are the side lengths of a triangle. By starting at an arbitrary position on the necklace and counting off three beads at a time, partition the 6n beads into 2n trios of consecutive beads. Let S be the sum obtained by adding together the smallest two numbers from every trio. (Thus, S is a sum of 4n numbers.) Let T be the sum obtained by adding the largest number from every trio. By our assumption, within each trio, the sum of the two smallest numbers is less than or equal to the largest number. By adding these inequalities across all trios, we see that S ≤T. On the other hand, S can be no smaller than the sum of the 4n smallest numbers. Using the formula for the sum of an arithmetic progression, we have S ≥(n+1)+(n+2)+···+(5n) = (4n)(6n+1) 2 = 12n2 +2n. Similarly, T can be no larger than the sum of the 2n largest numbers: T ≤(5n+1)+(5n+2)+···+(7n) = (2n)(12n+1) 2 = 12n2 +n. Thus T ≤12n2 +n < 12n2 +2n ≤S, which contradicts our earlier claim that S ≤T. We have arrived at a contradiction, so there must in fact be three beads in a row whose labels are the side lengths of a triangle. 5 A lattice point in the plane is a point with integer coordinates. Let T be a triangle in the plane whose vertices are lattice points, but with no other lattice points on its sides. Furthermore, suppose T contains exactly four lattice points in its interior. Prove that these four points lie on a straight line. Solution: Let us begin with some preliminaries. In the solution to follow, we treat points freely as vectors, e.g. writing nA to mean the point whose coordinates are n times the coordinates of A, or A+B to mean the point which is the coordinate-wise sum of A and B. A basic result from vector geometry, which we will assume, states that given any three noncollinear points A,B,C in the plane, every point Q may be represented in the form rA + sB +tC for unique r,s,t satisfying r +s+t = 1. Moreover, Q is in the interior of △ABC if and only if such r,s,t are all positive. In similar fashion, any point on the line through A and B can be expressed as rA+sB with r +s = 1, and lies between A and B if and only if r,s > 0. We will also make repeated use of Pick’s Theorem, which we now state without proof. This theorem asserts that a lattice polygon (a polygon whose vertices are lattice points) has area equal to i+ 1 2b−1, where i and b are the number of lattice points on the polygon’s interior and boundary, respectively. Thus (for instance), the triangle T described in the problem must have area 4+ 1 2(3)−1 = 9 2. Now we are ready to begin the solution. With no loss of generality, let us assume T has one vertex at the origin O, which we identify with the zero vector. Call the other two vertices A and B. Of the four lattice points in the interior of T, let P be the point closest to line OA. It follows that there are no lattice points lying inside △OPA or on its boundary, other than O,P,A themselves, since any such point would be closer than P to line OA. Therefore, by Pick’s Theorem, △OPA has area 1 2. Lemma: Every lattice point Q can be expressed in the form nP+kA for some pair of integers (n,k). Moreover, when Q is expressed in such form, we have n = 2[OQA]. (The brackets represent area.) Proof. Let Q be a lattice point. By Pick’s Theorem, [OQA] = n 2 for some integer n. Thus [OQA] = n·[OPA]. By the base–height formula for triangle area, it follows that Q is on the line parallel to line OA that passes through the point nP. Thus Q = nP+kA for some real k, where kA is a lattice point. We assert that k is an integer. Indeed, if {k} denotes the fractional part of k, then {k}A = kA−⌊k⌋A is a lattice point which lies on segment OA, part of the boundary of T. Since T has no lattice points on its boundary other than its vertices, we must have {k} = 0. This completes the proof of the lemma. Let us return to the main problem. As already noted, [T] = [OBA] = 9 2. Thus by the lemma, B = 9P− kA for some integer k (the minus sign in the expression is not a typo, but a deliberate convenience for what follows). Rearranging, and using the fact that O is the zero vector, we have P = k 9A+ 1 9B+ 8−k 9 O. Since P is in the interior of T, we have 0 < k < 8. We will consider the possible values of k in turn. If k ≡0 (mod 3), then 1 3B = 3P −k 3A is a lattice point lying on segment OB. This contradicts the speciﬁcation of T as having no lattice points on its sides. If k ≡2 (mod 3), then 1 3B + 2 3A = 3P −k−2 3 A is a lattice point lying on AB, similarly yielding a contradiction. The remaining possibilities are k = 1,4,7. If k = 1, then the interior of T contains in its interior the four collinear lattice points P = 1 9A+ 1 9B+ 7 9O, 2P = 2 9A+ 2 9B+ 5 9O, 3P = 3 9A+ 3 9B+ 3 9O, 4P = 4 9A+ 4 9B+ 1 9O. If k = 4, then the interior of T contains in its interior the four collinear lattice points P = 4 9A+ 1 9B+ 4 9O, 3P−A = 3 9A+ 3 9B+ 3 9O, 5P−2A = 2 9A+ 5 9B+ 2 9O, 7P−3A = 1 9A+ 7 9B+ 1 9O. If k = 7, then the interior of T contains in its interior the four collinear lattice points P = 7 9A+ 1 9B+ 1 9O, 2P−A = 5 9A+ 2 9B+ 2 9O, 3P−2A = 3 9A+ 3 9B+ 3 9O, 4P−3A = 1 9A+ 4 9B+ 4 9O. Thus, the four lattice points inside T are collinear in every case, as desired. Solution 2: We assume the same basic facts about vectors as in the previous solution, as well as Pick’s Theorem and the determinant formula for the area of a parallelogram. Let T have vertices A = (x1,y1), B = (x2,y2), and C = (x3,y3). We know that (1) x2 −x1 x3 −x1 y2 −y1 y3 −y1 = 2[ABC] = 9. Consider equation (1) modulo 3, that is, over the ﬁeld Z/3Z. In this setting, the determinant is zero, so the vectors u = (x2 −x1,y2 −y1) and v = (x3 −x1,y3 −y1) are linearly dependent. If either of these vectors is zero (mod 3, that is), or if they are equal, then the trisection points of a side of T are lattice points, which contradicts the problem statement. Thus u,v ̸= 0 and u = −v. An immediate consequence is that (x1 +x2 +x3,y1 +y2 +y3) = u+v+3(x3,y3) = 0 over Z/3Z, with the result that the centroid, G = 1 3(A+B+C), is a lattice point. Now consider △ABG, whose area is 1 3[ABC] = 3 2. By Pick’s Theorem, △ABG has either • one lattice point in its interior and none on its boundary (besides vertices), or • two lattice points on its boundary. Case 1: △ABG has a lattice point in its interior and none on its boundary. In this case, a repetition of the preceding (mod 3) argument shows that the centroid G1 of △ABG is a lattice point. In this case, G1 +k(G−G1) for k = 0,1,2,3 are four collinear lattice points inside T. Case 2: △ABG has two lattice points on its boundary. Note that if at least two lattice points occur on a line, then the lattice points on that line occur at regular intervals. Thus the two lattice points on the boundary of △ABG are either the midpoints of AG and BG or the trisection points of one of these sides (say, AG). In the two cases, if we extend side AG beyond G, the next lattice point occurring on the extension is respectively either on T (at the midpoint of side BC), which is a contradiction, or inside T, being then the fourth collinear lattice point inside T. So we are ﬁnished. Solution 3 (sketch): Linus Tang, who was the only participant to receive a full 7 points, used the simplifying idea of an afﬁne transformation. Start by assuming the that one vertex of the triangle is at the origin (0,0) and another at the point (m,n), where m and n are relatively prime (why?). Use this and Pick’s theorem to show that we can ﬁnd integers p,q so that the linear transformation that maps (x,y) to (nx−my, px+qy) will transform our triangle to one with vertices at (0,0) and (0,1), with the third vertex having x-coordinate of 9. We now have a (fairly) simple triangle with a number of relatively simple cases to examine.
8742	https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick?srsltid=AfmBOor3yxxIE1CtdSEnJHIS3RXJetNFFPhHk8HfBzmOP5zpvipEQR1F	Art of Problem Solving Simon's Favorite Factoring Trick - 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 Simon's Favorite Factoring Trick Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Simon's Favorite Factoring Trick Simon's Favorite Factoring Trick (SFFT) is often used in Diophantine equations where factoring is needed, applicable for Diophantine equations of the form , for integer constants , , and , where there is a constant on one side of the equation and on the other side and a product of variables with each of those variables in linear terms. Simon's Favorite Factoring Trick is named after AoPS user ComplexZeta, or Dr. Simon Rubinstein-Salzedo. Contents [hide] 1 Statement 2 Applications 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also 4.1 External Links Statement Let's put it in general terms. We have an equation , where , , and are integer constants. Simon's Favorite Factoring Trick states that this equation can be factored into the equation For example, is the same as: Here is another way to look at it. Consider the equation .Let's start to factor the first group out: . How do we group the last term so we can factor by grouping? Notice that we can add to both sides. This yields . Now, we can factor as . This is important because this keeps showing up in number theory problems. Let's look at this problem below: Determine all possible ordered pairs of positive integers that are solutions to the equation . (2021 CEMC Galois #4b) Let's remove the denominators: . Then . Take out the : (notice how I artificially grouped up the terms by adding ). Now, (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs to solve this problem. Look at all factor pairs of 20: . The first factor is for , the second is for . Solving for each of the equations, we have the solutions as . Applications This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually and are variables and are known constants. Sometimes, you have to notice that the variables are not in the form and . Additionally, you almost always have to subtract or add the and terms to one side so you can isolate the constant and make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. When coefficient of is not , you can sometimes achieve an equation that can be factored by dividing the coefficient off of the equation. Problems Introductory Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? (Source) Intermediate AoPS Online If has a remainder of when divided by , and has a remainder of when divided by , find the value of the remainder when is divided by . are integers such that . Find . (Source) A rectangular floor measures by feet, where and are positive integers with . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair ? (Source) Olympiad The integer is positive. There are exactly ordered pairs of positive integers satisfying: Prove that is a perfect square. (Source) See Also Number Theory Diophantine equation Algebra Factoring External Links Video on AoPS Retrieved from " Categories: Number theory Theorems 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.
8743	https://artofproblemsolving.com/wiki/index.php/Angle_addition_identities?srsltid=AfmBOoqeqtMshUBqiGdBTnE5NzXR3cj0NjvatmfQMJgxS2FMuSJX9bR3	Art of Problem Solving Angle addition identities - 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 Angle addition identities Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Angle addition identities The trigonometric angle addition identities state the following identities: This article is a stub. Help us out by expanding it. Contents [hide] 1 Proofs 2 Proof 1 3 Proof 2 4 See Also Proofs Proof 1 Proof 2 Let point and point be two points on the unit circle such that . By the law of cosine, we know that: Apply the distance formula to obtain the length of : \|A B\|=(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2=(cos 2⁡α+sin 2⁡α)+(cos 2⁡β+sin 2⁡β)−2 cos⁡α cos⁡β−2 sin⁡α sin⁡β=2−2 cos⁡α cos⁡β−2 sin⁡α sin⁡β Substituting and rearranging to get: See that the identity holds true (and makes sense geometrically) when due to the fact that . Then, let and substitute it into the identity and the angle addition identity for cosine follows. ~Bloggish See Also Trigonometric identities Retrieved from " Category: Stubs 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.
8744	https://www.6sigma.us/six-sigma-in-focus/left-skewed-histogram/	Lean Six Sigma Training Certification FacebookInstagramTwitterLinkedInYouTube (877) 497-4462 Email Us My Account \| FacebookInstagramTwitterLinkedInYouTube (877) 497-4462 \| Email Us \| My Account\|Logout\| Left Skewed Histogram: A Comprehensive Guide to Understanding, Interpreting, and Applying Skewed Data Distributions One pattern that consistently challenges even seasoned professionals is the left-skewed histogram. In my years of consulting and teaching, I’ve seen how understanding these asymmetrical distributions — a key skill often honed through six sigma certification — can be a game-changer for businesses and project managers alike. Here’s what you’ll learn The fundamentals of left-skewed histograms and how to identify them The critical relationship between mean, median, and mode in left-skewed distributions Real-world examples and applications across various industries Step-by-step techniques for analyzing and interpreting left-skewed data Common pitfalls to avoid and advanced strategies for handling skewed distributions What is a Left Skewed Histogram? A left-skewed histogram, also known as a negatively skewed histogram, is a graphical representation of data where the tail of the distribution extends towards the left side, with the bulk of the data concentrated on the right. In simpler terms, when you look at a left-skewed histogram, you’ll notice that the “peak” of the data is shifted towards the right, while there’s a longer tail stretching to the left. Key characteristics to identify in a left-skewed histogram include: The peak (mode) is on the right side of the center. The tail extends longer on the left side. The mean is typically less than the median. To help visualize this, imagine you’re looking at a dataset of employee salaries in a company. You might see a large group of employees with higher salaries on the right, and a smaller number of employees with much lower salaries creating that left tail. Comparing Left Skewed Histogram with Other Distribution Types To truly understand what a left-skewed histogram looks like, it’s helpful to compare it with other common distribution types I’ve encountered in my consulting work: Normal Distribution: This is the classic “bell curve” shape, symmetrical on both sides. Unlike a left-skewed histogram, the mean, median, and mode are all equal in a normal distribution. Right Skewed Distribution: This is the mirror image of a left-skewed histogram. The tail extends to the right, with the bulk of the data on the left side. Symmetrical Distribution: While similar to a normal distribution, not all symmetrical distributions are normal. The key is that they’re balanced on both sides, unlike our left-skewed histogram. Understanding these differences is crucial. In my work with companies like Intel and Motorola, I’ve seen how misinterpreting the type of distribution can lead to costly mistakes in process improvement initiatives – a risk mitigated by the rigorous analytical training provided in six sigma certification programs. Ready to deepen your understanding of data distributions and their impact on business decisions? Enroll in a short course on statistical analysis and data interpretation. Enroll Now —> Properties and Characteristics of a Left Skewed Histogram Understanding the properties and characteristics of left-skewed histograms is crucial for effective data analysis and decision-making. Let’s dive into these key aspects that I’ve seen make a significant difference in projects across various industries. Mean, Median, and Mode Relationship One of the most distinctive features of a left-skewed histogram is the relationship between its mean, median, and mode. In my workshops, I often use this relationship as a quick identifier for left skewness: Mean < Median < Mode This relationship is quite different from what we see in normal distributions, where these three measures of central tendency are typically equal. Let me break it down based on my experiences: Mean: In a left-skewed histogram, the mean is pulled towards the left tail by the smaller values. I’ve seen this cause confusion in salary analyses, where the average (mean) salary can be misleadingly low due to a few entry-level positions. Median: The median, being the middle value, is less affected by extreme values. It’s often a more reliable measure of center in skewed distributions. Mode: This is the peak of our distribution, representing the most common value. In left-skewed histograms, it’s the rightmost of these three measures. Tail and Peak Analysis Understanding the tail and peak of a left-skewed histogram can provide valuable insights: Long Left Tail: This extended tail represents the less frequent, smaller values in our dataset. In my work with manufacturing processes, I’ve seen this tail represent rare defects or outlier performance issues. Peak Position: The peak (mode) is shifted towards the right side of the distribution. This indicates that the most common values in our dataset are on the higher end of the range. Skewness Measurement To quantify the degree of skewness, we use the skewness coefficient: Negative Skewness: A left-skewed histogram will have a negative skewness coefficient. The more negative the value, the more pronounced the left skew. Interpreting Values: In my experience, values between -0.5 and -1 indicate moderate skewness, while values below -1 suggest high skewness. However, these are general guidelines and can vary depending on the specific context of your data. Key Properties of Left-Skewed Histograms The mean is less than the median, which is less than the mode The long tail extends to the left Peak (mode) is positioned on the right side Negative skewness coefficient The majority of data points are on the right side Fewer, but more extreme, low values on the left Understanding these properties of left-skewed histograms is essential for accurate data interpretation. Interpreting a Left Skewed Histogram: A Step-by-Step Guide In my years of consulting and teaching Six Sigma methodologies, which forms the core of six sigma certification, I’ve found that interpreting left-skewed histograms is a crucial skill for data-driven decision-making. Identifying the Distribution of Shape The first step in interpreting a left-skewed histogram is recognizing its unique shape. Here’s what to look for: Longer tail on the left side Peak (mode) shifted to the right Gradual slope on the left, steeper slope on the right Remember, a left-skewed histogram doesn’t always mean there’s a problem. In some cases, it’s the expected distribution. For instance, in a project, we found that the distribution of time-to-failure for certain products was naturally left-skewed. Analyzing Central Tendency Measures of a Left Skewed Histogram Understanding the relationship between mean, median, and mode is crucial: Mean < Median < Mode Mean is pulled towards the left tail Median is often more representative of the typical value In a Six Sigma project, we used this relationship to identify discrepancies in customer satisfaction scores that weren’t apparent from the average alone. This type of practical data application is a core component of six sigma green belt certification, where participants learn to manage projects and utilize statistical tools effectively. Understanding Data Concentration and Outliers In a left-skewed histogram: Most data points cluster on the right side Outliers typically appear on the left side The left tail often represents rare but significant events During a process improvement initiative, we identified critical quality issues by focusing on these left-tail outliers. Understanding why these outliers occurred required diving deeper, a process thoroughly covered in root cause analysis training. Contextualizing the Skewness Always consider the context of your data: What does the left tail represent in your specific scenario? Is the skewness expected or a sign of an underlying issue? How does the skewness impact your business goals? Common Pitfalls of a Left Skewed Histogram and How to Avoid Them Be aware of these common mistakes: Focusing solely on the mean: Always consider median and mode too Ignoring outliers: They often hold valuable insights Assuming normality: Many real-world processes aren’t normally distributed In a project, we initially missed a crucial insight by falling into the “normality assumption” trap. Always question your assumptions! Want to apply these interpretation techniques to real-world scenarios? Get hands-on experience with advanced statistical techniques with our Lean Six Sigma Green Belt certification. Learn More —> Applications and Examples These asymmetrical distributions often reveal critical insights that can drive business decisions and process improvements. Let me share some examples from my experience working with global corporations and government institutions. Income Distribution Analysis One of the most common examples of a left-skewed histogram is income distribution. In a project I led for a large multinational corporation, we analyzed employee salaries across different departments. The resulting left-skewed histogram revealed: A large cluster of mid-level salaries on the right A long tail to the left represents lower-paid entry-level positions A few extremely high salaries pull the mean lower than the median This analysis helped the company identify potential pay equity issues and informed its compensation strategy. Age-Related Data During my work with healthcare organizations, I’ve often encountered left-skewed histograms in age-related data: Retirement Age: A slightly left-skewed histogram typically shows most people retiring around 65, with a tail extending to earlier retirements. Life Expectancy: When analyzing life expectancy data for a government health initiative, we found a left-skewed distribution with a peak around 80 years and a tail extending towards younger ages. Academic Performance Metrics with a Left Skewed Histogram In an educational consulting project, we analyzed student performance data: GPA distributions often show a left skew, with most students clustered around higher GPAs Standardized test scores can also exhibit left skewness, especially for advanced exams Understanding these distributions helped institutions develop more effective support programs for struggling students. Environmental Data Environmental metrics frequently display left skewness: Pollution Levels: In a project for an environmental agency, we found that daily air quality readings often produced a left-skewed histogram, with most days having low pollution and a few high-pollution days creating the left tail. Rainfall: Analyzing rainfall data for an agricultural client revealed a left-skewed distribution, with most months having moderate rainfall and occasional drought periods forming the left tail. Financial Markets Stock returns typically show a left-skewed distribution: Most returns cluster around a slightly positive value The left tail represents occasional significant losses Rare but extreme positive returns create a slight right skew Integrating lean fundamentals into financial analysis helps organizations streamline decision-making while accounting for skewed risk profiles. These examples demonstrate the ubiquity and importance of left-skewed histograms across various fields. By recognizing and correctly interpreting these distributions, you can uncover valuable insights that drive informed decision-making in your organization. Creating and Analyzing Left Skewed Histograms Let me share some insights I’ve gained from working with companies and government institutions over the years. Data Collection Considerations When dealing with potentially left-skewed data, keep these points in mind: Sample Size: Ensure your sample is large enough to reveal the true distribution. In a project at Motorola, we initially missed a left skew due to insufficient data. Data Quality: Be vigilant about data accuracy. Identifying these outliers often requires root cause analysis training, which equips teams to systematically address underlying process issues. Time Frame: For time-series data, consider whether the skewness is consistent or varies over time. This was crucial in a long-term process improvement project. Choosing Appropriate Bin Sizes for a Left Skewed Histogram Bin size can significantly impact the appearance of your left-skewed histogram: Too few bins can mask the skewness Too many bins can create noise and obscure the overall pattern Consider using unequal bin widths for highly skewed data I often use the square root of the sample size as a starting point for the number of bins, adjusting as needed based on the data’s characteristics. Tools and Software for Histogram Creation Over the years, I’ve used various tools to create left-skewed histograms: Excel: Great for quick visualizations. Use the Data Analysis ToolPak for basic histograms. R: My go-to for advanced statistical analysis. The ggplot2 package is excellent for customizable histograms. Python Libraries: Matplotlib and Seaborn are powerful for creating histograms, especially when dealing with large datasets. Minitab: A staple in Six Sigma projects, it offers robust histogram capabilities. Best Practices for Visual Presentation To effectively communicate your findings: Use clear labeling for axes and bins Include a legend if comparing multiple datasets Consider overlaying a normal distribution curve for comparison Use color strategically to highlight key areas of the distribution Remember, the goal isn’t just to create a left-skewed histogram, but to use it as a tool for insight and decision-making. In my workshops, I always emphasize that the real value comes from interpreting the histogram in the context of your specific business challenge. Additions to Left-Skewed Distributions These topics are crucial for anyone looking to master data analysis and drive meaningful business improvements. Transforming Left Skewed Histogram Data In many Six Sigma projects, I’ve encountered situations where we needed to transform left-skewed data for further analysis: Log Transformation: Often effective for slightly left-skewed histograms, especially in financial data. Box-Cox Transformation: A powerful method I’ve used in manufacturing processes to normalize skewed data. Square Root Transformation: Useful for counting data that shows left skewness. Remember, the goal of transformation is not just to achieve normality, but to make your data more amenable to statistical analysis while preserving its fundamental characteristics. Statistical Tests for Skewed Data When dealing with left-skewed distributions, traditional tests may not be appropriate. Here are some alternatives I frequently use: Mann-Whitney U Test: A non-parametric test I’ve applied in quality control projects. Kruskal-Wallis Test: Useful for comparing multiple groups with skewed data. Bootstrapping Methods: I’ve found these particularly valuable for estimating confidence intervals in skewed datasets. Implications for Predictive Modeling Left skewed data can significantly impact predictive models: In regression analysis, skewed residuals can violate assumptions and lead to unreliable predictions. For classification problems, skewed features may need special handling to prevent bias. Time series forecasting with left-skewed data often requires specialized models like GARCH. During a project, we had to redesign our entire predictive maintenance model due to overlooked left skewness in equipment failure times. Ethical Considerations in Interpreting Skewed Data As data scientists and business leaders, we have an ethical responsibility to interpret skewed data correctly: Avoid cherry-picking metrics that hide important information in the left tail. Be transparent about the limitations of your analysis when dealing with skewed distributions. Consider the real-world implications of decisions based on skewed data, especially in sensitive areas like healthcare or finance. Even a foundational awareness, often gained through introductory programs like six sigma white belt certification, helps emphasize the importance of careful interpretation and transparency. These advanced topics underscore the complexity and importance of understanding left-skewed histograms in real-world scenarios. By mastering these concepts, you’ll be better equipped to extract valuable insights from your data and make informed decisions. Participating in six sigma certification programs helps individuals learn these concepts and add value to their organization. Ready to dive deeper into advanced statistical methods and their applications? Focus on cutting-edge data analysis techniques for process improvement with a short course on Introduction to Graphical Analysis with Minitab. Get Started —> Historical Context and Future Trends As a Six Sigma Master Black Belt with over two decades of experience, I’ve witnessed firsthand the evolution of data analysis techniques, particularly in the realm of left-skewed histograms. Let’s take a journey through time and peek into the future of this crucial statistical tool. Brief History of Skewed Distribution Analysis The concept of skewed distributions dates back to the late 19th century, but its practical applications have grown exponentially: 1895: Karl Pearson introduced the concept of skewness in distributions. 1920s: R. Fisher developed methods for handling non-normal distributions. 1960s: Box-Cox transformations emerged, revolutionizing how we handle skewed data. 1980s: Robust statistical methods gained popularity, addressing challenges with skewed data. During the 1990s, I saw a shift from avoiding skewed data to embracing it for deeper insights. Emerging Applications in Big Data and Machine Learning The big data revolution has brought new relevance to left-skewed histograms: Anomaly Detection: In cybersecurity projects I’ve led, left-skewed distributions often indicate potential threats. Predictive Maintenance: At companies like GE, we’ve used skewed time-to-failure distributions to optimize maintenance schedules. Customer Behavior Analysis: E-commerce giants are leveraging skewed purchase patterns for personalized marketing. Machine learning algorithms are becoming more adept at handling skewed data, opening up new possibilities for automated analysis and decision-making. Predicted Developments of a Left Skewed Histogram in Data Visualization Techniques The future of left-skewed histogram analysis is exciting: Interactive 3D Visualizations: I’m seeing prototypes that allow users to “walk through” skewed distributions in virtual reality. Real-time Dynamic Histograms: Imagine dashboards that update left-skewed histograms as data streams, something we’re implementing in IoT projects. AI-assisted Interpretation: Tools that not only display left-skewed histograms but provide context-aware insights are on the horizon. As we move forward, the ability to understand and leverage left-skewed histograms will become increasingly crucial. I can confidently say that mastering these concepts will be a key differentiator in data-driven decision-making. Going Forward As we wrap up our deep dive into left-skewed histograms, I want to emphasize the critical role these statistical tools play in modern business and project management. I’ve seen firsthand how understanding left-skewed distributions can lead to breakthrough insights and drive meaningful improvements. Let’s recap the key points we’ve covered: We defined what a left-skewed histogram is and explored its unique characteristics. We examined the crucial relationship between mean, median, and mode in left-skewed distributions. We discussed real-world applications across various industries, from finance to manufacturing. We delved into advanced topics like data transformation and ethical considerations. Finally, we looked at the historical context and future trends in skewed data analysis. Understanding left-skewed histograms is not just about statistical knowledge – it’s about gaining a deeper understanding of your processes, customers, and business environment. If you’re ready to take your data analysis skills to the next level, consider exploring our advanced Six Sigma courses at SixSigma.us. Together, we can turn your data into actionable insights that drive real business results. SixSigma.us offers both Live Virtual classes as well as Online Self-Paced training. Most option includes access to the same great Master Black Belt instructors that teach our World Class in-person sessions. Sign-up today! Virtual Classroom Training Programs Self-Paced Online Training Programs SixSigma.us Accreditation & Affiliations
8745	https://math.unideb.hu/sites/default/files/inline-files/s-pol_revised.pdf	ON POLYNOMIALS WITH ONLY RATIONAL ROOTS LAJOS HAJDU, ROBERT TIJDEMAN, AND N ´ ORA VARGA Abstract. In this paper we study upper bounds for the degrees of polynomials with only rational roots. First we assume that the coeﬃcients are bounded. In the second theorem we suppose that the primes 2 and 3 do not divide any coeﬃcient. The third theorem concerns the case that all coeﬃcients are composed of primes from a ﬁxed ﬁnite set. 1. Introduction Polynomials in Z[x] with only rational roots are the simplest exam-ples of decomposable polynomials and forms. Such polynomials play an important role in the theory of Diophantine equations, see e.g. Ch. 9 of Evertse and Gy˝ ory . They cover norm forms which are crucial in Schmidt’s Subspace Theorem , and index forms and discriminant forms, see Evertse and Gy˝ ory . Many papers on Diophantine equa-tions deal with polynomials in Z[x] with only rational roots themselves, see e.g. Section 2 of Hajdu and Tijdeman . There is also an extensive literature on polynomials with restricted coeﬃcients, in particular, with coeﬃcients belonging to one of the sets {−1, 1}, {0, 1} or {−1, 0, 1}, see Hare and Jankauskas and the ref-erences there. In the ﬁrst case the polynomials are called Littlewood polynomials, in the second case (assuming that the constant term is non-zero) Newman polynomials. For examples of studies of the loca-tion of the roots of such polynomials, see Borwein et al. and Berend and Golan for Littlewood polynomials, Odlyzko and Poonen and Mercer for Newman polynomials, and Borwein and Pinner , Borwein and Erd´ elyi and Drungilas and Dubickas for polynomials with all coeﬃcients in {−1, 0, 1}. The set of polynomials f(x) ∈Z[x] with all coeﬃcients in {−1, 0, 1}, constant term non-zero and only rational roots is very restricted as Date: April 25, 2023. 2020 Mathematics Subject Classiﬁcation. 11R09. Key words and phrases. Polynomials with restricted coeﬃcients. Research supported in part by the E¨ otv¨ os Lor´ and Research Network (ELKH) and by the NKFIH grants 128088 and 130909. 1 2 LAJOS HAJDU, ROBERT TIJDEMAN, AND N ´ ORA VARGA can be simply checked: The only possible roots are 1 and −1. Hence f(x) = ±(x −1)a(x + 1)b for some a, b ∈Z≥0. The coeﬃcient of x is ±(b −a). Therefore \|b −a\| ≤1. It follows that f(x) = ±(x2 −1)k maybe multiplied with either x −1 or x + 1 where k = min(a, b). Since the coeﬃcients of f are in {−1, 0, 1}, we obtain k ∈{0, 1} and the degree of f is at most 3. An example of such a polynomial of degree 3 is (1) f(x) = x3 −x2 −x + 1 = (x −1)2(x + 1). In this paper we generalize this result in two ways. In the ﬁrst place we require that the coeﬃcients of f are bounded. By the height of a polynomial with integer coeﬃcients we mean the maximum of the absolute values of its coeﬃcients. We prove the following result. Theorem 1.1. Let f(x) ∈Z[x] be a polynomial of degree n with only non-zero rational roots and height bounded by H ≥2. Then we have both (2) n ≤ 2 log 2 + o(1) log H (H →∞) and (3) n ≤ 5 log 2 log H. Further, the constants 2/ log 2 and 5/ log 2 in (2) and (3), respectively, are best possible. Remark 1. Observe that for any f ∈Z[x] of degree n, the height of g := xmf(x) is the same as that of f, while deg(g) = m + n. So the assumption that the roots of f are non-zero is clearly necessary. The second generalization concerns the case that none of the coeﬃ-cients of f(x) is divisible by 2 or 3. We prove Theorem 1.2. Every polynomial f(x) ∈Z[x] with only rational roots of which no coeﬃcient is divisible by 2 or 3 has degree at most 3. Remark 2. Example (1) shows that degree 3 is possible. A further restriction is that the coeﬃcients of f are integral S-units, that is integers composed of primes from a ﬁnite set S. Such polyno-mials are called S-polynomials. The next theorem shows that for any n there are only ﬁnitely many families of S-polynomials of degree n having only rational roots. Theorem 1.3. Let S be a ﬁnite set of primes with \|S\| = s and n a positive integer. There exists an explicitly computable constant C = ON POLYNOMIALS WITH ONLY RATIONAL ROOTS 3 C(n, s) depending only on n and s and sets T1, T2 with max(\|T1\|, \|T2\|) ≤ C of n-tuples of S-units and (n −1)/2-tuples of S-units for n odd, respectively, such that if f(x) is an S-polynomial of degree n having only rational roots q1, . . . qn, then q1, . . . , qn satisfy one of the conditions (i) or (ii): (i) (q1, . . . , qn) = u(r1, . . . , rn) with some (r1, . . . , rn) ∈T1 and S-unit u, (ii) n = 2t + 1 is odd, and re-indexing q1, . . . , qn if necessary, we have q1 = u and (q2, . . . , qn) = v(r1, −r1, . . . , rt, −rt) with some (r1, . . . , rt) ∈T2 and S-units u, v. Further, the possibilities (i) and (ii) cannot be excluded. The proof of Theorem 1.1 is elementary. In the proof of Theorem 1.2 we use an old result of Fine that if all the coeﬃcients of the polynomial (x + 1)n are odd, then n is of the form 2α −1 for some α ∈Z≥0. We derive a corresponding result for the prime 3 in place of 2. Its proof is elementary. The proof of Theorem 1.3 is based on an estimate of Amoroso and Viada on the number of non-degenerate, non-proportional solutions of S-unit equations. We ﬁnish the paper with stating some open questions. 2. Proofs Observe that the rational roots of an S-polynomial f(x) are S-units, i.e. rational numbers whose numerators and denominators are com-posed exclusively of primes in S. This follows from the well-known fact that the denominator of a root of f(x) divides the leading coeﬃ-cient of f(x), while its numerator divides the constant term of f(x). In the sequel we shall use this fact without any further mentioning. Proof of Theorem 1.1. On the one hand, let f(x) = Pn j=0 ajxj. Then (4) \|f(i)\| ≤ X j is even \|aj\| + i X j is odd \|aj\| ≤ r 1 2n2 + n + 1 H. On the other hand, we may write f(x) = Qn j=1(qjx −pj) with pj, qj ∈ Z̸=0 for all j. Then (5) \|f(i)\| = n Y j=1 (qji −pj) = n Y j=1 q q2 j + p2 j ≥( √ 2)n. Therefore, (6) n log 2 ≤log 1 2n2 + n + 1 + 2 log H. 4 LAJOS HAJDU, ROBERT TIJDEMAN, AND N ´ ORA VARGA From this (2) easily follows. For the height H of the polynomial f(x) = (x2 −1)n/2 with even n ≥2 by Stirling’s formula we have log H = (1 + o(1))n log 2/2. This shows that the constant 2/ log 2 in (2) is best possible. To prove (3), observe that assuming (5/ log 2) log H < n from (6) we obtain n log 2 < log 1 2n2 + n + 1 + 2n log 2 5 . Hence we easily get n ≤9. Further, observe that if we assume that f has a root diﬀerent from ±1, then (5) can be sharpened to (7) \|f(i)\| ≥ √ 5( √ 2)n−1. Thus, in this case combining (5/ log 2) log H < n with (4) and (7), we get a contradiction for n ≥1. So to prove (3), we only need to check the polynomials of the shape f(x) = ±(x + 1)a(x −1)n−a with 0 ≤a ≤n for 1 ≤n ≤9. A simple calculation gives that for all these polynomials (3) holds. In particular, for n = 5 and a = 2, 3 we have equality. Thus e.g. the polynomial (x −1)3(x + 1)2 = x5 −x4 −2x3 + 2x2 + x −1 shows that the constant 5/ log 2 in (3) is best possible. So the theorem is proved. □ Remark 3. Several authors have considered upper bounds for the number r of real roots of f(x) ∈R[x]. Bloch and P´ olya proved r ≪H n log log n/ log n. This was improved by E. Schmidt (unpublished) and further by Schur . Schur proved (8) r2 < 4n log Q for n > 6, where Q = 1 \|a0an\|1/2(a2 0 + a2 1 + · · · + a2 n)1/2. Further he showed that the constant 4 in (8) cannot be improved. With r = n we obtain for polynomials f(x) ∈Z[x] with only real roots that (9) n ≤(4 + o(1)) log H (H →∞). (Here we used that Q ≤√n H in this case.) By Theorem 1.1 the constant 4 in (9) cannot be replaced by a constant less than 2/ log 2 ∼ ON POLYNOMIALS WITH ONLY RATIONAL ROOTS 5 2.885. For related results see Erd˝ os and Tur´ an , Littlewood and Oﬀord , and Borwein, Erd´ elyi and K´ os too. To prove Theorem 1.2, we need two lemmas. The ﬁrst one is a direct consequence of Theorem 4 of Fine . Lemma 2.1. Let n be a positive integer such that all the coeﬃcients of (x + 1)n are odd. Then n is of the shape 2α −1 with some α ∈Z≥0. The next lemma is new, and provides a similar result for prime 3. Lemma 2.2. Let a, b be non-negative integers. Put n := a+b. If none of the coeﬃcients of (x −1)a(x + 1)b is divisible by 3, then n is of the shape 3β −1, 2·3β −1, 3γ +3δ −1 or 2·3γ +3δ −1 with β ≥0, γ > δ ≥0. Proof. We call a pair of non-negative integers (a, b) good if none of the coeﬃcients of f(a,b)(x) := (x −1)a(x + 1)b is divisible by 3; otherwise we say that (a, b) is bad. Observe that this property is symmetric in a and b in view of the substitution x →−x. We distinguish between the residue classes of a and b modulo 3. CASE a ≡ε (mod 3), b ≡0 (mod 3), ε ∈{0, 1}. Letting a = 3u + ε, b = 3v we get that f(a,b)(x) ≡(x3 −1)u(x3 + 1)v(x −1)ε (mod 3). Hence (a, b) is good if and only if u = v = 0, i.e. n = 0 or 1. CASE b ≡ε (mod 3), a ≡0 (mod 3), ε ∈{0, 1}. By symmetry, this yields the same conclusion as in the previous case. CASE a ≡2 (mod 3), b ≡0 (mod 3). Writing a = 3u + 2, b = 3v we see that f(a,b)(x) ≡(x3 −1)u(x3 + 1)v(x2 + x + 1) (mod 3). This shows that (a, b) is good if and only if (u, v) is good. CASE b ≡2 (mod 3), a ≡0 (mod 3). By symmetry, this yields the same conclusion as in the previous case. CASE a ≡b ≡ε (mod 3), ε ∈{1, 2}. Putting a = 3u + ε, b = 3v + ε we see that f(a,b)(x) ≡(x3 −1)u(x3 + 1)v(x2 −1)ε (mod 3). Hence (a, b) is bad, as the coeﬃcient of x will be 0 (mod 3). CASE a ≡2 (mod 3), b ≡1 (mod 3). Letting a = 3u + 2, b = 3v + 1 we obtain that (10) f(a,b)(x) ≡(x3 −1)u(x3 + 1)v(x3 −x2 −x + 1) (mod 3). 6 LAJOS HAJDU, ROBERT TIJDEMAN, AND N ´ ORA VARGA From this we immediately see that if (u, v) is bad, then (a, b) is bad, too. Assume that (u, v) is good. Then we may write (11) (x3 −1)u(x3 + 1)v = u+v X i=0 cix3i with 3 ∤ci (i = 0, . . . , u + v); in particular, cu+v = 1. Then, combining (10) and (11), we obtain that (a, b) is good if and only if none of the integers cu+v, cu+v + cu+v−1, . . . , c1 + c0, c0 is divisible by 3. Since cu+v = 1, this gives ci ≡1 (mod 3) (i = 0, . . . , u + v). Hence we obtain, on replacing x3 by x1 in (11), that every coeﬃcient of (x1 −1)u(x1 + 1)v is 1 (mod 3). This is equivalent with (12) (x1 −1)u+1(x1 + 1)v ≡xu+v+1 1 −1 (mod 3). We show that (12) holds precisely for (13) (u, v) = (3ℓ−1, 0), (3ℓ−1, 3ℓ) (ℓ≥0). It is easy to check that (12) is valid for (u, v) given by (13). Assume that (12) holds for some (u, v) and write u + 1 = 3U + p, v = 3V + q with 0 ≤p, q ≤2 and U, V ≥0. Then (12) can be rewritten as (x3 1 −1)U(x3 1 + 1)V (x1 −1)p(x1 + 1)q ≡xu+v+1 1 −1 (mod 3). Hence we get two possibilities. If (p, q) ̸= (0, 0), then we must have (p, q) = (1, 0), (1, 1) and (U, V ) = (0, 0). So (u, v) = (0, 0), (0, 1) be-longing to (13) with ℓ= 0. If (p, q) = (0, 0) then we easily see that either V = 0 or U = V must be valid. Then (12) can be rewritten as (x3 1 −1)U ≡x3U 1 −1 (mod 3) or (x6 1 −1)U ≡x6U 1 −1 (mod 3), respectively. These clearly hold if and only if U is a power of 3, and our claim follows. Altogether, we see that in this case (a, b) is good if and only if (u, v) is good and (13) holds. CASE b ≡2 (mod 3), a ≡1 (mod 3). By symmetry, (a, b) is good if and only if setting a = 3u + 1 and b = 3v + 2, (u, v) is good and (14) (u, v) = (0, 3ℓ−1), (3ℓ, 3ℓ−1) (ℓ≥0). We conclude that (a, b) with a + b = n > 1 is good if and only if writing a = 3u + i, b = 3v + j with 0 ≤i, j ≤2, (u, v) is good and ([(a, b) (mod 3) equals (2, 0) or (0, 2)] or [(a, b) ≡(2, 1) (mod 3) and (13) holds] or [(a, b) ≡(1, 2) (mod 3) and (14) holds]). ON POLYNOMIALS WITH ONLY RATIONAL ROOTS 7 Suppose (a, b) ≡(2, 1) (mod 3). Then by (13) we have two options. If (u, v) = (3ℓ−1, 0) (ℓ≥0) then we have n = 3ℓ+1 = 3ℓ+1+30−1 and we are done. If (u, v) = (3ℓ−1, 3ℓ) (ℓ≥0) then n = 2·3ℓ+1 = 2·3ℓ+1+30−1, and our claim follows. By symmetry the case (a, b) ≡(1, 2) (mod 3) with (14) leads to the same values of n. It remains to deal with the case (a, b) (mod 3) equals (2, 0) or (0, 2). In both cases we have n = a+b = 3u+3v+2. Writing u = 3u1+u0, v = 3v1 + v0 with u0, v0 ∈{0, 1, 2} we have, by the above conclusion: (u, v) with u + v > 1 is good if and only if (u1, v1) is good and ([(u, v) (mod 3) equals (2, 0) or (0, 2)] or [(u, v) ≡(2, 1) (mod 3) and (13) holds] or [(u, v) ≡(1, 2) (mod 3) and (14) holds]), where in (13) and (14) (u, v) is replaced with (u1, v1). Thus, by induction, all possible degrees n are obtained by applying the substitution n →3n + 2 a number of times on the possible starting values 0, 1, 3ℓ, 2 · 3ℓfor any non-negative integer ℓ. By applying the substitution k times we ﬁnd 3k −1, 2 · 3k −1, 3ℓ+ 3k −1, 2 · 3ℓ+ 3k −1, respectively, with ℓ> k, for the only possible values of n. □ Remark 4. For all the mentioned values in Lemma 2.2 there are polynomials without coeﬃcients divisible by 3. We have modulo 3: (x−1)3ℓ−1 = (x−1)3ℓ/(x−1) ≡(x3ℓ−1)/(x−1) = x3ℓ−1+x3ℓ−2+. . .+1, (x−1)2·3ℓ−1 ≡(x3ℓ−1)2/(x−1) = (x3ℓ−1 +x3ℓ−2 +. . .+1)(x3ℓ−1) = = x2·3ℓ−1 + x2·3ℓ−2 + . . . + x3ℓ−x3ℓ−1 −x3ℓ−2 −. . . −1, (x−1)3ℓ−1(x+1)3ℓ= (x−1)3ℓ(x+1)3ℓ/(x−1) ≡(x3ℓ−1)(x3ℓ+1)/(x−1) = (x3ℓ−1 + x3ℓ−2 + . . . + 1)(x3ℓ+ 1) = x2·3ℓ−1 + x2·3ℓ−2 + . . . + 1. The ﬁrst identity can be multiplied by (x + 1)3k ≡x3k + 1 for any k less than ℓand yields the solutions (3ℓ−1, 0), (3ℓ−1, 1), (3ℓ−1, 3), . . . , (3ℓ−1, 3ℓ−1) for (deg(x−1), deg(x+1)). This provides the total degrees 3ℓ−1, 3ℓ+ 3k −1. The second assertion provides the total degree 2 ·3ℓ−1. This degree is found in another way by the third formula. 8 LAJOS HAJDU, ROBERT TIJDEMAN, AND N ´ ORA VARGA The third identity can be multiplied by (x + 1)3k ≡x3k + 1 for any k less than ℓand yields the solutions (3ℓ−1, 3ℓ), (3ℓ−1, 3ℓ+ 1), (3ℓ−1, 3ℓ+ 3), . . . , (3ℓ−1, 3ℓ+ 3ℓ−1) for (deg(x −1), deg(x + 1)). This provides the total degrees 2 · 3ℓ−1 and 2 · 3ℓ+ 3k −1. Proof of Theorem 1.2. Let f be as in the statement. Since we argue modulo 2 and 3, and 2, 3 do not divide the leading coeﬃcient of f, we may assume that f is monic. Since the roots of f are odd, Lemma 2.1 shows that n + 1 is a power of 2. Further, since the roots of f are not divisible by 3, by Lemma 2.2 we get that n + 1 is of the shape 3β, 2 · 3β, 3γ + 3δ or 2 · 3γ + 3δ. The combination is possible only for n = 0, 1, 3, as a simple check reveals. □ For the proof of Theorem 1.3 we use the theory of S-unit equations. Let S be a ﬁnite set of primes, b1, . . . , bm non-zero rationals, and con-sider the equation (15) b1x1 + · · · + bmxm = 0 in S-units x1, . . . , xm. A solution (y1, . . . , ym) of (15) is called non-degenerate if X i∈I biyi ̸= 0 for each non-empty subset I of {1, . . . , m}. Further, two solutions (y1, . . . , ym) and (z1, . . . , zm) are called propor-tional, if there is an S-unit u such that (z1, . . . , zm) = u(y1, . . . , ym). The following result is due to Amoroso and Viada; see the paragraph after (1.7) on p. 412 of . (For an earlier version see , and for the case m = 2 see .) Note that in fact the original result of Amoroso and Viada concerns the inhomogeneous case, i.e. where the right hand side of (15) is 1. However, it is easy to transform their result into the shape of (15). Lemma 2.3. Equation (15) has at most (8m −8)4(m−1)4(m+s) non-degenerate, non-proportional solutions, where s = \|S\|. Proof of Theorem 1.3. Suppose that f(x) = Pn j=0 ajxj is an S-polynomial of degree n having only rational roots q1, . . . , qn. By our assumption, a0, a1, . . . , an are integral S-units. We have (16) Aj = σj(q1, . . . , qn) (1 ≤j ≤n) where Aj = (−1)jan−j/an and σj is the j-th elementary symmetric polynomial (of degree j) of q1, . . . , qn. Using (16) for j = 1, 2 we get (17) q2 1 + · · · + q2 n = A2 1 −2A2. ON POLYNOMIALS WITH ONLY RATIONAL ROOTS 9 This shows that (q2 1, . . . , q2 n, A2 1, A2) yields a solution to the S-unit equa-tion (18) x1 + · · · + xn −xn+1 + 2xn+2 = 0. If (q2 1, . . . , q2 n, A2 1, A2) is a solution with no vanishing subsums, then by Lemma 2.3 we can write q2 i = u0ℓi (i = 1, . . . , n), where (ℓ1, . . . , ℓn) comes from a ﬁnite set of cardinality bounded in terms of n and s, and u0 is an S-unit. Obviously, the squarefree parts of ℓ1, . . . , ℓn are the same, say ℓ0. Thus letting r2 i = ℓi/ℓ0 (i = 1, . . . , n) and u2 = u0ℓ0, we have qi = ±uri (i = 1, . . . , n) and we are done in this case. Hence we may assume that (q2 1, . . . , q2 n, A2 1, A2) contains a vanishing subsum. Since q2 i > 0 (1 ≤i ≤n), the only possibility is that (after re-indexing q1, . . . , qn if necessary) we have (19) q2 1 + · · · + q2 k −A2 1 = 0, (20) q2 k+1 + · · · + q2 n + 2A2 = 0 for some k with 1 ≤k < n. It is easy to see that (19) and (20) do not have a vanishing subsum. Thus, similarly as above, Lemma 2.3 yields that (q1, . . . , qk) = u(w1, . . . , wk), (qk+1, . . . , qn) = v(r1, . . . , rℓ), A1 = ut1 ̸= 0, A2 = v2t2 ̸= 0, where ℓ= n−k and both (w1, . . . , wk, t1) and (r1, . . . , rℓ, t2) come from ﬁnite sets of S-units of cardinalities bounded in terms of n and s, and u, v are S-units. Hence (16) for j = 1 yields that (21) u(w1 + · · · + wk) + v(r1 + · · · + rℓ) = ut1. If r1 + · · · + rℓ̸= 0 then the S-unit v/u comes from a set of cardinality bounded in terms of n and s, and we are in case (i). So we may suppose that w1 + · · · + wk = t1, r1 + · · · + rℓ = 0. As we have k ≥1, ℓ≥1 and, by (19) and (20), w2 1 + · · · + w2 k −t2 1 = 0, r2 1 + · · · + r2 ℓ+ 2t2 = 0, we obtain σ2(w1, . . . , wk) = 0, σ2(r1, . . . , rℓ) = t2. 10 LAJOS HAJDU, ROBERT TIJDEMAN, AND N ´ ORA VARGA We shall prove by contradiction that k = 1. Assume that k ≥2. If k = 2 then w1w2 = 0, which is not possible. So, k ≥3. Hence u3σ3(w1, . . . , wk) + u2vσ2(w1, . . . , wk)σ1(r1, . . . , rℓ)+ + uv2σ1(w1, . . . , wk)σ2(r1, . . . , rℓ) + v3σ3(r1, . . . , rℓ) −A3 = 0. Here σj(r1, . . . , rℓ) = 0 if ℓ< j. In view of the previously obtained assertions, we get (22) u3σ3(w1, . . . , wk) + uv2t1t2 + v3σ3(r1, . . . , rℓ) −A3 = 0. If σ3(w1, . . . , wk) ̸= 0 or σ3(r1, . . . , rℓ) ̸= 0 then (22) by Lemma 2.3 easily yields (both with or without vanishing subsums) that v/u belongs to a set of cardinality bounded in terms of n and s, and we are in case (i). So we may assume that σ3(w1, . . . , wk) = 0, σ3(r1, . . . , rℓ) = 0. Then we get w3 1 + · · · + w3 k = σ1(w1, . . . , wk)3 −3σ1(w1, . . . , wk)σ2(w1, . . . , wk)+ + 3σ3(w1, . . . , wk) = t3 1. We have obtained (23)      w1 + · · · + wk = t1, w2 1 + · · · + w2 k = t2 1, w3 1 + · · · + w3 k = t3 1. Note that (23) implies that there are indices i1, i2 with wi1 > 0 and wi2 < 0, thus \|t1\| = q w2 1 + · · · + w2 k = p \|w1\|2 + · · · + \|wk\|2 ≥ ≥ 3 p \|w1\|3 + · · · + \|wk\|3 > \|t1\|. This contradiction shows that k = 1 and ℓ= n −1. Recall that σ1(r1, . . . , rℓ) = 0, σ2(r1, . . . , rℓ) = t2, σ3(r1, . . . , rℓ) = 0. Hence (22) yields A3 = uv2t1t2. Further, by (16) and k = 1 we get (24) Aj = uvj−1w1σj−1(r1, . . . , rℓ) + vjσj(r1, . . . , rℓ) (4 ≤j ≤n). From this, taking j = 4 we obtain σ4(r1, . . . , rℓ) = A4/v4 ̸= 0. ON POLYNOMIALS WITH ONLY RATIONAL ROOTS 11 Now (24) for j = 5 by Lemma 2.3 yields that if σ5(r1, . . . , rℓ) ̸= 0 then v/u comes from a set of cardinality bounded by n and s, and we are in case (i). So we may assume that σ5(r1, . . . , rℓ) = 0. Now by repeating this argument, we may assume that ( σj(r1, . . . , rℓ) = Aj/vj ̸= 0 for j even, σj(r1, . . . , rℓ) = 0 for j odd. In particular, since σℓ(r1, . . . , rℓ) = r1 · · · rℓcannot be zero, ℓis even whence n = ℓ+ 1 is odd. Observing that (x + r1) · · · (x + rℓ) is an even polynomial, writing ℓ= 2t and re-indexing the S-units ri (1 ≤i ≤ℓ) such that rt+i = −ri (1 ≤i ≤t), we see that we are in case (ii). Finally, we show that the possibilities (i) and (ii) cannot be excluded. Indeed, if r1, . . . , rn is a set of rational roots of an S-polynomial of degree n, then clearly, the same is true for ur1, . . . , urn for any S-unit u, showing the necessity of (i). On the other hand, let r2 1, . . . , r2 t be the rational roots of the S-polynomial (x −r2 1) · · · (x −r2 t ). Then in the polynomial (x2 −r2 1) · · · (x2 −r2 t ), all the coeﬃcients of the even powers of x are S-units (while the coeﬃcients of the odd powers of x equal 0). Thus for any S-units u, v, all the coeﬃcients of the polynomial (x + u)(x −vr1)(x + vr1) · · · (x −vrt)(x + vrt) are S-units. This shows that (ii) cannot be excluded either. Note that it is easy to construct as many such non-proportional tuples as we like: Take arbitrary tuples of n integers (or t squares) that are non-proportional and deﬁne S as the set of prime factors of the product of their elementary symmetric polynomials. □ 3. Open problems We wonder whether the following statement is correct: Problem 1. Is it true that for any primes p and q there exists an n1 = n1(p, q) such that every polynomial f(x) ∈Z[x] with only rational roots of which no coeﬃcient is divisible by p or q has degree at most n1? Theorem 1.1 shows that the answer is ‘yes’ for the pair of primes (p, q) = (2, 3). A weaker statement is a restriction to S-polynomials. Problem 2. Is it true that for any ﬁnite set S of primes there exists an n2 = n2(S) such that every S-polynomial f(x) ∈Z[x] with only rational roots has degree at most n2? 12 LAJOS HAJDU, ROBERT TIJDEMAN, AND N ´ ORA VARGA Theorem 1.2 yields an aﬃrmative answer for sets S of primes with 2, 3 / ∈S. The last problem is raised by Lemmas 2.1 and 2.2. Problem 3. Is it true that for every prime p there exists a constant c(p) such that if f(x) ∈Z[x] has only rational roots and none of the coeﬃcients of f is divisible by p, then deg(f)+1 in its p-adic expansion has at most c(p) non-zero digits? In particular, can one take c(p) = p −1? Lemmas 2.1 and 2.2 show that the answer is ‘yes’ with c(p) = p −1 for p = 2, 3. Note that an aﬃrmative answer to Problem 3 through a deep result of Stewart would yield positive answers to Problems 1 and 2, as well. 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Hajdu, R. Tijdeman, N. Varga, Diophantine equations for Littlewood poly-nomials, Acta Arith. (to appear). L. Hajdu, N. Varga, Diophantine equations for polynomials with restricted co-eﬃcients, I (power values), Bull. Austral. Math. Soc. 106 (2022), 254–263. K.G. Hare, J. Jankauskas, On Newman and Littlewood polynomials with pre-scribed number of zeros inside the unit disk, Math. Comput. 90 (2021), 831– 870. J.E. Littlewood, A.C. Oﬀord, On the number of real roots of a random algebraic equation. II, Math. Proc. Cambridge Philos. Soc. 35 (1939), 133–148. I.D. Mercer, Newman polynomials not vanishing on the unit circle, Integers 12 (67) (2012), 7 pp. M. Odlyzko, B. Poonen, Zeros of polynomials with 0, 1 coeﬃcients, Enseign. Math. 39 (1993), 317–348. W.M. Schmidt, Diophantine Approximations and Diophantine Equations, Lect. Notes Math. 1467, Springer-Verlag, 1991. I. Schur, Untersuchungen ¨ uber algebraische Gleichungen, Sitz. Preuss. Akad. Wiss., Phys.-Math. Kl. (1933), 403–428. C.L. Stewart, On the representation of an integer in two diﬀerent bases, J. reine angew. Math 319 (1980), 63–72. G. Szeg˝ o, Bemerkungen zu einem Satz von E. Schmidt ¨ uber algebraische Glei-chungen, Sitz. Preuss. Akad. Wiss., Phys.-Math. Kl. (1934), 86–98. Institute of Mathematics, University of Debrecen, P. O. Box 400, H-4002 Debrecen, Hungary and ELKH-DE Equations, Functions, Curves and their Applications Research Group Email address: hajdul@science.unideb.hu Mathematical Institute, Leiden University, Postbus 9512, 2300 RA Leiden, The Netherlands Email address: tijdeman@ziggo.nl Institute of Mathematics, University of Debrecen, P. O. Box 400, H-4002 Debrecen, Hungary and ELKH-DE Equations, Functions, Curves and their Applications Research Group Email address: nvarga@science.unideb.hu
8746	https://math.stackexchange.com/questions/9933/why-is-negative-times-negative-positive?page=2&tab=scoredesc	abstract algebra - Why is negative times negative = positive? - 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 Why is negative times negative = positive? Ask Question Asked 14 years, 10 months ago Modified12 months ago Viewed 124k times This question shows research effort; it is useful and clear 155 Save this question. Show activity on this post. Someone recently asked me why a negative ×× a negative is positive, and why a negative ×× a positive is negative, etc. I went ahead and gave them a proof by contradiction like this: Assume (−x)⋅(−y)=−x y(−x)⋅(−y)=−x y Then divide both sides by (−x)(−x) and you get (−y)=y(−y)=y Since we have a contradiction, then our first assumption must be incorrect. I'm guessing I did something wrong here. Since the conclusion of (−x)⋅(−y)=(x y)(−x)⋅(−y)=(x y) is hard to derive from what I wrote. Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one? abstract-algebra algebra-precalculus arithmetic education faq Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Nov 20, 2023 at 11:54 calculatormathematical 2,324 2 2 gold badges 11 11 silver badges 28 28 bronze badges asked Nov 12, 2010 at 2:11 SevSev 2,151 5 5 gold badges 19 19 silver badges 19 19 bronze badges 21 41 You haven't proven that -xy = (-x)y.Qiaochu Yuan –Qiaochu Yuan 2010-11-12 02:24:58 +00:00 Commented Nov 12, 2010 at 2:24 74 It is not an uncommon question, and it's never not easy to show.J. M. ain't a mathematician –J. M. ain't a mathematician 2010-11-12 04:06:47 +00:00 Commented Nov 12, 2010 at 4:06 18 J.M. is pointing out that logical negation works the same way as multiplying negative numbers (two negatives make a positive), not belittling your question. You understood his double negative statements, and so you have another intuitive path to see that negativenegative=positive.Jonas Kibelbek –Jonas Kibelbek 2010-11-12 04:48:13 +00:00 Commented Nov 12, 2010 at 4:48 14 Does he understand why a negative times a positive is negative? And if so, how?Paul –Paul 2015-11-08 21:00:06 +00:00 Commented Nov 8, 2015 at 21:00 10 Have you shown him a sequence, to see a pattern? So for example 2×−3=−6 2×−3=−6, 1×−3=−3 1×−3=−3, 0×−3=0 0×−3=0, −1×−3=3−1×−3=3...?Paul –Paul 2015-11-08 21:03:43 +00:00 Commented Nov 8, 2015 at 21:03 \|Show 16 more comments 41 Answers 41 Sorted by: Reset to default Prev1 2 This answer is useful 2 Save this answer. Show activity on this post. Well, the way I think about it is this. We have the non-negative integers (0,1,2,3,4, etc.). We introduce the negative numbers and need to define multiplication with negative numbers so that we have internal consistency. We wish to keep the property that 0anything = 0, negative or positive. We also want to keep the distributive property. In order to keep the above two properties, we're forced to define the product of two negatives as a positive. 0(-3) = 0 (5 + (-5))(-3) = 0 (I'm plugging in 5+ (-5) for zero) 5(-3) + (-5)(-3) = 0 (distributive property) add 53 to both sides. 53 cancels with 5(-3) (-5)(-3) = 53 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 9, 2015 at 19:51 Ameet SharmaAmeet Sharma 3,007 15 15 silver badges 24 24 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. I think I have a pretty simple proof to why a negative times negative is a positive. We know that every x x has an inverse −x−x which when added together equals 0 0. Now as x+(−x)=0 x+(−x)=0 we can multiply this by some arbitrary −y−y to get −(y)x+(−y)(−x)=0−(y)x+(−y)(−x)=0. But we also know that x y+(−x y)=0 x y+(−x y)=0 so since the additive inverse is unique this means that x y=(−x)(−y)x y=(−x)(−y). Of course this assumes x(−y)=−(x y)x(−y)=−(x y) but that can be proved by multiplying the first equation with y y instead of −y−y. I am open to any suggestions and feedback. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jul 28, 2018 at 11:57 user500668user500668 345 3 3 silver badges 14 14 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Here's a simple proof that relies on the distributive property, the transitive property of equality, and cancellation. Theorem: If x,y∈N x,y∈N, then (−x)(−y)=x y(−x)(−y)=x y. Proof. First, notice that (−x)(−y)+x(−y)=((−x)+x)(−y)=0(−y)=0.(−x)(−y)+x(−y)=((−x)+x)(−y)=0(−y)=0. Similarly, x y+x(−y)=x(y+(−y))=x(0)=0.x y+x(−y)=x(y+(−y))=x(0)=0. By the transitive property of equality, (−x)(−y)+x(−y)=x y+x(−y)(−x)(−y)+x(−y)=x y+x(−y), and cancellation yields (−x)(−y)=x y(−x)(−y)=x y, as desired. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Apr 14, 2023 at 14:17 answered May 16, 2022 at 18:06 Pietro PaparellaPietro Paparella 3,690 1 1 gold badge 21 21 silver badges 31 31 bronze badges 1 2 Please don't duplicate prior answers, e.g. here (esp. very old answers).Bill Dubuque –Bill Dubuque 2023-03-13 22:23:54 +00:00 Commented Mar 13, 2023 at 22:23 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. If he already accepts that 1⋅−1=−1 1⋅−1=−1, then we can use some hand-waved algebra to show that 1=−1⋅−1 1=−1⋅−1 using symmetry, just like addition and subtraction. 2−1=¿w h a t?2−1=¿w h a t? , This is the question. 1+¿w h a t?=2 1+¿w h a t?=2 , But this is what we're really asking, in simpler terms. If 1 times -1 is -1, then -1 times what is 1? Because 1/1 is 1 we don't have to worry about reciprocal fractions, so even though it's a little misleading to use multiplication as the inverse of multiplication, I think it works well enough for now. 1⋅−1=−1 1⋅−1=−1 ¿w h a t?⋅−1=1¿w h a t?⋅−1=1 If we say ¿what? is 1, then we have a contradiction, because then 1⋅−1=−1 1⋅−1=−1 in the first equation, then 1⋅−1=1 1⋅−1=1 in the second equation. If we say ¿what? is -1, then there's no contradiction. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 9, 2015 at 3:31 MichaelSMichaelS 281 1 1 silver badge 8 8 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Why a negative times a negative can be reduced to the question of why -1 x -1 = 1. The reason for that is because it is forced upon you by the other rules of arithmetic. 1 + (-1) = 0 because of the definition of -1 as the additive inverse of 1 Now multiple both sides by -1 to get -1(1+(-1)) = 0 because 0 times anything is 0 Use distributive law to get: -1 1 + (-1)x(-1) = 0 Now -1 1 = -1 because 1 is multiplicative identity. So we have -1 + (-1)x (-1) = 0 Put -1 on the other side by adding 1 to both sides to get (-1) x (-1) = 1 So -1 x -1 = 1. Now for any other negative numbers x, y we have x = (-1) \|x\| and y= (-1) \|y\| So x y = (-1) \|x\| (-1) \|y\| = (-1) (-1) \|x\| \|y\| = \|x y\| is positive. Now that you know the reason it really doesn't make much difference in understanding. This question is not really that important. It's like asking why is 1 raised to the 0 power equal to 1? Because that's forced upon you by other rules of exponents,etc. A lot of time is wasted on this. This is not the kind of problem kids should be thinking about. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 9, 2015 at 6:37 larrylarry 63 1 1 bronze badge 8 5 I don't think this would be that accessible to a child who is only starting to learn how to multiply, so it does not really answer the question.Dylan –Dylan 2015-11-09 06:44:49 +00:00 Commented Nov 9, 2015 at 6:44 1 The short answer is forget about it kid. Just accept it as a rule of arithmetic. If a child doesn't understand this. then NO answer is going to satisfy him. There is no physical reason why it is so. Math is a man made. The rules can be anything. For instance why 1+1 = 2 and not 0? Well in other number systems it can be. 1+1 =0 . No reason to ask why. it just happens to be true. There is no universal truth in math. It is all in the axioms. Or why is parallel postulate true? It isn't? You can use other postulates.larry –larry 2015-11-09 06:50:18 +00:00 Commented Nov 9, 2015 at 6:50 3 I'm not planning to have children any time soon, but if I were, I certainly wouldn't want them to grow up just accepting things without questioning them or asking "why?". Also, that still doesn't change the fact this this is not an answer to the question which was asked. You could post it as a comment, if you like, but it is not an answer in my opinion.Dylan –Dylan 2015-11-09 06:54:32 +00:00 Commented Nov 9, 2015 at 6:54 4 But a lot of the time, we made the rules to model certain phenomenon or to capture certain intuitions, and it is perfectly valid to ask why the rules are the way that they are. The idea of the integers was around long before we started taking an axiomatic approach to them, and historically there were certainly other reasons to consider the product of two negative integers to be positive than just "them's the rules, kid".Dylan –Dylan 2015-11-09 07:10:30 +00:00 Commented Nov 9, 2015 at 7:10 1 Perhaps it is difficult to explain to a layperson, but that is precisely why this question is being asked: to gather input on what some approaches might be to explain the concept to a layperson, or--ideally--an eight-year-old.Dylan –Dylan 2015-11-09 07:13:49 +00:00 Commented Nov 9, 2015 at 7:13 \|Show 3 more comments This answer is useful 1 Save this answer. Show activity on this post. This is a sketch of an explanation that can easily be made more or less formal. My job now is to explain to you (who have a clue) my way to explain to somebody else (who has almost no clue). That's why I use formal language (like define) and notation (like f(x)f(x)) in order to keep my answer neat. At the same time I stick to integers, as if this "somebody else" was a kid. Adjust the form and scope to your interlocutor's level. Let's define f a(x)≡a x f a(x)≡a x The explanation goes as follows: On X Y X Y plane plot y=f 0(x),y=f 1(x),y=f 2(x),…y=f 0(x),y=f 1(x),y=f 2(x),… for x∈{0,1,2,3,…}x∈{0,1,2,3,…}; use different symbols/colors to distinguish the functions. Notice (or better let the interlocutor notice) that the points of f 0 f 0 lay along a straight line, the points of f 1 f 1 lay along another straight line etc. There is no reason this rule shouldn't apply when we consider x∈{…,−2,−1,0,1,2,…}x∈{…,−2,−1,0,1,2,…}; expand the plot. Notice ∀a f a(0)=0∀a f a(0)=0. Why? Notice ∀a f a(1)=a∀a f a(1)=a. Why? There is no reason the three rules shouldn't apply for negative a a; plot y=f−1(x),y=f−2(x),…y=f−1(x),y=f−2(x),… according to the rules. It will look like this: Only the blue points were obtained by actual multiplication. The rest of them were placed thanks to the rules found. And here you go. The result of (−1)⋅(−1)(−1)⋅(−1) is there among few others. You can replace "there is no reason the rules shouldn't apply" with some formal proofs if your interlocutor can understand them. While explaining to a kid it should be enough to point out this is the way mathematics works – coherence, no unnecessary exceptions, rules as broad as possible. I think it can be quite reassuring on an early stage of education. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 1, 2016 at 0:09 Kamil MaciorowskiKamil Maciorowski 2,908 1 1 gold badge 17 17 silver badges 24 24 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The following is based on this Reddit post that I edited for grammar and readability. As you've probably heard, multiplication is just a fancy word for repeated addition. If you wanted to repeatedly add 5 5 three times (5+5+5 5+5+5), multiplication allows you to rewrite that as 3×5 3×5. What if you wanted to repeatedly add −5−5? You would write (−5)+(−5)+(−5)(−5)+(−5)+(−5) as 3 x(−5)3 x(−5). Now what if I asked you to subtract (−5)(−5) three times? Well, you can write that as −(−5)−(−5)−(−5)−(−5)−(−5)−(−5), or (−3)×(−5)(−3)×(−5). This is just 5+5+5 5+5+5: a positive number. So why is subtracting a negative the same as adding a positive? Because as explained in the previous paragraph, when you subtract debt, you get money. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Dec 19, 2017 at 7:04 user53259 user53259 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. There is an abundance of intuitive answers to this question so it came as a surprise that a somewhat trivial (by the time someone finishes highschool he should have learned it..) but also fundamental example from physics is absent. Long story short: Opposites attract, Same repel When two particles with the same electromagentic charge interact, either positive or negative, they produce the same result-they repel each other. While if one has a negative charge and the other a positive, they attract each other. Let's symbolize the interaction with I, a positively charged particle with (+)(+), a negatively charged one with (−)(−), attraction with A A and repulsion with R R. So (+)I(+)=R(+)I(−)=A(−)I(+)=A(−)I(−)=R(+)I(+)=R(+)I(−)=A(−)I(+)=A(−)I(−)=R For me this looks like the best "natural" example, suitable to students relatively quickly. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Nov 12, 2018 at 22:52 MathematicianByMistakeMathematicianByMistake 5,935 2 2 gold badges 19 19 silver badges 35 35 bronze badges 2 Interesting viewpoint, but how does the story translate to magnetism and magnetic poles?Frenzy Li –Frenzy Li 2018-11-12 22:56:24 +00:00 Commented Nov 12, 2018 at 22:56 @FrenzyLi Well, the "hard math" answer involves the uniqueness of the inverse element in a ring, which obviously cannot be taught to highschool students the same time they learn about negatives. So instructors tend to look for examples from the physical world but end up talking about artificial creations like economics, running competitions etc. Electromagnetism is an integral part of the universe-one of the fundamental forces in particular..What could be more natural?MathematicianByMistake –MathematicianByMistake 2018-11-12 23:00:08 +00:00 Commented Nov 12, 2018 at 23:00 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. I like the most upvoted answer from Reddit much more than the 2nd most upvoted answer, but I reformat it with MathJax beneath. sjets3 14.1k points 7 months ago Imagine you are watching a movie. The first number is how the person in the movie is moving. The second number is how you are watching the film (normal or in reverse). 1×1 1×1 is a person walking forward, you watch it normal. Answer is you see a person walking forward, which is 1. 1×−1 1×−1 is a person walking forward, you watch it in reverse. You see a person walking backwards. -1 −1×1−1×1 is a person walking backward, you watch it normal. You see a person walking backwards. -1 −1×−1−1×−1 is a person walking backwards, but you watch it in reverse. What you will see is a person that looks like they are walking forward. 1 As a funny example of the last para. overhead, compare (original video) with (film person walking backward then play backward). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Jan 8, 2019 at 0:16 user53259 user53259 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. This is an explanation by example (in this case, a recipe for multiplying using the number line). It is not a proof. As I was reading Euler's explanation of the product of negatives, I was baffled and came looking to math stack exchange for help. The answers were very interesting, ranging from movies running forward and backward to vectors and rotations. I was particularly intrigued by the answer quoting Ruth McNeil as her experience mirrored my own. So, I worked through the answers given to see which, if any, satisfied my intuition, and while there were some that were appealing, I eventually settled on the following, drawn from different answers: Here is a recipe for doing multiplication using the traditional number line with origin 0, and a default direction of positive movement (to the right): Determine the direction to move - a) if the sign is positive, it indicates continuation of direction b) if the sign is negative, it indicates reversal of direction Determine the amount of movement - a) repeated addition of the quantities involved is fine Place the point at the indicated distance along the determined direction There are four cases to consider when multiplying two numbers (3 and 2, for example): postive times positive - (+3) (+2) the first sign is positive - continue forward, the second is also positive, again continue forward, the direction to move is forward 3 added twice is 6, as is 2 added thrice, the amount of movement is 6 units A point is placed at +6 negative times positive - (-3) (+2) the first sign is negative - reverse direction, the second is positive, again continue forward (in the reverse direction), the direction to move is reversed (to the left) 3 added twice is 6, as is 2 added thrice, the amount of movement is 6 units A point is placed at -6 positive times negative - (+3) (-2) the first sign is positive - continue forward, the second is negative, so reverse direction, the direction to move is reversed (to the left) 3 added twice is 6, as is 2 added thrice, the amount of movement is 6 units A point is placed at -6 negative times negative - (-3) (-2) the first sign is negative - reverse direction, the second is negative, reverse again, the direction to move is forward 3 added twice is 6, as is 2 added thrice, the amount of movement is 6 units A point is placed at +6 I don't think this is completely satisfying and YMMV, but I hope it helps some. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 12, 2023 at 17:14 answered Dec 12, 2023 at 16:59 decuserdecuser 133 6 6 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. We know that R⊂C R⊂C. Let x,y∈R x,y∈R and x<0,y<0 x<0,y<0 Any number n n can be written in polar form as n=\|n\|e i θ n=\|n\|e i θ where θ θ is the angle made by the line joining origin and n n with positive x x axis. Therefore x=\|x\|e i π x=\|x\|e i π and y=\|y\|e i π y=\|y\|e i π x y=x=\|x\|e i π\|y\|e i π=\|x\|\|y\|e i 2 π=\|x\|\|y\|x y=x=\|x\|e i π\|y\|e i π=\|x\|\|y\|e i 2 π=\|x\|\|y\| x x and y y being negative but there product is \|x\|\|y\|\|x\|\|y\| which is positive! NOTE e i m=cos m+ι sin m e i m=cos⁡m+ι sin⁡m Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jun 13, 2015 at 3:06 answered Jun 12, 2015 at 7:38 SinghSingh 2,178 1 1 gold badge 24 24 silver badges 47 47 bronze badges 3 6 -1: This is an incredibly over-complicated argument about real numbers that actually holds in far more generality anyway.Zev Chonoles –Zev Chonoles 2015-06-12 07:42:04 +00:00 Commented Jun 12, 2015 at 7:42 @ZevChonoles Sir though it may feel complicated it is technically correct and provides and alternative approach to the given problem. I request you to kindly elaborate your comment.Singh –Singh 2015-06-12 07:56:01 +00:00 Commented Jun 12, 2015 at 7:56 3 @Zev I agree ,but not only this, proving it's true in the complex plane does NOT necessarily imply it's true for real numbers only. To give a simple counterexample, the square roots of negative numbers can be multiplied in C, which does not carry over by inclusion to R.Mathemagician1234 –Mathemagician1234 2015-06-13 03:17:48 +00:00 Commented Jun 13, 2015 at 3:17 Add a comment\| Prev1 2 You must log in to answer this question. Protected question. To answer this question, you need to have at least 10 reputation on this site (not counting the association bonus). The reputation requirement helps protect this question from spam and non-answer activity. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions abstract-algebra algebra-precalculus arithmetic education faq 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 6Why minus times minus needs to be plus? 4Product of two negative numbers is positive 4Why is −a×−b=a b−a×−b=a b? 4Product of '-' and '-' is '+'! 39Explaining to a kid why a negative × negative = positive? 1Confused about (−1)×(−1)=(+1)(−1)×(−1)=(+1) 1How can a negative multiplied by a negative give positive? 3how can I prove negative times negative is positive. 1Why (−1)⋅(−1)=1(−1)⋅(−1)=1 0Why is (-1)(-1) = 1? 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Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. #### STANDARDS > US-GA Math Grade 1 Numerical Reasoning – Counting, Numbers, Equality, Place Value, Addition, Subtraction Patterning & Algebraic Reasoning – Repeating Patterns, Growing, Patterns, and Shrinking Patterns Geometric & Spatial Reasoning – Shapes, Attributes, Partitions Of Circles and Rectangles Numerical Reasoning – Base Ten Structure, Addition and Subtraction Within 100 Numerical Reasoning – Base Ten Structure, Addition and Subtraction Within 101 Numerical Reasoning – Base Ten Structure, Addition and Subtraction Within 102 Measurement & Data Reasoning – Length, Time, Money Grade 2 Numerical Reasoning – Counting Within 1000, Place Value, Addition and Subtraction, Fluency To 20, Developing Multiplication Through Arrays Patterning & Algebraic Reasoning – Patterns Up To 20 and Addition and Subtraction Within 1,000 Patterning & Algebraic Reasoning – Patterns Up To 20 and Addition and Subtraction Within 1,001 Measurement & Data Reasoning – Length, Distance, Time, and Money Geometric & Spatial Reasoning – Sorting Shapes, Lines Of Symmetry, Partitioning Circles and Rectangles Grade 3 Numerical Reasoning – Base Ten Numerals and Place Value Up To 10,000, and Rounding Up To 1,000 Patterning & Algebraic Reasoning – Fluency, Addition and Subtraction Within 10,000, Multiplication and Division Within 100, Equality, Properties Of Operations Numerical Reasoning – Unit Fractions, Equivalent Fractions, Fractions Greater Than 1 Measurement & Data Reasoning – Elapsed Time, Liquid Volume, Mass, Lengths In Half and Fourth Of An Inch, Data Geometric & Spatial Reasoning – Polygons, Parallel Line Segments, Perpendicular Line Segments, Right Angles, Lines Of Symmetry, Area, Perimeter Grade 4 Numerical Reasoning – Place Value, Rounding, Comparisons With Multi-Digit Numbers, Addition and Subtraction, Multiplicative Comparisons, Multiplication, and Division Involving Whole Numbers Patterning & Algebraic Reasoning – Patterns, Input-Output Tables, Factors, Multiples, Composite Numbers, Prime Numbers Numerical Reasoning – Fraction Equivalence, Comparison Of Fractions, and Addition and Subtraction Of Fractions With Like Denominators Measurement & Data Reasoning – Time, Metric Measurements, Distance, Elapsed Time, Liquid Volume, Mass, and Length Geometric & Spatial Reasoning – Polygons, Points, Lines, Line Segments, Rays, Angles, Perpendicular Lines, Area, Perimeter Grade 5 Numerical Reasoning – Place Value, Multiplying By Powers Of 10, Multiplication and Division Of Multi-Digit Numbers, Fractions, Decimal Numbers, Numerical Expressions Patterning & Algebraic Reasoning – Generating Patterns, Plotting Ordered Pairs In The First Quadrant Measurement & Data Reasoning – Measurements Within The Metric System, Measurement Conversions and Time As A Unit Of Measurement Geometric & Spatial Reasoning – Properties Of Polygons and Rectangular Prisms, Classify Polygons Grade 6 Numerical Reasoning – Multiplication and Division Of Whole Numbers and Fractions, and All Four Operations With Decimal Numbers Geometric & Spatial Reasoning – Area Of Polygons, Volume Of Right Rectangular Prisms, Surface Area Of 3-D Figures Patterning & Algebraic Reasoning – Numerical and Algebraic Expressions, Factors, Multiples, Algebraic Expressions, Plotting Points In All Four Quadrants, Rational Numbers On A Number Line, Polygons In The Coordinate Plane Grade 7 Numerical Reasoning – Integers, Percentages, Fractions, Decimal Numbers Patterning & Algebraic Reasoning – Linear Expressions With Rational Coefficients, Complex Unit Rates, Proportional Relationships Geometric & Spatial Reasoning – Vertical, Adjacent, Complementary, and Supplementary Angles, Circumference and Area Of Circles, Area and Surface Area, Volume Of Cubes, Right Prisms, and Cylinders Probability Reasoning – Likelihood, Theoretical and Experimental Probability Grade 8 Numerical Reasoning – Rational and Irrational Numbers, Decimal Expansion, Integer Exponents, Square and Cube Roots, Scientific Notation Patterning & Algebraic Reasoning – Expressions, Linear Equations, and Inequalities Functional & Graphical Reasoning – Relate Domain To Linear Functions, Rate Of Change, Linear Vs. Nonlinear Relationships, Graphing Linear Functions, Systems Of Linear Equations, Parallel and Perpendicular Lines Geometric & Spatial Reasoning – Pythagorean Theorem and Volume Of Triangles, Rectangles, Cones, Cylinders, and Spheres Algebra: Concepts & Connections Mathematical Modeling Functional & Graphical Reasoning – Function Notation, Modeling Linear Functions, Linear Vs. Nonlinear Comparisons Geometric & Spatial Reasoning – Distance, Midpoint, Slope, Area, and Perimeter Patterning & Algebraic Reasoning – Linear Inequalities and Systems Of Linear Inequalities Numerical Reasoning - Rational and Irrational Numbers, Square Roots and Cube Roots Patterning & Algebraic Reasoning – Quadratic Expressions & Equations Functional & Graphical Reasoning – Quadratic Functions Patterning & Algebraic Reasoning – Exponential Expressions and Equations Functional & Graphical Reasoning – Exponential Functions Data & Statistical Reasoning – Univariate Data and Single Quantitative Variables; Bivariate Data Advanced Algebra (Algebra II): Concepts and Connections Mathematical Modeling Data & Statistical Reasoning – Descriptive and Inferential Statistics Functional & Graphical Reasoning – Exponential and Logarithmic Functions Functional & Graphical Reasoning – Radical Functions Functional & Graphical Reasoning – Polynomial Functions Patterning & Algebraic Reasoning – Linear Algebra and Matrices Geometric & Spatial Reasoning – Trigonometry and The Unit Circle Functional & Graphical Reasoning – Rational Functions Advanced Financial Algebra Mathematical Modeling Numerical (Quantitative) Reasoning – Fractions, Decimals, Percents, and Ratios Functional & Graphical Reasoning – Linear, Exponential, Quadratic, Cubic, Rational, Square Root, Greatest Integer, and Piecewise Functions Patterning & Algebraic Reasoning – Formulas Patterning & Algebraic Reasoning – Systems Of Equations and Inequalities Geometric & Spatial Reasoning – Polygons, Circles, and Trigonometry Data & Statistical Reasoning – Data Displays Data & Statistical Reasoning – Investigative Research Linear Algebra With Computer Science Applications Mathematical Modeling Abstract & Digital Reasoning – Programming Geometric & Spatial Reasoning – Vectors Patterning & Algebraic Reasoning – Matrices Geometric & Spatial Reasoning – Matrices Probabilistic Reasoning – Markov Chains Patterning & Algebraic Reasoning – Vector Spaces Patterning & Algebraic Reasoning – Eigenvalues and Eigenvectors Geometry: Concepts & Connections Mathematical Modeling Patterning & Algebraic Reasoning – Polynomial Expressions Geometric & Spatial Reasoning – Congruence Geometric & Spatial Reasoning – Geometric Foundations, Constructions, and Proof Geometric & Spatial Reasoning – Similarity Geometric & Spatial Reasoning – Right Triangle Trigonometry Geometric & Spatial Reasoning – Trigonometry and The Unit Circle Geometric & Spatial Reasoning – Circles Geometric & Spatial Reasoning – Equations and Measurement Probabilistic Reasoning – Compound Events and Expected Values Data & Statistical Reasoning; Probablistic Reasoning – Categorical Data In Two-Way Frequency Tables; Conditional Probability Advanced Finite Mathematics Mathematical Modeling Logical Reasoning – Methods Of Proof Logical Reasoning – Logical Symbolism and Binary Numerical Reasoning – Number Theory Abstract & Quantitative Reasoning – Set Theory Abstract & Quantitative Reasoning – Probabilities and Combinatorics Abstract & Quantitative Reasoning – Graph Theory Advanced Mathematical Decision Making Mathematical Modeling Quantitative & Proportional Reasoning – Ratios, Rates, & Percents Quantitative & Proportional Reasoning – Averages & Indices Patterning & Algebraic Reasoning – Identification Numbers, Voting, & Algorithms Probabilistic Reasoning – Conditional Probabilities & Compound Events Probabilistic Reasoning – Mathematical Assumptions & Expected Value Data & Statistical Reasoning – Investigative Research Patterning & Algebraic Reasoning – Financial Models Functional & Graphical Reasoning – Modeling With Functions Geometric & Spatial Reasoning – Deterministic Models Patterning & Algebraic Reasoning – Vectors & Matrices Patterning & Algebraic Reasoning – Networks Precalculus Mathematical Modeling Functional & Graphical Reasoning – Functions and Their Characteristics Functional & Graphical Reasoning – Trigonometric Relationships and Functions Algebraic & Geometric Reasoning – Trigonometric Identities and Equations Geometric & Spatial Reasoning – Conic Sections and Polar Equations Algebraic & Graphical Reasoning – Vectors and Parametric Equations Patterning & Algebraic Reasoning – Sequences and Series Multivariable Calculus Mathematical Modeling Patterning & Algebraic Reasoning – Vectors, Functions, and Analytic Geometry In Three Dimensions Abstract & Quantitative Reasoning – Partial Differentiation Abstract & Quantitative Reasoning – Integration Differential Equations Mathematical Modeling Abstract Reasoning – First Order Differential Equations Abstract Reasoning – Higher Order Differential Equations Abstract Reasoning – Systems Of Differential Of Equations Abstract Reasoning – Laplace Transforms Engineering Calculus Mathematical Modeling Abstract Reasoning – Impact Of Engineering In Mathematics Abstract Reasoning – Algebra In Engineering Abstract Reasoning – Partial Differentiation In Engineering Abstract Reasoning – Integration In Engineering College Readiness Mathematics Mathematical Modeling Numerical & Quantitative Reasoning – Real Number System Patterning & Algebraic Reasoning – Expressions, Equations, & Inequalities Functional & Graphical Reasoning – Building & Interpreting Functions Geometric & Spatial Reasoning – Measurement, Surface Area, Volume, & Right Triangle Trigonometry Data & Statistical Reasoning – Interpreting Data & Calculating Probabilities Of Compound Events Mathematics Of Industry and Government Mathematical Modeling Abstract Reasoning & Deterministic Decision-Making – Linear Programming Abstract Reasoning & Deterministic Decision-Making – Optimal Locations Abstract Reasoning & Deterministic Decision-Making – Optimal Paths Abstract Reasoning & Probabilistic Decision-Making – Normal Distributions Abstract Reasoning & Probabilistic Decision-Making – Binomial, Geometric, and Poisson Distributions Probabilistic Reasoning – Probabilistic Models Abstract Reasoning & Probabilistic Decision-Making – Simulations Abstract Reasoning & Probabilistic Decision-Making – Fair Representation Statistical Reasoning Mathematical Modeling Data & Statistical Reasoning – Formulate Statistical Investigative Questions Data & Statistical Reasoning – Collect Data Data & Statistical Reasoning – Analyze Data Data & Statistical Reasoning – Interpret Results History Of Mathematics Mathematical Modeling Numerical Reasoning – Origins Of Mathematics Logical, Mathematical & Investigative Reasoning – Ancient Greek Mathematics Logical, Mathematical & Investigative Reasoning – Mathematics In The Middles Ages Logical, Mathematical & Investigative Reasoning – Mathematics Of The Classical Era Logical, Mathematical & Investigative Reasoning – Modern Mathematics Logical, Mathematical & Investigative Reasoning – 20Th Century Mathematics Georgia Math Advanced Algebra (Algebra II): Concepts and Connections: Functional & Graphical Reasoning – Polynomial Functions AA.FGRc.5 --------- Extend exploration of quadratic solutions to include real and non-real numbers and explore how these numbers behave under familiar operations and within real-world situations; create polynomial expressions, solve polynomial equations, graph polynomial functions, and model real-world phenomena. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ AA.FGR.5.1 Not covered Graph and analyze quadratic functions in contextual situations and include analysis of data sets with regressions. (Content unavailable) AA.FGR.5.2 Fully covered Define complex numbers i such that i^2 = -1 and show that every complex number has the form a + bi where a and b are real numbers and that the complex conjugate is a - bi. Classify complex numbers Classifying complex numbers i as the principal root of -1 Intro to complex numbers Intro to complex numbers Intro to the imaginary numbers Intro to the imaginary numbers Parts of complex numbers Powers of the imaginary unit Powers of the imaginary unit Powers of the imaginary unit Simplify roots of negative numbers Simplifying roots of negative numbers AA.FGR.5.3 Fully covered Use the relation i^2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. Add & subtract complex numbers Adding complex numbers Complex number conjugates Complex number equations: x³=1 Complex number operations review Complex number polar form review Dividing complex numbers in polar form Graphically add & subtract complex numbers Graphically multiply complex numbers Intro to complex number conjugates Multiply & divide complex numbers in polar form Multiply complex numbers Multiplying complex numbers Multiplying complex numbers Multiplying complex numbers graphically example: -1-i Multiplying complex numbers graphically example: -3i Multiplying complex numbers in polar form Powers of complex numbers Powers of the imaginary unit Powers of the imaginary unit Powers of the imaginary unit Subtracting complex numbers Taking and visualizing powers of a complex number Visualizing complex number multiplication Visualizing complex number powers AA.FGR.5.4 Fully covered Use the structure of an expression to factor quadratics. Comparing features of quadratic functions Difference of squares Features of quadratic functions Features of quadratic functions: strategy Finding features of quadratic functions Finding the vertex of a parabola in standard form Forms & features of quadratic functions Graph parabolas in all forms Graph quadratics in standard form Graphing quadratics review Graphing quadratics: standard form Interpret quadratic models Interpret quadratic models: Factored form Quadratic equations word problem: box dimensions Quadratic equations word problem: triangle dimensions Quadratic word problems (standard form) Quadratics by factoring Quadratics by factoring (intro) Solve equations using structure Solving quadratics using structure Worked examples: Forms & features of quadratic functions AA.FGR.5.5 Partially covered Write and solve quadratic equations and inequalities with real coefficients and use the solution to explain a mathematical, applicable situation. Completing the square Completing the square Completing the square (intermediate) Completing the square review Features of quadratic functions Finding the vertex of a parabola in standard form Interpret quadratic models Interpret quadratic models: Factored form Quadratic formula Quadratic word problem: ball Quadratic word problems (standard form) Quadratics by factoring Quadratics by factoring (intro) Quadratics by taking square roots Quadratics by taking square roots: strategy Solve by completing the square: Integer solutions Solve by completing the square: Non-integer solutions Solve equations using structure Solving quadratics by completing the square Solving quadratics by completing the square: no solution Solving quadratics by factoring Solving quadratics by factoring Solving quadratics by factoring review Solving quadratics by factoring: leading coefficient ≠ 1 Solving quadratics by taking square roots Solving quadratics by taking square roots Solving quadratics by taking square roots examples Solving quadratics by taking square roots: with steps Solving quadratics using structure Solving simple quadratics review Strategy in solving quadratic equations Strategy in solving quadratics The quadratic formula Understanding the quadratic formula Using the quadratic formula: number of solutions Worked example: Completing the square (intro) Worked example: completing the square (leading coefficient ≠ 1) Worked example: quadratic formula (example 2) Worked example: quadratic formula (negative coefficients) Worked example: Rewriting & solving equations by completing the square Zero product property Zero product property AA.FGR.5.6 Fully covered Solve systems of quadratic and linear functions to determine points of intersection. Quadratic system with no solutions Quadratic systems Quadratic systems: a line and a circle Quadratic systems: a line and a parabola AA.FGR.5.7 Mostly covered Create and analyze quadratic equations to represent relationships between quantities as a model for contextual situations. Completing the square Features of quadratic functions Finding the vertex of a parabola in standard form Graphing quadratics: vertex form Quadratic word problem: ball Quadratic word problems (standard form) Quadratics by factoring Quadratics by factoring (intro) Solve equations using structure Solving quadratics by completing the square Solving quadratics by completing the square: no solution Solving quadratics by factoring Solving quadratics by factoring Solving quadratics by factoring review Solving quadratics by factoring: leading coefficient ≠ 1 Solving quadratics by taking square roots Solving quadratics by taking square roots Solving quadratics by taking square roots examples Solving quadratics by taking square roots: with steps Solving quadratics using structure Solving simple quadratics review Strategy in solving quadratic equations Strategy in solving quadratics Worked example: Completing the square (intro) Worked example: completing the square (leading coefficient ≠ 1) Worked examples: Forms & features of quadratic functions AA.FGR.5.8 Partially covered Identify the number of zeros that exist for any polynomial based upon the greatest degree of the polynomial and the end behavior of the polynomial by observing the sign of the leading coefficient. Graphs of polynomials Graphs of polynomials: Challenge problems Multiplicity of zeros of polynomials Positive & negative intervals of polynomials Positive & negative intervals of polynomials Vertex form introduction Zeros of polynomials (factored form) Zeros of polynomials (multiplicity) Zeros of polynomials (multiplicity) Zeros of polynomials (with factoring) Zeros of polynomials (with factoring): common factor Zeros of polynomials (with factoring): grouping Zeros of polynomials & their graphs Zeros of polynomials introduction Zeros of polynomials: matching equation to graph Zeros of polynomials: matching equation to zeros Zeros of polynomials: plotting zeros AA.FGR.5.9 Fully covered Identify zeros of polynomial functions using technology or pre-factored polynomials and use the zeros to construct a graph of the function defined by the polynomial function. Analyze identify key features of these polynomial functions. End behavior of polynomials End behavior of polynomials Graphs of polynomials Graphs of polynomials: Challenge problems Increasing and decreasing intervals Intro to end behavior of polynomials Multiplicity of zeros of polynomials Positive & negative intervals of polynomials Positive & negative intervals of polynomials Vertex form introduction Zeros of polynomials (factored form) Zeros of polynomials (multiplicity) Zeros of polynomials (multiplicity) Zeros of polynomials (with factoring) Zeros of polynomials (with factoring): common factor Zeros of polynomials (with factoring): grouping Zeros of polynomials & their graphs Zeros of polynomials introduction Zeros of polynomials: matching equation to graph Zeros of polynomials: matching equation to zeros Zeros of polynomials: plotting zeros AA.FGR.5.10 Mostly covered Use the structure of an expression to factor polynomials, including the sum of cubes, the difference of cubes, and higher-order polynomials that may be expressed as a quadratic within a quadratic. Comparing features of quadratic functions Difference of squares Factor quadratics by grouping Factoring by grouping Features of quadratic functions Features of quadratic functions: strategy Finding features of quadratic functions Finding the vertex of a parabola in standard form Forms & features of quadratic functions Graph parabolas in all forms Graph quadratics in standard form Graphing quadratics review Graphing quadratics: standard form Interpret quadratic models Interpret quadratic models: Factored form Quadratic equations word problem: box dimensions Quadratic equations word problem: triangle dimensions Quadratic word problems (standard form) Quadratics by factoring Quadratics by factoring (intro) Solve equations using structure Solving quadratics using structure Worked examples: Forms & features of quadratic functions AA.FGR.5.11 Fully covered Using all the zeros of a polynomial function, list all the factors and multiply to write a multiple of the polynomial function in standard form. Factor quadratics by grouping Factoring by grouping Graphs of polynomials Graphs of polynomials: Challenge problems Multiplicity of zeros of polynomials Positive & negative intervals of polynomials Positive & negative intervals of polynomials Vertex form introduction Zeros of polynomials (factored form) Zeros of polynomials (multiplicity) Zeros of polynomials (multiplicity) Zeros of polynomials (with factoring) Zeros of polynomials (with factoring): common factor Zeros of polynomials (with factoring): grouping Zeros of polynomials & their graphs Zeros of polynomials introduction Zeros of polynomials: matching equation to graph Zeros of polynomials: matching equation to zeros Zeros of polynomials: plotting zeros Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization. Donate or volunteer today! 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8748	https://en.wikipedia.org/wiki/Logical_reasoning	Jump to content Search Contents (Top) 1 Definition 2 Basic concepts 3 Deductive reasoning 4 Non-deductive reasoning 4.1 Inductive 4.2 Abductive 4.3 Analogical 5 Fallacies 6 As a skill 7 See also 8 References 8.1 Citations 8.2 Sources Logical reasoning العربية Deutsch Eesti فارسی Français Bahasa Indonesia Қазақша Nederlands 日本語 پنجابی Português Русский Српски / srpski Türkçe اردو 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 Wikidata item Appearance From Wikipedia, the free encyclopedia Process of drawing correct inferences \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| Argument \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| \| Deductive \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| Valid \| \| \| \| --- \| \| \| Sound \| \| \| \| \| Unsound \| \| \| \| \| \| \| Invalid \| \| \| \| --- \| \| \| Unsound \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- \| Non‑deductive \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| Strong \| \| \| \| --- \| \| \| Cogent \| \| \| \| \| Uncogent \| \| \| \| \| \| \| Weak \| \| \| \| --- \| \| \| Uncogent \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Logical reasoning is concerned with the correctness of arguments. A key distinction is between deductive and non-deductive arguments. \| Logical reasoning is a mental activity that aims to arrive at a conclusion in a rigorous way. It happens in the form of inferences or arguments by starting from a set of premises and reasoning to a conclusion supported by these premises. The premises and the conclusion are propositions, i.e. true or false claims about what is the case. Together, they form an argument. Logical reasoning is norm-governed in the sense that it aims to formulate correct arguments that any rational person would find convincing. The main discipline studying logical reasoning is logic. Distinct types of logical reasoning differ from each other concerning the norms they employ and the certainty of the conclusion they arrive at. Deductive reasoning offers the strongest support: the premises ensure the conclusion, meaning that it is impossible for the conclusion to be false if all the premises are true. Such an argument is called a valid argument, for example: all men are mortal; Socrates is a man; therefore, Socrates is mortal. For valid arguments, it is not important whether the premises are actually true but only that, if they were true, the conclusion could not be false. Valid arguments follow a rule of inference, such as modus ponens or modus tollens. Deductive reasoning plays a central role in formal logic and mathematics. For non-deductive logical reasoning, the premises make their conclusion rationally convincing without ensuring its truth. This is often understood in terms of probability: the premises make it more likely that the conclusion is true and strong inferences make it very likely. Some uncertainty remains because the conclusion introduces new information not already found in the premises. Non-deductive reasoning plays a central role in everyday life and in most sciences. Often-discussed types are inductive, abductive, and analogical reasoning. Inductive reasoning is a form of generalization that infers a universal law from a pattern found in many individual cases. It can be used to conclude that "all ravens are black" based on many individual observations of black ravens. Abductive reasoning, also known as "inference to the best explanation", starts from an observation and reasons to the fact explaining this observation. An example is a doctor who examines the symptoms of their patient to make a diagnosis of the underlying cause. Analogical reasoning compares two similar systems. It observes that one of them has a feature and concludes that the other one also has this feature. Arguments that fall short of the standards of logical reasoning are called fallacies. For formal fallacies, like affirming the consequent, the error lies in the logical form of the argument. For informal fallacies, like false dilemmas, the source of the faulty reasoning is usually found in the content or the context of the argument. Some theorists understand logical reasoning in a wide sense that is roughly equivalent to critical thinking. In this regard, it encompasses cognitive skills besides the ability to draw conclusions from premises. Examples are skills to generate and evaluate reasons and to assess the reliability of information. Further factors are to seek new information, to avoid inconsistencies, and to consider the advantages and disadvantages of different courses of action before making a decision. Definition [edit] Logical reasoning is a form of thinking that is concerned with arriving at a conclusion in a rigorous way. This happens in the form of inferences by transforming the information present in a set of premises to reach a conclusion. It can be defined as "selecting and interpreting information from a given context, making connections, and verifying and drawing conclusions based on provided and interpreted information and the associated rules and processes." Logical reasoning is rigorous in the sense that it does not generate any conclusion but ensures that the premises support the conclusion and act as reasons for believing it. One central aspect is that this support is not restricted to a specific reasoner but that any rational person would find the conclusion convincing based on the premises. This way, logical reasoning plays a role in expanding knowledge. The main discipline studying logical reasoning is called logic. It is divided into formal and informal logic, which study formal and informal logical reasoning. studied by formal logic. But in a wider sense, it also includes forms of non-deductive reasoning, such as inductive, abductive, and analogical reasoning. The forms of logical reasoning have in common that they use premises to make inferences in a norm-governed way. As norm-governed practices, they aim at inter-subjective agreement about the application of the norms, i.e. agreement about whether and to what degree the premises support their conclusion. The types of logical reasoning differ concerning the exact norms they use as well as the certainty of the conclusion they arrive at. Deductive reasoning offers the strongest support and implies its conclusion with certainty, like mathematical proofs. For non-deductive reasoning, the premises make the conclusion more likely but do not ensure it. This support comes in degrees: strong arguments make the conclusion very likely, as is the case for well-researched issues in the empirical sciences. Some theorists give a very wide definition of logical reasoning that includes its role as a cognitive skill responsible for high-quality thinking. In this regard, it has roughly the same meaning as critical thinking. Basic concepts [edit] A variety of basic concepts is used in the study and analysis of logical reasoning. Logical reasoning happens by inferring a conclusion from a set of premises. Premises and conclusions are normally seen as propositions. A proposition is a statement that makes a claim about what is the case. In this regard, propositions act as truth-bearers: they are either true or false. For example, the sentence "The water is boiling." expresses a proposition since it can be true or false. The sentences "Is the water boiling?" or "Boil the water!", on the other hand, express no propositions since they are neither true nor false. The propositions used as the starting point of logical reasoning are called the premises. The proposition inferred from them is called the conclusion. For example, in the argument "all puppies are dogs; all dogs are animals; therefore all puppies are animals", the propositions "all puppies are dogs" and "all dogs are animals" act as premises while the proposition "all puppies are animals" is the conclusion. A set of premises together with a conclusion is called an argument. An inference is the mental process of reasoning that starts from the premises and arrives at the conclusion. But the terms "argument" and "inference" are often used interchangeably in logic. The purpose of arguments is to convince a person that something is the case by providing reasons for this belief. Many arguments in natural language do not explicitly state all the premises. Instead, the premises are often implicitly assumed, especially if they seem obvious and belong to common sense. Some theorists distinguish between simple and complex arguments. A complex argument is made up of many sub-arguments. This way, a chain is formed in which the conclusions of earlier arguments act as premises for later arguments. Each link in this chain has to be successful for a complex argument to succeed. An argument is correct or incorrect depending on whether the premises offer support for the conclusion. This is often understood in terms of probability: if the premises of a correct argument are true, it raises the probability that its conclusion is also true. Forms of logical reasoning can be distinguished based on how the premises support the conclusion. Deductive arguments offer the strongest possible support. Non-deductive arguments are weaker but are nonetheless correct forms of reasoning. The term "proof" is often used for deductive arguments or very strong non-deductive arguments. Incorrect arguments offer no or not sufficient support and are called fallacies, although the use of incorrect arguments does not mean their conclusions are incorrect. Deductive reasoning [edit] Main article: Deductive reasoning Deductive reasoning is the mental process of drawing deductive inferences. Deductively valid inferences are the most reliable form of inference: it is impossible for their conclusion to be false if all the premises are true. This means that the truth of the premises ensures the truth of the conclusion. A deductive argument is sound if it is valid and all its premises are true. For example, inferring the conclusion "no cats are frogs" from the premises "all frogs are amphibians" and "no cats are amphibians" is a sound argument. But even arguments with false premises can be deductively valid, like inferring that "no cats are frogs" from the premises "all frogs are mammals" and "no cats are mammals". In this regard, it only matters that the conclusion could not be false if the premises are true and not whether they actually are true. Deductively valid arguments follow a rule of inference. A rule of inference is a scheme of drawing conclusions that depends only on the logical form of the premises and the conclusion but not on their specific content. The most-discussed rule of inference is the modus ponens. It has the following form: p; if p then q; therefore q. This scheme is deductively valid no matter what p and q stand for. For example, the argument "today is Sunday; if today is Sunday then I don't have to go to work today; therefore I don't have to go to work today" is deductively valid because it has the form of modus ponens. Other popular rules of inference include modus tollens (not q; if p then q; therefore not p) and the disjunctive syllogism (p or q; not p; therefore q). The rules governing deductive reasoning are often expressed formally as logical systems for assessing the correctness of deductive arguments. Aristotelian logic is one of the earliest systems and was treated as the canon of logic in the Western world for over two thousand years. It is based on syllogisms, like concluding that "Socrates is a mortal" from the premises "Socrates is a man" and "all men are mortal". The currently dominant system is known as classical logic and covers many additional forms of inferences besides syllogisms. So-called extended logics are based on classical logic and introduce additional rules of inference for specific domains. For example, modal logic can be used to reason about what is possible and what is necessary. Temporal logic can be used to draw inferences about what happened before, during, and after an event. Classical logic and its extensions rest on a set of basic logical intuitions accepted by most logicians. They include the law of excluded middle, the double negation elimination, the principle of explosion, and the bivalence of truth. So-called deviant logics reject some of these basic intuitions and propose alternative rules governing the validity of arguments. reject the law of excluded middle and the double negation elimination while paraconsistent logics reject the principle of explosion. Deductive reasoning plays a central role in formal logic and mathematics. In mathematics, it is used to prove mathematical theorems based on a set of premises, usually called axioms. For example, Peano arithmetic is based on a small set of axioms from which all essential properties of natural numbers can be inferred using deductive reasoning. Non-deductive reasoning [edit] Non-deductive reasoning is an important form of logical reasoning besides deductive reasoning. It happens in the form of inferences drawn from premises to reach and support a conclusion, just like its deductive counterpart. The hallmark of non-deductive reasoning is that this support is fallible. This means that if the premises are true, it makes it more likely but not certain that the conclusion is also true. So for a non-deductive argument, it is possible for all its premises to be true while its conclusion is still false. There are various types of non-deductive reasoning, including inductive, abductive, analogical and others. Non-deductive reasoning is more common in everyday life than deductive reasoning. Non-deductive reasoning is ampliative and defeasible. Sometimes, the terms non-deductive reasoning, ampliative reasoning, and defeasible reasoning are used synonymously even though there are slight differences in their meaning. Non-deductive reasoning is ampliative in the sense that it arrives at information not already present in the premises. Deductive reasoning, by contrast, is non-ampliative since it only extracts information already present in the premises without adding any additional information. So with non-deductive reasoning, one can learn something new that one did not know before. But the fact that new information is added means that this additional information may be false. This is why non-deductive reasoning is not as secure as deductive reasoning. A closely related aspect is that non-deductive reasoning is defeasible or non-monotonic. This means that one may have to withdraw a conclusion upon learning new information. For example, if all birds a person has seen so far can fly, this person is justified in reaching the inductive conclusion that all birds fly. This conclusion is defeasible because the reasoner may have to revise it upon learning that penguins are birds that do not fly. Inductive [edit] Main article: Inductive reasoning Inductive reasoning starts from a set of individual instances and uses generalization to arrive at a universal law governing all cases. Some theorists use the term in a very wide sense to include any form of non-deductive reasoning, even if no generalization is involved. In the more narrow sense, it can be defined as "the process of inferring a general law or principle from the observations of particular instances." For example, starting from the empirical observation that "all ravens I have seen so far are black", inductive reasoning can be used to infer that "all ravens are black". In a slightly weaker form, induction can also be used to infer an individual conclusion about a single case, for example, that "the next raven I will see is black". Inductive reasoning is closely related to statistical reasoning and probabilistic reasoning. Like other forms of non-deductive reasoning, induction is not certain. This means that the premises support the conclusion by making it more probable but do not ensure its truth. In this regard, the conclusion of an inductive inference contains new information not already found in the premises. Various aspects of the premises are important to ensure that they offer significant support to the conclusion. In this regard, the sample size should be large to guarantee that many individual cases were considered before drawing the conclusion. An intimately connected factor is that the sample is random and representative. This means that it includes a fair and balanced selection of individuals with different key characteristics. For example, when making a generalization about human beings, the sample should include members of different races, genders, and age groups. A lot of reasoning in everyday life is inductive. For example, when predicting how a person will react to a situation, inductive reasoning can be employed based on how the person reacted previously in similar circumstances. It plays an equally central role in the sciences, which often start with many particular observations and then apply the process of generalization to arrive at a universal law. A well-known issue in the field of inductive reasoning is the so-called problem of induction. It concerns the question of whether or why anyone is justified in believing the conclusions of inductive inferences. This problem was initially raised by David Hume, who holds that future events need not resemble past observations. In this regard, inductive reasoning about future events seems to rest on the assumption that nature remains uniform. Abductive [edit] Main article: Abductive reasoning Abductive reasoning is usually understood as an inference from an observation to a fact explaining this observation. Inferring that it has rained after seeing that the streets are wet is one example. Often, the expression "inference to the best explanation" is used as a synonym. This expression underlines that there are usually many possible explanations of the same fact and that the reasoner should only infer the best explanation. For example, a tsunami could also explain why the streets are wet but this is usually not the best explanation. As a form of non-deductive reasoning, abduction does not guarantee the truth of the conclusion even if the premises are true. The more plausible the explanation is, the stronger it is supported by the premises. In this regard, it matters that the explanation is simple, i.e. does not include any unnecessary claims, and that it is consistent with established knowledge. Other central criteria for a good explanation are that it fits observed and commonly known facts and that it is relevant, precise, and not circular. Ideally, the explanation should be verifiable by empirical evidence. If the explanation involves extraordinary claims then it requires very strong evidence. Abductive reasoning plays a central role in science when researchers discover unexplained phenomena. In this case, they often resort to a form of guessing to come up with general principles that could explain the observations. The hypotheses are then tested and compared to discover which one provides the best explanation. This pertains particularly to cases of causal reasoning that try to discover the relation between causes and effects. Abduction is also very common in everyday life. It is used there in a similar but less systematic form. This relates, for example, to the trust people put in what other people say. The best explanation of why a person asserts a claim is usually that they believe it and have evidence for it. This form of abductive reasoning is relevant to why one normally trusts what other people say even though this inference is usually not drawn in an explicit way. Something similar happens when the speaker's statement is ambiguous and the audience tries to discover and explain what the speaker could have meant. Abductive reasoning is also common in medicine when a doctor examines the symptoms of their patient in order to arrive at a diagnosis of their underlying cause. Analogical [edit] Analogical reasoning involves the comparison of two systems in relation to their similarity. It starts from information about one system and infers information about another system based on the resemblance between the two systems. Expressed schematically, arguments from analogy have the following form: (1) a is similar to b; (2) a has feature F; (3) therefore b probably also has feature F. Analogical reasoning can be used, for example, to infer information about humans from medical experiments on animals: (1) rats are similar to humans; (2) birth control pills affect the brain development of rats; (3) therefore they may also affect the brain development of humans. Through analogical reasoning, knowledge can be transferred from one situation or domain to another. Arguments from analogy provide support for their conclusion but do not guarantee its truth. Their strength depends on various factors. The more similar the systems are, the more likely it is that a given feature of one object also characterizes the other object. Another factor concerns not just the degree of similarity but also its relevance. For example, an artificial strawberry made of plastic may be similar to a real strawberry in many respects, including its shape, color, and surface structure. But these similarities are irrelevant to whether the artificial strawberry tastes as sweet as the real one. Analogical reasoning plays a central role in problem-solving, decision-making, and learning. It can be used both for simple physical characteristics and complex abstract ideas. In science, analogies are often used in models to understand complex phenomena in a simple way. For example, the Bohr model explains the interactions of sub-atomic particles in analogy to how planets revolve around the sun. Fallacies [edit] Main article: Fallacy A fallacy is an incorrect argument or a faulty form of reasoning. This means that the premises provide no or not sufficient support for the conclusion. Fallacies often appear to be correct on the first impression and thereby seduce people into accepting and using them. In logic, the term "fallacy" does not mean that the conclusion is false. Instead, it only means that some kind of error was committed on the way to reaching the conclusion. An argument can be a fallacy even if, by a fortuitous accident, the conclusion is true. Outside the field of logic, the term "fallacy" is sometimes used in a slightly different sense for a false belief or theory and not for an argument. Fallacies are usually divided into formal and informal fallacies. Formal fallacies are expressed in a formal language and usually belong to deductive reasoning. Their fault lies in the logical form of the argument, i.e. that it does not follow a valid rule of inference. A well-known formal fallacy is affirming the consequent. It has the following form: (1) q; (2) if p then q; (3) therefore p. This fallacy is committed, for example, when a person argues that "the burglars entered by the front door" based on the premises "the burglars forced the lock" and "if the burglars entered by the front door, then they forced the lock". This fallacy is similar to the valid rule of inference known as modus ponens. It is faulty because the first premise and the conclusion are switched around. Other well-known formal fallacies are denying the antecedent, affirming a disjunct, denying a conjunct, and the fallacy of the undistributed middle. Informal fallacies are expressed in natural language. Their main fault usually lies not in the form of the argument but has other sources, like its content or context. Some informal fallacies, like some instances of false dilemmas and strawman fallacies, even involve correct deductive reasoning on the formal level. The content of an argument is the idea that is expressed in it. For example, a false dilemma is an informal fallacy that is based on an error in one of the premises. The faulty premise oversimplifies reality: it states that things are either one way or another way but ignore many other viable alternatives. False dilemmas are often used by politicians when they claim that either their proposal is accepted or there will be dire consequences. Such claims usually ignore that various alternatives exist to avoid those consequences, i.e. that their proposal is not the only viable solution. The strawman fallacy is another informal fallacy. Its error happens on the level of the context. It consists in misrepresenting the view of an opponent and then refuting this view. The refutation itself is often correct but the error lies in the false assumption that the opponent actually defends this view. For example, an alcohol lobbyist may respond to the suggestion to ban alcohol advertisements on television by claiming that it is impossible to make people give up drinking alcohol. This is a strawman fallacy since the suggestion was merely to ban advertisements and not to stop all alcohol consumption. Ambiguous and vague expressions in natural language are often responsible for the faulty reasoning in informal fallacies. For example, this is the case for fallacies of ambiguity, like the argument "(1) feathers are light; (2) light is opposed to darkness; (3) therefore feathers are opposed to darkness". The error is found in the ambiguous term "light", which has one meaning in the first premise ("not heavy") and a different meaning in the second premise ("visible electromagnetic radiation"). As a skill [edit] Some theorists discuss logical reasoning in a very wide sense that includes its role as a broad skill responsible for high-quality thinking. In this sense, it is roughly equivalent to critical thinking and includes the capacity to select and apply the appropriate rules of logic to specific situations. It encompasses a great variety of abilities besides drawing conclusions from premises. Examples are to understand a position, to generate and evaluate reasons for and against it as well as to critically assess whether to accept or reject certain information. It is about making judgments and drawing conclusions after careful evaluation and contrasts in this regard with uncritical snap judgments and gut feelings. Other core skills linked to logical reasoning are to assess reasons before accepting a claim and to search for new information if more is needed to reach a reliable conclusion. It also includes the ability to consider different courses of action and compare the advantages and disadvantages of their consequences, to use common sense, and to avoid inconsistencies. The skills responsible for logical reasoning can be learned, trained, and improved. Logical reasoning is relevant both on the theoretical and practical level. On the theoretical level, it helps decrease the number of false beliefs. A central aspect concerns the abilities used to distinguish facts from mere opinions, like the process of finding and evaluating reasons for and against a position to come to one's own conclusion. This includes being able to differentiate between reliable and unreliable sources of information. This matters for effective reasoning since it is often necessary to rely on information provided by other people instead of checking every single fact for oneself. This way, logical reasoning can help the person avoid the effects of propaganda or being manipulated by others. When important information is missing, it is often better to suspend judgment than to jump to conclusions. In this regard, logical reasoning should be skeptical and open-minded at the same time. On the practical level, logical reasoning concerns the issue of making rational and effective decisions. For many real-life decisions, various courses of action are available to the agent. For each possible action, there can be conflicting reasons, some in favor of it and others opposed to it. In such cases, logical reasoning includes weighing the potential benefits and drawbacks as well as considering their likelihood in order to arrive at a balanced all-things-considered decision. For example, when a person runs out of drinking water in the middle of a hiking trip, they could employ the skills associated with logical reasoning to decide whether to boil and drink water from a stream that might contain dangerous microorganisms rather than break off the trip and hike back to the parking lot. This could include considering factors like assessing how dangerous the microorganisms are and the likelihood that they survive the boiling procedure. It may also involve gathering relevant information to make these assessments, for example, by asking other hikers. Time also plays a central role in logical reasoning. If one lacks important information, it is often better to delay a decision and look for new information before coming to a conclusion. If the decision is time-sensitive, on the other hand, logical reasoning may imply making a fast decision based on the currently available evidence even if it is very limited. For example, if a friend yells "Duck!" during a baseball game the most logical response may be to blindly trust them and duck instead of demanding an explanation or investigating what might have prompted their exclamation. Generally speaking, the less time there is, the more significant it is to trust intuitions and gut feelings. If there is more time, on the other hand, it becomes important to examine ambiguities and assess contradictory information. 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ISBN 9789814619981. \| Authority control databases \| \| National \| United States Israel \| \| Other \| Yale LUX \| Retrieved from " Categories: Reasoning Concepts in logic Hidden categories: Good articles Articles with short description Short description is different from Wikidata CS1 maint: publisher location All articles with dead external links Articles with dead external links from August 2024 Articles with permanently dead external links Logical reasoning Add topic
8749	https://www.youtube.com/watch?v=-olHtXugp_w	Library of Functions Part 3: Graphs of Reciprocal Functions Ms. Hearn 10000 subscribers 4 likes Description 133 views Posted: 28 Jun 2017 In College Algebra, we must memorize a set of graphs of basic functions. This video introduces the names and graphs of a class of functions called reciprocal functions. The first function presented is y=1/x. For my students, I have labeled the 2 "Key Points" and the 2 Asymptotes that must be labeled on tests, both when you are asked to graph a reciprocal function and when you are asked to graph a related function using transformations. Transcript: hi I'm Miss Hearn let's get started now there's a another class of functions that we're going are going to be the last part of our library of functions and they're all related to this one this one is called the reciprocal function remember reciprocal means take one over a number so the reciprocal function is FX = 1/x take the reciprocal of the number so F of two is2 F of 3 is 1/3 F of 1/4 would be four the reciprocal of the number what ends up happening is you get a piece of the graph in quadrant 1 remember this is quadrant 2 quadrant 3 Quadrant 4 you get a piece of the graph in quadrant 1 and a piece in quadrant 3 because when you take the reciprocal of a positive number you get a positive number and when you take the reciprocal of a negative number you get a negative number the key points I want you to label are 1 one and 1 one 1 but there's something extra on these types of functions this is called a rational function we need to know that rational functions have these things called ASM tootes most of the time and for example this one has an ASM toote at xals Z why don't we get any function values at xal Z why doesn't this function cross this line what would happen if you tried to plug zero into X you'd get F of 0 = 1 /0 what's wrong with that it's not zero 0 over one is zero because what this is saying is how many ones does it take to add up to zero and none okay but one divided by 0 is asking how many zeros does it take to add up to one and you can add up an infinite number of zeros right and nothing happens so division by Zer is undefined yet so the function is undefined at zero whenever you have a value where the function is undefined you're either going to have a hole in the graph a little hole or an ASM toote in this case you'll have an ASM toote and then another interesting thing happens and we end up with a horizontal ASM toote along the line Y equals 0 the xais and the reason why that happens you can understand best if you try plugging in really big numbers so watch what happens if I make a little x f ofx chart right and let's say I plug in 10 for example what's F of 10 oh 1 over 10 1110th good what's F of 100 1 over 100 or .1 what's F of a th000 1 over a th000 or 0.1 what's 1 over a million 1 one millionth or 0.0000001 so what's happening is I'm looking at points way out here on this part of the graph and the graph is getting closer and closer and closer to what the Y values are getting closer and closer to zero aren't they but can you ever take the reciprocal of a large enough number that it actually gives you zero no that's not going to happen so we put in this imaginary boundary line to show to other people that the graph is going to come very very close to it and and run right alongside of it it's going to get so close that you won't be able to see the difference but it never actually touches and that happens in the negative direction as well it stays just below the axis so we see that we have these two asmm tootes so when I ask you to graph the reciprocal function I want to see not only the two key points but also the two ASM tootes in addition to the shape of the actual function now I said that this is the first in a whole family of related functions let's look at the next one this is 1 /x essentially to the first Power I don't ever write the first power but that's what it is if you take 1X to the 2 power you get a very similar graph it's just that instead of having pieces of it in quadrants one and three we end up with pieces in one and two why would it be that over here we got pieces in quadrants one and three but over here we're getting pieces in quadrants 1 and two because X is squ has what effect on the sign always positive that's right so we're not going to ever get anything in quadrant 3 or quadrant four because those have negative y values so it makes sense that both of these pieces are going to be up in quadrants one or two we still have two key points to plot one one now and one positive 1 and we still have two asmp tootes because you still can't plug in x equals z you'll get undefined still even if you Square Z it's still zero and you still find that as you plug in very large x values you're going to get tiny little fractions close to the axis but never touching it notice that the domain is everything from negative Infinity to zero not including zero and then everything from zero to Infinity The Domain we're allowed to plug in any x value we want except for zero is what that's saying and now the range is going to be 0 to Infinity not including zero again that's why it's parentheses because it never actually hits the X AIS now these two functions are representative of any function of the form 1 /x to a power if the power is Odd as it was in the first function then you get both positive and negative results so you're in quadrants 1 and three if the power is even you'll always get two positive quadrants one and two so in general we actually have a whole family of functions of this reciprocal form we call them reciprocals of power functions and they all look like this if the power is even you get quadrants one and two and you have to plot the key Point 1 one and the key Point -1 one if the power's odd you have quadrants 1 and three and you have to plot the key Point 1 one and the key point1 1 and in both cases you have to label the Asm toote what's the equation of the Y AIS is it xal 0 or yal 0 everybody always wants to say y equals 0 because it's the y axis that's not true because what do all these points have in common an x value of zero and then the xais is not xals 0 it's y equals 0 I see that error on test a lot so okay and that's the last of our library of functions I hope you found this video useful if you did please remember to like it
8750	https://www.youtube.com/watch?v=vKxCPFsvaXk	Find area between curve y = 3-x^2 and line y =-1 by integrating with respect to x and to y Ms Shaws Math Class 51100 subscribers 30 likes Description 3231 views Posted: 30 May 2021 1 comments Transcript: Intro hi everyone we're going to find the area the region between the curve and the line we have y equals 3 minus x squared y equals negative 1 and we're going to do this by integrating with both respect to x and with y so the first one we're going Area with respect to x to do is with respect to x and we already have it set up with y so we set these equal to each other and solve for look for intercepts so we have 3 minus x squared minus 1 equals negative 1 so we have x squared equals 4 so x equals plus or minus 2. if i if i just graph this a little bit i'll show you what it means a little sketch so we have y equals negative 1. all right our um x intercepts are two i'll put it here negative two and two and our y intercept here is three so basically we want this part right in here all right so basically what all we have to do is subtract these two areas you see it's symmetric to the y-axis here so we can just go from 0 to 2 here and then multiply it by 2 to get this side as well so we're going to go multiply our integral by 2 and just go from 0 to 2 for our limits and then we're going to have 3 minus x squared minus a negative 1 dx this is going to turn to positive 1 so i'll put that there so this is going to give you 2 times the integral from 0 to 2 of 4 minus x squared dx all right now let's go ahead and integrate this this is going to give us 2 times 4x minus one third x cubed and we're evaluating from zero to two so we don't really need to deal with that zero that's gonna zero out so we have two and then we have uh four times 2 minus 1 3 times 2 cubed this is going to give us 2 times 8 minus eight thirds and this is going to be uh 16 this is going to be minus 16 thirds and this is going to be equal to 32 divided by three all right so that's the area with uh by integrating with respect to x we should get the same thing when we integrate with respect to y now the only thing we have to do for y Area with y let's uh i'm going to graph this really quickly so y equals negative one and our y-intercept here is three and then we had twos here so it was like i just want to do a quick sketch right there all right and the main reason i want to do that is first of all let's look at this we already have y equals negative 1. but here let's solve for x so if you solve for x you're going to get [Music] x squared equals 3 minus y so x equals plus or minus square root of 3 minus y also our intercepts that we have to look at are our y-intercepts so our y-intercepts occur at y equals 1 and or negative 1 sorry y equals negative 1 and y equals 3 here and that's just by the graph it's a little easier to do it that way all right so we're going to set up we're going to do the same thing we did before we have the symmetry so we're only going to use the positive this part would be the negative so we're going to go 2 and our limits are going to go from negative 1 to 3 and i'm just going to write this as 3 minus y to the power of one half and this is d y now all right so when you integrate this you're going to get 2 times negative 2 thirds um three minus y to the power of three halves all right and then from there we're integrating from negative 1 to 3. so substituting all that in when you plug in substitute in 3 this whole thing is going to be 0. when you substitute a negative one you're going to get negative 16 thirds so this is going to give you um 2 times 16 thirds which equals 32 divided by three same answer this this one seems a little harder right all right thank you have a nice day bye-bye [Music] you
8751	https://pmc.ncbi.nlm.nih.gov/articles/PMC4445188/	Lhermitte's Sign: The Current Status - 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. 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 Ann Indian Acad Neurol . 2015 Apr-Jun;18(2):154–156. doi: 10.4103/0972-2327.150622 Search in PMC Search in PubMed View in NLM Catalog Add to search Lhermitte's Sign: The Current Status Supreet Khare Supreet Khare 1 Armed Forces Medical College, Pune, India Find articles by Supreet Khare 1,✉, Deeksha Seth Deeksha Seth 1 Kasturba Medical College, Mangalore (Manipal University), India Find articles by Deeksha Seth 1 Author information Article notes Copyright and License information 1 Armed Forces Medical College, Pune, India 1 Kasturba Medical College, Mangalore (Manipal University), India ✉ For correspondence: Dr. Supreet Khare, Virat Khand, Gomtinagar, Lucknow - 226 010, Uttar Pradesh, India. E-mail: drsupreet.khare@gmail.com Received 2014 Oct 17; Revised 2014 Nov 20; Accepted 2014 Nov 26. Copyright: © Annals of Indian Academy of Neurology This is an open-access article distributed under the terms of the Creative Commons Attribution-Noncommercial-Share Alike 3.0 Unported, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC4445188 PMID: 26019410 Abstract Lhermitte's sign was described by Marie and Chatelin and named after Jean Lhermitte. This sign is mostly described as an electric shock like condition by some patients of multiple sclerosis. This sensation occurs when the neck is moved in a wrong way or rather flexed. It can also travel down to the spine, arms, and legs, and sometimes the trunk. Demyelination and hyperexcitability are the main pathophysiological reasons depicted for the Lhermitte's sign. Other causes for Lhermitte's sign include transverse myelitis, behçet's disease, trauma, etc. This article reviews the Lhermitte's sign, its history, and its etiopathophysiology. Very few studies are available on Lermitte's sign and more research need to be done on the same to ensure its sensitivity and specificity. Keywords: Demyelination, Lhermitte's sign, shock like sensation Introduction Lhermitte's sign (also known as Lhermitte's phenomenon also referred to as the barber chair phenomenon is the name which describes an electric shock-like sensation that occurs on flexion of the neck. This sensation radiates down the spine, often into the legs, arms, and sometimes to the trunk. Lhermitte's sign was first described by Marie and Chatelin in 1917. The first reference to the symptom was described by Beriel and Devic in 1918 in multiple sclerosis (MS). In 1924, Lhermitte et al. described in detail a patient with MS and electric dysesthesias. Recently, Lhermitte's sign has been associated with the intensity-modulated radiotherapy (IMRT) for head and neck cancer patients. It is one of the late term effects following IMRT in these patients.[2,3,4] Pathophysiology and eliciting the sign Lhermitte's sign is caused by miscommunication between the nerves that have become demyelinated. The pathophysiology of Lhermitte's sign was described as the stretching of the hyper excitable demyelinated dorsal column of the spinal cord, particularly at the cervical level, thus triggering an electric shock-like sensation. Till date, hyper excitability is considered as the main pathophysiological mechanism for the occurrence of Lhermitte's sign. Also, from an etiological point of view, in the original paper by Lhermitte, the shock-like sensations were supposed to be caused by medullary lesions due to demyelination or trauma of the dorsal column. Neck movements are said to exacerbate Lhermitte's sign where Lhermitte's phenomenon is said to be induced by neck flexion while the reverse Lhermitte's phenomena is defined when symptoms are induced by neck extension. Reverse Lhermitte's phenomenon is said to be induced by extrinsic compression of the cervical cord and neck collar immobilization. A relatively rare form called inverse Lhermitte phenomenon can be described by upward moving paresthesia with neck flexion which can be a sign of myelopathy. Causes The causes for Lhermitte's sign are shown in the Table 1.[8,9] Table 1. The causes of Lhermitte's sign are listed below Open in a new tab Validity Two studies have measured the diagnostic accuracy of Lhermitte's sign which is found to be sensitive ranging from 3 to 17%, which is poor. One of these studies also found out that it had good specificity (97%) for non-specific compressive myelopathy. Review A study reported that Lhermitte's sign was experienced by 33.3% out of 114 patients of MS; and in 16%, it was reported to have been occurred in the first episode of MS. One out of eleven patients with subacute combined degeneration of the cord due to pernicious anemia also reported the presence of Lhermitte's sign. The sign is said to occur commonly in conditions such as subacute combined degeneration of the cord, neck trauma, prolapsed cervical disc, and radiation myelitis. Lhermitte's sign has also been reported in cavernous angioma of the cervical spinal cord. A case reported that a 49-year-old woman diagnosed as having breast cancer and on treatment with cisplatin presented with Lhermitte's sign. An 80-year-old man previously operated on for adenocarcinoma colon, with no further treatment also presented with Lhermitte's sign. A 54-year-old man who was being treated for laryngeal cancer by radiotherapy presented with Lhermitte's sign. Lhermitte's sign is a non-specific sign, although in oncological patients a detailed history and clinical examination should be done for data regarding radiotherapy, chemotherapy, and spinal compression. A strong association between Lhermitte's sign and abnormalities of the cervical spinal cord has been seen on magnetic resonance imaging. The study presumed that Lhermitte's sign in MS is the result of a lesion in the cervical spinal cord and it was confirmed that a lesion in the posterior columns of the cervical spinal cord is the cause of Lhermitte's sign in MS. A case was reported where Lhermitte's sign occurred during yawning and was associated with congenital partial aplasia of the posterior arch of the atlas. Computed tomography (CT) myelography during yawning showed compression of the upper cervical cord due to the inward mobility of the isolated posterior tubercle. The symptoms completely disappeared following removal of the isolated posterior tubercle. Another case of a 34-year-old man suffering from herpes zoster was accompanied by Lhermitte's sign. A study was done to investigate the pathophysiology of the radiation-induced, chronic Lhermitte's sign on the basis of long-standing case histories with partial functional recovery. Positron Emission Tomography (PET) demonstrated increased fluorodeoxyglucose (FDG) accumulation and butanol perfusion, but negligible methionine uptake in the irradiated spinal cord segments in the patients. A 29-year-old boy with an intrinsic, fusiform mass extending from C5 to C7, identified as low-grade ependymoma; developed Lhermitte's sign. Lhermitte's sign was most likely caused by tumor-induced distortion and demyelination of cervical dorsal column sensory axons. Another case reports of two patients who developed intrinsic cervical spinal cord damage as permanent complication of cervical epidural steroid injections which were administered while the patients were sedated. The patients were found to develop Lhermitte's sign. Response to treatment A study conducted treatment with extra-cranial picotesla range pulsed electromagnetic fields(EMFs) which was found to be effective in the management of various MS symptoms. Three MS patients in whom two brief applications of EMFs were done resulted in resolution of the Lhermitte's sign which emerged during a period of exacerbation of symptoms in one patient and during a prolonged phase of symptom deterioration in the other two patients with MS. As Lhermitte's sign is thought to result from the spread of ectopic excitation in demyelinated plaques in the cervical and thoracic regions of the spinal cord, it was hypothesized that the effects of EMFs were related to the reduction of axonal excitability via a mechanism involving changes in ionic membrane permeability. Neck movements are said to exacerbate Lhermitte's sign therefore, a brace can keep the patient from bending his neck too much which may be prescribed by a physical therapist to help with posture and positioning of the head in such cases. If a neck brace or collar is used, periodic monitoring is required to ensure that strength and range of motion is not compromised. An occupational therapist may offer progressive muscle relaxation techniques, deep breathing exercises, and active or passive stretching. Conclusion Lhermitte's sign is popularly described as a shock-like sensation by MS patients commonly. But recently several other causes for the Lhermitte's sign have been found where the most recent one includes the IRMT therapy for head and neck cancer patients. Lhermitte's sign has variants such as inverse and reverse Lhermitte's phenomena in which it can present itself. Diagnosis at the right time with proper management can help the patient cope up with the sign. Footnotes Source of Support: Nil Conflict of Interest: None declared. References 1.Walton J.N., John N. (MD), editors. 8th ed. Oxford: Oxford University Press; 1977. Brain's Diseases of the Nervous System. pp. xvi + 1277:526-568. [Google Scholar] 2.Studer G, Linsenmeier C, Riesterer O, Najafi Y, Brown M, Yousefi B, et al. Late term tolerance in head neck cancer patients irradiated in the IMRT era. Radiat Oncol. 2013;8:259. doi: 10.1186/1748-717X-8-259. [DOI] [PMC free article] [PubMed] [Google Scholar] 3.Lim DC, Gagnon PJ, Meranvil S, Kaurin D, Lipp L, Holland JM. Lhermitte's sign developing after imrt for head and neck cancer. Int J Otolaryngol 2010. 2010 doi: 10.1155/2010/907960. 907960. [DOI] [PMC free article] [PubMed] [Google Scholar] 4.Pak D, Vineberg K, Feng F, Ten Haken RK, Eisbruch A. Lhermitte sign after chemo-IMRT of head-and-neck cancer: Incidence, doses, and potential mechanisms. Int J Radiat Oncol Biol Phys. 2012;83:1528–33. doi: 10.1016/j.ijrobp.2011.10.052. 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[DOI] [PubMed] [Google Scholar] 9.Teive HA, Haratz S, Zavala J, Munhoz RP, Scola RH, Werneck LC. Lhermitte's sign and vitamin B12 deficiency: Case report. Sao Paulo Med J. 2009;127:171–3. doi: 10.1590/S1516-31802009000300011. [DOI] [PMC free article] [PubMed] [Google Scholar] 10.Uchihara T, Furukawa T, Tsukagoshi H. Compression of brachial plexus as a diagnostic test of cervical cord lesion. Spine (Phila Pa 1976) 1994;19:2170–3. doi: 10.1097/00007632-199410000-00007. [DOI] [PubMed] [Google Scholar] 11.Murphy KD, Gutrecht JA. Lhermitte's sign in cavernous angioma of the cervical spinal cord. J Neurol Neurosurg Psychiatry. 1998;65:954–5. doi: 10.1136/jnnp.65.6.954a. [DOI] [PMC free article] [PubMed] [Google Scholar] 12.Porta-Etessam J, Martínez-Salio A, Berbel A, Balsalobre-Aznar J, Esteban J, Benito-León J, et al. Lhermitte's sign in three oncological patients. Rev Neurol. 2000;30:649–51. [PubMed] [Google Scholar] 13.Gutrecht JA, Zamani AA, Slagado ED. 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8752	https://www.khanacademy.org/science/ap-college-physics-1/xf557a762645cccc5:force-and-translational-dynamics/xf557a762645cccc5:gravitational-force/e/gravitational-forces	Gravitational forces (practice) \| 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 AP®︎/College Physics 1 Course: AP®︎/College Physics 1>Unit 2 Lesson 3: Gravitational force Newton's law of gravitation Gravitational force The concept of gravitational field Gravitational fields Inertial and gravitational mass Inertial and gravitational mass Science> AP®︎/College Physics 1> Force and translational dynamics> Gravitational force © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Gravitational force Google ClassroomMicrosoft Teams Problem Planet A‍ interacts gravitationally with Planets B‍ and C.‍ Planet B‍ is half as massive as Planet C.‍ The distance between the centers of mass of Planets A‍ and B‍ is half the distance between the centers of mass of Planets A‍ and C,‍ as modeled below. (Planetary sizes and distances are not shown to scale.) 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8753	https://www.cureepilepsy.org/understanding-epilepsy/treatments/epilepsy-medications/phenobarbital/	Phenobarbital (Luminal, Solfoton) - Epilepsy Medication Skip to main content About Us Image 2: Our Mission #### Our Mission Learn More #### Why We Exist Our Mission Our Story Research Impact Annual Impact Reports #### Who We Are Our Team CURE Epilepsy in the News Careers Financials Contact Us Our Research #### Our Research #### Our Research Strategy Learn More #### Overview Research Strategy Areas of Focus Grants Awarded #### Progress Research Impact Annual Impact Reports Discoveries Understanding Epilepsy #### Educational Resources Epilepsy Explained Videos Webinars Treatment Talks Medication Access Epilepsy Research News #### Epilepsy Basics What is Epilepsy? What Causes Epilepsy? Seizures 101 Seizure Action Plans SUDEP Other Epilepsy Risks #### Diagnosing Epilepsy How is Epilepsy Diagnosed? 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Based on their judgment and knowledge, a drug may be prescribed for other epilepsy types not included in the indications. For more information, please see the prescribing information. How can you take phenobarbital? (Available formulations) Phenobarbital is available as a tablet, injection solution, or oral solution. Who should not take phenobarbital? If you are allergic to phenobarbital, barbiturates, or any of the inactive ingredients, then you should not take it. If you have a personal or familial history of acute intermittent porphyria should not take phenobarbital. If you have marked impairment of liver function, respiratory disease in which shortness of breath or obstruction is evident, or have known previous addiction to the sedative/hypnotic drugs, then you should not take phenobarbital. Other considerations may influence whether you should take phenobarbital. Tell your healthcare provider if you: have or have had depression, mood problems, or suicidal thoughts or behavior. have borderline hypoadrenal function (underactive adrenal gland). have liver problems. have abused or been dependent on prescription medicines, street drugs, or alcohol. are pregnant or plan to become pregnant. are breastfeeding or plan to breastfeed. What is important to know about taking phenobarbital? Do not stop taking phenobarbital suddenly unless directed to do so by your healthcare provider. As with all antiseizure medications, phenobarbital should be withdrawn gradually to minimize the risk of causing or worsening seizures or status epilepticus. You should not stop using phenobarbital suddenly unless your healthcare provider tells you to stop the medicine because of a serious side effect. Tell your healthcare provider about all the medicines you take, including prescription and over-the-counter medicines, vitamins, and herbal supplements. Taking phenobarbital with certain other medicines may cause side effects or affect how well they work. Do not start or stop other medicines without talking to your healthcare provider. Especially tell your healthcare provider if you take: coumarin anticoagulants. Women or those who are/plan to become pregnant Use in Pregnancy At this time, there is not enough evidence regarding developmental risks associated with the use of phenobarbital in pregnant people. In animal studies, there were instances of developmental issues at clinically relevant doses. However, having a seizure during pregnancy could harm both the pregnant individual and the baby. Tell your healthcare provider right away if you become pregnant. Do not start or stop taking seizure medication during pregnancy without your healthcare provider’s advice. If you become pregnant while taking phenobarbital, talk to your healthcare provider about registering with the North American Antiepileptic Drug (NAAED) Pregnancy Registry. The purpose of this registry is to collect information about the safety of antiseizure medicine during pregnancy. You can enroll in this registry by calling 1-888-233-2334. Use during breastfeeding It is unknown if there are effects on the breastfed infant, or if phenobarbital impacts milk production. Talk to your healthcare provider about the risks. Your healthcare provider will consider the developmental and health benefits of breastfeeding along with your need for phenobarbital and the potential effect on the infant from phenobarbital or from your epilepsy. Effect on birth control and fertility Phenobarbital may decrease the effectiveness of hormonal contraceptives, including birth control pills, injections, implants, skin patches, and vaginal rings. To prevent pregnancy while using phenobarbital, use a barrier form of birth control: condom, diaphragm, cervical cap, or contraceptive sponge. Phenobarbital side effects Phenobarbital is not approved by the FDA. There are risks associated with all medicines. Some side effects caused by phenobarbital can be very serious, and even life-threatening. It is important to be informed about these serious reactions and to be aware of their symptoms. Common side effects The most common side effects that were reported in studies of phenobarbital are central nervous system depression, respiratory depression, apnea (temporarily stopped breathing during sleep), circulatory collapse, hypersensitivities, nausea and vomiting, headache. Rare, but serious side effects Abuse, misuse, and addiction Phenobarbital is a federally controlled substance (Schedule IV) as it has a risk of abuse, misuse, and addiction, which can lead to overdose or death. Using phenobarbital more frequently than recommended, even at recommended dosages, may increase these risks, especially when used in combination with other medications (e.g., opioids), alcohol, and/or illicit substances. Talk to your healthcare provider about the signs and symptoms of phenobarbital addiction. Pain In small doses, the barbiturates may increase the reaction to painful stimuli. Taken by themselves, the barbiturates cannot be relied upon to relieve pain or even to produce sedation or sleep in the presence of severe pain. On a mission to find a cure Our Mission Financials Careers Contact Us Shop Privacy Policy ©2024 CURE Epilepsy. All rights reserved. CURE Epilepsy is a qualified 501(c)3 tax-exempt organization. (EIN: 36-4253176) Privacy Policy Notifications
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Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA HMH Careers Exploring a career at HMH? Do work that matters. Learn about our culture, benefits, and available job opportunities. Accessibility Explore HMH’s approach to designing affirming and accessible curriculum materials and learning tools for students and teachers. Shop Support 0 Log in Back to Shaped Math Teaching Linear Equations in Math Richard Blankman August 15, 2024 10 Min Read For many students in Grades 8 and up, many of the numbers and shapes they’ve learned about start to come together once they are creating and solving linear equations. This topic connects ideas about algebra, geometry, and functions and can be difficult for many children—and adults!—to wrap their heads around. This article explains what a linear equation is and walks through different examples. Then it offers lesson ideas for introducing and developing the concept of linear equations in one variable to your students. What is a linear equation? Just like any other equation, a linear equation is made up of two expressions set equal to each other. There are some key features common to all linear equations: A linear equation only has one or two variables. No variable in a linear equation is raised to a power greater than 1 or is used in the denominator of a fraction. When you find pairs of values that make a linear equation true and plot those pairs on a coordinate grid, all of the points lie on the same line. The graph of a linear equation is a straight line. A linear equation in two variables can be described as a linear relationship between x and y, that is, two variables in which the value of one of them (usually y) depends on the value of the other one (usually x). In this case, x is the independent variable, and y depends on it, so y is called the dependent variable. Whether or not it’s labeled x, the independent variable is usually plotted along the horizontal axis. Most linear equations are functions. In other words, for every value of x, there is only one corresponding value of y. When you assign a value to the independent variable, x, you can compute the unique value of the dependent variable, y. You can then plot the points named by each (x,y) pair on a coordinate grid. Describing linear relationships Students should already know that any two points determine a line. So graphing a linear equation only requires finding two pairs of values and drawing a line through the points they describe. All other points on the line will provide values for x and y that satisfy the equation. The graphs of linear equations are always lines. However, math models but does not always perfectly describe the real world. Not every point on the line that an equation describes will necessarily be a solution to the problem that the equation models. For example, the problem may not make sense for negative numbers (say, if the independent variable is time) or very large numbers (say, numbers over 100 if the dependent variable is grade in class). What does a linear equation look like? Example 1: Distance = rate × time Given a steady rate, the relationship between distance and time will be linear. However, both distance and time are usually expressed as positive numbers or zero, so most graphs of this relationship will only show points in the first quadrant. Notice that the direction of the line in the graph below is from bottom left to top right. Lines that tend in this direction have positive slope. A positive slope indicates that as values on the horizontal axis increases, so do the values on the vertical axis. Example 2: Amount of water in a leaky bucket = rate of leak × time In this equation, since you can’t have a negative amount of water in the bucket, the graph will show points only in the first quadrant. Notice that the direction of the line in this graph is top left to bottom right. Lines that tend in this direction have negative slope. A negative slope indicates that the values on the y- axis are decreasing as the values on the x- axis are increasing. Example 3: Number of angles of a polygon = number of sides of that polygon Again in this graph, we are relating values that only make sense if they are positive, so we show points only in the first quadrant. Moreover, in this case, since no polygon has fewer than 3 sides or angles and the number of sides or angles of a polygon must be a whole number, we show the graph starting at (3,3) and indicate with a dashed line that points between those plotted are not solutions to the problem. Example 4: Degrees Celsius = 5/9 × (degrees Fahrenheit – 32) Since it’s reasonable to have both positive and negative temperatures, we plot the points on this graph on the full coordinate grid. (Although it’s not visible on the graph, the lowest possible physical temperature is around –460° Fahrenheit, so not every solution on the graph is useful.) Slope The slope of a line tells a couple things: first, how steep the line is with respect to the y- axis, and second, whether the line slopes up or down when you look at it from left to right. That is to say, the slope tells you the rate at which the dependent variable changes with respect to a change in the independent variable. Calculating slope Pick any two points on the line. To find how fast y is changing, subtract the y value of the first point from the y value of the second point: (y 2 – y 1). To find how fast x is changing, subtract the x value of the first point from the x value of the second point: (x 2 – x 1). To find the rate at which y is changing with respect to the change in x, calculate the ratio: (y 2 – y 1)/(x 2 – x 1). If we designate Point A as the first point and Point B as the second point, the slope of the line is (–2–4)/(–1–2) = –6/–3, or 2. It does not matter which points along the line you designate as A and B, just as long as we’re consistent with which point is the “first” (x 1,y 1) and which is the “second” (x 2,y 2). If we designate Point B as the first point and Point A as the second point, the value of the slope is the same: (4 – -2)/(2 – -1) = 6/3 or 2. It would remain the same value no matter what pair of points on the line you choose to compute slope. Linear equation formula The equation of a line can be written in a form that makes the slope obvious and allows you to draw the line without having to do arithmetic. If students are comfortable with solving a simple two-step linear equation, they can write linear equations in slope-intercept form. The slope-intercept form of a linear equation isy=mx+b. In the equation, x and y are the variables. The numbers m and b give the slope of the line (m) and the value of y when x is 0 (b). The number b corresponds to the y-intercept because (0,y) is the point at which the line crosses the y-axis. You can draw the line for an equation matching this linear formula by plotting (0,b), then using m to find another point. For example, if m is 1/2, you can interpret that as a difference in 1 among y coordinates for every difference in 2 among x coordinates. That is, (y 2 – y 1)/(x 2 – x 1) = 1/2. Count +2 on the x- axis, then +1 on the y- axis to get to another point: (2, b + 1). The equation for this line is y+3=2 x. In slope-intercept form, the equation is y=2 x–3. In this form you can easily see that the slope m = 2. Looking at the graph, the slope can be visualized as 2 since for every +2 change in y, there is a +1 change in x. Now look at b in the equation: –3 is where the line intercepts the y-axis. Positive slope When a line slopes up from left to right, it has a positive slope. This means that a positive change in y is associated with a positive change in x. The steeper the slope, the greater the rate of change in y in relation to the change in x. A slope of 6 is steeper than a slope of 1, which is in turn steeper than a slope of 1/6. When the line represents real-world data points plotted on a coordinate plane, a positive slope indicates a positive correlation, and the steeper the slope, the more dramatic the correlation. Consider a linear equation where the independent variable g is gallons of gas used and the dependent variable d is the distance traveled in miles. If you drive a car with poor gas mileage, the amount of miles traveled is low relative to the amount of gas consumed, in which case the value m is a low number. The incline of the line is fairly gradual. If instead you drive a car with better gas mileage, you travel more miles relative to the same amount of gas consumed, so the value of m is greater and the line is steeper. Both rates are positive because you still travel a positive number of miles for every gallon of gas you consume. Negative slope When a line slopes down from left to right, it has a negative slope. This means that a negative change in y is associated with a positive change in x. When the line represents real-world data points plotted on a coordinate plane, a negative slope indicates a negative correlation. Consider a line that represents the number of peppers left to plant after minutes spent gardening. Notice that as you spend more time gardening, there are fewer plants left to go. If the garden can fit 18 pepper plants, and you plant 1 pepper plant every 2 minutes, the value of m is –1/2 and the equation representing plants remaining, p, is given by p=18–(1/2)t. Contrast that with being able to plant 1 pepper plant every 1 minute. The faster rate results in a steeper line, shown as p =18–t. Zero slope When there is no change in y as x changes, the graph of the line is horizontal. A horizontal line has a slope of zero. Undefined slope When there is no change in x as y changes, the graph of the line is vertical. You cannot compute the slope of this line, because you would need to divide by 0. Notice how you can think of these lines as “infinitely steep,” either positively or negatively. The slope of a vertical line is not defined. Lines with the same slope Two lines with the same slope have identical steepness. This means one of two things: either the lines are parallel or they’re the same line. In all three of these lines, every 1-unit change in y is associated with a 1-unit change in x. All three have a slope of 1. Solving two-step linear equations When a linear equation has two variables (as it usually does), it has an infinite number of solutions. Each solution is a pair of numbers (x,y) that makes the equation true. Solving a linear equation often means finding the value of y for a given value of x. When the equation is already in slope-intercept form If the equation is already in the form y = mx + b, with x and y variables and m and b real numbers, solving for specific values is straightforward. Choose a value for x, and compute the corresponding value for y. An easy value to choose for x is 0 because, in that case, y = b. Students may be asked to make tables of values for linear equations. These are simply T-tables that lists of values for x with corresponding values for y. Two-step equations refer to equations requiring two operations to solve: both adding or subtracting and multiplying or dividing. Two-step equations will contain more than one term. A term can be a number, a variable, or numbers and variables multiplied together. The terms in an expression are separated by addition or subtraction symbols. 2 x is an expression with one term. 2 x + 6 has two terms. To find a value for y given a value for x, substitute the x-value into the expression. Consider the equation y = 2 x+ 6. Here is how you can find the value for y when x = 5: Substitute the value for x into the equation.y = 2(5) + 6 Multiply.y = 10 + 6 Add.y = 16 When the equation is not in slope-intercept form When a linear equation is not in slope-intercept form (that is, not written as y=mx+b), students can still make a table of values to find solutions for the equation, but it may be simpler to put the equation in slope-intercept form first. This requires mirroring operations on each side of the equation until y is by itself on the one side of the equation, set equal to a linear expression involving x. You can manipulate the equation in this way because of equality properties: If a = b, then a + c = b + c. If a = b, then a – c = b – c. If a = b, then ac = bc. If a = b, then a ÷ c = b ÷ c(as long as c≠ 0). Consider 2 x + y – 6 = 0. This equation is not in slope-intercept form, but you can use equality properties to get y on one side of the equation. You can subtract y from both sides of the equation to get 2 x – 6 = –y. Then multiply both sides of the equation by –1 to get –2 x + 6 = y. Alternatively, you can subtract 2 x and add 6 to both sides of the equation to get y = –2 x + 6. The two equations, –2 x + 6 = y and y = 6 – 2 x are equivalent. You can turn one into the other by using the Commutative Property of Addition, which states that a + b = b + a, and the Symmetric Property of Equality, which states that if a = b, then b = a. Commutative Property of Addition–2x + 6 = y is equivalent to 6 – 2 x = y. Symmetric Property of Equality 6 – 2 x = y is equivalent to y = 6 – 2 x. Lesson 1 – Linear equations: Introducing the concept Materials: Coordinate grid that all students can see (grid should go at least from at least –10 to +10 on both axes), tool to mark up the grid with points and lines Preparation: Students will be reading points from graphs and graphing lines from lists of points. Unless doing it digitally, they (and you) will need to be prepared to use a straightedge to generate accurate straight lines. Prerequisite skills and concepts: Students need to be able to plot points on a coordinate plane and should be familiar with the various ways to indicate multiplication and division in an equation. They should also be familiar with the order of operations and the properties of equality. Carefully draw a line through (0,0) and (2,2) on the grid. Be sure to extend it in both directions so there are plenty of easy-to-name points on it. Say:Name some points on this line. Students should come up with a list of the points named by integer coordinates. If not, spend some time reviewing how to name points on a grid before continuing with this lesson. If students name non-integer points, such as (1.5,1.5), spend time explaining why those are on the line too. Ask:Can you give me a rule for how to find y when we know x on this line? Discuss how the coordinates are related, then ask students to write the rule in equation form. The equation for this line is y = x. Say:That was the line for the equation y = x. How would you draw the line for the equation y = x + 3? Have students independently try to draw the line. If feasible, have them pair up and compare their lines. Facilitate a discussion around the different lines students drew, highlighting similarities and differences. Then show one way of drawing the line: substitute some values for x into the equation, find the corresponding values for y, and then plot those coordinate pairs. Two points give you enough information to draw the line, but because mistakes are possible and human drawing isn’t perfect, it is wise to generate at least three points. Display a T-table with the related x- and y-values, and draw the graph of the line. Say:Now how would you draw the line for the equation y = 2x + 3? Students are likely to use the strategy of making a T-table and computing points. If they forget to multiply their x-values by 2 before adding 3, remind them about the order of operations (multiply or divide left to right, then add or subtract left to right). Ask different students to generate different points, talking through their reasoning as they go. Ask:Can someone give me a number between –5 and 5? Now, how about a number between –10 and 10? Use these numbers to generate linear equations. The first number will be the coefficient of x, and the second will be added to the x term. Generate equations, find points, then draw the lines. You can make this activity more gamelike by having students roll real or virtual number cubes. If you’ve been working with slope, these problems will give you a chance to reinforce that concept as well. (Ask: Do you think the slope of this line will be positive or negative? Do you think it will be very steep or not so steep? Will this line go through the origin?) Lesson 2 – Linear equations: Developing the concept Materials: Coordinate grid that all students can see (grid should go at least from –10 to +10 on both axes), tool to mark up the grid with points and lines SayWhen we generated points for lines in our last lesson, our equations always looked the same. In other words, they were always in the same form. Today, we'll look at different ways they might look. SayCan someone describe how to find some coordinate pairs for the linear equation, 2x + y = 15? This is a two-step equation. Solutions involve assigning a value to x, then multiplying this value by 2 before trying to figure out what value of y would satisfy the equation. Students can use trial and error, or they can transform the equation using the equality properties: Write the equation.2 x + y = 15 Assign a value to x.2(3) + y = 15 Multiply.6 +y = 15 Subtract 6 from each side.6 – 6 + y = 15 – 6 Subtract.y= 9 This solution gives us the point (3,9). Continue finding solutions, or coordinate pairs, for this equation until you are satisfied that students are comfortable with the process. Then, plot the points on your grid and draw the line. SayCan someone describe how to find some points on the line described by the equation y + x/3 = 5? Write the equation.y+ x/3 = 5 Assign a value to x.y + 3/3 = 5 Divide.y + 1 = 5 Subtract 1 from each side.y + 1 – 1 = 5 – 1 Subtract.y= 4 This solution gives us the point (3,4). Students may notice that this doesn't look like previous linear equations. Explain that because x/3 is the same thing as 1/3 times x, x/3 is still an ordinary term. Continue finding solutions, or coordinate pairs, for this equation until you are satisfied that students are comfortable with the process. Then, plot the points on your grid and draw the line. SayCan someone describe how to find some points on the line described by the equation y – 6 = 2x? Write the equation.y – 6 = 2 x Assign a value to x.y– 6 = 2(3) Multiply.y – 6 = 6 Add 6 to each side.y – 6 + 6 = 6 + 6 Add.y= 12 This solution gives us the point (3,12). By now, students may have noticed that an easy substitution for x is 0. This substitution will give you the point where the line crosses the y-axis. Prompt students to come to this realization if they don’t do so independently. Assessment hints When students are solving multistep equations, pay particular attention to whether they follow the order of operations. This is an important concept of arithmetic and algebra. Also, watch for whether students truly understand the equality properties: if they do something to one side of an equation, they MUST do the same thing to the other side of the equation. What they do will depend on what problem they are trying to solve. If a number is subtracted from y and they want y to be by itself, add that number to each side of the equation and its opposite appears to ‘move’ to the other side. Similarly, if y is multiplied by a number, division will help them get y by itself. Looking for a solution for students in Grades 3 and up that can help unlock learning for linear relationship equations, formulas, and beyond? ExploreMath 180, a revolutionary approach to math intervention. 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8756	https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.01%3A_Properties_of_Transition_Metals_and_Their_Compounds	19.1.1 19.1.1 19.1.2 19.1.2 19.1.3 19.1.3 Co(s)+2HCl⟶H2+CoCl2(aq) Skip to main content 19.1: Properties of Transition Metals and Their Compounds Last updated : Sep 12, 2022 Save as PDF 19: Transition Metals and Coordination Chemistry 19.2: Coordination Chemistry of Transition Metals Page ID : 38321 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Outline the general approach for the isolation of transition metals from natural sources Describe typical physical and chemical properties of the transition metals Identify simple compound classes for transition metals and describe their chemical properties We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing. Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. As shown in Figure 19.1.219.1.2, the d-block elements in groups 3–11 are transition elements. The f-block elements, also called inner transition metals (the lanthanides and actinides), also meet this criterion because the d orbital is partially occupied before the f orbitals. The d orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals. The d-block elements are divided into the first transition series (the elements Sc through Cu), the second transition series (the elements Y through Ag), and the third transition series (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the fourth transition series, which also includes Rf through Rg. The f-block elements are the elements Ce through Lu, which constitute the lanthanide series (or lanthanoid series), and the elements Th through Lr, which constitute the actinide series (or actinoid series). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series. Example 19.1.119.1.1: Valence Electrons in Transition Metals Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the s electrons before the d or f electrons. Then, for each ion, give the electron configuration: cerium(III) lead(II) Ti2+ Am3+ Pd2+ For the examples that are transition metals, determine to which series they belong. Solution For ions, the s-valence electrons are lost prior to the d or f electrons. Ce3+[Xe]4f1; Ce3+ is an inner transition element in the lanthanide series. Pb2+[Xe]6s25d104f14; the electrons are lost from the p orbital. This is a main group element. titanium(II) [Ar]3d2; first transition series americium(III) [Rn]5f6; actinide palladium(II) [Kr]4d8; second transition series Exercise 19.1.119.1.1 Check Your Learning Give an example of an ion from the first transition series with no d electrons. Answer : V5+ is one possibility. Other examples include Sc3+, Ti4+, Cr6+, and Mn7+. Uses of Lanthanides in Devices Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as rare earth elements. Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver (4.5 × 10−5% versus 0.79 × 10−5% by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together. The commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure 19.1.319.1.3). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines. As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to $470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials. The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable coordination compounds, in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series. Properties of the Transition Elements Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials (Table P1), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be2+ and Mg2+. On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry. Ions of the lighter d-block elements, such as Cr3+, Fe3+, and Co2+, form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo3+, Ru3+, and Ir2+) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier d-block elements are oxyanions such as MoO2−4MoO2−4 and ReO−4ReO−4. Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the platinum metals. With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions. Both the d- and f-block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The f-block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts. Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure 19.1.419.1.4. As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+). For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the s and d orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3d and two 4s electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air. The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr3+ ions in water and, in the absence of air, less stable Cr2+ ions. The sulfide with the highest oxidation state for chromium is Cr2S3, which contains the Cr3+ ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+. Example 19.1.219.1.2: Activity of the Transition Metals Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)? Solution First, we need to look up the reduction half reactions (Table P1) for each oxide in the specified oxidation state: Cr2O2−7+14H++6e−⟶2Cr3++7H2O+1.33V Cr2O2−7+14H++6e−⟶2Cr3++7H2O+1.33V MnO−4+8H++5e−⟶Mn2++H2O+1.51V MnO−4+8H++5e−⟶Mn2++H2O+1.51V TiO2+4H++2e−⟶Ti2++2H2O−0.50V TiO2+4H++2e−⟶Ti2++2H2O−0.50V A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set. Exercise 19.1.219.1.2 Predict what reaction (if any) will occur between HCl and Co(s), and between HBr and Pt(s). You will need to use the standard reduction potentials from (Table P1). Answer : Co(s)+2HCl⟶H2+CoCl2(aq); no reaction because Pt(s) will not be oxidized by H+ Preparation of the Transition Elements Ancient civilizations knew about iron, copper, silver, and gold. The time periods in human history known as the Bronze Age and Iron Age mark the advancements in which societies learned to isolate certain metals and use them to make tools and goods. Naturally occurring ores of copper, silver, and gold can contain high concentrations of these metals in elemental form (Figure 19.1.519.1.5). Iron, on the other hand, occurs on earth almost exclusively in oxidized forms, such as rust (Fe2O3). The earliest known iron implements were made from iron meteorites. Surviving iron artifacts dating from approximately 4000 to 2500 BC are rare, but all known examples contain specific alloys of iron and nickel that occur only in extraterrestrial objects, not on earth. It took thousands of years of technological advances before civilizations developed iron smelting, the ability to extract a pure element from its naturally occurring ores and for iron tools to become common. Generally, the transition elements are extracted from minerals found in a variety of ores. However, the ease of their recovery varies widely, depending on the concentration of the element in the ore, the identity of the other elements present, and the difficulty of reducing the element to the free metal. In general, it is not difficult to reduce ions of the d-block elements to the free element. Carbon is a sufficiently strong reducing agent in most cases. However, like the ions of the more active main group metals, ions of the f-block elements must be isolated by electrolysis or by reduction with an active metal such as calcium. We shall discuss the processes used for the isolation of iron, copper, and silver because these three processes illustrate the principal means of isolating most of the d-block metals. In general, each of these processes involves three principal steps: preliminary treatment, smelting, and refining. Preliminary treatment. In general, there is an initial treatment of the ores to make them suitable for the extraction of the metals. This usually involves crushing or grinding the ore, concentrating the metal-bearing components, and sometimes treating these substances chemically to convert them into compounds that are easier to reduce to the metal. Smelting. The next step is the extraction of the metal in the molten state, a process called smelting, which includes reduction of the metallic compound to the metal. Impurities may be removed by the addition of a compound that forms a slag—a substance with a low melting point that can be readily separated from the molten metal. Refining. The final step in the recovery of a metal is refining the metal. Low boiling metals such as zinc and mercury can be refined by distillation. When fused on an inclined table, low melting metals like tin flow away from higher-melting impurities. Electrolysis is another common method for refining metals. Isolation of Iron The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities. The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure 19.1.619.1.6) in which the roasted ore, coke, and limestone (impure CaCO3) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons. Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide: CO2(g)+C(s)⟶2CO(g) CO2(g)+C(s)⟶2CO(g) The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure 19.1.619.1.6. The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore: CaO(s)+SiO2(s)⟶CaSiO3(l) CaO(s)+SiO2(s)⟶CaSiO3(l) Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure 19.1.719.1.7). Much of the iron produced is refined and converted into steel. Steel is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle. Isolation of Copper The most important ores of copper contain copper sulfides (such as covellite, CuS), although copper oxides (such as tenorite, CuO) and copper hydroxycarbonates [such as malachite, Cu2(OH)2CO3] are sometimes found. In the production of copper metal, the concentrated sulfide ore is roasted to remove part of the sulfur as sulfur dioxide. The remaining mixture, which consists of Cu2S, FeS, FeO, and SiO2, is mixed with limestone, which serves as a flux (a material that aids in the removal of impurities), and heated. Molten slag forms as the iron and silica are removed by Lewis acid-base reactions: CaCO3(s)+SiO2(s)⟶CaSiO3(l)+CO2(g) CaCO3(s)+SiO2(s)⟶CaSiO3(l)+CO2(g) FeO(s)+SiO2(s)⟶FeSiO3(l) FeO(s)+SiO2(s)⟶FeSiO3(l) In these reactions, the silicon dioxide behaves as a Lewis acid, which accepts a pair of electrons from the Lewis base (the oxide ion). Reduction of the Cu2S that remains after smelting is accomplished by blowing air through the molten material. The air converts part of the Cu2S into Cu2O. As soon as copper(I) oxide is formed, it is reduced by the remaining copper(I) sulfide to metallic copper: 2Cu2S(l)+3O2(g)⟶2Cu2O(l)+2SO2(g) 2Cu2S(l)+3O2(g)⟶2Cu2O(l)+2SO2(g) 2Cu2O(l)+Cu2S(l)⟶6Cu(l)+SO2(g) 2Cu2O(l)+Cu2S(l)⟶6Cu(l)+SO2(g) The copper obtained in this way is called blister copper because of its characteristic appearance, which is due to the air blisters it contains (Figure 19.1.819.1.8). This impure copper is cast into large plates, which are used as anodes in the electrolytic refining of the metal (which is described in the chapter on electrochemistry). Isolation of Silver Silver sometimes occurs in large nuggets (Figure 19.1.919.1.9) but more frequently in veins and related deposits. At one time, panning was an effective method of isolating both silver and gold nuggets. Due to their low reactivity, these metals, and a few others, occur in deposits as nuggets. The discovery of platinum was due to Spanish explorers in Central America mistaking platinum nuggets for silver. When the metal is not in the form of nuggets, it often useful to employ a process called hydrometallurgy to separate silver from its ores. Hydrology involves the separation of a metal from a mixture by first converting it into soluble ions and then extracting and reducing them to precipitate the pure metal. In the presence of air, alkali metal cyanides readily form the soluble dicyanoargentate(I) ion, [Ag(CN)2]−[Ag(CN)2]−, from silver metal or silver-containing compounds such as Ag2S and AgCl. Representative equations are: 4Ag(s)+8CN−(aq)+O2(g)+2H2O(l)⟶4[Ag(CN)2]−(aq)+4OH−(aq) 4Ag(s)+8CN−(aq)+O2(g)+2H2O(l)⟶4[Ag(CN)2]−(aq)+4OH−(aq) 2Ag2S(s)+8CN−(aq)+O2(g)+2H2O(l)⟶4[Ag(CN)2]−(aq)+2S(s)+4OH−(aq) 2Ag2S(s)+8CN−(aq)+O2(g)+2H2O(l)⟶4[Ag(CN)2]−(aq)+2S(s)+4OH−(aq) AgCl(s)+2CN−(aq)⟶[Ag(CN)2]−(aq)+Cl−(aq) AgCl(s)+2CN−(aq)⟶[Ag(CN)2]−(aq)+Cl−(aq) The silver is precipitated from the cyanide solution by the addition of either zinc or iron(II) ions, which serves as the reducing agent: 2[Ag(CN)2]−(aq)+Zn(s)⟶2Ag(s)+[Zn(CN)4]2−(aq) 2[Ag(CN)2]−(aq)+Zn(s)⟶2Ag(s)+[Zn(CN)4]2−(aq) Example 19.1.319.1.3: Refining Redox One of the steps for refining silver involves converting silver into dicyanoargenate(I) ions: 4Ag(s)+8CN−(aq)+O2(g)+2H2O(l)⟶4[Ag(CN)2]−(aq)+4OH−(aq) 4Ag(s)+8CN−(aq)+O2(g)+2H2O(l)⟶4[Ag(CN)2]−(aq)+4OH−(aq) Explain why oxygen must be present to carry out the reaction. Why does the reaction not occur as: 4Ag(s)+8CN−(aq)⟶4[Ag(CN)2]−(aq)? 4Ag(s)+8CN−(aq)⟶4[Ag(CN)2]−(aq)? Solution The charges, as well as the atoms, must balance in reactions. The silver atom is being oxidized from the 0 oxidation state to the 1+ state. Whenever something loses electrons, something must also gain electrons (be reduced) to balance the equation. Oxygen is a good oxidizing agent for these reactions because it can gain electrons to go from the 0 oxidation state to the 2− state. Exercise 19.1.319.1.3 During the refining of iron, carbon must be present in the blast furnace. Why is carbon necessary to convert iron oxide into iron? Answer : The carbon is converted into CO, which is the reducing agent that accepts electrons so that iron(III) can be reduced to iron(0). Transition Metal Compounds The bonding in the simple compounds of the transition elements ranges from ionic to covalent. In their lower oxidation states, the transition elements form ionic compounds; in their higher oxidation states, they form covalent compounds or polyatomic ions. The variation in oxidation states exhibited by the transition elements gives these compounds a metal-based, oxidation-reduction chemistry. The chemistry of several classes of compounds containing elements of the transition series follows. Halides Anhydrous halides of each of the transition elements can be prepared by the direct reaction of the metal with halogens. For example: 2Fe(s)+3Cl2(g)⟶2FeCl3(s) 2Fe(s)+3Cl2(g)⟶2FeCl3(s) Heating a metal halide with additional metal can be used to form a halide of the metal with a lower oxidation state: Fe(s)+2FeCl3(s)⟶3FeCl2(s) Fe(s)+2FeCl3(s)⟶3FeCl2(s) The stoichiometry of the metal halide that results from the reaction of the metal with a halogen is determined by the relative amounts of metal and halogen and by the strength of the halogen as an oxidizing agent. Generally, fluorine forms fluoride-containing metals in their highest oxidation states. The other halogens may not form analogous compounds. In general, the preparation of stable water solutions of the halides of the metals of the first transition series is by the addition of a hydrohalic acid to carbonates, hydroxides, oxides, or other compounds that contain basic anions. Sample reactions are: NiCO3(s)+2HF(aq)⟶NiF2(aq)+H2O(l)+CO2(g) NiCO3(s)+2HF(aq)⟶NiF2(aq)+H2O(l)+CO2(g) Co(OH)2(s)+2HBr(aq)⟶CoBr2(aq)+2H2O(l) Co(OH)2(s)+2HBr(aq)⟶CoBr2(aq)+2H2O(l) Most of the first transition series metals also dissolve in acids, forming a solution of the salt and hydrogen gas. For example: Cr(s)+2HCl(aq)⟶CrCl2(aq)+H2(g) Cr(s)+2HCl(aq)⟶CrCl2(aq)+H2(g) The polarity of bonds with transition metals varies based not only upon the electronegativities of the atoms involved but also upon the oxidation state of the transition metal. Remember that bond polarity is a continuous spectrum with electrons being shared evenly (covalent bonds) at one extreme and electrons being transferred completely (ionic bonds) at the other. No bond is ever 100% ionic, and the degree to which the electrons are evenly distributed determines many properties of the compound. Transition metal halides with low oxidation numbers form more ionic bonds. For example, titanium(II) chloride and titanium(III) chloride (TiCl2 and TiCl3) have high melting points that are characteristic of ionic compounds, but titanium(IV) chloride (TiCl4) is a volatile liquid, consistent with having covalent titanium-chlorine bonds. All halides of the heavier d-block elements have significant covalent characteristics. The covalent behavior of the transition metals with higher oxidation states is exemplified by the reaction of the metal tetrahalides with water. Like covalent silicon tetrachloride, both the titanium and vanadium tetrahalides react with water to give solutions containing the corresponding hydrohalic acids and the metal oxides: SiCl4(l)+2H2O(l)⟶SiO2(s)+4HCl(aq) SiCl4(l)+2H2O(l)⟶SiO2(s)+4HCl(aq) TiCl4(l)+2H2O(l)⟶TiO2(s)+4HCl(aq) TiCl4(l)+2H2O(l)⟶TiO2(s)+4HCl(aq) Oxides As with the halides, the nature of bonding in oxides of the transition elements is determined by the oxidation state of the metal. Oxides with low oxidation states tend to be more ionic, whereas those with higher oxidation states are more covalent. These variations in bonding are because the electronegativities of the elements are not fixed values. The electronegativity of an element increases with increasing oxidation state. Transition metals in low oxidation states have lower electronegativity values than oxygen; therefore, these metal oxides are ionic. Transition metals in very high oxidation states have electronegativity values close to that of oxygen, which leads to these oxides being covalent. The oxides of the first transition series can be prepared by heating the metals in air. These oxides are Sc2O3, TiO2, V2O5, Cr2O3, Mn3O4, Fe3O4, Co3O4, NiO, and CuO. Alternatively, these oxides and other oxides (with the metals in different oxidation states) can be produced by heating the corresponding hydroxides, carbonates, or oxalates in an inert atmosphere. Iron(II) oxide can be prepared by heating iron(II) oxalate, and cobalt(II) oxide is produced by heating cobalt(II) hydroxide: FeC2O4(s)⟶FeO(s)+CO(g)+CO2(g) FeC2O4(s)⟶FeO(s)+CO(g)+CO2(g) Co(OH)2(s)⟶CoO(s)+H2O(g) Co(OH)2(s)⟶CoO(s)+H2O(g) With the exception of CrO3 and Mn2O7, transition metal oxides are not soluble in water. They can react with acids and, in a few cases, with bases. Overall, oxides of transition metals with the lowest oxidation states are basic (and react with acids), the intermediate ones are amphoteric, and the highest oxidation states are primarily acidic. Basic metal oxides at a low oxidation state react with aqueous acids to form solutions of salts and water. Examples include the reaction of cobalt(II) oxide accepting protons from nitric acid, and scandium(III) oxide accepting protons from hydrochloric acid: CoO(s)+2HNO3(aq)⟶Co(NO3)2(aq)+H2O(l) Sc2O3(s)+6HCl(aq)⟶2ScCl3(aq)+3H2O(l) The oxides of metals with oxidation states of 4+ are amphoteric, and most are not soluble in either acids or bases. Vanadium(V) oxide, chromium(VI) oxide, and manganese(VII) oxide are acidic. They react with solutions of hydroxides to form salts of the oxyanions VO3−4, CrO2−4, and MnO−4. For example, the complete ionic equation for the reaction of chromium(VI) oxide with a strong base is given by: CrO3(s)+2Na+(aq)+2OH−(aq)⟶2Na+(aq)+CrO2−4(aq)+H2O(l) Chromium(VI) oxide and manganese(VII) oxide react with water to form the acids H2CrO4 and HMnO4, respectively. Hydroxides When a soluble hydroxide is added to an aqueous solution of a salt of a transition metal of the first transition series, a gelatinous precipitate forms. For example, adding a solution of sodium hydroxide to a solution of cobalt sulfate produces a gelatinous pink or blue precipitate of cobalt(II) hydroxide. The net ionic equation is: Co2+(aq)+2OH−(aq)⟶Co(OH)2(s) In this and many other cases, these precipitates are hydroxides containing the transition metal ion, hydroxide ions, and water coordinated to the transition metal. In other cases, the precipitates are hydrated oxides composed of the metal ion, oxide ions, and water of hydration: 4Fe3+(aq)+6OH−(aq)+nH2O(l)⟶2Fe2O3⋅(n+3)H2O(s) These substances do not contain hydroxide ions. However, both the hydroxides and the hydrated oxides react with acids to form salts and water. When precipitating a metal from solution, it is necessary to avoid an excess of hydroxide ion, as this may lead to complex ion formation as discussed later in this chapter. The precipitated metal hydroxides can be separated for further processing or for waste disposal. Carbonates Many of the elements of the first transition series form insoluble carbonates. It is possible to prepare these carbonates by the addition of a soluble carbonate salt to a solution of a transition metal salt. For example, nickel carbonate can be prepared from solutions of nickel nitrate and sodium carbonate according to the following net ionic equation: Ni2+(aq)+CO2−3⟶NiCO3(s) The reactions of the transition metal carbonates are similar to those of the active metal carbonates. They react with acids to form metals salts, carbon dioxide, and water. Upon heating, they decompose, forming the transition metal oxides. Other Salts In many respects, the chemical behavior of the elements of the first transition series is very similar to that of the main group metals. In particular, the same types of reactions that are used to prepare salts of the main group metals can be used to prepare simple ionic salts of these elements. A variety of salts can be prepared from metals that are more active than hydrogen by reaction with the corresponding acids: Scandium metal reacts with hydrobromic acid to form a solution of scandium bromide: 2Sc(s)+6HBr(aq)⟶2ScBr3(aq)+3H2(g) The common compounds that we have just discussed can also be used to prepare salts. The reactions involved include the reactions of oxides, hydroxides, or carbonates with acids. For example: Ni(OH)2(s)+2H3O+(aq)+2ClO−4(aq)⟶Ni2+(aq)+2ClO−4(aq)+4H2O(l) Substitution reactions involving soluble salts may be used to prepare insoluble salts. For example: Ba2+(aq)+2Cl−(aq)+2K+(aq)+CrO2−4(aq)⟶BaCrO4(s)+2K+(aq)+2Cl−(aq) In our discussion of oxides in this section, we have seen that reactions of the covalent oxides of the transition elements with hydroxides form salts that contain oxyanions of the transition elements. High Temperature Superconductors A superconductor is a substance that conducts electricity with no resistance. This lack of resistance means that there is no energy loss during the transmission of electricity. This would lead to a significant reduction in the cost of electricity. Most currently used, commercial superconducting materials, such as NbTi and Nb3Sn, do not become superconducting until they are cooled below 23 K (−250 °C). This requires the use of liquid helium, which has a boiling temperature of 4 K and is expensive and difficult to handle. The cost of liquid helium has deterred the widespread application of superconductors. One of the most exciting scientific discoveries of the 1980s was the characterization of compounds that exhibit superconductivity at temperatures above 90 K. (Compared to liquid helium, 90 K is a high temperature.) Typical among the high-temperature superconducting materials are oxides containing yttrium (or one of several rare earth elements), barium, and copper in a 1:2:3 ratio. The formula of the ionic yttrium compound is YBa2Cu3O7. The new materials become superconducting at temperatures close to 90 K (Figure 19.1.10), temperatures that can be reached by cooling with liquid nitrogen (boiling temperature of 77 K). Not only are liquid nitrogen-cooled materials easier to handle, but the cooling costs are also about 1000 times lower than for liquid helium. Although the brittle, fragile nature of these materials presently hampers their commercial applications, they have tremendous potential that researchers are hard at work improving their processes to help realize. Superconducting transmission lines would carry current for hundreds of miles with no loss of power due to resistance in the wires. This could allow generating stations to be located in areas remote from population centers and near the natural resources necessary for power production. The first project demonstrating the viability of high-temperature superconductor power transmission was established in New York in 2008. Researchers are also working on using this technology to develop other applications, such as smaller and more powerful microchips. In addition, high-temperature superconductors can be used to generate magnetic fields for applications such as medical devices, magnetic levitation trains, and containment fields for nuclear fusion reactors (Figure 19.1.11). Summary The transition metals are elements with partially filled d orbitals, located in the d-block of the periodic table. The reactivity of the transition elements varies widely from very active metals such as scandium and iron to almost inert elements, such as the platinum metals. The type of chemistry used in the isolation of the elements from their ores depends upon the concentration of the element in its ore and the difficulty of reducing ions of the elements to the metals. Metals that are more active are more difficult to reduce. Transition metals exhibit chemical behavior typical of metals. For example, they oxidize in air upon heating and react with elemental halogens to form halides. Those elements that lie above hydrogen in the activity series react with acids, producing salts and hydrogen gas. Oxides, hydroxides, and carbonates of transition metal compounds in low oxidation states are basic. Halides and other salts are generally stable in water, although oxygen must be excluded in some cases. Most transition metals form a variety of stable oxidation states, allowing them to demonstrate a wide range of chemical reactivity. Glossary actinide series : (also, actinoid series) actinium and the elements in the second row or the f-block, atomic numbers 89–103 coordination compound : stable compound in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons d-block element : one of the elements in groups 3–11 with valence electrons in d orbitals f-block element : (also, inner transition element) one of the elements with atomic numbers 58–71 or 90–103 that have valence electrons in f orbitals; they are frequently shown offset below the periodic table first transition series : transition elements in the fourth period of the periodic table (first row of the d-block), atomic numbers 21–29 fourth transition series : transition elements in the seventh period of the periodic table (fourth row of the d-block), atomic numbers 89 and 104–111 hydrometallurgy : process in which a metal is separated from a mixture by first converting it into soluble ions, extracting the ions, and then reducing the ions to precipitate the pure metal lanthanide series : (also, lanthanoid series) lanthanum and the elements in the first row or the f-block, atomic numbers 57–71 platinum metals : group of six transition metals consisting of ruthenium, osmium, rhodium, iridium, palladium, and platinum that tend to occur in the same minerals and demonstrate similar chemical properties rare earth element : collection of 17 elements including the lanthanides, scandium, and yttrium that often occur together and have similar chemical properties, making separation difficult second transition series : transition elements in the fifth period of the periodic table (second row of the d-block), atomic numbers 39–47 smelting : process of extracting a pure metal from a molten ore steel : material made from iron by removing impurities in the iron and adding substances that produce alloys with properties suitable for specific uses superconductor : material that conducts electricity with no resistance third transition series : transition elements in the sixth period of the periodic table (third row of the d-block), atomic numbers 57 and 72–79 19: Transition Metals and Coordination Chemistry 19.2: Coordination Chemistry of Transition Metals
8757	https://www.escardio.org/Journals/E-Journal-of-Cardiology-Practice/Volume-18/physical-examination-in-aortic-valve-disease-do-we-still-need-it-in-the-modern	Physical examination in aortic valve disease: do we still need it in the modern era? In order to bring you the best possible user experience, this site uses Javascript. If you are seeing this message, it is likely that the Javascript option in your browser is disabled. For optimal viewing of this site, please ensure that Javascript is enabled for your browser. Did you know that your browser is out of date? To get the best experience using our website we recommend that you upgrade to a newer version. Learn more. 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Vol. 18, N° 12 - 26 Feb 2020 Dr. Roxana-Nicoleta Siliste Assoc. Prof Calin Siliste Compared to echocardiography, auscultation is of low sensitivity in detecting aortic valve lesions (56.6-73%), but has a high agreement rate between examiners and a high specificity (92-98%). This is why, even in the era of high technology, physical examination still plays a crucial role in the screening, diagnosis and the severity assessment of aortic valve disease. The classic signs such as a "parvus et tardus" carotid pulse, a loud late-peaking systolic murmur in the aortic area or a diminished A2 correlate with the severity of aortic stenosis, while wide pulse pressure and a holodiastolic regurgitant murmur are the hallmark findings in significant chronic aortic regurgitation. Topic(s): Valvular Heart Disease Introduction Aortic valve disease, especially aortic stenosis (AS), has an increasing prevalence worldwide, so a good screening tool for the selection of patients being referred for further testing and surgical or interventional procedure is always needed. For many years, the physical examination based mostly on cardiac auscultation represented the most useful bedside diagnostic tools for the diagnosis of these diseases. Unfortunately, expertise and proficiency in auscultation has been diminishing in the modern era, since new technologies have been developed. Increasingly, over the years, cardiac auscultation has tended to be replaced by sophisticated high technology - electronic stethoscopes, handheld ultrasound or echocardiographic evaluation. Until now, there has been no evidence to support the role of auscultatory devices in increasing the sensitivity of auscultation. There were several studies that showed some benefit of handheld ultrasound (HHU) versus physical examination for the diagnosis of different cardiac diseases, but there are no substantial data concerning the role of HHU in the screening of asymptomatic aortic valve lesions. A systematic review by Stanger et al. showed that the sensitivity and specificity of insonation in identifying AS ranged from 62% to 94% and 85% to 98%, respectively, and that these ranges were similar to auscultation . Compared to echocardiography, auscultation is of low sensitivity (56.6-73%) in detecting aortic valve lesions, but with a high agreement rate between examiners and a high specificity (92-98%) [2,3]. The lowest sensitivity was found in subjects with aortic regurgitation (AR), mild valve lesions, as well as in patients with significant lesions and concomitant left ventricle systolic dysfunction . Regarding the value of the diastolic murmur for the diagnosis of AR, older studies have found a sensitivity of 0% to 38% for mild AR and 60% to 80% for moderate or greater AR . With a certain degree of limitation, cardiac auscultation may have a prognostic value in AS. For 35 years, Bodegard et al have followed up a cohort of apparently healthy middle-aged men, including 23.4% of subjects with systolic ejection murmurs. In patients with a low-grade murmur, there was a 4.7-fold, age-adjusted increase in the risk of aortic valve replacement (AVR) compared to an 89-fold increased risk of AVR for those with increased risk of AVR for those with moderate-grade murmur. The physical examination has also an important role in the preoperative assessment of patients before non-cardiac surgery. The preoperative screening using cardiac auscultation identified 908 out of 3,997 hip fracture patients with an undiagnosed heart murmur, of whom almost a third were diagnosed by echocardiography with significant aortic stenosis, this finding being important for the further perioperative management . Physical Examination in Valvular Aortic Stenosis Physical examination often provides the first clue to the presence of AS and helps in assessing the severity of the lesion. Precordial palpation, cardiac auscultation and examination of the carotid pulse are very valuable when evaluating a patient with suspicion of AS. In patients with isolated AS, the precordial apical thrust is accentuated and initially normal in location. In the left lateral decubitus, a bifid apical impulse is sometimes felt: the first impact comes from the left atrial contraction (corresponding to the presystolic gallop) and the second one from the LV contraction. A low-intensity and/or displaced systolic apical impulse could be a sign of low-flow AS or may be caused by other associated cardiac conditions. A thrill may be palpable over the aortic area in significant AS. Thecarotid pulse is characterised in normal subjects by a relatively rapid upstroke and a smooth, more gradual downstroke, interrupted only briefly at the pulse peak. These palpable pulsatile changes in the carotid arterial diameter are virtually identical to the intraluminal pressure pulse. In patients with significant AS, the carotid pulse is weak (“pulsus parvus”), rises slowly and has a delayed systolic peak (“pulsus tardus”). The absence of this finding, particularly in an elderly patient with non-compliant vasculature or in patients in a hyperkinetic state, does not exclude severe AS . Cardiac auscultation reveals many important findings that suggest both the diagnosis and the severity of AS. The most important auscultatory findings in AS are: a normal first sound (S1), a soft and single second sound (S2), the presence of an aortic ejection click and the typical basal systolic murmur. Normally, the murmur is a systolic ejection murmur with onset a short interval after the S1 and the end before S2. When the valve is still flexible the aortic ejection click precedes the murmur. It is commonly heard in the 2nd right intercostal space (the aortic area), but it could also be audible along the left sternal border in the 3rd and 4th interspaces. The murmur is usually harsh and medium-pitched, so it is audible with either the bell or the diaphragm of the stethoscope. It has a characteristic “diamond shape” in the phonocardiogram. Generally, the murmur is loudest in the aortic area, but in 15% of cases it can be better heard at the apex . The murmur is transmitted well and equally to the carotid arteries and both clavicles. Etchells et al showed that the absence of the transmission of the murmur over the right clavicle effectively rules out AS . The murmur may also radiate to the apex, sometimes with a musical quality due to high frequency vibrations, the so-called Gallavardin phenomenon (usually encountered in degenerative AS). The presence of three or four associated findings - slow carotid artery upstroke, reduced carotid artery volume, maximal murmur intensity at the second right intercostal space, and reduced or absent second heart sound - effectively rules in AS . The AS murmur has to be differentiated from other conditions that could associate a basal systolic murmurs: Functional murmurs are generally faint, medium-pitched, very short (proto-, mid- or late-systolic); the S2 is always normal. Patients with hypertension or aortic sclerosis could have a similar harsh, medium-pitched murmur, but with a normal or even loud S2. In hypertrophic obstructive cardiomyopathy a systolic ejection murmur is heard along the left sternal border and the apex. Typically the murmur is variable, being amplified by exercise, squatting, Valsalva manoeuvre or administration of vasodilator or positive inotropic drugs. Supravalvular aortic stenosis produces most of the signs of valvular stenosis, but the systolic click is absent, the S2 is accentuated and the carotid murmurs are very loud. The murmur of pulmonic stenosis has similar intensity, configuration and pitch to AS, but it is loudest in the pulmonary area, the S2 is widely split and the murmur extends beyond the S2. A similar murmur to that of pulmonic stenosis can be heard in atrial septal defect, but the S2 is wide and with a fixed split. When the AS murmur radiates to the apex, there are some hints to differentiate it from mitral regurgitation: the apical impulse is normal in location, the S1 is normal, the murmur starts after the S1, it is not holosystolic and it is not transmitted to the axilla. Clinical Signs Predictive of Severe Aortic Stenosis The following signs help in the diagnosis of severe AS (Table 1), but no combination of physical findings has both a high sensitivity and specificity, particularly in asymptomatic patients. Table 1. Clinical Signs Predictive of Severe Aortic Stenosis. Clinical Signs of Severe Aortic Stenosis A loud systolic murmur (grade 4 or greater) Mechanism The intensity of the murmur reflects the velocity and turbulence of blood flow across the valve Pitfalls↓- LV dysfunction ↑-Hyperdynamic states (anaemia, fever, hyperthyroid) Clinical Signs of Severe Aortic Stenosis Mid- or late- peaking murmur Mechanism It takes longer for blood to eject through a very reduced AV area Pitfalls Difficult to be assessed clinically, especially in patients with tachycardia, atrial fibrillation, low grade murmur Clinical Signs of Severe Aortic Stenosis A diminished or absent S2 Mechanism The aortic cusps are immobile, so the A2 is faint or even not audible Pitfalls S2 is increased in associated pulmonary hypertension, other valve heart diseases, hypertensive heart Clinical Signs of Severe Aortic Stenosis Paradoxical splitting of S2 Mechanism LV ejection time is prolonged and AV closes after PV Pitfalls In LV dysfunction the ejection time does not correlate to the AV area Clinical Signs of Severe Aortic Stenosis The disappearance of an ejection click Mechanism It occurs at the moment of maximal opening of the AV, when the valve is bicuspid, still flexible Pitfalls Usually it is not audible Clinical Signs of Severe Aortic Stenosis Presystolic gallop (S4) Mechanism Forceful atrial contraction into a hypertrophied, non-compliant left ventricle Pitfalls Other causes of LV hypertrophy Absent in atrial fibrillation Clinical Signs of Severe Aortic Stenosis The “parvus et tardus” carotid pulse Mechanism Occlusion of more than 75% of the AV orifice produces a plateau pulse and diminished pulse pressure Pitfalls The pulse pressure increases in patients with non-compliant vasculature or hyperdynamic states \| Clinical Signs of Severe Aortic Stenosis \| Mechanism \| Pitfalls \| --- \| A loud systolic murmur (grade 4 or greater) \| The intensity of the murmur reflects the velocity and turbulence of blood flow across the valve \| ↓- LV dysfunction ↑-Hyperdynamic states (anaemia, fever, hyperthyroid) \| \| Mid- or late- peaking murmur \| It takes longer for blood to eject through a very reduced AV area \| Difficult to be assessed clinically, especially in patients with tachycardia, atrial fibrillation, low grade murmur \| \| A diminished or absent S2 \| The aortic cusps are immobile, so the A2 is faint or even not audible \| S2 is increased in associated pulmonary hypertension, other valve heart diseases, hypertensive heart \| \| Paradoxical splitting of S2 \| LV ejection time is prolonged and AV closes after PV \| In LV dysfunction the ejection time does not correlate to the AV area \| \| The disappearance of an ejection click \| It occurs at the moment of maximal opening of the AV, when the valve is bicuspid, still flexible \| Usually it is not audible \| \| Presystolic gallop (S4) \| Forceful atrial contraction into a hypertrophied, non-compliant left ventricle \| Other causes of LV hypertrophy Absent in atrial fibrillation \| \| The “parvus et tardus” carotid pulse \| Occlusion of more than 75% of the AV orifice produces a plateau pulse and diminished pulse pressure \| The pulse pressure increases in patients with non-compliant vasculature or hyperdynamic states \| AV: aortic valve; A2: aortic valve closer; LV: left ventricle; PV: pulmonary valve; S2: the second heart sound The Intensity of the Murmur Haemodynamic studies have shown that the intensity of the murmur reflects the velocity of blood flow across the valve, so a very loud murmur (grade 4 or greater) has a high specificity for severe AS. In one study, the echocardiographically derived peak gradient, mean gradient and aortic valve area were highly predictive for correctly identifying the degree of severity of the murmur (87%, 81%, and 87%, respectively) . Despite the high specificity, the intensity of the murmur has a low sensitivity in diagnosing severe AS, being dependent on the hemodynamic status. In patients with a low left ventricular (LV) ejection fraction or low stroke volume, the murmur’s intensity decreases and in hyperdynamic states it is frequently augmented. A Mid- or Late-Peaking Murmur As aortic stenosis worsens, it takes longer for blood to be ejected through the valve. Thus, mild AS would have an early peaking murmur, while in severe AS the murmur peaks later in systole. Even if it is a highly specific clinical sign for severe AS, the evaluation of the timing of the murmur has a very low reproducibility among examiners. Derived from this clinical finding, time-to-peak-velocity (TPV) is a new echocardiographic variable that is easily measured, reproducible and useful to evaluate AS severity (Figure 1). In the study of Kamimura et al, a TPV cut-off value of 99 msec had the highest sensitivity, speciﬁcity, and positive and negative predictive values in detecting severe AS (irrespective of ejection fraction) and also for predicting a poor prognosis in AS patients . Figure 1. Time-to-Peak Velocity Measured by Continuous-Wave Doppler Echocardiography in a Patient with Severe AS and a Late-Peaking Systolic Murmur. A Diminished or Absent S2 In severe AS, the aortic cusps are immobile, so the aortic component of the second sound is faint or even not audible. The real proportion of this feature was not extensively studied. In a series of 397 patients with AS at their first haemodynamic evaluation, only 9% had an absent S2 . Paradoxical Splitting of the S2 The paradoxical splitting of S2 is present when the transaortic pressure gradient is very high and the aortic valve closes very late, after the pulmonic valve (more obvious during expiration). Haemodynamic studies showed that, in patients with AS, the direct linear relationship between ejection time and stroke volume is totally obliterated and the degree of prolongation of left ventricular ejection time above that predicted from stroke volume is closely correlated with aortic valve area. On the other hand, in patients with failing ventricles, the ejection time was less prolonged, and the duration of ejection was unrelated to the valve area . The Disappearance of the Ejection Click The ejection click is a high-pitched sound, best heard at the apex that occurs at the moment of maximal opening of the aortic valve (AV), shortly after the S1. The sound occurs in the presence of a dilated aorta or in the presence of a bicuspid or flexible stenotic AV. It disappears in patients with a severe calcific, immobile aortic valve . The Presence of Presystolic Gallop (The Fourth Sound or So-Called Atrial Gallop) A prominent S4 can be audible and palpable due to forceful atrial contraction into a hypertrophied, non-compliant left ventricle . The presence of an S4 in a young patient with AS indicates a significant AV lesion, but in an elderly or hypertensive person this is not necessarily true because it can be related to the very common diastolic dysfunction. The “Parvus et Tardus” Carotid Pulse Occlusion of more than 75% of the aortic orifice produces a plateau pulse and diminished pulse pressure that can be objectified by carotid or peripheral pulse palpation. Of its two components, “pulsus tardus” is the better discriminator, detecting severe AS with a sensitivity of 31% to 90%, and a specificity of 68% to 93% . A prolonged carotid upstroke time was found in 58% to 68% of patients with severe AS, in 33% of patients with moderate AS and only in 3% of patients with mild AS . A recent study demonstrated that the invasively measured time difference between LV and aortic systolic pressure peaks was significantly associated with the severity of AS and AV calcification calculated on multidetector computed tomography (MDCT) . Physical Examination in Valvular Aortic Regurgitation In patients with AR, the clinical examination focuses on precordial and peripheral pulse inspection and palpation, as well as cardiac auscultation. The typical early diastolic murmur has a high sensitivity and specificity for the diagnosis of AR (76% and 96%, respectively) , though its absence does not exclude the diagnosis. An isolated mid-systolic murmur is a more common auscultatory finding by a non-cardiologist in patients with mild to moderate AR . A displaced (laterally and inferior), diffuse and hyperdynamic apical thrust is a frequent finding in significant chronic AR. The pulse is characterised by a very brisk upstroke, large amplitude, and rapid collapse (known as Corrigan’s pulse or Watson’s water hammer pulse). It is an extreme form of the hyperkinetic pulse, so we have to differentiate it from other hyperdynamic states . Besides the Corrigan Pulse, there is a long list of classic findings related to wide pulse pressure, but their sensibility and specificity is not well established - dancing carotid arteries (also described by Corrigan), de Musset’s sign (head bobbing with systolic pulse), Mueller’s sign (systolic pulsations of the uvula), Becker’s sign (pulsation of the retinal arteries), Landolfi’s sign (pupillary hippus), Traube’s sign (systolic and diastolic “pistol shot” sounds heard while auscultating the femoral artery), Duroziez’s sign (a systolic and diastolic bruit heard when the femoral artery is partially compressed), Quincke’s pulse (systolic pulsations seen upon light compression of the nail bed) and Hill’s sign (the lower limb systolic blood pressure exceeds the upper limb systolic blood pressure by more than 20 mmHg). Cardiac auscultation in chronic AR reveals a normal S1, a soft, single or accentuated S2, the typical basal diastolic murmur and a faint mid-systolic murmur. Normally the murmur is an early diastolic regurgitant murmur that immediately follows the 2nd sound. It is heard in the 2nd right or 3rd left intercostal space (the Erb area), but it could commonly be audible along the left sternal border in the 3rd and 4th interspaces in patients with valvular AR. This is a high-pitched, blowing murmur, best heard with the diaphragm of the stethoscope . Occasionally, the murmur can be musical in quality, the so-called diastolic whoop. It is usually protodiastolic and decrescendo in the phonocardiogram, but in severe AR the murmur is holodiastolic. The murmur is transmitted towards the cardiac apex. The intensity is usually low and does not correlate to the severity of the valve lesion. When faint, the murmur can be better heard end-expiratory or with the patient leaning forward. A mid-systolic murmur is frequent in isolated AR and is caused either by a large ejection volume through the AV or by calcification of the cusps. All these findings described in chronic AR may not be present in the acute setting of a severe AR. Usually the murmur with acute AR is a low-pitched early diastolic murmur beginning after the S2, and the signs of acute heart failure dominate the clinical picture. The AR murmur has to be differentiated from other conditions that could associate basal diastolic murmurs: Pulmonic regurgitation is indistinguishable in location, timing and quality from the AR murmur, so the distinction is made only by precordial palpation and the lack or presence of peripheral signs. Coronary artery stenosis (Dock’s murmur) is a high-pitched early diastolic murmur, best heard in the 2nd or 3rd left space, near the left border; it is not as widespread as AR. The following clinical findings suggest a severe AR: 1. A Holodiastolic Murmur The duration and quality of murmur are directly proportional to the severity of AR. In mild AR the murmur is blowing and only early diastolic. In moderate to severe AR the murmur is holodiastolic (>2/3 diastole) and rougher in quality. However, it becomes very soft or even disappears in patients with LV dysfunction. 2. Marked Peripheral Signs There are only few data about the predictive value of peripheral signs in diagnosing the severity of AR. Sensitivity reported in the literature for Hill’s sign and Duroziez’s murmur is 69% and 88%, respectively . A difference in blood pressures between lower and upper limb (ankle-brachial difference =ABD) >60 mmHg is highly suggestive of severe AR. In a recent study, the ABD determined by using the cardio ankle vascular index was correlated with LV dimensions and the vena contracta of AR measured by Color echo-Doppler . 3. The Austin Flint Murmur The murmur typically begins in mid-diastole, often has a presystolic accentuation, and terminates at the end of diastole. It is low-pitched, with a rough and rumbling quality, and is best heard at the apex. The proposed mechanisms for the presence of this murmur are fluttering of the anterior mitral leaflet due to the regurgitant jet, increased mitral inflow velocity due to narrowing of the valve orifice by the jet, early mitral valve closure caused by high left ventricular end-diastolic (LVED) pressure, and diastolic mitral regurgitation. The sensitivity reported in various studies is 25% to 100%; the specificity is irrelevant since the Austin Flint murmur is by definition only found in AR . 4. Signs of LV Dilation and Dysfunction The systolic apical impulse is laterally and inferiorly displaced, the intensity of the S1 is decreased (due to the elevated LVED pressure and the early closure of the mitral valve) and a protodiastolic gallop (S3 gallop) is usually heard at the apex. Conclusion If we are aware of its limitations and strengths and we succeed in keeping our expertise and proficiency in cardiac auscultation, then clinical examination remains a valuable and cost-effective tool that often enables a rapid, integrative, accurate and patient-orientated diagnosis of aortic valve disease. Although advanced technologies have become part of our daily lives as clinicians, physical examination still plays a crucial role in the screening, diagnosis and evaluation of the severity of aortic valve disease, especially in AS. References Stanger D, Wan D, Moghaddam N, Elahi N, Argulian E, Narula J, Ahmadi A. Insonation versus Auscultation in Valvular Disorders: Is Aortic Stenosis the Exception? A Systematic Review. Ann Glob Health. 2019;85:104. Patel A, Tomar NS, Bharani A. Utility of physical examination and comparison to echocardiography for cardiac diagnosis.Indian Heart J. 2017;69:141-5. Thoenes M, Bramlage P, Zamorano P, Messika-Zeitoun D, Wendt D, Kasel M, Kurucova J, Steeds RP. Patient screening for early detection of aortic stenosis (AS)-review of current practice and future perspectives.J Thorac Dis. 2018;10:5584-94. Choudhry NK, Etchells EE. Does This Patient Have Aortic Regurgitation?JAMA. 1999;281:2231-8. Bodegard J, Skretteberg PT, Gjesdal K, Pyörälä K, Kjeldsen SE, Liestøl K, Erikssen G, Erikssen J. Low-grade systolic murmurs in healthy middle-aged individuals: innocent or clinically significant? A 35-year follow-up study of 2014 Norwegian men.J Intern Med. 2012;271:581-8. McBrien ME, Heyburn G, Stevenson M, McDonald S, Johnston NJ, Elliott JR, Beringer TR. Previously undiagnosed aortic stenosis revealed by auscultation in the hip fracture population--echocardiographic findings, management and outcome.Anaesthesia. 2009;64:863-70. Morris DC. The Carotid Pulse. In: Walker HK, Hall WD, Hurst JW, editors .Clinical Methods: The History, Physical, and Laboratory Examinations.3rd edition. Boston: Butterworths; 1990. Chapter 20, pp.312. Le Blond R, Brown D. Cardiovascular signs. In: DeGowin’s Diagnostic Examination, 10th Edition. New York: McGraw Hill Education; 2015. Chapter 6, pp 308-333. Etchells E, Glenns V, Shadowitz S, Bell C, Siu S. A bedside clinical prediction rule for detecting moderate or severe aortic stenosis.J Gen Intern Med. 1998;13:699-704. Rama BN, Mohiuddin SM, Esterbrooks DJ, Lynch JD, Holmberg MJ, Mooss AN, Hilleman DE. Correlation of intensity of aortic stenosis murmur by auscultation with echocardiographically determined transvalvular gradients and valve area.Journal of Noninvasive Cardiology. 1999;3:25-31. Kamimura D, Hans S, Suzuki T, Fox ER, Hall ME, Musani SK, McMullan MR, Little WC. Delayed Time to Peak Velocity Is Useful for Detecting Severe Aortic Stenosis.J Am Heart Assoc. 2016;5:3002907. Lombard JT, Selzer A. Valvular aortic stenosis. A clinical and hemodynamic profile of patients.Ann Intern Med. 1987;106:292-8. Kligfield P, Okin P.Effect of ventricular function on left ventricular ejection time in aortic stenosis. Br Heart J. 1979;42:438-41. Steven McGee MD. Pulse rate and contour. In: Evidence-Based Physical Diagnosis (Fourth Edition). Amsterdam, the Netherlands: Elsevier; 2018. Chapter 15, pp. 95-108. Munt B, Legget ME, Kraft CD, Miyake-Hull CY, Fujioka M, Otto CM. Physical examination in valvular aortic stenosis: correlation with stenosis severity and prediction of clinical outcome.Am Heart J. 1999;137:298-306. Sato K, Kumar A, Jobanputra Y, Betancor J, Halane M, George R, Menon V, Krishnaswamy A, Tuzcu EM, Harb S, Jaber WA, Mick S, Svensson LG, Kapadia SR. Association of Time Between Left Ventricular and Aortic Systolic Pressure Peaks With Severity of Aortic Stenosis and Calcification of Aortic Valve.JAMA Cardiol. 2019;4:549-55. Heidenreich PA, Schnittger I, Hancock SL, Atwood JE. A systolic murmur is a common presentation of aortic regurgitation detected by echocardiography.Clin Cardiol. 2004;27:502-6. Babu AN, Kymes SM, Carpenter Fryer SM. Eponyms and the Diagnosis of Aortic Regurgitation: What Says the Evidence?Ann Intern Med. 2003;138:736-742. Shiraishi H, Shirayama T, Maruyama N, Kaimoto S, Otakara A, Kurimoto R, Nakanishi N, Nakamura T, Yamano T, Matsumuro A, Doi K, Yaku H, Matoba S. Usefulness of peripheral arterial signs in the evaluation of aortic regurgitation.J Cardiol. 2017;69:769-73. Notes to editor Authors: Roxana-Nicoleta Siliste 1,2, MD, PhD, Assist. Prof.; Calin Siliste 1,3, MD, PhD, Assoc. Prof. University of Medicine and Pharmacy "Carol Davila", Bucharest, Romania; Department of Internal Medicine and Cardiology, Coltea Clinical Hospital, Bucharest, Romania; Department of Cardiology, University Emergency Hospital, Bucharest, Romania Address for correspondence: Assist. Prof. Roxana-Nicoleta Siliste; Department of Internal Medicine and Cardiology, Coltea Clinical Hospital, Bulevardul Ion C. Brătianu 1, Bucharest, Romania E-mail: silisterox@gmail.com Tel: +40 7 24001158 Author disclosures: The authors have no conflicts of interest to declare. The content of this article reflects the personal opinion of the author/s and is not necessarily the official position of the European Society of Cardiology. Our mission: To reduce the burden of cardiovascular disease. About the ESC Who We Are What We Do ESC Board and Committees ESC Policies Statutes & Reports Press and Media ESC Press Office Press Releases ESC Congress ESC TV ESC Cardio Talk Information Our Offices Conference Facilities Jobs in Cardiology Terms & Conditions Update your cookie settings Follow us Need help? Help centre Contact us © 2025 European Society of Cardiology. All rights reserved. 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8758	https://ahdictionary.com/word/search.html?q=trowel	American Heritage Dictionary Entry: trowel HOW TO USE THE DICTIONARY To look up an entry in The American Heritage Dictionary of the English Language, use the search window above. For best results, after typing in the word, click on the “Search” button instead of using the “enter” key. Some compound words (like bus rapid transit, dog whistle, or identity theft) don’t appear on the drop-down list when you type them in the search bar. For best results with compound words, place a quotation mark before the compound word in the search window. guide to the dictionary THE USAGE PANEL The Usage Panel is a group of nearly 200 prominent scholars, creative writers, journalists, diplomats, and others in occupations requiring mastery of language. Annual surveys have gauged the acceptability of particular usages and grammatical constructions. The Panelists AMERICAN HERITAGE DICTIONARY APP The new American Heritage Dictionary app is now available for iOS and Android. THE AMERICAN HERITAGE DICTIONARY BLOG The articles in our blog examine new words, revised definitions, interesting images from the fifth edition, discussions of usage, and more. See word lists from the best-selling 100 Words Series! Find out more! INTERESTED IN DICTIONARIES? Check out the Dictionary Society of North America at trow·el (trouə l) Share: Tweet n. 1.A flat-bladed hand tool for leveling, spreading, or shaping substances such as concrete or mortar. 2.A small implement with a pointed, scoop-shaped blade used for digging, as in setting plants. tr.v.trow·eled, trow·el·ing, trow·elsortrow·elled or trow·el·ling To spread, smooth, form, or scoop with a trowel. [Middle English trowell, from Old French truele, from Late Latin truella, diminutive of Latin trua, ladle.] trowel·er, trowel·lern. (click for a larger image) trowel top to bottom: inside corner, pointing, and finishing trowels The American Heritage® Dictionary of the English Language, Fifth Edition copyright ©2022 by HarperCollins Publishers. All rights reserved. Indo-European & Semitic Roots Appendices Thousands of entries in the dictionary include etymologies that trace their origins back to reconstructed proto-languages. You can obtain more information about these forms in our online appendices: Indo-European Roots Semitic Roots The Indo-European appendix covers nearly half of the Indo-European roots that have left their mark on English words. A more complete treatment of Indo-European roots and the English words derived from them is available in our Dictionary of Indo-European Roots. American Heritage Dictionary Products The American Heritage Dictionary, 5th Edition The American Heritage Dictionary of Idioms The American Heritage Roget's Thesaurus Curious George's Dictionary The American Heritage Children's Dictionary CONTACT US Customer Service Make Me An Author Ebooks Help with Glose Reader ABOUT US Company Profile Leadership Team Corporate Social Responsibility HarperCollins Careers HarperCollins Imprints HarperGreen Social Media Directory Accessibility FOR READERS Browse Reading Guides FOR AUTHORS Submit a Manuscript Report Piracy Agent Portal MEDIA Publicity Contacts Press Room SERVICES HarperCollins Speakers Bureau Library Services Academic Services Desk & Exam Copies Review Copies OpenBook API Marketing Partnerships COVID-19 RESOURCES & PERMISSIONS Permissions for Adult Online Readings Permissions for Kids Online Readings SALES & RIGHTS Booksellers & Retailer Ordering HarperCollins Catalogs Permissions Subsidiary Rights Media Rights and Content Development GLOSE APP iPhone Android GLOBAL DIVISIONS HarperCollins US HarperCollins Canada HarperCollins Christian HarperCollins Australia HarperCollins India HarperCollins UK Terms of Use•Terms of Sale•Your Ad Choices•Privacy Policy•California Privacy Policy Do Not Sell My Personal Information Copyright 2022 HarperCollins Publishers All rights reserved. This website is best viewed in Chrome, Firefox, Microsoft Edge, or Safari. Some characters in pronunciations and etymologies cannot be displayed properly in Internet Explorer.
8759	https://www.droracle.ai/articles/239775/how-do-you-know-if-labs-show-agranulocytosis	How do you diagnose agranulocytosis based on lab results? ▼ How do you diagnose agranulocytosis based on lab results? Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: August 2, 2025 • View editorial policy Diagnosing Agranulocytosis Based on Laboratory Results Agranulocytosis is diagnosed when the absolute neutrophil count (ANC) falls below 500 cells/μL, which represents a severe and potentially life-threatening condition requiring immediate medical attention.1, 2 Key Laboratory Parameters for Diagnosis Absolute Neutrophil Count (ANC) Severe neutropenia/agranulocytosis: ANC < 500 cells/μL 1 Moderate neutropenia: ANC 500-999 cells/μL 1 Mild neutropenia: ANC 1000-1499 cells/μL 1 Normal range: ANC ≥ 1500 cells/μL 1 Calculation of ANC The ANC can be calculated using the following formula: ANC = Total WBC count × (% neutrophil "segs" + % neutrophil "bands") 1 Pattern of Neutrophil Decline In drug-induced agranulocytosis, particularly with medications like clozapine, a distinctive pattern may be observed: Continuous and rapid neutrophil count decline Mean time for neutrophils to fall from ANC > 2000 to ANC < 500 cells/μL is approximately 8.4 days (range 2-15 days) Characterized by a prolonged nadir and delayed recovery (4-16 days) 3 Differential Diagnosis When evaluating abnormal neutrophil counts, consider: Drug-induced agranulocytosis: Most common culprits include: Clozapine Trimethoprim-sulfamethoxazole Methimazole Dipyrone Diclofenac Antithyroid drugs Carbamazepine β-lactam antibiotics 2, 4 Infectious causes: Viral infections (e.g., Epstein-Barr virus) 5 Bacterial sepsis Hematologic disorders: Myelodysplastic syndromes Leukemia Aplastic anemia Additional Laboratory Tests to Consider When agranulocytosis is identified, additional testing should include: Complete blood count with differential to assess other cell lines Peripheral blood smear examination Bone marrow examination (if etiology remains unclear) Specific drug-dependent antibody testing (though rarely performed due to complexity and limited availability) 2 Monitoring Recommendations For patients on medications known to cause agranulocytosis (e.g., clozapine): Obtain baseline ANC before initiating treatment Regular monitoring of ANC according to medication-specific protocols For clozapine: weekly monitoring for first 6 months, then biweekly for next 6 months, then monthly after 12 months if ANC remains stable 6, 1 Clinical Implications and Management When agranulocytosis is detected: Immediately discontinue the suspected causative medication Initiate broad-spectrum antibiotics if fever or signs of infection are present Consider granulocyte colony-stimulating factor (G-CSF) in high-risk patients Implement infection control measures 4 Common Pitfalls in Diagnosis Failing to calculate ANC correctly (relying only on total WBC count) Not recognizing medication-induced neutropenia patterns Overlooking mild neutropenia that may rapidly progress to agranulocytosis Misinterpreting transient neutropenia as agranulocytosis Remember that drug-induced agranulocytosis carries a mortality rate of approximately 5% if not promptly recognized and managed, making early detection through laboratory monitoring crucial 4. References 1 Drug Official FDA Drug Label For clozapine (PO) FDA, 2025 2 Research Drug-induced immune neutropenia/agranulocytosis. Immunohematology, 2014 3 Research Distinctive pattern of neutrophil count change in clozapine-associated, life-threatening agranulocytosis. Schizophrenia (Heidelberg, Germany), 2022 4 Research Nonchemotherapy drug-induced neutropenia and agranulocytosis: could medications be the culprit? Journal of pharmacy practice, 2014 5 Research [Agranulocytosis as a complication of acute infectious mononucleosis]. Medicinski pregled, 1998 6 Guideline Clozapine Management Guidelines Praxis Medical Insights: Practical Summaries of Clinical Guidelines, 2025 Related Questions What is the diagnosis for a patient with leukopenia (white blood cell count of 2.43), normal red blood cell count (4.07), normal hemoglobin and hematocrit, normal platelets, neutropenia (neutrophil count of 1.37), lymphocytopenia (lymphocyte count of 0.79), and monocytopenia (monocyte count of 0.14)?What is agranulocytosis?What are the causes of neutropenia (low neutrophil count)?What is the management for antithyroid drug-induced neutropenia?What are the causes of leukopenia (low white blood cell count)?Does a common cold pose additional risks in a child with thalassemia?What are we monitoring for with regular complete blood counts (CBC) in patients on clozapine?What is the recommended dosage and treatment duration for cyclobenzaprine in treating acute skeletal muscle spasms?What is the comprehensive approach to soft tissue infections, including definition, natural history, epidemiology, pathophysiology, classification, clinical manifestations, diagnosis, prevention, and treatment?What is the starting dose for a continuous infusion of morphine for analgesia?What is the recommended approach for managing cyclobenzaprine (a muscle relaxant) withdrawal? 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8760	https://en.wiktionary.org/wiki/back-to-back	back-to-back - Wiktionary, the free dictionary Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main Page Community portal Requested entries Recent changes Random entry Help Glossary Contact us Special pages Feedback If you have time, leave us a note. 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. Donations Preferences Create account Log in [x] Personal tools Donations Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 EnglishToggle English subsection 1.1 Pronunciation 1.2 Adjective 1.2.1 Translations 1.3 Adverb 1.4 Noun back-to-back [x] 12 languages Eesti Ελληνικά 한국어 Italiano Lietuvių Malagasy မြန်မာဘာသာ Polski Português Suomi தமிழ் 中文 Entry Discussion Citations [x] English Read Edit View history [x] Tools Tools move to sidebar hide 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 Create a book Download as PDF Printable version In other projects Visibility Show translations Show quotations From Wiktionary, the free dictionary English [edit] Pronunciation [edit] IPA(key): /ˌbæk.təˈbæk/ Audio (Southern England):Duration: 2 seconds.0:02(file) Adjective [edit] back-to-back (not comparable) Sequential or consecutive. quotations▼Ruth and Gehrig hit back-to-back home runs. They sat through two back-to-back movies. 2011 February 12, Les Roopanarine, “Birmingham 1 - 0 Stoke”, in BBC‎:An injury-time goal from Nikola Zigic against an obdurate Stoke side gave Birmingham back-to back Premier League wins for the first time in 14 months. With one's back facing that of somebody else. (physics, by extension)Emerging in exactly opposite directions. quotations▼ 2009, R. Fiore, Alessandro Papa, Igor Ivanov, Jacques Soffer, Diffraction 2008: International Workshop on Diffraction in High Energy Physics, Amer Inst of Physics:Seen in laboratory frame the photon-jet pair is not any longer back-to-back and the energy balance is distorted. 2000, S. Söldner-Rembold, Photon '99: Proceedings of the International Conference on the Structure and Interactions of the Photon, Including the 12th International Workshop on Photon-Photon Collisions, Freiburg, Germany, 23-27 May 1999:This produces two back-to-back jets of hadrons that dominate the inclusive high energy cross-section. 2015, Thomas Schörner-Sadenius, The Large Hadron Collider: Harvest of Run 1, Springer, →ISBN, page 392:Photons at leading order are produced back-to-back with an associated parton with nearly identical transverse momenta. 1992, David Shaw Brown, A Comparison of High Transverse Momentum Direct Photon and Neutral Pion Events in Negative Pion and Proton-nucleus Collisions at 31.5 GeV Center of Mass Energy:In the rest frame of the π 0{\displaystyle \pi ^{0}} the photons emerge back-to-back isotropically. (of a house) Having a party wall at the rear. We lived in a row of back-to-back houses. (pokerslang)Synonym of wired(“being a pair in seven card stud with one face up and one face down”). Translations [edit] show ▼±sequential or consecutive [Select preferred languages] [Clear all] Finnish: peräkkäinen(fi), perättäinen(fi) Italian: consecutivo(it) Russian: оди́н за одним(ru)(odín za odnim), друг за другом(ru)(drug za drugom), поочерёдно(ru)(poočerjódno) Spanish: consecutivo(es) Turkish: arka arkaya(tr) Ukrainian: послідо́вний(poslidóvnyj) Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) show ▼±with one's back facing somebody else's back [Select preferred languages] [Clear all] Dutch: rug-aan-rug Finnish: seläkkäinen French: dos à dos(fr) Italian: dorso a dorso Navajo: ghąąjįʼ Ukrainian: спина́ в спи́ну(spyná v spýnu) Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) show ▼±(physics) emerging in exactly opposite directions [Select preferred languages] [Clear all] Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) show ▼±(house) having a party wall at the rear [Select preferred languages] [Clear all] Spanish: espalderaf Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) The translations below need to be checked and inserted above into the appropriate translation tables. See instructions at Wiktionary:Entry layout §Translations. show ▼±Translations to be checked‌: "adverbs or adjectives?" [Select preferred languages] [Clear all] Romanian: (please verify)spate în spate Turkish: (please verify)sırt sırta(tr) Add translation: More [x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective Noun class: Plural class: Transliteration: (e.g. zìmǔ for 字母) Literal translation: Raw page name: (e.g. 疲れる for 疲れた) Qualifier: (e.g. literally, formally, slang) Script code: (e.g. Cyrl for Cyrillic, Latn for Latin) Nesting: (e.g. Serbo-Croatian/Cyrillic) Adverb [edit] back-to-back (not comparable) Alternative form of back to back. quotations▼ 2021 December 15, Paul Clifton, “There is nothing you can do”, in RAIL, number 946, page 37:A ScotRail Driver: [...] A good friend of mine overshot two stations back-to-back a couple of years ago. He tried to stop at one station and slid by it. Tried to stop at the next station. He slid by that, too. Noun [edit] back-to-back (pluralback-to-backs) A house with a party wall at the rear. (oil industry) One of a pair of rig workers who are rostered on alternately. quotations▼ 2005, Paul Carter, Don't Tell Mum I Work on the Rigs, Crows Nest: Allen and Unwin, page 101:"He's been standing-by in town for the last week, talk to his back-to-back on the rig in the morning." ("Back-to-back" is the man on the rig who does your job when you're not there. Retrieved from " Categories: English 3-syllable words English terms with IPA pronunciation English terms with audio pronunciation English lemmas English adjectives English uncomparable adjectives English multiword terms English terms with usage examples English terms with quotations en:Physics en:Poker English adverbs English uncomparable adverbs English nouns English countable nouns en:Oil industry English reduplicated coordinated pairs Hidden categories: Pages with entries Pages with 1 entry Quotation templates to be cleaned Entries with translation boxes Terms with Finnish translations Terms with Italian translations Terms with Russian translations Terms with Spanish translations Terms with Turkish translations Terms with Ukrainian translations Terms with Dutch translations Terms with French translations Terms with Navajo translations Terms with Romanian translations Requests for review of Romanian translations Requests for review of Turkish translations This page was last edited on 24 June 2025, at 06:26. 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8761	https://www.math.ualberta.ca/~xinweiyu/314.A1.13f/314%20Sets%20and%20Functions.pdf	Sets and Functions Table of contents 1. Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.1. Relations between sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2. Operations that create new sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.3. Operations on arbitrary number of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2. Sets of real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.1. Subsets of R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Open sets and closed sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2. Properties of sets of real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Sup and Inf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Nested sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.3. Extended real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3. Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.1. Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.2. When functions meet sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Image and pre-image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Eﬀects of functions on set relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Eﬀects of functions on set operations . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.3. Composite functions and inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . 16 Composite function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Inverse function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 3.4. Functions with range R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3.5. Important functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Absolute value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Max and Min . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Odd and Even functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Monotone functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Periodic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 A few more words about the deﬁnition of A B . . . . . . . . . . . . . . . . . . . . . . . . . 20 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1 1. Sets. Deﬁnition 1. (Set) A set is a collection of objects. These objects are called “elements” or “members” of the set. If an object x is an element of a set A, we write x ∈A, or equivalently A ∋x. If an object is not an element of a set A, we write x A. There are two ways of deﬁning a set. • The ﬁrst is to list all elements: A {1, 2, 3}; B {1, {2, 3}} (1) Note that A and B are diﬀerent! Exercise 1. Explain why A and B are diﬀerent. On the other hand, we have {1, 2, 3} = {3, 2, 1} = {1, 2, 3, 3, 1} (2) that is the order of elements and repetition of elements do not matter. • The second is to deﬁne a set to be all objects satisfying one or more properties: ◦ Let A = {All the real numbers that satisfy \|x −1\| < 3}. Then 1 ∈B, 4 B. (3) ◦ Let B = {All natural numbers n such that there are integers solving xn + yn = zn}. Thanks to Andrew Wiles we now know that 1, 2 ∈C, 3, 4, C. (4) ◦ Let C = {x ∈B: x2 = 9}. Then 3 ∈D, −3 D. In general, a set deﬁned this way is written as A = {x O P(x)} (5) which reads “A is the set of all x such that the statement P(x) is true”. Exercise 2. Write the above sets A, B, C in the form {x O P(x)}. Make P(x) explicit. As seen above in the deﬁnition of C, we also use the notation A = {x ∈B O P(x)} (6) to mean “A is the set of all x in B such that P(x) is true.” This is useful when the set B is previously deﬁned and most of the x’s under consideration are elements of B. For example, in real analysis we often write deﬁnitions like A = {x ∈R O .}. Exercise 3. Sometimes a set is deﬁned as A = {f(x)O P(x)} (7) where f is a previously deﬁned function. Explain what this means. Remark 2. In mathematics sets usually have inﬁnitely many members. For such sets we have to take the second way. However, see the next remark. Remark 3. The above two methods of writing a set are based on the following two assumptions: 1. Axiom of extensionality: A set is determined by its elements, that is, two sets with same elements are the same set. 2 Sets and Functions 2. Axiom of intentionality: A set can be determined by one or more properties. That is we can write A = {x O P(x) is true} or simply A = {x O P(x)} (8) where P(x) is a mathematical statement involving x. These two axioms seem naturally true but unfortunately they lead to paradoxes. Fortunately, at the level of analysis, such paradoxes have little eﬀect. Exercise 4. (Russel’s paradox) Let A = {S O S S}. Show why this is a paradox. Example 4. (Empty set) There is a special set called “empty set”, denoted ∅, which is deﬁned as a set with no element. In other words, there is no object a satisfying a ∈∅, or equivalently, every object a satisﬁes a ∅. Exercise 5. Prove that there is exactly one “empty set”. Which axiom are you using in your argument? Example 5. There are several special sets of numbers that are so important that they have special letters assigned to them. • The set of natural numbers is denoted N; So N= {0, 1, 2, }. • The set of integers is denoted Z; So Z = { , −1, 0, 1, 2, }. • The set of rational numbers is denoted Q; • The set of real numbers is denoted R; • The set of complex numbers is denoted C. Note that in the following of this course we will use these standard symbols. 1.1. Relations between sets. Two sets A, B can have the following possible relations: • Subset. If every element of A is also an element of B, we say A is a subset of B, denoted A ⊆B. Using the logical statements in the last subsection, A ⊆B is deﬁned as x ∈A x ∈B. (9) For example N ⊆Z. • Equal. If A, B have exactly the same elements, we say A = B. For example, if A {x ∈R: x2 + 1 = 0}, then A = ∅. On the other hand, if B {x ∈C: x2 + 1 = 0}, then B = {i, −i}. A = B can be deﬁned as x ∈A x ∈B (10) which immediately gives A = B is equivalent to [(A ⊆B) and (B ⊆A)]. (11) When the relation between A, B are not obvious, showing A ⊆B and then B ⊆A is most likely the only way to prove A = B. • Proper subset. From the above deﬁnitions we know that if A = B, then A ⊆B. However, when we talk about subsets, often we do not mean this trivial situation. Thus we deﬁne “A is a proper subset of B”, denoted A ⊂B, if A ⊆B but A B. Sometimes it is also denoted as A ⊊B to emphasize these. How to prove set relations: • A ⊆B. To prove A ⊆B, we show every x ∈A also ∈B. More speciﬁcally, we show that an element x ∈A, taken arbitrarily, must also be in B. Example 6. Let A = {x > 0}, B = {x > −1}. Prove A ⊆B. 3 Proof. Let x ∈A. Then x > 0. This gives x > −1 which means x ∈B. Thus any x ∈A also ∈B which means A ⊆B. □ • A = B. To show this we need to show A ⊆B and B ⊆A. Example 7. Let A = {x O ex > 1} and B = {x > 0}, prove A = B. Proof. We ﬁrst show A ⊆B and then B ⊆A. ◦ A ⊆B. Take arbitrary x ∈A, we have ex > 1 = e0. This gives x > 0 so x ∈B. ◦ B ⊆A. Take arbitrary x ∈B, we have ex > e0 = 1. This gives ex > 1 so x ∈A. □ • A ⊊B. Show A ⊆B and A B. More speciﬁcally, ﬁrst prove “any x ∈A also ∈B”, then prove “there is y ∈B such that y A”. This is usually done through explicitly ﬁnding this element y. Example 8. Let A = {x > 0}, B = {x > −1}. Prove A ⊊B. Proof. We have already proved A ⊆B, therefore all we need to show is A B. To show this we need to ◦ either ﬁnd an element of A which is not an element of B, ◦ or ﬁnd an element of B which is not an element of A. Since A ⊆B, all elements of A are elements of B, so we try to ﬁnd an element of B which is not an element of A. This is easy: 0 ∈B but 0 A. □ Proposition 9. We have • ∅⊆A for any set A. • Let A be a set. If A ⊆∅, then A = ∅. • A ⊆B and B ⊆C then A ⊆C. Proof. • This is true because there is no element in ∅.1 • We need to show that A has no element. Assume the contrary, let x ∈A. Then x ∅. Therefore A⊆φ, contradiction. • Take arbitrary x ∈A. As A ⊆B, x ∈B. As B ⊆C, x ∈C. Thus we have: for any x ∈A, x ∈C which is exactly A ⊆C. □ 1.2. Operations that create new sets. Given two sets A, B, the following operations create new sets from A and B. • Union. The union of two sets A, B is the new set obtained from putting all elements of A and all elements of B together. We denote A ∪B {x O (x ∈A) ∨(x ∈B)}. (12) • Intersection. The intersection of A, B is the set of all common elements of A, B: A ∩B {x O (x ∈A) ∧(x ∈B)}. (13) • Set diﬀerence. The “set diﬀerence of B from A” is the set of all elements that are in A but not in B: A −B {x O (x ∈A) ∧(x B)}. (14) 1. W e need to show {x ∈∅} {x ∈A}. But x ∈∅is always false so the whole statement is always true no matter what A is. 4 Sets and Functions It is also called the “complement of B relative to A”. In some textbooks the notation A\B is also used. • Complement. Often all the sets relevent to our discussion are subsets of an “ambient set” X. For example in 314 almost all the sets we discuss are subsets of the ambient set R. In this case X −B is often denoted as Bc and called “complement of B”. Exercise 6. Prove the following: Let A, B be any set. A −A = ∅; A −∅= A; ∅−A = ∅; (15) A −B = A −(A ∩B); (16) Ac ∩Bc = (A ∪B)c; Ac ∪Bc = (A ∩B)c. (17) Proposition 10. The following are very useful in proving set relations. a) A ⊆C, B ⊆C A ∪B ⊆C; b) C ⊆A, C ⊆B C ⊆A ∩B; c) A ⊆B A ∩C ⊆B ∩C; d) A ⊆B A ∪C ⊆B ∪C. e) A ⊆B C −B ⊆C −A. Proof. a) We are given x ∈A x ∈C and x ∈B x ∈C (18) and need to show (x ∈A or x ∈B) x ∈C. (19) We only need to deal with the case x ∈A or x ∈B is true. Thus at least one of x ∈A,x ∈B is true. Say x ∈A is true. Then x ∈A x ∈C gives x ∈C is true. Similarly if x ∈B is true we also conclude x ∈C is true. Consequently if x ∈A or x ∈B is true then x ∈C is true, which is exactly (x ∈A or x ∈B) x ∈C. (20) b)–e) can be proved similarly and is omitted. □ Exercise 7. Prove b) – e). Theorem 11. (Properties of set operations) Let A, B, C be sets a) A ∩B ⊆A ⊆A ∪B. b) A ∪A = A; A ∩A = A. c) A ∪B = B ∪A; A ∩B = B ∩A. d) (A ∪B) ∪C = A ∪(B ∪C); (A ∩B) ∩C = A ∩(B ∩C). e) A ∩(B ∪C) = (A ∩B) ∪(A ∩C); A ∪(B ∩C) = (A ∪B) ∩(A ∪C). Proof. The proofs are quite similar, thus we will not give all the details but only prove the ﬁrst half of e). Recall that to prove A = B, all we need are A ⊆B and B ⊆A. • [(A ∩B) ∪(A ∩C)] ⊆A ∩(B ∪C). Since B ⊆B ∪C, application of Proposition 10 gives A∩B ⊆A∩(B ∪C). Similarly we have A∩C ⊆A∩(B ∪C). Applying Proposition 10 again, we have [(A ∩B) ∪(A ∩C)] ⊆A ∩(B ∪C). (21) 5 • A∩(B ∪C)⊆[(A∩B)∪(A∩C)]. This is a bit tricky. First consider the case A∩(B ∪C)=∅. In this case A ∩(B ∪C) ⊆any other set, so the claim holds. Otherwise, for every x ∈A ∩(B ∪C), by deﬁnition x ∈A and x ∈B or C. We discuss the two cases separately. If x ∈B, then x ∈A ∩B ⊆(A ∩B) ∪(A ∩C); If x ∈C, then x ∈A ∩C ⊆(A ∩B) ∪(A ∩C). □ Exercise 8. Prove the remaining claims of the theorem. Remark 12. From property d) in the above theorem we see that the intersection of three sets: A ∩B ∩C is well-deﬁned. Remark 13. A good tool to understand relations of sets is the Venn graph, or Venn diagram (the wiki page is a good enough reference). However, drawing a Venn graph is NOT A PROOF of set relations. It is just a help of visualizing what is going on and may inspire how the proof could be constructed. 1.3. Operations on arbitrary number of sets. The operations ∪, ∩can be generalized naturally to involve more than two or even inﬁnitely many sets. For example we can deﬁne A ∩B ∩C = {x: x ∈A and x ∈B and x ∈C}. (22) More speciﬁcally, let E = {Eα}α∈A be a collection of sets. Then i. The union of E is ∪α∈AEα {x: x ∈Eα for at least one α ∈A}; (23) ii. The intersection of E is ∩α∈AEα {x: x ∈Eα for all α ∈A}. (24) Example 14. Let Ea = {x ∈R: x < 1/a} for a ∈A = {a ∈R: a > 1}. Calculate (meaning: give the simplest description possible, preferably one single formula) ∩a∈AEa and ∪a∈AEa. Solution. • ∩a∈AEa. Solving such problems usually involve three steps. 1. Determine the answer. If a > 1, then 0 < 1/a. So {x: x ⩽0} ⊆Ea for all a. As a gets larger, 1/a gets smaller so we are pretty sure E {x: x ⩽0} should be the answer. What we need to do now is to show ∩a∈AEa = E. Remember our only method of showing equality of two sets? 2. ∩a∈AEa ⊆E. We need to show that if x ∈R satisﬁes x < 1/a for all a > 1, then x ⩽0. We prove by contradiction. Assume there is x > 0 such that x < 1/a for all a > 1. Set a = 1 + x x > 1. Then 1/a = x 1 + x < x as x > 0, contradiction.2 3. E ⊆∩a∈AEa. We need to show that every x ⩽0 satisﬁes x < 1/a for every a > 1. This is obvious as x ⩽0 < 1/a. • ∪a∈AEa. The procedure is similar, we get ∪a∈AEa = {x: x < 1}. 2. Think: Why not just set a = 1/x? Then 1/a = x ⩾x, contradiction! 6 Sets and Functions 2. Sets of real numbers. 2.1. Subsets of R. The most important sets to us are subsets of the set of real numbers, that is E ⊆R. Intervals. One special class of subsets of R is intervals: Deﬁnition 15. Let a, b be real numbers. A closed interval is a set of the form [a, b] {x ∈R: a ⩽x ⩽b}, [a, ∞) {x ∈R: a ⩽x} (25) (−∞, b] {x ∈R: x ⩽b}; (−∞, ∞) R. (26) An open interval is a set of the form (a, b) {x ∈R: a < x < b}, (a, ∞) {x ∈R: a < x} (27) (−∞, b) {x ∈R: x < b}; (−∞, ∞) R. (28) One can also deﬁne half-open, half-closed intervals: [a, b): ={x ∈R: a ⩽x < b}, (a, b] {x ∈R: a < x ⩽b}. (29) Remark 16. Note that R is both an open interval and a closed interval. Example 17. Write the following in interval notation: a) A {x ∈R: \|x −3\| ⩽1}; b) B {x ∈R: \|x −3\| > 5}. c) (1, 2)c. Solution. a) For A we have A = {x ∈R: 2 ⩽x ⩽4} so A = [2, 4]; b) B = {x ∈R: x > 8 or x < −2} so B = (−∞, −2) ∪(8, ∞). c) (1,2) = {x ∈R:1 < x < 2} so (1, 2)c = {x ∈R:x ⩽1 or x ⩾2}= {x ∈R: x ⩽1} ∪{x ∈R:x ⩾2} which equals (−∞, 1] ∪[2, ∞). Exercise 9. Prove the following: a) (a, b) ⊆(c, d) [(a ⩾c) ∧(b ⩽d)]; b) (a, b) ⊆[c, d] [(a ⩾c) ∧(b ⩽d)]; c) [a, b] ⊆[c, d] [(a ⩾c) ∧(b ⩽d)]; d) [a, b] ⊆(c, d) [(a > c) ∧(b < d)]. Remember: To prove you need to prove both and ! Open sets and closed sets. Using intervals we can deﬁne open and closed sets, which are crucial in real analysis. Deﬁnition 18. A set E ⊆R is open if for every x ∈E, there is an open interval (a, b) ⊆E such that x ∈(a, b). A set E ⊆R is closed if its complement Ec R −E is open. Remark 19. Traditionally, we say R and ∅are both open and closed. 7 Lemma 20. Open intervals are open, closed intervals are closed. Half-open, half-closed intervals are neither open nor closed. Proof. Let I be an open interval. For every x ∈I, we have x ∈I ⊆I. Therefore I is open. Next let a, b ∈R and [a, b] be an closed interval. Then we have [a, b]c = (−∞, a) ∪(b, ∞). Take any x ∈[a, b]c. Then there are two cases: • x ∈(−∞, a). Since (−∞, a) is an open interval, we have x ∈(−∞, a) ⊆[a, b]c. • x ∈(b, ∞). Similarly, (b, ∞) is an open interval so we have x ∈(b, ∞) ⊆[a, b]c. The other three cases are easier: • [a, ∞): We have [a, ∞)c = (−∞, b) open; • (−∞, b]: We have (−∞, b]c = (b, ∞) open; • (−∞, ∞): We have (−∞, ∞)c = ∅open. Finally we show that [a,b) and (a,b] are neither open nor closed. For [a,b), to see that it is not open, take x = a ∈[a, b). Then for every open interval (c, d) containing x = a, we have c < a and therefore c + a 2 ∈(c,d) but c + a 2 [a,b). To see that it is not closed, we consider [a,b)c=(−∞,a)∪[b,∞). Take x =b and argue similarly, we see that [a,b)c is not open. Therefore [a,b) is neither open nor closed. The proof for (a, b] is similar. □ Exercise 10. Prove by deﬁnition: a) (0, 1) ∪(1, 2) is open; b) [1, 2] ∪[3, 4] is closed. Lemma 21. If E is open, then Ec is closed; If E is closed then Ec is open. Proof. The second part is by deﬁnion. For the ﬁrst part, because (Ec)c = E, Ec is closed if E is open. □ Theorem 22. We have the following results about intersection and union of sets: a) The intersection of ﬁnitely many open sets is open; The union of open sets is open. b) The intersection of closed sets is closed; The union of ﬁnitely many closed sets is closed. Proof. We prove a). b) follows from a), Lemma 21, and De Morgan’s rule of set operations: (∩α∈AEα)c = ∪α∈AEα c; (∪α∈AEα)c = ∩α∈AEα c. (30) • Intersection of ﬁnitely many open sets. Denote these sets by E1, , En. We show that for every x ∈∩i=1 n Ei, there is (a, b) ⊆∩i=1 n Ei such that x ∈(a, b). As E1 is open, there is (a1, b1) ⊆E1 with x ∈(a1, b1); As E2 is open, there is (a2, b2) ⊆E2 with x ∈(a2, b2); Doing this for all Ei, we obtain (ai, bi) ⊆Ei such that x ∈(ai, bi). Now set a = max {a1, , an} and b = min {b1, , bn}. We claim that a < x < b. Since x∈(ai,bi), we have ai<x0. Then we know ∪x>0(1/x, ∞) is open. On the other hand, the intersection of inﬁnitely many open sets may be closed, for example ∩n∈N(−1/n, 1/n); The union of inﬁnitely many closed sets may be open, for example ∪n∈N h 1 n, 1 −1 n i . Of course it may also be half-open-half-closed. Remark 24. From the above theorem we see why it is a good idea to say R and ∅are both open and closed. Theorem 25. (Structure of open sets) Let E ⊆R be open. Then there are ai,bi ∈R, i∈N such that E = ∪i∈N(ai, bi). Proof. The proof is beyond the level of this course and is omitted. □ 2.2. Properties of sets of real numbers. Sup and Inf. For a set of ﬁnitely many real numbers, we often talk about its “largest” and “smallest” elements: its maximum and minimum. Deﬁnition 26. (max and min) Let A be a nonempty set of numbers. Then the maximum of A is an element x ∈A such that ∀y ∈A y ⩽x. (31) Similarly, the minimum of A is an element z ∈A such that ∀y ∈A y ⩾z. (32) The maximum and minimum of a set give us a rough idea of how “spread out” the set is. However, when the set is inﬁnite, maximum or minimum may not exist. Example 27. (max/min may not exist) Let A= n 1 −1 n:n ∈N o . Then minA=0, while maxA does not exist. • min A = 0. Checking deﬁnition of min we see that we need to prove 1. 0 ∈A; This is true as 0 = 1 −1 1 ∈A. 2. For any x∈A, x⩾0. Let x∈A be arbitrary. Then there is n∈N such that x=1−1 n ⩾0. Thus we have proved min A = 0. • max A does not exist. Assume the contrary, then there is amax ∈A. Then there is n0 ∈N such that amax= 1 −1 n0. Taking n > n0 we have amax< 1 −1 n ∈A, contradiction. Exercise 11. Let S ⊂R be a ﬁnite set. Prove that max S and min S exist. Exercise 12. Give an example of a inﬁnite set A ⊂R whose maximum and minimum both exist. Justify your answer. 9 To ﬁx this situation, we introduce the notion of supreme and inﬁmum, two numbers characterize a how spread out a set of (possibly inﬁnite) real numbers is. Deﬁnition 28. (sup and inf) Let A be a nonempty set of numbers. The supreme of A is deﬁned as sup A = min {b ∈R: b ⩾a for every a ∈A}. (33) If {b ∈R: b ⩾a for every a ∈A} = ∅, we write sup A = ∞; (34) The inﬁmum of A is deﬁned as inf A = max {b ∈R: b ⩽a for every a ∈A}. (35) If {b ∈R: b ⩾a for every a ∈A} = ∅, write inf A = −∞. (36) Example 29. (max/min may not exist) Let A = n 1 −1 n: n ∈N o . Then sup A = 1, inf A = min A = 0, while max A does not exist. • sup A = 1. We show two things: 1. ∀a ∈A, 1 ⩾a. Take any a ∈A. Then there is n ∈N such that a = 1 −1 n < 1. 2. ∀b ∈R such that b ⩾a for all a ∈A, b ⩾1. Since b ⩾a for all a ∈A, b ⩾1 −1 n for all n ∈N. Assume b < 1. Taking n > 1 1 −b leads to contradiction. • inf A, min A = 0. Omitted. • max A does not exist. Assume the contrary, then there is amax ∈A. Then there is n0 ∈N such that amax= 1 −1 n0. Taking n > n0 we have amax< 1 −1 n ∈A, contradiction. Deﬁnition 30. (Upper/lower bound) Let A ⊆R be nonempty. a∈R is said to be a upper bound of A if and only if ∀x ∈A x ⩽a; (37) b ∈R is said to be a lower bound of A if and only if ∀x ∈A x ⩾b. (38) Remark 31. Thus supA is “least upper bound” of A. Similarly, inf A is “greatest lower bound” of A. Remark 32. The advantage of supreme and inﬁmum is that they always exist. Exercise 13. Try to prove the above claim. Do you encounter any diﬃculty? Remark 33. In fact, the existence of sup and inf is part of the deﬁnition of the set of real numbers R. Exercise 14. More precisely, in the deﬁnition of R we only need to put in existence of one of the two. Prove a) If sup A exists for any A ⊆R, then inf B exists for any B ⊆R. b) If inf A exists for any A ⊆R, then sup B exists for any B ⊆R. 10 Sets and Functions sup and inf are generalizations of max and min . Proposition 34. Let A ⊆R. If maxA exists, then supA= maxA; Similarly, if minA exists, then inf A = min A. Proof. Let amax=maxA. Set B ={b∈R:b⩾a for every a∈A}. We need to show that amax=minB, that is 1. amax∈B. As amax= max A, we have amax⩾a for all a ∈A. Therefore a1 ∈B; 2. ∀b ∈B, amax⩽b. Take any b ∈B. Then b ⩾a for all a ∈A. In particular b ⩾amax. The proof for the inf/min part is similar. □ Exercise 15. Identify a real number r ∈R with the set Mr {x ∈R O x < r}. a) Prove that the sets correspond to sup A and inf A are ∪r∈AMr, ∩r∈A Mr respectively. (39) b) What sets correspond to r1 ± r2? Justify your answer. c) What set corresponds to the product r1 r2? Justify your answer. (Be careful about sign!) d) What set corresponds to 1/r? Theorem 35. (Monotone property) Suppose A ⊆B are nonempty subsets of R. Then a) sup B ⩾sup A. b) inf B ⩽inf A. Proof. We prove a) and leave b) as exercise. If sup B = ∞, then sup B ⩾sup A holds; If sup B ∈R, then by deﬁnition we have sup B ⩾b for every b ∈B sup B ⩾a for every a ∈A (40) because A ⊆B. By deﬁnition of sup A we conclude that sup B ⩾sup A. □ Exercise 16. Are there simple formulas of “sup” and “inf” for A ∩B, A ∪B, A −B? Justify your answers. Nested sets. Deﬁnition 36. (Nested sets) A sequence of sets {In}n∈N is said to be nested if I1 ⊇I2 ⊇ (41) Theorem 37. (Nested interval) If In = [an, bn] with an, bn ∈R is nested, then ∩n=1 ∞In is not empty. • Discussion. First understand what is the implication of I1 ⊇I2 ⊇ . This means a1 ⩽a2 ⩽ ⩽an ⩽ ⩽bn ⩽ ⩽b2 ⩽b1. (42) Next to understand the proof, we think about the case where there are only ﬁnitely many intervals [a1,b1]⊇[a2,b2]⊇ ⊇[an,bn]. Then clearly any x ∈[an,bn] would belong to ∩k=1 n In, for the following reason: an = max {a1, , an} ⩾any ak; bn = min {b1, , bn} ⩽any bk. (43) 11 Now return to the proof of the theorem. We have inﬁnitely many intervals so it may not be possible to pick the largest an or the smallest bn. But we always have sup an and inf bn. Proof. Let A = {an} and B = {bn}. Consider a = sup A and b = inf B. We prove • a ⩽b. Assume the contrary, that is b < a. Since a = sup A, by deﬁnition b is not an upper bound of A, there is there is n0 ∈N such that an0 > b. As the intervals are nested, we have an0 ⩽bn for all n ∈N that is an0 is a lower bound of B. Now b = inf B by deﬁnition implies b ⩾an0. Contradiction. • [a,b]⊆∩n=1 ∞In. We show that for any n∈N, [a,b]⊆[an,bn]. To show this we only need a⩾an and b ⩽bn. Both follow directly from a = sup A and b = inf B. □ Exercise 17. a) Give a proof using lim an and lim bn after we have discussed limits. b) If furthermore limn ∞bn −an = 0, show that ∩n=1 ∞ In consists of a single point. Remark 38. • an, bn ∈R is necessary. Otherwise we can take In = [an, ∞) with an ∞which leads to ∩n=1 ∞In ∅. • It is also necessary that the intervals are closed. Counter-examples are In = (0, 1/n), or In = (0, 1/n]. Theorem 39. (Approximation of sup and inf) Let A ⊆R with supA,inf A∈R. Then for every ε > 0, there are a, b ∈A such that sup A −a < ε; b −inf A < ε (44) Proof. We prove the sup case and left the inf case as exercise. Assume the contrary. Then there is ε0 > 0 such that for all a ∈A, sup A −a ⩾ε0. Now set asup sup A −ε0/2. We have asup> a for all a ∈A but asup< sup A. Contradiction. □ Remark 40. By setting ε=1/n, we can obtain a sequence {xn} with xn ∈A such that xn supA (or inf A). However note that xn may not be diﬀerent elements from one another. For example when A is ﬁnite, we basically will have to take the sequence xn = amax. Exercise 18. Can we deﬁne “sup”, “inf” for sets in Q, Z or N? Do “sup”, “inf” always exist in subsets of these ambient sets? 2.3. Extended real numbers. It is often convenient to add to R two elements +∞, −∞to form the set of “extended real numbers”. Most of the arithmetics on R can be carried over: x + ∞= ∞, x −∞= −∞, x ∈R (45) x · ∞= ∞, x · (−∞) = −∞, x > 0 (46) x · ∞= −∞, x · (−∞) = ∞, x < 0 (47) ∞+ ∞= ∞, −∞−∞= −∞ (48) ∞· ∞= (−∞) · (−∞) = ∞∞· (−∞) = (−∞) · (∞) = −∞ (49) and the following are not involved: ∞−∞or 0 · (+∞) or 0 · (−∞). 12 Sets and Functions 3. Functions. Functions used to be deﬁned through formulas. However, as people gain more understanding of sciences and mathematics, it became necessary to study functions that cannot be described by formulas. The current understanding is to deﬁne functions through what they do: How they map the input value to the output value. 3.1. Deﬁnitions. Deﬁnition 41. (Functions) A function f:A B is a rule assigning to each element a∈A exactly one element in B, this element is denoted f(a). We call A the domain of the function and B the range of the function. Example 42. Let A = {1, 2, 3, 4, 5}, B = {a, b, c, d, e}. Let the rule be given by 1 a (50) 2 c (51) 3 b (52) 4 d (53) 5 d (54) then this is a function with domain A and range B. On the other hand, the rule 1 a (55) 1 b (56) 2 c (57) 3 e (58) fails to be a function because • It does not assign an element in B to every a ∈A; • For some a ∈A it assigns more than one element in B. Remark 43. It is important to keep in mind that a function is a triplet: Domain A, Range B, Rule f. Changing any one of the three leads to a diﬀerent function. Rigorously speaking, sinx over (−1, 1) and sin x over (−2, 2) are two diﬀerent function, they just happen to coincide over (−1, 1). In real analysis this may seem like some annoying triviality, but this understanding is extremely important when studying Complex Analysis, Functional Analysis, and many other higher level analysis courses. Deﬁnition 44. (Sequence) A sequence is a function with domain N. Deﬁnition 45. (Restriction of functions) Let f:A B be a function. Let C ⊂A. Then we can deﬁne a new function, called “restriction of f on C” by keeping the same rule but change the domain to C. Example 46. Consider the function f: R R deﬁned through x sin x. Then its restriction on N is the sequence {sin n}. 3.2. When functions meet sets. Image and pre-image. 13 We can consider the eﬀect of a function on subsets of either the domain or the range. Deﬁnition 47. (Image) Let A, B be sets and let f: A B be a function. The image of a subset S ⊆A under f is deﬁned as f(S) {b ∈B: ∃a ∈S such that f(a) = b} or simply f(S) {f(a) O a ∈S} (59) The image of f is deﬁned as the special case Image(f) f(A). (60) Example 48. Let A = B = R, S = {a ∈R: 0 < a < π}, f = sin . Then we have Image(f) = {x ∈R: −1 ⩽x ⩽1}. (61) f(S) = {x ∈R: 0 < x ⩽1}. (62) From this example we see that the image of f may only be a proper subset of B. When studying functions, it is often important to study those a ∈A such that f(a) has certain property. More precisely, we need a notation for those a ∈A such that f(a) belongs to a certain subset S ⊆B. Deﬁnition 49. (Pre-image) Let A, B be sets and let f: A B be a function. The pre-image of a subet S ⊆B is deﬁned as f −1(S) {a ∈A O f(a) ∈S}. (63) Remark 50. When S ={s} has exactly one element, f −1(S) is called a “level set” of the function f. Example 51. Let A = B = R, f = sin , S = {1}. Then f −1(S) = 2 k π + π 2, k ∈Z. (64) If we let S = {2}, then clearly f −1(S) = ∅. (65) We see that the pre-image of a single element may not be a single element, it may also be empty or contain more than one element. Exercise 19. (Functions and Sets) Let X be an ambient space. For any A ⊆X deﬁne a function fA: X R by fA(x) = 1 if x ∈A and 0 otherwise. a) Prove that fA∩B = fA · fB. (66) b) Use a) to prove A ∩(B ∩C) = (A ∩B) ∩C. (67) c) Find similar formulas for fA∪B, fA−B, fAc. (68) d) Use the above to prove other set relations. Eﬀects of functions on set relations. Let f: X Y be a function. Let A, B be two subsets of the domain X. Then there are three possible relations: A ⊆B, A = B, A ⊂B. (69) 14 Sets and Functions We can discuss what we can conclude for the relation between f(A) and f(B). Similarly we can study what the eﬀect of f −1 is on subsets S, T ⊆Y , in each of the following situations: S ⊆T , S = T , S ⊂T. (70) Lemma 52. Let f: X Y be a function. Let A, B ⊆X and S, T ⊆Y. Then the following holds. a) If A ⊆B then f(A) ⊆f(B). b) If S ⊆T then f −1(S) ⊆f −1(T). Proof. Left as exercise. □ Exercise 20. Prove Lemma 52. Remark 53. Note that A ⊂B (S ⊂T) does not imply f(A) ⊂f(B) (f −1(S) ⊂f −1(T)). The construction of counterexamples are left as exercises. Eﬀects of functions on set operations. Theorem 54. Let f: X Y be a function. Let A, B ⊆X and S, T ⊆Y. Then a) f(A ∩B) ⊆f(A) ∩f(B). b) f(A ∪B) = f(A) ∪f(B). c) f(A −B) ⊇f(A) −f(B). d) f −1(S ∩T) = f −1(S) ∩f −1(T). e) f −1(S ∪T) = f −1(S) ∪f −1(T). f ) f −1(S −T) = f −1(S) −f −1(T). Proof. a) Since A ∩B ⊆A, Lemma 52 gives f(A ∩B) ⊆f(A); Application of the same lemma to A ∩B ⊆B gives f(A ∩B) ⊆f(B). Therefore f(A ∩B) ⊆f(A) ∩f(B). b) We need to show f(A ∪B) ⊆f(A) ∪f(B) and f(A) ∪f(B) ⊆f(A ∪B). • f(A ∪B) ⊆f(A) ∪f(B). Take any y ∈f(A ∪B). Then there is x ∈A ∪B such that y = f(x). Now x ∈A ∪B has two cases: x ∈A and x ∈B. In the ﬁrst case we have y ∈f(A) and in the second we have y ∈f(B). Therefore x ∈A ∪B implies y ∈f(A) ∪f(B). So f(A ∪B) ⊆f(A) ∪f(B). • f(A) ∪f(B) ⊆f(A ∪B). Since A ⊆A ∪B, application of Lemma 52 gives f(A) ⊆ f(A ∪B). Application of the same lemma to B ⊆A ∪B gives f(B) ⊆f(A ∪B). Therefore f(A) ∪f(B) ⊆f(A ∪B). c) Left as exercise. d) We need to show f −1(S ∩T) ⊆f −1(S) ∩f −1(T) and f −1(S) ∩f −1(T) ⊆f −1(S ∩T). • f −1(S ∩T) ⊆f −1(S) ∩f −1(T). As S ∩T ⊆S, application of Lemma 52 gives f −1(S ∩T) ⊆f −1(S). The same lemma applied to S ∩T ⊆T givesf −1(S ∩T) ⊆ f −1(T). Therefore f −1(S ∩T) ⊆f −1(S) ∩f −1(T). 15 • f −1(S)∩f −1(T)⊆f −1(S ∩T). Take any x∈f −1(S)∩f −1(T). As f −1(S)∩f −1(T)⊆ f −1(S), we have x ∈f −1(S) so f(x) ∈S. On the other hand, f −1(S) ∩f −1(T) ⊆ f −1(T) gives f(x) ∈T. Therefore f(x) ∈S ∩T which means x ∈f −1 (S ∩T). Thus ends the proof for f −1(S) ∩f −1(T) ⊆f −1(S ∩T). e) Left as exercise. f) Left as exercise. □ Exercise 21. Prove c), e), f) of the above theorem. 3.3. Composite functions and inverse functions. Composite function. Deﬁnition 55. (Composite function) Let f: A B and g: C D. If B ⊆C, we can deﬁne the composite function g ◦f: A D through ∀x ∈A (g ◦f)(x) g(f(x)). (71) Exercise 22. Let f(x) = sin x, g(x) = x2, show that f ◦g g ◦f. Exercise 23. Prove that if f ◦(g ◦h) is well-deﬁned, so is (f ◦g) ◦h and furthermore f ◦(g ◦h) = (f ◦g) ◦h. Inverse function. Given a function f: A B, it is often necessary to understand the rule relating f(a) back to a, that is is we would like to ﬁnd a function g: B A such that g(f(a)) = a for every a ∈A; f(g(b)) = b for every b ∈B. (72) Such a function is called an inverse function of f. However as we have seen, for general function this is not possible. Now let’s see what extra conditions we need. Remark 56. Note that in this course f −1 always denotes the pre-image, not the inverse function. More precisely, given f:A B, f −1 is a function from the set of subsets of B to the set of subsets of A, while the inverse function g, if it exists, is a function from B to A. Deﬁnition 57. (one-to-one,onto,bijection) Let A, B be sets and f: A B a function. We say f is one-to-one if whenever f(a1) = f(a2), we have a1 = a2. We say that f is onto if for every b ∈B there exists a ∈A such that f(a) = b. We say f is a bijection if it is both one-to-one and onto. Example 58. Consider the following functions: A = B = R, f1(x) = 2 x + 4; f2(x) = arctan x; f3(x) = sin x; f4(x) = 2 x3 + x2 + 12 x + 4. (73) Then f1 is one-to-one and onto, f2 is one-to-one but not onto, f3 is neither, f4 is onto but not one-to-one. We give proof to the claim about f2 and leave others as exercise. Proof of f2 being one-to-one but not onto. To prove that f2 is one-to-one, we need to show that whenever f2(a1) = f2(a2), we must have a1 = a2. One way to show this is through Fundamental Theorem of Calculus: f2(a2) −f2(a1) = Z a1 a2 f2 ′(x) dx = Z a1 a2 1 1 + x2 dx. (74) Now if a2>a1, from the above we have f2(a2)> f2(a1); If a2<a1, we have f2(a2)< f2 (a1). Therefore if f2(a1) = f2(a2), we must have a1 = a2. 16 Sets and Functions To show that f2 is not onto, all we need is a counter-example. That is all we need is one b∈B =R such that there is no a ∈A = R such that f2(a) = b. This is easy. For example b = 3. Note 59. Is the following “proof” of f2 being one-to-one correct? If arctan a1 = arctan a2, taking tan gives a1 = a2. Theorem 60. f has an inverse function if and only if f is a bijection. Proof. We need to prove f has an inverse function f is a bijection. (75) Recall that we need to prove and . • . Let g be an inverse function of f. As for every b ∈B, f(g(b)) = b, f is onto. Now we show f is one-to-one. Let f(a1)= f(a2). As g is a function, it maps f(a1)= f(a2) to a single element in a: a1 = g(f(a1)) = g(f(a2)) = a2. (76) • . Since f is a bijection, for every b∈B there is a unique a∈A such that f(a)=b. We deﬁne g(b) = a. (77) Thus automatically g(f(a)) = a. On the other hand, f(g(b)) = f(a) = b. So g is an inverse function. □ Exercise 24. Let f: A B be a function. a) Assume there is a function g: B A such that (g ◦f)(x) = x for all x ∈A. Prove that f has an inverse function and furthermore it is exactly g.3 b) Formulate a similar claim assuming existence of g with (f ◦g)(y) = y for all y ∈B, and prove your claim. 3.4. Functions with range R. For functions with range R, that is for functions taking real values, we can talk about how “spread out” the image is through the notion of supreme and inﬁmum of sets of real numbers. Deﬁnition 61. (Sup and Inf of functions) Let f: E R be a function. We deﬁne sup x∈E f(x) sup f(E); inf x∈Ef(x) inf f(E). (78) Example 62. supx∈(−1,1)x2 = 1, infx∈(−1,1)x2 = 0. Theorem 63. (sup and inf under operations of functions) Let f , g be functions with domains containing E ⊆R. Let c ∈R be a positive number. Then a) supx∈E (c f) = c supx∈E f; infx∈E (c f) = c infx∈Ef; b) supx∈E (−f) = −infx∈Ef; infx∈E (−f) = −supx∈Ef; c) supx∈E (f + g) ⩽supx∈Ef + supx∈Eg; d) infx∈E (f + g) ⩾infx∈Ef + infx∈Eg. The above holds even when the supreme/inﬁmum is ∞or −∞. 3. This is an example of the following fact in group theory: if a has a left (or right) inverse, then it has an inverse. 17 Proof. We only prove b), c) here. Other cases are left as exercise. • b). We show supx∈E (−f)= −infx∈Ef. Once this is done setting g =−f gives the other half. Let a = infx∈Ef (maybe −∞). We need to show 1. −a ⩾−f(x) for all x ∈E. Since a = infx∈Ef, by deﬁnition a ⩽f(x) for all x ∈E. Therefore −a ⩾−f(x) for all x ∈E. 2. For any b ∈R satisfying b ⩾−f(x) for all x ∈E, we have b ⩾−a. Since b ⩾−f(x) for all x∈E, we have −b⩽f(x) for all x∈E. Since a=infx∈Ef, a⩾−b. Therefore b⩾−a. Note that the above argument still holds when a = −∞. • c). Denote a = supx∈Ef , b = supx∈Eg, we need to show that a + b ⩾f(x) + g(x) for all x ∈E. If one of a, b is ∞, then we have a + b = ∞⩾f(x) + g(x) for all x ∈E.4 If both a, b ∈R, take any x ∈E. We have a = supx∈Ef ⩾f(x) and b = supx∈Eg ⩾g(x). Consequently a + b ⩾f(x) + g(x). □ Exercise 25. Prove a), d). Example 64. The inequalities in c),d) may hold strictly. The reason is that the “peak” of f and the “peak” of g may not be at the same location. For example, take f = 1 0 < x < 1 0 elsewhere and g = 1 1 ⩽x < 2 0 elsewhere , we have supx∈Rf = supx∈Rg = 1, but supx∈R(f + g) = 1 < 1 + 1 = 2. Remark 65. For more on functions, see [Sib09] Chapters 3, 4. 3.5. Important functions. Absolute value. Deﬁnition 66. (Absolute value function) The function \|·\|: R R is deﬁned through \|x\| x if x ⩾0 −x if x < 0 . (79) Exercise 26. Let x, y ∈R. Prove the following a) \|x\| = 0 x = 0; b) \|x y\| = \|x\| \|y\|; c) \|\|x\| −\|y\|\| ⩽\|x + y\| ⩽\|x\| + \|y\|. Exercise 27. Let x1, , xn ∈R. Prove \|x1 + + xn\| ⩽\|x1\| + \|xn\|. (80) Max and Min. Exercise 28. Let x ∈R. Prove that \|x\| = max {x, −x}. (81) Exercise 29. Let x, y ∈R. Prove that max {x, y} = x + y 2 + \|x −y\| 2 ; min{x, y} = x + y 2 −\|x −y\| 2 . (82) Odd and Even functions. 4. Note that by deﬁnition sup can only be real number or ∞, while inf can only be real number or −∞. 18 Sets and Functions Deﬁnition 67. A function f: E ⊆R R is odd if and only if ∀x ∈E f(x) = −f(−x) (83) It is even if and only if ∀x ∈E f(x) = f(−x). (84) Remark 68. From the above it is clear that E cannot be arbitrary, it has to satisfy (x ∈E) (−x ∈E). (85) Monotone functions. Deﬁnition 69. A function f: E ⊆R R is • increasing if and only if x, y ∈E, x < y f(x) ⩽f(y); (86) • strictly increasing if and only if x, y ∈E, x < y f(x) < f(y); (87) • decreasing if and only if x, y ∈E, x < y f(x) ⩾f(y); (88) • strictly decreasing if and only if x, y ∈E, x < y f(x) > f(y); (89) • monotone if it is either increasing or decreasing. Exercise 30. Prove that if f is strictly increasing, then it is one-to-one. What if f is only increasing? Periodic functions. Deﬁnition 70. A function f: R R is said to be periodic, if and only if there is L > 0 such that f(x) = f(x + L) (90) holds for all x ∈R. Such L is called a “period” of f. Exercise 31. Prove that a periodic function f always have inﬁnitely many periods. Remark 71. Often, among the inﬁnitely many periods of f, there is a smallest one. In this case, when we talk about “the period of the function f”, we are referring to this smallest period. Exercise 32. What are the periods for sin x, cos 3 x, tan 5 x? Exercise 33. Let f: R R be a periodic function and deﬁne the set A to be its periods, that is A {L > 0 O f(x) = f(x + L) for all x ∈R}. (91) Let T inf A. a) Prove that if T > 0, then A = {n T O n ∈N}, that is any other period is a multiple of T. b) Consider the Dirichlet function D(x): = 1 x ∈Q 0 x Q . (92) Prove that it is periodic but without smallest period. 19 A few more words about the deﬁnition of A B. The deﬁnition for the logical statement A B as false only when A is true and B is false is a bit puzzling. The following may help.5 Consider the claim If it rains, then I will bring an umbrella. So here A = it rains, and B = I bring an umbrella. It is natural to write the above as A B. Now we consider, among the following four situations, which makes the claim false: 1. A true and B true: It rains, and I brought an umbrella; 2. A true and B false: It rains, and I didn’t bring an umbrella; 3. A false and B true: It didn’t rain, but I still brought an umbrella; 4. A false and B false: It didn’t rain, and I didn’t bring an umbrella. I hope everyone agrees that the only situation that makes the original claim false is 2. Bibliography [Sib09] Thomas Q. Sibley. Foundations of Mathematics. John Wiley & Sons, 2009. 5. If anyone has better way of making sense of this deﬁnition, please either let me know or post on piazza or tell other students. 20 Sets and Functions
8762	https://community.coda.io/t/how-does-and-and-or-logic-actually-work-when-used-together-in-a-formula/34632	Skip to main content How does AND and OR logic actually work when used together in a formula? Ask the Community You have selected 0 posts. select all cancel selecting 3 Oct 2022 1 / 14 Oct 2022 Jan 2023 Fran_Vidicek Oct 2022 This is what I thought is going on but it can’t be true: I thought the conditions (red circles with numbers) that are surrounding the OR function need to have at leas one true. so for example, either condition 1 or 2 need to be true in order to show rows that the true condition is true for. And if the condition 1 is true, it wont even check if the condition 2 is true or not, it won’t matter. It will just move on to the next condition. In this example the next is another set of conditions: 3 and 4. The same goes for them, but between 1 and 2, and 3 and 4 is AND function. And AND function outputs true only if both sets of conditions output true. Therefore Set 1 (condition 1 & 2) = true AND = true Set 2 (condition 3 & 4) = true but if: Set 1 (condition 1 & 2) = true AND = false Set 2 (condition 3 & 4) = false Now when we come to the next set and the next AND function, does that AND function require this entire thing… Set 1 (condition 1 & 2) = true AND = true Set 2 (condition 3 & 4) = true …to be true and the next set of conditions (5 & 6) to be true. Or how does it work? image1265×1165 165 KB If the condition 9 is false, it won’t show any tasks at all no matter if everything before is true? I’m sure someone can explain this to me in very simple terms but I can’t figure it out, I just wan to know how a program would go trough this step by step. I would appreciate it very much. 6 3 3 Christian_Antonsen Oct 2022 Line with number 7 should start with a parantheses? (_Team responsible,…,.,. That way 7 and 8 would be correct… I might be wrong, but that’s a first guess from me… Fran_Vidicek Oct 2022 That is a good observation, I didn’t think about parentheses, but if I put it there it gives an error “missing close parenthesis” . So the formula itself works fine, but I have trouble understanding it because the number 9 should be false for a lot of rows but they still show up. So how do they show up if the expression/condition (I don’t know how to call it) is outputs false and it’s right after the AND function. So My guess is that both 9 and the set of 7 and 8 need to output True for a row to be shown. Christian_Antonsen Oct 2022 Then try removing the parentheses at the start (beginning) of 8. I believe you still need the one at the beginning of 7. Fran_Vidicek Oct 2022 If I remove it from 8 it looks fine and outputs the same number of rows. But that’s not even my issue. I just want to understand how this actually works. How can a row be displayed if the condition is false… Christian_Antonsen Oct 2022 I am new to coda.io, but have some (just a littlebit) experience with programming in general. Going after the parentheses was my first idea, simply by experience, those things are so easy to overlook. My best suggestion for you is to first copy the entire formula, and paste it in a texteditor. Then start by adding one layer at a time to the formula, making each step work, one by one, until you managed to fix the entire formula. By first hand, your formula does seem slightly complicated, so doing it this way may help both in understanding, and also to troubleshoot for errors. In simple basic terms: AND → Both criterias must be met, otherwise it gives false. OR → One or the other or both must be met, otherwise it gives false. When you look closely in each filter pair that you group together with the logical operator OR / AND, I believe you will manage to see how it goes. This way, checking your filterpair one by one, you will see weather or not you got the logic correct in your formula. I hope this can be of some help for you. louca.developer Oct 2022 Is there something that leads you to believe the formula is not working as expected? From glancing it seems correct. What you are probably looking for is the associativity of operators in CFL, i.e. whether an expression like a AND b AND b is parsed as (a AND b) AND c, or as a AND (b AND c). The former is called left-associative, the latter is right-associative. I would imagine that CFL, like most other established programming grammars, uses left-associativity. In any case, it shouldn’t matter at all since both are equivalent. The only case it would matter is if you were depending on a, b, and c to be evaluated in some specific order, in which case you shouldn’t. Fran_Vidicek Oct 2022 No there is nothing wrong with the formula. That’s exactly what I’m looking for; the associativity of operators. So based on that, would this be a correct order of how things go: image2031×573 54 KB This should be the exact replica of my filter formula but in a simpler format. There should be 3 bigger sets of conditions: a - d, e - g, h - i and j Is that right? If it is, now everything makes sense. Thank you! Paul_Danyliuk Community Champion Oct 2022 Hey — without reading too much here’s the overall answer on the associativity of operators. AND — it’s like multiplication (actually it’s called logical multiplication). Just like 0x0=0, 0x1=0, 1x0=0, and only 1x1=1, AND is only true if both are true. OR — it’s called logical addition. Only 0+0=0, otherwise 1+0=1, 0+1=1, and logically 1+1 also gives 1. The order of operation applies just the same way as with arithmetic multiplication and addition. Unless there are brackets, AND is evaluated first, OR second: a AND b OR c AND d equals (a AND b) OR (c AND d) There’s also short-circuiting: a AND b AND c AND d... will stop calculating as soon as it finds the first false. Analogically, a OR b OR c OR d will stop calculating as soon as it finds the first true.\| P.P.S. Wait, the formula in the original post looks a lot like something in my starter template (: P.S. Each of the a b c etc can be a boolean expression on its own. Also when unsure, use brackets (). Fran_Vidicek Oct 2022 Is the following true? (a OR b) AND (c OR d) Equals a OR b AND c OR d Also this: (a OR b) AND c) OR (d AND e) ADN f Equals a OR b AND c OR d AND e AND f Let’s see if I get this right: It would calculate whether… b AND c = true d AND e = true e AND f = true Then it if it’s all true, it would calculate OR operators: a OR b = true c OR d = true ← This is confusing because it would make more sense if it checked whether instead of c it checked everything that is in brackets before it, and if that’s false then check if d is true. ← If this is true, then how would you type (a OR b) AND c) OR (d AND e) ADN f without any brackets P.S. Only learning from the best Paul_Danyliuk Community Champion Oct 2022 No. Just like (1 + 2) (3 + 4) is not equal 1 + 2 3 + 4 and ((1 + 2) 3) + (5 6) 7 is not equal to 1 + 2 3 + 5 6 7 It’s not the associativity in play but the order of operations. Fran_Vidicek Oct 2022 Got it! So It cannot be written without brackets because it’s nested. Paul_Danyliuk Community Champion Oct 2022 Yes, you use brackets to control the order of operations Here’s the full documentation for JS. Coda is written in JS so it follows pretty much the same order (it’s only that there are fewer operators in Coda, e.g. no bitwise operators) Here you can learn that e.g. equality has precedence over logical operators, i.e. a = b AND c = d is evaluated as (a = b) AND (c = d), not a = (b AND c) = d 3 months later Closed on Jan 16, 2023 This topic was automatically closed 90 days after the last reply. New replies are no longer allowed. Related topics Topic list, column headers with buttons are sortable. \| Topic \| Replies \| Views \| Activity \| \| Changing Filter Order & interaction between “and” and “or” Ask the Community \| 5 \| 480 \| Jul 2022 \| \| Filter Help - Lookup with Condition Ask the Community \| 7 \| 961 \| Nov 2021 \| \| Issue with filtering because of full evaluation of AND operator Ask the Community \| 4 \| 387 \| Sep 2021 \| \| Adding Filter formula is confusing \| 1 \| 606 \| Jun 2020 \| \| Filtering using OR logic? Ask the Community \| 3 \| 660 \| Jan 2019 \|
8763	https://www.keyence.com/products/3d-measure/cmm/applications/in-process-inspection.jsp	To use all available functions on this website, JavaScript must be enabled in your browser. View More Products CMM (Coordinate Measuring Machine) In-Process Inspection Before any product leaves manufacturing and enters the hands of a consumer, it must be inspected. Without inspections, quality control wouldn’t exist and unreliable products would be rampant. Inspections ensure that parts are measured correctly for assembly, that no defects would harm the whole product or consumer, and that products are processed uniformly. To achieve optimal cost-efficiency and effectiveness, manufacturers rely on in-process inspection. In-process quality inspection involves examining components during the production process rather than waiting until the final stage before distribution. 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8764	https://physics.stackexchange.com/questions/191949/velocity-of-rocket-exhaust	momentum - Velocity of Rocket Exhaust - Physics Stack Exchange Join Physics 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 Velocity of Rocket Exhaust Ask Question Asked 10 years, 3 months ago Modified10 years, 3 months ago Viewed 12k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. i recently learned a bit of rocket propulsion.It wasn't much complex but was explained in simple terms.The only problem i had understanding it was that in calculating the thrust of rocket the velocity of the exhaust was taken relative to the rocket.My problem is : Shouldn't the velocity of the exhaust be taken as the relative to earth.In all the previous examples we had done so , why don't we use the velocity relative to the rocket.Thanks PS:A simple explanation would be much appreciated momentum rocket-science propulsion Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications asked Jun 30, 2015 at 7:16 BatwayneBatwayne 313 3 3 gold badges 6 6 silver badges 11 11 bronze badges 3 The fuel is moving with the rocket and the engine exhausts the fuel with a constant velocity relative to the rocket. There is, of course, also a relative velocity of the plume to Earth, which changes as the rocket accelerates.CuriousOne –CuriousOne 2015-06-30 07:40:08 +00:00 Commented Jun 30, 2015 at 7:40 @CuriousOne but if were to calculate the thrust at the start of launch ,then the both the gas and the rocket will be at rest and there would not yet be any exhaust ,so how do we work out the thrust for that particular instant when the launch just starts ....???Batwayne –Batwayne 2015-06-30 07:55:44 +00:00 Commented Jun 30, 2015 at 7:55 At launch the gas will be at −v e x h a u s t−v e x h a u s t relative to earth, in general it will move at v r o c k e t−v e x h a u s t v r o c k e t−v e x h a u s t.CuriousOne –CuriousOne 2015-06-30 08:03:46 +00:00 Commented Jun 30, 2015 at 8:03 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Imagine the rocket before and after throwing a small ("infinitessimal") amount of fuel out its exhaust. You apply the momentum conservation notion by equating the increase in the rocket's forwards momentum with the momentum of the fuel thrown backwards. The easiest inertial frame to do one's analysis in is that of rocket immediately before the increment d m d m of fuel is thrown: in this frame, before the fuel is thrown, the rocket has a momentum of nought; after the fuel is thrown, we have a mass d m d m flying backwards at velocity v e v e and the rocket is moving forwards a speed d v d v: the increment in speed wrought by the fuel increment. Momentum conservation is then v e d m=(m−d m)d v⇔v e m=d v d m v e d m=(m−d m)d v⇔v e m=d v d m which is the Tsiolkovsky equation. It should be very clear that the exhaust velocity v e v e to use is the velocity relative to the rocket. Note that there are modifications to this equation to account for the pressures on the rocket when it is steeped in an atmosphere: see my answer here. The rocket's or exhaust's motion relative to Earth or to anything else is irrelevant: this is simply Galileo's Relativity postulate (oka the "First" postulate of special relativity) - the rocket and its fuel are to be thought of as the below-decks protagonists in the Allegory of Salviati's Ship. I'm not one generally for Hollywood depictions of science, but I must say I rather liked the way Newton's third law was stated by the Matthew McConaughey character "Joseph Cooper" in the film "Interstellar" when he said "Newton's Third Law: [One] Can't get anywhere without leaving something behind"! This is a great thought to recall when analyzing a problem such as this one Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:39 CommunityBot 1 answered Jun 30, 2015 at 7:58 Selena BallerinaSelena Ballerina 90.4k 7 7 gold badges 198 198 silver badges 444 444 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. To see why the exhaust speed is important, let's do a calculation. Let's start with a rocket of mass m m going at speed u u. (We measure all speeds with respect to some inertial reference frame.) Now, suppose it exhausts a tiny amount of propellant of mass δ m δ m and the propellant is traveling at speed u P u P. After it exhausts that fuel, the rocket now has mass m−δ m m−δ m and is now going at speed u+δ u u+δ u. Conservation of momentum says: (m−δ m)(u+δ u)+δ m u P=m u(m−δ m)(u+δ u)+δ m u P=m u Simplifying the above and keeping only first order terms, we obtain: m δ u=δ m(u−u P)m δ u=δ m(u−u P) In other words, for a given amount of mass, the increase in speed of the rocket, δ u δ u, depends only on the difference between the speed of the rocket, u u, and the speed of the propellant, u P u P. The absolute value of either speed is irrelevant. The difference between the rocket speed and the propellant speed is called the exhaust speed. For the best chemical rockets, the exhaust speed is around 3,000 meters per second. When electric propulsion is used, exhaust speeds can be up to 20,000 meters per second or more. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jun 30, 2015 at 8:07 answered Jun 30, 2015 at 7:58 John1024John1024 3,025 19 19 silver badges 14 14 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Physics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. 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8765	https://blog.scaledcode.com/blog/effective-java/effective_java_override_tostring/	Effective Java! Override toString Skip to main contentScaled Code Effective Java! Override toString 03 February 2020 java, effective java review, design, architecture Today's topic is a lot less prescriptive than our previous couple chapters where there were very mathematical sounding principles behind the contracts we were to follow. Today's is much simpler. The toString() method. As Effective Java points out, Object gives us a default implementation of this method but it's rather not useful, especially when you are trying to debug and are simply greeted with something like Animal@23a5b2. The documentation for toString() even says that "It is recommended that all subclasses override this method." Well all classes are subclasses of Object so that must mean it is recommended that all classes should override this method. One of the main uses of the toString method in my experience is to assist in debugging. Whether you override the toString method or not developers will try using the toString method for debugging. There is an art to knowing how much of the objects data to expose via the toString method. You should try to expose as much information as possible via the method but obviously a large object may not be able to practically expose all that information. In these cases we must just use our judgement to determine which fields are the most useful. A decision that must be made is whether to document the format with which the method will return. With such a contract the users of your class can expect what will be returned and can use it in expected ways. If you do define the contract it can be a good idea to also provide a static factory method that takes the string representation and creates the object (this of course would be impossible if not all the data was exposed via the toString method, see above). Of course the downside of specifying the format is that you will then be stuck with that format for life and thus end up with a loss of flexibility. Another item to keep in mind as well is to expose all the data in the toString method via regular getters as well. What we want to avoid is forcing a developer to parse the output of the toString method in order to get at the information they need. Not only will this not be performant but also error prone. There are some times that overriding the toString method can be skipped such as in enums and utility classes. Once again we find a place where Lombok can help us overcome the boilerplate. Lombok's @ToString annotation has a very specific format. It's a fairly solid format that gets the information out in a simple manner. There definitely can be better formats for a particular context but it uses a good default. Thankfully this chapter was a little simpler than our previous chapters. Overriding the toString method is one of those little things that doesn't get much attention but it's the little things that matter. Previous: Effective Java! Obey the hashcode contract Next: Effective Java! Override clone judiciously Written By: Kyle Carter Home Archive About Me Search RSS Feed
8766	https://jcbpr.org/storage/upload/pdfs/1740727491-en.pdf	61 Bilişsel Davranışçı Psikoterapi ve Araştırmalar Dergisi Journal of Cognitive-Behavioral Psychotherapy and Research DOI: 10.14744/JCBPR.2024.62723 J Cogn Behav Psychother Res 2025;14(1):61–72 A Review of Panic Disorder and Cognitive Behavioral Therapy Simay Yılmaz,1 Merve Kabadayı,2 Derin Kubilay2 1Department of Psychology, İstanbul Gelişim University, İstanbul, Türkiye 2Department of Psychology, Private Practice, İstanbul, Türkiye One mental health issue that impacts people’s social and professional functioning is panic disorder (PD). Numerous studies have demonstrated the efficacy of cognitive behavioral therapy (CBT) in treating PD, making it a well-known intervention technique in the field of mental health. CBT integrates cognitive strategies and behavioral interventions in the treatment of PD and fosters more adaptive cognitive patterns by challenging negative automatic thoughts, thereby enhancing emotional regulation and interpersonal functioning. At the same time, behavioral techniques such as exposure and systematic desensitization equip individuals to confront and manage panic-inducing situations. Third-wave cognitive behavioral therapies, such as Acceptance and Commitment therapy, Metacognitive Therapy, Mindfulness-Based Cognitive Therapy, and Schema therapy, assist with symptom management through various methods. This clearly demonstrates that these therapies are effective interventions for managing PD, reducing its severity, and ensuring that the benefits persist after the treatment period. Keywords: Panic disorder, panic attack, cognitive behavioral therapy. Panik Bozukluk ve Bilişsel Davranışçı Terapi Üzerine Derleme Panik bozukluk, bireylerin sosyal ve mesleki işlevselliğinde ciddi düşüşlere neden olan ruhsal bir bo-zukluktur. Panik bozukluk tedavisinde, bilişsel davranışçı terapi yaygın olarak kullanılan ve etkinliği kanıtlanmış bir yöntemdir. Bu tedavi hem bilişsel süreçleri hem de davranışsal müdahaleleri içerir; bu nedenle panik bozukluğu tedavi etmede bütünsel bir yaklaşım sunar. Bilişsel davranışçı terapi, kişinin olumsuz otomatik düşüncelerini daha işlevsel ve gerçekçi olanlarla değiştirmelerine yardımcı olurken, maruz bırakma ve sistematik duyarsızlaştırma teknikleri kişinin panik atakları tetikleyen du-rumlarla yüzleşmelerini sağlar. Üçüncü dalga bilişsel davranışçı terapiler de panik bozukluk tedavi-sinde önemli bir rol oynamaktadır. Kabul ve Kararlılık Terapisi, Metakognitif Terapi, Farkındalık Temelli Bilişsel Terapi, Şema Terapi gibi üçüncü dalga bilişsel davranışçı terapi yaklaşımları, danışanların psi-kolojik esneklik kazanmalarına ve panik bozukluk semptomlarını yönetmede daha etkili olmalarına yardımcı olur. Yaptığımız derleme çalışması panik bozukluğun tedavisinde bilişsel davranışçı terapi-nin ve üçüncü dalga bilişsel davranışçı terapilerin etkili olduğunu ve bu etkilerin tedaviden sonraki süreçte de korunduğunu göstermektedir. Anahtar Kelimeler: Panik bozukluk, panik atak, bilişsel davranışçı terapi. Cite this article as: Yılmaz S, Kabadayı M, Kubilay D. A Review of Panic Disorder and Cognitive Behavioral Therapy. J Cogn Behav Psychother Res 2025; 14(1): 61–72. Address for correspondence: Simay Yılmaz. İstanbul Gelişim Üniversitesi, Psikoloji Bölümü, İstanbul, Türkiye Phone: +90 212 422 70 00 E-mail: simyilmaz@gelisim.edu.tr Submitted: 29.04.2024 Revised: 24.06.2024 Accepted: 06.11.2024 Available Online: 26.02.2025 JCBPR, Available online at This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. ABSTRACT ÖZ Review / Derleme 62 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy J Cogn Behav Psychother Res 2025;14(1):61–72 INTRODUCTION Panic Disorder (PD) is a condition wherein a person is in constant fear of future panic attacks and employs various behavioral changes to prevent triggers that may lead to a panic attack (APA, 2013). A panic episode can be precipitated either before its occurrence or suddenly, as triggered by certain situations (Kessler et al, 2012). According to recent research, the prevalence of PD in the last 12 months is approximately 3% (Walker et al, 2021). Individuals suffering from PD are believed to have a lower quality of life than people who do not have anxiety disorders (Barrera & Norton 2009). Moreover, it has also been mentioned that PD produces quite a pronounced limitation in the functioning of people (Alonso et al, 2004). Cognitive behavioral therapy (CBT) has been found to be effective in several clinical studies (Barlow et al, 2000; Manfro et al, 2008; Pompoli et al, 2016). Numerous studies have confirmed that CBT is effective in treating PD (Gould et al, 1995; Oei et al, 1999; Otto et al, 2001; Butler et al, 2006; Otto et al, 2012; Bruinsma et al, 2016; Otto et al, 2016; Ateş & Arcan 2018). This review comprehensively examined the existing literature on CBT’s techniques and effectiveness in treating PD. In this context, the study evaluates the effects of basic CBT techniques, such as cognitive restructuring, exposure, breathing, and relaxation, with a focus on the effect of CBT on PD. The study also addresses the limited research on third-wave cognitive behavioral therapies (e.g., Acceptance and Commitment Therapy, Schema Therapy, Metacognitive Therapy, and Mindfulness-Based Cognitive Therapy) that may enhance the effectiveness of CBT. Additionally, it examines the application of these therapies in the treatment of PD and evaluates their clinical effectiveness. Panic Disorder PD is an anxiety disorder characterized by the experience of sudden and intense fear accompanying several bodily sensations and/or cognitive symptoms without a specific external factor, as well as the recurrent experience of unexpected panic attacks (APA, 2013). According to DSM-5, a person is diagnosed with PD if, after experiencing a panic attack, they remain constantly worried about having another one and avoid triggers that may lead to panic attacks to prevent this situation (APA, 2013). Experiencing an important event months before the onset of PD is an important reason for the development of PD. Approximately 80% of people diagnosed with PD may have experienced one or more stressful life events in their lives before the first panic attack (Uhde et al, 1985). PD can result in reduced social and occupational functionality, difficulty in performing daily tasks, diminished quality of life, decreased well-being, and personal stagnation (Fava et al, 2001). Modern theories on Panic Disorder suggest that biological and psychosocial vulnerabilities, along with an insufficient sense of personal control, are majorly responsible for the incurrence and maintenance of attacks (Suárez et al, 2008). Such models provide an explanation for the catastrophic interpretation of bodily sensations and identify that anxiety sensitivity and intolerance of uncertainty trigger these interpretations. Individuals with Panic Disorder frequently develop heightened anxiety and panic attacks due to a combination of factors; particularly intolerance to uncertainty, which is accompanied by misinterpretations of bodily sensations, thus increasing the frequency and intensity of panic attacks (Carleton et al, 2013). With the DSM-5, agoraphobia has been considered a separate disorder from PD. People with PD may develop agoraphobia. In the event of a panic attack, individuals may avoid being in environments or situations where help may not be available, or avoidance may be difficult (APA, 2013). The lifetime prevalence rate of PD with agoraphobia as a comorbidity is 1.1%. The lifetime prevalence rate of PD without agoraphobia is 3.7% (Kessler et al, 2006). The presence of comorbid agoraphobia is associated with greater loss of functioning, and a person’s level of agoraphobic avoidance can predict their degree of disability (Wittchen et al, 2010). Cognitive Behavioral Therapy CBT refers to a set of interventions based on scientific foundations that aim to directly change dysfunctional thinking styles and behavioral patterns to alleviate psychological distress (Hofmann et al, 2013). The main idea underlying CBT developed by Beck (1970) and Ellis (1962) is that maladaptive thoughts cause emotional distress and behavioral problems to persist. According to Beck’s model, these maladaptive thoughts form general beliefs or schemas about the world, the self, and the future, with automatic thoughts emerging in certain situations (Beck, 1970; Ellis, 1962). According to one of the basic assumptions of the CBT school, it is possible to create positive changes in emotions and behaviors by reconsidering and reorganizing one’s thinking style and cognitive processes (Beck, 2014). Cognitive techniques used in CBT include questioning evidence, exploring options, considering the worst-case scenario, restructuring, and monitoring emotions and thoughts. Behavioral techniques include systematic desensitization, exposure, response prevention, and relaxation (Beck, 2014). In CBT, the client and therapist seek to understand problems by examining the relationship between thoughts, feelings, and behaviors, and the focus of therapy is on the present moment. CBT directly targets symptoms, reduces distress, reappraises thoughts, and promotes helpful behavioral responses. The therapist helps the client manage problems using their resources and teaches specific psychological and practical skills (Leichsenring et al, 2006). 63 J Cogn Behav Psychother Res 2025;14(1):61–72 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy Cognitive Behavioral Therapy for Panic Disorder CBT interventions for treating anxiety attempt to change maladaptive beliefs about the likelihood and actual effects of future harm. Various cognitive (e.g., thought restructuring) and behavioral (e.g., exposure) techniques are used in this process (Hofmann, 2008; Smits et al, 2012). Several studies have shown the effectiveness of CBT as a treatment method for PD (Telch et al, 1995; Barlow et al, 2000; Otto & Deveney, 2005; Efron & Wootton, 2021). According to the Cognitive Model of Panic developed by Clark (1986) and Beck (1988), panic attacks arise from the misinterpretation of bodily sensations as frightening, dangerous, or fatal. This catastrophizing interpretation leads to a rapid increase in anxiety, followed by heightened arousal and, ultimately, panic attacks (Clark, 1986; Beck, 1988). The therapy process begins with psychoeducation about the nature and symptoms of PD, aiming to correct misconceptions that these symptoms are harmful. After psychoeducation, a self-monitoring process is initiated to help clients develop objective self-awareness. The participants were asked to record when they experienced panic attacks and during daily periods of anxiety and depression. Cognitive restructuring directly addresses the misinterpretation of bodily sensations during therapy. In this process, therapists guide clients to develop alternative evaluations that are not based on fear or schema but based on their references and evidence, using Socratic questioning (Craske & Barlow, 2007). The treatment process and stages of PD with CBT are described below. Psychoeducation and Self-Monitoring Treatment for PD begins with explaining the general appearance and nature of PD, how it starts, why the panic state is caused, and the physical, cognitive, and behavioral cycle in PD to the client, while providing psychoeducation on these issues (Craske & Barlow, 2007). During psychoeducation, the aim is to explain the components that cause anxiety to the client and address myths and false beliefs about panic symptoms. During the psychoeducation process, clients are encouraged to evaluate their beliefs as assumptions rather than absolute truths (Taylor, 2000). Self-monitoring is used to increase objective self-awareness and improve the accuracy of observation. The participants were asked to keep three types of records: noting the level of distress, symptoms, and thoughts after each panic attack; noting daily levels of anxiety and depression; and tracking progress by updating the number of panic attacks and anxiety levels weekly (Craske & Barlow, 2007). Working with Automatic Thoughts in Panic Disorder Automatic thoughts occur spontaneously and often quite rapidly. Even if one is rarely aware of these thoughts, one is more likely to become aware of the behaviors and emotions that follow such thoughts (Beck, 2014). According to Wenzel and Cochran (2006), individuals with PD often recall autobiographical memories that are closely related to their automatic thoughts. These thoughts are usually negative and threat-oriented. The recall process is faster and more intense in individuals with PD than in individuals who do not experience anxiety. According to Ottaviani and Beck (1987), clients with PD often misinterpret physical sensations as signs of catastrophic events such as “having a heart attack” or “going crazy.” These automatic thoughts are closely linked to the fear of physical or mental catastrophe and can lead to an escalating cycle of panic. According to Goldberg (2001), individuals with PD often experience a cognitive bias that causes them to focus on negative and personally important information, which reinforces their fear and leads to panic attacks. CBT is effective in teaching clients to identify these automatic thoughts, replacing their catastrophic interpretations with more realistic and less threatening thoughts. This process plays an important role in reducing the frequency and severity of panic attacks and helping clients control their emotional reactions. Role-playing or imagery techniques can be applied to individuals during sessions to help them attain warmer cognitions. In this context, when working on automatic thoughts in PD, individuals may be asked to visualize the last panic attack they have experienced and then find the automatic thoughts that went through their mind while experiencing a panic attack (Beck, 2014). Distraction Distraction techniques play an important role in the cognitive behavioral treatment of PD and help individuals to distract from anxiety-provoking thoughts and feelings (Barlow, 2002). Studies have shown that distraction techniques combined with CBT contribute to individuals coping with their symptoms more effectively by reducing hypersensitivity to bodily symptoms and catastrophizing tendencies (Schmidt et al, 2000). Restructuring Automatic Thoughts Cognitive restructuring helps clients identify and replace their automatic negative thoughts with more realistic and functional ones. For example, a client experiencing palpitations may perceive them as a heart attack; with this technique, the accuracy of this thought is questioned, and the client is made to realize that palpitations are a symptom of anxiety. The therapist offers restructuring strategies by teaching how these distorted thoughts are related to panic attacks. Thus, the anxiety levels of the clients decrease and their severity of panic attacks decreases, enabling them to better understand PD (Strauss et al, 2019). 64 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy J Cogn Behav Psychother Res 2025;14(1):61–72 Cognitive restructuring aims to enable a person to realize erroneous thoughts in the form of catastrophizing to question the validity of these thoughts and to produce more realistic thoughts (Craske & Barlow, 2014). This section aims to identify a person’s negative automatic thoughts, and various methods are used for this. With the direct question method, a person can be asked about their thoughts during a panic attack (e.g., “What were you thinking when your heart was beating fast?”). In addition, the behavioral experiments allow the participants to record their thoughts in the presence of similar symptoms. In this process, information about automatic thoughts is provided, and an automatic thought registration form is used. The reality of these thoughts is then assessed. The purpose of CBT is to shake up negative thoughts and develop functional alternatives. The decrease in the person’s belief in these thoughts contributes to a decrease in discomfort (Hocaoğlu et al, 2023). Various techniques are used to evaluate negative automatic thoughts. These techniques include examining the evidence, finding alternative explanations, the double-standard technique, finding the worst and best thing that can happen, questioning the effect of believing in automatic thinking on the person, and problem-solving (Beck, 2014). The double-standard technique recognizes and alters negative automatic thoughts. By asking questions such as “What would a friend of yours think if they were going through this?” the person is made to approach the situation more objectively. When viewed from someone else’s perspective, the subjectivity of the emotional reaction decreases, and the person can evaluate the event more objectively (Türkçapar, 2018). Breathing Exercises and Progressive Muscle Relaxation When a person experiences anxiety, their breathing may accelerate. This may cause mild discomfort or may be intense and severe enough to cause panic attacks. Breathing exercises can be taught to individuals to control their breathing during a panic attack (Hocaoğlu et al, 2023). Progressive muscle relaxation is a technique that allows the whole body to relax through voluntary and progressive relaxation of the muscles in the body. It is effective for individuals with high anxiety levels. This technique demonstrates the difference between tension and relaxation and teaches the awareness of muscle tension and the release of these muscles. This technique induces relaxation in the whole body by eliminating muscle contractions (McCallie, 2006; Essa et al, 2017). According to the results of a meta-analysis study conducted by Mitte (2005), for PD without agoraphobia, treatment using CBT combined with applied relaxation was found to be as effective as using each therapy separately. Additionally, both methods, whether used separately or together, were found to be superior to medication. In Barlow et al.’s (2000) treatment model, the exposure of the client to internal sensations plays a central role. This exposure triggers feared sensations through exercises, such as visualization of anxiety scenes, excessive breathing, and spinning. Treatment includes education about panic and its factors and cognitive therapy to change false beliefs. The program also includes progressive muscle relaxation training and homework assignments at different stages (Barlow et al, 2000). Behavioral Interventions Behavioral techniques play an important role in treating PD. Instead of avoiding situations that trigger panic attacks, these techniques help clients confront these situations and become desensitized to the stimuli over time. In particular, real-life exposure therapy allows clients to directly confront anxiety-provoking stimuli. Thus, it restructures individuals’ reactions to these stimuli and reduce panic symptoms (Ham et al, 2005). Rapid breathing and hyperventilation are the most commonly used behavioral experiments in patients diagnosed with PD. In the treatment, panic attacks are experienced with rapid breathing exercises, and it is shown that this is a natural reaction of the body and not a sign of a major disaster. The client was shown that an attack can be controlled through breathing exercises (Bouchard et al, 1996). Exposure Exposure techniques applied within the framework of PD are based on exposing the individual to the bodily sensations that trigger panic attacks and contribute to the maintenance of these attacks (Durna, 2016). The exposure technique eliminates fear and anxiety by reducing avoidance behavior and destructive thoughts. Research has shown that direct exposure to situations that cause intense anxiety can be beneficial; however, gradual exposure is more effective (Fiegenbaum, 1988). In CBT, exposure and cognitive restructuring techniques are commonly used together. However, studies have shown no significant difference between CBT and behavioral therapy in which only the exposure technique is used (Mitte, 2005; Ougrin, 2011). The most frequently used techniques in these therapies include exposure, cognitive restructuring, and breathing and relaxation training. In a study conducted by Aydın Yeral (2024), 15 meta-analyses investigating the effectiveness of these treatment methods were analyzed, and it was shown that the exposure technique was the most effective factor in the treatment process. Mechanisms of Cognitive and Behavioral Change in Cognitive Behavioral Therapy for Panic Disorder Cognitive and behavioral factors affect the persistence of PD. These factors include the individual’s level of anxiety sensitivity, panic self-efficacy, and catastrophizing interpretations of events. These factors are addressed as targets in the CBT process to ensure changes in individuals (Sandin et al, 2015). 65 J Cogn Behav Psychother Res 2025;14(1):61–72 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy Level of Anxiety Sensitivity In general, anxiety sensitivity is the fear of anxiety-related sensations, and anxiety sensitivity is thought to predispose patients to PD. A high level of anxiety sensitivity is considered an effective factor for increasing the likelihood of experiencing more panic attacks and relapse of PD. Therefore, the level of anxiety sensitivity is considered a target in CBT to prevent the recurrence of panic attacks (Scholten et al, 2013). Panic Self-Efficacy Within the scope of PD, the perception that a person can control their bodily sensations, emotions, thoughts, or situations related to their panic attack experienced is considered as panic self-efficacy. If a person believes they cannot cope with perceived danger, this situation causes the person’s anxiety to continue (Kılıç & Yalçınkaya-Alkar, 2021). Increasing the panic self-efficacy level of the patient during therapy can predict a decrease in panic attacks. At the same time, it has been revealed that the elements adopted by CBT also increase the person’s panic self-efficacy (Casey et al, 2005). Catastrophizing According to Clark’s PD model, symptoms of PD are sustained by catastrophizing interpretations of uncertain physical or mental states, which are defined as catastrophizing (Clark, 1994; Clark, 1996). Catastrophizing interpretations of a person are considered a basic mechanism of change in treatment. In the treatment process, the change in the person’s catastrophizing interpretations are examined as factors that reduce the severity of panic symptoms (Hofmann et al, 2007). Safety-Providing Behaviors Safety-providing behaviors refer to behavioral factors developed to prevent the threat perceived by the person with a panic attack. The presence of safety-providing behaviors causes persistent anxiety symptoms. People think that catastrophizing thoughts do not occur because they perform safety-providing behaviors (Salkovskis, 1991). While these behaviors include avoiding environments and situations that the person thinks are likely to cause panic attacks (Salkovskis et al, 1996), they also include internal or external measures to prevent panic attacks and seeking reassurance from those around them (Helbig-Lang et al, 2014). It has been shown that the frequent use of reassuring behaviors by individuals in daily life negatively affects the long-term effectiveness of CBT (Beesdo-Baum et al, 2012). Effectiveness of Cognitive Behavioral Therapy in the Treatment of Panic Disorder In a meta-analysis study conducted by Bandelow et al. (2018), it was investigated whether the persistence of psychotherapy methods (CBT and other psychotherapy methods) used in the treatment of anxiety disorders, including PD, makes a difference compared with the control groups (drug treatment and placebo). The meta-analysis included 93 studies with 185 study arms (CBT, n=120; other psychotherapies, n=32; medication, n=16; placebo effect, n=17) that included follow-up evaluations of psychological treatment for PD, social anxiety disorder, and generalized anxiety disorder. Because of this study; it was stated that the gains obtained after psychotherapy treatment were preserved for up to 24 months. In addition, the clients who received CBT treatment showed improvement compared with the values at the end of the treatment. It was stated that the clients in the drug group remained stable in the untreated period and that there was no significant difference compared with those who received psychotherapy, whereas the clients in the placebo group showed significantly worse results compared with the clients receiving CBT. In a meta-analysis study conducted by Cuijpers et al. (2016), the effectiveness of CBT in the treatment of major depression, generalized anxiety disorder, PD, and social anxiety disorder was evaluated. This meta-analysis included studies comparing CBT with a control group, such as a waiting list or placebo. In the 144 included studies, the overall effects were large, and specifically for PD, the effect size was calculated as 0.81. It was concluded that CBT was effective for the treatment of PD. In a randomized controlled study conducted by Pincus et al. (2010) with adolescents, the effectiveness of Panic Control Treatment in Adolescents was examined. Panic Control Treatment is a Cognitive Behavioral treatment consisting of 11 sessions including psychoeducation, cognitive restructuring, breathing exercise training, exposure to internal perception, and exposure to situations that the individual is afraid of and avoids, aiming to correct the individual’s misinformation about panic attacks. In the study, 13 adolescents aged 14–17 years were included in the 11-session Panic Control Treatment in Adolescents group, and 13 adolescents were randomized to the control group. According to the results of the study, there was a significant decrease in the severity of PD, self-reported anxiety, anxiety sensitivity, and depression levels of the adolescents in the treatment group. These treatment gains were maintained at 3- and 6-month follow-ups. A research study was carried out by Hendriks et al. (2012), which attempted to establish the use of paroxetine along with a CBT program for elderly clients suffering from PD, irrespective of whether they have agoraphobia or not. The study included 49 66 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy J Cogn Behav Psychother Res 2025;14(1):61–72 participants aged above 60 years. The participants of the study were placed into three different categories. The first cohort received 40 mg of paroxetine, the second group received individual CBT sessions, and the third category was assigned to the waiting group over a span of 14 weeks. According to the findings of the study, most of the participants profiled to be using paroxetine and CBT have progressed significantly more than the ones in the control group. According to the study findings, paroxetine and CBT were found to have a positive outcome in elderly patients with PD. Pompoli et al (2016) conducted a meta-analysis of 60 studies comparing the effect of eight psychological treatments and three control groups for the treatment of PD. Psychoeducation, psychodynamic therapy, supportive psychotherapy, physiology therapy, cognitive therapy, behavior therapy, CBT, and third-wave cognitive behavior therapy people diagnosed with PD with agoraphobia and those without it and face-to-face therapies were examined. The most efficient, reliable, and long-lasting therapies were CBT and psychodynamic therapy. With technological advances, it became possible to develop internet-based CBT. As indicated by the research conducted by Domhardt et al. 2020, internet-based CBT reduced the symptoms of PD in the same efficient manner as face-to-face CBT. A meta-analysis study by Efron and Wootton (2021) investigated the effectiveness of remote CBT in the diagnosis and treatment of PD. Findings of this study suggest that remote CBT generates appreciable treatment effects, with improvement observed after the intervention. In a study conducted by Rabasco et al. (2022), studies effective in treating PD were analyzed, and 38 meta-analyses and systematic reviews were discussed. Most of these studies have addressed the effectiveness of CBT. According to the study results, face-to-face CBT was effective in treating PD in adults, especially compared with the control conditions. In addition, with the development of technology, it was also revealed that internet-based CBT or other computer-based CBT applications are suitable alternatives to face-to-face CBT. In the CBT treatment of patients with PD, combinations of different techniques included in the therapy framework can greatly affect the treatment outcome (Pompoli et al, 2016; Pompoli et al, 2018). Pompoli et al. (2018) conducted a meta-analysis of 72 studies (4064 participants) to evaluate the use of CBT in the treatment of PD. The study showed that interoceptive exposure and face-to-face therapy increased treatment efficacy and acceptability; relaxation techniques and virtual reality therapy resulted in lower efficacy; breathing exercise and real-life exposure increased acceptability but had little effect on efficacy. The highest remission rates were associated with CBT, including face-to-face therapy, placebo effect, psychoeducation, psychosocial support, cognitive restructuring, and interoceptive exposure. However, the least effective CBT included the placebo effect, psychoeducation, psychosocial support, breathing exercises, relaxation techniques, real-life exposure, and virtual reality exposure. In the same study, relaxation techniques and virtual reality exposure were found to reduce the likelihood of treatment response (55 studies, 3275 participants). In terms of treatment discontinuation, the placebo effect was determined as the most significant component (68 studies, 3705 participants) (Pompoli et al, 2018). CBT uses structured techniques to reduce the physical and cognitive symptoms of panic attacks, anticipatory anxiety, and avoidance behaviors in the treatment of PD and generally improves the functionality of individuals (Kılıç & Yalçınkaya-Alkar, 2021). However, not all individuals showed the same level of improvement. According to Kılıç and Yalçınkaya-Alkar’s (2021) review, pretreatment factors include disorder-related characteristics such as age of onset, duration, symptom severity, and impairment in functionality; co-diagnoses such as cardiac and respiratory clients, depression, anxiety, and personality disorders; and the person’s social environment conditions. In the treatment process, cognitive and behavioral mechanisms, such as the individual’s expectations of therapy, commitment, therapeutic alliance, anxiety sensitivity, and safety behaviors, affect success. The efficacy of CBT in the treatment of PD can vary according to various factors. For example, early-onset PD is associated with more negative outcomes in terms of panic symptoms (Chambless et al, 2017), whereas late-onset PD may yield more positive results (Hendriks et al, 2012). While it has been observed that symptom severity decreases as the duration of the disorder increases (Nakano et al, 2008), it has been reported that clients with high pretreatment symptom severity initially respond less to CBT, but may have difficulty in maintaining their gains in the long term (Brown & Barlow, 1995). Although experts working in the clinical field generally consider comorbid diagnoses as a factor that prevents the benefit of therapy (Wolf & Goldfried, 2014), experimental studies have revealed that these diagnoses do not always have a negative effect on the course of PD, and that CBT used in the treatment of the disorder can reduce the symptoms of these diagnoses as well as the symptoms of the disorder (Brown et al, 1995). Social environmental conditions are also an important factor affecting the CBT process. Stressful events may cause symptoms to reappear after treatment, and lack of social support may negatively affect the effectiveness of therapy (Heldt et al, 2011; Wolf & Goldfried, 2014). Therefore, to evaluate the effectiveness of CBT, it is important to consider the factors before and during the treatment of the individual. 67 J Cogn Behav Psychother Res 2025;14(1):61–72 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy Third-wave Cognitive Behavioral Therapies and Panic Disorder Acceptance and Commitment Therapy in Panic Disorder Acceptance and Commitment Therapy (ACT) aims to increase psychological resilience through six processes of change: acceptance, cognitive dissociation, staying in the moment, seeing the self as context, values, and behavioral commitments. Each of these processes has received considerable support from outside the treatment programs (Hayes et al, 2006; Ruiz, 2010). The systematic review study conducted by Swain et al. (2013) examined the effectiveness of ACT in the treatment of anxiety. Among these studies, two research articles specific to PD were identified, and these two studies were analyzed. In the first study, among 22 participants, only the group receiving panic control treatment and the group receiving ACT-supported panic control treatment were compared. According to the results of the study, more agoraphobia severity was observed in the group receiving only panic control treatment than in the group receiving ACT, and the participants experienced a higher level of avoidance (Karekla, 2005). The second study is a case study conducted by López (2000). In this study, 12 sessions of ACT were administered. According to the results of the study, the anxiety, worry, and fear levels of the participants decreased significantly. Use of Schema Therapy in Panic Disorder According to Martin and Young (2010), Schema Therapy (ST) is a therapy method and theoretical framework based on three main structures consisting of “schemas,” “coping styles” and “modes” (Young & Brown., 2003), which is used in the treatment of people with personality disorders, some chronic axis 1 diagnoses, as well as other individual and couple problems (Young, 1990). Gude et al. (2001) examined the effectiveness of a combined cognitive therapy and ST treatment. The study was conducted with 47 participants who had cluster C personality traits and PD and/or agoraphobia. The first 5 weeks of therapy were based on the cognitive model of panic and agoraphobia (Clark et al, 1994). The second part was based on Young’s (1990) personality-focused ST and was planned for 6 weeks. All participants were evaluated in terms of their agoraphobic avoidance before, during, after, and 1-year follow-up. According to the results, the agoraphobic avoidance scores were higher before than during treatment and higher after than during treatment. Hoffart et al. (2002) conducted a more comprehensive study. Participants consisted of 35 individuals with cluster C personality disorder, PD, and/ or agoraphobia. All participants received at least nine sessions of ST. Participants were evaluated twice before starting treatment, during treatment, after treatment, and at 1-year follow-up after treatment. According to the results of the study, the PD and agoraphobia measurements of the participants decreased significantly after treatment compared with pretreatment. Use of Mindfulness-Based Cognitive Therapy in Patients with Panic Disorder Mindfulness-Based Cognitive Therapy (MBCT) is a therapy method that combines elements of the Mindfulness-Based Stress Reduction Program and CBT. Although MBCT mainly focuses on encouraging people to relate to their feelings and thoughts and to adopt a new way of being, it also includes, albeit to a lesser extent, changing or challenging certain cognitions during therapy (Sipe & Eisendrat, 2012). Kim et al. (2013) conducted a study with 65 patients with PD, examining the results of MBCT treatment in terms of potential determinants. The patients were compared according to completion, response to treatment, and remission rates during the 8-week treatment period. The results of the study revealed a significant relationship between the improvement in anxiety sensitivity and the response rate of the participants who completed the treatment program. In addition, improvement in anxiety sensitivity has been reported to reduce symptoms and may predict remission at 1-year follow-up. Metacognitive Therapy in Panic Disorder Metacognitive Therapy (MCT) is a transdiagnostic treatment method that aims to change the processes leading to cognitive and emotional dysregulation in individuals with mental disorders (McEvoy, 2019). The metacognitive approach is the center of a person’s approach to mastering their interactions with thoughts and managing the mechanisms that control these thoughts, as described by Wells (2009). In their meta-analysis, Normann et al. (2014) analyzed a total of 16 studies focusing on the effect of MCT on depression and anxiety disorders. The effectiveness of MCT in treating depression and anxiety disorders was found to be very high. Furthermore, large effect sizes between pre- and posttreatment were reported in 16 different studies, and based on the follow-up periods, the treatment results were maintained. Based on several studies, MCT showed significantly stronger treatment effects than CBT. In another study, Afshari et al. (2010) examined the effect of MCT on the panic beliefs of women with PD and found that the panic beliefs of participants receiving MCT decreased. In the treatment of PD, therapies such as Dialectical Behavior therapy, Emotional regulation therapy, and Integrative Emotional regulation therapy, known as third-wave therapies, are also included (Vatan, 2016). 68 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy J Cogn Behav Psychother Res 2025;14(1):61–72 DISCUSSION In this review study, CBT was examined in terms of its role in the treatment protocol for PD. A number of studies have investigated the effectiveness of CBT in the treatment of PD and have concluded that it is indeed effective (Gould et al, 1995; Butler et al, 2006; Bruinsma et al, 2016). For example, CBT for PD as a method of psychological therapy that is more productive for panic attacks than other psychotherapeutic techniques is now a known fact (Pompoli et al, 2016; Bandelow et al, 2018) in addition to the nontemporary (Bandelow et al, 2018; Pincus et al, 2010; Pompoli et al, 2016) nature of its outcomes. As reported by Bandelow et al. (2007) in the results of a meta-analysis, CBT maintains the gains of patients after therapy completion. According to Deacon and Abramowitz (2005), patients regard CBT as a more useful and practical form of therapy than long-term medication. These results support the conclusions regarding the benefits of CBT for patients relative to drug use. For treating PD, CBT helps patients by providing them with lifelong coping skills through intervening with automatic thoughts and core beliefs, thereby showing prolonged results. While the return of symptoms is observed in drug treatment, gains are also preserved in CBT treatment (Bandelow et al, 2018). In addition, research has shown that CBT can be made more effective using different techniques and combinations. In particular, methods such as face-to-face work, psychoeducation, psychosocial support, exposure to bodily sensations, and cognitive restructuring positively affect treatment outcomes (Pompoli et al, 2018). This demonstrates the potential benefits of combining different techniques in the CBT process. The use of different cognitive behavioral techniques together may help develop a deeper understanding of the therapy process and more permanent changes in the client. However, internet-based CBT applications have recently come to the fore as an important innovation. Studies have shown that internet-based CBT is as effective as face-to-face CBT (Domhardt et al, 2020; Rabasco et al, 2022). This can be considered a new opportunity for clients who have difficulty obtaining therapy in person. The integration of technology into therapeutic processes indicates that treatment options have recently expanded, and future studies should examine the long-term effects and effectiveness of digital therapies in more detail. In addition, third-wave cognitive behavioral therapies such as ACT (Lopez, 2000; Karekla, 2004), ST (Hoffart, 2002), MCT (Afshari et al, 2010; Normann et al, 2014), and MBCT (Kim et al, 2013) have been shown to be effective in the treatment of PD. Based on this evaluation, the role of third-wave therapies for treating PD emerges as a field of study that needs to be addressed in more depth. In particular, there is a need for comparative studies on whether these therapies contribute to the long-term recovery processes of patients and studies in which their effectiveness is compared with that of CBT. These studies will provide important findings that will expand treatment options. However, it can be mentioned that studies on therapy methods other than ST, ACT, MCT, and MBCT are not sufficient in the literature on CBT of PD (Lopez & Salas, 2009; Hawke & Provencher, 2011; McEvoy, 2019). Therefore, it is thought that more studies on third-wave therapies in the treatment of PD will contribute to the literature. The strengths of this review study include the fact that the current studies in the literature are included, the findings of meta-analyses and systematic review studies, and the effect of CBT on PD are discussed in detail, and the limited number of studies on the effect of third-generation cognitive behavioral therapies on PD are also included. Based on the findings of this review study, existing studies on the effectiveness and long-term effects of CBT for treating PD generally show that CBT is an effective treatment method. CONCLUSION This review discusses CBT, a therapy method with proven efficacy in the treatment of PD. It has been mentioned that CBT significantly reduces PD symptoms in individuals (Pincus et al, 2010; Hendriks et al, 2012), and there are findings that the gains of CBT in the treatment of PD are maintained for up to 24 months (Bandelow et al, 2018). Although studies have shown that third-wave CBT approaches are effective in treating PD, additional research is needed in this field. The results of this study indicate that CBT is highly accepted in the treatment of PD and clearly demonstrate its widespread application and proven effectiveness. However, it can be mentioned that case studies in the existing literature and research on relapse rates after treatment can be increased, and in-depth investigations and more scientific studies are needed in this field. In particular, increasing the number of studies evaluating the clinical efficacy and long-term results of third-wave CBT approaches may allow these therapies to find a wider clinical application area; thus, more effective and permanent solutions can be offered in the treatment of PD. These studies may contribute to optimizing the approaches used in the treatment process and improve the treatment responses of individuals. Author Contributions: Concept – SY, MK; Design – SY, MK; Supervision – SY, MK, DK; Resource – SY, MK, DK; Materials – SY, MK, DK; Data Collection and/or Processing – SY, MK, DK; Analysis and/or Interpretation – SY, MK, DK; Literature Review – SY, MK, DK; Writing – SY, MK, DK; Critical Review – SY, MK. Conflict of Interest: The authors have no conflict of interest to declare. Use of AI for Writing Assistance: Not declared. Financial Disclosure: The authors declared that this study has received no financial support. Peer-review: Externally peer-reviewed. 69 J Cogn Behav Psychother Res 2025;14(1):61–72 Yılmaz et al. Panic Disorder and Cognitive Behavioral Therapy REFERENCES Afshari, M., Neshat Doost, H. T., Bahrami, F., & Afshar, H. (2010). Efficacy of metacognitive-behavioral therapy on panic beliefs in female panic patients. J. Arak Uni Med Sci, 13(1), 9–16. Alonso, J., Angermeyer, M. C., Bernert, S., Bruffaerts, R., Brugha, T. S., Bryson, H., de Girolamo, G., Graaf, R., Demyttenaere, K., Gasquet, I., Haro, J. M., Katz, S. J., Kessler, R. C., Kovess, V., Lépine, J. P., Ormel, J., Polidori, G., Russo, L. J., Vilagut, G., Almansa, J., Arbabzadeh-Bouchez, S., Autonell, J., Bernal, M., Buist-Bouwman, M. A., Codony, M., Domingo-Salvany, A., Ferrer, M., Joo, S. 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8767	https://www.ncbi.nlm.nih.gov/books/NBK279170/	46,XY Differences of Sexual Development - Endotext - 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. Feingold KR, Ahmed SF, Anawalt B, et al., editors. Endotext [Internet]. South Dartmouth (MA): MDText.com, Inc.; 2000-. Endotext [Internet]. Show details Feingold KR, Ahmed SF, Anawalt B, et al., editors. South Dartmouth (MA): MDText.com, Inc.; 2000-. Contents www.endotext.org Search term < PrevNext > 46,XY Differences of Sexual Development Sorahia Domenice, MD, Rafael Loch Batista, MD, Ivo J, P. Arnhold, MD, Maria Helena Sircili, MD, Elaine M. F. Costa, MD, and Berenice Bilharinho Mendonca, MD. Author Information and Affiliations Sorahia Domenice, MD Assistant Professor of Endocrinology, Unidade de Endocrinologia do Desenvolvimento, Laboratório de Hormônios e Genética Molecular, LIM/42, Disciplina de Endocrinologia, Hospital das Clínicas da Universidade de São Paulo, Brasil, Developmental Endocrinology Unit, Laboratory of Hormones and Molecular Genetics LIM/42, Division of Endocrinology, Clinical Hospital, Medical School, Sao Paulo University, Sao Paulo, Brazil. Email: moc.liamg@daiharos Rafael Loch Batista, MD Assistant Professor of Endocrinology, Unidade de Endocrinologia do Desenvolvimento, Laboratório de Hormônios e Genética Molecular, LIM/42, Disciplina de Endocrinologia, Hospital das Clínicas da Universidade de São Paulo, Brasil, Developmental Endocrinology Unit, Laboratory of Hormones and Molecular Genetics LIM/42, Division of Endocrinology, Clinical Hospital, Medical School, Sao Paulo University, Sao Paulo, Brazil. Ivo J, P. Arnhold, MD Assistant Professor of Endocrinology, Unidade de Endocrinologia do Desenvolvimento, Laboratório de Hormônios e Genética Molecular, LIM/42, Disciplina de Endocrinologia, Hospital das Clínicas da Universidade de São Paulo, Brasil, Developmental Endocrinology Unit, Laboratory of Hormones and Molecular Genetics LIM/42, Division of Endocrinology, Clinical Hospital, Medical School, Sao Paulo University, Sao Paulo, Brazil. Maria Helena Sircili, MD Assistant Professor of Urology, Unidade de Endocrinologia do Desenvolvimento, Laboratório de Hormônios e Genética Molecular, LIM/42, Disciplina de Endocrinologia, Hospital das Clínicas da Universidade de São Paulo, Brasil, Developmental Endocrinology Unit, Laboratory of Hormones and Molecular Genetics LIM/42, Division of Endocrinology, Clinical Hospital, Medical School, Sao Paulo University, Sao Paulo, Brazil Elaine M. F. Costa, MD Assistant Professor of Endocrinology, Unidade de Endocrinologia do Desenvolvimento, Laboratório de Hormônios e Genética Molecular, LIM/42, Disciplina de Endocrinologia, Hospital das Clínicas da Universidade de São Paulo, Brasil, Developmental Endocrinology Unit, Laboratory of Hormones and Molecular Genetics LIM/42, Division of Endocrinology, Clinical Hospital, Medical School, Sao Paulo University, Sao Paulo, Brazil. Berenice Bilharinho Mendonca, MD Head Professor of Endocrinology, Unidade de Endocrinologia do Desenvolvimento, Laboratório de Hormônios e Genética Molecular, LIM/42, Disciplina de Endocrinologia, Hospital das Clínicas da Universidade de São Paulo, Brasil, Developmental Endocrinology Unit, Laboratory of Hormones and Molecular Genetics LIM/42, Division of Endocrinology, Clinical Hospital, Medical School, Sao Paulo University, Sao Paulo, Brazil. Email: rb.psu@nemereb Corresponding author. Last Update: August 21, 2022. Go to: ABSTRACT The 46,XY differences of sex development (46,XY DSD) can result either from decreased synthesis of testosterone and/or DHT or from impairment of androgen action. 46,XY DSD are characterized by micropenis, atypical or female external genitalia, caused by incomplete intrauterine masculinization with or without the presence of Müllerian structures. Male gonads are identified in the majority of 46,XY DSD patients, but in some of them no gonadal tissue is found. Complete absence of virilization results in normal female external genitalia and these patients generally seek medical attention at pubertal age, due to the absence of breast development and/or primary amenorrhea. A careful clinical evaluation of the neonate is essential because most DSD patients could be recognized in this period and prompt diagnosis allows a better therapeutic approach. Family and prenatal history, complete physical examination and assessment of genital anatomy are the first steps for a correct diagnosis. The diagnostic evaluation of DSD includes hormone measurements (assessment of Leydig and Sertoli cell function), imaging (ultrasonography is always the first and often the most valuable imaging modality in DSD patients’ investigation), cytogenetic, and molecular studies. Endoscopic and laparoscopic exploitation and/or gonadal biopsy are required in very few cases. Psychological evaluation is of crucial importance to treat DSD patients. Every couple that has a child with atypical genitalia must be assessed and receive counseling by an experienced psychologist, specialized in gender identity, who must act as soon as the diagnosis is suspected, and then follow the family periodically, more frequently during the periods before and after genitoplasty. Parents must be well informed by the physician and psychologist about normal sexual development. A simple, detailed, and comprehensive explanation about what to expect regarding integration in social life, sexual activity, need of hormonal and surgical treatment and the likely possibility or not of fertility according to the sex of rearing, should also be discussed with the parents before the assignment of final social sex. Optimal care of DSD patients begins in the newborn period and sometimes in prenatal life and requires a multidisciplinary team. Most of the well-treated DSD patients present a normal quality of life in adulthood. For complete coverage of all related areas of Endocrinology, please visit our on-line FREE web-text, WWW.ENDOTEXT.ORG. Go to: INTRODUCTION Male phenotypic development is a 2-step process: 1) testis formation from the primitive gonad (sexual determination) and 2) internal and external genitalia differentiation by action of factors secreted by the fetal testis (sexual differentiation). The first step is very complex and involves interplay of several transcription factors and signaling cells (1-3). Dosage imbalances in genes involved in DSD (deletions or duplication) have also been identified as a cause of these developmental differences (Fig. 1). Figure 1. Summary of the molecular events in sex determination indicating the genes in which molecular defects can cause gonadal disorders in animal models. Some of these disorders were confirmed in humans. Nr5a1, Wnt4 and Wt1 are expressed in the urogenital ridge whose development results in formation of the gonads, kidneys, and adrenal cortex. Several genes, Wt1, Nr5a1, M33 (CBX2 mouse homologue), Lhx9, Lim1, Gata4/Fog2, Dmrt1, Emx2 and Cited are expressed in the bipotential gonad. Nr5a1 up-regulates Cbx2 expression that is required for upregulation of the Sry gene. Nr5a1 and Wt1 up-regulate Sry expression in pre-Sertoli cells and Sry initiates male gonad development. Sry strongly up-regulates Sox9 in Sertoli cells. Sox9 up-regulates Fgf9 and Fgf9 maintains Sox9 expression, forming a positive feed-forward loop in XY gonads. The balance between Fgf9 and Rspo1/Wnt4 signals is shifted in favor of Fgf9, establishing the male pathway. If Wnt4/Rspo1 is overexpressed activating the β-catenin pathway, this system blocks Fgf9 and disrupts the feed-forward loop between Sox9 and Fgf9. Pdg2 signaling up-regulates Sox9 and Sox9 activate Ptgds. Sox9 establishes a feed-forward loop with the Pgd2. Sox9 inhibits beta-catenin-mediated Wnt signaling. Overexpression in either Dax1 (locus DSS) or Rspo1/Wnt4 antagonizes testis formation. On the other hand, Dax1 regulates the development of peritubular myoid cells and the formation of testicular cords. Dmrt1 has recently been shown to be required for the maintenance of gonadal sex and to prevent female reprogramming in postnatal testis, Cbx2 directly or indirectly repress ovarian development. Dhx37 has critical roles in early human testis determination and also in the maintenance of testicular tissue. The second step, male sex differentiation, is a more straightforward process. Mesonephric (Wolffian) and paramesonephric (Mullerian) ducts are present in both, male and female fetuses, and originate from the anterolateral epithelium of the urogenital ridge. Anti Müllerian hormone (AMH) secreted by the testicular Sertoli cells acts on its receptor in the Müllerian ducts to cause their regression. Testosterone secreted by the testicular Leydig cells acts on the androgen receptor in the Wolffian ducts to induce the formation of epididymis, deferent ducts and seminal vesicles (Fig. 2) (4). The external genitalia of the fetus derive from mesenchyme cells from the primitive streak. Under androgen stimuli male fetal urethral folds, genital tubercle and genital swellings give rise to corpus spongiosum and primitive urethra, phallus, and scrotal swellings respectively. This process is mediated by testosterone and its further reduced dihydrotestosterone (DHT), which acts on the androgen receptor of the prostate and external genitalia leading to their masculinization (5,6) (Figs. 3 - 8). Figure 2. Summary of the molecular events in sex differentiation indicating the genes in which molecular defects cause 46,XY DSD in humans. After testis determination, hormones produced by the male gonad induce the differentiation of internal and external genitalia acting on their specific receptor. The regulation of AMH gene requires cooperative interaction between SOX9 and NR5A1, WT1, GATA4 and HSP70 at the AMH promoter. Combined expression of DHH, MAMLD1 and NR5A1 is required for Leydig cell development. NR5A1 regulates gonadal steroidogenesis. The Leydig cells also produce INSL3, which causes the testes to descend to the scrotum. Figure 3. The development of male internal genitalia in the human embryo. The 6-wk-end embryo is equipped with both male and female genital ducts derived from the mesonephrons. Figure 4. The development of male internal genitalia in the human embryo. The regression of the Müllerian ducts is mediated by the action of AMH secreted by the fetal Sertoli cells. Figure 5. The development of male internal genitalia in the human embryo. The stabilization and differentiation of the Wolffian ducts are mediated by testosterone synthesized by the fetal Leydig cells. The enzyme 5α-reductase 2 converts testosterone to dihydrotestosterone (DHT). The Wolffian ducts differentiate into epididymis, vas deferens and seminal vesicles. DHT contributes to prostate differentiation. Figure 6. Development of male external genitalia in the human embryo. At the 8-wk-end embryo the external genitalia of both sexes are identical and have the capacity to differentiate in both directions: male or female. DHT stimulates growth of the genital tubercle and induces fusion of urethral folds and labioscrotal swellings. It also inhibits growth of the vesicovaginal septum, preventing development of the vagina. Figure 7. Development of male external genitalia in the human embryo. At the 12-week-end embryo the male external genitalia are entirely formed. Figure 8. Development of male internal and external genitalia in the human embryo. At the 12-week-end embryo both internal and external genitalia are completely formed. The term differences of sex development (DSD) include- congenital conditions in which development of chromosomal, or gonadal or anatomical sex is atypical. This nomenclature has been proposed to replace terms such as intersex, pseudo-hermaphroditism and sex reversal (6,7). These terms, previously used to describe the differences of sex development, are potentially offensive to the patients and the consensus on the management of intersex disorders recommends a new nomenclature that will be followed in this chapter (6). The proposed changes in terminology aim to integrate upcoming advances in molecular genetics in the most recent DSD classification (8). The 46,XY DSDs are characterized by micro-penis, atypical or female external genitalia, caused by incomplete intrauterine masculinization with or without the presence of Müllerian structures. Male gonads are identified in the majority of 46,XY DSD patients, but in some of them no gonadal tissue is found. Complete absence of virilization results in normal female external genitalia and these patients generally seek medical attention at pubertal age, due to the absence of breast development and/or primary amenorrhea. 46,XY DSD can result either from decreased synthesis of testosterone or DHT or from impairment of androgen action (9,10). Our proposal classification of 46,XY DSD is displayed in Table 1. Table 1. Classification of 46,XY DSD View in own window 46,XY DSD DUE TO ABNORMALITIES OF GONADAL DEVELOPMENT Gonadal agenesis Gonadal dysgenesis - complete and partial forms 46,XY DSD ASSOCIATED WITH CHOLESTEROL SYNTHESIS DEFECTS Smith-Lemli-Opitz syndrome 46,XY DSD DUE TO TESTOSTERONE PRODUCTION DEFECTS Impaired Leydig cell differentiation (LHCGR defects) Complete and partial forms Enzymatic defects in testosterone synthesis Defects in adrenal and testicular steroidogenesis STAR deficiency P450scc deficiency 3-β-hydroxysteroid dehydrogenase II deficiency 17α-hydroxylase and 17,20 lyase deficiency P450 oxidoreductase defect (electron transfer disruption) Defects in testicular steroidogenesis Isolated 17,20-lyase deficiency Cytochrome b5 defect (allosteric factor for P450c17 and POR interaction) 17β-hydroxysteroid dehydrogenase III deficiency Alternative pathway to DHT 3α- hydroxysteroid dehydrogenase deficiency due to AKR1C2 and AKR1C4 defects 46,XY DSD DUE TO DEFECTS IN TESTOSTERONE METABOLISM 5α-reductase type 2 deficiency 46,XY DSD DUE TO DEFECTS IN ANDROGEN ACTION Androgen insensitivity syndrome Complete and partial forms 46,XY DSD DUE TO PERSISTENCE OF MÜLLERIAN DUCTS Defect in AMH synthesis Defect in AMH receptor CONGENITAL NON-GENETIC 46,XY DSD Maternal intake of endocrine disruptors Associated with impaired prenatal growth 46,XY OVOTESTICULAR DSD NON-CLASSIFIED FORMS Hypospadias 46,XY gender dysphoria Go to: INVESTIGATION OF DSD PATIENTS Optimal care of patients with DSD requires a multidisciplinary team and begins in the newborn period. A careful clinical evaluation of the neonate is essential because most DSD patients could be recognized in this period and prompt diagnosis allows a better therapeutic approach. Family and prenatal history, complete physical examination and assessment of genital anatomy are the first steps for a correct diagnosis. The diagnostic evaluation of DSD includes hormone measurements, imaging, cytogenetic, and molecular studies (11). In very few cases, endoscopic and laparoscopic exploitation and/or gonadal biopsy are required (12). The endocrinological evaluation of 46,XY DSD infants includes assessment of testicular function by basal measurements of LH, FSH, inhibin B, anti-Mullerian hormone (AMH), and steroids. AMH and inhibin B are useful markers of the presence of Sertoli cells and their assessment could help in the diagnosis of testis determination disorders. In boys with bilateral cryptorchidism serum AMH and inhibin B correlate with the presence of testicular tissue and undetectable values are highly suggestive of absence of testicular tissue (13,14). In minipuberty and in postpubertal patients with testosterone synthesis defects, the diagnosis is made through basal steroid levels. Testosterone levels are low and steroids upstream from the enzymatic blockage are elevated. This pattern can be confirmed by an hCG stimulation test, which increases the accumulation of steroids before the enzymatic blockage, with a slight elevation of testosterone. In prepubertal individuals, an hCG stimulation test is essential for the diagnosis, since basal levels are not altered. There are several hCG stimulation protocols and normative data must be established for each of them. We established a normal testosterone response 72 and 96 hours after the last of 4 doses of hCG, 50-100 U/kg body weight, given via intramuscular every 4 days in boys with cryptorchidism but otherwise normal external genitalia: testosterone peak levels reached 391 ± 129 ng/dL and we consider a subnormal response a value <130 ng/dL (equivalent to -2 SD) (15). Imaging evaluation is indicated in the neonatal period when atypical genitalia are identified. If apparent female genitalia with clitoral hypertrophy, posterior labial fusion, foreshortened vulva with single opening or inguinal/labial mass is present, imaging studies may also be performed. A family history of DSD and later presentations as abnormal puberty or primary amenorrhea, cyclic hematuria in a male, and inguinal hernia in a female also require an imaging evaluation. Ultrasonography is always the first and often the most valuable imaging modality in DSD patients’ investigation. Ultrasound shows the presence or absence of Müllerian structures at all ages and can locate the gonads and characterize their echo texture. This exam can also identify associated malformations such as kidney abnormalities (16). Genitography and cystourethrography can display the type of urethra, the presence of vagina, cervix, and urogenital sinus. MRI contributes to accurate morphologic evaluation of Mullerian duct structures, the gonads, and the development of the phallus, all of which are essential for appropriate gender assignment and planning of surgical reconstruction (17). Go to: CYTOGENETIC AND MOLECULAR INVESTIGATION The routine use of genetic testing for reaching a diagnosis in XY DSD is increasingly playing an important role in the diagnostic process. A wide range of techniques may be used, each one having a different investigative application and genetic resolution (18,19). More than 75 genes involved in gonadal development and/or sex hormone biosynthesis/action are known causes of DSDs and the molecular methodologies have contributed to identify already known as well as novel causes of DSD. These results have led to the adoption of molecular tests into clinical practice for diagnosis and genetic counseling, reducing the need of hormonal and imaging tests to reach the correct diagnosis (20). Advances in molecular biological techniques for diagnosing DSD are reviewed in recent publications (18,19). Chromosomal Analysis Early identification of chromosomal regions and candidate-genes involved in the DSD etiology were established by finding microscopically visible structural changes in the karyotype, using conventional cytogenetic techniques. Many of them were achieved by positional cloning and linkage analysis, which are not widely used tools. Although conventional karyotyping is still used frequently in routine clinical diagnosis, faster molecular cytogenetic techniques that do not require cell culture can be employed. Array techniques [array comparative genomic hybridization (aCGH) and single-nucleotide polymorphism (SNP) array] are all capable to identify submicroscopic genome imbalance / copy number variation (CNV), as small as 10 KB (CNVs between 10 kb and 5 Mb in size), and which may affect several genes, in patients with an apparently normal karyotype (21,22). CNV affecting coding sequences or regulatory elements of critical dosage-sensitive genes are known causes of DSDs (23-26). Novel DSD candidate chromosomic regions and genes, with potential roles in sex determination and DSD, such as SUPT3H, C2ORF80, KANK1, ADCY2, VAMP7 and ZEB2, have been also identified by array studies, many of them waiting for further validation (25,27). Array techniques can diagnose pathogenic CNV in almost 30% of syndromic DSD patients as a single method (27,28). Thus, a CGH or SNP-array was proposed as the first genomic test to investigate this group of DSD patients. Sequencing Analysis Among the genetic tests, many use a candidate-gene approach (Sanger sequencing), while targeted DSD gene panels, wider whole-exome (WES) and whole-genome (WGS) scale are high-throughput screening technologies, in which multiple short DNA target sequences are analyzed to identified the presence of allelic variants (29). Sanger sequencing is often the method of choice if a specific genetic condition is highly suspected by an established clinical and hormonal diagnosis. AR and SRD5A2, in addition to almost all testosterone synthesis defects, are the most requested genes in 46,XY DSD to be sequenced using this approach (20). The superiority of targeted DSD gene panel tests, that can evaluate simultaneously several and non-standard sets of genes, over single-gene testing approaches is well established, especially considering time and cost-effectiveness (26,30). Whole-Exome Sequencing (WES) and Whole-Genome Sequencing (WGS) are also based on short-read sequencing. They present a clear improvement over single-gene testing in providing clinical diagnosis for DSD. The advantage of WES/WGS is the potential to identify new DSD-related genes in the research setting. On the other hand, WGS has more consistent coverage of gene sequences throughout the genome, including the non-coding regions and so it has the potential to provide a much higher diagnostic yield than WES (25,31). Nevertheless, WES and WGS require significant bioinformatic resources and are expensive strategies; consequently, their application for first-line diagnostic investigation in many clinical settings are still limited (18,32). The target DNA can also be read in longer fragments (several kb). The main advantage of using long-reads during the process is that repetitive elements and complex structural variants can often be resolved to a greater extent than in assemblies generated from short-read sequencing (18). Long-read sequencing offers a potential solution to genome-wide short tandem repeats analysis, which are highly variable elements, which play a pivotal role in the regulation of gene expression. Studies in animal models have suggested the involvement of epigenetic regulation in the process of gonadal formation, reinforcing a probable role of epigenetic variation in the etiology of DSD (33). Careful selection of the genetic test indicated for each condition remains important for a good clinical practice (Figure 9). Figure 9. Algorithm for 46,XY DSD diagnosis Go to: 46,XY DSD DUE TO ABNORMALITIES IN GONADAL DEVELOPMENT Uncountable allelic variants identified in several genes involved in the process of human gonadal determination have been associated with 46,XY gonadal dysgenesis. They will be described according to the period of gene expression in gonadal determination. Gonadal Determination and Differentiation The intermediate mesoderm is the primary embryonic tissue at gastrulation that gives rise to the urogenital ridge. This, in turns, is going to derive the primitive gonad from a condensation of the medioventral region of the urogenital ridge. The primitive gonad separates from the adrenal primordium at about 5 weeks but remains bipotential until the 6 th week after conception. Mammals sex determination is a complex process, which involves many genes acting in networks. Several genes have been involved in the development of the urogenital ridge, including Emx2, Lim1, Lhx9, Wt1, Gata-4/Fog2, Nr5a1/Sf1. Although knockout models of these genes produce abnormal gonads in mice, not all of them have been implicated in the human gonadal dysgenesis etiology. To date, Emx2 null mice have absent kidneys, ureters, gonads and genital tracts and have developmental abnormalities of the brain (34). In humans, variants in EMX2 have been found in patients with schizencephaly (a rare condition in which a person is born with clefts in the brain that are filled with liquor) but no gonadal phenotypes have been described. WT1, NR5A1 and NR0B1/DAX1 are well known genes that are critical for the formation of the urogenital ridge in humans. The products of WT1 are essential for both gonadal and renal formation (35) whereas NR5A1/SF1 protein is essential for gonadal and adrenal formation (36,37). NR0B1/DAX1 is also essential for gonadal and adrenal differentiation and when mutated, results in congenital adrenal hypoplasia and hypogonadotropic hypogonadism (38). After the formation of the bipotential gonad, by the 6 th week after conception, in 46, XY individuals, the expression of the testis-determining gene Sry, which is transcriptionally regulated by the expression of Wt1 (39) and GATA Binding Protein 4 (Gata4), its cofactor the Friend-of-GATA (Fog2) (40) and chromobox protein homolog 2 (Cbx2) (41) trigger the gonadal masculinizing fate process. In the mammalian male embryo, the first molecular signal of sex determination is the expression of Sry within a subpopulation of somatic cells of the indifferent genital ridge (42). The transient expression of Sry drives the initial differentiation of pre-Sertoli cells that would otherwise follow a female pathway, becoming granulosa cells. Once Sry expression begins, it initiates the cascade of gene interactions and cellular events that direct the formation of a testis from the undifferentiated fetal gonad. So, pre-Sertoli cells proliferate, polarize and aggregate around the germ cells to define the testes cords. Migration of cells into the gonad from the mesonephros or the coelomic epithelium is subsequently induced by signals emanating from the pre-Sertoli cells. Peritubular myoid cells surround the testes cords and cooperate with pre-Sertoli cells to deposit the basal lamina and further define the testis cords. Signaling molecules produced by the pre-Sertoli cells promote the differentiation of somatic cells, found outside the cords, into fetal Leydig cells, thus ultimately allowing the production of testosterone. Endothelial cells are associated to form the coelomic vessel, which promotes efficient export of testosterone into plasma. The gene Sox9 is up-regulated immediately after Sry expression and is involved in the initiation and maintenance of Sertoli cell differentiation during the early phases of testis differentiation. The mechanism by which NR5A1 and SRY increase endogenous SOX9 expression was clearly demonstrated in human embryonal carcinoma cell line NT2/D1 (43). Extracellular signaling pathways (Fgf9 and Igf1r/Irr/Ir) play a significant role in Sox9 expression. A model has been suggested in that the fate of the bipotential gonad is controlled by mutually antagonistic signals between Fgf9 and Wnt4/Rspo1. In this model Sox9 up-regulates Fgf9-Fgfr2 and Fgf9 maintains Sox9 expression, forming a positive feed-forward loop in XY gonads. The balance between Fgf9 and Wnt4/Rspo1 signals is shifted in favor of Fgf9, establishing the male pathway. In addition, Sry inhibits β-catenin-mediated Wnt signaling (44). In the absence of this feed-forward loop between Sox9 and Fgf9, Wnt4/Rspo1, the activated β-catenin pathway, blocks Fgf9 and promotes the ovarian fate (45,46). Furthermore, Sox9 directly binds to the promoter of the Ptgds gene which encodes prostaglandin D synthase that mediates the production of PGD2 (47) which, in turn, promotes nuclear translocation of Sox9, facilitating Sertoli cell differentiation (48). Antagonism between Dmrt1 and Foxl2 comprises another step for sex-determining decision. Dmrt1 has been described as essential to maintain mammalian testis determination, preventing female reprogramming in the postnatal mammalian testis (49). MAP3K1 has been described to be important to the balance between SOX9/FGF9 to WNT/beta-catenin signaling in functional studies (50,51). However, the role of MAP3K1 in human sex-determination remains unknown as the downstream effectors of MAP3K1 in the human developing testis have not been identified (52). Similarly, the precise mechanism by which DHX37 interferes with testis determination/maintenance remains to be elucidated (53). Abnormalities in the expression (underexpression or overexpression or timing of expression) of genes involved in the cascade of testis determination can cause anomalies of gonadal development and consequently, 46,XY DSD. The absence, regression, or the presence of dysgenetic testes results in abnormal development of the genital ducts and/or external genitalia in those patients. 46,XY Gonadal Agenesis Total absence of gonadal tissue confirmed by laparoscopy has rarely been described in XY subjects with female external and internal genitalia indicating the absence of testicular determination (54). Mendonca et al described a pair of siblings, one XY and the other XX, born to a consanguineous marriage, with normal female external and internal genitalia associated with gonadal agenesis (55). Pathogenic allelic variants in NR5A1 and LHX9 were later ruled out in these siblings (56). The origin of this disorder remains to be determined, but a defect in another gene essential for bipotential gonad development is the most likely cause of this disorder. 46,XY Gonadal Dysgenesis - Complete and Partial Forms 46,XY gonadal dysgenesis consists of a variety of clinical conditions, in which the development of the fetal gonad is abnormal and encompasses both a complete and a partial form. The complete form of gonadal dysgenesis was first described by Swyer et al. (57) and is characterized by female external and internal genitalia, lack of secondary sexual characteristics, normal or tall stature without somatic stigmata of Turner syndrome, eunuchoid habitus and the presence of bilateral dysgenetic gonads in XY subjects. Mild clitoromegaly is present in some cases. The partial form of this syndrome is characterized by variable degrees of impaired testicular development and testicular function. These patients present a spectrum of atypical genitalia with or without Müllerian structures. Similar phenotypes can also result from a 45,X/46,XY karyotype. Serum gonadotropin levels are elevated in both the complete and partial forms, mainly FSH levels, which predominate over LH levels. Testosterone levels are at the prepubertal range in the complete form. Meanwhile, in the partial form, it can range from prepubertal levels to normal adult male levels. The clinical condition named embryonic testicular regression syndrome (ETRS) has been considered part of the clinical spectrum of partial 46,XY gonadal dysgenesis (58). In this syndrome, most of the patients present atypical genitalia or micropenis associated with complete regression of testicular tissue in one or both sides. Pathogenic/likely pathogenic variants in DHX37 were reported in patients with 46,XY GD at a frequency of 14%. Considering only the ETRS phenotype (micropenis and absence of uni- or bilateral testicular tissue), this frequency increases to 50% (53). The masculinization degrees of internal and external genitalia presented are related to the time and duration of the hormonal secretion, prior to cessation of testicular function. The dysgenetic testes showed disorganized seminiferous tubules and stroma with occasional primitive sex cords without germ cells (59). Familial cases of gonadal dysgenesis with variable degrees of genital atypia have been reported, and the nature of the underlying genetic defect is still unknown in several families, despite new genetic investigation methodologies available (58). Regarding the genetic etiology, 46,XY gonadal dysgenesis is heterogeneous and can result from defects of any gene involved in the process of gonadal formation. The following review will focus on the main genes causing gonadal dysgenesis in humans, presenting as an isolated or syndromic phenotype. Dysgenetic 46,XY DSD Due to Under Expression of GATA4 and FOG2/ ZFPM2 Genes Gata4 (GATA-binding factor 4 gene) cooperatively interacts with several proteins to regulate the expression of genes involved in testis determination and differentiation as SRY, SOX9, NR5A1, AMH, DMRT1, STAR, CYP19A1, and others (60). In humans, GATA4 variants were first described in patients with congenital heart defects without genital abnormalities (61). However, genitourinary anomalies, such as hypospadias and cryptorchidism, were described in 46,XY patients with deletion of the 8p23.1 region, in which GATA4 is located (62). The p.G221R GATA4 pathogenic variant was identified in five members of a French family, three 46,XY DSD patients, two of them with cardiac anomalies, and in their apparently unaffected mothers (63). The role of FOG2 in human testis development was corroborated by the identification of a balanced translocation (8;10) (q23.1;q21.1) in a patient with partial gonadal dysgenesis and congenital heart abnormalities (64). Bashamboo et al. identified missense FOG2 variants, using exome sequencing, in two patients with 46,XY gonadal dysgenesis. One patient carried the non-synonymous p.S402R heterozygous variant. The second patient carried the inherited homozygous p.M544I variant and the de novo heterozygous p.R260Q variant. The p.M544I variant by itself has little effect on the biological activity of FOG2 protein in transactivation of the gonadal promoters, but it shows reduced binding with GATA4. In the in vitro assays, a combination of both the p.R260Q and the p.M544I variants altered the biological activity of the FOG2 protein on specific downstream targets, as well as obliterated its interaction with GATA4. In the patient, the two variants together may result in an imbalance of the delicate equilibrium between antagonistic male and female pathways leading ultimately to gonadal dysgenesis (65). Although several GATA4 and FOG2/ZFPM2 variants have been identified in 46,XY DSD patients, the real role of the majority of them in the etiology of gonadal disease is still unclear. The re-study of seven GATA4 and ten FOG2/ZFPM2 variants previously identified by Eggers et al. (26) in a cohort of 46,XY DSD patients, using updated tools and testing their molecular activity in the context of gonadal signaling by in vitro assays, support that the majority of them are benign in their contribution to 46,XY DSD. Only one variant (p. W228C) located in the conserved N-terminal zinc finger of GATA4, was considered pathogenic, with functional analysis confirming differences in its ability to regulate Sox9 and AMH, and in protein interaction with ZFPM2 (66). Dysgenetic 46,XY DSD Due to Under Expression of the CBX2 Gene In humans, variants in both CBX2.1 and CBX2.2 isoforms were associated with 46,XY DSD (67,68). The compound heterozygous CBX2.1 variants, c.C293T (p.P98L) and c.G1370C (p.R443P), inherited from the father and the mother respectively, was identified in a 46,XY patient, who was born with a completely normal female phenotype. The patient had uterus and histologically normal ovaries (67) and high serum FSH levels. Her phenotype resembles the Cbx2 knock-out XY mice phenotype (41). Cbx2 (M33) knockout mice present hypoplastic gonads in both sexes, but a small or absent ovaries are observed in the XY Cbx2 knockout, consequently to the reduced expression of Sry and Sox9 in the gonadal tissue (41). Functional studies demonstrated that these variants do not bind to, or adequately regulate the expression of target genes important for gonadal development, such as NR5A1 (67). Mutated CBX2.2 isoforms were also implicated in the etiology of partial 46,XY gonadal dysgenesis in two other patients. Each patient carried a distinct variant, the p. C132R (c.394T>C) and the C154fs (c.460delT). These CBX2.2 variants were shown to be related to a defective expression of EMX2 in the developing gonad (68). However, analysis of populational genetics data indicates that p.C154fs is present in general populations at high frequency, inconsistent with causing gonadal dysgenesis (69). 46,XY DSD Due to Under Expression of the WT1 Gene The Wilms’ tumor suppressor gene (WT1) is located on 11p13, and encodes a zinc-finger transcription factor involved in the development and function of the kidneys and gonads. The WT1 contains 10 exons, of which exons 1–6 encode a proline/glutamine-rich transcriptional-regulation region and exons 7–10 encode the four zinc fingers of the DNA-binding domain. Four major species of RNA with conserved relative amounts, different binding specificities, and different subnuclear localizations are generated by two alternative splicing regions (70). Splicing at the first site results in either inclusion or exclusion of exon 5. The second alternative splicing site is in the 3’ end of exon 9 and allows the inclusion or exclusion of three amino acids lysine, threonine and serine between the third and fourth zinc fingers, resulting in either KTS-positive or negative isoforms. Isoforms that only differ by the presence or absence of the KTS amino acids have different affinities for DNA and, therefore, possibly different regulatory functions (70). A precise balance between WT1 isoforms is necessary for its normal function (71). WT1 presents a complex network of interaction with several protein systems so that abnormalities in it can determine a wide phenotypic spectrum in XY and XX individuals (72). Several syndromes are associated with WT1 pathogenic variants, including: WAGR, Denys-Drash and Frasier syndrome. WAGR syndrome is characterized by Wilms’ tumor, aniridia, genitourinary abnormalities and mental retardation. Genitourinary anomalies are frequently observed, such as renal agenesis or horseshoe kidney, urethral atresia, hypospadias, cryptorchidism and more rarely atypical genitalia (73). Heterozygous deletions of WT1 and other contiguous genes are the cause of this syndrome (74). Deletions of PAX6 gene are related to the presence of aniridia in these patients. Severe obesity is present in a substantial proportion of subjects with the WAGR syndrome, and the acronym WAGRO has been suggested for this condition (75). The phenotype of obesity and hyperphagia in WAGRO syndrome is attributable to deletions that determine haploinsufficiency of the BDNF gene (Brain-Derived Neurotrophic Factor) (76). Le Caignec et al. described a 46,XY patient with an interstitial deletion of approximately 10 Mb located on 11p13, encompassing WT1 and PAX6 , who presented a female external and internal genitalia, an unusual phenotype of WAGR syndrome (77). This report demonstrated an overlap of clinical and molecular features in WAGR, Frasier and Denys-Drash syndromes that confirms these conditions as a spectrum of disease due to WT1 alterations. Denys-Drash syndrome is characterized by dysgenetic 46,XY DSD associated with early-onset renal failure (steroid-resistant nephrotic syndrome with diffuse mesangial sclerosis and progression to end-stage kidney disease) and Wilms´ tumor development in the first decade of life (78). Müllerian duct differentiation varies according to the Sertoli cells' function. The molecular defect of this syndrome is the presence of heterozygous missense allelic variants in the zinc finger encoding exons (DNA-binding domain) of WT1 gene (79). Gonadal development is impaired to variable degrees, resulting in a spectrum of 46,XY DSD (80). Frasier syndrome is characterized by a female to atypical external genitalia phenotype in 46,XY patients, streak or dysgenetic gonads, which are at high risk for gonadoblastoma development, and renal failure (early steroid-resistant nephrotic syndrome with focal and segmental glomerular sclerosis). In these patients the progression to end-stage renal disease often occurs in adolescence, although the late-onset nephropathy has also been described in Frasier syndrome (81), reinforcing that patients carrying WT1 pathogenic variants should have the renal function carefully monitored (82). Constitutional heterozygous variants of the WT1, almost all located at intron 9, are found in patients with Frasier syndrome, leading to a change in splicing that results in reversal of the normal KTS positive/negative ratio from 2:1 to 1:2 (78,83). Exonic variants have been also associated with Frasier syndrome (84). The report of atypical external genitalia (84) , the presence of Wilms’ tumor (85), and the description of exonic variants in the DNA binding domain of WT1 gene (84) in patients with Frasier syndrome indicate an overlap of clinical and molecular features between Denys Drash and Frasier syndromes. 46,XY DSD Due to Under Expression of the NR5A1/SF1 Gene NR5A1 was originally identified as a master-regulator of steroidogenic enzymes in the early 1990s following the Keith L. Parker and Kenichirou Morohashi inspiring work (86-88). NR5A1 has since been shown to control many aspects of adrenal and gonadal function (37,89), NR5A1, together with several signaling molecules, are also involved in adrenal stem cell maintenance, proliferation and differentiation inducing adrenal zonation, probably acting in the progenitor cells (90). Homozygous 46,XY null mice (−/−) have adrenal agenesis, complete testicular dysgenesis, persistent Müllerian structures, partial hypogonadotropic hypogonadism, and other features such as late-onset obesity (91). Therefore, it was clear demonstrated that Nr5a1 is an essential factor in sexual and adrenal differentiation, and a key regulator of adrenal and gonadal steroidogenesis and also of the hypothalamic-pituitary-gonadal axis. The first reported human case of NR5A1 pathogenic variant, the heterozygous p.G35E, was a 46,XY patient who presented female external genitalia and Müllerian duct derivatives, indicating the absence of male gonadal development, associated with adrenal insufficiency. This patient presented with salt-losing adrenal failure in early infancy and was thought to have a high block in steroidogenesis (e.g., in CYP11A1, STAR) affecting both adrenal and testicular functions. However, the identification of streak-like gonad and Müllerian structures was consistent with testicular dysgenesis, thereby, a disruption of a common developmental regulator such as NR5A1 was hypothesized. The patient was found to have a de novo heterozygous p.G35E change in the P-bo of NR5A1, which is important in dictating DNA binding specificity through its interaction with DNA response elements in the regulatory regions of target genes (92). The second report of NR5A1 defects in humans was described by Biason-Lauber and Schoenle, in a 14-month-old 46,XX girl who had presented with primary adrenal insufficiency and seizures (93) . She had a de novo heterozygous NR5A1 change resulting in the p.R255L variant into the proximal part of the ligand-like binding domain of the protein. The ovaries were detected by MRI scan and Inhibin A levels were normal for her age, suggesting that NR5A1 change had not disrupted ovarian function. The follow up of this girl until 16.5 years old showed a normal puberty and regular menstruation showing that phenotypic variant of NR5A1 allelic variant in a 46, XX affected person includes adrenocortical insufficiency but no ovarian dysfunction at pubertal age (94). The third report of NR5A1 defects in humans was found in an infant with a similar phenotype of the first case: primary adrenal failure and 46,XY DSD. However, this child had inherited the homozygous p.R92Q alteration in a recessive manner (95). The change lies within the A-box of NR5A1, which interferes with monomeric DNA binding stability, but in vitro functional activity was in the order of 30–40% of the wild type (95-97). Carrier parents showed normal adrenal function suggesting that the loss of both alleles is required for the phenotype development when disrupted protein keeps this level of functional activity. In addition, another family has been reported with a homozygous missense variant (p.D293N) in the LBD of NR5A1 (98). This change also showed partial loss-of-function (50%) in gene transcription assays. In 2004, we reported the fourth NR5A1 deleterious variant in humans, which brought two novel variables to the NR5A1 phenotype: it was the first frameshift variant and it was identified in a 34-year-old 46,XY DSD female with normal adrenal function (99). Another interesting aspect in this patient was the absence of gonadal tissue at laparoscopy. Since she had atypical genitalia and absence of Müllerian derivatives, we assumed that testicular tissue regressed completely late in fetal life. NR5A1 changes associated with 46,XY DSD are usually frameshift, nonsense or missense changes that affect DNA-binding and gene transcription (96). Most of the point variants identified in NR5A1 are located in the DNA-binding domain of the protein. The p.L437Q variant, the first located in the ligand-binding region, was identified in a patient with a mild phenotype, a penoscrotal hypospadias. This protein retained partial function in several NR5A1-expressing cell lines and its location points to the existence of a ligand for NR5A1, considered an orphan receptor so far (97). NR5A1 is bound to sphingosine (SPH) and lyso-sphingomyelin (lysoSM) under basal conditions (100,101). Progressive androgen production and virilization in adolescence has been observed in several XY patients with NR5A1 variants, in contrast to the severe under virilized external genitalia found in most patients (101,102). The almost normal testosterone levels after hCG stimulation test or at pubertal age suggest that NR5A1 action might be less implicated in pubertal steroidogenesis than during fetal life. In contrast, fetal Sertoli cell function seems to be preserved in most patients with heterozygous NR5A1 variants based on the common observation of absent Müllerian derivatives and primitive seminiferous tubules in histology. The reviewed data of seventy-two 46,XY DSD patients with NR5A1 pathogenic variants reported in the literature, for whom information on the presence or absence of Müllerian derivatives was available, suggested that Müllerian derivatives are present in about 24% of the cases. However, persistently elevated FSH levels after puberty found in all patients studied suggest an impairment of Sertoli cells function in post pubertal age. More than 180 different NR5A1 variants, distributed across the full length of the protein, have been described and the majority are nonsynonymous variants (103-105). Most of these variants are in the DNA binding domain and are in heterozygous state or compound heterozygous state with the p.G46A (rs1110061) variant. A clear correlation between the location of a variant, it’s in vitro functional performance and the associated phenotype is not observed. Indeed, family members bearing the same NR5A1 variant may present with different phenotypes (106). The contribution of other genetic modifiers has been suggested to explain phenotypic variability. Exome sequencing analyses of DSD patients have identified pathogenic variants or variants of uncertain significance in several genes involved in sexual development (42). In a 46,XY patient with atypical external genitalia, palpable inguinal gonads, absent uterus in pelvic ultrasonography and poor testosterone response to hCG stimulation, Mazen and colleagues identified, by exome sequencing, the previously described p.Arg313Cys NR5A1 variant in compound heterozygous state with a p.Gln237Arg MAP3K1 variant 27169744 (107). This NR5A1 variant was previously reported in association with mild hypospadias (108), and a possible digenic inheritance was proposed to explain the phenotypic heterogeneity (107). In several cohort studies, NR5A1 changes have been reported in approximately 10–15% of the individuals with gonadal dysgenesis (89,96). Although many of the heterozygous changes are de novo, about one-third of these changes have been shown to be inherited from the mother in a sex-limited dominant manner (96). These women are at potential risk of primary ovarian insufficiency but while fertile they can pass NR5A1 heterozygous changes to their children. This mode of transmission can mimic X-linked inheritance (96). The features in different affected family members can be variable. A different role of NR5A1 in human reproductive function was described by Bashamboo and co-workers (109). They investigated whether changes in NR5A1 could be found in a cohort of 315 men with normal external genitalia and non-obstructive male factor infertility where the underlying cause was unknown (109). Analysis of NR5A1 in this cohort identified heterozygous changes in seven individuals; all of them were located within the hinge region of the NR5A1 protein. The men who harbored NR5A1 changes had more severe forms of infertility (azoospermia, severe oligozoospermia) and in several cases low testosterone and elevated gonadotropins were found. A serial decrease in sperm count was found in one-studied men, raising the possibility that heterozygous changes in NR5A1 might be transmitted to offspring, especially if fatherhood occurs in young adulthood rather than later in life (110). As progressive gonadal dysgenesis is likely, gonadal function should be monitored in adolescence and adulthood, and early sperm cryopreservation considered in male patients, if possible. In conclusion, this study shows that changes in NR5A1 may be found in a small subset of phenotypically normal men with non- obstructive male factor infertility where the cause is currently unknown. These individuals may be at risk of low testosterone in adult life and may represent part of the adult testicular dysgenesis syndrome (110,111). A novel heterozygous missense variant (p.V355M) in NR5A1 was identified in one boy with a micropenis and testicular regression syndrome (112). NR5A1 variants have also been identified in familial and sporadic forms of 46,XX primary ovarian insufficiency not associated with adrenal failure (98,113). Most of these women harbored heterozygous alterations in NR5A1 and had been identified in families with histories of 46,XY DSD and 46,XX POI. Heterozygous NR5A1 changes were also found in two girls with sporadic forms of POI (98). In one large kindred, a partial loss-of-function NR5A1 change (p.D293N) was inherited in an autosomal recessive manner. These 46,XX women with p.D293N NR5A1 variant presented with either primary or secondary amenorrhea and with a variable age of features onset. The detection of NR5A1 alterations in 46,XX ovarian failure shows that NR5A1 is also a key factor in ovarian development and function in humans. Thus, some 46,XX women with NR5A1 variants have normal ovarian function and can transmit the variant in a sex-limited dominant pattern. Therefore, the inheritance patterns associated with NR5A1 changes can be autosomal dominant, autosomal recessive or sex-limited dominant. NR5A1 defects can be found in association with a wide range of human reproductive phenotypes such as 46,XY and 46,XX disorders of sex development (DSD) associated or not with primary adrenal insufficiency, male infertility, primary ovarian insufficiency and finally testicular or ovo-testicular 46,XX DSD (101) (103) (Table 4). Spleen development anomalies have been described in patients with NR5A1 variants (103). Table 2. Spectrum of Phenotypes Caused by NR5A1 Defects View in own window \| Karyotype \| Phenotypes \| Number of reported patients \| Reference \| --- --- \| \| 46,XY \| DSD and adrenal insufficiency \| 2 \| (92,95) \| \| DSD without adrenal insufficiency \| 69 \| (63,89,96,98,100,101,103,106) \| \| Male infertility \| 10 \| (63) \| \| Ovotesticular DSD and genitopatellar syndrome \| 1 \| (114) \| \| 46,XX \| Adrenal insufficiency \| 2 \| (92,93) \| \| Female infertility, POI \| 14 \| (98,101,103) \| \| (Ovo) testicular DSD without adrenal insufficiency \| 11 \| (98,115) \| 46,XY DSD Due to Under Expression of the SRY Gene Most of the authors reported pathogenic allelic variants in SRY gene in less than 20% of the patients with complete 46,XY gonadal dysgenesis (116-118). In the partial form, the frequency of SRY variants is even lower than in the complete form. To date, most of the SRY variants are located in the HMG box, showing the critical role of this domain, and are predominantly de novo variants. However, some cases of fertile fathers and their XY affected children, sharing the same altered SRY sequence, have been reported (116,119). In a few of these cases, the father’s somatic mosaicism for the normal and mutant SRY gene has been proven (120) The variable penetrance of SRY variants in familial cases have been described in SRY mutant proteins with relatively well preserved in vitro activity (121). Dysgenetic 46,XY DSD Associated with Campomelic Dysplasia (Under Expression of the SOX9 Gene) SRY-related HMG-box gene 9 (SOX9) is a transcription factor involved in chondrogenesis and sex determination.SOX9 gene, located on human chromosome 17, is a highly conserved HMG family member and it is also implicated in the male sex-determining pathway (122,123). Pathogenic allelic variants in SOX9 have been identified in heterozygous state in patients with Campomelic dysplasia (122). This syndrome is characterized by severe skeletal malformations associated with dysgenetic 46,XY DSD. These patients have variable external genitalia ranging from that of normal male with cryptorchidism to atypical or female genitalia, and the internal genitalia may include vagina, uterus, and fallopian tubes (124). Intact SOX9 were also reported in patients with Campomelic dysplasia and 46,XY gonadal dysgenesis. The genomic analysis of the SOX9 locus in these patients identified a key regulatory element termed RevSex, located approximately 600 kb upstream from SOX9. RevSex is duplicated in individuals with 46,XX (ovo)testicular DSD and deleted in individuals with 46,XY GD (125,126). Moreover, structural changes involving multiple regions both upstream and downstream of the SOX9 gene have been associated with non-syndromic XY DSD (127,128). These findings indicate that variants located in the regulatory elements of SOX9 should be routinely screened in a DSD diagnostic setting (69). Dysgenetic 46,XY DSD Due to Under Expression of the FGF9/FGFR2 Genes The importance of Fgf9/Fgfr2 signaling pathway in mouse testis determination is well known (129,130). In the developing testis occurs a positive feedback loop among Fgf9/Fgfr2/Sox9;Fgf9 is upregulated by Sox9 and signals through Fgfr2 maintain Sox9 expression (129) and this loop represses Wnt4(131). Mice homozygous for a null variant in Fgf9 or Fgfr2 exhibit male-to-female sex reversal, with all testis-specific cellular events being disrupted, including cell proliferation, mesonephric cell migration, Sertoli cell differentiation, and testis cord formation (129,130,132). However, in human sex development the role of FGF9 and FGFR2 remains unclear. In humans, the only reported pathogenic variants in FGF9 are associated with craniosynostosis or multiple synostosis phenotypes, and no FGF9 variants were identified in 46,XY GD patients (133). Human FGFR2 variants have been related with some syndromes as lacrimo-auriculo-dento-digital, characterized by tear tract, ear, teeth and digit abnormalities (133) and craniosynostosis syndromes including Crouzon, Pfeiffer, Apert and Antley-Bixler syndromes (134-136). FGFR2 variants can lead to loss (LAAD syndrome) or gain (craniosynostosis syndromes) of function in these disorders (137). No gonadal defects were described in patients with LADD or craniosynostosis syndromes. A single 46,XY patient with gonadal dysgenesis and craniosynostosis was described by Bagheri-Fam et al (138). This patient had abnormalities which are identified in several craniosynostosis syndromes (short stature, brachycephaly, proptosis, down slanting palpebral fissures, low-set dorsally rotated ears, reduced extension at the elbows but absence of hand and feet anomalies). She also presented female external genitalia, primary amenorrhea and gonadal dysgenesis with dysgerminoma. DNA sequencing revealed a cysteine-to-serine substitution at position 342 in the FGFR2c isoform (p.C342S). Cys342 substitutions by Ser or other amino acids (Arg/Phe/Trp/Tyr) occur frequently in the craniosynostosis syndromes Crouzon and Pfeiffer but these patients do not present gonadal abnormalities. Variants in the 2c isoform of FGFR2 is in agreement with knockout data showing that FGFR2c is the critical isoform during sex determination in the mouse. Taken together, these data suggest that the FGFR2c c.1025G>C (p.C342S) variant might contribute to 46,XY DSD in this patient. The authors proposed that this heterozygous variant leads to gain of function in the skull, but to loss of function in the developing gonads and that she might harbor a unique set of modifier genes, which exacerbate this testicular phenotype (138). The authors proposed that the p.C342S heterozygous variant in FGFR2c leads to gain of function in the skull, but loss of function in the developing gonads; and that the presence of modifier genes would exacerbate the testicular phenotype in this patient (138). However, the presence of a pathogenic variant involving other DSD genes, cannot be completely excluded in this patient. Dysgenetic 46,XY DSD Due to Disruption in Hedgehog Signaling DESERT HEDGEHOG (DHH) GENE It is a member of the hedgehog family of signaling proteins, is located in chromosome 12-q13.1 and is one of the genes involved in the testis-determining pathway (139). Dhh seems to be necessary for Nr5a1 up-regulation in Leydig cells in mice (140). To date, six homozygous variants have been described in DHH gene in 46,XY patients conferring phenotypes ranging from partial to complete gonadal dysgenesis, associated or not with polyneuropathy. The first one, the homozygous missense variant (p.M1T) is located at the initiation codon of exon 1 and was found in a 46,XY patient with partial gonadal dysgenesis associated with polyneuropathy (141). Two other variants, one the p.L162P located at exon 2 and the other the p.L363CfsX4 located in exon 3 were identified in three patients with complete gonadal dysgenesis without polyneuropathy; two of them harbored gonadal tumors (bilateral gonadoblastoma and dysgerminoma, respectively) (142). Later, the c.1086delG variant was identified in heterozygous state in two patients with partial gonadal dysgenesis (143). In addition, two novel homozygous variants were described in two patients with complete 46,XY gonadal dysgenesis without clinically overt polyneuropathy (144). In both sisters, clinical neurological examination revealed signs of a glove and stocking like polyneuropathy. The first defect, the c.271_273delGAC resulted in deletion of one amino acid (p.D90del) and the second one, a duplication c.57_60dupAGCC resulted in a premature termination of DHH protein (144) . The p.R124Q variant was identified by exome sequencing in two sisters of a consanguineous family with 46, XY gonadal dysgenesis and testicular seminoma (145). HEDGEHOG ACETYL-TRANSFERASE (HHAT) GENE The HHAT protein is a member of the MBOAT family of membrane-bound acyl-transferases which catalyzes amino-terminal palmitoylation of Hh proteins. The novel variant (p.G287V) in the HHAT gene was found in a syndromic 46,XY DSD patient with complete gonadal dysgenesis and skeletal malformation by exome sequencing. This variant disrupted the ability of HHAT protein to palmitoylated Hh proteins including DHH and SHH (146) In mice, the absence of Hhat in the XY gonad did not affect testis-determination, but impaired fetal Leydig cells and testis cords development (146). The phenotype of the girl carrying the homozygous p.G287V variant is a rare combination of gonadal dysgenesis and chondrodysplasia. Moreover, a de novo dominant variant in the MBOAT domain of HHAT was reported in association with intellectual disability and apparently normal testis development (147). 46,XY DSD Due to Under Expression of the DMRT1 Gene Raymond et al identified both DNA-binding Motif (DM) domain genes expressed in testis (DMRT1 and DMRT2) located in chromosome 9p24.3, a region associated with gonadal dysgenesis and 46,XY DSD (148-150). The human 9p monosomy syndrome is characterized by variable degrees of 46,XY DSD, from female genitalia to male external genitalia with cryptorchidism associated to agonadism, streak gonads or hypoplastic testes and internal genitalia disclosing normal Müllerian or Wolffian ducts, mental retardation and craniofacial abnormalities (151). Gonadal function varies from insufficient to near normal testicular production. It is inferred that haploinsufficiency of DMRT1 and DMRT2 primarily impairs the formation of the undifferentiated gonad, leading to various degrees of testis or ovary formation defects (151). Although 9p24 deletions are a relatively common cause of syndromic 46,XY gonadal dysgenesis, the pathogenic variants within DMRT1 are rarely identified (152). Genomic–wide copy number variation screening revealed that DMRT1 deletions were associated with isolated 46,XY gonadal dysgenesis in addition to inactivation variants (133,148).In vitro studies to analyze the functional activity of the DMRT1 (p.R111G) variant identified by exome sequencing in a patient with 46,XY complete gonadal dysgenesis, indicated that this protein had reduced DNA affinity and altered sequence specificity. This mutant DMRT1, when mixed with the wild-type protein bound as a tetramer complex to an in vitro Sox9 DMRT1-binding site, differs from the wild-type DMRT1 that is usually bound as a trimer. This suggests that a combination of haploinsufficiency and a dominant disruption of the normal DMRT1 target binding site is the cause of the abnormal process of testis-determination seen in this patient (153). Matson et al. (2011) have shown in mice that Dmrt1 and Foxl2 create another regulatory network necessary for maintenance of the testis during adulthood. Loss of Dmrt1 in mouse Sertoli cells induces the reprogramming of those into granulosa cells, due to Foxl2 upregulation. Consequently, theca cells are formed, estrogens are produced, and germ cells appear feminized (49). ATR-X Syndrome (X-linked α-Thalassemia and Mental Retardation) ATR-X syndrome results from variants in the gene that encodes for X-linked helicase-2, implicating ATR-X in the development of the human testis (154). Genital anomalies leading to a female sex of rearing were reported in several affected 46,XY patients with ATR-X syndrome (155). ATR-X syndrome is characterized by severe mental retardation, alpha thalassemia and a range of genital abnormalities in 80% of cases (154). In addition to these definitive phenotypes, patients also present with typical facial anomalies comprising a carp-like mouth and a small triangular nose, skeletal deformities and a range of lung, kidney, and digestive problems. A variety of phenotypically overlapping conditions (Carpenter-Waziri syndrome, Holmes-Gang syndrome, Jubert-Marsidi syndrome, Smith-Fineman-Myers syndrome, Chudley-Lowry syndrome and X-linked mental retardation with spastic paraplegia without thalassemia) have also been associated with ATRX variants (154). ATRX lies on the X chromosome (Xq13) and the disease has been confined to males; in female carriers of an ATRX variant, the X-inactivating pattern is skewed against the X chromosome carrying the mutant allele. Urogenital abnormalities associated with variants in human ATRX range from undescended testes to testicular dysgenesis with female or atypical genitalia. Duplication of Xq12.2-Xq21.31 that encompasses ATRX along with other genes has been described in a male patient with bilateral cryptorchidism and severe mental retardation. The patient entered spontaneous puberty by the age of 12 and developed bilateral gynecomastia (156). There are two major functional domains in ATRX protein: 1- the ATRX-DNMT3-DNMT3L (ADD) domain at the N-terminus and 2- the helicase/ATPase domain at the C-terminal half of the protein, both acting as chromatin remodeling. variants in the ADD domain have been related to severe psychomotor impairment associated with urogenital abnormalities. On the other hand, variants in the C-terminus region have been related with mild psychomotor impairment without severe urogenital abnormalities (157,158). Although all cases of severe genital abnormality reported in ATRX syndrome have been associated with severe mental retardation, this is not true for alpha-thalassemia. The role of ATRX in the sexual development cascade is poorly understood and it is suggested that it could be involved in the development of the Leydig cells (159). Dysgenetic 46,XY DSD Due to Under Expression of the MAP3K1 Gene MAPK signaling pathway role in mammalian sex-determination is still poorly understood. In mice, it has been shown that the Map3k4 gene is essential for testicular determination, since the lack of activity of this protein leads to failure of testicular cord development and disorganization of gonadal tissue in development (160). In mice, the reduction of the Gadd45/Map3k4/p38 pathway activity is associated with a reduction in the Sry expression in the XY mice gonad at sex-determination causing sex-reversal in these animals (161). Studies with knock-in animals for the Map3k1 gene demonstrated a lower repercussion in the testicular tissue, which present a reduction in the Leydig cells number (162,163). However, in patients with 46, XY gonadal dysgenesis, different non-synonymous allelic variants were identified in the MAP3K1 gene. The first variant described was identified for mapping by linkage analysis of an autosomal sex-determining gene locus at the long arm of chromosome 5 in two families with 46,XY DSD, including patients with complete and partial gonadal dysgenesis. The splice-acceptor variant c.634-8T>A in the MAP3K1 disrupted RNA splicing and was segregated with the phenotype in the first family. Variants in the MAP3K1 were also demonstrated in the second family (p.G616R) and in two of 11 sporadic 46,XY DSD patients (p.L189P, p.L189R) studied (51,164). Subsequently, the two novel variants p.P153L and c.2180- 2A>G in the MAP3K1 were identified in non-syndromic patients with 46,XY gonadal dysgenesis. Functional studies of mutated MAP3K1 proteins identified change in phosphorylation targets in subsequent steps of the cascade of MAP3K1, p38 and ERK1/2 and enhanced the binding of the Ras homolog gene family, member A (RHOA) to the MAP3K1 complex (51). In normal male gonadal development, the binding of MAP3K1 to the RHOA protein promotes a normal phosphorylation of p38 and ERK1/2, and a blockade of the β-catenin pathway is determined by MAP3K4. In the female development, hyperphosphorylation of p38 and ERK1/2 occurs and the presence of p38 and ERK1/2 hyperphosphorylated determine the activation of the β-catenin pathway, that result in a block of the positive feedback pathway of SOX9 and the testicular development (51) . Cohorts of patients with 46,XY DSD studied by a targeted gene panel have found several new potentially deleterious variants and uncertain significance variants in the MAP3K1(26). Although the findings strongly indicate the participation of the MAP3K1 variants in the etiology of testicular development abnormalities, a better understanding of the mechanisms of MAPK pathway in the gene regulatory networks of the human testicular determination process is still necessary (52,107). 46,XY DSD Due to Over Expression of the NR0B1/DAX1 Gene Male patients with female or atypical external and internal genitalia due to partial duplications of Xp in the presence of an intact SRY gene have been described (28). These patients present with dysgenetic or absent gonads associated or not with mental retardation, cleft palate, and dysmorphic face. Bardoni et al. identified in these patients, a common 160-kb region of Xp2 containing NR0B1/DAX1 gene named dosage sensitive sex locus which, when duplicated, resulted in 46,XY DSD (164). The large duplications of Xp21 reported prior to array-CGH and MLPA techniques were identified by conventional karyotyping. Patients carried large genomic rearrangements involving several genes. In these patients, the presence of XY gonadal dysgenesis was part of a more complex phenotype, which also included dysmorphic features and/or mental retardation (165). Interestingly, in all cases with isolated 46,XY gonadal dysgenesis, the IL1RAPL1 gene located immediately to the duplication containing NR0B1/DAX1, is not disrupted. Deletions or variants of this gene have been identified in patients with mental retardation (166). Disruption of this gene could explain the mental retardation previously described in patients with larger Xp21 duplications (167). Several patients with isolated 46,XY gonadal dysgenesis due to duplications of Xp21 have been described. The first report identified a 637 kb tandem duplication on Xp21.2 that in addition to NR0B1/DAX1 includes the four MAGEB genes in two sisters with isolated 46,XY gonadal dysgenesis and gonadoblastoma (168). The second case exhibited a duplication with approximately 800 kb in size and, in addition to NR0B1/DAX1, contains the four MAGEB, Cxorf21 and GK genes. The healthy mother was a carrier of the duplication (169). Smyk et al. described a 21-years-old 46,XY patient manifesting primary amenorrhea, a small immature uterus, gonadal dysgenesis and absence of adrenal insufficiency with a submicroscopic deletion (257 kb) upstream of NR0B1/DAX1. The authors hypothesized that loss of regulatory sequences may have resulted in up-regulation of DAX1 expression, consistent with phenotypic consequences of NR0B1/DAX1 duplication (170). By using array-CGH and MLPA techniques, additional NR0B1/DAX1 locus duplications have been identified in patients with isolated 46,XY gonadal dysgenesis (28,169,171). Barbaro et al. identified a relatively small NR0B1/DAX1 locus duplication responsible for isolated complete 46,XY gonadal dysgenesis in a large English family (28). The duplication extends from the MAGEB genes to part of the MAP3K7IP3 gene, including NR0B1, CXorf21, and GK genes. Unfortunately, the authors were unable to set up the rearrangement mechanism and distinguish between a nonallelic homologous recombination or a nonhomologous end joining mechanism. Therefore, until now, there is not a direct proof that an isolated NR0B1/DAX1 duplication is sufficient to cause 46,XY gonadal dysgenesis in humans, suggesting that other contiguous genes located in the DSS locus, should be involved in dosage-sensitive 46,XY DSD. X-inactivation patterns in fertile female carriers of each of the three small NR0B1 locus duplications were analyzed (169). They established that female carrier of macroscopic Xp21 duplications are healthy and fertile due to the preferential inactivating of the duplicated chromosome and thereby protecting them from increased NR0B1 expression (169). 46,XY DSD Due to the Over Expression of WNT4 Gene The WNT4(wingless-type mouse mammary tumor virus integration site member 4) gene belongs to a family that consists of structurally related genes that encode cysteine-rich secreted glycoproteins that act as extracellular signaling factors (172). Overexpression of the WNT4 and RSPO1 may be a cause of 46,XY DSD. A 46,XY newborn infant, with multiple congenital anomalies including bilateral cleft lips and palate, intrauterine growth retardation, microcephaly, tetralogy of Fallot, atypical external and internal genitalia, and undescended gonads consisted of rete testes and rudimentary seminiferous tubules, who carried a duplication of 1p31-p35, including both WNT4 and RSPO1 genes, was reported (173).In vitro functional studies showed that Wnt4 up-regulates Nr0b1/Dax1 in Sertoli cells, suggesting that Nr0b1/Dax1 overexpression was the cause of 46,XY DSD in this infant (174). Table 3. Phenotypic Spectrum of Defects in the Genes Involved in Human Male Sex Determination View in own window \| Genes \| Chromosome position \| Molecular defect \| External genitalia \| Müllerian ducts derivatives \| Testes \| Associated anomalies \| Associated Syndrome \| --- --- --- --- \| \| ARX \| Xp22 \| Deletion/ Inactivating variants \| Atypical/ micropenis with cryptorchidism Dysgenetic \| Abnormal psychomotor development, epilepsy, spasticity, and intellectual disability \| X-linked lissencephaly, Proud syndrome, Ohtahara syndrome \| \| ATRX \| Xq13 \| Inactivating variants \| Atypical / Male with cryptorchidism Dysgenetic \| Severe psycho-motor retardation, dysmorphic face, cardiac and skeletal abnormalities, thalassemia \| Alpha thalassemia and mental retardation X-linked \| \| CBX2 \| 17q25 \| Inactivating variants \| Female \| + \| Normal Ovary \| No \| No \| \| DHH \| 12q12 \| Inactivating variants \| Female/Atypical \| +/- \| Dysgenetic / Testis \| Minifascicular neuropathy \| No \| \| DHX37 \| 12q24.31 \| Inactivating variants \| Male with cryptorchidism and micropenis, Atypical \| +/- \| Dysgenetic/ Absent \| No \| No \| \| DMRT1 \| 9p24 \| Deletion/Inactivating variants \| Female/ Atypical/ Male with cryptorchidism \| +/- \| Dysgenetic/Absent/ Hypoplastic \| Craniofacial Abnormalities, microcephaly, mental retardation \| No \| \| DSS locus (DAX-1 /MAGEB) \| Xp21 \| Gene duplication \| Female/ Atypical/ male \| +/- \| Dysgenetic/ Absent \| Mental retardation, cleft palate, dysmorphic face \| No \| \| FGFR2 \| 10q26 \| Inactivating variants \| Female \| ND \| Dysgenetic \| Short stature, craniofacial abnormalities, elbow and knee contractures \| Craniosynostosis syndrome \| \| FOG2/ZFPM2 \| 8q23 \| Balanced translocation, inactivating variants \| Male Probable dysgenetic \| Heart defects \| No \| \| GATA4 \| 8p23 \| Inactivating variants \| Atypical / male with micropenis Normal/ Dysgenetic \| Heart defects \| No \| \| HHAT \| 1q32 \| Inactivating variants \| Female \| + \| Dysgenetic \| Chondrodysplasia \| Nivelon-Nivelon-Mabille syndrome \| \| MAP3K1 \| 5q11.2 \| Inactivating mutation \| Female/Atypical \| + \| Dysgenetic \| No \| No \| \| MYRF \| 11q12.2 \| Inactivating variants \| Female/Atypical ND \| Congenital heart defects, urogenital anomalies, congenital diaphragmatic hernia, and pulmonary hypoplasia \| Cardiac urogenital syndrome \| \| NR5A1 \| 9q33 \| Inactivating variants \| Female/Atypical/ Male with cryptorchidism Male with spermatogenic failure \| +/- \| Normal/ Dysgenetic/ Absent \| Adrenal Insufficiency \| No \| \| PPP1R12A \| 12q21.2- q21.31 \| Inactivating variants \| Female/Atypical \| +/- \| Dysgenetic \| Genitourinary and/or brain malformations \| No \| \| SOX9 \| 17q24.3-25.1 \| Inactivating variants, 5’ and 3’ Rearrangements \| Female/ Atypical Male \| +/- \| Dysgenetic \| Severe skeletal defects \| Campomelic displasia \| \| SRY \| Yp11.3 \| Inactivating variants \| Female/ Atypical \| + \| Dysgenetic \| No \| No \| \| WNT4 /RSPO1 locus \| 1p34.3-p35 \| Gene duplication \| Atypical \| + \| Dysgenetic \| Cleft lips and palate, tetralogy of Fallot, intrauterine growth retardation, microcephaly \| No \| \| WT1 \| 11p13 \| Inactivating variants \| Female/ Atypical \| +/- \| Dysgenetic \| Late-onset renal failure Gonadoblastoma \| Frasier \| \| Inactivating variants \| Atypical \| +/- \| Dysgenetic \| Early-onset renal failure, Wilm's tumor \| Denys-Drash \| \| Inactivating variants \| Female/ Atypical / Male with cryptorchidism Dysgenetic \| Mental retardation, Wilm's tumor, Aniridia, renal agenesis or horseshoe kidney \| WAGR \| \| WWOX \| 16q23 \| Multi-exons deletion \| Atypical \| + \| Dysgenetic \| No ND: data not described Go to: 46,XY DSD ASSOCIATED WITH CHOLESTEROL SYNTHESIS DEFECTS Smith-Lemli-Opitz Syndrome (SLOS) This syndrome,caused by a deficiency of 7-dehydrocholesterol reductase, is the first true metabolic syndrome leading to multiple congenital malformations (179,180). This disorder is caused by variants in the sterol delta-7-reductase (DHCR7) gene, which maps to 11q12-q13. Typical facial appearance is characterized by short nose with anteverted nostrils, blepharoptosis, microcephaly, photosensitivity, mental retardation, syndactyly of toes 2 and 3, hypotonia, and atypical genitalia. Adrenal insufficiency may be present or evolve with time. Atypical external genitalia are a frequent feature of males (71%) and ranges from hypospadias to female external genitalia despite normal 46,XY karyotype and SRY sequences. Müllerian derivative ducts can also be present (181-183). The etiology of masculinization failure in SLOS remains unclear. However, the description of patients with SLOS who present with hyponatremia, hyperkalemia, and decreased aldosterone-to-renin ratio suggest that the lack of substrate to produce adrenal and testicular steroids is the cause of adrenal insufficiency and atypical genitalia (184), although, a revision of HPA axis in these patients showed normal HPA axis function (185). Affected children present elevations of 7-dehydrocholesterol (7DHC) in plasma or tissues. 7DHC is best assayed using Gas Chromatography/Mass Spectroscopy (GC/MS). Considering the relative high frequency of Smith-Lemli-Opitz syndrome, approximately 1 in 20,000 to 60,000 births, we suggest that at least cholesterol levels should be routinely measured in patients with 46,XY DSD. However, although frequently low, plasma cholesterol levels can be within normal limits in affected patients. DHCR7 variant analysis can confirm a diagnosis of SLOS. The human DHCR7 gene is localized on chromosome 11q13 and contains nine exons encoding a 425 amino-acid protein (64). More than 130 different variants of DHCR7 have been identified and the great majority of them are located at the exons 6 to 9 (186,187). However, the genotype-phenotype correlation in SLOS is relatively poor (188). Currently, most SLOS patients are treated with cholesterol supplementation that can be achieved by including high cholesterol foods and/or suspensions of pharmaceutical grade cholesterol. Data suggests that early intervention may be of benefit to SLOS patients (189). Observational studies report improved growth and muscle tone and strength, increased socialization, decreased irritability and aggression in SLOS patients treated with cholesterol supplementation. However, in a group of SLOS patients’ treatment with a high cholesterol diet did not improve developmental scores (190). Treatment with simvastatin, an HMG-CoA reductase inhibitor, aiming to block the cholesterol synthesis pathway avoiding the formation of large amounts of 7DHC/8DHC, and in this manner limiting exposure to potentially toxic metabolites in SLOS patients has been proposed. Simvastatin can also cross the blood–brain barrier and may provide a means to treat the biochemical defect present in the CNS of SLOS patients (191). A major effect of statin therapy is the transcriptional upregulation of genes controlled by the transcriptional factor SREBP, including DHCR7. Thus, if any residual activity is present in the mutant DHCR7, its upregulation could increase intracellular cholesterol synthesis. Simvastatin use in SLOS patients resulted in a paradoxical increase in serum and cerebrospinal fluid cholesterol levels (191). Randomized controlled-placebo trials were performed with simvastatin in SLOS showing significant reduction in plasmatic 7DHC associated with improvement in irritability symptoms (192). Determination of residual DHCR7 enzymatic activity may be helpful in selecting SLOS patients to be considered for a beneficial response of statins (187). Recently, promising gene therapy using an adeno-associated virus vector carrying a functional copy of the DHCR7 gene was administered by intrathecal injection in a mouse model with improvement of cholesterol levels in the central nervous system (193). Table 4. Phenotype of 46,XY Subjects with Smith-Lemli-Optiz Syndrome View in own window Inheritance Autosomal recessive External genitalia Micropenis and/or hypospadias, hypoplastic or bifid scrotum; female Müllerian duct derivatives May be present Wolffian duct derivatives Absent to male Testes Scrotum, inguinal or intra-abdominal region Clinical features Facial and bone abnormalities. Heart and pulmonary defects. Renal agenesis. Mental retardation, Seizures, hypotonia, syndactyly of second and third toes. Puberty Apparently normal Hormonal diagnosis Low cholesterol, elevated 7-dehydrocholesterol. Decreased aldosterone-to-renin ratio Gender role Male DHCR7 gene location 11q12-q13 Molecular defect variants in DHCR7 gene Treatment Dietary cholesterol supplies accompanied by ursodeoxycholic acid, and statins Outcome Severe mental retardation Dysgenetic 46,XY DSD Due to Under Expression of the DHX37 Gene 46,XY gonadal dysgenesis (GD) is a heterogeneous group of disorders with a wide phenotypic spectrum, including embryonic testicular regression syndrome (ETRS) (Table 5). Screening of 87 patients with 46,XY DSD (17 familial cases from 8 unrelated families and 70 sporadic cases) using whole-exome sequencing and target gene-panel sequencing identified a new player in the complex cascade of male gonadal differentiation and maintenance - the Asp-Glu-Ala-His-box (DHX) helicase 37 (DHX37) gene (53). The variants were especially associated with ETRS (7/14 index cases; 50%). The frequency of rare, predicted-to-be-deleterious DHX37 variants in this cohort (14%) is significantly higher than that observed in the Genome Aggregation Database (0.4%; P < 0.001). Immunohistochemistry analysis in human testis showed that DHX37 is mainly expressed in germ cells at different stages of testis maturation, in Leydig cells, and rarely in Sertoli cells. Other papers confirmed these findings, associating 46,XY gonadal dysgenesis with defects in DHX37 gene (152,175). Table 5. Phenotype of 46,XY Subjects with Gonadal Dysgenesis Due to DHX37 Defects View in own window Inheritance Autosomal dominant External genitalia Micropenis, atypical genitalia or typical female Müllerian duct derivatives Absent uterus, Fallopian tubes may be present Wolffian duct derivatives Present Testes Abdominal region or absent Histological analysis Dysgenetic, no gonadal tissue Puberty Hypergonadotropic hypogonadism Hormonal diagnosis Elevated serum levels of LH and FSH; very low levels of testosterone and normal testosterone precursors levels Gender role Male, female, male to female DHX37 gene location 12q24.31 Molecular defect Heterozygous variants in DHX37 gene Treatment Repair of atypical genitalia; estrogen or testosterone replacement according to social sex Outcome Most patients keep the male social sex; some change to female social sex Different modes of inheritance have been reported in familial cases of 46,XY gonadal dysgenesis, including autosomal dominant, autosomal recessive, X-linked and multifactorial inheritance (polygenic) (107,176-178). Oligogenic mode of inheritance might explain genotype/phenotype variability observed in 46,XY gonadal formation patients. Pathogenic allelic variants in NR5A1, DHX37, MAP3K1 and SRY are the most frequent molecular causes of 46,XY gonadal dysgenesis (20). Go to: 46,XY DSD DUE TO TESTOSTERONE PRODUCTION DEFECTS 46,XY DSD Due to Impaired Leydig Cell Differentiation (Complete and Partial Forms) Inactivating variants of human LHCG receptor (LHCGR) have been described in 46,XY individuals with a rare form of disorder of sex development, termed Leydig cell hypoplasia. These inactivating variants in the LHCGR prevent LH and hCG signal transduction and thus testosterone production both pre- and postnatally in genetic males (194). Both hCG and LH act by stimulating a common transmembrane receptor, the LHCGR (195) LHCGR is a member of G protein-coupled receptors, which are characterized by the canonical serpentine region, composed of seven transmembrane helices interconnected by three extracellular and three intracellular loops (196,197). The large amino-terminal extracellular domain, rich in leucine-repeats, mediates the high affinity binding of pituitary LH or placental human chorionic gonadotropin (hCG) (197). LHCGR activates the Gs protein, which determines an increase in intracellular cAMP and a subsequent stimulation of steroidogenesis in gonadal cells such as testicular Leydig cells, ovarian theca cells and differentiated granulosa cells (195,198) A secondary mechanism of LHCGR stimulation is through G q/11 protein activation and the inositol phosphate signaling pathway (197). The LHCGR gene is located on the short arm of chromosome 2 (2p21). It spans nearly 80 kb and has been thought to be composed of 11 exons and 10 introns. Exon 11 of the LHCGR gene encodes the entire serpentine domain as well as the carboxy-terminal portion of the hinge region (NCBI GeneID 3973; The amino-terminal portion of the hinge region is encoded by exon 10 and the signal peptide and remaining portion of the extracellular domain are encoded by exons 1-9 (194,196). A novel primate-specific exon (termed exon 6A) was identified within intron 6 of the LHCGR gene. This exon is not used by the wild-type full-length receptor. It displays composite characteristics of an internal/terminal exon and possesses stop codons triggering nonsense-mediated mRNA decay in LHCGR. When exon 6A is utilized, it results in a truncated LHCGR protein (199). In 1976, Berthezene et al. (200) described the first patient with Leydig cell hypoplasia and subsequently several cases have been reported (201-203). The clinical features are heterogeneous and result from a failure of intrauterine and pubertal virilization. A review of the literature allowed delineation of the characteristics of 46,XY DSD due to the complete form of Leydig cell hypoplasia as: 1) female external genitalia leading to female sex assignment 2) no development of sexual characteristics at puberty, 3) undescended testes slightly smaller than normal with relatively preserved seminiferous tubules and absence of mature Leydig cells, 4) presence of rudimentary epididymis and vas deferens and absence of uterus and fallopian tubes, 5) low testosterone levels despite elevated gonadotropin levels, with elevated LH levels predominant over FSH levels, 6) testicular unresponsiveness to hCG stimulation, and 7) no abnormal step up in testosterone biosynthesis precursors (194,204) (table 6). Several different variants in the LHCGR gene were reported in patients with Leydig cell hypoplasia in both sexes (194,205). Table 6. Phenotype of 46,XY Subjects with the Complete Form of Leydig Cell Hypoplasia View in own window Inheritance Autosomal recessive External genitalia Female, occasionally mild clitoromegaly or labial fusion Müllerian derivatives Absent Wolffian ducts derivatives Absent or vestigial Testes Inguinal or intra-abdominal, slightly subnormal size Puberty Absence of spontaneous virilization or feminization Hormonal diagnosis Elevated serum LH, normal or slightly elevated FSH and very low testosterone levels with normal levels of testosterone precursors Gender role Female LHCGR gene location 2p21 Molecular defect Pathogenic variants in LHCGR gene (complete inactivation) and in the internal exon 6A LHCGR (increase of nonfunctional isoform); defects in LHCGR were not identified in several families Treatment Estrogen replacement at pubertal age, bilateral orchiectomy and vaginal dilation Outcome Female gender role and behavior, infertility In contrast to the homogenous phenotype of the complete form of Leydig cell hypoplasia, the partial form features a broad spectrum, ranging from incomplete male sexual differentiation characterized by micropenis and/or hypospadias to hypergonadotropic hypogonadism without ambiguity of the male external genitalia (194,195,206,207). Testes are cryptorchidic or in the scrotum and during puberty, partial virilization occurs and testicular size is normal or only slightly reduced, while penile growth is significantly impaired. Spontaneous gynecomastia does not occur. Before puberty, the testosterone response to the hCG test is subnormal without accumulation of testosterone precursors. After puberty, LH levels are elevated as a result of insufficient negative feedback of gonadal steroid hormones on the anterior pituitary and testosterone levels are intermediate between those of children and normal males. Several mutations in the LHCGR gene have also been identified in patients with the partial form of Leydig cell hypoplasia. Latronico et al. reported the first homozygous mutation in the LHCGR (p.Ser616Tyr) in a boy with micropenis (207). Subsequently, other milder mutations were identified in further patients with the partial form of Leydig cell hypoplasia (194,195,207). In vitro studies showed that cells transfected with LHCGR gene containing these mutations had an impaired hCG-stimulated cAMP production (195,207). Leydig cell hypoplasia was found to be a genetic heterogenous disorder since Zenteno et al. (197) ruled out, by segregation analysis of a known polymorphism in exon 11 of the LHCG receptor gene, molecular defects in the LHCG receptor as being responsible for Leydig cell hypoplasia in three siblings with 46,XY DSD. Most inactivating mutations of the LHCGR are missense mutations that result in a single amino acid substitution in the LHCGR. In addition, mutations causing amino acid deletions, amino acid insertions, splice acceptor mutation or premature truncations of the receptor have also been reported (208). LHCGR mutations are usually located in the coding sequence, resulting in impairment of either LH/CG binding or signal transduction. Although it is well known that hCG and LH act by stimulating a common receptor, a differential action of them in the LHCGR has been suggested. The identification of a deletion of exon 10 of the LHCGR in a patient with normal male genitalia at birth, but no pubertal development indicated that the mutant LHCGR was responsive to fetal hCG, but resistant to pituitary LH. The binding affinity of hCG for LHCGR was normal in vitro analysis, suggesting that exon 10 is necessary for LH, but not for hCG action (199). The identification and characterization of a novel, primate-specific bona fide exon (exon 6A) within the LHCGR determined a new regulatory element within the genomic organization of this receptor and a new potential mechanism of this disorder. Kossack et al analyzing the exon 6A in 16 patients with 46,XY DSD due to Leydig cells hypoplasia without molecular diagnosis, detected mutations (p.A557C or p.G558C) in three patients. Functional studies revealed a dramatic increase in expression of the mutated internal exon 6A transcripts, resulting in the generation of predominantly nonfunctional isoforms of the LHCGR, thereby preventing its proper expression and functioning (209). A new compound heterozygous mutation of the LHCGR, constituted by a previously described missense mutation (p.Cys13Arg) and a large deletion of the paternal chromosome 2 was identified by array-Comparative Genomic Hybridization (array-CGH) in a 46,XY infant with sexual ambiguity and low hCG-stimulated testosterone levels associated with high LH and FSH levels (200). In addition, causative mutations in LHCGR were absent in around 50% of the patients strongly suspected to have Leydig cell hypoplasia. These findings supported the idea that other genes must be implicated in the molecular basis of this disorder. We observed that 46,XX sisters of the patients with 46,XY DSD due to Leydig cell hypoplasia, carrying the same homozygous mutation in the LHCGR, have primary or secondary amenorrhea, spontaneous breast development, infertility, normal or enlarged cystic ovaries with elevated LH and LH/FSH ratio, normal estradiol and progesterone levels for early to mid-follicular phase, but not for luteal phase levels, confirming lack of ovulation (198,207,210). Our findings were subsequently confirmed by other authors who studied 46,XX sisters of 46,XY DSD patients with Leydig cell hypoplasia (201,202,211). Subsequently, a novel homozygous missense mutation, p.N400S, has been identified by whole genome sequencing in two sisters with empty follicle syndrome (204). Table 7. Phenotype of 46,XY Subjects with Partial Leydig Cells Hypoplasia View in own window Inheritance Autosomal recessive External genitalia Atypical to male Müllerian derivatives Absent Wolffian ducts derivatives Rudimentary to male Testes Scrotum, labial folds, or inguinal regions, normal or only slightly subnormal size Puberty Partial virilization without gynecomastia, discrepancy between reduced penis size and normal testicular growth Hormonal diagnosis Elevated serum LH levels, normal or slightly elevated FSH and low T levels with normal levels of T precursors in relation to T Gender role Male LHCGR gene location 2p21 Molecular defect Variants which confer partial inactivation of LHCGR Treatment Repair of the hypospadias, testosterone replacement at pubertal age Outcome Male gender role and behavior, possible fertility under treatment 46,XY DSD Due to Enzymatic Defects in Testosterone Synthesis Six enzymatic defects that alter the normal synthesis of testosterone have been described to date (Figure 10). Three of them are associated with defects in cortisol synthesis leading to congenital adrenal hyperplasia. All of them present an autosomal recessive mode of inheritance and genetic counseling is mandatory since the chance of recurring synthesis defects among siblings is 25%. Figure 10. Standard steroidogenesis and alternative pathway to DHT synthesis. DEFECTS IN ADRENAL AND TESTICULAR STEROIDOGENESIS Adrenal hyperplasia syndromes are examples of hypoadrenocorticism or mixed hypo- and hyper cortico-adrenal steroid secretion. Synthesis of cortisol or both cortisol and aldosterone are impaired. When cortisol production is impaired, there is a compensatory increase in ACTH secretion. If mineralocorticoid production is impeded, there is a compensatory increase in renin-angiotensin production. These compensatory mechanisms may return cortisol or aldosterone production to normal or near normal levels, but at the expense of excessive production of precursors that can cause undesirable hormonal effects. Lipoid Congenital Adrenal Hyperplasia due to Deficiency of the Steroidogenic Acute Regulatory Protein (StAR) StAR is a mitochondrial phosphoprotein which facilitates the influx of cholesterol from the outer to the inner mitochondrial membrane for the subsequent action of the P450scc enzyme (212). StAR is encoded by the STAR gene and its deficiency leads to congenital lipoid adrenal hyperplasia (CLAH), the most severe form of congenital adrenal hyperplasia (213). Lipoid adrenal hyperplasia is rare in Europe and America, but it is thought to be the second most common form of adrenal hyperplasia in Japan where 1 in 300 individuals carries the p.Q258X variant (214). Affected subjects are phenotypic females irrespective of gonadal sex or sometimes have slightly virilized external genitalia with or without cryptorchidism, underdeveloped internal male organs and an enlarged adrenal cortex, engorged with cholesterol and cholesterol esters (215). Adrenal steroidogenesis deficiency leads to salt wasting, hyponatremia, hyperkalemia, hypovolemia, acidosis, and death in infancy, although patients can survive to adulthood with appropriate mineralocorticoid- and glucocorticoid-replacement therapy (215). Hormonal diagnosis is based on high ACTH and renin levels and the presence of low levels of all glucocorticoids, mineralocorticoids, and androgens. The disease was firstly attributed to P450scc deficiency, but most of the cases studied through molecular analysis showed an intact P45011A gene and its RNA (216). Since StAR is also required for the conversion of cholesterol to pregnenolone, molecular studies were performed in StAR gene and variants were found in most of the affected patients (217) Congenital lipoid adrenal hyperplasia (LCAH) in most Palestinian cases is caused by a founder c.201_202delCT variant causing premature termination of the StAR protein (217) Histopathological findings of excised XY gonads included accumulation of fat in Leydig cells since 1 yr. of age, positive placental alkaline phosphatase and octamer binding transcription factor (OCT4) staining indicating a neoplastic potential (217). A two-hit model has been proposed by Bose et al. (216) as the pathophysiological explanation for LCAH. In response to a stimulus (e.g., ACTH), the normal steroidogenic cell recruits cholesterol from endogenous synthesis, stored lipid droplets or low-density lipoprotein-receptor mediated endocytosis. Subsequently StAR promotes the cholesterol transport from the outer to the inner mitochondrial membrane in which cholesterol is further processed to pregnenolone. In cells with mutant StAR (first hit), there is no rapid steroid synthesis, but still some StAR-independent cholesterol flows into the mitochondria, resulting in a low level of steroidogenesis. Due to increased steroidogenic stimuli in response to inadequately low steroid levels, additional cholesterol accumulates. Massive cholesterol storage and resulting biochemical reactions eventually destroy all steroidogenic capacity (second hit) (217). This two-hit model has been confirmed by clinical studies (218) as well as StAR knockout mice research (219). The human STAR gene is localized on chromosome 8p11.2 and consists of seven exons (220). It is translated as a 285-amino acid protein including a mitochondrial target sequence (N terminal 62 amino acids), which guides StAR to the outer mitochondrial membrane and a cholesterol binding site, which is located at the C-terminal region. In vitro studies revealed that StAR protein lacking the N terminal targeting sequence (N-62 StAR) can still stimulate steroidogenesis in transfected COS-1 cells, whereas variants in the C-terminal region led to severely diminished or absent function (221-223). Most of the STAR gene variants associated with LCAH are located in the C-terminal coding region between exons 5 and 7 StAR related lipid transfer (START) domain (224). Mild phenotype of lipoid CAH is a recognized disorder caused by StAR variants that retain partial activity (225). Affected males can present with adrenal insufficiency resembling autoimmune Addison disease with micropenis or normal development with hypergonadotropic hypogonadism (224,225). More than 40 StAR variants causing classic lipoid CAH have been described (217,226,227), but very few partial loss-of-function variants have been reported (224-226). Therefore, there is a broad clinical spectrum of StAR variants, however, the StAR activities in vitro correlate well with clinical phenotypes (228). Three 46,XY patients with the homozygous p.R188C STAR variant causing primary adrenocortical insufficiency without atypical genitalia were reported (229). Patients with nonclassical lipoid CAH may present with male genitalia and preserved testicular function (230). Table 8. Phenotype of 46,XY Subjects with StAR Deficiency View in own window Inheritance Autosomal recessive External genitalia Female Micropenis (mild form) Müllerian duct derivatives Absent Wolffian duct derivatives Absent -> hypoplastic Testes Small size Clinical Features Early adrenal insufficiency; no pubertal development; hypergonadotropic hypogonadism Hormonal diagnosis Elevated ACTH and renin levels; low levels of all glucocorticoids, mineralocorticoids, and androgens Gender role Female Male (mild form) STAR gene location 8p11.2 Molecular defect Inactivating variants in STAR Treatment Early gluco- and mineralocorticoid replacement; estrogen replacement at pubertal age Outcome Infertile, female or male gender role and behavior Deficiency of P450 Side Chain Cleavage Enzyme (P450scc) Due to Variants in CYP11A1 The first step in the conversion of cholesterol to hormonal steroids is hydroxylation at carbon 20, with subsequent cleavage of the 20-22 side chain to form pregnenolone. In steroidogenic tissues, such as adrenal cortex, testis, ovary, and placenta, this is the initial and rate-limiting step in steroidogenesis. This reaction, known as cholesterol side-chain cleavage, is catalyzed by a specific cytochrome P450 called P450scc or CYP11A1 encoded by the CYP11A1 gene (231). A number of patients with CYP11A1 variants have now been described (232-235), including late-onset non-classical forms secondary to variants that retain partial enzyme activity (236,237). Clinically, these patients are indistinguishable from those with lipoid CAH, but none of them present enlarged adrenals that characterize lipoid CAH. Analyzing infants with adrenal failure and disorder of sexual differentiation compound heterozygous variants in CYP11A1 have been identified, recognizing that this disorder may be more frequent than originally thought. The phenotypic spectrum of P450scc deficiency ranges from severe loss-of-function variants associated with prematurity, complete under androgenization, and severe early-onset adrenal failure, to partial deficiencies found in children born at term with mild masculinization and later-onset adrenal failure (236,237). 3β-Hydroxysteroid Dehydrogenase type II Deficiency 3β-hydroxysteroid dehydrogenase 2 (3βHSD2) deficiency is a rare form of congenital adrenal hyperplasia (CAH), with fewer than 200 cases reported in the world literature. 3β-HSD converts 3β-hydroxy 5 steroids to 3-keto 4 steroids and is essential for the biosynthesis of mineralocorticoids, glucocorticoids and sex steroids Two forms of the enzyme have been described in man: the type I enzyme which is expressed in placenta and peripheral tissues such as the liver and skin, and type II that is the major form expressed in the adrenals and gonads (238). The two forms are very closely related in structure and substrate specificity, though the type I enzyme has higher substrate affinities and a 5-fold greater enzymatic activity than type II (238). Male patients with 3β-HSD type II deficiency present with atypical external genitalia, characterized by microphallus, proximal hypospadias, bifid scrotum and a blind vaginal pouch associated or not with salt loss (239,240). Precocious pubarche and gynecomastia at pubertal stage is a common phenotype in 3β-HSD type II deficiency (241). Serum levels of Δ-5 steroids such as pregnenolone, 17OHpregnenolone (17OHPreg), DHEA, DHEAS are elevated and basal levels of 17OHPreg and 17OHPreg/17OHP ratio are the best markers of this deficiency in both prepubertal and postpubertal stage. Δ-4 steroids are slightly increased due to the peripheral action of 3β-HSD type I enzyme but the ratio of Δ-5/Δ-4 steroids is elevated. Cortisol secretion is reduced but the response to exogenous ACTH stimulation varies from decreased (more severe deficiency) to normal. At adult age, affected males can reach normal or almost normal levels of testosterone due to the peripheral conversion of elevated Δ-5 steroids by 3β-HSD type I enzyme and also due to testicular stimulation by the high LH levels (242). The human genome encodes two functional 3βHSD genes on chromosome 1p13.1. The HSD3B2 gene is expressed in adrenal and gonads and consists of four exons coding for a 372 amino acid protein (243). To date, around 40 variants in the HSD3B2 gene have been described. Most of them are base substitutions, and they are located especially at the N-terminal region of the protein. The amino acids A10, A82, P222 and T259 could be considered a hotspot since different variants were reported in these HSD3B2 positions. Variants abolishing 3β-HSD type II activity lead to congenital adrenal hyperplasia (CAH) with severe salt-loss (244). Variants that reduce, but do not abolish type II activity ( > 5% of wild type 3βHSD2 activity in vitro) lead to CAH with mild or no salt-loss, which in males is associated with 46,XY DSD due to the reduction in androgen synthesis (241,242,245,246). Male subjects with 46,XY DSD due 3β-HSD type II deficiency without salt loss showed clinical features in common with the deficiencies of 17β-HSD3 and 5α-reductase 2. Most of the patients were raised as males and kept the male social sex at puberty. In one Brazilian family, two cousins with 46,XY DSD due to 3β-HSD type II deficiency were reared as females; one of them was underwent orchiectomy in childhood and kept the female social sex; the other did not undergo orchiectomy at childhood and changed to male social sex at puberty (246). There is little data on the outcomes of 3β-HSD type II deficiency. A mixed longitudinal and cross-sectional study from a single Algerian center reported 14 affected subjects (8 females) with pathogenic variants in HSD3B2 gene (247). Premature pubarche was observed in four patients (3F:1M). Six patients (5F:1M) entered puberty spontaneously, aged 11 (5-13) years in 5 girls and 11.5 years in one boy. Testicular adrenal rest tumors were found in three boys. Four girls reached menarche at 14.3 (11-14.5) years, with three developing adrenal masses and polycystic ovary syndrome (PCOS), with radiological evidence of ovarian adrenal rest tumor in one. The median IQ was 90 (43-105), >100 in only two patients and <70 in three of them (247). Table 9. Phenotype of 46,XY Subjects with 3β-HSD Type II Deficiency View in own window Inheritance Autosomal recessive External genitalia Atypical (proximal hypospadias, bifid scrotum, urogenital sinus), precocious pubarche Müllerian derivatives Absent Wolffian duct derivatives Normal Testes Well developed; generally topic Clinical features Adrenal insufficiency or not in infancy; virilization at puberty with or without gynecomastia Hormonal diagnosis Elevated basal and ACTH-stimulated 17OHPreg and 17OHPreg/17OHP ratio Gender role Male; female to male HSD3B2 gene location 1p13.1 Molecular defect Inactivating variants in HSD3B2 Treatment Glucocorticoid replacement along with mineralocorticoids in salt-losing form; at puberty variable necessity for testosterone replacement Outcome Variable spermatogenesis; fertility possible by in vitro fertilization Combined 17-Hydroxylase and C-17-20 lyase deficiency CYP17 is a steroidogenic enzyme that has dual functions: hydroxylation and lyase. It is located in the fasciculata and reticularis zone of the adrenal cortex and gonadal tissues. The first activity results in hydroxylation of pregnenolone and progesterone at the C(17) position to generate 17α-hydroxypregnenolone and 17α-hydroxyprogesterone, while the second enzyme activity cleaves the C(17)-C(20) bond of 17α-hydroxypregnenolone and 17α-hydroxyprogesterone to form dehydroepiandrosterone and androstenedione, respectively. The modulation of these two activities occurs through cytochrome b5, necessary for lyase activity (248). Deficiency of adrenal 17-hydroxylation activity was first demonstrated by Biglieri et al. (249). The phenotype of 17-hydroxylase deficiency in most of the male patients described is a female-like or slightly virilized external genitalia with blind vaginal pouch, cryptorchidism and high blood pressure, usually associated with hypokalemia. New in 1970, reported the first affected patient with atypical genitalia which was assigned to the male sex (250). The 17-hydroxylase deficiency is the second most common cause of CAH in Brazil (251). At puberty, patients usually present sparse axillary and pubic hair. Male internal genitalia are hypoplastic and gynecomastia can appear at puberty. Most of the male patients were reared as females and sought treatment due to primary amenorrhea or lack of breast development. Genetic female patients may also be affected and present normal development of internal and external genitalia at birth and hypergonadotropic hypogonadism and amenorrhea at post pubertal age; enlarged ovaries at adult age and infarction from twisting can occur (252,253). These patients do not present signs of glucocorticoid insufficiency, due to the elevated levels of corticosterone, which has a glucocorticoid effect. The phenotype is similar to 46,XX or 46,XY complete gonadal dysgenesis and the presence of systemic hypertension and absence of pubic hair in post pubertal patients suggests the diagnosis of 17-hydroxylase deficiency (254). Serum levels of progesterone, corticosterone, and 18-OH-corticosterone are elevated, while aldosterone, 17-OH-progesterone, cortisol, androgens and estrogens are decreased. Martin et al, performed a clinical, hormonal, and molecular study of 11 patients from 6 Brazilian families with the combined 17-alpha-hydroxylase/17,20-lyase deficiency phenotype (255). All patients had elevated basal serum levels of progesterone and suppressed plasma renin activity. The authors concluded that basal progesterone measurement is a useful marker of P450c17 deficiency and suggest that its use should reduce the misdiagnosis of this deficiency in patients presenting with male DSD, primary or secondary amenorrhea, and mineralocorticoid excess syndrome. Excessive production of deoxycorticosterone and corticosterone results in systemic hypertension, suppression of renin levels and inhibition of aldosterone synthesis. The CYP17A1 gene, which encodes the enzymes 17-hydroxylase and 17-20 lyase, is a member of a gene family within the P450 supergene family and is mapped at 10q24.3 (254) (256). Several variants in the CYP17A1 gene have been identified in patients with both 17-hydroxylase and 17,20 lyase deficiencies (252,253,257). Four homozygous variants, p.A302P, p.K327del, p.E331del and p.R416H, were identified by direct sequencing of the CYP17A1 gene. Both P450c17 activities were abolished in all the mutant proteins but the mutant proteins were normally expressed, suggesting that the loss of enzymatic activity is not due to defects of synthesis, stability, or localization of P450c17 proteins (257). Glucocorticoid replacement for hypertension management, gonadectomy and estrogen replacement at puberty for patients reared in the female social sex are indicated. In male patients, androgen replacement is usually necessary since they present very low levels of testosterone. These patients are very sensitive to glucocorticoids and low doses of dexamethasone (0.125-0.5 mg at night) are sufficient to control blood pressure. In some patients, however, estrogens might aggravate hypertension. The control of blood pressure can be initially achieved by salt restriction although mineralocorticoid antagonists might be necessary (257). Table 10. Phenotype of 46,XY Subjects with 17a-Hydroxylase and 17,20-Lyase Deficiency View in own window Inheritance Autosomal recessive External genitalia Female like --> atypical Müllerian duct derivatives Absent Wolffian duct derivatives Hypoplastic --> normal Testes Intra-abdominal or inguinal Clinical features Low renin hypertension; absent or slight virilization at puberty; gynecomastia Hormonal diagnosis Elevated progesterone, DOC, corticosterone; low plasma renin activity low cortisol not stimulated by ACTH Gender role Female in most patients CYP17 gene location 10q24.3 Molecular defect Variants in CYP17A1 gene Treatment Repair of sexual ambiguity; glucocorticoid and estrogen or testosterone replacement according to social sex Outcome Female behavior, infertility Cytochrome P450 Reductase (POR) Deficiency (Electron Transfer Disruption) The apparent combined P450C17 and P450C21 deficiency is a rare variant of congenital adrenal hyperplasia, first reported by Peterson et al in 1985 (258). Affected girls and boys are born with atypical genitalia, indicating intrauterine androgen excess in females and androgen deficiency in males. Boys and girls can also present with skeletal malformations, which in some cases resemble a pattern seen in patients with Antley-Bixler syndrome. Findings of biochemical investigations of urinary steroid excretion in affected patients have shown accumulation of steroid metabolites, indicating impaired C17 and C21 hydroxylation, suggesting concurrent partial deficiencies of the 2 steroidogenic enzymes, P450C17 and P450C21. However, sequencing of the genes encoding these enzymes showed no variants, suggesting a defect in a cofactor that interacts with both enzymes. POR is a flavoprotein that donates electrons to all microsomal P450 enzymes, including the steroidogenic enzymes P450c17, P450c21 and P450aro (259). Shephard et al. (1989) isolated and sequenced cDNA clones that encode the rat and human NADPH-dependent cytochrome P-450 reductase and located the human gene at 7q11.2 (260). The underlying molecular basis of congenital adrenal hyperplasia with apparent combined P450C17 and P450C21 deficiency was defined in 3 patients, who were compound heterozygotes for variants in POR (259,260). Antley-Bixler syndrome is characterized by craniosynostosis, severe midface hypoplasia, proptosis, choanal atresia/stenosis, frontal bossing, dysplastic ears, depressed nasal bridge, radio-humeral synostosis, long bone fractures, femoral bowing, phalangeal malformation (arachno-/campto-/clinodactyly, brachy-tele-phalanges, rocker bottom feet) and urogenital abnormalities (259). The occurrence of genital abnormalities in patients with Antley-Bixler syndrome, especially females was reported in 2000 (261). In a recent large survey of patients with Antley-Bixler syndrome, it was demonstrated that individuals with an Antley-Bixler-like phenotype and normal steroidogenesis have FGFR2 variants, whereas those with atypical genitalia and altered steroidogenesis have POR deficiency (262). The skeletal malformations observed in many, but not all patients with POR deficiency, are thought to be due to disruption of enzymes involved in sterol synthesis, 14α-lanosterol demethylase (CYP51A1) and squalene epoxidase, and disruption of retinoic acid metabolism catalyzed by CYP26 isoenzymes that depend on electron transfer from POR (263). Pubertal presentations in females with congenital POR deficiency were described. Incomplete pubertal development and large ovarian cysts prone to spontaneous rupture were the predominant findings in females (264).The ovarian cysts may be driven not only by high gonadotropins but possibly also by impaired CYP51A1-mediated production of meiosis-activating sterols due to mutant POR. In the two boys evaluated, pubertal development was more mildly affected, with some spontaneous progression. These findings may suggest that testicular steroidogenesis may be less dependent on POR than adrenal and ovarian steroidogenesis (265). Table 11. Phenotype of 46,XY Patients with POR Deficiency View in own window Inheritance Autosomal recessive External genitalia Atypical Müllerian duct derivatives Normally developed Wolffian duct derivatives Normally developed Testes Well developed, frequent cryptorchidism Hormonal diagnosis Low T and cortisol and elevated basal ACTH, Prog and 17OHP POR gene location 7q11.2 Molecular defect Inactivating variants of POR gene Puberty Spontaneous pubertal development in males Gender role Male Treatment Repair of sexual ambiguity; glucocorticoid replacement and estrogen or testosterone replacement according to social sex Outcome Puberty development, fertility? DEFECTS IN TESTICULAR STEROIDOGENESIS Three defects in testosterone synthesis that are not associated with adrenal insufficiency have been described: CYP17A1 deficiency, cytochrome B5 deficiency and 17-β-HSD3 deficiency. CYP17A1 Deficiency Human male sexual differentiation requires production of fetal testicular testosterone, whose biosynthesis requires steroid 17,20-lyase activity. The existence of true isolated 17,20-lyase deficiency has been questioned because 17-α-hydroxylase and 17,20-lyase activities are catalyzed by a single enzyme and because combined deficiencies of both activities were found in functional studies of the variant found in a patient thought to have had isolated 17,20-lyase deficiency (266,267). Later, clear molecular evidence of the existence of isolated 17,20 desmolase deficiency was demonstrated (268). The patients present atypical genitalia with micropenis, proximal hypospadias and cryptorchidism. Gynecomastia Tanner stage V can occur at puberty (268). Elevated serum levels of 17-OHP and 17-OHPreg, with low levels of androstenedione, dehydroepiandrosterone and testosterone, are described. The hCG stimulation test results in a slight stimulation in androstenedione and testosterone secretion with an accumulation of 17-OHP and 17-OHPreg. The CYP17A1 gene of two Brazilian 46,XY DSD patients with clinical and hormonal findings indicative of isolated 17,20-lyase deficiency, since they produce cortisol normally, were studied. Both were homozygous for missense variants in CYP17A1 (268). When expressed in COS-1 cells, the mutants retained 17α-hydroxylase activity and had minimal 17,20-lyase activity. Both variants alter the electrostatic charge distribution in the redox-partner binding site, so that the electron transfer for the 17,20-lyase reaction is selectively lost (268). Table 12. Phenotype of 46,XY Subjects with 17,20 Lyase Deficiency View in own window Inheritance Autosomal recessive External genitalia Atypical (proximal hypospadias, bifid scrotum, urogenital sinus) Müllerian derivatives Absent Wolffian ducts derivatives Hypoplastic --> normal Testes In the inguinal region, small size Clinical features Gynecomastia variable; poor virilization at puberty Hormonal diagnosis Elevated 17OHP and 17OHP/A ratio after hCG stimulation and decreased A and T levels; Gender role Male or female CYP17 gene location 10q24.3 Molecular defect Variants in the redox partner binding site of CYP17A1 enzyme Treatment Repair of hypospadias and gynecomastia; testosterone replacement at pubertal age Outcome Male or female behavior Cytochrome B5 deficiency (Allosteric Factor for P450c17 and POR Interaction) In 1986, Hegesh et al described a 46,XY DSD patient with type IV hereditary methemoglobinemia (269). The patient had a 16-bp deletion in the cytochrome b5 mRNA leading to a new in-frame termination codon and a truncated protein. The etiology of 46,XY DSD in this patient was attributed to the cytochrome b5 defect since cytochrome b5 acts as an allosteric factor, promoting the interaction of. P450c17 and POR favoring 17,20 lyase reactions (270). Two homozygous variants in CYB5 in 46,XY DSD patients with elevated methemoglobin levels but without clinical phenotype of methemoglobinemia were reported (269). 46,XY DSD due to 17β-HSD 3 Deficiency This disorder consists of a defect in the last phase of steroidogenesis when androstenedione is converted to testosterone and estrone to estradiol. This disorder was described by Saez and his colleagues (271) and is the most common disorder of androgen synthesis, reported from several parts of the world (272,273). There are 5 steroid 17β-HSD enzymes that catalyze this reaction (274) and 46,XY DSD results from variants in the gene encoding the 17β-HSD3 isoenzyme (275). Patients present female-like or atypical genitalia at birth, with the presence of a blind vaginal pouch, intra-abdominal or inguinal testes and epididymis, vasa deferentia, seminal vesicles and ejaculatory ducts. Most affected males are raised as females (276,277), but some have less severe defects in virilization and are raised as males (274). Virilization in subjects with 17β-HSD3 deficiency occurs at the time of expected puberty. This late virilization is usually a consequence of the presence of testosterone in the circulation because of the conversion of androstenedione to testosterone by some other 17β-HSD isoenzyme (presumably 17β-HSD 5) in extra-gonadal tissue and, occasionally, of the secretion of testosterone by the testes when levels of LH are elevated in subjects with some residual 17β-HSD3 function (274,277). However, the discrepancy between the failure of intrauterine masculinization and the virilization that occurs at the time of expected puberty is poorly understood. A limited capacity to convert androstenedione into testosterone in the fetal extragonadal tissues may explain the impairment of virilization of the external genitalia in the newborn. Bilateral orchiectomy resulted in a clear reduction of androstenedione levels indicating that the main origin of this androgen is the testis (278). 46,XY DSD phenotype is sufficiently variable in 17β-HSD3 deficiency to cause problems in accurate diagnosis, particularly in distinguishing it from partial androgen insensitivity syndrome (276,279). Laboratory diagnosis is based on elevated serum levels of androstenedione and estrone and low levels of testosterone and estradiol resulting in elevated androstenedione/testosterone and estrone/ estradiol ratios or low (or low testosterone/androstenedione and estradiol/estrone ratios) indicating impairment in the conversion of 17-keto into 17-hydroxysteroids. Testosterone/Androstenedione ratio of 0.4±0.2 was found in prepubertal patients with 17β-HSD3 deficiency after hCG stimulation. Based on these data, a T/A ratio below <0.8 is suggestive of 17β-HSD3 deficiency (272). At the time of expected puberty, serum LH and testosterone levels rise in all affected males and testosterone levels may reach the normal adult male range (277,278). Pitfalls in the hormonal diagnosis of 17β-HSD3 deficiency had been reported in the literature. Two of the fourteen cases of 17β-HSD3 deficiency reported from the UK database had a T/A ratio > 0.8 (276). Both patients were from a consanguineous pedigree, with two affected sisters (both assigned in the female gender) and one nephew. The former patient had atypical genitalia with proximal hypospadias and was assigned as male. The hCG test was performed at 2 years and 2 months of age, respectively, resulting in a T/A ratio of 3.4 and 1.5. Two other patients with atypical genitalia, who were also assigned in the female social sex, were evaluated at 5 months and 9.2 year of age, respectively (280). After the hCG stimulation test, there was a clear elevation of serum testosterone (measured by HPLC tandem mass spectrometry) with a small increase of the androstenedione levels resulting in a high T/A ratio (2.47 and 2.27 respectively). Sequencing of the HSD17B3 gene identified deleterious molecular defects in both alleles in both patients. The possible explanation for the normal T/A ratio in these 4 children is the individual and temporal variability in the HSD17B isoenzymes activity (280). The disorder is due to homozygous or compound heterozygous variants in the HSD17B3 gene which encodes the 17β-HSD3 isoenzyme. Up to now, almost 40 variants in the HSD17B3 gene have been reported. These include missense, nonsense, exonic deletion, duplication and intronic splice site variants (274,277). Although allelic variants have been described throughout HSD17B3, a variant cluster region was identified in the exon 9. The 17β-HSD3 activity was completely eliminated in the majority of the HSD17B3 variants (276). Outside exon 9, the most frequent site of variant in HSD17B3 gene is the R80 in exon 3, which primarily disrupts the binding of the NADPH cofactor to the protein. The p.R80Q variant has been found in Palestinian, Brazilian, and Turkish families (281). Most patients are raised as girls during childhood. Change to male gender role behavior at puberty has been frequently described in individuals with this disorder who were reared as females (282-285), including members of a large consanguineous family in the Gaza strip (286). In a review of all adult patients with 46,XY DSD due to 17β-HSD3 deficiency reared as female and not castrated during childhood reported until now, we found that 30 of them (61%) kept the female gender and 19 of them (39%) changed to male gender (277). After a histological analysis of testicular tissue stained with hematoxylin-eosin from 40 reported cases of 46,XY patients with 17β-HSD3 deficiency, the prevalence of germ cell tumor was 5%, which is lower than the estimated GCT risk for some 46,XY DSD etiologies (287-289). However, the maintenance of the testes in male patients is safe if the testes can be positioned into the scrotum (277,290). Table 13. Phenotype of 46,XY Patients with 17β-HSD 3 Deficiency View in own window Inheritance Autosomal recessive External genitalia Atypical, frequently female-like at birth Müllerian duct derivatives Absent Wolffian duct derivatives Normally developed Testes Well developed, frequent cryptorchidism Hormonal diagnosis Low T and elevated basal and hCG-stimulated A and A/T ratio HSD17B3 gene location 9q22 Molecular defect Inactivating variants of HSD17B3 Puberty Virilization at puberty; variable gynecomastia Gender role Most patients keep the female social sex; some change to male social sex Treatment Repair of sexual ambiguity; estrogen or testosterone replacement according to social sex Outcome Male or female gender identity; in males’ fertility possible by in vitro fertilization ALTERNATIVE PATHWAY TO DHT SYNTHESIS 46,XY DSD Due to 3α-Hydroxysteroid Dehydrogenase Deficiency (AKR1C2 and AKR1C4 Defects) Back in 1972, the molecular analysis of 46,XY DSD due to isolated 17,20-lyase deficiency patients failed to find variants in the CYP17A1 (248). However, the hormonal data was inconsistent with other adrenal enzymatic deficiencies. Therefore, the alternative or backdoor pathway was considered to explain the etiology of the DSD in these patients. The backdoor pathway was firstly described in marsupials and is remarkable for having both reductive and oxidative 3α-HSD steps: the reductive reaction converts 17-OH-dihydroprogesterone (17OH-DHP) into 17OH-allopregnanolone (17OH-Allo), and the oxidative reaction converts androstanediol into DHT (291,292) (Figure 6). Therefore, synthesis of dihydrotestosterone (DHT) occurs without the intermediacy of DHEA, androstenedione or testosterone (291). All the human genes participating in the backdoor pathway have not been identified, however it has been thought that the reductive 3α-HSD activity can be catalyzed by an aldo-keto reductase called AKR1C2 (293), as well as by other enzyme, such as the oxidative 3α-HSD activity by 17β-HSD6, also called as RoDH (294) and possibly by AKR1C4 (295). The first reported cases with isolated 17,20 lyase deficiency from 1972 (266) were found to carry variants in two aldo-keto reductases, AKR1C2 and AKR1C4 which catalyze 3α-hydroxysteroid dehydrogenase activity. The two affected 46,XY females were compound heterozygotes for AKR1C2 variants, the p.I79V/H90Q and p.I79V/N300T. However, the mutant AKR1C2 enzymes retained 22-82% of wild-type activity in vitro analysis suggesting that another gene might be involved (248). Analysis of AKR1C cDNA found that AKR1C4 was spliced incorrectly and gene sequencing displayed an intronic variant 106 bases upstream from exon 2 that caused this exon skipping. So, in this family, a digenetic inheritance was found to impair testicular synthesis of DHT during prenatal life (296). AKR1C2 is abundantly expressed in the fetal testis, but minimally expressed in the adult testis; on the other hand, the AKR1C4 was found in fetal and adult testes at lower levels (293). Therefore, it appears that both AKR1C2 and AKR1C4 participate in the backdoor pathway to DHT in the fetal testis, and that molecular defects in these genes appear to cause incomplete male genital development (297). However, the relative roles of these two AKR1C enzymes remain unclear and testosterone levels at adult age are not available in these patients (298). All findings described above, which substantially advanced our understanding of the underlying mechanisms of male sexual differentiation, illustrate the importance of detailed studies of rare 17,20 lyase deficiency patients. Go to: 46,XY DSD DUE TO DEFECTS IN TESTOSTERONE METABOLISM 5α-Reductase Type 2 Deficiency A condition named pseudo-vaginal perineo-scrotal hypospadias in 46,XY individuals was reported in 1961, in which the phenotype included female-like external genitalia, bilateral testes, and male urogenital tracts with a blind-ending vagina (299). Thereafter, experimental studies showed that the male external genitalia virilization depended on the conversion of testosterone into dihydrotestosterone (DHT), an enzymatic reaction catalyzed by the 5α-reductase enzyme. Further, that enzymatic deficiency was biochemically and clinically reported in 24 individuals from the Dominican Republic and two siblings from North America (300,301). Typically, affected individuals are born with female-like external genitalia but develop clinical and psychological virilization at puberty with no gynecomastia (300). Both studies characterized this syndrome as a genetic condition with an autosomal recessive pattern of inheritance, resulting from the inability to convert testosterone into DHT. Later, two different genes encoding two 5α-reductase isoenzymes were isolated by cloning technology: the 5α-reductase type 1 and 2 (SRD5A1 and SRD5A2) (302). Allelic variants in the SRD5A2 gene were found in two individuals from Papua New Guinea with clinical features of 5α-reductase type 2 deficiency, whereas controls did not have variants in this gene, suggesting that variants in the SRD5A2 were the molecular basis of this condition (303). Further, the SRD5A2 gene was mapped at chromosome 2 (2p23), containing 5 exons and 4 introns, and encoding a 254 amino-acids protein (304). Since then, several SRD5A2 allelic variants have been reported across the whole gene in individuals presenting this particular 46,XY DSD (305). We recently reviewed all 5α-reductase type 2 deficiency cases reported in the literature. We identified 451 cases of 5α-reductase type 2 deficiency from several countries, harboring 121 different SRD5A2 allelic variants (306). These variants have been reported in all exons of this gene, but mainly are located at exons 1 (33%) and at exon 4 (25%). Among the 254 amino acids that make up the SR5A2 protein, we found allelic variants in the SRD5A2 gene in 76 of them (306). Regarding the SRD5A2 allelic variants, most are missense variants, but small deletions, variants at splicing sites, stop codons, small indels (n = 20) and large deletions have also been described. We also identified homozygosity in 70% of the SRD5A2 allelic variants causing 5α-reductase type 2 deficiency (306). Neonatal diagnosis was carried out in 29.7%, whereas the remaining had the 5α-reductase type 2 deficiency diagnosis later in life. Most cases were assigned as female (69.4%), and an association between higher scores of external genitalia virilization (less virilization) and female sex assignment was identified. However, when we divided the cases into those who were diagnosed after and before 1999, the percentage of male sex assignment rose from 26.8% to 42.8%, suggesting a temporal trend pointing toward an increased likelihood of 5α-reductase type 2 deficiency patients being raised as boys (305). Intriguingly, 5α-reductase type 2 deficiency is a condition with no genotype-phenotype correlation (307-309). This observation is based on several 5α-reductase type 2 deficiency families carrying the same genotype but presenting a broad range of external genitalia virilization. However, some SRD5A2 variants are consistent in the way they affect phenotype. It is the case of the p.Arg246Gln variant, which is associated with more external genitalia virilization (304,310-312), and also the case of both p.Gly183Ser and p.Gln126Arg variants, that are consistently reported with more severe external genitalia under-virilization (304,313-315). The diagnosis is usually made at birth, infancy or at puberty. In the newborn, the features of 46,XY DSD due to 5a-reductase type 2 deficiency overlap with other forms of male DSD such as androgen insensitivity syndrome (partial form) and testosterone synthesis defects (304,316). At puberty or in young adult men, the basal hormonal evaluation demonstrates normal male serum testosterone levels, low or low/normal dihydrotestosterone levels, and elevated or normal serum testosterone to dihydrotestosterone ratio (307). For appropriated use of this ratio, the testosterone levels should be in the post puberal range. Likewise, in prepubertal children, a hCG stimulation to increase serum testosterone levels is necessary (313). The biggest challenge is the diagnosis in newborns. This difficult largely arises from the fact that even when serum testosterone has undergone a neonatal surge, the ratio of serum testosterone to dihydrotestosterone may be normal, because expression of the 5a-reductase type 1 enzyme can occasionally be higher than expected (317,318). Measurement of dihydrotestosterone is difficult because this steroid is present in very low concentrations and has a high rate of cross-reactions (319). To obtain an accurate dihydrotestosterone measurement, a precise assay must be utilized since serum testosterone levels are higher than dihydrotestosterone levels (about 10-fold). Consequently, the separation of testosterone from dihydrotestosterone is necessary to provide and accurate dihydrotestosterone measurement. Using such methodologies, the testosterone/dihydrotestosterone ratio for 5a-reductase type 2 deficiency hormonal diagnosis is generally over 18 in most cases (320,321). However, the testosterone/dihydrotestosterone ratio for 5a-reductase type 2 deficiency hormonal diagnosis has been debatable. Another approach to 5a-reductase type 2 deficiency diagnosis is the measurement of urinary steroids by gas chromatography– mass spectrometry (GC–MS) to determine the ratio of 5a- to 5ß- reduced steroids in urine. This evaluation is very helpful for the diagnosis in subjects at prepubertal age and in orchiectomized adults. In one review, extremely low ratios of 5a- to 5ß-reduced steroid metabolites in urine were pathognomonic for 5a-reductase type 2 deficiency (322). Based on the challenges on the hormonal diagnosis, genetic analysis of the SRD5A2 gene is recommended to confirm the diagnosis [39,40]. The management in subjects with female social sex includes a careful psychological evaluation of gender identity (323). Subsequent management is similar to that in women with other forms of 46,XY DSD (324). Treatment must simulate a normal puberty pattern and low to normal estrogen doses, considering the height, and it should be administered at the age of expected puberty (10 – 12 years old). After complete breast development, adult estrogen doses are maintained continuously. Progesterone replacement is not necessary because these patients do not have a uterus (11). For women with this condition, feminizing genitoplasty is often necessary to provide an adequate vaginal opening, a functional vaginal introitus, full separation between urethral and vaginal orifice and phallic erectile tissue removal (11). Vaginal dilatation to promote vaginal length with acrylic molds is recommended when the patients decide to initiate sexual activity (325). Laparoscopic orchiectomy is recommended for all female patients to avoid virilization and gonadal malignancies. Usually, testosterone replacement is not often necessary for male patients since most retain testes and present adequate testicular function towards puberty (307). However, since the degree of virilization is usually unsatisfactory in male patients, a limited use of intramuscular testosterone or transdermal dihydrotestosterone may be helpful to improve virilization (307,319). Maximum penile length is obtained after 6 months of high dose testosterone therapy (e.g., 500 mg of testosterone cypionate per week) (319). The therapeutic penile response does not result in normal penile length in all individuals, even when initiated during childhood, and the final penile length is still below -2 SD in most patients (326). Surgical treatment consists of orthophalloplasty, scrotoplasty, resection of the vaginal pouch and proximal and distal urethroplasty. Correction of hypospadias is indicated in early childhood (up to two years old) (326). Gender change (from female to male) is common among 5α-reductase type 2 deficiency individuals (327). It occurs in over 50% among those assigned as girls in some series (328). It may change since there is growing evidence suggesting male sex assignment for 5α-reductase type 2 deficiency newborns to avoid gender incongruence and gender dysphoria (329-331). Regarding long-term follow up in the males from the Sao Paulo cohort, most of these subjects were satisfied with the appearance of the external genitalia and sexual life, although a small penile length made sexual intercourse difficult for some of them (326). Most of the adult males patients get married, and those reared as male report a more satisfactory quality of life than the female patients (332). Among female individuals, most describe a satisfactory sexual life, but none are married or have adopted children (333). Table 14. Phenotype of 46,XY Subjects with 5α-Reductase 2 Deficiency View in own window Inheritance Autosomal recessive External genitalia Atypical, small phallus, proximal hypospadias, bifid scrotum, blind vaginal pouch Müllerian duct derivatives Absent Wolffian duct derivatives Normal Testes Normal size at inguinal, intra-abdominal region or topic Puberty Virilization at puberty, absence of gynecomastia Hormonal diagnosis Increased T/DHT ratio in basal and hCG-stimulation conditions in postpubertal patients and after hCG-stimulation in pre-pubertal subjects. Elevated 5β/5α C 21 and C 19 steroids in urine in all ages SRD5A2 gene location 2 p23 Molecular defect Inactivating variants in 5RD5A2 Gender role Female → male in 60% of the cases Treatment High doses of T and/or DHT for 6 months to increase penis size Outcome Maximum penis size in males after treatment = 9 cm; fertility is possible by in vitro fertilization Go to: 46,XY DSD DUE TO DEFECTS IN ANDROGEN ACTION Androgen Insensitivity Syndrome Androgen insensitivity syndrome (AIS) is the most frequent etiology of 46,XY DSD individuals (334). The underlying molecular basis of AIS is variants in the androgen receptor gene (AR), which is located on the long arm of the X chromosome (Xq11-12) (335). AIS is an X-linked inherited condition, and up to 30% of AIS cases present de novo variants (334,336). Due to disruptive variants in the AR gene, affected individuals present a broad spectrum of under-virilization, which will depend upon the residual activity of the mutant AR. There are three phenotypes of AIS: complete (typically female external genitalia; CAIS), partial (a wide spectrum of external genitalia under virilization; PAIS) or mild (typically male external genitalia with further gynecomastia and/or infertility; MAIS) (337,338). The AR contains eight exons and encodes a 920 amino-acids protein (10). The AR is composed of three major functional domains: the N-terminal transactivation domain (NTD), a central DNA-binding domain (DBD), a C-terminal ligand-binding domain (LBD), and a hinge region connecting the DBD and LBD (339,340). The main difference between the AR and other steroid receptors is the presence of a longer NTD (341). Exon 1 encodes for the NTD, while exons 2 and 3 encode for the DBD and exons 4-8 encode for the LDB (342). In the presence of androgens, the AR recruits multiple epigenetic coregulators. These co-regulators can be either co-activators or co-repressors and act upon AR influencing DNA binding, nuclear translocation, chromatin remodeling, AR stability and bridging AR with transcriptional machinery (343,344). AR coding region has two polymorphic trinucleotide repeat regions, located at exon 1, the CAG and GGC repeats (345). The length of these repeats can cause human diseases. In general, longer CAG repeats may lead to AR transactivation impairment whereas shorter CAG repeats may enhance the AR transactivation (346). A high number of CAG repeats (>38) is the molecular cause of Spinal and Bulbar Muscular Atrophy (Kennedy’s disease) (347). This condition is characterized by severe muscular atrophy and a mild AIS phenotype, including gynecomastia. On the other hand, shorter CAG repeats are related with increased risk for prostate cancer (348). There are more than 800 variants in the AR gene reported in AIS patients (www.androgendb.mcgill.ca/; HGMD). Most of them are missense variants leading to amino acid substitutions (349). However, small indels, variants at splicing sites, premature stop codons and large deletions were also reported, most of them related to CAIS (350). Despite a well-characterized monogenic condition, AR variants are identified in 90-95% of CAIS, but only in 28-50% of PAIS (334,350). Therefore, the molecular diagnosis of PAIS individuals remains challenging. Advances in molecular biology have been helpful to clarify unusual molecular mechanisms or deep DNA alterations related to AIS. Alterations immediately upstream of the AR were identified in AIS patients without variants in the coding region of the AR, either by promoting aberrant AR transcripts, or disrupting AR expression by the insertion of a large portion of a long-interspersed element retrotransposon, which were proven to cause AIS (351,352). Rare synonymous variants within the encoding region of the AR gene were proven to play a role in AIS by disrupting splicing (353). The sequencing of intronic regions of the AR was able to identify a deep intronic variant leading to pseudo-exon activation in AIS (354). Additionally, studies involving AR variant-negative individuals with AIS revealed the deficiency of the androgen-responsive apolipoprotein D, indicating functional AIS, and epigenetic repression of the AR transcription was reported in a group of AIS variant-negative individuals, a condition defined as AIS type II (355). However, a specific role of certain coregulators in the pathophysiology of AIS is not established yet and the contribution of AR-associated coregulators in AIS remains poorly understood. COMPLETE ANDROGEN INSENSITIVITY SYNDROME Prenatal diagnosis of CAIS is possible and can be suspected based on the discordance between 46,XY karyotype on prenatal fetal sex determination and the identification of a female genitalia at prenatal ultrasound (356). At birth, CAIS individuals present typically female external genitalia. In childhood, the identification of an inguinal hernia in a girl may be a clinical indication of CAIS, since inguinal hernias in girls are rare (357). At puberty, CAIS patients present with complete breast development and primary amenorrhea (358). Pubic hair and axillar hair are sparse in most of them, and Mullerian ducts are often absent in CAIS patients (337). The endocrine evaluation after puberty shows normal or elevated serum testosterone levels and slightly elevated LH levels, whereas FSH levels can be slightly elevated, with normal presence of testosterone precursors (334). Patients with CAIS are assigned and raised as girls and usually present a female gender identity (328,331). Estrogen replacement is recommended to induce puberty if bilateral gonadectomy has been performed before puberty. There is gonadal tumor risk in CAIS patients, but this risk is very low before puberty (359). Therefore, gonadectomy can be postponed because after puberty is complete in CAIS patients (360,361) . An increasing number of adult women with CAIS prefer to decline or delay gonadectomy for several reasons, such as fear of surgery, to avoid estrogen replacement, and expectations for future fertility (362). Table 15. Phenotype of 46,XY Subjects with Complete Androgen Insensitivity Syndrome View in own window Inheritance X-linked recessive External genitalia Female Müllerian duct derivatives Absent Wolffian duct derivatives Absent or vestigial Testes Inguinal or intra-abdominal, slightly subnormal size Puberty Complete breast development Hormonal diagnosis High or normal serum LH and T levels, normal or slightly elevated FSH levels Gender role Female AR gene location Xq11-12 Molecular defect Pathogenic allelic variants in AR gene Treatment Psychological support Estrogen replacement after gonadectomy. Vaginal dilation for sexual intercourse Outcome Female identity, infertility PARTIAL ANDROGEN INSENSITIVITY SYNDROME Patients with PAIS have a broad spectrum of virilization impairment (337). The external genitalia ranged from predominantly female with clitoromegaly and labial fusion to predominantly male with micropenis and hypospadias. Testes are in the inguinal canal or labioscrotal folds or, less frequently, intraabdominal. At puberty, under-virilization and gynecomastia are observed (334). The final height of PAIS individuals is intermediate between the average height for control males and females. In addition, PAIS individuals presented decreased bone mineral density in the lumbar spine compared to controls (363). In male PAIS, gynecomastia is common at puberty which is helpful in the differential diagnosis from other 46,XY DSD etiologies (364). In the endocrine analysis, serum LH levels are in the normal upper range or slightly elevated, and testosterone levels are normal or slightly elevated (334). A definitive diagnosis of PAIS is established by identifying a variant in the AR gene, but AR variants are found only in about 40% of PAIS (350). The sex of rearing is female in half of the cases, and gender change is uncommon in PAIS patients either raised as female or male (328). Estrogen replacement is necessary for female patients to induce adequate puberty since most female PAIS patients undergo gonadectomy in childhood (334,365). For male patients, androgen replacement, either to induce puberty or to enhance virilization post-puberty, is commonly required (11). High doses of intramuscular testosterone preparations or topical DHT can be tried for six months to improve virilization, but it is unnecessary after that (11). If the testes are at the scrotum, gonadectomy is unnecessary in male PAIS individuals. However, bilateral gonadectomy is still recommended for female PAIS due to avoid partial virilization and due to the gonadal malignancy risk (288,337). Table 16. Phenotype of 46,XY Subjects with Partial Androgen Insensitivity Syndrome View in own window Inheritance X-linked recessive External genitalia Broad spectrum from female with mild clitoromegaly to male with micropenis and/or hypospadias Müllerian duct derivatives Absent Wolffian duct derivatives Broad spectrum from absent or male Testes Eutopic, inguinal or intra-abdominal, normal or slightly subnormal size Puberty Gynecomastia Hormonal diagnosis High or normal serum LH and T levels, normal or slightly elevated FSH levels Gender role Female or male AR gene location Xq11-12 Molecular defect Pathogenic variants in AR gene Treatment Females: surgical feminization, gonadectomy, replacement with estrogens at the time of puberty, vaginal dilation (if necessary) Males: hypospadias repair, bifid scrotum; high doses of T or DHT to increase penis size Outcome Infertility, female or male gender role Go to: 46,XY DSD DUE TO PERSISTENT MÜLLERIAN DUCT Defect in AMH Synthesis or AMH Receptor The development of female internal genitalia in a male individual is due to the incapacity of Sertoli cells to synthesize or secrete anti-Mullerian hormone (AMH) or to alterations in the hormone receptor. Persistent Müllerian duct syndrome (PMDS) phenotype can be produced by a variant in the gene encoding anti-Müllerian hormone or by a variant in the AMH receptor. These two forms result in the same phenotype and are referred to as type I and type II, respectively (366). AMH is a 145,000 MW glycoprotein homodimer produced by Sertoli cells not only during the period when it is responsible for regression of the Müllerian ducts but also in late pregnancy, after birth, and even, albeit at a reduced rate, in adulthood (13,367,368). AMH is a small gene containing 5 exons, located in chromosome19p.13.3 (367) and its protein product acts through its specific receptor type 2 (AMHR2) a serine/threonine kinase, member of the family of type II receptors for TGF-β-related proteins (369). Affected patients present a male phenotype, usually along with bilateral cryptorchidism and inguinal hernia (368). Leydig cell function is preserved, but azoospermia is common due to the malformation of ductus deferens or agenesis of epididymis. When the hernia is surgically corrected, the presence of a uterus, fallopian tubes and the superior part of the vagina can be verified. PMDS is a heterogeneous disorder that is inherited in a sex-limited autosomal recessive manner. variants in AMH gene or AMH receptor 2 gene in similar proportions are the cause of approximately 85% of the cases of PMDS (370,371). In the remaining cases the cause of the persistent Mullerian duct syndrome is unknown (368). Normally, AMH levels are measurable during childhood and decrease at puberty. Patients with AMH gene defects have low AMH levels since birth whereas patients with variants in AMH receptor gene have elevated AMH levels (372). Treatment is directed toward an attempt to assure fertility in males. Early orchiopexy, proximal salpingectomy (preserving the epididymis), and a complete hysterectomy with dissection of the vas deferens from the lateral walls of the uterus are indicated (368,373). Go to: CONGENITAL NON-GENETIC 46,XY DSD Maternal Intake of Endocrine Disruptors The use of synthetic progesterone or its analogs during the gestational period has been implicated in the etiology of 46,XY DSD (374). Some hypotheses have been proposed to explain the effect of progesterone in the development of male external genitalia, such as reduction of testosterone synthesis by the fetal testes or a decrease in the conversion of testosterone to DHT due to competition with progesterone (also a substrate for 5α-reductase 2 action). The effect of estrogen use during gestation in the etiology of 46,XY DSD has not been confirmed to date (375). Recently, a study in Japanese subjects supports the hypothesis that homozygosis for the specific estrogen receptor alpha 'AGATA' haplotype may increase the susceptibility to the development of male genital abnormalities in response to estrogenic effects of environmental endocrine disruptors (376). Environmental chemicals that display anti-androgenic activity via multiple mechanisms of action have been identified. They are pesticides, fungicides, insecticides, plasticizers and herbicides. They can work as androgen receptor antagonists like pesticides, or they can decrease mRNA expression of key steroidogenic enzymes and also the peptide hormone insl3 from the fetal Leydig cells, like plasticizers and fungicides (377). Daily exposure to residues of a fungicide (vinclozolin), either alone or in association with a phytoestrogen genistein (present in soy products), induce hypospadias in 41% of mice, supporting the idea that exposure to environmental endocrine disruptors during gestation could contribute to the development of hypospadias (378). Supporting the idea that exposure to a mixture of chemicals can produce greater incidences of genital malformations, Rider et al examined the effects of exposure to a mixture of two chemicals that act as androgen receptor antagonists. They observed that the exposure to vinclozolin (fungicide) alone resulted in a 10% incidence of hypospadias and no vaginal pouch development in male rats, whereas procymidone, another fungicide exposure, failed to generate malformations. However, the combined exposure resulted in a 96% incidence of hypospadias and 54% incidence of vaginal pouch in treated animals. Similar results were observed in phthalate (plasticizer) mixture studies (377). Given that severe alterations of sexual differentiation can be produced in animal laboratory studies, the question arises of what would be expected in exposed humans given that humans are exposed to mixtures of compounds in their environment. Congenital Non-Genetic 46,XY DSD Associated With Impaired Prenatal Growth Despite the multiple genetic causes of 46,XY DSD, around 30-40% of cases remain without diagnosis. Currently, there is a frequent, non-genetic variant of 46,XY DSD characterized by reduced prenatal growth and lack of evidence for any associated malformation or endocrinopathy (379,380). Using the model of monozygotic twins, hypospadias has now been linked to low birth weight (379). We have identified a pair of 46,XY monozygotic twins (identical for 13 informative DNA loci) born at term after an uneventful pregnancy sustained by one placenta who were discordant for genital development (perineal hypospadias versus normal male genitalia) and postnatal growth (low birth weight versus normal birth weight). No evidence for uniparental disomy was found (381). The most plausible cause of incomplete male differentiation associated with early-onset growth failure is a post-zygotic, micro-environmental factor since different DNA methylation patterns associated with silencing of genes important for sex differentiation has been shown (382). Additionally, three cohorts of undetermined 46,XY DSD report around 30% of cases as associated with low birth weight, indicating that adverse events in early pregnancy are frequent causes of congenital non-genetic 46,XY DSD (383-385). A genetic defect that clarifies the etiology of hypospadias was not found in 41 non-syndromic SGA children, supporting the hypothesis that multifactorial causes, new genes, and/or unidentified epigenetic defects may have an influence in this condition (385). Go to: 46,XY OVOTESTICULAR DSD There are rare descriptions of 46,XY DSD patients with well characterized ovarian tissue with primordial follicles and testicular tissue, a condition that is histologically characterized 46,XY ovo-testicular DSD (386). The differential diagnosis of 46,XY ovo-testicular DSD with partial 46,XY gonadal dysgenesis should be performed considering that in the latter condition there are descriptions of dysgenetic testes with disorganized seminiferous tubules and ovarian stroma with occasional primordial follicles in the first years of life (46). To our knowledge there are no descriptions of an adult patient with 46,XY ovo-testicular DSD with functioning ovarian tissue, as occurs in all 46,XX ovo-testicular DSD. Therefore, the diagnosis of 46,XY ovo-testicular DSD is debatable. Go to: NON-CLASSIFIED FORMS Hypospadias Hypospadias is one of the most frequent genital malformations in the male newborn and 40% of the cases are associated with other defects of the urogenital system (387). Hypospadias results from an abnormal penile and urethral development that appears to be a consequence of various mechanisms including genetic and environmental factors. It is usually a sporadic phenomenon, but familial cases can be observed, with several affected members (388,389). The presence of hypospadias indicates an intra uterus interference in the correct genetic program of the complex tissue interactions and hormonal action through enzymatic activities or transduction signals. MAMLD1 (mastermind-like domain containing 1) has been reported to be a causative gene for hypospadias (390). It appears to play a supportive role in testosterone production around the critical period for sex development. To date, microdeletions involving MAMLD1 and nonsense and frameshift variants in the gene have been found in 46, XY DSD patients, suggesting that MAMLD1 variants cause 46,XY DSD primarily because of compromised fetal testosterone production, however, its role in the molecular network involved in fetal testosterone production is not known so far (391). The activating transcription factor 3 (ATF3) expression was identified in the developing male urethra. Apparently ATF3 variants may influence the risk of hypospadias (392). By definition, hypospadias is a form of 46,XY DSD and although most of the patients maintain fertility and masculinization at puberty, their testicular function should be assessed to rule out causes such as defects in testosterone synthesis and action, which require hormonal treatment and genetic counseling in addition to surgical treatment. Go to: GONADAL TUMOR DEVELOPMENT IN 46,XY DSD PATIENTS Any disturbance in the gonadal development, including testicular descent, increases the risk of developing gonadal malignancies (288). For inherited disturbances in gonadal development or endocrine alterations, patients with 46,XY DSD are at increased risk of developing type II germ cell tumors (GCT) (289). In testicular tissues, GCTs comprise both premalignant conditions, such as germ cell neoplasia in situ (GCNIS) and malignant invasive germ cell tumors, including seminomas and non-seminomas (393). The term GCNIS was introduced in the 2016 WHO classification of urological tumors to define precursor lesions of invasive GCTs, since GCNIS has the potential to develop into several types of GCTs (394,395). GCNIS cells are fetal gonocytes present in the seminiferous tubules arrested during gonadal development that failed to mature into spermatogonia (396). GCNIS are often detected in testicular tissues from 46,XY DSD subjects (397). It is estimated that 50% of GCNIS progress to an invasive GCT in five years (396,398,399). A high risk of GCT is found when sex determination is disrupted at an early stage of Sertoli cell differentiation (due to abnormalities in SRY, WT1, SOX9) (289,397,400). For that reason, specific etiologies of 46,XY DSD (401) have a significant risk factor for GCT development (393). Early Sertoli cell development is also disturbed in patients with 45X/46,XY mosaicism (402). The presence of the well-defined Y chromosome region, known as the gonadoblastoma Y locus (GBY), is a prerequisite for malignant transformation. Among the genes located in the GBY region the testis-specific protein Y (TSPY) seems to be the most significant candidate gene for the tumor-promoting process (288,403). The presence of undifferentiated gonadal tissue containing germ cells that abundantly express TSPY has also been identified as a gonadal differentiation pattern bearing a high risk for GCT development (404). Prolonged expression of OCT3/4 (POU5F1) and the stem cell factor KITL after one year of age are also estimated to play a role in GCNIS/GCT development. Other factors implicated in that risk include MAP3K1 variants in 46,XY patients with gonadal dysgenesis due to MAPK signaling pathway upregulation and loss of androgen receptor function in patients with androgen insensitivity syndrome (289,405). Additionally, gonads at the abdominal region are at higher risk of GCNIS/GCT development than those appropriately positioned (393,406). Unfortunately, GCNIS/GCT screening is challenging due to a lack of a predictive factor or a biomarker with adequate sensitivity and specificity (407). As far as imaging is concerned, ultrasound (US) is more sensitive than MRI at identifying dysgenetic gonads, but MRI showed better sensitivity and specificity than US at localizing non-palpable gonads (408). However, both imaging techniques are poor at identifying GCNIS/GCT, since MRI failed to identify GCNIS in patients with CAIS and the US only identified one out of ten malignant lesions in 46,XY DSD people (409). There are serum markers that are associated with GCT in non-DSD people, such as alpha-fetoprotein, beta-hCG, and lactate dehydrogenase, but there is poor evidence about how useful they are for GCT screening in 46,XY DSD individuals (410). An interesting perspective for GCT screening are microRNAs (miRNA), since some miRNA clusters are expressed in the presence of GCT (411). For non-DSD people, microRNAs are more sensitive than serum markers and imaging to detect GCT. Noteworthy, GCNIS also expresses some embryonic-type miRNAs (miR-371-3, miR-302, and miR-367) that are also expressed by GCTs (410,412). Therefore, they have the potential to serve as a biomarker even for GCNIS(407). Overall, neoplastic transformation of germ cells in dysgenetic gonads (gonadoblastoma and/or an invasive germ cell tumor) occurs in 20-30% of 46,XY DSD individuals, but the risk varies among 46,XY DSD etiologies (413). Individuals with Denys-Drash syndrome (40%), Frasier syndrome (60%), and gonadal dysgenesis (15 - 35%) have the highest risk of GCNIS/GCT among 46,XY DSD etiologies (413). On the other hand, individuals with CAIS (at prepubertal age) and ovotestis DSD have lower risk of GCNIS/GCT (414). The age matters in the estimation of GCNIS/GCT risk. For example, it is as low as 1.3% in CAIS individuals before puberty, but it can reach 33% thereafter (287,415). For 46,XY DSD subjects, gonadectomy is classically recommended to avoid GCNIS/GCT development, preventing additional therapies and related risks (290). Despite a very effective strategy to avoid GCNIS/GCT, gonadectomy leads to hypogonadism and infertility. Regarding the time for gonadectomy, bilateral gonadectomy should be performed in early childhood in 46,XY DSD patients with gonadal dysgenesis, females with Y chromosome material, and patients with androgen biosynthesis defect, unless the gonad is functional and easily accessible to palpation and imaging studies, which should be performed yearly (11,289). Although data are limited, in the androgen insensitivity syndrome the risk seems to be markedly lower in the complete form before puberty than in the other 46,XY DSD (416). Therefore, gonadectomy can be postponed until puberty is complete in CAIS individuals (417). Unfortunately, the GCNIS/GCT risk for other causes of 46,XY DSD patients, such as Leydig Cell Hypoplasia and 5 alpha reductase type 2 deficiency has not been estimated yet. Rarely, gonadal tumors can produce sexual steroids (418). In those cases that are able to produce estrogens, spontaneous breast development may be a clinical sign that suggests the presence of an estrogen-secreting gonadal tumor, and bilateral gonadectomy is indicated even at early childhood, regardless of the 46,XY DSD etiology. Overall, 46,XY DSD patients are at increased risk for gonadal malignancy which seems to be related to 46,XY DSD etiology. While it is clear that prepubertal CAIS patients are at low risk for GCT and 46,XY DSD individuals harboring WT1 variants present a high risk for GCT development, the real GCT risk for other 46,XY DSD etiologies is not that clear (413). In the absence of a reliable predictive factor or biomarker of GCNIS/GCT as well as appropriate recommendations for GCT screening, bilateral gonadectomy will still be recommended for most 46,XY DSD etiologies. Go to: FERTILITY IN PATIENTS WITH 46,XY DSD Most 46,XY DSD individuals face infertility due to abnormal gonadal development, endocrine disturbances, anatomical issues, or prophylactic gonadectomy for malignancy risk (419). However, there has been growing evidence showing that fertility is relevant for several 46, XY DSD people, in addition to the possibility of delaying gonadectomy in some 46,XY DSD etiologies (420). In parallel, fertility preservation technologies have been improved in recent years along with a better social perception of non-traditional family structures (421,422). In 46,XY DSD, the fertility potential varies depending on the underlying etiology as well as the severity of the condition (421). In this sense, all options for fertility should be discussed considering the 46,XY DSD etiology or the gonadal structure and internal genitalia in those in whom the 46,XY etiology is unknown (420). For example, individuals with complete gonadal dysgenesis possess uterus, despite lacking gametes (423). Therefore, pregnancy by oocyte donation is an alternative for these patients. On the other hand, male individuals with partial and mild androgen insensitivity often present oligospermia, but biological fertility is possible (334,338). Overall, there is limited literature about fertility potential among 46,XY DSD people. Successful biological fertility was obtained in a man with PAIS after prolonged high-dose testosterone therapy followed by intracytoplasmic sperm injection (424). The possibility of fertility seems to be more frequent among MAIS since there are six cases of successful fertility (338). There are few reported cases of successful pregnancies and live births in men with 5RD2 deficiency, both spontaneous and with assisted reproductive technology (319,425-427). Biological fertility has also been documented in individuals with nonclassical congenital lipoid adrenal hyperplasia, 3b-HSD2 deficiency, and LHCG receptor defect (420). Conversely, there are no reported cases of biological fertility in individuals with classic congenital lipoid adrenal hyperplasia, cytochrome p450 oxidoreductase deficiency, complete CYP17A1 deficiency, 17b-HSD3 deficiency, and CAIS (419,428). To estimate fertility potential, a pilot study evaluated the presence of germ cells and the germ cell density in individuals with several 46,XY DSD etiologies (429). In six patients with CAIS, all presented Sertoli Cell nodules and hyperplasia, but germ cells were detected in areas between nodules. All six patients with mixed gonadal dysgenesis and two with ovo-testicular DSD presented germ cells, and ten out of twelve 46,XY DSD patients with unknown etiology presented germ cells in their gonads. On the other hand, germ cells were not found in any of the patients with either complete or partial gonadal dysgenesis. However, the number of germ cells was inversely correlated with age, suggesting that the gonadectomy delay may decrease fertility potential. It needs to be confirmed by more extensive studies, but it indicates that 46,XY DSD fertility potential may be greater than previously thought. As far as desire for fertility is concerned, a large follow-up study included 1,040 DSD individuals to investigate their fertility preferences (430). The authors reported that 55% of patients expressed a desire to have had fertility treatments in the past or have it in the future, and 40% mentioned that they would like to try new fertility treatment techniques. Additionally, CAIS women reported the possibility of future fertility as one of the reasons to keep their gonads (362). Indeed, fertility preservation has been primarily assessed in oncology to preserve patients' fertility under gonadotoxic treatments (431). In this sense, cryopreservation of postpubertal testicular tissue is helpful to keep fertility potential in patients having gonadectomy or those before gonadotoxic treatment. As an alternative, cryopreservation of immature testicular tissue containing spermatogonial cells or spermatogonial stem cells can be offered to prepubertal patients (432). These techniques could also be considered for 46,XY DSD patients. In summary, addressing fertility is essential in 46,XY DSD management. The fertility potential must be discussed considering the 46,XY DSD etiology and the patient’s desire. As assisted fertility and preservation techniques improve, these advancements should be offered and accessible to all 46,XY individuals. Go to: 46,XY GENDER IDENTITY DISORDERS Transgender Women are characterized by the wish to live as members of the female sex with conviction and consistently and progressively efforts to achieve such state. 46,XY gender identity disorders are more frequent among the male sex, although it also occurs in the female sex. Its first manifestations usually start during childhood. If it has a biological basis is still unknown, but some hormonal alterations during intrauterine life and familial factors before and after birth cannot be ruled out (433). The term used to name men and women who live a relevant incongruence between their gender identity and their inborn physical phenotype has changed over time. The term “trans-sexualism” was coined by Hirschfeld in 1923 and was adopted by the International Classification of Diseases – version 10 (ICD-10). The American Psychiatric Association, in its 4 th edition, adopted “gender identity disorder” to define persons who are not satisfied with their biological gender (Association, American Psychiatric. "Diagnostic and statistical manual of mental disorders(2000). Finally, the current classification system of the American Psychiatric Association (DSM-5) replaced the term “gender identity disorder” with “gender dysphoria” and the upcoming version of International Classification of Diseases – version 11 (ICD-11) has proposed the term “gender incongruence” (434). In this chapter we will use the current DSM-5 term, “gender dysphoria”. To refer to male to female gender-dysphoric persons we will use the term transgender woman (American Psychiatric Association. Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition - DSM-5; 2013). Therefore, the term “transgender woman” refers to all 46, XY individuals with typical male phenotype who wish to live and be accepted as a female. Higher prevalence of addictions and suicidal thoughts or suicide attempt than those observed in the general population, revealed the need for early care of these patients by health professionals. Among transgender women, total mortality was 51% higher than in the general population, mainly from increased mortality rates due to suicide, acquired immunodeficiency syndrome, cardiovascular disease, drug abuse, and unknown cause (435). Based on these data, supervised gender-affirming treatment for gender dysphoric persons is crucial because they are at increased risk of committing suicide and self-harm (436). Management of Adult Transgender Women As proposed by the Harry Benjamin International Gender Dysphoria Association, now known as World Professional Association for Transgender Health (WPATH), the process of gender-affirming treatment should be given by a multi and interdisciplinary team, in which the endocrinologist has a key role (437). The interdisciplinary team should consist of a psychologist, a psychiatrist, an endocrinologist, and a surgeon, at least (438). It would be ideal that they all participate in an integrated and consistent way across all the steps of the treatment (439). The mental health professionals (psychologist and psychiatrist) make a distinction between gender dysphoria and conditions with similar features (body dysmorphic disorder and body identity integrity disorder), decide whether the individuals fulfill ICD-11 and DSM-5 criteria, recommend the appropriate treatment and follow-up before, during and after gender-affirming treatment. The endocrinologist will inform about the possibilities and limitations of all sorts of treatment, initiate and monitor the cross-sex hormonal treatment and participate in the indication of gender-affirming surgery. At the final step, the surgeon performs surgical procedures of the treatment (439). DIAGNOSTIC ASSESSMENT AND MENTAL HEALTH CARE Psychological evaluation of persons with gender dysphoria should consider the evolution of the individual as whole, using psychological assessment instruments, such as: freely structured interviews and patterned psychological assessment instruments. For the structured interview, we use a specific questionnaire developed by our mental health professionals that covers childhood, adolescence, and adulthood aspects. During the psychotherapeutic follow up, besides offering an ideal condition for elaborating conflicts and issues regarding gender identity, other variables should be considered, such as individual general state of mental health, ability and manner of conflict resolution, quality of interpersonal relationships, ability to deal with frustrations and limitations, particularly regarding to surgery’s esthetic and functional results idealization. It is recommended that the relatives and/or spouses are invited for interviews to clear them up upon the offered treatment. HORMONAL THERAPY FOR ADULT TRANSGENDER WOMEN Endocrinologists have the responsibility to confirm that persons fulfill criteria for hormonal treatment and clarify the consequences, risks, and benefits of treatment. Hormone therapy must follow well-defined criteria. The person with gender dysphoria has to: 1) demonstrate knowledge and understanding of the expected and side effects of cross-sex hormone use; 2) complete a real life experience in the gender identity for at least three months, or psychotherapy for a period determined by the mental health professional to consolidate gender identity; and 3) be likely to take hormones appropriately (439). There are two major goals of hormonal therapy: 1) to replace endogenous sex hormone levels and, thus, induce the appearance of sexual characteristics compatible with female gender identity; 2) to reduce endogenous sex hormone levels and, thereby, the secondary male sexual characteristics and 3) to establish the ideal hormones dosage which allows physiological hormone serum levels compatible with female gender identity by adopting the principles of hormone replacement treatment of hypogonadal patients (439,440). Hormone therapy provides a strong relief from the suffering caused by the incongruence of the phenotype with the gender identity. In our clinical practice, we observe that most transgender women consume very high doses of female sex hormones, guided by their wish to obtain fast breast development, and reduce facial hair. However, high doses of hormones are not necessary to achieve the desired effects and are frequently associated with undesirable side effects. The chosen hormone to induce female secondary sexual characteristics are the estrogens. Several pharmaceutical estrogen preparations, including oral, injectable, transdermal, and intravaginal forms associated or not with progesterone are available. Due to the higher cost of the transdermal preparations, the oral route is the most widely used. Nevertheless, the transdermal route is recommended for transgender women over 40 years of age due to the lower association of transdermal 17β-estradiol replacement with thromboembolic events (441). Anti-androgens are used as adjuvants to estrogen, especially in the reduction of male sexual characteristics and the suppression of testosterone to levels compatible with the female sex. Cyproterone acetate blocks testosterone binding to its receptor, and in a dose of 50-100 mg/day associated with estrogen can maintain testosterone in female levels in transgender women (442). At the time, most of the patients in our clinic used conjugated equine estrogens at a dose of 0.625-1.25 mg/day associated with 50 mg/day of cyproterone acetate for an average period of 11 years. At clinical examination we observed satisfactory breast development, decrease of spontaneous erections, thinning of facial and body hair (especially after association with cyproterone acetate), body fat redistribution, enlargement of the areola and nipple and reduction of testicular volume (440). Testosterone levels remained at pre- or intra-pubertal female range (< 14-99 ng/dL) in all patients; LH levels were pre-pubertal (<0.6-0.7 U/L) in 72% of the cases, and the FSH levels were suppressed (<1.0 U/L) in 40% of cases. Therefore, daily use of oral conjugated estrogens at low doses in association with cyproterone acetate is effective in suppressing the hypothalamic-pituitary-testicular axis in transgender women (440). Venous thromboembolism may be a serious complication related to estrogen therapy, particularly during the first year of treatment, when the incidence of this event is 2-6% falling to 0.4% in the second year, significantly higher when compared to the overall young population (0.005 to 0.01%/year). This high incidence of thromboembolic events in transgender women seems to be more associated with ethinyl estradiol than natural oral or transdermal estrogens (441). All patients on estrogen therapy have a mild prolactin level increase. However, a small percentage of these subjects have galactorrhea. In our cohort, two patients had macroprolactinoma, which totally regressed with dopamine agonist treatment. Both had previously used high doses of estrogen (443). Endocrinologists should monitor weight, blood pressure, breast enlargement, body hair involution, body fat redistribution and testicular size every six months. The laboratory evaluation should include measurement of LH, FSH, testosterone, estradiol, prolactin, liver enzymes, complete blood count, coagulation factors, and lipid profile. Bone densitometry and breast ultrasound should be performed yearly. After surgery in patients over 50 years old, the measurement of PSA should be conducted yearly (440). The current key issues include avoiding supraphysiological doses of estrogen and the use of ethinyl estradiol. The preference should be given to conjugated estrogens or transdermal natural estrogen, especially in patients over 40 years of age (444). Hormone therapy provides a strong relief from the suffering caused by gender dysphoria. (440). Go to: MANAGEMENT OF PATIENTS WITH 46,XY DSD It is important to stress that the treatment of 46,XY DSD patients requires an appropriately trained multi-disciplinary team. Early diagnosis is important for better outcomes and should start with a careful examination of the newborn’s genitalia at birth (445-447). Psychological Evaluation It is of crucial importance to treat DSD patients (448). Every couple that has a child with atypical genitalia must be assessed and receive counseling by an experienced psychologist, specialized in gender identity, who must be act as soon as the diagnosis is suspected, and then follow the family periodically, more frequently during the periods before and after genitoplasty (449,450). Parents must be well informed by the physician and psychologist about sexual development (451). A simple, detailed, and comprehensive explanation about what to expect regarding integration in social life, sexual activity, need of hormonal replacement and surgical treatment and fertility issues should also be discussed with the parents, before sex assignment (11). The sex assignment must consider the etiological diagnosis, external genitalia, cultural and social aspects, sexual identity and the acceptance of the assigned gender by the parents (452). In case parents and health care providers disagree over the sex of rearing, the parents’ choice must be respected. The affected child and his/her family must be followed throughout life to ascertain the patient’s adjustment to his/her social sex. Hormonal Therapy Sex steroid replacement is an important component of management for some types of 46,XY DSD (453,454). The goals of replacement include induction and maintenance of secondary sex characteristics as well as other aspects of pubertal development including growth. Bone mineral optimization and promotion of uterine development may also be helped by treatment with sex steroids for some patients. Hormone replacement can also impact psychosocial and psychosexual development, as well as general wellbeing, in positive ways for some people (455,456). Induction and maintenance of pubertal development is necessary in most patients affected by 46,XY DSD regardless of male or female rearing; however, specific indications depend on the underlying etiology of the condition. FEMALE SOCIAL SEX The purpose of the hormonal therapy is the development of female sexual characteristics and menses in the patients with uterus. There are several options available for estrogen replacement as well as different combinations and doses of progestins (457) however, 17β-estradiol (oral or transdermal) is preferred. Estrogen therapy should be initiated at a low dose (1/6 to 1/4 of the adult dose) to avoid excessive bone maturation in short children and increase gradually at intervals of 6 months. Doses can then be adjusted to the response (Tanner stage, bone age), with the aim of completing feminization gradually over a period of 2–3 years. In 46,XY females with tall stature, adult estrogen dosage is recommended to avoid high final stature. Transdermal delivery avoids hepatic first-pass metabolism resulting in less thrombogenicity and more neutral effects on lipids (458,459). It is also easier to administer small doses of estrogen by cutting up a patch or by using a metered-dose gel dispenser. An initial recommended dose of oral 17-β estradiol is 5 μg/kg daily, titrated every 6–12 months to an additional 5 μg/kg daily until an adult dose of 1–2 mg daily is achieved (459) . In case of transdermal replacement, the initial recommended dose for the 17-β estradiol patch is 3.1–6.2 mg/24h overnight (1/8–1/4 of 25 mg/24h patch). Transdermal doses can increase 3.1–6.2 mg/24h every 6 months until an adult dose of 50-100 mg/24 h twice a week is achieved (460) Once breast development is complete, an adult dose can be maintained continuously (11). For patients who do not have a uterus, estrogen alone is indicated (458,461). Progesterone is needed to induce endometrial cycling and menses in patients with a uterus. For the latter group, medroxyprogesterone acetate (5 to 10 mg/day) or micronized progesterone (200 mg/day from the 1 st to the 12 th day of each month) are appropriate to maintain uterine health. Some females with CAIS report decreased psychological wellbeing and sexual dissatisfaction following bilateral gonadectomy and subsequent estrogen replacement (462,463). Testosterone treatment has been proposed as an alternative to estrogen for hormone replacement in these women and such treatment improves sexual desire (464). However, long-term follow-up studies on the impact of T replacement on additional psychological measures, as well as on bone metabolism and cardiovascular outcomes, are needed (465). The dilation of the blind vaginal pouch with acrylic molds (325) or exceptionally surgical neovagina promote development of a vagina adequate for sexual intercourse after 6-10 months of treatment when patients desire to initiate sexual activity (466). MALE SOCIAL SEX For those raised male, T replacement should strive to mimic masculine pubertal induction between 10 and 12 years of age, provided the child’s predicted height and growth are normal and he indicates a desire and readiness for puberty (5). Intramuscular, short-acting injections of T esters are the most suitable formulation to induce male puberty, although other options include oral T undecanoate and transdermal preparations (467,468).. The initial dose of short-acting T esters is 25–50 mg/month intramuscularly, with further increments of 50- 100 mg every 6–12 months, thereafter. After reaching a replacement dose of 100–150 mg/month, the delivery interval can decrease to every two weeks. An adult dose of 200-250 mg every two weeks (short-acting T esters), 1000 mg every 10-14 weeks (long-acting T esters), or 50-100 mg for T gel or other transdermal preparations applied topically are effective to maintain male secondary sex characteristics (12,468). Monitoring of T levels should be performed on the day preceding the next hormone administration, and serum levels should fall just above the lower limit of the normal range for eugonadal men. In male patients with androgen insensitivity, higher doses of testosterone esters (250-500 mg twice a week) are used to increase penile length and male secondary characteristics. Maximum penis enlargement is obtained after 6 months of high doses and after that, the normal dosage is re-instituted (272,313). The use of topical DHT gel is also useful to increase penile length with the advantage of not causing gynecomastia. Glucocorticoid Replacement It is necessary for 46,XY DSD patients with classical forms of congenital lipoid adrenal hyperplasia, POR, 3β-HSD type II deficiency to receive glucocorticoid replacement for adrenal insufficiency and in 17α-hydroxylase/17,20-lyase deficiency for hypertension management (469) (267). Mineralocorticoid replacement is also required for 46,XY DSD salt-losing patients (470). Surgical Treatment Surgical approach for 46,XY DSD patients includes: gonadal management, removal of internal structures that disagree with the social sex and reconstruction of the atypical external genitalia. Genital reconstruction involves the feminization or masculinization of external genitalia; these procedures are being widely discussed and controversy continues over the ideal age for genital surgery (471,472). There is a lack of data concerning this issue: a survey with 459 individuals (≥ 16 years) with a DSD diagnosis concerning patients desire about timing of genital surgery was published (473). A total of 66% of individuals with CAH and 60% of those with 46,XY DSD thought that infancy or childhood were the appropriate age for genital surgery. This report concluded that case-by-case decision-making is the best approach (473). In our experience, patients submitted to surgery in adulthood, preferred surgery in infancy and none of the patients operated during childhood regretted the surgery at that age (474). Laparoscopy is the ideal method of surgical treatment of the internal genital organs in patients with 46,XY DSD (475). In these patients, the indications for laparoscopy are the removal of gonads and ductal structures that are contrary to the assigned gender and the removal of dysgenetic gonads, which are nonfunctional and present potential for malignancy. In addition to being a minimally invasive surgery, one of the main advantages of this method is the lack of scars. Feminizing genitoplasty includes the reduction of enlarged clitoral size, opening the urogenital sinus to separate the urethra from the vaginal introitus, and constructing labioscrotal folds. Feminizing techniques have evolved over time to achieve better cosmetic outcomes (476,477). Many techniques have been proposed to separate the urethra from the vaginal introitus and bring both to the surface of the perineum. Fortunoff and Latimer in 1964 described the most commonly used technique until the present day, using an inverted U-shaped perineal skin flap to enlarge the vaginal introitus allowing adequate menstrual flow and future sexual activity (478). Failure to interposing an adequate flap will result in persistent urogenital sinus or vaginal introitus stenosis, requiring later revision (479). Vaginal dilation with acrylic molds in patients with short vagina or introitus stenosis showed to be a good treatment choice when these patients wished to start sexual intercourse, resulting in better outcomes (325). To reduce clitoral enlargement a number of techniques were proposed during the years (480). Kogan described preserving the neuro-vascular bundle attached to the dorsal portion of the tunica albuginea to protect the nerves and blood supply (481) and this is the technique of choice. The redundant clitoral skin obtained during clitoroplasty is used to create the labia minora; this skin is divided longitudinally and then sutured along either side of the vagina. When necessary, the reduction of labioscrotal folds is performed to create the labia majora, often using a Y-V plasty technique (482). The most common surgical complications, in feminizing genitoplasty, includes: clitoral ischemia or necrosis that can rarely occur in patients with high grade of virilization; introitus stenosis or vaginal stenosis particularly when the confluence of the vagina and urethra is far from the perineum surface and urinary infections mostly observed in patients with persistence of urogenital sinus (471,479,480). In order to minimize surgical complications and dissatisfaction in adulthood, only skilled surgeons with specific training should perform these procedures in the DSD patients (8). In our experience, the single-stage feminizing genitoplasty consisting of clitoroplasty with the preservation of dorsal nerves and vessels and ventral mucosa, vulvoplasty and Y-V perineal flap, followed by vaginal dilation with acrylic molds, allowed good cosmetic and functional results (483). For the males, masculinizing procedures aims to allow the patient to have micturition standing up without effort with a straight and wide stream and to have a satisfactory sexual life with straight erections. The genital surgery consists in correction of hypospadias and scrotal abnormalities, relocation of the testes to the scrotum or removal when dysgenetic, and resection of Mullerian remnants (326,484). Correction of hypospadias includes correction of phallic curvature (orthophalloplasty) and construction of a urethra to the tip of the glans (urethroplasty). Preoperative administration of testosterone is indicated for patients with a small penis (485). Usually, multistage procedures are preferred for male genital reconstruction in DSD patients, due to the severe under virilization represented by proximal hypospadias with severe curvature. The first stage repair consists in ortho-phaloplasty and scrotoplasty (486) (487). The second stage is performed 6-9 months later and consists in urethroplasty). The two-stage approach typically results in better cosmetic outcomes and fewer postoperative complications for patients with severe hypospadias and significant chordee (326,487-489). The most frequent complication in correction of hypospadias is urethral fistula (23%) followed by urethral strictures (9%) and diverticula (4%) (490). This frequency is highly variable in the literature (490). Fistula can be observed just after surgery or months later, but urethral stenosis in some cases can occur several years after surgery. Reoperations are necessary to correct fistula, diverticula and particularly to treat severe urethral strictures. The buccal mucosa graft is commonly used to enlarge the urethra in these cases (490). For patients with undescended testes, simultaneous orchidopexy may be performed. The surgical treatment of gonads of 46, XY DSD patients aims to preserve testicular function (production of testosterone and sperm) and prevent malignancy (288,360,491). Finally, gonadectomy is recommended for patients at risk for neoplastic transformation of germ cells (gonadoblastomas and/or an invasive germ cell tumor) in dysgenetic gonads (287). When gonadectomy is recommended, patients may then choose to have a testicular prosthesis placed in the scrotum (492). Müllerian structures are rudimentary in some patients and present as a cystic prostatic utricle. These utricles may be left in situ when asymptomatic, but in cases of recurrent urinary tract infection, stones, or significant post-void urethral dribbling due to urinary pooling, they can be removed either laparoscopically or through a sagittal posterior incision of the perineum (475). With either approach, great care must be taken to prevent injury to the vas deferens, seminal vesicles and pelvic nerves so as to avoid subsequent infertility, erectile dysfunction and urinary incontinence (488,493). Late evaluation of 46,XY DSD patients operated in childhood due to proximal hypospadias reveals that many felt that their genitals had an unusual appearance or presented some degree of urinary or sexual dysfunction (494). Objectively, most DSD patients have a penile length below the -2.0 SD (5.2 ± 2.0 cm) (326). Dysfunctional voiding and lower urinary tract symptoms are also more frequent in these patients than in controls (495). However, between 55.6 and 91% of these patients after genitoplasty were satisfied with their overall sexual function after genitoplasty, when considering sexual contacts, libido, erections, orgasm, as well as size of the penis and volume of ejaculation (326),(494),(496),(497),(498). The long-term outcomes were evaluated for a long time concerning functional and cosmetic results that could be analyzed by objective criteria. The subjective long-term evaluation analyzing psychological and sexual implications in quality of life were often neglected in the past, but is being currently explored (332) (499) (500). Jones et al reported that 81% were satisfied with their genital appearance and that 90% were satisfied with their overall body image (500). Most of our patients were satisfied with their genital appearance and present satisfactory sexual performance as long as they present a penis size of at least 6 cm (326). 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[PubMed: 15076297] ABSTRACT INTRODUCTION INVESTIGATION OF DSD PATIENTS CYTOGENETIC AND MOLECULAR INVESTIGATION 46,XY DSD DUE TO ABNORMALITIES IN GONADAL DEVELOPMENT 46,XY DSD ASSOCIATED WITH CHOLESTEROL SYNTHESIS DEFECTS 46,XY DSD DUE TO TESTOSTERONE PRODUCTION DEFECTS 46,XY DSD DUE TO DEFECTS IN TESTOSTERONE METABOLISM 46,XY DSD DUE TO DEFECTS IN ANDROGEN ACTION 46,XY DSD DUE TO PERSISTENT MÜLLERIAN DUCT CONGENITAL NON-GENETIC 46,XY DSD 46,XY OVOTESTICULAR DSD NON-CLASSIFIED FORMS GONADAL TUMOR DEVELOPMENT IN 46,XY DSD PATIENTS FERTILITY IN PATIENTS WITH 46,XY DSD 46,XY GENDER IDENTITY DISORDERS MANAGEMENT OF PATIENTS WITH 46,XY DSD ACKNOWLEDGMENT REFERENCES Copyright © 2000-2025, MDText.com, Inc. This electronic version has been made freely available under a Creative Commons (CC-BY-NC-ND) license. A copy of the license can be viewed at Bookshelf ID: NBK279170 PMID: 25905393 Contents < PrevNext > Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page ABSTRACT INTRODUCTION INVESTIGATION OF DSD PATIENTS CYTOGENETIC AND MOLECULAR INVESTIGATION 46,XY DSD DUE TO ABNORMALITIES IN GONADAL DEVELOPMENT 46,XY DSD ASSOCIATED WITH CHOLESTEROL SYNTHESIS DEFECTS 46,XY DSD DUE TO TESTOSTERONE PRODUCTION DEFECTS 46,XY DSD DUE TO DEFECTS IN TESTOSTERONE METABOLISM 46,XY DSD DUE TO DEFECTS IN ANDROGEN ACTION 46,XY DSD DUE TO PERSISTENT MÜLLERIAN DUCT CONGENITAL NON-GENETIC 46,XY DSD 46,XY OVOTESTICULAR DSD NON-CLASSIFIED FORMS GONADAL TUMOR DEVELOPMENT IN 46,XY DSD PATIENTS FERTILITY IN PATIENTS WITH 46,XY DSD 46,XY GENDER IDENTITY DISORDERS MANAGEMENT OF PATIENTS WITH 46,XY DSD ACKNOWLEDGMENT REFERENCES LINKS TO WWW.ENDOTEXT.ORG View this chapter in Endotext.org Related information PMCPubMed Central citations PubMedLinks to PubMed Recent Activity Clear)Turn Off)Turn On) 46,XY Differences of Sexual Development - Endotext46,XY Differences of Sexual Development - Endotext 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 Domenice S, Batista RL, Arnhold IJP, et al. 46,XY Differences of Sexual Development. [Updated 2022 Aug 21]. In: Feingold KR, Ahmed SF, Anawalt B, et al., editors. Endotext [Internet]. South Dartmouth (MA): MDText.com, Inc.; 2000-. Available from: Making content easier to read in Bookshelf Close We are experimenting with display styles that make it easier to read books and documents in Bookshelf. Our first effort uses ebook readers, which have several "ease of reading" features already built in. The content is best viewed in the iBooks reader. You may notice problems with the display of some features of books or documents in other eReaders. 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8768	https://www.e-dmj.org/journal/view.php?doi=10.4093/dmj.2020.0004	Skip Navigation : Skip to contents KDA Contact us E-SUBMISSION E-SUBMISSION (Before 2025) Current Current issue Ahead-of print Browse All issues Article by category Article by topic Article by Category 2023-2024 Best paper of the year Most view Most cited Funded articles Diabetes Metab J Search Author index Collections Guidelines in DMJ Fact sheets in DMJ COVID-19 in DMJ For contributors For Authors Instructions for authors Article processing charge For Reviewers Instructions for reviewers How to become a reviewer Best reviewers For Readers Readership Subscription Permission guidelines About Aims and scope About the journal Editorial board Management team Best practice Metrics Contact us Editorial policy Research and publication ethics Peer review policy Copyright and open access policy Article sharing (author self-archiving) policy Archiving policy Data sharing policy Preprint policy Advertising policy Articles Page Path : HOME > Diabetes Metab J > Volume 44(1); 2020 > Article Guideline/Fact Sheet Metformin Treatment for Patients with Diabetes and Chronic Kidney Disease: A Korean Diabetes Association and Korean Society of Nephrology Consensus Statement : Kyu Yeon Hur1, Mee Kyoung Kim2, Seung Hyun Ko3, Miyeun Han4, Dong Won Lee5, Hyuk-Sang Kwon2, Committee of Clinical Practice Guidelines, Korean Diabetes Association, Committee of the Cooperative Studies, Korean Society of Nephrology : Diabetes & Metabolism Journal 2020;44(1):3-10. DOI: Published online: February 21, 2020 20,585 Views 512 Download 16 Web of Science 16 Crossref 22 Scopus Author information Article notes Copyright and License information 1Division of Endocrinology and Metabolism, Department of Medicine, Samsung Medical Center, Sungkyunkwan University School of Medicine, Seoul, Korea. 2Division of Endocrinology and Metabolism, Department of Internal Medicine, Yeouido St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea. 3Division of Endocrinology and Metabolism, Department of Internal Medicine, St. Vincent's Hospital, College of Medicine, The Catholic University of Korea, Suwon, Korea. 4Division of Nephrology, Department of Internal Medicine, Pusan National University Hospital, Pusan National University School of Medicine, Busan, Korea. 5Division of Nephrology, Department of Internal Medicine, Pusan National University School of Medicine, Yangsan, Korea. : Corresponding author: Dong Won Lee. Division of Nephrology, Department of Internal Medicine, Pusan National University School of Medicine, 20 Geumo-ro, Mulgeum-eup, Yangsan 50612, Korea. dongwonlee@pusan.ac.kr : Corresponding author: Hyuk-Sang Kwon. Division of Endocrinology and Metabolism, Department of Internal Medicine, Yeouido St. Mary's Hospital, College of Medicine, The Catholic University of Korea, 10 63-ro, Yeongdeungpo-gu, Seoul 07345, Korea. drkwon@catholic.ac.kr : Kyu Yeon Hur and Mee Kyoung Kim contributed equally to this study as first authors. • Received: January 7, 2020 • Accepted: January 23, 2020 Copyright © 2020 Korean Diabetes Association This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. prev next Full Article : Full Article Figure & data Reference Citations Metrics Download PDF Twitter Facebook Linked ABSTRACT INTRODUCTION CURRENT STATUS OF METFORMIN USE IN PATIENTS WITH TYPE 2 DIABETES MELLITUS IN KOREA LACTIC ACIDOSIS ASSOCIATED WITH METFORMIN THERAPY RECENT RECOMMENDATIONS ON THE USE OF METFORMIN IN CHRONIC KIDNEY DISEASE IODINATED CONTRAST MEDIA-ASSOCIATED RENAL DAMAGE IODINATED CONTRAST MEDIA-ASSOCIATED RENAL DAMAGE WHEN USING METFORMIN IN PATIENTS WITH TYPE 2 DIABETES MELLITUS-RELATED CHRONIC KIDNEY DISEASE RECENT RECOMMENDATIONS REGARDING METFORMIN DISCONTINU-ATION PRIOR TO CONTRAST PROCEDURES ACKNOWLEDGMENTS NOTES SUPPLEMENTARY MATERIALS REFERENCES About this article ABSTRACT : The safety of metformin use for patients with type 2 diabetes mellitus (T2DM) and advanced kidney disease is controversial, and more recent guidelines have suggested that metformin be used cautiously in this group until more definitive evidence concerning its safety is available. The Korean Diabetes Association and the Korean Society of Nephrology have agreed on consensus statements concerning metformin use for patients with T2DM and renal dysfunction, particularly when these patients undergo imaging studies using iodinated contrast media (ICM). Metformin can be used safely when the estimated glomerular filtration rate (eGFR) is ≥45 mL/min/1.73 m2. If the eGFR is between 30 and 44 mL/min/1.73 m2, metformin treatment should not be started. If metformin is already in use, a daily dose of ≤1,000 mg is recommended. Metformin is contraindicated when the eGFR is <30 mL/min/1.73 m2. Renal function should be evaluated prior to any ICM-related procedures. During procedures involving intravenous administration of ICM, metformin should be discontinued starting the day of the procedures and up to 48 hours post-procedures if the eGFR is <60 mL/min/1.73 m2. Keywords: Contrast media; Diabetes mellitus; Metformin; Renal insufficiency INTRODUCTION : Chronic kidney disease (CKD) is a global public health problem, and its prevalence is gradually increasing, mainly due to an increase in the number of patients with type 2 diabetes mellitus (T2DM) . CKD develops in approximately 35% of patients with T2DM and is associated with increased mortality . CKD stages are defined as follows : CKD stage 1 (an estimated glomerular filtration rate [eGFR] ≥90 mL/min/1.73 m2, normal or high), stage 2 (eGFR 60–89 mL/min/1.73 m2, mildly decreased), stage 3a (eGFR 45–59 mL/min/1.73 m2, mildly to moderately decreased), stage 3b (eGFR 30–44 mL/min/1.73 m2, moderately to severely decreased), stage 4 (eGFR 15–29 mL/min/1.73 m2, severely decreased), and stage 5 (eGFR <15 mL/min/1.73 m2, kidney failure). According to a study conducted among elderly Canadian patients with diabetes mellitus (DM) , metformin continues to be prescribed to patients with advanced CKD stages 4 to 5, although many guidelines and recommendations have suggested that metformin therapy be avoided because of its potential association with lactic acidosis and all-cause mortality in patients with CKD stage 5 . The safety of metformin in advanced kidney disease is controversial, and more recent guidelines have suggested cautious use in this patient group until more definitive evidence concerning its safety is made available . For several decades, one of the most important limitations imposed by regulatory authorities on metformin use was related to renal function, for which a creatinine limit at 1.4 mg/dL for women and 1.5 mg/dL for men was established to contraindicate its use. In subsequent years, clinical studies and reviews were published that justified extending the use of metformin to patients with CKD stage 3 (i.e., an eGFR between 30 and 59 mL/min/1.73 m2) . In 2016, the U.S. Food and Drug Administration (FDA) revised metformin's indication for use in patients with CKD stage 3 , whereas the indications for metformin use set by the Korean Ministry of Food and Drug Safety remain unchanged. In this report, we present the opinions of experts on the use of metformin according to renal function. The Korean Diabetes Association (KDA) and the Korean Society of Nephrology (KSN) have agreed on consensus statements for the use of metformin for patients with T2DM and renal dysfunction or patients undergoing imaging studies using iodinated contrast media (ICM). CURRENT STATUS OF METFORMIN USE IN PATIENTS WITH TYPE 2 DIABETES MELLITUS IN KOREA : We investigated trends in the prescription of antidiabetic medications for patients with T2DM, focusing on renal function. Retrospective data involving patients with T2DM aged ≥30 years were analyzed using information from the National Health Information Database as collected by the National Health Insurance Service in Korea, from January 2009 to December 2015. This study was approved by the Institutional Review Board of Catholic University of Korea (No. SC19ZCSI0094). Anonymized and de-identified information was used for analyses and, therefore, informed consent was waived. We identified patients with T2DM who had at least one service claim in each year during the study period. Among antidiabetic medications in 2009, sulfonylurea (SU) was the most commonly used agent (75.9%), followed by metformin (74.7%). However, in 2015, the use of metformin increased to 90.3% of all antidiabetic prescriptions (Fig. 1). The use of the dipeptidyl peptidase-4 (DPP-4) inhibitor began immediately after its release late in 2008 and dramatically increased to 54.3% in 2015. In contrast, the use of SU declined dramatically from 75.9% in 2009 to 48.4% in 2015. Among patients with an eGFR between 30 and 45 mL/min/1.73 m2, metformin was the most commonly used agent (68.6%) in 2015, SU was the second-most commonly used agent (59.7%), and DPP-4 inhibitor was the third-most commonly used agent (58.6%). Therefore, we found that even among patients classified with CKD stage 3b, the use of metformin was relatively common in a real-world clinical setting. Fig. 1 ### Secular trends in antidiabetic medication prescriptions in adults with type 2 diabetes mellitus, according to kidney function: (A) in patients with an estimated glomerular filtration rate (eGFR) ≥60 mL/min/1.73 m2, (B) in patients with eGFR between 45 and 59 mL/min/1.73 m2, and (C) in patients with an eGFR between 30 and 44 mL/min/1.73 m2. SU, sulfonylurea; MFOM, metformin; GLN, glinide; TZD, thiazolidinedione; DPP-4i, dipeptidyl peptidase-4 inhibitor; α-Gi, α-glucosidase inhibitor. : Because of the potential risk of fatal lactic acidosis, metformin has long been under-prescribed, especially for patients with heart failure (HF) . This situation has changed considerably following recent epidemiological studies that showed the risk of lactic acidosis associated with metformin use was not greater than that of other antidiabetic medications, and that metformin use in patients with DM and HF did not necessarily increase the risk of lactic acidosis . We also found that the use of metformin was not associated with an increased incidence of HF in a case-control study of patients with T2DM who were first diagnosed with HF between 2013 and 2015 (Supplementary Table 1). We analyzed retrospective data on patients with T2DM from January 2013 to December 2015. We excluded those with a history of HF before the index year. Between 2013 and 2015, there were 97,178 incident HF cases in patients with T2DM. Potential case subjects were identified by International Classification of Diseases, 10th Revision (ICD-10) code I50 (HF). Control subjects were matched to each case subject according to age (±3 years) and sex. The index date for the control corresponded to the incidence date of the matched HF subject. The use of metformin was associated with a lower risk of HF (odds ratio [OR], 0.92; 95% confidence interval [CI], 0.87 to 0.97), but the use of insulin was associated with an increased risk of HF (OR, 3.60; 95% CI, 3.33 to 3.89). Similar results were obtained when the analysis was performed on a subset consisting of those who had received ≥2 prescriptions of antidiabetic medication (Supplementary Table 1). A recent population-based retrospective cohort study in Taiwan reported a reduced risk of HF associated with metformin use in patients with T2DM . LACTIC ACIDOSIS ASSOCIATED WITH METFORMIN THERAPY : Lactic acidosis, a well-known complication of biguanide therapy that usually occurs when lactic acid production exceeds lactic acid clearance, is generally defined as a blood pH <7.35 and a blood lactate level >45.0 mg/dL or >5 mmol/L. Lactic acidosis is a rare but potentially life-threatening complication, with a reported mortality risk approaching 50% when the serum lactate level is >23 mg/dL . : Phenformin induces conversion of glucose to lactate through the intestinal mucosa, enhances anaerobic metabolism, suppresses hepatic gluconeogenesis, and impairs renal excretion of lactate . Since metformin is a type of biguanide with a chemical structure similar to that of phenformin, concerns have been raised regarding lactic acidosis. However, in contrast to phenformin, metformin does not inhibit glucose oxidation or interfere with lactate turnover or hydroxylation polymorphism and is not metabolized and excreted in the urine. Since the underlying mechanisms of these two drugs are different, the incidence of lactic acidosis in metformin therapy is theoretically considered to be lower than in phenformin therapy. Lactic acidosis is a rare event, with an estimated incidence of 4.3 cases per 100,000 person-years among metformin users . Furthermore, evidence indicating that metformin increases the risk of lactic acidosis is lacking, and the frequency of lactic acidosis due to metformin is lower than expected. Most cases of lactic acidosis among metformin users were reported to have occurred when metformin was used inappropriately, such as in patients with severely decreased renal function . : The term metformin-associated lactic acidosis (MALA) has been used to describe almost all cases of lactic acidosis observed in a metformin-treated patient. However, use of the word ‘associated’ is ambiguous. It is difficult to distinguish between lactic acidosis due to metformin accumulation (i.e., acute kidney injury [AKI] and voluntary intoxication) and systemic conditions (sepsis, cardiac failure, and bleeding) in patients taking metformin. One recent review article emphasized that metformin therapy does not necessarily induce metformin accumulation, just as metformin accumulation does not necessarily induce hyperlactatemia, and hyperlactatemia does not necessarily induce lactic acidosis. Therefore, it is important to distinguish between metformin-unrelated lactic acidosis (MULA) and metformin-induced lactic acidosis (MILA). True MALA, which falls between MULA and MILA, is very rare . RECENT RECOMMENDATIONS ON THE USE OF METFORMIN IN CHRONIC KIDNEY DISEASE : In 2016, the FDA and the European Medicines Agency (EMA) removed the contraindications concerning the use of metformin in patients with CKD stages 3a and 3b (eGFR 30–44 and 45–59 mL/min/1.73 m2, respectively). The FDA recommends metformin use in patients with T2DM and CKD as follows: : (1) Before starting metformin, obtain the patient's eGFR. (2) Metformin is contraindicated in patients with an eGFR <30 mL/min/1.73 m2. (3) Starting metformin in patients with an eGFR between 30 and 45 mL/min/1.73 m2 is not recommended. (4) Obtain an eGFR at least annually in all patients taking metformin. (5) For patients at increased risk for the development of renal impairment, such as elderly patients, renal function should be assessed more frequently. (6) For patients taking metformin whose eGFR later falls below 45 mL/min/1.73 m2, assess the benefits and risks of continuing treatment. (7) Discontinue metformin if a patient's eGFR later falls below 30 mL/min/1.73 m2. : In addition, the EMA recommends dose reduction for patients with a moderate reduction in kidney function. Lalau et al. recently published three complementary studies concerning the use of metformin for patients with T2DM and CKD stages 3a, 3b, or 4, namely, a dose-finding study, a chronic metformin treatment study, and a pharmacokinetic study. On the basis of the dose-finding study results, they selected a chronic dosage regimen of 1,500 mg/day for patients with T2DM and CKD stage 3a, 1,000 mg/day for patients with T2DM and CKD stage 3b, and 500 mg/day for patients with T2DM and CKD stage 4 (off-label use) . : The Korean Ministry of Food and Drug Safety continues to contraindicate the use of metformin for patients with T2DM and CKD stage 3b (creatinine clearance <45 mL/min or an eGFR <45 mL/min/1.73 m2). However, in 2019, the KDA revised its recommendations as follows : Metformin is contraindicated in patients with an eGFR <30 mL/min/1.73 m2; Metformin can be used cautiously in patients with an eGFR between 30 and 45 mL/min/1.73 m2. IODINATED CONTRAST MEDIA-ASSOCIATED RENAL DAMAGE : When ICM is used for patients with T2DM-related CKD, a patient's renal function should be considered based on CKD staging, the eGFR, and the risk of AKI . The FDA has suggested the following risk factors that may warrant renal function assessment prior to contrast procedures: age >60 years, history of renal disease including dialysis, kidney transplantation, single kidney, renal cancer, renal surgery, history of hypertension requiring medical therapy, history of DM, and metformin or metformin-containing drug combinations . : Post-contrast AKI (PC-AKI) is a general term used to describe a sudden decline in renal function that occurs within 48 hours following intravascular administration of ICM; therefore, PC-AKI is a correlative diagnosis. However, contrast-induced nephropathy (CIN) is a more specific and causative diagnostic term in cases of sudden renal function decline due to the intravascular administration of ICM . : The exact pathophysiologic mechanism of CIN is not fully understood; however, some etiologic factors have been suggested, such as renal hemodynamic changes, direct tubular toxicity, or agent-specific chemotoxicity . : In Korea, contrast media was listed third among the most common causes of adverse drug reactions in 2018, following anticancer and nonsteroidal anti-inflammatory drugs. Moreover, 25 fatal cases have been reported in the past 5 years . Therefore, the appropriate management for adverse reactions involving ICM is of growing importance; however, its current institutional management has not been assessed. IODINATED CONTRAST MEDIA-ASSOCIATED RENAL DAMAGE WHEN USING METFORMIN IN PATIENTS WITH TYPE 2 DIABETES MELLITUS-RELATED CHRONIC KIDNEY DISEASE : Many concerns have been raised about MALA following ICM exposure, especially regarding patients who develop AKI while taking metformin. However, metformin does not confer an increased risk of CIN, and there have been no documented cases among properly selected patients . The American College of Radiology (ACR) and the European Society of Urogenital Radiology (ESUR) have proposed more relaxed guidelines in recent years . However, 57.1% (32/56) of Korean hospitals and 53.3% (8/15) of hospitals outside Korea have been reported to discontinue metformin prior to contrast-enhanced computed tomography (CT) scans, regardless of a patient's renal function, and have failed to take into account up-to-date guidelines . The median cut-off eGFR for stopping metformin prior to contrast-enhanced CT scans has been reported to be 60 mL/min/1.73 m2 in Korean hospitals, which was significantly higher than that reported in hospitals outside Korea (30 or 44 mL/min/1.73 m2) . The KSN conducted a simple survey on metformin use in 35 Korean university hospitals. The survey results indicated that metformin was usually stopped prior to contrast exposure, regardless of the patient's eGFR. : Metformin inhibits pyruvate carboxylase and induces excessive lactate production, inhibits the mitochondrial electron transport chain, and blocks the entry of pyruvate into the tricarboxylic acid cycle . As mentioned earlier, MALA is very rare, and in almost all cases, lactic acidosis was reported to have occurred because of overlooked contraindications and several comorbid factors, such as cardiovascular, hepatic, or renal diseases . RECENT RECOMMENDATIONS REGARDING METFORMIN DISCONTINU-ATION PRIOR TO CONTRAST PROCEDURES : In 2016, the FDA issued the following revised recommendation concerning the use of metformin for certain patients with reduced renal function: : (1) Discontinue metformin at the time of or prior to an iodinated contrast imaging procedure in patients with an eGFR between 30 and 60 mL/min/1.73 m2; in patients with a history of liver disease, alcoholism, or HF; or in patients who will be administered intra-arterial ICM. : (2) Re-evaluate eGFR 48 hours after the imaging procedure; restart metformin if renal function is stable . : In contrast, the ACR proposed more specific guidelines in 2018. The ACR guidelines recommend that patients taking metformin be classified into one of two categories, based on a patient's renal function, measured by the eGFR, as follows: : (1) Category I: In patients with no evidence of AKI and with an eGFR ≥30 mL/min/1.73 m2, there is no need to discontinue metformin either prior to or following intravenous administration of ICM, nor is there an obligatory need to reassess the patient's renal function following the test or procedure. : (2) Category II: In patients with AKI or severe CKD (stage 4 or 5; eGFR <30 mL/min/1.73 m2) or patients who are undergoing intra-arterial contrast studies, metformin should be temporarily discontinued at the time of or prior to the procedure, withheld for 48 hours and reinstituted only after renal function has been re-evaluated and found to be normal . : The ESUR also focused on patients with an eGFR <30 mL/min/1.73 m2, intra-arterial administration, and AKI. They distinguished between the risks of first or second pass renal exposure in intra-arterial administration. Patients with an eGFR >30 mL/min/1.73 m2 and no evidence of AKI, receiving either intravenous ICM or intra-arterial ICM with second pass renal exposure, could continue taking metformin . : However, according to the Korean Ministry of Food and Drug Safety, even patients with an eGFR >60 mL/min/1.73 m2 should discontinue metformin at the time of or prior to a contrast procedure. Moreover, patients with an eGFR between 45 and 60 mL/min/1.73 m2 should discontinue metformin 48 hours prior to and after the procedure, for a total of 96 hours . : In 2019, the KDA revised its recommendation that patients with an eGFR <60 mL/min/1.73 m2 should discontinue metformin on the day of the procedure for up to 48 hours, and reinstituted metformin therapy if renal function has not declined following the procedures . KOREAN DIABETES ASSOCIATION AND KOREAN SOCIETY OF NEPHROLOGY JOINT CONSENSUS STATEMENTS ON THE USE OF METFORMIN IN DIABETES : (1) Metformin can be used when the eGFR is ≥45 mL/min/1.73 m2. (2) If the eGFR is 30–44 mL/min/1.73 m2, do not start metformin treatment. If metformin is already in use, administer a daily dose of ≤1,000 mg. (3) Metformin is contraindicated when the eGFR is <30 mL/min/1.73 m2. (4) Evaluate renal function before any procedures involving the use of ICM. (5) Any decision to use ICM and whether to stop metformin should be based on renal function test results. (6) During procedures involving intra-arterial administration of ICM, metformin should be discontinued starting the day of the procedures and up to 48 hours after the procedures. (7) During procedures involving intravenous administration of ICM, metformin should be discontinued starting the day of the procedures and up to 48 hours after the procedures if eGFR is <60 mL/min/1.73 m2. (8) Re-evaluate renal function after procedures involving the use of ICM, and re-administer metformin if renal function has not declined after the procedures. Acknowledgements ACKNOWLEDGMENTS : This work was performed through the cooperation of the National Health Insurance Service (NHIS) and the Korean Diabetes Association. The results do not necessarily represent the opinion of the National Health Insurance Corporation. NOTES : This manuscript is simultaneously published in the Diabetes Metabolism Journal and in Kidney Research and Clinical Practice by the Korean Diabetes Association and the Korean Society of Nephrology. : CONFLICTS OF INTEREST: No potential conflict of interest relevant to this article was reported. SUPPLEMENTARY MATERIALS : Supplementary materials related to this article can be found online at Supplementary Table 1 : The risk of congestive heart failure according to antidiabetic medication dmj-44-3-s001.pdf REFERENCES Kim KS, Park SW, Cho YW, Kim SK. 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Management of adverse reactions to iodinated contrast media for computed tomography in Korean referral hospitals: a survey investigation. Korean J Radiol 2019;20:148-157. ArticlePubMedPDF Corremans R, Vervaet BA, D'Haese PC, Neven E, Verhulst A. Metformin: a candidate drug for renal diseases. Int J Mol Sci 2018;20:E42Article Korea Ministry of Food and Drug Safety: Instructions on the change of permit for metformin hydrochloride (tablet) (uniform adjustment) updated 2018 Jan 3. Available from: Figure & Data Fig. 1 Secular trends in antidiabetic medication prescriptions in adults with type 2 diabetes mellitus, according to kidney function: (A) in patients with an estimated glomerular filtration rate (eGFR) ≥60 mL/min/1.73 m2, (B) in patients with eGFR between 45 and 59 mL/min/1.73 m2, and (C) in patients with an eGFR between 30 and 44 mL/min/1.73 m2. SU, sulfonylurea; MFOM, metformin; GLN, glinide; TZD, thiazolidinedione; DPP-4i, dipeptidyl peptidase-4 inhibitor; α-Gi, α-glucosidase inhibitor. Supplementary Table 1 : The risk of congestive heart failure according to antidiabetic medication dmj-44-3-s001.pdf References Kim KS, Park SW, Cho YW, Kim SK. Higher prevalence and progression rate of chronic kidney disease in elderly patients with type 2 diabetes mellitus. Diabetes Metab J 2018;42:224-232. ArticlePubMedPMCPDF Bae JC. Trends of diabetes epidemic in Korea. Diabetes Metab J 2018;42:377-379. ArticlePubMedPMCPDF Kim KJ, Kwon TY, Yu S, Seo JA, Kim NH, Choi KM, Baik SH, Choi DS, Kim SG, Park Y, Kim NH. Ten-year mortality trends for adults with and without diabetes mellitus in South Korea, 2003 to 2013. Diabetes Metab J 2018;42:394-401. ArticlePubMedPMCPDF Kim BY, Won JC, Lee JH, Kim HS, Park JH, Ha KH, Won KC, Kim DJ, Park KS. Diabetes fact sheets in Korea, 2018: an appraisal of current status. Diabetes Metab J 2019;43:487-494. ArticlePubMedPMCPDF Inker LA, Astor BC, Fox CH, Isakova T, Lash JP, Peralta CA, Kurella Tamura M, Feldman HI. KDOQI US commentary on the 2012 KDIGO clinical practice guideline for the evaluation and management of CKD. Am J Kidney Dis 2014;63:713-735. ArticlePubMed Clemens KK, Liu K, Shariff S, Schernthaner G, Tangri N, Garg AX. Secular trends in antihyperglycaemic medication prescriptions in older adults with diabetes and chronic kidney disease: 2004-2013. Diabetes Obes Metab 2016;18:607-614. ArticlePubMedPDF Hung SC, Chang YK, Liu JS, Kuo KL, Chen YH, Hsu CC, Tarng DC. Metformin use and mortality in patients with advanced chronic kidney disease: national, retrospective, observational, cohort study. Lancet Diabetes Endocrinol 2015;3:605-614. ArticlePubMed American Diabetes Association. 9. Pharmacologic approaches to glycemic treatment: standards of medical care in diabetes-2019. Diabetes Care 2019;42(Suppl 1):S90-S102. ArticlePubMedPDF U.S. Food and Drug Administration: FDA Drug Safety Communication: FDA revises warnings regarding use of the diabetes medicine metformin in certain patients with reduced kidney function cited 2020 Jan 28. Available from: Packer M. Is metformin beneficial for heart failure in patients with type 2 diabetes? Diabetes Res Clin Pract 2018;136:168-170. ArticlePubMed Aharaz A, Pottegard A, Henriksen DP, Hallas J, Beck-Nielsen H, Lassen AT. Risk of lactic acidosis in type 2 diabetes patients using metformin: a case control study. PLoS One 2018;13:e0196122. ArticlePubMedPMC Lee EY, Hwang S, Lee YH, Lee SH, Lee YM, Kang HP, Han E, Lee W, Lee BW, Kang ES, Cha BS, Lee HC. Association between metformin use and risk of lactic acidosis or elevated lactate concentration in type 2 diabetes. Yonsei Med J 2017;58:312-318. ArticlePubMedPMCPDF Tseng CH. Metformin use is associated with a lower risk of hospitalization for heart failure in patients with type 2 diabetes mellitus: a retrospective cohort analysis. J Am Heart Assoc 2019;8:e011640. ArticlePubMedPMC Chan NN, Brain HP, Feher MD. Metformin-associated lactic acidosis: a rare or very rare clinical entity? Diabet Med 1999;16:273-281. ArticlePubMedPDF Oates NS, Shah RR, Idle JR, Smith RL. Influence of oxidation polymorphism on phenformin kinetics and dynamics. Clin Pharmacol Ther 1983;34:827-834. ArticlePubMed Richy FF, Sabido-Espin M, Guedes S, Corvino FA, Gottwald-Hostalek U. Incidence of lactic acidosis in patients with type 2 diabetes with and without renal impairment treated with metformin: a retrospective cohort study. Diabetes Care 2014;37:2291-2295. ArticlePubMedPDF Luft D, Schmulling RM, Eggstein M. Lactic acidosis in biguanide-treated diabetics: a review of 330 cases. Diabetologia 1978;14:75-87. ArticlePubMedPDF Lalau JD. Lactic acidosis induced by metformin: incidence, management and prevention. Drug Saf 2010;33:727-740. PubMed Lalau JD, Kajbaf F, Protti A, Christensen MM, De Broe ME, Wiernsperger N. Metformin-associated lactic acidosis (MALA): moving towards a new paradigm. Diabetes Obes Metab 2017;19:1502-1512. ArticlePubMedPDF European Medicines Agency: Note for guidance on the evaluation of the pharmacokinetics of medicinal products in patients with impaired renal function cited 2020 Jan 28. Available from: Lalau JD, Kajbaf F, Bennis Y, Hurtel-Lemaire AS, Belpaire F, De Broe ME. Metformin treatment in patients with type 2 diabetes and chronic kidney disease stages 3A, 3B, or 4. Diabetes Care 2018;41:547-553. ArticlePubMedPDF Kim MK, Ko SH, Kim BY, Kang ES, Noh J, Kim SK, Park SO, Hur KY, Chon S, Moon MK, Kim NH, Kim SY, Rhee SY, Lee KW, Kim JH, Rhee EJ, Chun S, Yu SH, Kim DJ, Kwon HS, Park KS. 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Available from: Citations Citations to this article as recorded by Metformin and risk factors for chronic kidney disease in a European population based on Mendelian randomization Xiaopei Liu, Xingyao Li, Peng An, Qi Gao, Yanhong Zhao, Xingmin Shi, Xili Wu Renal Failure.2025;[Epub] CrossRef Distribution and elimination kinetics of midazolam and metabolites after post-resuscitation care: a prospective observational study Wonjoon Jeong, Jung Sunwoo, Yeonho You, Jung Soo Park, Jin Hong Min, Yong Nam In, Hong Joon Ahn, So Young Jeon, Jang Hee Hong, Ji Hye Song, Hyein Kang, My Tuyen Thi Nguyen, Jaehan Kim, Changshin Kang Scientific Reports.2024;[Epub] CrossRef Prediction of glycosylated hemoglobin level in patients with cardiovascular diseases and type 2 diabetes mellitus with respect to anti-diabetic medication Alisher Ikramov, Shakhnoza Mukhtarova, Raisa Trigulova, Dilnoza Alimova, Saodat Abdullaeva Frontiers in Endocrinology.2024;[Epub] CrossRef Clinical Management of Type II DM in patients Showing Progressive Increase in the Creatinine Level – A Cross-sectional Study Prabhudatta Mohapatra, Durga Madhab Kar, Karmajeet Rath, Abhisek Pal Research Journal of Pharmacy and Technology.2024; : 2719. CrossRef Trends in Antidiabetic Drug Use and Safety of Metformin in Diabetic Patients with Varying Degrees of Chronic Kidney Disease from 2010 to 2021 in Korea: Retrospective Cohort Study Using the Common Data Model Sung Hwan Joo, Seungwon Yang, Suhyun Lee, Seok Jun Park, Taemin Park, Sang Youl Rhee, Jae Myung Cha, Sandy Jeong Rhie, Hyeon Seok Hwang, Yang Gyun Kim, Eun Kyoung Chung Pharmaceuticals.2024; 17(10): 1369. CrossRef Determinants of vitamin B12 deficiency in patients with type-2 diabetes mellitus — A primary-care retrospective cohort study Andrew Kien Han Wee, Rehena Sultana BMC Primary Care.2023;[Epub] CrossRef Risk factors for post-contrast acute kidney injury in patients sequentially administered iodine- and gadolinium-based contrast media on the same visit to the emergency department: a retrospective study Changshin Kang, Soo Hyun Han, Jung Soo Park, Dae Eun Choi Kidney Research and Clinical Practice.2023; 42(3): 358. CrossRef Guideline for the diagnosis and treatment of diabetes mellitus in patients with transfusion-dependent thalassemia Mohammad E. Khamseh, Mojtaba Malek, Nahid Hashemi-madani, Fariba Ghassemi, Neda Rahimian, Amir Ziaee, Mohammad Reza Foroughi-Gilvaee, Pooya Faranoush, Negin Sadighnia, Ali Elahinia, Mohammad Reza Rezvany, Mohammad Faranoush Iranian Journal of Blood and Cancer.2023; 15(4): 293. CrossRef Glucagon-Like Peptide-1 Receptor Agonists in Type 2 Diabetes Mellitus and Cardiovascular Disease: The Past, Present, and Future Filipe Ferrari, Rafael S. Scheffel, Vítor M. Martins, Raul D. Santos, Ricardo Stein American Journal of Cardiovascular Drugs.2022; 22(4): 363. CrossRef Comparative Efficacy of Lobeglitazone Versus Pioglitazone on Albuminuria in Patients with Type 2 Diabetes Mellitus Kyung-Soo Kim, Sangmo Hong, Hong-Yup Ahn, Cheol-Young Park Diabetes Therapy.2021; 12(1): 171. CrossRef Treatment Patterns of Type 2 Diabetes Assessed Using a Common Data Model Based on Electronic Health Records of 2000–2019 Kyung Ae Lee, Heung Yong Jin, Yu Ji Kim, Yong-Jin Im, Eun-Young Kim, Tae Sun Park Journal of Korean Medical Science.2021;[Epub] CrossRef Continuous use of metformin in patients receiving contrast medium: what is the evidence? A systematic review and meta-analysis Ting-Wan Kao, Kuo-Hua Lee, Wing P. Chan, Kang-Chih Fan, Che-Wei Liu, Yu-Chen Huang European Radiology.2021; 32(5): 3045. CrossRef Albuminuria Is Associated with Steatosis Burden in Patients with Type 2 Diabetes Mellitus and Nonalcoholic Fatty Liver Disease (Diabetes Metab J 2021;45:698-707) Eugene Han, Hye Soon Kim Diabetes & Metabolism Journal.2021; 45(6): 972. CrossRef KRCP's past and future path Tae-Hyun Yoo Kidney Research and Clinical Practice.2020; 39(3): 233. CrossRef Metformin Use and Risk of All-Cause Mortality and Cardiovascular Events in Patients With Chronic Kidney Disease—A Systematic Review and Meta-Analysis Yao Hu, Min Lei, Guibao Ke, Xin Huang, Xuan Peng, Lihui Zhong, Ping Fu Frontiers in Endocrinology.2020;[Epub] CrossRef Pharmacotherapy for patients with diabetes mellitus Joon Ho Moon, Soo Lim Journal of the Korean Medical Association.2020; 63(12): 766. CrossRef \| \| \| \| \| \| \| \| PubReader ePub Link Cite this Article Cite this Article : Close Download Citation : Close XML Download;) See more details Patents (1) Mendeley (63) Figure Related articles : Evaluation and Management of Patients with Diabetes and Heart Failure: A Korean Diabetes Association and Korean Society of Heart Failure Consensus Statement Lipid Management in Korean People with Type 2 Diabetes Mellitus: Korean Diabetes Association and Korean Society of Lipid and Atherosclerosis Consensus Statement Metformin Treatment for Patients with Diabetes and Chronic Kidney Disease: A Korean Diabetes Association and Korean Society of Nephrology Consensus Statement Fig. 1 Secular trends in antidiabetic medication prescriptions in adults with type 2 diabetes mellitus, according to kidney function: (A) in patients with an estimated glomerular filtration rate (eGFR) ≥60 mL/min/1.73 m2, (B) in patients with eGFR between 45 and 59 mL/min/1.73 m2, and (C) in patients with an eGFR between 30 and 44 mL/min/1.73 m2. SU, sulfonylurea; MFOM, metformin; GLN, glinide; TZD, thiazolidinedione; DPP-4i, dipeptidyl peptidase-4 inhibitor; α-Gi, α-glucosidase inhibitor. Fig. 1 Metformin Treatment for Patients with Diabetes and Chronic Kidney Disease: A Korean Diabetes Association and Korean Society of Nephrology Consensus Statement About this article : Hur KY, Kim MK, Ko SH, Han M, Lee DW, Kwon HS, , . Metformin Treatment for Patients with Diabetes and Chronic Kidney Disease: A Korean Diabetes Association and Korean Society of Nephrology Consensus Statement. Diabetes Metab J. 2020;44(1):3-10. : Received: Jan 07, 2020; Accepted: Jan 23, 2020 : DOI: : - Download citation ↓ : ABOUT + Aims and scope + About the journal + Editorial board + Management team + Best practice + Metrics + Contact us BROWSE ARTICLES + All issues + Article by category + Article by topic + Article by Category 2023-2024 + Best paper of the year + Most view + Most cited + Funded articles + Diabetes Metab J Search + Author index EDITORIAL POLICY + Research and publication ethics + Peer review policy + Copyright and open access policy + Article sharing (author self-archiving) policy + Archiving policy + Data sharing policy + Preprint policy + Advertising policy FOR CONTRIBUTORS + For Authors + Instructions for authors + Article processing charge + For Reviewers + Instructions for reviewers + How to become a reviewer + Best reviewers + For Readers + Readership + Subscription + Permission guidelines \| \| \| \| \| \| TOP
8769	https://next-gen.materialsproject.org/materials/mp-8039	Updating... The Materials Project Apps About About News and UpdatesPeoplePartners and SupportTestimonialsHow to CiteOpen Source SoftwarePublicationsPressTerms of Use Community Community Getting InvolvedSeminarHelp MLAPI Apps About AboutNews and UpdatesPeoplePartners and SupportTestimonialsHow to CiteOpen Source SoftwarePublicationsPressTerms of Use Community CommunityGetting InvolvedSeminarHelp ML API Materials Project Apps Overview About Community Machine Learning API Dashboard About News and Updates People Partners and Support How to Cite Open Source Software Publications Press Terms of Use MPContribs Terms of Use Testimonials Community Seminar Documentation Forum The Materials Project is powered by open-source software. Our data is licensed under a Creative Commons Attribution 4.0 International License. Contributed data is owned by the respective contributors. The current website version is 68cd16dd and database version is v2025.09.25. Complete your Materials Project registration Before continuing, we would like some information to complete your account registration. You can make changes to this information at any time by going to your dashboard page. What institution are you affiliated with? (Optional) Which sector most closely describes your institution? (Optional) Select... Which role most closely describes you? (Optional) Select... The Materials Project is a non-profit organization and does not share your information with any outside organizations. The information you choose to provide is used solely for improving the Materials Project. Our full privacy policy can be read here. We have recently updated our terms of use, please review them. [x] I agree to the terms of MP. Save
8770	https://www.sohu.com/a/246999678_100182824	2019高考数学备战，函数图像关于直线x＝a对称，掌握了特点特简单_f(x) 新闻体育汽车房产旅游教育时尚科技财经娱乐更多母婴健康历史军事美食文化星座专题游戏搞笑动漫宠物无障碍关怀版登录 2019高考数学备战，函数图像关于直线x＝a对称，掌握了特点特简单 2018-08-14 07:47 来源: 幽默笑话萌萌萌链接复制成功一般来说，对于函数f(x)（x∈R），若满足f(x)＝f(2a－x)或者f(a＋x)＝f(a－x)，则函数f(x)的图像关于直线x＝a对称；两个点关于直线x＝a对称，则这两个点的横坐标之和等于2a，且纵坐标相等，根据这一特点可以设计出丰富的题型，但不管什么题型，只要理解了这一特点，都可以很顺利地做出来。第1题分析：由f(x)＝f(2－x)可知函数f(x)的图像关于直线x＝1对称，明显第二个函数的图像的对称轴也是直线x＝1，则它俩的交点也是关于直线x＝1对称，那么任意一对关于直线x＝1对称的交点的横坐标之和都等于2，详细的过程如下：第2题分析：本题是一道选择题，函数的表达式给了出来，只需对选项一一进行判断即可；对于函数问题，首先要确定定义域，f(x)的定义域为(0,2)；对于A、B选项，只需求出导函数，然后判断导函数的符号即可；对于C选项，只需判断等式f(2－x)＝f(x)是否成立；对于D选项只需判断等式f(2－x)＋f(x)＝0是否成立即可；详细过程如下：初中、高中、基础、提高、中考、高考；你想要的，这里都有！禁止转载！孙老师微信公众号：slsh2018返回搜狐，查看更多平台声明：该文观点仅代表作者本人，搜狐号系信息发布平台，搜狐仅提供信息存储空间服务。首赞 +1 点赞失败阅读 () 我来说两句 0 人参与， 0 条评论搜狐“我来说两句” 用户公约点击登录搜狐小编发布推荐阅读刷新中.. 幽默笑话萌萌萌文章 0 总阅读 0 24小时热文搜狐号返回首页举报/反馈邮箱账号登录忘记密码请输入正确的登录账号或密码手机号验证码登录获取验证码我已阅读并同意搜狐网用户服务协议和隐私政策需阅读并勾选同意其他方式微信登录 qq登录微博登录账号密码登录手机号验证码登录安全提示为保证您的账户安全，建议您绑定手机号码请输入正确的登录账号或密码获取动态码收不到短信验证码？点击获取语音验证码安全提示系统出于安全考虑，在点击“发送语音验证码”后，您将会收到一条来自950开头号码的语音验证码，请注意接听。暂不发送发送语音验证码手机注册邮箱注册获取验证码同意《搜狐服务协议》使用已有账号登录
8771	https://www.jci.org/articles/view/10305	JCI - Antibodies against keratinocyte antigens other than desmogleins 1 and 3 can induce pemphigus vulgaris–like lesions Go to JCI Insight About Editors Consulting Editors For authors Publication ethics Publication alerts by email Advertising Job board Contact Clinical Research and Public Health Current issue Past issues By specialty Back COVID-19 Cardiology Gastroenterology Immunology Metabolism Nephrology Neuroscience Oncology Pulmonology Vascular biology All ... Videos Back Videos Conversations with Giants in Medicine Video Abstracts Reviews Back Reviews Reviews View all reviews ... Review Series Pancreatic Cancer (Jul 2025) Complement Biology and Therapeutics (May 2025) Evolving insights into MASLD and MASH pathogenesis and treatment (Apr 2025) Microbiome in Health and Disease (Feb 2025) Substance Use Disorders (Oct 2024) Clonal Hematopoiesis (Oct 2024) Sex Differences in Medicine (Sep 2024) View all review series ... 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Email the journal Go to Top Abstract Introduction Methods Results Discussion Acknowledgments References Version history PDF Metrics Article usage Citations to this article (40) Advertisement Article Free access \| 10.1172/JCI10305 ShareXFacebookLinkedInWeChatBlueskyEmailCopy Link Antibodies against keratinocyte antigens other than desmogleins 1 and 3 can induce pemphigus vulgaris–like lesions Vu Thuong Nguyen,1Assane Ndoye,1Leonard D. Shultz,2Mark R. Pittelkow,3 andSergei A. Grando 1 1 Department of Dermatology, University of California at Davis, School of Medicine, Davis, California, USA 2 The Jackson Laboratory, Bar Harbor, Maine, USA 3 Department of Dermatology, Mayo Clinic, Rochester, Minnesota, USA Address correspondence to: Sergei A. Grando, 4860 Y Street, Suite 3400, Sacramento, California 95817, USA. Phone: (916) 734-6057; Fax: (916) 734-6793; E-mail: sagrando@ucdavis.edu. Find articles by Nguyen, V. in: PubMed \| Google Scholar 1 Department of Dermatology, University of California at Davis, School of Medicine, Davis, California, USA 2 The Jackson Laboratory, Bar Harbor, Maine, USA 3 Department of Dermatology, Mayo Clinic, Rochester, Minnesota, USA Address correspondence to: Sergei A. Grando, 4860 Y Street, Suite 3400, Sacramento, California 95817, USA. Phone: (916) 734-6057; Fax: (916) 734-6793; E-mail: sagrando@ucdavis.edu. Find articles by Ndoye, A. in: PubMed \| Google Scholar 1 Department of Dermatology, University of California at Davis, School of Medicine, Davis, California, USA 2 The Jackson Laboratory, Bar Harbor, Maine, USA 3 Department of Dermatology, Mayo Clinic, Rochester, Minnesota, USA Address correspondence to: Sergei A. Grando, 4860 Y Street, Suite 3400, Sacramento, California 95817, USA. Phone: (916) 734-6057; Fax: (916) 734-6793; E-mail: sagrando@ucdavis.edu. Find articles by Shultz, L. in: PubMed \| Google Scholar 1 Department of Dermatology, University of California at Davis, School of Medicine, Davis, California, USA 2 The Jackson Laboratory, Bar Harbor, Maine, USA 3 Department of Dermatology, Mayo Clinic, Rochester, Minnesota, USA Address correspondence to: Sergei A. Grando, 4860 Y Street, Suite 3400, Sacramento, California 95817, USA. Phone: (916) 734-6057; Fax: (916) 734-6793; E-mail: sagrando@ucdavis.edu. Find articles by Pittelkow, M. in: PubMed \| Google Scholar 1 Department of Dermatology, University of California at Davis, School of Medicine, Davis, California, USA 2 The Jackson Laboratory, Bar Harbor, Maine, USA 3 Department of Dermatology, Mayo Clinic, Rochester, Minnesota, USA Address correspondence to: Sergei A. Grando, 4860 Y Street, Suite 3400, Sacramento, California 95817, USA. Phone: (916) 734-6057; Fax: (916) 734-6793; E-mail: sagrando@ucdavis.edu. Find articles by Grando, S. in: PubMed \| Google Scholar Published December 15, 2000 - More info Published in Volume 106, Issue 12 on December 15, 2000 J Clin Invest. 2000;106(12):1467–1479. © 2000 The American Society for Clinical Investigation Published December 15, 2000 - Version history Received: May 11, 2000; Accepted: November 6, 2000 View PDF Abstract Pemphigus is an autoimmune disease of skin adhesion associated with autoantibodies against a number of keratinocyte antigens, such as the adhesion molecules desmoglein (Dsg) 1 and 3 and acetylcholine receptors. The notion that anti-Dsg antibodies alone are responsible for blisters in patients with pemphigus vulgaris (PV) stems from the ability of rDsg1 and rDsg3 to absorb antibodies that cause PV-like skin blisters in neonatal mice. Here, we demonstrate that PV IgGs eluted from rDsg1-Ig-His and rDsg3-Ig-His show similar antigenic profiles, including the 38-, 43-, 115-, and 190-kDa keratinocyte proteins and a non–Dsg 3 130-kDa polypeptide present in keratinocytes from Dsg 3 knockout mouse. We injected into Dsg 3–lacking mice the PV IgGs that did not cross-react with the 160-kDa Dsg 1 or its 45-kDa immunoreactive fragment and that showed no reactivity with recombinant Dsg 1. We used both the Dsg3 null mice with a targeted mutation of the Dsg3 gene and the “balding” Dsg3 bal/Dsg3 bal mice that carry a spontaneous null mutation in Dsg3. These PV IgGs caused gross skin blisters with PV-like suprabasal acantholysis and stained perilesional epidermis in a fishnet-like pattern, indicating that the PV phenotype can be induced without anti–Dsg 3 antibody. The anti–Dsg 1 antibody also was not required, as its presence in PV IgG does not alter the PV-like phenotype in skin organ cultures and because pemphigus foliaceus IgGs produce a distinct phenotype in Dsg3 null mice. Therefore, mucocutaneous lesions in PV patients could be caused by non-Dsg antibodies. Introduction In pemphigus vulgaris (PV), blisters develop on oral mucosa. Mucosal lesions are often followed by skin involvement. The deep intraepidermal cleft occurs between the basal cells and the overlaying spinous keratinocytes. In pemphigus foliaceus (PF), the oral mucosa is usually not involved, and cutaneous erosive lesions develop owing to a superficial epidermal split localized to the stratum granulosum. The pathophysiological mechanism causing autoimmune pemphigus is unknown and still being intensively investigated. To date, a catalogue of self-antigens, demonstrated by various authors and detection techniques to react uniquely with pemphigus IgGs, includes approximately 20 molecules with different relative molecular masses, namely: 12, 18, 33, 47, 50, 52, 55, 59, 66, 67, 68, 75, 78, 80, 85, 102, 105, 160, 180, 185/190, and 210 kDa (reviewed in ref. 1). The number of detectable target molecules varies from patient to patient and depends on the sensitivity of the detection technique, i.e., immunoblotting versus immunoprecipitation. Hypothetically, some of these bands may represent degradation products of pemphigus antigens having higher native molecular weights. The number of detectable pemphigus antigens can be substantially reduced by altering the sensitivity of the technique, as is performed when the keratinocyte protein suspension, the source of antigens, is first preabsorbed with normal human serum and then used in an immunoprecipitation assay with PV and PF sera (2, 3). Only a few protein bands remain, including the pairs of 85/130 and 85/160 kDa that were considered to represent the pathophysiologically important targets of PV and PF autoimmunity, respectively (4). Likewise, the number of clones detected in a λgt 11 keratinocyte cDNA library by PV IgG was reduced by substituting the whole PV IgG fraction with the affinity-purified IgG from a single band, the 130-kDa keratinocyte polypeptide (5). The 130-kDa PV antigen was reported to be a novel keratinocyte desmoglein (Dsg) 3 (5); the 160-kDa PF antigen, Dsg 1 (6); and the 85-kDa antigen, recognized by both PV and PF IgGs, was identified as the adhesion molecule plakoglobin (7). Autoantibodies to these adhesion molecules in pemphigus were interpreted as direct, cause-and-effect pathogenesis with autoantibody binding to an adhesion molecule inducing a disease of skin dyshesion (8–10). The broad spectrum variety of clinical and histological manifestations of autoimmune pemphigus has been explained by some investigators as an interplay between Dsg 1 and Dsg 3 antibodies. These antibodies are proposed to cause pemphigus directly by disrupting desmosomal junctions within the upper and lower epidermal compartments (11), based on the predominant differential expression of the Dsg1 and Dsg3 genes in the upper and lower epidermis, respectively (12, 13). However, a deletion mutation in the NH 2-terminal extracellular domain of Dsg 1 results in the dominantly inherited condition of striate palmoplantar keratoderma without any intraepidermal dyshesion (14). Similar thickening of the stratum corneum without skin blistering also occurs in Dsg 3-truncated transgenic mice (15). Furthermore, no spontaneous gross skin blistering is observed in the Dsg3 null mouse with a targeted mutation of the Dsg3 gene or the “balding” Dsg3 bal/Dsg3 bal mouse with a spontaneous null mutation in the Dsg3 gene (16, 17). Interestingly, even in P-cadherin/Dsg 3 double knockout mice neither spontaneous nor trauma (skin rubbing)–induced gross and microscopic alterations of the integrity of the epidermis can be found (18). Preabsorption of patients’ sera with recombinant Dsg1-Ig and Dsg3-Ig chimeric baculoproteins eliminate disease-causing activities of PF and PV IgG fractions, respectively. Further, IgGs eluted from recombinant Dsg molecules elicit acantholysis and gross skin blisters in neonatal BALB/c mice (19–23). Surprisingly, although both the recombinant protein representing the extracellular epitope of Dsg 3 and a chimeric baculoprotein combining sequences of the extracellular epitope of Dsg 3 and of the Fc portion of human IgG1 effectively absorb anti–Dsg 3 antibody from PV sera, only absorption on the chimeric baculoprotein can eliminate the antibodies capable of causing gross skin blisters in neonatal mice. The explanation for this phenomenon is that addition of the Fc IgG1 portion assists the Dsg 3 portion to fold properly and acquire an “active” conformation; however, confirmation that rDsg3-Ig-His absorbs only antibodies to Dsg 3 was not provided. Using Dsg 3–deficient mice, we have recently demonstrated that the PV IgGs lacking anti–Dsg 1 antibody can (a) produce a PV-like, fishnet staining pattern of the epidermis when used in indirect immunofluorescence (IIF) reactions; and (b) induce clinical and histological signs of PV when injected intraperitoneally into Dsg3 null neonates (1). Furthermore, we have shown rDsg3-Ig-His can absorb the non–Dsg 3 antibodies that are pathogenic in Dsg3 null neonates (24). These results indicated that non–Dsg 3 PV antibodies mediate acantholysis but could not eliminate a possibility that the undetected anti–Dsg 1 antibody contributed to acantholytic activity of test PV IgGs in Dsg3 null pups, because (a) 25–60% of patients with PV have both anti–Dsg 1 and anti–Dsg 3 antibodies, as demonstrated by immunoprecipitation (25, 26), immunoblotting (27–29), and immunohistochemistry (30); (b) spread of lesions from the oral mucosa to the skin in one patient with PV was reported to be associated with the appearance of anti–Dsg 1 antibody (31); and, most importantly, (c) absorption of PV IgGs with rDsg1 was reported to eliminate the disease-causing activity, whereas the anti–Dsg 1 PV IgG was suggested to play a critical role in induction of PV-like deep suprabasilar acantholysis in Dsg3 null neonates (32). In this study, we attempt to resolve the controversy about the role of anti-Dsg antibodies in pemphigus. First , we demonstrate that antibodies targeting Dsg 1 and Dsg 3 are not essential for induction of PV-like lesions in neonatal mice and that PV antibodies to other keratinocyte self-antigens are sufficient for causing gross skin blisters. Second, we reveal the lack of monospecificity of the rDsg1-Ig-His and rDsg3-Ig-His chimeric baculoproteins for anti–Dsg 1 and anti–Dsg 3 antibodies, respectively. Third, we document that PV IgG fraction containing anti–Dsg 3 antibody without anti–Dsg 1 antibody or PV IgG fraction containing both anti–Dsg 1 and anti–Dsg 3 antibodies produce deep suprabasilar acantholysis in either wild-type or Dsg3 null mice, whereas PF IgG fraction containing anti–Dsg 1 antibody without anti–Dsg 3 antibody produces subcorneal acantholysis localized to the granular layer. The cumulative results obtained in this study indicate that non-Dsg antibodies are sufficient to induce PV-like pathology in neonatal mice. Lesions in patients with PV may therefore result from cumulative and/or synergistic effects of a set of autoantibodies targeting different specific self-antigens on the keratinocyte cell surface. Methods Pemphigus and control sera, and IgG fractions. The results reported here were obtained in experiments using sera and IgG fractions from patients with well-established PV and PF, and healthy volunteers. This study had been approved by the University of California Davis Human Subjects Review Committee. The diagnosis of PV or PF was made based on the results of both comprehensive clinical and histological examinations and immunological studies, including direct immunofluorescence (DIF), IIF on various epithelial substrates, immunoblotting, and immunoprecipitation, following standard protocols (33). The serum IgG fractions were isolated using 40% ammonium sulfate followed by dialysis against PBS (Life Technologies Inc., Gaithersburg, Maryland, USA), lyophilized and reconstituted in PBS as detailed elsewhere (1). The protein concentration was determined using the Micro BCA kit (Pierce Chemical Co., Rockford, Illinois, USA). The serum level of rheumatoid factor was measured by a standard nephelometric technique using a commercially available antigen (Kit P/N 465320; Beckman Instruments Inc., Brea, California, USA). According to the reference range used in the clinical laboratory of University of California Davis Medical Center, normal values do not exceed 20 IU/ml. Enhanced sensitivity immunoprecipitation assay. Cultures of epidermal keratinocytes used for metabolic radioactive labeling were established from normal human neonatal foreskins and grown to approximately 95% confluence in 150 cm 2 flasks (Corning Costar, Cambridge, Massachusetts, USA) in serum-free keratinocyte growth medium (KGM; Life Technologies Inc.) containing 0.09 mM Ca 2+ at 37°C in a humid 5% CO 2 incubator, as described previously (34). Cultured keratinocyte monolayers were metabolically labeled for 16 hours at 37°C with 100 μCi/mL [35 S]methionine (1,000 Ci/mmol; Amersham Life Science Inc., Arlington Heights, Illinois, USA) in 1.8 mM Ca 2+ labeling medium, and the [35 S]methionine-labeled proteins were separated by centrifugation and used as a source of naturally folded keratinocyte proteins in an immunoprecipitation assay (25). The immune complexes were precipitated with a protein A-Sepharose suspension (Sigma Chemical Co., St. Louis, Missouri, USA), washed, and resolved on SDS-PAGE gels. The gels were fixed, enhanced, and the radioactivity was analyzed using the storage phosphor autoradiography feature of the Storm system (Molecular Dynamics, Mountain View, California, USA), as described by us previously (35). The presence of Dsg 1 antibody in test sera was investigated using the 45-kDa tryptic fragment of Dsg 1 representing its immunoreactive epitope (36). Briefly, [35 S]methionine-labeled keratinocytes were homogenized in 10 mM Tris-buffered saline containing 0.025% NaN 3, and 20 mM Ca 2+ (TBS-Ca) buffer containing 0.25% bovine trypsin (Sigma Chemical Co.) and then incubated for 1 hour at 37°C. The trypsinized homogenate was treated with 1 mM PMSF to stop the reaction, centrifuged at 40,000 g, and the supernatant was passed through a Concanavalin A Sepharose column (Con-A) (EY Laboratories Inc., San Mateo, California, USA). The Con-A–bound glycoproteins containing the 45-kDa immunoreactive epitope of Dsg 1 were eluted with 0.2 M α-methylmannoside (Sigma Chemical Co.), dialyzed against TBS-Ca, concentrated by lyophilization and used in enhanced sensitivity immunoprecipitation assay (ESIA) as a source of antigen. ELISA. The reactivity of test sera with the extracellular portions of the Dsg 1 or Dsg 3 peptide sequences that reportedly represent pathogenic epitopes of the PV or PF antigens (37–39) was determined using Dsg 1– and Dsg 3–coated ELISA plates purchased from MBL (Nagoya, Japan), including the positive and the negative controls, and following the protocols provided by the manufacturer. All serum samples were tested in duplicate or triplicate on at least two different occasions, using different lots of ELISA plates, and the results were expressed as ELISA scores, or index values (IVs), calculated as follows: IV = (OD sample – OD negative control)/(OD positive control – OD negative control) × 100. As recommended by the manufacturer, a serum sample was considered to be positive for Dsg 1 or Dsg 3 antibody if the IV exceeded 20. The samples of the PV, PF and normal human control sera used in this study were also independently tested in the laboratory of L.A. Diaz at Medical College of Wisconsin using the Dsg 1 ELISA developed in this laboratory. In vitro model of pemphigus. The standard protocols for treatment of normal human skin organ cultures with pemphigus serum (40, 41) were modified to eliminate known shortcomings. The samples of normal facial skin freshly removed from adult Caucasians undergoing surgical procedures were freed of fat and clotted blood, and 3-mm-diameter samples were obtained using a standard skin biopsy punch. These skin samples were rinsed in PBS, placed in plastic tubes containing ice-cold KGM without supplements, transported to the laboratory, and used in experiments immediately. The skin samples, containing epidermis and papillary and reticular dermis, were placed into the wells of the standard U-bottom 96-well cell-and-tissue culture plates (Falcon-Becton Dickinson Labware, Lincoln Park, New Jersey, USA), and 200 μL of test serum was added to each well. Each serum was tested in triplicate skin samples. The dishes were incubated at 37°C in a humid 5% CO 2-containing atmosphere for 24, 36, 48, 72, and 96 hours. After each incubation period, the skin samples were harvested and examined by both light microscopy and DIF. In vivo model of pemphigus. The skin blisters were induced in neonatal mice by passive transfer of serum IgG fraction (42). This study had been approved by the University of California Davis Review Committee on the Use of Animals in Research. The IgGs were injected intraperitoneally through a 30-gauge needle into 10- to 12-hour-old pups at the doses specified in Results. The neonates in each progeny always received the same amount of test or control IgG. The latter was isolated from normal human serum purchased from Sigma Chemical Co. Both Dsg 3–containing BALB/c mice, and Dsg 3–deficient Dsg3 null and C57BL/6 Dsg3 bal/Dsg3 bal mice were used (17). The integument of mice lacking Dsg 3 is normal at birth, making these neonates a suitable model for induction of pemphigus phenotype in passive transfer experiments (1, 32). In each litter, the lack of Dsg 3 in pups with induced blisters was confirmed by genotyping at the end of each passive transfer experiment. PCR was used to amplify the genomic sequences from the DNA isolated from pieces of mouse tails. The genotyping of Dsg3 null mice was performed using PCR primers and reaction conditions described by us previously (1). The genotyping of Dsg3 bal/Dsg3 bal mice was performed by sequencing the PCR product that amplified a portion of the exon 14 of the Dsg3 gene containing the 2275insT functional null mutation. The sense 5′-gccatagcatgaactgttag-3′ and antisense 5′-gttggcttgtcttgtgagtt-3′ primers (Bio-Synthesis Inc., Lewisville, Texas, USA) were used in PCR and the amplification condition included preheating at 94°C for 4 minutes; hot start with Taq DNA polymerase (Promega Corp., Madison, Wisconsin, USA), then 35 cycles at 94°C for 1 minute, 60°C for 1 minute, and 72°C for 3 minutes. The knockout Dsg3 null mice were homozygous for a targeted mutation of the Dsg3 gene, and “balding” Dsg3 bal/Dsg3 bal mice were homozygous for a spontaneous null mutation of the Dsg3 gene (17). Immunoaffinity absorption experiments. For preabsorption of antibodies to Dsg 1 and 3, we used the baculoproteins Dsg1-Ig-His, Dsg3-His, and Dsg3-Ig-His produced in High Five cells grown in Sf-900 II SFM medium (Life Technologies Inc.). The baculoviruses provided by M. Amagai (Keio University, Japan) were amplified in Sf9 insect cell (Life Technologies Inc.), and the preabsorption experiments were performed exactly as described by Amagai et al. (23). Briefly, supernatant containing each recombinant protein was incubated with TALON metal affinity resin overnight at 4°C with gently rocking. After washing ten times with PBS containing 1 mM CaCl 2, the TALON resins carrying immobilized baculoproteins were incubated with PV serum overnight. The resins with bound antibodies were washed 15 times with PBS, and the antibodies were eluted with an IgG gentle elution buffer (Pierce Chemical Co.). The eluants were cleared on D-Salt Excellulose Plastic Desalting Column (Pierce Chemical Co.) and diluted in TBS containing 2% nonfat milk and 10 mM CaCl 2. For preabsorption of antibodies to the Fc portion of human IgG, we used the Ig-immunosorbent column containing Sepharose conjugated with the Fc fragment of human IgG (Bethyl Laboratories, Inc., Montgomery, Texas, USA), and the incubation and elution procedures already described here. The efficiency of adsorption of anti-Fc IgG antibodies on this column was tested by the ability of the preabsorbed FITC-conjugated goat anti-human Fc IgG antibody (Sigma Chemical Co.) to visualize PV IgG bound to monkey esophagus in a standard IIF assay. The amount of adsorbed anti-Fc IgG antibodies was monitored by measuring the protein concentration in test IgG fraction before and after its absorption and comparing the differences with the protein concentration of the eluant. Characterization by immunoblotting of antigenic reactivities of adsorbed PV IgGs. The PV IgGs eluted from Dsg 1 and Dsg 3 constructs and from the Ig-immunosorbent column were characterized by immunoblotting using as substrate protein the extracts of normal human keratinocytes and keratinocytes from Dsg 3–positive BALB/c or Dsg 3–negative Dsg3 null mice that were resolved by SDS-PAGE and transferred to nitrocellulose membrane (Millipore Corp., Bedford, Massachusetts, USA), as detailed previously (24). The immunoblotting membranes were blocked with 5% nonfat milk in 20 mM Ca 2+ -containing TBS (pH 7.4) for 1 hour at 37°C. The blots were cut into 4-mm–wide vertical strips, and each strip was exposed overnight at 4°C to an affinity-purified PV IgG diluted in TBS containing 1% normal goat serum, 3% nonfat milk, and 0.05% Tween 20 (Sigma Chemical Co.). Binding of primary antibody was visualized using horseradish peroxidase (HRP) conjugated goat anti-human IgG (Pierce Chemical Co.) with an HRP color developer (Bio-Rad Laboratories Inc., Hercules, California, USA). The specificity of binding was determined in negative control experiments, in which the primary antibody was omitted. In a separate set of immunoblotting experiments, we examined antigenic reactivities of PV IgGs eluted from distinct keratinocyte protein bands, in accordance with the procedure for immunoaffinity antibody purification (43). Briefly, approximately 3-mm–wide horizontal strips carrying a keratinocyte protein with a specific M r ± 3 kDa were cut out from the immunoblotting membrane and incubated overnight with PV IgG serum fraction diluted 1:5 in TBS containing 20 mM CaCl 2, 0.05% Tween 20 (Sigma Chemical Co.) and 1% nonfat milk to allow antibody binding. The strips were then washed thoroughly, and the IgGs were eluted by a 3-minute incubation at 37°C in a solution containing 500 μL of 20 mM sodium citrate, 1% milk, and 0.05% Tween 20 (pH 3.2), and immediately neutralized by adjusting the pH to 7.4 with the 2 M Tris base. Results Characterization of Dsg 1 and Dsg 3 antibody profiles of PV and PF sera. The profile of anti-keratinocyte antibodies present in the sera of patients with PV was characterized by ESIA, using radiolabeled, naturally folded Dsg 1 and Dsg 3, and ELISA, using recombinant proteins representing the extracellular domains of Dsg 1 and Dsg 3 (Figure 1). By both ESIA and ELISA, all PV sera, but not normal human or PF sera, contained autoantibody to the 130-kDa Dsg 3. In ESIA, the PV1 and PV2 sera did not precipitate the 160-kDa Dsg 1 nor its 45-kDa tryptic fragment (Figure 1), which constitutes the major antigenic region for anti–Dsg 1 PF autoantibody and has been shown to be a very sensitive and specific probe for the anti–Dsg 1 antibody developed by patients with PV (31). Both PV1 and PV2 sera were also found negative for Dsg 1 antibody in ELISA assays (Figure 1). The commercial Dsg 1 ELISA kit used in our laboratory contained recombinant Dsg 1 developed by Amagai et al. (37). These negative results were confirmed in ELISA experiments performed in Diaz’s laboratory (data not shown), using recombinant Dsg 1 developed by Ding et al. (44). Figure 1 Characterization of PV and control sera. To identify the profiles of anti-keratinocyte antibodies, whole-cell (top) or 0.25% trypsin-digested (bottom) protein extracts of [35 S]methionine-labeled human keratinocytes were immunoprecipitated with three different PV sera, one PF serum, and two sera from healthy volunteers, as detailed in Methods. The immune complexes were resolved on 5% (top) or 15% (bottom) SDS-PAGE, and the radiolabeled antigens visualized using the Storm system’s phosphor-imager feature. The positions of the relative molecular mass markers in kDa are shown on the left. The titer of intercellular antibodies, determined by IIF on monkey esophagus, and the ELISA scores (IV) are shown on the bottom. MW, molecular weight. To eliminate the possibility that the autoantibodies to the conformational epitope of Dsg 1 remained undetectable in ESIA and ELISA due to limitations of these assays, we included PV3 serum and the PF serum as positive controls. These sera reacted with the 160-kDa Dsg 1 that was present in the keratinocyte protein solution used in ESIA (Figure 1). These two positive control sera also immunoprecipitated the 45-kDa fragment of Dsg 1 in ESIA and showed high titer of anti–Dsg 1 antibodies in ELISA (Figure 1). As negative controls, we used the sera from two healthy subjects without history of any blistering skin condition. These sera immunoprecipitated none of 130- or 160- or 45-kDa keratinocyte polypeptides in ESIA and were negative for Dsg 1 and Dsg 3 antibodies in ELISA (Figure 1). Thus, the four different approaches to detect Dsg 1 antibody in PV1 and PV2 sera produced negative results. Is anti–Dsg 3 antibody the sole mediator of PV lesions? Mice lacking Dsg 3 do not spontaneously develop gross skin blisters (1, 16, 17). If Dsg 3 is indeed essential for normal keratinocyte cell-to-cell adhesion in normal epidermis, the lack of skin lesions in Dsg3 null and Dsg3 bal/ Dsg3 bal mice may be explained by a compensation mechanism in which upregulation of other adhesion molecules occurs. Therefore, these animals provide a unique Dsg 3–free system for probing PV sera for the presence of non–Dsg 3 autoantibodies targeting the molecules that mediate and regulate keratinocyte cell-to-cell adhesion in the epidermis (1, 32). To determine whether non–Dsg 1/non–Dsg 3 PV antibodies are sufficient at causing skin blisters, we selected the two PV sera (PV1 and PV2) that did not contain Dsg 1 antibody, isolated IgG fractions and injected them intraperitoneally through a 30-gauge needle at a dose of 20 mg/1 g body weight into 10- to 12-hour-old offsprings bred from homozygous female and male Dsg3 null mice, as well as litters resulting from breeding pairs of homozygous Dsg3 bal/ Dsg3 bal mice. Control littermates received the same doses of IgG isolated from pooled normal human serum (control). All neonates injected with PV IgGs (Figure 2; n = 16), but not with normal human IgG (n = 12; data not shown), developed extensive, flaccid skin blisters approximately 18–24 hours after a single injection. The blisters resulted from suprabasilar acantholysis associated with binding of non–Dsg 1/non–Dsg 3 antibodies to the cell surface of murine epidermal keratinocytes (Figure 2). Identical clinical and microscopic features were produced in normal BALB/c mice by PV1 (n = 4) or PV2 (n = 6) IgGs (data not shown). Figure 2 Induction of pemphigus in neonatal mice lacking Dsg 3 by passive transfer of IgGs from sera of PV patients that lack Dsg 1 antibody. Neonatal knockout Dsg3 null mice homozygous for a targeted mutation of the Dsg3 gene (a–d) and “balding” Dsg3 bal/Dsg3 bal mice homozygous for a spontaneous null mutation in the Dsg3 gene (e–h) were injected intraperitoneally with 20 mg/g body weight of IgGs from the PV1 serum that lacked Dsg 1 antibody. Approximately 18–24 hours after injection, large, flaccid blisters filled with serous fluid were seen on the skin of both Dsg3 null (a) and Dsg3 bal/Dsg3 bal (e) mice. At this time, the Nikolsky sign could be elicited on the skin of Dsg3 null (b) and Dsg3 bal/Dsg3 bal (f) mice by mechanical extension of a large erosion after spontaneous rupture of the blister. The blisters developed due to intraepidermal separation showing suprabasilar acantholysis and prominent tombstone appearance of the basal cell layer (c and g; hematoxylin and eosin [H&E]). DIF with FITC-conjugated anti-human IgG antibody revealed intercellular staining of epidermis due to deposition of injected PV IgGs (d and h). All neonates of the progeny from mating of homozygous Dsg3 null mice showed the PCR product of 280 bp, representing the sequence of the neomycin resistance gene used for the targeted disruption of the Dsg3 gene (17), and no 500-bp product, representing normal Dsg3 gene amplified from DNA extracted from a normal, Dsg 3–positive mouse (positive control). Likewise, all pups in the progeny from mating homozygous Dsg3 bal/Dsg3 bal mice lacked Dsg 3, as illustrated by finding a nonfunctional homozygous mutational insertion, 2275insT, upon direct nucleotide sequencing of the PCR product representing a portion of exon 14 of the mouse Dsg3 gene (data not shown). Scale bars = 50 μm. To eliminate the possibility that pups with blisters developed a revertant mutation that restored expression of the Dsg3 gene, the expected genotypes of all pups was confirmed at the end of each experiment (data not shown). Thus, PV-like skin lesions in neonatal mice can be produced by PV autoantibodies to keratinocyte proteins other than Dsg 1 and Dsg 3. Does Dsg 1 antibody contribute to acantholytic activity of PV serum? The results discussed here indicated that anti-Dsg 3 antibody may not be critical for skin blister formation in patients with PV, in agreement with the conclusion of Mahoney et al. (32) that anti–Dsg 3 alone is not sufficient at causing skin blistering. However, the possibility that in our passive transfer experiments the skin lesions were caused by a “hidden” anti-Dsg 1 antibody could not be excluded because, theoretically, this antibody might not recognize the conformational epitopes of any Dsg 1 molecules present in either antigenic substrate used by four different antibody detection assays (Figure 1). Although patients with PV never develop the superficial epidermal acantholysis seen with anti–Dsg 1 (44), this antibody still may be pathogenic. If so, the lack of PF-like pathology in patients with PV might be explained by prevention of anti–Dsg 1 antibody entry of the granular cell layer by the deep suprabasilar split in the epidermis of these patients. We tested this hypothetical mechanism using an in vitro model of PV in which all epidermal layers are equally accessible to antibodies. The normal human skin specimens were bathed in the medium containing pemphigus antibodies for different time periods, and penetration of the antibodies throughout all epidermal layers with their subsequent deposition on the cell membrane of keratinocytes was confirmed by DIF. We tested (a) PV serum containing anti–Dsg 3 antibody without anti–Dsg 1 antibody (PV2 serum); (b) PV serum with both anti–Dsg 1 and anti–Dsg 3 antibodies (PV3 serum); (c) PF serum featuring anti–Dsg 1 antibody without anti–Dsg 3 antibody; and (d) a healthy volunteer serum (N2, i.e., normal serum 2 in Figure 1). The first signs of acantholysis in skin organ cultures treated with pemphigus sera could be observed as early as 24 hours of incubation (data not shown). Extensive acantholysis with wide intraepidermal clefting was observed by the end of the second day of incubation. After the 4th day, nonspecific changes, such as keratinocyte apoptosis and vacuolar and ballooning alterations, could be observed in the epidermis of all skin organ cultures. Therefore, all experiments were terminated at 72 hours of incubation, and the changes produced by test sera by that time point were compared. As seen in Figure 3, the PV2 serum containing the 130-kDa Dsg 3 antibody caused deep suprabasilar acantholysis. The basal cells remained attached to the basement membrane, rendering the tombstone appearance, which is the histopathological hallmark of PV (45). The PF serum with the 160-kDa Dsg 1 antibody produced acantholysis in the superficial layer of the epidermis (Figure 3). The split occurred within the granular cell layer. In marked contrast, the PV3 serum that, in addition to anti–Dsg 3 antibody, also contained anti–Dsg 1 antibody caused only suprabasilar acantholysis, but no superficial split that would be expected if anti–Dsg 1 PV antibodies were pathogenic (Figure 3). When these skin organ cultures were examined by DIF, the IgGs present in all three test pemphigus sera stained the entire epidermis, including the granular cell layer, in a similar intercellular, pemphigus-like pattern, which is consistent with the general observations that the DIF patterns of PV and PF sera are indistinguishable in most cases (46). Figure 3 Characterization of acantholytic activities of pemphigus sera with different combinations of Dsg 1 and Dsg 3 antibodies in skin organ culture. The anti-keratinocyte antibody profiles of two Dsg 3 antibody–positive PV sera, one with (PV3) and one without (PV2) Dsg 1 antibody, Dsg 1 antibody–positive PF serum, and normal human serum (N2) were characterized by ESIA using 7.5% SDS-PAGE (IP), and tested in fresh cultures of normal human skin as detailed in Methods. The morphological changes and antibody binding were examined by light microscopy (H&E) and DIF, respectively. The PV2 serum that contained Dsg 3 antibody but no Dsg 1 antibody and the PV3 serum that contained both Dsg 1 and Dsg 3 antibodies produced identical changes: 36 hours after addition of each serum, deep suprabasilar, but not superficial, acantholysis and intercellular antibody binding throughout the entire epidermis could be seen in these skin specimens. In marked contrast, the PF serum had induced by this time point distinct subcorneal acantholysis accompanied by intercellular antibody binding throughout the entire epidermis. As can be seen in the left upper corner of the biopsy stained with H&E, the split occurred within the granular cell layer. No alterations of the integrity of the epidermis or epidermal deposits of IgGs could be found in the cultures exposed for 96 hours to a healthy subject’s serum. In the column labeled IP, the relative molecular mass of the bands precipitated by each pemphigus serum is designated with arrows, and the positions of standard relative molecular mass markers are shown in the lowermost row. Scale bars = 50 μm. To distinguish pemphigus antibody–induced effects from nonspecific or degenerative changes in the cultured epidermis, parallel skin samples from the same skin donor were incubated with normal human sera. The bottom row of Figure 3 shows that no alterations of the integrity of the epidermis or any deposits of IgG could be detected in the cultures incubated with the N2 serum for 4 days. Thus, anti–Dsg 1 antibody present in PV serum does not exhibit acantholytic activity at its normal serum concentration, in agreement with the findings of Ding et al. (44). Can anti–Dsg 1 antibody cause PV-like skin lesions in Dsg3 null mice? A role for anti–Dsg 1 antibody in PV was proposed in the “desmoglein compensation” hypothesis (47), which is based on a report that both PV and PF IgGs produce identical suprabasilar split in Dsg3 null mice (32). The hypothesis attempts to reconcile the discrepancy between the known correlation of anti–Dsg 1 antibody with subcorneal epidermal split of PF and suprabasilar location of acantholysis in PV. However, insufficient thickness of murine epidermis may have hampered localization of the split. Therefore, in this study we compared the epidermal morphology at different anatomical regions of wild-type and Dsg3 null mice. The IgG fractions of PF sera and PV3 sera, both of which contained anti–Dsg 1 antibody (Figure 1), were injected intraperitoneally at a dose of 10 mg/g body weight into 10- to 12-hour-old Dsg3 null mice (16 pups received PF IgG; seven pups received PV3 IgG) or BALB/c mice (nine pups received PF IgG; five pups received PV3 IgG), and skin samples were obtained 24 hour thereafter. Both strains of mice injected with PF IgG developed superficial skin peeling, or wrinkling, with a positive Nikolsky sign (Figure 4a). Light microscopic examination revealed superficial, or subcorneal, epidermal acantholysis, as in patients with PF (Figure 4b). However, in locations where the basal cell layer was the only nucleated cell layer of the epidermis (Figure 4, b and c; arrows), the subcorneal split appeared as a suprabasilar split. The morphological pattern shown in Figure 4b was seen in the specimens obtained from the head/neck and axillary areas, where the epidermis is relatively thick, whereas the pattern seen in Figure 4c could be observed in skin of the trunk and extremities, where epidermis is thinner. In marked contrast, all pups injected with PV3 IgGs (Figure 4d) showed clinical and morphological acantholysis that was indistinguishable from that seen in mice injected with the PV IgGs that did not contain anti–Dsg 1 antibody. The deep intraepidermal localization of the acantholytic cleft, which was obvious in the skin areas with thick epidermis (Figure 4d), confirmed that the split induced by PV IgGs in the abdominal skin of Dsg 3–deficient mice, where the epidermis is thin (Figure 2, c and g), was also suprabasilar rather then subcorneal. The subcorneal split was never found in skin specimens from mice injected with the PV IgGs that were either positive or negative for anti–Dsg 1 antibody. Figure 4 Passive transfer of PV and PF IgGs to neonatal Dsg3 null mouse produce two distinct phenotypes in the epidermis. The PF IgG that immunorecognized Dsg 1 and did not cross-react with Dsg 3 and the PV3 IgG that contained antibodies to both Dsg 1 and Dsg 3 were injected at a dose of 10 mg/g body weight into 10- to 12-hour-old neonates of the litter that resulted from breeding homozygous female and male Dsg3 null mice. Wrinkling of the skin of a Dsg3 null mouse was noticed approximately 20 hours after a single injection of PF IgG (a). Approximately 22 hours after injection, the pups were euthanized, and two skin specimens, one from the neck/head area and one from abdominal skin, were obtained, stained with H&E, and examined by light microscopy. In both anatomical regions, the epidermal split induced by PF IgGs is subcorneal, rather than suprabasilar (b and c). The subcorneal location of the split can be easily misinterpreted as being suprabasilar in locations where the basal cell layer is the only nucleated cell layer of the epidermis (arrows). The accurate conclusions, however, should be drawn based on the phenotype found in the skin regions where the epidermis is thick, such as in the most parts of the specimen from the head/neck area shown in the b. In this skin region, the suprabasilar split caused by PV3 IgG is seen in the deep epidermis (d). The Dsg 3–/– genotype of these mice was confirmed by PCR amplification of the sequences of genomic DNA extracted from the tail (1). Scale bar = 50 μm. Thus, upon passive transfer to neonatal Dsg3 null mice, PV or PF IgGs produce two distinct, PV-like or PF-like phenotypes, respectively, eliminating the possibility that gross skin blisters and deep suprabasilar acantholysis in Dsg 3–lacking mice injected with PV IgGs (Figure 2) were caused by any unidentified anti–Dsg 1 antibody. The PV-like skin lesions were therefore induced by non–Dsg 1/non–Dsg 3 antibodies. Do Dsg 1 and Dsg 3 chimeric baculoproteins absorb similar disease-causing non-Dsg PV antibodies? These results provided evidence that non–Dsg 1/non–Dsg 3 antibodies can cause skin blisters in patients with PV. How then could Dsg 1 and Dsg 3 constructs absorb out all disease-causing PV antibodies in previous studies (21, 23, 32, 48)? Because the antigenic profile of the eluted PV IgGs was not characterized in those studies, the issue of monospecificity of the recombinant Dsg proteins used in absorption experiments required examination. We therefore produced the recombinant baculoproteins using the baculoviruses provided by M. Amagai, repeated the absorption experiments exactly as described in the original studies (21, 23), and characterized the immunoreactivities of the eluted, disease-causing PV IgG by immunoblotting using different keratinocyte substrates. Preliminary results indicated that the antigenic profiles of PV IgG eluted from rDsg3-His and rDsg3-Ig-His are different, suggesting that addition of the Fc IgG portion to the extracellular portion of Dsg 3 created a new epitope recognized by pathogenic non–Dsg 3 PV IgG (24). Alternatively or additionally, rDsg3-Ig-His might absorb anti-Fc IgG antibody (also known as rheumatoid factor). To test this hypothesis, we in this study compared the antigenic profiles of PV IgGs eluted from the rDsg3-His versus rDsg3-Ig-His versus rDsg1-Ig-His versus Fc IgG columns. The authenticity of the Dsg3-His, Dsg3-Ig-His and Dsg1-Ig-His baculoviruses was verified by PCR (Figure 5a), and the reactivities of produced baculoproteins with PV antibodies were confirmed by immunoblotting (Figure 5b). The efficiency of the Fc IgG column in absorbing ant-Fc IgG antibodies was demonstrated by the ability of the FITC-conjugated goat anti-human Fc IgG antibody that was first preabsorbed on and then eluted from the column to visualize PV IgG bound to the monkey esophagus substrate in a standard IIF assay (data not shown). Figure 5 Characterization of rDsg1-Ig-His, rDsg3-His, and rDsg3-Ig-His baculoviruses and baculoproteins. (a) PCR amplifications of DNA isolated from High Five insect cells infected with the rDsg3-His (lanes 2 and 3), rDsg3-Ig-His (lanes 5 and 6), or rDsg1-Ig-His (lanes 7 and 8). Lanes 1, 4, and 9 show positions of 100 bp DNA markers. The 380-bp Dsg3-His product (lane 3) was amplified using a pair of forward and reverse primers matching the sequences of the Dsg3 gene. In the negative control experiment (lane 2), the same template and a sense primer were used, but the reverse primer matched the sequence of the Fc portion of human IgG. The latter set of primers amplified the 420-bp product from the DNA extracted from the insect cells infected with the rDsg3-Ig-His baculovirus (lane 5), but not from noninfected cells (lane 6). Similarly, rDsg1-Ig-His was amplified using the forward Dsg 1 primer and the reverse Fc IgG1 primer, giving a 530-bp PCR product (lane 7), whereas no product was obtained in the control experiment using DNA from noninfected cells (lane 8). (b) The reactivity of PV3 IgGs with recombinant Dsg 1 and Dsg 3 baculoproteins. Western blots of the SDS-PAGE–resolved baculoproteins rDsg3-His (left), rDsg3-Ig-His (middle), and rDsg1-Ig-His (right) purified and concentrated on the TALON affinity metal resin and probed with PV3 serum, as detailed in Methods. Each visualized protein migrated with the expected relative molecular mass (left lane of each panel). The right lane of each panel shows absence of staining due to omission of the primary antibody. The antigenic profile of PV IgGs adsorbed on each recombinant protein was characterized by immunoblotting (Figure 6). The PV3 IgG eluted from rDsg3-His recognized mainly the 130-kDa human keratinocyte protein but also stained weakly a few bands with lower relative molecular mass. The PV3 IgG eluted from rDsg3-Ig-His reacted with a mixture of protein bands, including a 130-kDa polypeptide present in Dsg3 null keratinocytes (Figure 6). Most of protein bands visualized by PV3 IgG eluted from rDsg1-Ig-His showed molecular weights that were very similar to those visualized by PV3 IgG eluted from rDsg3-Ig-His (Figure 6). The PV3 IgG eluted from the Fc IgG column (∼12 mg total) did not recognize murine epidermal proteins in either Western blotting (Figure 6) or IIF (data not shown) assays. However, given that PV3 serum contained a rheumatoid factor (62 IU/ml), to rule out even a remote possibility that this antibody could induce skin changes in vivo, we tested PV3 IgG eluted from the Fc IgG column in passive transfer experiments with Dsg 3–deficient neonatal mice. We could not find any gross or microscopic skin changes, elicit Nikolsky sign, or detect any specific antibody binding to the epidermis in any of the four Dsg3 bal/ Dsg3 bal mice that were injected subcutaneously with 1.5 mg/g body weight of affinity-purified anti-Fc IgG antibody and observed for at least 24 hours after injection after which the animals were sacrificed. These results showed that the anti-Fc IgG antibody present in the PV3 IgG fraction does not cross react with keratinocyte proteins by immunoblotting and IIF, does not bind to the epidermis of Dsg 3 deficient mice in vivo and does not cause any skin changes in passive transfer experiments. This antibody, therefore, could not contribute to the staining of murine epidermal proteins and intraepidermal blisters produced by PV3 IgG eluted from the rDsg3-Ig-His construct. Figure 6 The profiles of the PV IgGs absorbed by recombinant Dsg 1 and Dsg 3 baculoproteins. Western blots of protein extracts of normal human keratinocytes resolved by 7.5% SDS-PAGE and stained with PV3 IgG affinity-purified on rDsg3-His (lane 1), and that of Dsg3 null keratinocytes stained with PV3 IgG affinity purified on rDsg3-Ig-His (lane 5), rDsg1-Ig-His (lane 4) or on the Fc IgG column (lane 9). Binding of PV3 IgGs to the immunoblotting membranes was visualized using secondary, HRP-conjugated goat anti-human IgG antibodies. Lane 6 shows a single protein band with apparent M r of 190 kDa among SDS-PAGE–resolved Dsg 3–positive keratinocyte proteins from BALB/c mouse. This band was visualized by the PV3 IgG that was eluted from the 190-kDa area of the Western blot of Dsg3 null keratinocyte proteins stained with the PV3 IgG adsorbed on rDsg3-Ig-His. Note: Only a 190-kDa but not 130- or 160-kDa band or any other keratinocyte protein was visualized, indicating that the antibody targeting the 190-kDa protein is a unique one, as it does not recognize the 130-kDa Dsg 3 or 160-kDa Dsg 1. Lanes 2, 3, and 7 are the negative controls omitting primary antibody. The Dsg3 null keratinocyte protein extract in lane 8 was blotted with normal human IgG-affinity purified on rDsg3-Ig-His. The lack of multiple bands in lane 8 as well as complete absence of specific staining in lanes 3 and 7 indicate that there were no nonspecific cross reactivities of human or goat IgGs with murine epidermal proteins. The positions of relative molecular mass markers run in parallel lanes of each blot are shown to the left of the respective blot. The apparent relative molecular mass of keratinocyte protein bands visualized due to PV antibody binding is shown to the right of lanes 2 and 5 in the columns designated M r. Most likely, addition of the Fc IgG1 sequence to the Dsg 1 and Dsg 3 extracellular epitopes rendered novel secondary or tertiary structures to these adhesion molecules. If the newly formed antigenic epitope represented conformational epitope of Dsg 3, then multiple bands visualized by PV3 IgGs eluted from rDsg1-Ig-His or rDsg3-Ig-His would result from random cross-reactivities of anti-Dsg 3 antibodies with other keratinocyte proteins. If these cross reacting anti-Dsg 3 autoantibodies were polyclonal, then anti-Dsg 3 IgG eluted from any single band should have been able to produce the same or very similar multiple-band staining pattern as the whole PV3 IgG fraction did. If the autoantibodies were monoclonal, then at least two bands, the 130-kDa band and the one from which that specific IgG was eluted, should have been visualized upon re-staining. To examine these intriguing possibilities, we characterized the antigenic profile of PV3 IgG eluted from a horizontal 190-kDa band strip cut from a wide immunoblotting membrane. We selected this band based on the following two main reasons: (a) it had the highest relative molecular mass compared with other bands visualized by PV3 IgGs eluted from rDsg3-Ig-His, which excluded a possibility that it represented a degradation product of any desmosomal cadherin; and (b) it was located at the edge of the antigenic “ladder” visualized on the immunoblotting membrane, which diminished to the minimum a chance that the horizontal strip cut at this area of the membrane could carry any other pemphigus antigens. The experiments showed that only 190-kDa band, but not 130- or 160-kDa band or any other band with a low relative molecular mass, could be stained with the antibody eluted from 190-kDa area of the immunoblotting membrane (Figure 6). These results indicate that the disease-causing PV IgGs absorbed with rDsg3-Ig-His represent a mixture of autoantibodies to an array of new keratinocyte self-antigens, including yet unidentified non–Dsg 3 polypeptides with apparent M r of 190 and 130 kDa. These results also indicate that the protein bands with lower relative molecular mass recognized by PV IgG eluted from the chimeric baculoproteins do not represent degradation products of the 190-kDa pemphigus antigen. Thus, Dsg 3 chimeric baculoprotein absorbs out all disease-causing PV antibodies nonspecifically, perhaps due to creation of new secondary or tertiary epitopes as a result of unique folding of the chimeric baculoprotein composed of two unrelated polypeptide sequences. Discussion In this study, we demonstrate for the first time that, in addition to Dsg 1 and Dsg 3, other keratinocyte proteins can be targeted by disease-causing PV antibodies and that neither anti–Dsg 1 nor anti–Dsg 3 antibody is critical for induction of PV-like skin blisters. These results were obtained in in vitro and in vivo models of pemphigus, using a novel, high sensitivity-and-specificity metabolic uptake assay, ESIA, in combination with the ELISA that uses supposedly pathogenic recombinant extracellular epitopes of Dsg 1 and Dsg 3. The discrepancy of these results with the postulated exclusive roles of anti–Dsg 1 and anti–Dsg 3 antibodies in mediating clinical and microscopic signs of pemphigus can be explained by our findings that both rDsg1-Ig-His and rDsg3-Ig-His chimeric baculoproteins, in addition to absorbing the purportedly unique autoantibodies against the adhesion molecules Dsg 1 and Dsg 3, also absorb a heterogeneous group of autoantibodies targeting other keratinocytes proteins. Blisters in pemphigus occur by intraepidermal clefting that results from breakdown of cell-to-cell adhesion. Intercellular spaces enlarge; desmosomes decrease in number and eventually disappear; and the cells round up and detach from one another without cell death (acantholysis) (45, 49–51). Blisters in pemphigus are associated with the binding of IgG autoantibodies to the keratinocyte cell surface (52). The pathophysiological importance of anti-Dsg autoimmunity in pemphigus has been suggested (8–10) by the results of experiments showing the ability of recombinant Dsg 1 and Dsg 3 constructs to absorb disease-causing antibodies from PF and PV sera, respectively (21–23). Validity of this theory was diminished when marked epidermal acantholysis and extensive skin blistering was induced in otherwise normal Dsg3 null neonatal mice by passive transfer of PV IgGs (1). As was determined by immunoblotting and standard immunoprecipitation techniques, the PV IgG fractions injected into Dsg 3–deficient mice did not contain anti–Dsg 1 antibody (1). However, it has been recently reported that absorption of PV sera with recombinant Dsg 1 could eliminate all disease-causing activity in passive transfer experiments with Dsg3 null mice, suggesting that anti–Dsg 1 antibody is responsible for the PV-like phenotype in these pups (32). Therefore, in our previous study (1) the disease-causing anti–Dsg 1 antibodies might have been undetectable due to destruction of conformational epitope of Dsg 1 in the course of immunoblotting and/or relatively low sensitivity of the standard immunoprecipitation assay used. In this study, the anti-keratinocyte antibodies developed by patients with pemphigus were characterized using naturally folded keratinocyte proteins as antigens. We used the enhanced sensitivity metabolic labeling assay, ESIA, in which the sensitivity and resolution is enhanced by the ability of the Storm system’s phosphor-imager feature to capture the image from both strong and weak signals in a single exposure (53, 54). To increase the sensitivity of our detection technique even further, we probed the sera with a 45-kDa immunoreactive fragment of Dsg 1 purified by Con A affinity chromatography (36). The 45-kDa polypeptide is a fragment of Dsg 1, as can be judged from the sequencing data reported by Abreu-Velez et al. (55), and is believed to be a very sensitive probe for anti–Dsg 1 antibody produced by patients with PV (31). Finally, to validate the sensitivity of our approach, we performed series of ELISA experiments using as antigens the extracellular epitopes of Dsg 1 and Dsg 3 that have been reported to be highly sensitive and specific targets for pathogenic anti-Dsg antibodies produced by patients with different forms of autoimmune pemphigus (23, 37, 39, 44, 56). This combination of highly sensitive and specific immunoassays was required to identify the PV sera that were free of anti-Dsg 1 antibody in order to test their acantholytic activities in vivo and in vitro. The IgG fractions of Dsg 1 antibody–negative PV sera caused gross skin blisters with suprabasilar acantholysis and produced a fishnet-like, intercellular staining pattern of perilesional epidermis in two different strains of mice lacking Dsg 3. By ESIA, in addition to the 130-kDa Dsg 3 target that was absent in these mice, the non–Dsg 1 pathogenic PV IgGs uniquely recognized keratinocyte proteins of approximately 38, 43, 115, and 190 kDa. The possibility that the anti–Dsg 3 antibody cross-reacted with Dsg 1 on the cell surfaces of murine keratinocytes, leading to acantholysis via this pathway, is unlikely because Dsg 3 antibody produced by patients with PV does not cross-react with Dsg 1 (44, 56). This was confirmed in this study by lack of protein bands with apparent M r of 160 kDa in Western blots stained with the Dsg 1 antibody–positive PV3 IgG eluted from rDsg3-His or rDsg3-Ig-His. Furthermore, a semiquantitative IIF assay shows that the epidermis of Dsg3 null mice immunoreacts with only 58% of the relative amounts of PV IgGs bound in the epidermis of the Dsg3+/+ BALB/c mice, defined as 100%, indicating that anti–Dsg 3 antibody present in PV IgG fractions does not bind to the Dsg 3–negative keratinocytes (1). Indeed, commercially available antibody to Dsg 3 also does not stain the epidermis of Dsg3 null and Dsg3 bal/Dsg3 bal mice (1). Furthermore, the results of our organ culture experiments demonstrate that anti–Dsg 1 antibody present in PV sera is unable to induce superficial acantholysis characteristic of PF. Thus, PV-like skin lesions in Dsg 3–deficient mice injected with PV IgG mice seemed to be induced through a pathophysiological mechanism exclusive of any Dsg 1 and Dsg 3 antibody–mediated pathways. However, the ultimate conclusion could not be made until after resolution of the controversy about microscopic localization of skin blisters in Dsg3 null neonatal mice injected with anti–Dsg 1 antibody (32, 48). To address a hypothetical possibility that in our passive transfer experiments (Figure 2) the skin blisters were caused by some hidden anti–Dsg 1 antibody, we investigated thoroughly the pathology of the skin of Dsg3 null neonatal mice injected with the Dsg 1 antibody-containing PF IgG. We found the superficial intraepidermal split. However, the location of the split could be also called suprabasilar in anatomical regions where the basal cell layer was the only nucleated cell layer of the epidermis. Identical gross and microscopic changes were observed in wild-type Dsg 3–positive mice injected with PF IgGs. Therefore, as either Dsg3 null or Dsg3 bal/Dsg3 bal or wild-type mice injected with PV IgG never develop any evidence of PF-like subcorneal split, but always demonstrate deep, suprabasal acantholysis, it is extremely unlikely that the PV-like phenotype in Dsg 3–negative neonates injected with PV IgG was caused by some hidden anti–Dsg 1 antibody. The keratinocytes in Dsg 3-negative mice, however, may be more sensitive to non-Dsg antibodies than in wild-type mice. Having established that non–Dsg 1/non–Dsg 3 PV IgGs are capable of inducing PV-like skin blisters, we inquired how could the recombinant Dsg 1 and Dsg 3 constructs that also carry Fc IgG1 portion absorb out all disease-causing antibodies from PV sera (21–23, 32)? Furthermore, in our own experiments, absorption of PV sera with rDsg3-Ig-His eliminated the ability of IgGs purified from these sera to cause acantholysis in Dsg 3-negative Dsg3 null mice, whereas the PV IgG eluted from rDsg3-Ig-His induced skin blisters in these mice (24). Therefore, because the Dsg 3 antigen was not present in the epidermis of Dsg3 null mice, the interpretations that rDsg3-Ig-His absorbs disease-causing PV IgGs due to a permissive conformation mimicking the authentic epitope, in contrast to rDsg3-His , which lacks this ability, and that its conformation is important in the pathogenesis of PV become untenable. To clarify the relevant mechanisms, we compared the antigenic profiles of PV antibodies absorbed by rDsg1-Ig-His and rDsg3-Ig-His . Both chimeric baculoproteins absorbed a series of autoantibodies, recognizing, in addition to the 160-kDa Dsg 1 and 130-kDa Dsg 3, respectively, keratinocyte proteins with apparent M r of 38, 43, 115, and 190 kDa. These results strongly suggested that both chimeric baculoproteins absorb identical non-Dsg antibodies, and that addition of Fc portion of IgG1 to the extracellular epitopes of Dsg 1 and Dsg 3 accounts for similar antibody reactivities of both chimeras. Thus, the chimeric baculoproteins used in the past to remove disease-causing activity of PV and PF sera, which is the only evidence directly supporting the notion that anti–Dsg 1 and anti–Dsg 3 antibodies are pathogenic, actually absorbed out autoantibodies directed to multiple keratinocyte cell membrane proteins. Hence, although the conformational epitopes of these baculoproteins are indeed recognizable by disease-causing PV and PF antibodies, the autoantibodies are heterogeneous. These chimeras, therefore, may become a useful tool for elimination and further characterization of antibodies causing PV and PF. Further investigation of PV and PF disease-causing antibodies may take advantage of recent discoveries of two new human keratinocyte molecules targeted by disease-causing PV IgGs, namely: (a) the α9 acetylcholine receptor with dual muscarinic-and-nicotinic pharmacology (24), and (b) pemphaxin, a novel annexin (also known as annexin 31 or ANXA9) that can act as an acetylcholine receptor (35). The pathophysiological importance of these new molecules in pemphigus is illustrated by the ability of rabbit monoepitopic anti-α9 antibody to cause pemphigus-like acantholysis in keratinocyte monolayers, and by elimination of the disease-causing activity of PV IgGs by preabsorption with recombinant pemphaxin. These observations are in keeping with earlier findings showing that 85% of patients with PV and of those with PF develop autoantibodies to keratinocyte acetylcholine receptors (1), and a large body of evidence demonstrating that pemphigus antibody binding to the keratinocyte cell membrane evokes a cascade of intracellular biochemical reactions (reviewed in refs. 1, 57), including those that directly alter expression and function of Dsg 3 (58, 59) and other adhesion molecules, such as vinculin and connexin 43 (60). Acetylcholine receptors and annexins are functionally coupled to regulation of intracellular Ca 2+ and control cell adhesion and motility (reviewed in refs. 61, 62). Recent studies of molecular mechanisms of epithelial adhesion have revealed that Ca 2+-mediated actin-polymerization is the major driving force for keratinocyte cell-to-cell attachment and that desmosome assembly is a passive process that occurs once the stable cell-cell contact has been made (63). It is therefore worth noting that acantholysis and a loss of skin adhesion in Hailey-Hailey disease and in Darier disease result from defects in genes encoding an ATP-powered calcium pump that sequesters calcium in the Golgi apparatus (64) and a sarco/endoplasmic reticulum-Golgi calcium pump (65), respectively. Altogether, these findings elucidate, on the one hand, the morphological similarities between keratinocyte acantholysis induced by either cholinolytic drugs or pemphigus antibody, and, on the other hand, the ability of the cholinomimetic drugs to abolish PV IgG-induced acantholysis (reviewed in ref. 66). In summary, the results of this comprehensive study reveal, for the first time to our knowledge, the following facts which will be critical in formulating further conclusions about the pathophysiology of PV: (a) PV-like skin lesions can be induced without participation of either anti-Dsg 1 or anti-Dsg 3 antibodies. (b) PV and PF IgGs produce different disease phenotypes in Dsg3 null mice. (c) Absorption of similar non-Dsg antibodies with rDsg1-Ig-His and rDsg3-Ig-His reconciles the controversy about the role of anti-Dsg 1 and anti-Dsg 3 antibodies in pemphigus. We conclude that acantholysis in PV is mediated by an active process triggered by a yet incompletely understood pathogenic mechanism, including autoantibodies to different types of keratinocyte cell membrane proteins. Our “multiple hit” hypothesis assimilates diverse postulates on the roles of different autoantibodies in pemphigus (66). We propose that antibody binding to multiple cell-surface target antigens induce acantholysis. The severity of disease and the specific clinical presentation in each particular patient depend on the contributions and relative concentrations of multiple anti-keratinocyte autoantibodies. Future studies, therefore, should be directed toward identification of the keratinocyte membrane proteins targeted by disease-causing antibodies in each patient with PV or PF, and elucidation of the immunopharmacological actions of novel disease-causing pemphigus IgGs. Acknowledgments We thank Peter M. Yau from the Department of Biochemistry, University of California Davis Medical School, for help with analysis of the radioactivity of immunoprecipitation gels on the Storm system. The colony of mice with targeted disruption of the Dsg3 gene was established from animals generously donated to The Induced Mutant Resource of The Jackson Laboratory by John R. Stanley (University of Pennsylvania). 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Udey, MC, Stanley, JR. Pemphigus: diseases of antidesmosomal autoimmunity. JAMA 1999. 282:572-576. View this article via: PubMedCrossRefGoogle Scholar 48. Mahoney, MG, et al. Explanation for localization of blisters in pemphigus patients. J Invest Dermatol 1998. 110:499. (Abstr.) 49. Bellone, AG, Leone, V. Ricerche sull’influenza esercitata da sieri di soggetti sani o affetti da pemfigo su pelle umana normale e pemfigosa coltivata “in vitro”. G Ital Dermatol Minerva Dermatol 1956. 97:97-109. 50. Farb, RM, Dykes, R, Lazarus, GS. Anti-epidermal-cell-surface pemphigus antibody detaches viable epidermal cells from culture plates by activation of proteinase. Proc Natl Acad Sci USA 1978. 75:459-463. View this article via: PubMedCrossRefGoogle Scholar 51. Patel, HP, Diaz, LA, Anhalt, GJ, Labib, RS, Takahashi, Y. Demonstration of pemphigus antibodies on the cell surface of murine epidermal cell monolayers and their internalization. J Invest Dermatol 1984. 83:409-415. View this article via: PubMedCrossRefGoogle Scholar 52. Beutner, E, Jordon, R. Demonstration of skin antibodies in sera of pemphigus vulgaris patients by indirect immunofluorescent staining. Proc Soc Exp Biol Med 1964. 117:505-510. View this article via: PubMedGoogle Scholar 53. Johnston, RF, Pickett, SC, Barker, DL. Autoradiography using storage phosphor technology. Electrophoresis 1990. 11:355-360. View this article via: PubMedCrossRefGoogle Scholar 54. Zouboulis, CC, Tavakkol, A. Storage phosphor imaging technique improves the accuracy of RNA quantitation using 32 P-labeled cDNA probes. Biotechniques 1994. 16:290-292, 294. View this article via: PubMedGoogle Scholar 55. Abreu-Velez, AM, et al. Characterization of a 45 kD epidermal tryptic peptide recognized by pemphigus foliaceus sera. J Invest Dermatol 1997. 108:541. (Abstr.) 56. Ishii, K, et al. Characterization of autoantibodies in pemphigus using antigen-specific enzyme-linked immunosorbent assays with baculovirus-expressed recombinant desmogleins. J Immunol 1997. 159:2010-2017. View this article via: PubMedGoogle Scholar 57. Kitajima, Y, Aoyama, Y, Seishima, M. Transmembrane signaling for adhesive regulation of desmosomes and hemidesmosomes, and for cell-cell detachment induced by pemphigus IgG in cultured keratinocytes: involvement of protein kinase C. J Invest Dermatol Symp Proc 1999. 4:137-144. View this article via: CrossRefGoogle Scholar 58. Aoyama, Y, Kitajima, Y. Pemphigus vulgaris-IgG causes a rapid depletion of desmoglein 3 (Dsg3) from the triton X-100 soluble pools, leading to the formation of Dsg3-depleted desmosomes in a human squamous carcinoma cell line, DJM-1 cells. J Invest Dermatol 1999. 112:67-71. View this article via: PubMedCrossRefGoogle Scholar 59. Aoyama, Y, Owada, MK, Kitajima, Y. A pathogenic autoantibody, pemphigus vulgaris-IgG, induces phosphorylation of desmoglein 3, and its dissociation from plakoglobin in cultured keratinocytes. Eur J Immunol 1999. 29:2233-2240. View this article via: PubMedCrossRefGoogle Scholar 60. Xue, W. Pemphigus vulgaris IgG regulates expression of urokinase receptor and junctional proteins that may contribute to acantholysis. J Am Acad Dermatol 1999. 41:462-463. View this article via: PubMedGoogle Scholar 61. Benz, J, Hofmann, A. Annexins: from structure to function. Biol Chem 1997. 378:177-183. View this article via: PubMedGoogle Scholar 62. Grando, SA. Biological functions of keratinocyte cholinergic receptors. J Invest Dermatol Symp Proc 1997. 2:41-48. 63. Vasioukhin, V, Bauer, C, Yin, M, Fuchs, E. Directed actin polymerization is the driving force for epithelial cell-cell adhesion. Cell 2000. 100:209-219. View this article via: PubMedCrossRefGoogle Scholar 64. Hu, Z, et al. Mutations in ATP2C1, encoding a calcium pump, cause Hailey-Hailey disease. Nat Genet 2000. 24:61-65. View this article via: PubMedCrossRefGoogle Scholar 65. Sakuntabhai, A, et al. Mutations in ATP2A2, encoding a Ca 2+ pump, cause Darier disease. Nat Genet 1999. 21:271-277. View this article via: PubMedCrossRefGoogle Scholar 66. Grando, SA. Autoimmunity to keratinocyte acetylcholine receptors in pemphigus. Dermatology (Basel) 2000. 201:290-295. View this article via: PubMedCrossRefGoogle Scholar Version history Version 1 (December 15, 2000): No description Article tools View PDF Download citation information Send a comment Terms of use Standard abbreviations Need help? 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8772	http://www.cs.toronto.edu/~jacobson/computer-graphics-csc418/slides/lecture2+3.pdf	Today’s Topics 3. Transformations in 2D 4. Coordinate-free geometry 5. 3D Objects (curves & surfaces) 6. Transformations in 3D Topic 3: 2D Transformations • Simple Transformations • Homogeneous coordinates • Homogeneous 2D transformations • Affine transformations & restrictions Transformations Transformation/Deformation in Graphics: A function f, mapping points to points. simple transformations are usually invertible. [x y] T [x’ y’] T Applications: • Placing objects in a scene. • Composing an object from parts. • Animating objects. Processing Tree Demo! f f-1 Lets start out simple… Translate a point [x y]T by [tx ty]T : x’ = x + tx y’ = y + ty Rotate a point [x y]T by an angle t: x’ = x cost - y sint y’ = x sint + y cost Scale a point [x y]T by a factor [sx sy]T x’ = x sx y’ = y sy Representing 2D transforms as a 2x2 matrix Rotate a point [x y]T by an angle t: x’ = cost -sint 0 x y’ sint cost 0 y 1 0 0 1 1 Scale a point [x y]T by a factor [sx sy]T x’ = sx 0 0 x y’ 0 sy 0 y 1 0 0 1 1 Translate? Points as Homogeneous 2D Point Coords [1 0]T [0 1]T p=[x y 1]T p= x[1 0 0]T + y[0 1 0]T +[0 0 1]T basis vectors Cartesian Homogeneous 2D Points Cartesian [x y]T => Homogeneous [x y 1]T Homogeneous [x y w]T => Cartesian [x/w y/w 1]T What about w=0? Homogeneous points are equal if they represent the same Cartesian point. For eg. [4 -6 2] T = [-6 9 -3] T. Points at ∞ in Homogeneous Coordinates [x y w] T with w=0 represent points at infinity, though with direction [x y] T and thus provide a natural representation for vectors, distinct from points in Homogeneous coordinates. Line Equations in Homogeneous Coordinates A line given by the equation ax+by+c=0 can be represented in Homogeneous coordinates as: l=[a b c] , making the line equation l.p= [a b c][x y 1] T =0. Aside: cross product as a matrix [ 0 -c b] [x y 1] T [ c 0 -a] [-b a 0] The Line Passing Through 2 Points For a line l that passes through two points p0, p1 we have l.p0 = l.p1 = 0. In other words we can write l using a cross product as: l= p0 X p1 p0 p1 Point of intersection of 2 lines For a point that is the intersection of two lines l0, l1 we have p.l0 = p.l1 = 0. In other words we can write p using a cross product as: p= l0 X l1 l1 l0 p What happens when the lines are parallel? Representing 2D transforms as a 3x3 matrix Translate a point [x y]T by [tx ty]T : x’ = 1 0 tx x y’ 0 1 ty y 1 0 0 1 1 Rotate a point [x y]T by an angle t: x’ = cost -sint 0 x y’ sint cost 0 y 1 0 0 1 1 Scale a point [x y]T by a factor [sx sy]T x’ = sx 0 0 x y’ 0 sy 0 y 1 0 0 1 1 Properties of 2D transforms …these 3x3 transforms have a variety of properties. most generally they map lines to lines. Such invertible Linear transforms are also called Homographies. …a more restricted set of transformations also preserve parallelism in lines. These are called Affine transforms. …transforms that further preserve the angle between lines are called Conformal. …transforms that additionally preserve the lengths of line segments are called Rigid. Where do translate, rotate and scale fit into these? Properties of 2D transforms Homography, Linear (preserve lines) Affine (preserve parallelism) shear, scale Conformal (preserve angles) uniform scale Rigid (preserve lengths) rotate, translate Homography: mapping four points How does the mapping of 4 points uniquely define the 3x3 Homography matrix? Homography: preserving lines Show that if points p lie on some line l, then their transformed points p’ also lie on some line l’. Proof: We are given that l.p = 0 and p’=Hp. Since H is invertible, p=H-1p’. Thus l.(H-1p’)=0 => (lH-1).p’=0, or p’ lies on a line l’= lH-1. QED Affine: preserving parallel lines What restriction does the Affine property impose on H? If two lines are parallel their intersection point at infinity, is of the form [x y 0]T. If these lines map to lines that are still parallel, then [x y 0]T transformed must continue to map to a point at infinity or [x’ y’ 0]T i.e. [x’ y’ 0]T = [x y 0]T ? ? ? Affine: preserving parallel lines What restriction does the Affine property impose on H? If two lines are parallel their intersection point at infinity, is of the form [x y 0]T. If these lines map to lines that are still parallel, then [x y 0]T transformed must continue to map to a point at infinity or [x’ y’ 0]T i.e. [x’ y’ 0]T = A t [x y 0]T 0 0 1 In Cartesian co-ordinates Affine transforms can be written as: p’ = Ap + t Affine properties: composition Affine transforms are closed under composition. i.e. Applying transform (A1,t1) (A2,t2) in sequence results in an overall Affine transform. p’= A2 (A1p+t1) + t2 => (A2 A1)p+ (A2t1 + t2) Affine properties: inverse The inverse of an Affine transform is Affine. - Prove it! Affine transform: geometric interpretation A change of basis vectors and translation of the origin t A a1 a2 a1 a2 t p 0 0 1 p point p in the local coordinates of a reference frame defined by is -1 Affine transform: change of reference frame How can we transform a point p from one reference frame , to another frame ? a1 b1 p a2 o1 b2 o2 Composing Transformations Any sequence of linear transforms can be collapsed into a single 3x3 matrix by concatenating the transforms in the sequence. In general transforms DO NOT commute, however certain combinations of transformations are commutative… try out various combinations of translate, rotate, scale. Rotation about a fixed point The typical rotation matrix, rotates points about the origin. To rotate about specific point q, use the ability to compose transforms… Tq R T-q Topic 4: Coordinate-Free Geometry (CFG) • A brief introduction & basic ideas CFG: dimension free geometric reasoning Points p [ … 1] Vectors v [ … 0] Lines l [ ….. ] Dot products, Cross products, Length of vectors, Weighted average of points… How do you find the angle between 2 vectors? Topic 5: 3D Objects • General curves & surfaces in 3D • Normal vectors, surface curves & tangent planes • Implicit surface representations • Example surfaces: surfaces of revolution, bilinear patches, quadrics 3D parametric curves p(t)=(fx(t),fy(t),fz(t))) 3D parametric surfaces p(t,s)=(fx(t,s),fy(t,s),fz(t,s))) 3D parametric plane p(s,t)= q + as +tb q a b Topic 5: 3D Objects • General curves & surfaces in 3D • Normal vectors, surface curves & tangent planes • Implicit surface representations • Example surfaces: surfaces of revolution, bilinear patches, quadrics Tangent / Normal vectors of 2D curves Explicit: y=f(x). Tangent is dy/dx. Parametric: x=fx(t) Tangent is (dx/dt, dy/dt) y=fy(t) Implicit: f(x,y) = 0 Normal is gradient(f). direction of max. change Given a tangent or normal vector in 2D how do we compute the other? What about in 3D? Normal vector of a plane q a b p(s,t)= q + as +tb Normal vector of a plane q a b n n=aXb Normal vector of a parametric surface [f(u0,v0)] f(u0,v) f(u,v0) Normal vector of a parametric surface [f(u0,v0)] f(u0,v) f(u,v0) n=f’(u0,v) X f’(u,v0) n Topic 5: 3D Objects • General curves & surfaces in 3D • Normal vectors, surface curves & tangent planes • Implicit surface representations • Example surfaces: surfaces of revolution, bilinear patches, quadrics Implicit function of a plane q a b n f(p) = (p-q).n=0 Implicit function: level sets Topic 5: 3D Objects • General curves & surfaces in 3D • Normal vectors, surface curves & tangent planes • Implicit surface representations • Example surfaces: surfaces of revolution, bilinear patches, quadrics 3D parametric surfaces • Extrude • Revolve • Loft • Square Maya Live Demo… 3D parametric surfaces: Coons interpolation b0 b3 b2 b1 interpolate(b0,b2) interpolate(b1,b3) bilinear interpolation
8773	http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/twin.html	Twin paradox and other special relativity topics Time Dilation Experiments ========================= The abandonment of the concept of universal time embodied in the time dilation expression is so counter-intuitive that one must look at the experiments to confirm this extraordinary prediction of special relativity. These are some of the experiments which confirm time dilation. Muon experimentHafele-Keating experiment Atomic fine structureKaivole atomic beamIndex HyperPhysics RelativityR NaveGo Back Twin Paradox ============ The story is that one of a pair of twins leaves on a high speed space journey during which he travels at a large fraction of the speed of light while the other remains on the Earth. Because of time dilation, time is running more slowly in the spacecraft as seen by the earthbound twin and the traveling twin will find that the earthbound twin will be older upon return from the journey. The common question: Is this real? Would one twin really be younger? The basic question about whether time dilation is real is settled by the muon experiment. The clear implication is that the traveling twin would indeed be younger, but the scenario is complicated by the fact that the traveling twin must be accelerated up to traveling speed, turned around, and decelerated again upon return to Earth. Accelerations are outside the realm of special relativity and require general relativity. Despite the experimental difficulties, an experiment on a commercial airline confirms the existence of a time difference between ground observers and a reference frame moving with respect to them. The Pole-Barn Paradox The Bug-Rivet ParadoxIndex HyperPhysics RelativityR NaveGo Back Atomic Fine Structure One of the great successes of the quantum theory was the prediction of the energy levels of the hydrogen atom. When attempts were made to explain the fine structure of the hydrogen spectral lines, it was found that the splitting of the lines was in error by a factor of two. It was realized that relativistic time dilation must be used in calculating the frequencies, and calculations showed that this relativistic correction, called Thomas precession, was the factor of two which was needed for agreement with experiment.Index HyperPhysics RelativityR NaveGo Back Kaivola Time Dilation Experiment A precise measurement of time dilation was made in a double-photon experiment by Kaviola et al. in 1985. An atomic beam of neon atoms at v= 0.004c was excited by two lasers which were colinear with the beam and incident upon the atoms from opposite directions. The absorption frequencies for neon are shifted by both the Doppler effect and time dilation, with the Doppler shift being much larger. By measuring a double photon transition with the two lasers from opposite directions, the Doppler shift was canceled and the time dilation shift measured precisely . The measurement was made by measuring the beat frequency between the two tunable lasers. This experiment confirmed the expected time dilation within 4 parts in 100,000.Index Kaivola, et al. HyperPhysics RelativityR NaveGo Back
8774	https://www.youtube.com/watch?v=eq8APJ6qf2M	Ex: Solve a One Step Equation by Multiplying by Reciprocal (a/b)x=-c/d Mathispower4u 330000 subscribers 1 likes Description 915 views Posted: 31 Oct 2015 This video explains how to solve a one step equation with fractions by multiplying by the reciprocal. The solution is checked. Transcript: We want to solve the equation three-fourths, x equals negative nine-eighths. Three-fourths, x means three-fourths times x, and therefore, the solution to the equation is the value of x that makes the product on the left equal to negative nine-eighths. In order to solve for x we need to isolate x on one side of the equation. And because the x is on the left side, the solution will be in the form x equals some value, again, that makes this product equal to negative nine-eighths. And, again, because three-fourths, x means three-fourths times x, in order to solve for x, we need to undo this multiplication by performing the inverse, or opposite operation of multiplying by three-fourths. So to undo multiplying by three-fourths, we would divide by three-fourths, but remember instead of dividing by a fraction, we almost always multiply by the reciprocal. So instead of dividing both sides by three-fourths, let's multiply both side by the reciprocal of three-fourths, which would be four-thirds. So we'll multiply the left side by four-thirds, and multiply the right side by four-thirds. Remember, what ever operation we perform to one side, we must also perform to the other, to maintain the equality. Now looking at the left side of the equation, notice how we have a product of two reciprocals, and therefore four-thirds times three-fourths is equal to one. Of course we can also show the simplification. There's a common factor of four here, the fours simplify to ones, and the threes also simplify to ones. So the left side is now just one times x, which is just x. And now let's find the product on the right. First notice how we have a positive times a negative, so the product will be negative, but before multiplying let's simplify. Notice four and eight share a common factor of four. There's one four in four, and two fours in eight. Three and nine share a common factor of three. There's one three in three, and three threes in nine. Multiplying the numerators we have one times negative three, that's negative three. In the denominator we have one times two, which equals two. So our solution is x equals negative three-halves. Let's take a look at a second method for solving this equation. Beginning with the original equation, again, we have three-fourths, x equals negative nine-eighths. Whenever we have an equation that contains fractions we always have the option of clearing the fractions from the equation. Looking at our notes below, to clear the fractions from an equation, we multiply both sides of the equation by the least common denominator. So notice here we have a denominator of four, and here we have a denominator of eight. The least common denominator is the least common multiple of four and eight, which would be eight. So let's solve this again, by clearing the fractions from the equation by multiplying both sides of the equation by eight. So multiply the left side by eight, and the right side by eight. If we want we can write eight as a fraction with a denominator of one, if that's helpful. Now looking at the left of the equation here, we can simplify. Four and eight share a common factor of four, there's one four in four and two fours in eight. So on the left side we have two times three times x, which equals six, x, equals on the right side, eight over eight simplifies to one over one. So notice on the right side we just have one over one, times negative nine, which equals negative nine. Now we need to solve the equation six, x equals nine, and again because six, x means six times x, in order to undo the multiplication and solve for x, we need to divide both sides of the equation by six. Six divided by six is equal to one. One times x equals x. So we have x equals negative nine-sixths, but the fraction does simplify. Six and nine share a common factor of three. Dividing the numerator and denominator by three, notice how we do get negative three-halves. So of course our solution is the same using the method of clearing the fractions. We have x equals negative three-halves. Before we go, let's check our solution by substituting negative three-halves for x in the original equation to verify that three-fourths times negative three-halves does equals negative nine-eighths. So we'd have three-fourths times negative three-halves equals negative nine-eighths. No simplifying required here. A positive times a negative is negative. Three times three is nine. Four times two is eight. So we have negative nine-eighths equals negative nine-eighths, which is true, verifying our solution is correct. I hope you found this helpful.
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8776	https://www.sciencedirect.com/topics/engineering/ferromagnet	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Supplement 2 19.3.1.5 Ferrimagnets In these crystalline materials, magnetic dipoles are arranged in antiparallel directions as in antiferromagnets, though always with only partial reciprocal compensation, and thus with a nonzero vector sum of moments. Both at the microscopic and the macroscopic level, ferrimagnetism57–59 is a magnetic behavior very similar to ferromagnetism. Ferrimagnets, indeed, are characterized by a Weiss domain microstructure, magnetization mechanisms and hysteresis loops, high values of susceptibility and permeability (μs, i of the order of 102 to 104), and a Curie transition to paramagnetic materials, as in ferromagnets. Major differences are: (i) ferrimagnets are nonconductors or semiconductors (with respect to ferromagnetic materials, which are, with a few exceptions such as CrO2, metals or metal alloys); (ii) above TC, χm, s decays with increasing temperature more sharply than according to the Curie–Weiss law, but approaches asymptotically this behavior as the temperature becomes higher and higher. Ferrimagnets87, 88, 91, 92 are usually classified as: (a) spinels, i.e. γ-Fe2O3 and compounds of general formula MO · Fe2O3, with M a divalent cation such as Fe2+ (magnetite), Ni2+, CO2+, Zn2+, Mg2+, etc.; (b) magnetoplumbites MO · 6Fe2O3, where M is a divalent ion such as Ba2+, Sr2+, Pb2+; (c) magnetic garnets 3M2O3 · 5Fe2O3, with M a trivalent rare-earth ion. Very fine grinding of ferri- and ferromagnets fractionates the Weiss domains and cancels the ferro- or ferrimagnetic properties, thus yielding peculiar paramagnetic powders (useless, e.g. for magnetic information storage93) which have outstandingly high values of susceptibility (superparamagnets).57, 88, 93 View chapterExplore book Read full chapter URL: Chapter Meso- and Nanostructures Ferromagnets Have Only One Domain and a Low Anisotropy Energy Macroscopic ferromagnets acquire domains in order to minimize the energy due to the magnetic field. When a magnetic field is applied, the atomic magnetic moments align through movement of the domain walls. If the size of the ferromagnet is smaller than that of one domain, then alignment with the magnetic field can only occur by the concerted rotation of all the moments. There is an energy penalty for this (anisotropy energy). However, the smaller the ferromagnet, the lower the anisotropy energy. Thus, if the magnetic particle is small enough, then a collection of them can behave as a paramagnet with a very large susceptibility (superparamagnetism). If the particle is elongated, it forms a single domain, while if it is circular it tends to form a vortex state. View chapterExplore book Read full chapter URL: Chapter Rare-earth magnets: properties, processing and applications 2012, Functional Materials for Sustainable Energy ApplicationsI.R. Harris, G.W. Jewell 19.2.1 What is a ferromagnetic material? A useful starting point in understanding the behaviour of permanent magnets is to consider the fundamental behaviour of ferromagnetic materials.1,2 The very special feature of these materials is the spontaneous alignment of the magnetic moments of the individual atoms. In the ferromagnetic metals, Fe, Co, and Ni, these arise predominantly from the spin contribution of the 3d-type electrons. However, under equilibrium conditions, the bulk material does not exhibit an external magnetic field because the total energy of the system is minimised by the formation of ‘magnetic domains’ which form a closed magnetic circuit and hence eliminate any external field. This is illustrated in Fig. 19.1 and the response to the application of an external magnetic field (H) is to begin to align these domains with a large increase in the total field and this contribution is called the magnetisation (M) which reaches saturation (Ms) when all the domains are aligned in the direction of the applied field. The value of Ms is proportional to the number of unpaired 3d-electron spins and therefore decreases in going from Fe to Co to Ni. The value of Ms also decreases with increasing temperature due to progressive thermal disorder and the ferromagnetic state disappears completely at a particular temperature called the Curie point (Tc). This effect is totally reversible and the ferromagnetic state reappears on cooling below Tc. The Curie points for Fe, Co and Ni are 1043, 1393 and 631 K, respectively, and these indicate that the ferromagnetic coupling (exchange energy) is strongest in Co. The Ms and Tc are intrinsic properties of a ferromagnet. Another intrinsic property is the magneto-crystalline anisotropy (MCA) which, in the case of Fe, Co and Ni is a consequence of the relatively weak coupling between the spin and orbital moments of the 3d-electrons. In the case of the rare-earth metals on the other hand, the 4f spin-orbit coupling is much more pronounced giving rise to much stronger MCAs. The consequence of this spin-orbit coupling can be that, within a particular crystalline structure, there is an easy direction of magnetisation and this is illustrated schematically for a hexagonal metal such as Co in Fig. 19.2, where the easy magnetisation direction is along the c-axis and is an example of uniaxial MCA (UMCA). This effect is much more pronounced in a crystal structure where there is a combination of ferromagnetic transition metal and a rare-earth metal, for example, SmCo5, Sm2(Co,Fe,Cu,Zr)17 and Nd2Fe14B. This essential characteristic for permanent magnets based on these compounds will be discussed later in the chapter. To become a permanent magnet the domains have to remain aligned when the external field is removed or applied in the opposite direction. This leads to the phenomenon of hysteresis (from the Greek ‘to lag’). This gives rise to the properties of remanence (Br) when the external field is at zero and the intrinsic coercivity (Hcj) when M is reduced to zero. Fortunately, these are extrinsic properties and can be controlled by modifying the microstructure of the material. The resulting bulk magnetic behaviour of permanent magnets which leads to their utility for many electromagnetic devices can be illustrated by reference to its magnetic hysteresis loop, which captures both magnetisation and demagnetisation characteristics. View chapterExplore book Read full chapter URL: Book2012, Functional Materials for Sustainable Energy ApplicationsI.R. Harris, G.W. Jewell Chapter Supplement 2 1989, Comprehensive Polymer Science and SupplementsFabrizio Parodi 19.3.1.3 Ferroelectrics and ferromagnets Conceptually, ferroelectricity and ferromagnetism are two almost equivalent types of response to electric and magnetic fields, respectively. Ferroelectrics82–86 and ferromag-nets57–59, 87–89 are crystalline materials achieving strong, partially to almost entirely permanent electrization or magnetization under the influence of electric or magnetic fields, respectively, even of modest intensity. Such materials exhibit abnormally high values of χe, s and χm, s (and thus of εs and μs) which depend in a complicated manner on temperature and on electric or magnetic field strength, as well as on their previous electrization or magnetization history, according to their hysteretic behavior, expressed by well-known electric or magnetic polarization-depolarization-reverse polarization loops. Polarization phenomena in ferroelectrics and ferromagnets are based on crystal domains of isooriented permanent atomic, ionic, or molecular dipoles (i.e. of spontaneously and permanently polarized material) called Weiss domains in ferromagnets, with linear dimensions of the order of microns, up to millimeters in ferroelectrics. Two polarization mechanisms may be involved: (a) enlargement of domains already aligned with the external field, by displacement of domain boundaries (or Bloch walls), in weak fields; (b) rotation of domain dipoles in the field direction, in strong fields. Due to such mechanisms, the polarization of ferroelectrics and ferromagnets shows marked saturation effects at certain field strength values, above which P or M cannot increase further. Starting from the depolarized state, D or B increases with the applied field strength according to a sigmoidal curve, which is the locus of the E–D or E–D limits of hysteresis loops, up to saturation. Accordingly, εs or μs increases from an initial value for E or H → 0 (εs, i and μs, i, respectively), reaches a maximum, and then decreases again to level off at a constant value at saturation. In ferromagnets, whose μs normally decreases with temperature, the dependence of B and μs on the magnetic field strength at constant temperature is expressed, for weak fields, by the simple Rayleigh law (equations (30)). (30a) (30b) Inorganic ferroelectrics display εs, i values from 102 up to 104 and over, while μs, i of ferromagnets, commonly in the range 102 to 104, can reach values of the order of 105. Values of μs at saturation are of the order of 103 to 106. As the temperature is increased, at the critical Curie point TC, variable from material to material, any ferroelectric or ferromagnet undergoes a sharp transition to paraelectric or paramagnetic behavior. Above TC, εs or μs decays rapidly with temperature. Whereas ferromagnetic behavior pertains almost exclusively to inorganic substances, ferroelectric properties are exhibited by such a rather common organic material as poly(vinylidene fluoride) (PVDF), a semicrystalline polymer whose ferroelectric characteristics have been studied since 1969.76, 90 At room temperature it shows an εs, i value (in the range 10–20) 2.5–5-fold the values known for normal paraelectric polymers, and maximum εs values of 50–60; its TC (above the melting point) is extrapolated to be ∼200 °C. Copolymers of vinylidene fluoride with up to about 50% mol of tri- or tetrafluoroethylene are also ferroelectric, with TC values ranging between room temperature and their melting points. View chapterExplore book Read full chapter URL: Reference work1989, Comprehensive Polymer Science and SupplementsFabrizio Parodi Chapter Ferromagnetism 2003, Encyclopedia of Physical Science and Technology (Third Edition)H.R. Khan VI.A Ferromagnetic Order In a ferromagnetic material, the individual magnetic moments are ordered in parallel arangement, as shown in Fig. 3. A ferromagnet possesses a magnetic moment even in the absence of an externally applied magnetic field, and this spontaneous magnetic moment is also called the saturation moment. When a ferromagnet is heated, the parallel arrangement disappears above the Curie temperature. The magnetic susceptibility of a ferromagnetic material at temperatures close to the Curie temperature is related to the temperature by which is also shown in Fig. 9. View chapterExplore book Read full chapter URL: Reference work2003, Encyclopedia of Physical Science and Technology (Third Edition)H.R. Khan Chapter Theoretical studies of implanted muons in organic magnets 1996, C,H,N and O in Si and Characterization and Simulation of Materials and ProcessesR.M. Valladares, ... A.J. Fisher Abstract Wholly organic ferromagnets are a new class of materials of great potential interest, but the early examples exhibit only very weak magnetic order at very low temperatures (M. Tamura et al., Chem. Phys. Lett., 186 (1991) 401). In these circumstances, muon spin rotation (μSR) is among the most sensitive sources of information about the magnetic structure and interactions on a molecular scale (F.L. Pratt et al., Synthetic Metals, 61 (1993) 171). However, interpretation of the μSR data is complicated because of the large number of binding sites made possible by the relatively complicated molecular structure. We have performed semi-empirical and ab initio density functional calculations of the electronic and molecular structure of muonium incorporated into the organic ferromagnets para-nitrophenyl α-nitronyl nitroxide (p-NPNN), 3-quinolyl nitronyl nitroxide (3-QNNN) and para-pyridyl nitronyl nitroxide (p-PYNN). We find evidence for a wide variety of possible muon binding sites including sites in the conjugated ring systems of the materials as well as in the nitronyl nitroxide group. Our calculations also suggest the formation of local triplet electronic states near the muon within the spin-1/2 magnetic system. View chapterExplore book Read full chapter URL: Book1996, C,H,N and O in Si and Characterization and Simulation of Materials and ProcessesR.M. Valladares, ... A.J. Fisher Chapter Ferrimagnetic Heterostructures for Applications in Magnetic Recording 2018, Novel Magnetic NanostructuresFlorin Radu, Jaime Sánchez-Barriga 9.2 Ferrimagnetism The ferromagnetic, antiferromagnetic, and ferrimagnetic traits of thin films are part of the same class of magnetic ordering based on the exchange interaction, which couples the magnetic spins of the lattice leading to preferred orientations of individual magnetic moments along certain direction, which defines the anisotropy axis. While all three cases of magnetic ordering are characterized by an onset of spontaneous sublattice magnetization at a certain temperature, that is, Curie temperature TC for ferromagnets and ferrimagnets and Néel temperature for antiferromagnets, they differ in some essential general aspects as depicted in Fig. 9.9. For instance, a ferromagnet exhibits a macroscopic magnetization in all the temperature range below the ordering temperature, whereas the magnetization of the sublattices in an antiferromagnet is fully compensated for all temperatures below the Néel temperature. For the ferrimagnetic state, the total magnetization can be uncompensated as similar to the ferromagnets but it can also be fully compensated for one certain temperature, depending on the stoichiometry of the constituent elements. Moreover, the ferromagnet can be easily aligned with external magnetic fields, whereas for aligning the magnetic moments of an antiferromagnet, very large magnetic fields are required, generally of the order of up to hundreds of Teslas. In these very high external magnetic fields, the antiferromagnet can be driven through spin-flip and spin-flop transitions when the magnetic field is applied along the anisotropy axis, or its total magnetization can be linearly increased as a function of the external magnetic field when applied perpendicular to the anisotropy axis. A ferrimagnet can be easily aligned with external field similar to the ferromagnet, and for higher magnetic fields spin-flip and spin-flop transitions can also occur leading to a boost of the total magnetization value. Moreover, differently from antiferromagnets, a linear change of magnetization as a function of the magnetic field cannot be achieved outside the magnetic compensation. The ferromagnetic materials are extensively used for applications due to their high magnetization, low anisotropy, relatively high polarization at the Fermi energy, and abundance in nature. Nevertheless, optimization and control of their properties are by now somewhat limited. For instance, a perpendicular magnetic orientation is difficult to be maintained for single films because the demagnetization fields which are directed against the magnetization direction are much stronger in the perpendicular direction as compared to the ones which are virtually missing for the in-plane configuration. Stabilizing an out-of-plane anisotropy can be achieved for multilayers systems like Co/Pt or Co/Pd by using thin Co and Pd layers. This is also possible for ultrathin magnetic films with optimized interface anisotropy through Dzyaloshinskii-Moriya interaction, interlayer coupling, exchange bias effect, and strain-induced interactions. Also, the magnetization can be tuned at room temperature by the right choice of materials or by doping. The strength of the anisotropy is generally rather low because the orbital moments are nearly quenched, leaving the crystal field splitting as the major source of anisotropy. As such, an intrinsic control of anisotropy is rather limited. In Table 9.1 we show the orbital and the spin moment of Fe, Co, and Ni. The orbital moments are below 0.2 μB which amounts to about 10% of the full moment. Table 9.1. Spin and orbital moments for Fe, Co, and Ni \| \| \| \| --- \| Empty Cell \| Spin Moment (μB) \| Orbital Moment (μB) \| \| Fe \| 2.2 \| 0.07 \| \| Co \| 1.7 \| 0.11 \| \| Ni \| 0.6 \| 0.05 \| The antiferromagnetic materials, instead, exhibit a vanishing macroscopic magnetization and generally high anisotropy. Therefore, the control of the magnetic states requires very high magnetic fields of order of several tens of Teslas, field cooling through the ordering temperature in high magnetic fields, or field cooling in the proximity of ferromagnetic materials. Nevertheless, antiferromagnetic materials are an integral part of spintronic devices because they can mediate magnetic stability through induced anisotropies in ferromagnetic materials. More recently, it has been shown that antiferromagnets can also be functionalized by making use of spin currents . An alternative route for controlling magnetic properties of materials can be achieved by combining ferromagnetic transition metals (TM) with rare-earth (RE) elements to form ferrimagnetic alloys. The RE materials have specific characteristics which are of high interest in this respect. In Table 9.2 we show the magnetic moments of the lanthanide series compiled and updated based on Refs. [171, 172]. The magnetism of RE materials is largely given by the 4f shells. These shells are localized closer to the atom, at about 10% of the atomic radii, being outwards screened by the d and s shells. As such the magnetic moment of the RE is more localized , with its single-ion anisotropy caused by the electrostatic interaction of the nonspherical charge distribution of the 4f electron with the local electric field of the surrounding atoms. This implies that the SOC is the main source of anisotropy, with crystal fields considered as a perturbation . Table 9.2. Orbital, spin, total, and experimental moments for the trivalent rare-earth elements \| Ion 3+ \| GS \| NE \| L \| S \| J \| g \| μl (μB) \| μs (μB) \| μt (μB) \| μexp (μB) \| --- --- --- --- --- \| La \| 1S0 \| 0 \| 0 \| 0 \| 0 \| – \| – \| – \| – \| 0 \| \| Ce \| \| 1 \| 3 \| 1/2 \| 5/2 \| 6/7 \| 3.381 \| –0.845 \| 2.535 \| 2.3–2.5 \| \| Pr \| 3H4 \| 2 \| 5 \| 1 \| 4 \| 4/5 \| 5.367 \| –1.789 \| 3.578 \| 3.4–3.6 \| \| Nd \| \| 3 \| 6 \| 3/2 \| 9/2 \| 8/11 \| 6.332 \| –2.714 \| 3.618 \| 3.5–3.6 \| \| Pm \| 5I4 \| 4 \| 6 \| 2 \| 4 \| 3/5 \| 6.261 \| –3.578 \| 2.683 \| – \| \| Sm \| \| 5 \| 5 \| 5/2 \| 5/2 \| 2/7 \| 5.071 \| –4.226 \| 0.845 \| 1.4–1.7 \| \| Eu \| 7F0 \| 6 \| 3 \| 3 \| 0 \| – \| 3.464 \| –3.464 \| 0 \| 3.3–3.5 \| \| Gd \| \| 7 \| 0 \| 7/2 \| 7/2 \| 2 \| 0 \| 7.937 \| 7.937 \| 7.9–8.0 \| \| Tb \| 7F6 \| 8 \| 3 \| 3 \| 6 \| 3/2 \| 3.240 \| 6.481 \| 9.721 \| 9.5–9.8 \| \| Dy \| \| 9 \| 5 \| 5/2 \| 15/2 \| 4/3 \| 5.323 \| 5.323 \| 10.646 \| 10.4–10.6 \| \| Ho \| 5I8 \| 10 \| 6 \| 2 \| 8 \| 5/4 \| 6.364 \| 4.243 \| 10.607 \| 10.4–10.7 \| \| Er \| \| 11 \| 6 \| 3/2 \| 15/2 \| 6/5 \| 6.387 \| 3.194 \| 9.581 \| 9.4–9.6 \| \| Tm \| 3H6 \| 12 \| 5 \| 1 \| 6 \| 7/6 \| 5.401 \| 2.160 \| 7.561 \| 7.1–7.6 \| \| Yb \| \| 13 \| 3 \| 1/2 \| 7/2 \| 8/7 \| 3.402 \| 1.134 \| 4.536 \| 4.3–4.9 \| \| Lu \| 1S0 \| 14 \| 0 \| 0 \| 0 \| – \| – \| – \| – \| 0 \| Notes: The values have been calculated based on Hund's rules as described in Ref. . GS and NE refer to ground state and number of unpaired electrons, respectively. The experimental moments are reproduced from Lanthanide Magnetism, Available from: Accessed 24 July 2017. The exchange interaction between the localized moment of the RE with the 3d itinerant magnetic moments of the TM was suggested to be mediated by the d-shell of the RE. The electron spin on the f-shell creates a positive local moment on the d shell through an ordinary exchange which further couples to the d-shell of the TM. For Fe, Co, and Ni, the exchange interaction is negative leading to an antiferromagnetic ground state of these TM and RE ions. In Table 9.2 we show the calculated spin and orbital moments of the RE elements. This is instructive when choosing the material for achieving certain properties of the RE-TM ferrimagnets. For instance, the orbital moments are quite high in most cases except for the Gd ion which is expected to exhibit a vanishing orbital moment. As such, films with low anisotropy can be easily achieved by combining Gd with the TM. Note that divalent Eu (ground state: ) exhibits also a pure spin moment (L = 0, S = 7/2), therefore magnetic anisotropy of Eu2+-based ferrimagnets and antiferromagnets should exhibit soft magnetic properties. Instead, for films which are required to exhibit high anisotropy, the RE elements that have high orbital moments, like Dy, Tb, Sm, Nd, are preferred . An exception is the Terfenol-D (TbxDy1−xFe2), a material which in spite of containing Dy and Tb elements exhibits soft magnetic properties, probably due to a high degree of oxidation . Another interesting variable is the total moment of the RE. This can also be tuned by the right choice of elements. For the first part of the series (Ce to Eu), the spin and orbital moments are opposite oriented leading to a reduced total moment, whereas for the second part of the series (Gd to Yb), the orbital and the spin moments are parallel oriented leading to a higher moment for the RE ion. This is a consequence of the Hund's rules, which dictate the occupation of the available state in the 4f shell. One interesting element in this respect is trivalent Eu, which is expected to exhibit equal spin and orbital moments of opposite sign in the ground state. This leads to an interesting paradox, namely the existence of an ordered magnetic material, which exhibits a zero total magnetic moment, so macroscopically it appears as being nonmagnetic. This situation has been actually achieved on an Sm-based material, which behaves intrinsically similar to a ferrimagnet due to the compensation of the spin and orbital moments . The stoichiometry of the chosen material plays also a crucial role in opening an effective and easy way to tune the total magnetization of the RE-TM alloy for a certain temperature, usually close to room temperature. Due to the antiparallel orientation of the RE and TM ions, the total magnetization can vanish at temperatures below the ordering temperature when the relative stoichiometry of the RE-TM leads to an equal amount of net magnetic moments for each sublattice. This is called magnetization compensation temperature. View chapterExplore book Read full chapter URL: Book2018, Novel Magnetic NanostructuresFlorin Radu, Jaime Sánchez-Barriga Chapter Exchange and Anisotropy 1998, Hysteresis in MagnetismGiorgio Bertotti 5.1 EXCHANGE We begin this section with a brief discussion of paramagnetism, which is the natural basis for the description of ferromagnetism. In fact, a ferromagnet can be imagined as a paramagnet where a strong coupling, described by Weiss molecular field, exists between the magnetic moments. The molecular field approach is not able to accurately reproduce all the details of the behavior of ferromagnets below the Curie point, but it is of invaluable importance in giving a simple and at the same time deep physical interpretation of the existence of spontaneous magnetization, and in introducing one of the building blocks necessary to the study of magnetic domains and hysteresis effects. 5.1.1 Paramagnetism Some aspects of paramagnetism were briefly introduced in Section 1.1.1 and Section 4.1.3. The physical assumptions defining an ideal paramagnet (Fig. 5.1) can be summarized as follows. (i) : The paramagnet is described as an assembly of identical permanent magnetic moments mi of strength m, subject to the action of the external field Ha. Each moment can be treated as a classical elementary moment of the type discussed in Section 3.1.3, or as a quantum moment. The qualitative behavior is the same in both cases, though the quantitative details may differ. We shall use quantum rules for spin-½ moments to calculate the magnetization, mainly because in this way one obtains simpler mathematical expressions. This is a minor exception to our program of avoiding any consideration of quantum aspects, which, anyhow, will not affect the final results in any substantial way. (ii) : There is no interaction between the moments. Even the magnetostatic dipolar coupling is neglected. (iii) : Each moment has a potential energy –μ0mi · Ha in the external field (Eq. 3.77). Due to this energy, each moment tends to stay parallel to Ha. (iv) : Thermal agitation randomly shakes each moment and tends to destroy any preferential alignment to the field. The thermodynamic behavior of a system with these properties results from the competition between the field and the thermal coupling. Let us first consider the case where the moments are classical vectors. Because there is no interaction between the moments, one can study one moment at a time and then sum up the result over all moments. The thermodynamic information is all contained in the single-moment partition function (5.1) where the dimensionless parameter a is given by the expression (5.2) In Eq. (5.1), θ is the angle between the moment and the field, and the sum over all microstates amounts to an integral over all possible moment orientations with respect to the field. The total partition function Z associated with a certain number N of independent moments is just the product of the single-moment functions, Z = (Zm)N. According to statistical thermodynamics, the free energy is given by G(Ha,T) = –kBT lnZ = –N kBT lnZm, that is, (5.3) The average magnetization M is given by Eq. (4.43), (5.4) where ΔV is the volume occupied by the moments. By carrying out the field derivative, one finds (5.5) where (5.6) is the maximum possible magnetization, obtained when the moments are all perfectly aligned. The function L(a) is called the Langevin function. Its behavior as a function of Ha/T is shown in Fig. 5.2. Let us now consider the quantum description of the problem. Paramagnetism is a phenomenon in which the quantum nature of magnetic moments becomes clearly manifest, and it is only after quantum considerations are introduced that a satisfactory agreement with experiments is found. Nonetheless, we will abandon quantum ideas soon after this section. Once a suitable expression for the average magnetization M of the system is derived, we will forget about the classical or quantum origin of that result, and we will use it as the macroscopic starting point for the discussion of molecular field effects. Let us then consider the case where the moments are quantum moments, associated with spin ½. Only two states are then possible, in which the magnetic moment component along the field is either +m or –m. Equation (5.1) through Eq. (5.5) are modified as follows: (5.7) (5.8) (5.9) The behavior of Eq. (5.9) is shown in Fig. 5.2. Because th(a) ≅ a when a is small, we see that, at low fields, the magnetization is proportional to the field, M = χHa, and the susceptibility is given by (5.10) The inverse proportionality with temperature expresses the Curie law. The classical result (Eq. 5.5) is qualitatively similar, but the low field susceptibility is different, because L(a) ≅ a/3 for small a. Note, however, that the quantity m appearing in the two expressions has a different physical meaning, because in the classical case it represents the modulus of the magnetic moment, whereas in the quantum case it represents its component along the field. Saturation sets in when all moments begin to be substantially aligned to the field. This takes place at fields Ha ~ kBT/μ0m. It is interesting to make some order-of-magnitude estimate. The natural atomic unit of magnetic moments is the Bohr magneton, μB = hqe/4πme ≅ 9.27 10−24 Am2 (see Section 3.2.1). By taking m ~ μB and μ0M0 ~ 2 T, we find, at room temperature, where kBT ≅ 4 10− 21 J, that, χ ~ 5 10− 3 and kBT/μ0m ~ 4 108 Am−1. Thermal agitation is very effective at room temperature and substantial moment alignment can be obtained only by applying extremely high fields. An account of the quantum theory of paramagnetism is outside the scope of this book. We simply recall that, for spins J greater than ½, instead of Eq. (5.9) the more complicated expression M = M0BJ(a) is found, where BJ(a) is the Brillouin function (5.11) For J = ½, Eq. (5.9) is recovered. Conversely, the classical expression, Eq. (5.5), is approached in the limit J → ∞. Equation (5.9) will be assumed in the next section as the starting point for the macroscopic description of spontaneous magnetization. 5.1.2 Weiss molecular field We have seen that enormous fields, of the order of 108 Am−1, are required to produce substantial moment alignment in a paramagnet. This is in striking contradiction to the observation that certain materials, which are paramagnetic at high temperatures, at room temperature can exhibit a magnetization of the order of 1 T under fields as low as a few tens of ampères per meter. Weiss molecular field hypothesis resolves this difficulty by describing a ferromagnet as a paramagnet where one of the four assumptions listed in the previous section no longer holds. In a ferromagnet, the moments are by no means uncoupled. There exists a strong interaction field, which favors collective alignment of the moments along a common direction. As the mean degree of moment alignment is measured by the magnetization M, it seems reasonable, as a first approximation, to postulate that the interaction field should be proportional to M itself. This interaction field, named molecular field, thus takes the form (5.12) where the dimensionless constant NW measures the strength of the interaction. Under the presence of HW, each moment experiences an effective field that is the sum of the applied field and of the molecular field. This is described by modifying Eq. (5.9) as follows: (5.13) According to Eq. (5.13), the presence of the molecular field introduces a positive feedback mechanism, because a nonzero magnetization gives rise to a field that favors a further increase of the magnetization itself. We can use Eq. (5.13) to calculate, as before, the low field susceptibility. By expanding th(x) to first order, th(x) ≅ x, we again obtain M = χHa, but now (5.14) where the characteristic temperature (5.15) is the Curie temperature. Equation (5.14) is known as the Curie-Weiss law. The comparison with Eq. (5.10) shows that the only difference caused by the molecular field is the appearance of the additional temperature Tc in the denominator. When T → Tc from higher temperatures, the susceptibility tends to infinity. In addition, the law certainly fails for T < Tc as a large negative susceptibility is a sure indication of instability. Something important occurs at T = Tc. To analyze the problem in more detail, let us simplify the notation by introducing appropriate dimensionless variables: (5.16) Equation (5.13) becomes (5.17) This equation expresses in implicit form the equation of state x(ha, t) for the system. Let us first discuss its meaning when no applied field is present. The magnetization then follows the law (5.18) The solution of this equation for x as a function of t describes the state of the system under no external field. In order to visualize the solutions of Eq. (5.18), one can separately plot the two members of the equation and consider the points of intersection. The result is shown in Fig. 5.3. When t > 0, i.e., T > Tc, there is only one solution at x = 0. Conversely, three solutions, at x = 0 and at x = ± xs(t) (i.e., M = ± Ms(T)), exist for t < 0 (i.e., T < Tc). As we shall see shortly, the solution x = 0 is unstable and should not be considered. The other two solutions represent states where the system spontaneously acquires the nonzero magnetization Ms(T) even if no field is applied. Tc is just the temperature at which the positive feedback operated by the molecular field overcomes thermal agitation and drives the system to a globally ordered state. Ms(T) is called spontaneous magnetization or saturation magnetization. Its temperature dependence, as predicted by Eq. (5.18), is shown in Fig. 5.4. The molecular field approach operates on macroscopic, phenomenological grounds. The average magnetization is the important parameter, the existence of the molecular field is postulated, and nowhere in the theory is there direct reference to the microscopic configurations attained by magnetic moments. A microscopic interpretation of the molecular field can be worked out only in the framework of quantum mechanics, in terms of the so-called exchange interaction. The idea is that the overall antisymmetric character of electronic states gives rise to a coupling between the spin momentum and the wave function in real space. In other words, if two electrons carry parallel spins (which corresponds to a symmetric spin wave function), they cannot stay close to each other (which is the typical property of an antisymmetric two-electron wave function in real space). The fact that they never get close reduces the average energy associated with their electrostatic interaction. It is this reduction that favors the parallel spin configuration. Thus, in the end, exchange interaction, the basis of ferromagnetism, turns out to be an electrostatic effect. This schematic picture is nothing more than a starting point. The real situation is much more complex. One has to deal not with two, but with a macroscopic number of electrons, and one has to decide whether the electrons contributing to the magnetization are localized around single ions or itinerant over large distances. We shall not go into the details of these descriptions. We only recall the fact that these problems are often described in terms of the Heisenberg Hamiltonian (5.19) where Si is the spin angular momentum of the ion located at the ith site of some lattice, and the exchange integral Jij (positive in a ferromagnet) measures the strength of the exchange coupling between the moments i and j. Exchange is a short-range interaction, which falls off rapidly with distance, and it is often a perfectly acceptable approximation to truncate the summation in Eq. (5.19) to nearest neighbors. When Jij > 0, the minimum energy configuration is the one where all spins are aligned to each other. The energy increases above this ground level whenever some mis-aligmnent of neighboring spins occurs. A final point should be mentioned, which is not evident from the simplified description just given. Spin ordering at low temperatures does not necessarily reduce to parallel alignment. Depending on the lattice symmetry and the value of the exchange integral, more complex ordered structures may be formed, where one can identify several sublattices, such that the spins belonging to the same sublattice are aligned, but the magnetization of different sublattices points along different directions. In this context, the word ferromagnetism describes the case where Ms is due to the parallel alignment of identical moments on one lattice, and ferrimagnetism the one where Ms results from the vector sum of competing partial contributions coming from several sublattices. This competition can lead to perfect cancellation of the partial contributions. One then has antiferromagnetism, characterized by long-range spin ordering but no spontaneous magnetization. As already mentioned in Section 4.1.3, in this book we generically use the expression magnetic material to refer to any system displaying spontaneous magnetization and hysteresis, with no further distinction. 5.1.3 Energetic aspects We said that, of the three solutions of Eq. (5.18) for T < Tc, the one giving x = 0 is actually unstable and should not be considered. This aspect is better clarified by analyzing the problem from the energy viewpoint. Let us reconsider Eq. (5.17). By inverting the hyperbolic tangent, one can rewrite it in the form (5.20) According to the approach presented in Section 2.1.4, the stable or metastable states of a system are local minima of its Landau free energy GL, satisfying the condition ∂GL/∂X = 0. Considering that GL = F(X) – HX, this condition can also be written as H = ∂F/∂X. This is just the form taken by Eq. (5.20), with H → ha, X → x. By integrating Eq. (5.20) with respect to x, we can thus obtain the dimensionless energy of the system. We shall denote this energy by gEX to remind us of its exchange origin. The relation of gEX to the energy GEX expressed in physical units is gEX = GEX /μ0NWM02ΔV. The integration of Eq. (5.20) with respect to x gives (5.21) The first term, proportional to 1 + t = T/Tc, describes the effect of thermal agitation. It is a function of x with positive curvature, favoring the zero-magnetization state x = 0. Conversely, the second term, –x2/2, has negative curvature and favors states where x ≠ 0. This is the effect of the molecular field. We can appreciate the consequences of this competition by making a Taylor expansion of Eq. (5.21) with respect to x, and by keeping the lowest-order terms. One finds (5.22) We recognize the cusp catastrophe discussed in Section 2.2.2. t = 0, i.e. T = Tc, is the critical point where a qualitative change in the potential takes place, as shown in Fig. 5.5. At the Curie point, the x = 0 state becomes unstable and the system spontaneously acquires a nonzero magnetization. The behavior of gEX around the Curie point given by Eq. (5.22) is also interesting in another respect. Figure 5.4 shows the solutions of Eq. (5.17) under nonzero field. The presence of the field destroys the magnetization singularity at the Curie temperature. This raises the problem of how one should measure the Curie temperature in practice. In fact, Tc marks a qualitative change in the spontaneous magnetization Ms(T), but in order to measure Ms one needs to apply magnetic fields large enough to sweep away magnetic domains and this just destroys the singularity that one aims at determining. A possible solution is to carry out measurements under different fields, and then to attempt some extrapolation of the results to zero field. Here is where Eq. (5.22) comes into play. In fact, the equilibrium condition ∂gEX/∂x = 0, when applied to Eq. (5.22), gives (5.23) By dividing both members by x and by expressing the result in physical units through Eq. (5.16), one finds (5.24) Equation (5.24) shows that, if we measure the saturation magnetization Ms(Ha,T) under different fields and we represent Ha/Ms as a function of Ms2, we should get a straight line. In addition, the intersection of this line with the field axis should be positive if T > Tc, and negative in the opposite case. Through this change of sign, one has a sensitive way to estimate where the measurement temperature is with respect to the Curie point. Representations based on this idea are known as Arrott plots, and are of great value in Curie point studies. The energy per unit volume UW/ΔV associated with the molecular field is represented by the term -x2/2 of Eq. (5.21). By going back to physical quantities, one finds (5.25) An energy term analogous to Eq. (5.25) was encountered in Section 4.1.1 (Eq. 4.5), when we discussed magnetostatic energy. The molecular field energy is greater than the magnetostatic one by the factor 3NW. We can estimate NW from the experimental knowledge of the Curie point. From Eq. (5.15), we have that NW = kBTc/μ0mM0. In iron, where Tc ~ 103 K, by taking m ~ μB ~ 10−23 Am2 and μ0M0 ~ 2 T, we obtain 3NW ~ 2 103. The molecular field interaction is three orders of magnitude greater than the magnetostatic one. This confirms the fact that the M 2 term of Eq. (4.5) is in fact of negligible importance when ferromagnetic materials are considered. The free energy of Fig. 5.5 has the typical two-well structure discussed in Chapter 2. The presence of two equivalent minima points at the existence of two phases, a “magnetization up” and a “magnetization down” phase. All the considerations made in Section 2.2.3 about phase coexistence can be applied to the present case. But one should by no means jump to the conclusion that we have already at hand a realistic framework for the interpretation of magnetic domains. In this respect, the approach that we have discussed so far is inadequate and unrealistic. The point is that Ha and M have been treated everywhere just as numbers, whereas they are vectors. At the beginning, the reason for this was that we were considering the magnetization component along the field direction. However, later we focused attention on the case where no field is present, and this raises the question of what magnetization “up” or “down” might mean under such circumstances. The two-well potential of Fig. 5.5 describes a situation where, due to some anisotropy of unspecified origin, the magnetization is forced to lie along a certain fixed direction, and we are estimating the energy involved in passing from magnetization “up” to “down” along that direction. This energy is huge. If we apply the analysis of two-well potentials presented in Chapter 2 to the present case, we come to the conclusion that the field that we must apply to reverse the magnetization from “up” to “down” is of the order of the molecular field, HW = NWM. With NW ~ 103 and μ0M ~ 1 T, this gives fields of the order of 109 Am−1. Certainly, it is a prediction grossly out of scale with respect to the hysteresis loops observed in reality. The point is that the anisotropy just mentioned is an artifact of the description, useful to keep the discussion simple but unphysical. Exchange interactions, as described for instance by the Heisenberg Hamiltonian, Eq. (5.19), are isotropic in space. There is no a priori favored direction for the spontaneous magnetization, but only that particular direction selected by the system when the spontaneous magnetization sets in. To illustrate the consequences of this fact, it is sufficient to pass from one to two dimensions. The equation analogous to Eq. (5.21) is (5.26) where ha and x are now two-dimensional vectors and x = \|x\|. The behavior of the potential under zero field is shown in the upper part of Fig. 5.6. There is a continuous ring of equivalent ground states (it would be a spherical shell in three dimensions). This reflects the rotational symmetry of the problem. One says that the symmetry is continuous, because one can move through the whole set of ground states with continuity. When the system spontaneously acquires a magnetization in the way shown in Fig. 5.6, it selects one among the available ground states, thus destroying the original symmetry of the problem. In field and phase transition theory, this mechanism is often referred to as spontaneous symmetry breaking. When a nonzero external field is applied, the energy surface is modified as shown in the lower part of Fig. 5.6. The existence of continuous symmetry in the zero-field potential plays a crucial role, because it permits the system to adjust its state to the field without having to overcome any energy barrier. The path “around the hill” shown in the figure is just coherent and simultaneous rotation of all magnetic moments toward the field direction. This does not imply any change in the modulus of spontaneous magnetization and in exchange interactions, and it can take place under fields no matter how small. On the contrary, the path “above the hill,” which was the only one available in one dimension (Fig. 5.5), is a path where the magnetization modulus must change with continuity from its initial value to the opposite of it. This can only be obtained by finely mixing magnetic moments pointing along different directions, a process that entails a huge amount of exchange energy. This path is never followed. The curvature of the energy profile in the radial M direction, where the strength of the spontaneous magnetization changes, plays a role when one wants to estimate the joint field-temperature dependence Ms(Ha, T). Let us consider the spontaneous magnetization Ms(0, T) under zero field. x = xs(0, T) = Ms(0, T)/M0 is then the solution of Eq. (5.18). Now let us apply a field parallel to the magnetization. The field favors further alignment of the magnetic moments and thus produces a forced increase of Ms. This increase is evaluated by equating the external field force to the springlike counterforce arising from the curvature of the potential of Eq. (5.21) around the minimum at x = xs(0, T). We have (5.27) The curvature, calculated from Eq. (5.21), is (5.28) where we have taken into account that xs(0, t) satisfies Eq. (5.18). At temperatures T << Tc, xs ≅ 1 and xs2 + t ≅ 1 + t = T/Tc. By using this approximation when Eq. (5.28) is inserted into Eq. (5.27), and by taking into account Eq. (5.16), one finds (5.29) As T << Tc, the second term of Eq. (5.29) is exponentially small and the function Ms(Ha, T) is thus a very weak function of Ha. In the following chapters, we shall usually neglect this dependence, by assuming that there exists a spontaneous magnetization of modulus Ms(T) independent of field. This is a perfectly acceptable approximation in most cases, provided the temperature is not too close to the Curie point. View chapterExplore book Read full chapter URL: Book1998, Hysteresis in MagnetismGiorgio Bertotti Chapter Multiferroic Oxide Thin Films 2018, Encyclopedia of Interfacial ChemistryM. Gich, Z. Ma Ferroic Materials As its name suggests, multiferroic materials are characterized by the simultaneous presence of more than one ferroic order.2 The practical interest of coexisting ferroic orders is the possibility of exploiting an eventual cross coupling between them. Ferroics are materials composed of regions, named domains, which in equilibrium are found in at least two possible orientation states. A typical ferroic, showing domains of different orientation states, separated by domain walls, is depicted in Fig. 1. The appearance of a ferroic order is in fact a phase transition: on cooling, the symmetry of a high-temperature nonferroic state is broken by the emergence of an order parameter. In ferromagnetic and ferroelectric materials, the better-known examples of ferroics, the order parameter coincides with the orientation states that define the different domains. In ferromagnets, the order parameter is the magnetization, M. Typically, within a given domain, M is either parallel or antiparallel to some specific crystallographic directions but can be switched by a magnetic field, H. In ferroelectric materials, the order parameter is the electric polarization, P, switchable by an electric field, E. In these examples, the order parameters are described by vectors that are normally found in their “up” or “down” ground states of equal energy and can be switched reversibly between them by applying an external field, provided that it exceeds the so-called coercive field. The latter is a consequence of hysteresis, a characteristic feature of ferroics that is related to changes in the domain configuration to minimize the energy of the system in response to applied fields. For instance, Fig. 2 illustrates the typical evolution of ferromagnetic domains under different magnetic fields: from an initial state A in the absence of magnetic field, which has zero magnetization due to the compensation of an equal number of “up” and “down” domains, a positive magnetic field favors the growth of “up” domains at expenses of “down” domains until a saturation magnetization Ms is achieved at E. Upon decreasing the field from state E, the system does not respond reversibly but goes instead to states F, G, and finally H, a state of nonzero magnetization at zero field due to the larger fraction of “up” domains compared with “down” domains. From state H, a magnetic field applied in the opposite direction will not immediately result in an opposite magnetization, and M will only be reversed for fields above the coercive field Hc at which the number of “up” and “down” domains is again compensated and hence M is zero. Such behavior is analogous for ferroelectrics and also for a third type of ferroic materials known as ferroelastics. The latter presents domains of different spontaneous strain, also known as twins, which are indeed different orientations of the same crystal structure, and those can be switched by the application of a mechanical stress. These three ferroic orders have been known for quite a long time, and their orientation states break symmetries of time reversal (in ferromagnets), spatial inversion (in ferroelectrics), and rotation (in ferroelastics). A fourth type of ferroic, ferrotoroidics, was lately considered and linked to an order parameter that could simultaneously break time-reversal and spatial-inversion symmetries. Ferrotoroidics are associated to domains of magnetic vortices (i.e., circular arrangements of magnetic moments)3, which would be switched by a toroidal field defined as the cross product of a magnetic and an electric field, but this has not been experimentally achieved so far. The four types of ferroic materials described earlier constitute the so-called primary ferroics characterized by the response of their domains to a single external field or force. Even though in the following we will only consider primary ferroics excepting ferrotoroidics, let us mention the existence of secondary ferroic orders, characterized by their response to more than one external field or higher-order field terms. The interested reader is referred to the work of Newnham in the Further Reading section. View chapterExplore book Read full chapter URL: Reference work2018, Encyclopedia of Interfacial ChemistryM. Gich, Z. Ma Chapter Bulk high temperature superconductor (HTS) materials 2012, High Temperature Superconductors (HTS) for Energy ApplicationsT. Coombs Ferrimagnetism/antiferromagnetism Ferrimagnetism differs from ferromagnetism in that there are two distinct magnetic sublattices. These, when ordered, have antiparallel spins but the spins are not equal. This results in a net magnetic moment. Ferrimagnets, like ferromagnets, have a Curie point. In an antiferromagnet the spins in the two sublattices are equal, resulting in no magnetic moment. View chapterExplore book Read full chapter URL: Book2012, High Temperature Superconductors (HTS) for Energy ApplicationsT. Coombs Related terms: Energy Engineering Superconductivity Graphene Tunnel Construction Magnetic Moment Magnetic Field Binary Digit Room Temperature Magnetoelectronics External Magnetic Field View all Topics
8777	https://www.geeksforgeeks.org/dsa/sum-divisors-1-n/	Sum of all divisors from 1 to n Last Updated : 14 Aug, 2023 Suggest changes 31 Likes Given a positive integer n. Find the value of ∑i=1i=nF(i) where function F(i) for number i be defined as the sum of all divisors of 'i'. Examples : 4 15F(1) = 1F(2) = 1 + 2 = 3F(3) = 1 + 3 = 4F(4) = 1 + 2 + 4 = 7ans = F(1) + F(2) + F(3) + F(4) = 1 + 3 + 4 + 7 = 15 5 21 Naive approach is to traverse for every number(1 to n), find all divisors and keep updating the sum with that divisor. See this to understand more. Try it on GfG Practice C++ ```` // C++ program to find sum of all // divisor of number up to 'n' include using namespace std; // Utility function to find sum of // all divisor of number up to 'n' int divisorSum(int n) { int sum = 0; for(int i = 1; i <= n; ++i) { // Find all divisors of i and add them for(int j = 1; j j <= i; ++j) { if (i % j == 0) { if (i / j == j) sum += j; else sum += j + i / j; } } } return sum; } // Driver code int main() { int n = 4; cout << " " << divisorSum(n) << endl; n = 5; cout << " " << divisorSum(n); return 0; } ```` // C++ program to find sum of all // C++ program to find sum of all // divisor of number up to 'n' // divisor of number up to 'n' ``` include#include ``` using namespace std; using namespace std // Utility function to find sum of // Utility function to find sum of // all divisor of number up to 'n' // all divisor of number up to 'n' int divisorSum(int n) int divisorSum int n { int sum = 0; int sum = 0 for(int i = 1; i <= n; ++i) for int i = 1 i<= n ++ i { ``` ``` // Find all divisors of i and add them // Find all divisors of i and add them for(int j = 1; j j <= i; ++j) for int j = 1 j j<= i ++ j { if (i % j == 0) if i% j == 0 { if (i / j == j) if i/ j == j sum += j; sum += j else else sum += j + i / j; sum += j + i/ j } } } return sum; return sum } // Driver code // Driver code int main() int main { int n = 4; int n = 4 cout << " " << divisorSum(n) << endl; cout<< " "<< divisorSum n<< endl ``` ``` n = 5; n = 5 cout << " " << divisorSum(n); cout<< " "<< divisorSum n ``` ``` return 0; return 0 } Java ```` // JAVA program to find sum of all // divisor of number up to 'n' import java.io.; class GFG { // Utility function to find sum of // all divisor of number up to 'n' static int divisorSum(int n) { int sum = 0; for (int i = 1; i <= n; ++i) { // Find all divisors of i // and add them for (int j = 1; j j <= i; ++j) { if (i % j == 0) { if (i / j == j) sum += j; else sum += j + i / j; } } } return sum; } // Driver code public static void main(String args[]) { int n = 4; System.out.println(divisorSum(n)); n = 5; System.out.println(divisorSum(n)); } } /This code is contributed by Nikita tiwari./ ```` Python3 ```` Python3 code to find sum of all divisor of number up to 'n' Utility function to find sum of all divisor of number up to 'n' def divisorSum( n ): sum = 0 for i in range(1, n + 1): # Find all divisors of i # and add them j = 1 while j j <= i: if i % j == 0: if i / j == j: sum += j else: sum += j + i / j j = j + 1 return int(sum) Driver code n = 4 print( divisorSum(n)) n = 5 print( divisorSum(n)) This code is contributed by "Sharad_Bhardwaj". ```` C# ```` // C# program to find sum of all // divisor of number up to 'n' using System; class GFG { // Utility function to find sum of // all divisor of number up to 'n' static int divisorSum(int n) { int sum = 0; for (int i = 1; i <= n; ++i) { // Find all divisors of i // and add them for (int j = 1; j j <= i; ++j) { if (i % j == 0) { if (i / j == j) sum += j; else sum += j + i / j; } } } return sum; } // Driver code public static void Main() { int n = 4; Console.WriteLine(divisorSum(n)); n = 5; Console.WriteLine(divisorSum(n)); } } /This code is contributed by vt_m./ ```` JavaScript ```` // Javascript program to find sum of all // divisor of number up to 'n' // Utility function to find sum of // all divisor of number up to 'n' function divisorSum(n) { let sum = 0; for (let i = 1; i <= n; ++i) { // Find all divisors of i // and add them for (let j = 1; j j <= i; ++j) { if (i % j == 0) { if (i / j == j) sum += j; else sum += j + i / j; } } } return sum; } // Driver code let n = 4; document.write(divisorSum(n) + "<br>"); n = 5; document.write(divisorSum(n) + "<br>"); // This code is contributed by _saurabh_jaiswal ```` PHP ```` php // PHP program to find sum of all // divisor of number up to 'n' // Utility function to find sum of // all divisor of number up to 'n' function divisorSum($n) { $sum = 0; for ($i = 1; $i <= $n; ++$i) { // Find all divisors of i // and add them for ($j = 1; $j $j <= $i; ++$j) { if ($i % $j == 0) { if ($i / $j == $j) $sum += $j; else $sum += $j + $i / $j; } } } return $sum; } // Driver code $n = 4; echo "\n", divisorSum($n), "\n"; $n = 5; echo divisorSum($n), "\n"; // This code is contributed by aj_36 ? ```` Output 15 21 Time complexity: O(n√(n))) Auxiliary space: O(1) Efficient approach is to observe the function and co-relate the pattern. For a given number n, every number from 1 to n contributes its presence up to the highest multiple less than n. For instance, Let n = 6, => F(1) + F(2) + F(3) + F(4) + F(5) + F(6) => 1 will occurs 6 times in F(1), F(2), F(3), F(4), F(5) and F(6) => 2 will occurs 3 times in F(2), F(4) and F(6) => 3 will occur 2 times in F(3) and F(6) => 4 will occur 1 times in F(4) => 5 will occur 1 times in F(5) => 6 will occur 1 times in F(6) From the above observation, it can easily be observed that number i is occurring only in their multiples less than or equal to n. Thus, we just need to find the count of multiples and then multiply it with i for full contribution in the final sum. It can easily be done in O(1) time by taking the floor of (n / i) and then multiply it with i for the sum. C++ ```` // C++ program to find sum of all // divisor of number up to 'n' include using namespace std; // Utility function to find sum of // all divisor of number up to 'n' int divisorSum(int n) { int sum = 0; for (int i = 1; i <= n; ++i) sum += (n / i) i; return sum; } // Driver code int main() { int n = 4; cout <<" "<< divisorSum(n)<<endl; n = 5; cout <<" "<< divisorSum(n)<< endl; return 0; } // This code is contributed by shivanisinghss2110 ```` // C++ program to find sum of all // C++ program to find sum of all // divisor of number up to 'n' // divisor of number up to 'n' ``` include#include ``` using namespace std; using namespace std // Utility function to find sum of // Utility function to find sum of // all divisor of number up to 'n' // all divisor of number up to 'n' int divisorSum(int n) int divisorSum int n { int sum = 0; int sum = 0 for (int i = 1; i <= n; ++i) for int i = 1 i<= n ++ i sum += (n / i) i; sum += n/ i i return sum; return sum } // Driver code // Driver code int main() int main { int n = 4; int n = 4 cout <<" "<< divisorSum(n)<<endl; cout<< " "<< divisorSum n<< endl n = 5; n = 5 cout <<" "<< divisorSum(n)<< endl; cout<< " "<< divisorSum n<< endl return 0; return 0 } // This code is contributed by shivanisinghss2110 // This code is contributed by shivanisinghss2110 C ```` // C program to find sum of all // divisor of number up to 'n' include // Utility function to find sum of // all divisor of number up to 'n' int divisorSum(int n) { int sum = 0; for (int i = 1; i <= n; ++i) sum += (n / i) i; return sum; } // Driver code int main() { int n = 4; printf("%d\n", divisorSum(n)); n = 5; printf("%d", divisorSum(n)); return 0; } ```` Java ```` // Java program to find sum of all // divisor of number up to 'n' import java.io.; class GFG { // Utility function to find sum of // all divisor of number up to 'n' static int divisorSum(int n) { int sum = 0; for (int i = 1; i <= n; ++i) sum += (n / i) i; return sum; } // Driver code public static void main(String args[]) { int n = 4; System.out.println(divisorSum(n)); n = 5; System.out.println(divisorSum(n)); } } /This code is contributed by Nikita Tiwari./ ```` Python3 ```` Python3 code to find sum of all divisor of number up to 'n' Utility function to find sum of all divisor of number up to 'n' def divisorSum( n ): sum = 0 for i in range(1, n + 1): sum += int(n / i) i return int(sum) Driver code n = 4 print( divisorSum(n)) n = 5 print( divisorSum(n)) This code is contributed by "Sharad_Bhardwaj". ```` C# ```` // C# program to find sum of all // divisor of number up to 'n' using System; class GFG { // Utility function to find sum of // all divisor of number up to 'n' static int divisorSum(int n) { int sum = 0; for (int i = 1; i <= n; ++i) sum += (n / i) i; return sum; } // Driver code public static void Main() { int n = 4; Console.WriteLine(divisorSum(n)); n = 5; Console.WriteLine(divisorSum(n)); } } /This code is contributed by vt_m./ ```` JavaScript ```` // Javascript program to find sum of all // divisor of number up to 'n' // Utility function to find sum of // all divisor of number up to 'n' function divisorSum(n) { let sum = 0; for (let i = 1; i <= n; ++i) sum += Math.floor(n / i) i; return sum; } // Driver code let n = 4; document.write(divisorSum(n) + ""); n = 5; document.write(divisorSum(n) + ""); // This code is contributed by _saurabh_jaiswal. ```` PHP ```` php // PHP program to find sum of all // divisor of number up to 'n' // Utility function to find sum of // all divisor of number up to 'n' function divisorSum( $n) { $sum = 0; for ( $i = 1; $i <= $n; ++$i) $sum += floor($n / $i) $i; return $sum; } // Driver code $n = 4; echo divisorSum($n),"\n"; $n = 5; echo divisorSum($n),"\n"; // This code is contributed by anuj_67. ? ```` Output 15 21 Time complexity: O(n) Auxiliary space: O(1) More efficient solution: We need to calculate ∑i=1i=N(i∗⌊N/i⌋) To evaluate the above expression in O(sqrt(N)) we make use of The Harmonic Lemma. Consider the harmonic sequence on integer division: {N/1, N/2, N/3, ..... ,N/N} The lemma states that the above sequence is non-increasing, and there are at most 2sqrt(N) different elements. Consider floor(N/i) = k. Thus, k <= N/i < k+1. From this we get largest = floor(N/k). Therefore, we can find a range of values of i for which floor(N/i) is constant. And using The Harmonic Lemma we know that will be at most 2sqrt(N) terms, thus we can calculate it programmatically in O(sqrt(N)) complexity. Consider the following example for better clarification. C++ ```` // C++ program to calculate sum of divisors // of numbers from 1 to N in O(sqrt(N)) complexity include using namespace std; define ll long long define mod 1000000007 / Function to calculate x^y using Modular exponentiation Refer to modular-exponentiation-power-in-modular-arithmetic/ / ll power(ll x, ll y, ll p) { // re x^y if p not specified // else (x^y)%p ll res = 1; x = x % p; while (y > 0) { if (y & 1) res = (res x) % p; y = y >> 1; x = (x x) % p; } return (res + p) % p; } // Function to find modular // inverse of a under modulo m // Assumption: m is prime ll modinv(ll x) { return power(x, mod - 2, mod); } // Function to calculate sum from 1 to n ll sum(ll n) { // sum 1 to n = (n(n+1))/2 ll retval = ((((n % mod) ((n + 1) % mod)) % mod) modinv(2)) % mod; return retval; } ll divisorSum(ll n) { ll l = 1; ll ans = 0; while (l <= n) { ll k = n / l; ll r = n / k; k %= mod; // For i=l to i=r, floor(n/i) will be k ans += ((sum(r) - sum(l - 1) % mod) k) % mod; // Since values can be very large // we need to take mod at every step ans %= mod; l = r + 1; } ans = ans % mod; // ans can be negative // for example n = 831367 ans would be -534577982 if (ans < 0){ return ans+mod; }else{ return ans; } } / Driver program to test above function / int main() { int n = 5; cout << "The sum of divisors of all numbers from 1 to " << n << " is: " \ << divisorSum(n) << '\n'; n = 14; cout << "The sum of divisors of all numbers from 1 to " << n << " is: " \ << divisorSum(n) << '\n'; } ```` // C++ program to calculate sum of divisors // C++ program to calculate sum of divisors // of numbers from 1 to N in O(sqrt(N)) complexity // of numbers from 1 to N in O(sqrt(N)) complexity ``` include #include ``` using namespace std; using namespace std ``` define ll long long #define ll long long ``` ``` define mod 1000000007 #define mod 1000000007 ``` / / Function to calculate x^y using Function to calculate x^y using Modular exponentiation Modular exponentiation Refer to Refer to modular-exponentiation-power-in-modular-arithmetic/ modular-exponentiation-power-in-modular-arithmetic/ / / ll power(ll x, ll y, ll p) ll power ll x ll y ll p { ``` ``` // re x^y if p not specified // re x^y if p not specified // else (x^y)%p // else (x^y)%p ll res = 1; ll res = 1 x = x % p; x = x% p while (y > 0) while y> 0 { if (y & 1) if y& 1 res = (res x) % p; res = res x% p y = y >> 1; y = y>> 1 x = (x x) % p; x = x x% p } return (res + p) % p; return res + p% p } // Function to find modular // Function to find modular // inverse of a under modulo m // inverse of a under modulo m // Assumption: m is prime // Assumption: m is prime ll modinv(ll x) ll modinv ll x { return power(x, mod - 2, mod); return power x mod - 2 mod } // Function to calculate sum from 1 to n // Function to calculate sum from 1 to n ll sum(ll n) ll sum ll n { // sum 1 to n = (n(n+1))/2 // sum 1 to n = (n(n+1))/2 ll retval = ((((n % mod) ((n + 1) % ll retval = n% mod n + 1% mod)) % mod) modinv(2)) % mod; mod% mod modinv 2% mod return retval; return retval } ll divisorSum(ll n) ll divisorSum ll n { ll l = 1; ll l = 1 ll ans = 0; ll ans = 0 while (l <= n) while l<= n { ll k = n / l; ll k = n/ l ll r = n / k; ll r = n/ k k %= mod; k%= mod ``` ``` // For i=l to i=r, floor(n/i) will be k // For i=l to i=r, floor(n/i) will be k ans += ((sum(r) - sum(l - 1) % ans += sum r - sum l - 1% mod) k) % mod; mod k% mod ``` ``` // Since values can be very large // Since values can be very large // we need to take mod at every step // we need to take mod at every step ans %= mod; ans%= mod l = r + 1; l = r + 1 } ans = ans % mod; ans = ans% mod // ans can be negative // ans can be negative // for example n = 831367 ans would be -534577982 // for example n = 831367 ans would be -534577982 if (ans < 0){if ans< 0 return ans+mod; return ans + mod }else{else return ans; return ans } } / Driver program to test above function / / Driver program to test above function / int main() int main { int n = 5; int n = 5 cout << "The sum of divisors of all numbers from 1 to " << n << " is: " \ cout<< "The sum of divisors of all numbers from 1 to "<< n<<" is: " \ << divisorSum(n) << '\n';<< divisorSum n<< '\n' n = 14; n = 14 cout << "The sum of divisors of all numbers from 1 to " << n << " is: " \ cout<< "The sum of divisors of all numbers from 1 to "<< n<<" is: " \ << divisorSum(n) << '\n';<< divisorSum n<< '\n' } Java ```` // Java program to calculate // sum of divisors of numbers // from 1 to N in O(sqrt(N)) // complexity import java.util.; class Main{ static int mod = 1000000007; / Function to calculate x^y using Modular exponentiation Refer to modular-exponentiation-power-in- modular-arithmetic/ / public static long power(long x, long y, long p) { // re x^y if p not specified // else (x^y)%p long res = 1; x = x % p; while (y > 0) { if ((y & 1) != 0) res = (res x) % p; y = y >> 1; x = (x x) % p; } return (res + p) % p; } // Function to find modular // inverse of a under modulo m // Assumption: m is prime public static long modinv(long x) { return power(x, mod - 2, mod); } // Function to calculate sum // from 1 to n public static long sum(long n) { // sum 1 to n = (n(n+1))/2 long retval = ((((n % mod) ((n + 1) % mod)) % mod) modinv(2)) % mod; return retval; } public static long divisorSum(long n) { long l = 1; long ans = 0; while (l <= n) { long k = n / l; long r = n / k; k %= mod; // For i=l to i=r, // floor(n/i) will be k ans += ((sum(r) - sum(l - 1) % mod) k) % mod; // Since values can be very // large we need to take mod // at every step ans %= mod; l = r + 1; } ans = ans % mod; return ans; } // Driver code public static void main(String[] args) { int n = 5; System.out.println("The sum of divisors of" + " all numbers from 1 to " + n + " is: " + divisorSum(n)); n = 14; System.out.println("The sum of divisors of all" + " numbers from 1 to " + n + " is: " + divisorSum(n)); } } // This code is contributed by divyeshrabadiya07 ```` Python3 ```` Python program to calculate sum of divisors of numbers from 1 to N in O(sqrt(N)) complexity mod = 1000000007; Function to calculate x^y using Modular exponentiation Refer to https:#www.geeksforgeeks.org/ modular-exponentiation-power-in- modular-arithmetic/ def power(x, y, p): # re x^y if p not specified # else (x^y)%p res = 1; x = x % p; while (y > 0): if ((y & 1) != 0): res = (res x) % p; y = y >> 1; x = (x x) % p; return (res + p) % p; Function to find modular inverse of a under modulo m Assumption: m is prime def modinv(x): return power(x, mod - 2, mod); Function to calculate sum from 1 to n def sum(n): # sum 1 to n = (n(n+1))/2 retval = ((((n % mod) ((n + 1) % mod)) % mod) modinv(2)) % mod; return retval; def divisorSum(n): l = 1; ans = 0; while (l <= n): k = n // l; r = n // k; k %= mod; # For i=l to i=r, # floor(n/i) will be k ans += ((sum(r) - sum(l - 1) % mod) k) % mod; # Since values can be very # large we need to take mod # at every step ans %= mod; l = r + 1; ans = ans % mod; return ans; Driver code if name == 'main': n = 5; print("The sum of divisors of all numbers from 1 to " , n , " is: " ,int( divisorSum(n))); n = 14; print("The sum of divisors of all numbers from 1 to ", n ," is: " , int(divisorSum(n))); This code contributed by aashish1995 Write ```` C# ```` // C# program to calculate // sum of divisors of numbers // from 1 to N in O(sqrt(N)) // complexity using System; class GFG{ static int mod = 1000000007; / Function to calculate x^y using Modular exponentiation Refer to modular-exponentiation-power-in- modular-arithmetic/ / static long power(long x, long y, long p) { // re x^y if p not specified // else (x^y)%p long res = 1; x = x % p; while (y > 0) { if ((y & 1) != 0) res = (res x) % p; y = y >> 1; x = (x x) % p; } return (res + p) % p; } // Function to find modular // inverse of a under modulo m // Assumption: m is prime static long modinv(long x) { return power(x, mod - 2, mod); } // Function to calculate sum // from 1 to n static long sum(long n) { // sum 1 to n = (n(n+1))/2 long retval = ((((n % mod) ((n + 1) % mod)) % mod) modinv(2)) % mod; return retval; } static long divisorSum(long n) { long l = 1; long ans = 0; while (l <= n) { long k = n / l; long r = n / k; k %= mod; // For i=l to i=r, // floor(n/i) will be k ans += ((sum(r) - sum(l - 1) % mod) k) % mod; // Since values can be very // large we need to take mod // at every step ans %= mod; l = r + 1; } ans = ans % mod; return ans; } // Driver code static void Main() { int n = 5; Console.WriteLine("The sum of divisors of" + " all numbers from 1 to " + n + " is: " + divisorSum(n)); n = 14; Console.WriteLine("The sum of divisors of all" + " numbers from 1 to " + n + " is: " + divisorSum(n)); } } // This code is contributed by divyesh072019 ```` JavaScript ```` // Javascript program to calculate // sum of divisors of numbers // from 1 to N in O(sqrt(N)) // complexity var mod = 10007; / Function to calculate x^y using Modular exponentiation Refer to modular-exponentiation-power-in- modular-arithmetic/ / function power(x, y, p) { // re x^y if p not specified // else (x^y)%p var res = 1; x = x % p; while (y > 0) { if ((y & 1) != 0) res = (res x) % p; y = y >> 1; x = (x x) % p; } return (res + p) % p; } // Function to find modular // inverse of a under modulo m // Assumption: m is prime function modinv(x) { return power(x, mod - 2, mod); } // Function to calculate sum // from 1 to n function sum(n) { // sum 1 to n = (n(n+1))/2 var retval = Math.floor((((n % mod) ((n + 1) % mod)) % mod) modinv(2)) % mod; return retval; } function divisorSum(n) { var l = 1; var ans = 0; while (l <= n) { var k = n / l; var r = n / k; k %= mod; // For i=l to i=r, // floor(n/i) will be k ans += Math.floor((sum(r) - sum(l - 1) % mod) k) % mod; // Since values can be very // large we need to take mod // at every step ans %= mod; l = r + 1; } ans = ans % mod; return ans; } // Driver code var n = 5; document.write("The sum of divisors of" + " all numbers from 1 to " + n + " is: " + divisorSum(n) +"<br>"); n = 14; document.write("The sum of divisors of all" + " numbers from 1 to " + n + " is: " + divisorSum(n)); // This code is contributed by shivanisinghss2110 ```` Output The sum of divisors of all numbers from 1 to 5 is: 21 The sum of divisors of all numbers from 1 to 14 is: 165 Time complexity: O(sqrt(N)) Auxiliary space: O(1) Another sqrt(n) approach: Anywhere division is used in the below article, it means integer division. Let's start with an example assume that n = 20, now let's see how each number from 1 to 20 appears as the factor of some other number. 1 : 1 1, 1 2, 1 3, 1 4..., 1 (20 / 1) 2 : 2 1, 2 2, 2 3, 2 4,..., 2 (20 / 2) 3 : 3 1, 3 2, 3 3, 3 4...., 3 (20 / 3) our goal is to add every number each time it appears as the factor of some other number. For example 3 appears as the factor of (3 1), (3 2), (3 3)..., (3 (20 / 3)). Now let's start from 1 and add 1 to our sum each time it appears and also we will add all the numbers that appeared with 1, we'll do the same thing with 2 and when we reach 3 we have already added 3 to our sum when it appeared with 1 and 2 so now we will only add 3 when it appears with numbers greater than 2 i.e. 3, 4, 5, 6 also we will add the numbers that appeared with 3 so we'll add 4, 5 and 6 as well (notice here we will not add 3 twice because of 3 3). Similarly when we reach 4 we have already added 4 when it appeared with 1, 2 and 3 so we'll add it only when it appears with numbers >= itself and add the numbers that appear with 4. Finally we can say that when we are at a number i, we have already processed numbers from1 to i - 1 and hence we have added i every time it appears with numbers 1 to i - 1 so this time we only need to add i every time it appears with numbers >= i also we have to add all the numbers that appear together with i and they are > i. Therefore for every number i we want to add the following terms to our sum t1 : (add i each time it appears with numbers >= itself) -> i (num / i - (i - 1)) (recall i will appear with numbers 1 to num / i and we have already added i each time it appeared with a numbers less than itself)t2 : (add numbers that appear with i) -> (i + 1) + (i + 2) ... + (num / i) (numbers 1 to num / i will appear with i but we have already processed numbers 1 to i - 1 and added them when they appeared with i so now we only have to add the numbers that appear with i and are greater than i, here we will not add i itself because when i appears with itself it should be added only once and we have added it once in t1)we need to calculate t2 in O(1) time, here's how to do thatt2 = (i + 1) + (i + 2) + (i + 3) + ... + (num / i)add and subtract 1 + 2 + 3 ... + i => t2 = 1 + 2 + 3 + ... + i + (i + 1) + (i + 2) + ... + (num / i) - (1 + 2 + 3 + ... + i) => t2 = (1 + 2 + 3 + .. + (num / i)) - (1 + 2 + 3 .. + i) => t2 = ((num / i) (num / i + 1)) / 2 - (i (i + 1)) / 2 Finally, let's look at the numbers that are greater than sqrt(num). These numbers will only appear with numbers that are lesser than sqrt(num). Let's say x is a number greater than sqrt(num) we have,x > sqrt(num)multiply sqrt(num) on both sides => x sqrt(num) > sqrt(num) sqrt(num) => x sqrt(num) > num we want to add x each time it appears, from above proof we see that x multiplied by root of num itself is greater than num hence x will only appear with numbers less than root of num so if we process all the numbers from 1 to sqrt(num) we will add every time x appears. For example take n = 100 now consider 11, 11 10 > 100 so 11 appears only with 1 to 9 i.e. as a factor of 11, 22, 33,..., 99 same is true for rest of the numbers that are greater than 10 they will only appear with numbers lesser than 10 and hence we only need to process numbers from 1 to 10 to add the numbers greater than 10 for n = 100. Finally, our solution is this for each i in 1 to sqrt(num) //no need to visit numbers greater than the root add t1 and t2 to the sum below is the c++ code C++ ```` include using namespace std; long long sum_all_divisors(long long num) { long long sum = 0; for (long long i = 1; i <= sqrt(num); i++) { long long t1 = i (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself long long t2 = (((num / i) (num / i + 1)) / 2) - ((i (i + 1)) / 2); // adding numbers that appear with i and are greater than i sum += t1 + t2; } return sum; } int main() { int n; long long sum = sum_all_divisors(n); cout << sum << '\n'; return 0; } ```` ``` include #include ``` using namespace std; using namespace std long long sum_all_divisors(long long num) long long sum_all_divisors long long num { long long sum = 0; long long sum = 0 for (long long i = 1; i <= sqrt(num); i++) {for long long i = 1 i<= sqrt num i ++ long long t1 = i (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself long long t1 = i num/ i - i + 1 // adding i every time it appears with numbers greater than or equal to itself long long t2 = (((num / i) (num / i + 1)) / 2) - ((i (i + 1)) / 2); // adding numbers that appear with i and are greater than i long long t2 = num/ i num/ i + 1/ 2 - i i + 1/ 2 // adding numbers that appear with i and are greater than i sum += t1 + t2; sum += t1 + t2 } return sum; return sum } int main() int main { int n; int n long long sum = sum_all_divisors(n); long long sum = sum_all_divisors n cout << sum << '\n'; cout<< sum<< '\n' return 0; return 0 } Java ```` import java.io.; class GFG { public static int sum_all_divisors(int num) { int sum = 0; for (int i = 1; i <= Math.sqrt(num); i++) { int t1 = i (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself int t2 = (((num / i) (num / i + 1)) / 2) - ((i (i + 1)) / 2); // adding numbers that appear with i and are greater than i sum += t1 + t2; } return sum; } // Driver code public static void main (String[] args) { int n = 1; int sum = sum_all_divisors(n); System.out.println(sum); } } // This code is contributed by shivanisinghss2110 ```` Python3 ```` import math def sum_all_divisors(num): sum = 0; for i in range(1,math.floor(math.sqrt(num))+1): t1 = i (num / i - i + 1) # adding i every time it appears with numbers greater than or equal to itself t2 = (((num / i) (num / i + 1)) / 2) - ((i (i + 1)) / 2) # adding numbers that appear with i and are greater than i sum += t1 + t2; return sum; n = 1 sum = sum_all_divisors(n) print(sum) This code is contributed by shivanisinghss2110 ```` C# ```` using System; class GFG { public static int sum_all_divisors(int num) { int sum = 0; for (int i = 1; i <= Math.Sqrt(num); i++) { int t1 = i (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself int t2 = (((num / i) (num / i + 1)) / 2) - ((i (i + 1)) / 2); // adding numbers that appear with i and are greater than i sum += t1 + t2; } return sum; } // Driver code public static void Main (String[] args) { int n = 1; int sum = sum_all_divisors(n); Console.Write(sum); } } // This code is contributed by shivanisinghss2110 ```` JavaScript ```` function sum_all_divisors(num) { var sum = 0; for (var i = 1; i <= Math.sqrt(num); i++) { var t1 = i (num / i - i + 1); // adding i every time it appears with numbers greater than or equal to itself var t2 = (((num / i) (num / i + 1)) / 2) - ((i (i + 1)) / 2); // adding numbers that appear with i and are greater than i sum += t1 + t2; } return sum; } var n; var sum = sum_all_divisors(n); document.write( sum ); // This code is contributed by shivanisinghss2110 ```` Output 0 Time complexity: O(sqrt(N)) Auxiliary space: O(1) Example in c: Approach: Create an array divisor_sum of size n+1, initialized with 1 for each index. Loop through each prime p (starting from 2) and check if divisor_sum[p] is equal to 1. If so, update divisor_sum for each multiple of p using the formula: divisor_sum[i] = (1 - pow(p, exp+1)) / (1 - p), where exp is the highest power of p that divides i. For each prime p, add divisor_sum[p] to the variable sum. Return sum. C++ ```` include using namespace std; int sum_of_divisors(int n) { int sum = 1; int divisor_sum[n+1]; for (int i = 1; i <= n; i++) { divisor_sum[i] = 1; } for (int p = 2; p <= n; p++) { if (divisor_sum[p] == 1) { for (int i = p; i <= n; i += p) { int exp = 0; int q = i; while (q % p == 0) { exp++; q /= p; } divisor_sum[i] = (1 - pow(p, exp+1)) / (1 - p); } } sum += divisor_sum[p]; } return sum; } int main() { int n = 4; int sum = sum_of_divisors(n); cout<<"The sum of all divisors from 1 to "<<n<<" is "<<sum<<endl; return 0; } ```` ``` include #include ``` using namespace std; using namespace std int sum_of_divisors(int n) {int sum_of_divisors int n int sum = 1; int sum = 1 int divisor_sum[n+1]; int divisor_sum n + 1 for (int i = 1; i <= n; i++) {for int i = 1 i<= n i ++ divisor_sum[i] = 1; divisor_sum i = 1 } for (int p = 2; p <= n; p++) {for int p = 2 p<= n p ++ if (divisor_sum[p] == 1) {if divisor_sum p == 1 for (int i = p; i <= n; i += p) {for int i = p i<= n i += p int exp = 0; int exp = 0 int q = i; int q = i while (q % p == 0) {while q% p == 0 exp++; exp ++ q /= p; q/= p } divisor_sum[i] = (1 - pow(p, exp+1)) / (1 - p); divisor_sum i= 1 - pow p exp + 1/ 1 - p } } sum += divisor_sum[p]; sum += divisor_sum p } return sum; return sum } int main() {int main int n = 4; int n = 4 int sum = sum_of_divisors(n); int sum = sum_of_divisors n cout<<"The sum of all divisors from 1 to "<<n<<" is "<<sum<<endl; cout<< "The sum of all divisors from 1 to "<< n<< " is "<< sum<< endl return 0; return 0 } C ```` include include int power(int p,int exp) { int res=1; for(int i=0;i<exp;i++) { res=resp; } return res; } int sum_of_divisors(int n) { int sum = 1; int divisor_sum[n+1]; for (int i = 1; i <= n; i++) { divisor_sum[i] = 1; } for (int p = 2; p <= n; p++) { if (divisor_sum[p] == 1) { for (int i = p; i <= n; i += p) { int exp = 0; int q = i; while (q % p == 0) { exp++; q /= p; } divisor_sum[i] = (1 - power(p, exp+1)) / (1 - p); } } sum += divisor_sum[p]; } return sum; } int main() { int n = 4; int sum = sum_of_divisors(n); printf("The sum of all divisors from 1 to %d is %d\n", n, sum); return 0; } ```` Java ```` /package whatever //do not write package name here / import java.io.; class GFG { static int sum_of_divisors(int n) { int sum = 1; int divisor_sum[] = new int[n+1]; for (int i = 1; i <= n; i++) { divisor_sum[i] = 1; } for (int p = 2; p <= n; p++) { if (divisor_sum[p] == 1) { for (int i = p; i <= n; i += p) { int exp = 0; int q = i; while (q % p == 0) { exp++; q /= p; } int temp = (int)Math.pow(p,exp+1); divisor_sum[i]= (1 - temp) / (1 - p); } } sum += divisor_sum[p]; } return sum; } public static void main (String[] args) { int n = 4; int sum = sum_of_divisors(n); System.out.println("The sum of all divisors from 1 to "+n+" is "+sum); } } //this code is contributed by shubhamrajput6156 ```` Python3 ```` import math def sum_of_divisors(n): sum = 1 divisor_sum = (n+1) for i in range(2, n+1): if divisor_sum[i] == 1: for j in range(i, n+1, i): exp = 0 q = j while q % i == 0: exp += 1 q //= i divisor_sum[j] = (1 - math.pow(i,exp+1)) // (1 - i) sum += divisor_sum[i] return sum n = 4 sum = sum_of_divisors(n) print(f"The sum of all divisors from 1 to {n} is {sum}") ```` C# ```` using System; class Program { static int sum_of_divisors(int n) { int sum = 1; int[] divisor_sum = new int[n+1]; for (int i = 1; i <= n; i++) { divisor_sum[i] = 1; } for (int p = 2; p <= n; p++) { if (divisor_sum[p] == 1) { for (int i = p; i <= n; i += p) { int exp = 0; int q = i; while (q % p == 0) { exp++; q /= p; } divisor_sum[i] = (1 - (int)Math.Pow(p, exp+1)) / (1 - p); } } sum += divisor_sum[p]; } return sum; } static void Main(string[] args) { int n = 4; int sum = sum_of_divisors(n); Console.WriteLine("The sum of all divisors from 1 to " + n + " is " + sum); } } ```` JavaScript ```` function sum_of_divisors(n) { let sum = 1; let divisor_sum = new Array(n + 1).fill(1); for (let p = 2; p <= n; p++) { if (divisor_sum[p] === 1) { for (let i = p; i <= n; i += p) { let exp = 0; let q = i; while (q % p === 0) { exp++; q /= p; } let temp = Math.pow(p, exp + 1); divisor_sum[i] = (1 - temp) / (1 - p); } } sum += divisor_sum[p]; } return sum; } let n = 4; let sum = sum_of_divisors(n); console.log(The sum of all divisors from 1 to ${n} is ${sum}); ```` Output The sum of all divisors from 1 to 4 is 15 The time complexity of this approach is O(n log log n) . The space complexity is O(n) for the divisor_sum array. Approach: We initialize n to the maximum number till which we want to find the sum of divisors. In this example, we have taken n as 10. We initialize an array of size n+1 to store the sum of divisors for each number from 1 to n. We use two nested loops to iterate through all the numbers from 1 to n. The outer loop runs from 1 to n, while the inner loop runs from i to n in steps of i. For each number j that is a multiple of i, we add i to the sum of divisors for j. This is done by accessing the array element divisors[j] and incrementing it by i. Finally, we iterate through the array of sum of divisors and add up all the values to get the total sum of divisors from 1 to n. C++ ```` include using namespace std; define MAXN 1000000 int main() { int n = 10; // for example int i, j; int sum = 0; int divisors[MAXN+1] = {0}; // array to store the sum of divisors for (i = 1; i <= n; i++) { for (j = i; j <= n; j += i) { divisors[j] += i; } } for (i = 1; i <= n; i++) { sum += divisors[i]; } cout<<"The sum of all divisors from 1 to "<<n<<" is "<<sum<<endl; return 0; } ```` ``` include #include ``` using namespace std; using namespace std ``` define MAXN 1000000 #define MAXN 1000000 ``` int main() int main { int n = 10; // for example int n = 10 // for example int i, j; int i j int sum = 0; int sum = 0 int divisors[MAXN+1] = {0}; // array to store the sum of divisors int divisors MAXN + 1 = 0 // array to store the sum of divisors for (i = 1; i <= n; i++) {for i = 1 i<= n i ++ for (j = i; j <= n; j += i) {for j = i j<= n j += i divisors[j] += i; divisors j += i } } for (i = 1; i <= n; i++) {for i = 1 i<= n i ++ sum += divisors[i]; sum += divisors i } cout<<"The sum of all divisors from 1 to "<<n<<" is "<<sum<<endl; cout<< "The sum of all divisors from 1 to "<< n<< " is "<< sum<< endl return 0; return 0 } C ```` include define MAXN 1000000 int main() { int n = 10; // for example int i, j; int sum = 0; int divisors[MAXN+1] = {0}; // array to store the sum of divisors for (i = 1; i <= n; i++) { for (j = i; j <= n; j += i) { divisors[j] += i; } } for (i = 1; i <= n; i++) { sum += divisors[i]; } printf("The sum of all divisors from 1 to %d is %d\n", n, sum); return 0; } ```` Java ```` import java.util.; public class Main { public static void main(String[] args) { int n = 10; // for example int i, j; int sum = 0; int[] divisors = new int; // array to store the sum of divisors for (i = 1; i <= n; i++) { for (j = i; j <= n; j += i) { divisors[j] += i; } } for (i = 1; i <= n; i++) { sum += divisors[i]; } System.out.println("The sum of all divisors from 1 to " + n + " is " + sum); } } // This code is contributed by shivhack999 ```` Python3 ```` MAXN = 1000000 n = 10 # for example sum = 0 divisors = (MAXN+1) # list to store the sum of divisors for i in range(1, n+1): for j in range(i, n+1, i): divisors[j] += i for i in range(1, n+1): sum += divisors[i] print(f"The sum of all divisors from 1 to {n} is {sum}") ```` C# ```` using System; class Program { const int MAXN = 1000000; static void Main(string[] args) { int n = 10; // for example int i, j; int sum = 0; int[] divisors = new int[MAXN+1]; // array to store the sum of divisors for (i = 1; i <= n; i++) { for (j = i; j <= n; j += i) { divisors[j] += i; } } for (i = 1; i <= n; i++) { sum += divisors[i]; } Console.WriteLine("The sum of all divisors from 1 to {0} is {1}", n, sum); } } ```` JavaScript ```` function main() { const n = 10; // for example let i, j; let sum = 0; const divisors = new Array(1000001).fill(0); // array to store the sum of divisors for (i = 1; i <= n; i++) { for (j = i; j <= n; j += i) { divisors[j] += i; // add the divisor 'i' to the sum of divisors for 'j' } } for (i = 1; i <= n; i++) { sum += divisors[i]; // add the sum of divisors for 'i' to the overall sum } console.log("The sum of all divisors from 1 to " + n + " is " + sum); } main(); // calling the main function to start the computation //This code is contributed by uomkar369 ```` Output The sum of all divisors from 1 to 10 is 87 Time Complexity: O(nloglogn), where n is the maximum number till which we want to find the sum of divisors. 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8778	https://math.stackexchange.com/questions/42411/algebra-equation-for-percentage-increase-needed-to-get-the-current-value-as-a-20	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 Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Algebra equation for percentage increase needed to get the current value as a 20% discount Ask Question Asked Modified 9 years, 6 months ago Viewed 12k times 4 $\begingroup$ I have some products that I want to increase in value such that a 20% discount gives their current value. It's been ~25 years since college algebra and so I'm a bit rusty on setting up the equation. I've been trying to figure out how to solve for X being the percentage increase needed in order that 20% off would give the current value. For example a product worth 100. I know a 20% increase would make it 120, but 20% off of that would be 96 which isn't 100. I'd give a bounty for explaining the algebra and steps to figure it out, but I'm new to this exchange and am unable to award one - thanks if you spend the time to explain this to me! And if someone wouldn't mind tagging this appropriately for this exchange I'd appreciate it. algebra-precalculus Share edited Jan 12, 2012 at 15:48 Michael Chen 4,27911 gold badge3333 silver badges4545 bronze badges asked May 31, 2011 at 19:03 Chuck SavageChuck Savage 14311 gold badge11 silver badge55 bronze badges $\endgroup$ 3 $\begingroup$ You wouldn't be trying to mislead your customers, would you? Is that allowed where you come from? $\endgroup$ TonyK – TonyK 2011-05-31 19:36:42 +00:00 Commented May 31, 2011 at 19:36 $\begingroup$ Actually, I'm thinking of setting up my website so independent sales people can sell my products, and their customers enter a code that gives them a 20% discount but then I also know who's code it is and they get their commission. So, their customers get the actual price and someone that doesn't use the sales code gets jacked (the 25% increase we figured out below), aka I'm not undercutting the sales people. $\endgroup$ Chuck Savage – Chuck Savage 2011-05-31 20:00:01 +00:00 Commented May 31, 2011 at 20:00 $\begingroup$ OK, sorry I misinterpreted you. $\endgroup$ TonyK – TonyK 2011-05-31 20:22:16 +00:00 Commented May 31, 2011 at 20:22 Add a comment \| 2 Answers 2 Reset to default 3 $\begingroup$ Increasing an amount $A$ by $X$ percent means adding $A\cdot \frac{X}{100}$ to $A$, resulting in $A+A\cdot\frac{X}{100}=A\left(1+\frac{X}{100}\right)$. Decreasing an amount $B$ by $Y$ percent means subtracting $B\cdot\frac{Y}{100}$, resulting in $B-B\cdot\frac{Y}{100}=B\left(1-\frac{Y}{100}\right)$. To have an $X$ percent increase followed by a $20$ percent decrease with an initial amount $A$, you will first multiply by $1+\frac{X}{100}$ to obtain a new amount. If we call that amount $B$, then the next step is to decrease $B$ by $20$ percent by multiplying by $1-\frac{20}{100}$. At this point you will have $A\cdot \left(1+\frac{X}{100}\right)\cdot\left(1-\frac{20}{100}\right)$. For this to leave you where you started, you need to solve the equation $$A\cdot \left(1+\frac{X}{100}\right)\cdot\left(1-\frac{20}{100}\right) =A.$$ You can cancel $A$ from both sides, leaving an equation $$\frac{4}{5}\left(1+\frac{X}{100}\right)=1$$ with $X$ as the only unknown, which can then be solved by division, subtraction, and multiplication. Does that get you where you want to be? Share edited May 31, 2011 at 20:07 answered May 31, 2011 at 19:12 Jonas MeyerJonas Meyer 55.9k99 gold badges216216 silver badges310310 bronze badges $\endgroup$ 8 $\begingroup$ I'm reading over it trying to figure it out. How come it isn't A^2 in the first equation, since you had an A for both the plus and the minus? $\endgroup$ Chuck Savage – Chuck Savage 2011-05-31 19:21:01 +00:00 Commented May 31, 2011 at 19:21 $\begingroup$ @ChuckSavage: When decreasing by 20%, the amount that had been called "$A$" when decreasing by $Y$ percent won't be the same as the original $A$, but instead will be the amount that results from the $X$ percent increase, namely $\displaystyle{A\cdot\left(1+\frac{X}{100}\right)}$. For example, if you were to start with $A$, then increase by 7%, then decrease by 23%, then increase by 12%, you would go through the sequence $A$, $A\cdot 1.07$, $A\cdot 1.07\cdot 0.77$, and finally $A\cdot 1.07\cdot 0.77\cdot 1.12$. I will edit to clarify. $\endgroup$ Jonas Meyer – Jonas Meyer 2011-05-31 19:23:31 +00:00 Commented May 31, 2011 at 19:23 $\begingroup$ Yes - thank you. How do you write those equations... I came up for X = ((1/(1-20/100)-1)100) and my math is so rusty, I don't know what to do with that right side to reduce it. $\endgroup$ Chuck Savage – Chuck Savage 2011-05-31 19:28:13 +00:00 Commented May 31, 2011 at 19:28 $\begingroup$ @Chuck: In general, simplifying $\displaystyle{\frac{1}{\left(1-\frac{Y}{100}\right)}}$ by hand won't be that neat, depending on what $Y$ is. In this case, it is a little easier if you simplify as follows: $\displaystyle{1-\frac{20}{100}=1-\frac{1}{5}=\frac{5}{5}-\frac{1}{5}=\frac{4}{5}}$, and $\displaystyle{\frac{1}{\left(\frac{4}{5}\right)}=\frac{5}{4}}$. Then $\frac{5}{4}-1=\frac{5}{4}-\frac{4}{4}=\frac{1}{4}$, and finally $100\cdot\frac{1}{4}=25$. Working this out in decimal form also wouldn't be too bad. $\endgroup$ Jonas Meyer – Jonas Meyer 2011-05-31 19:36:11 +00:00 Commented May 31, 2011 at 19:36 1 $\begingroup$ @Jonas ty again, I worked through it on my own and came up with the same. These exchanges are great! If you ever have C#/asp.net coding questions head over to stackoverflow.com and I'd be happy to answer them! $\endgroup$ Chuck Savage – Chuck Savage 2011-05-31 19:52:47 +00:00 Commented May 31, 2011 at 19:52 \| Show 3 more comments 2 $\begingroup$ Let's work first with your particular numbers. Suppose that an item is originally priced at 100 (dollars). Imagine that you inrease the price by $x$ percent. Then the new price is $100+x$. You want to make sure that if you apply a 20 percent discount to this new price $100+x$, you end up with a price of exactly $100$ dollars. A $20$ percent discount on a price of $100+x$ means that the price will be $80$ percent of $100+x$. So the new price is $$(100+x)\frac{80}{100}, \qquad\text{or equivalently,} \qquad \frac{(100+x)(80)}{100}$$ It so happens that you want this new price to be $100$ dollars. This gives you the equation $$\frac{(100+x)(80)}{100}=100.$$ You would like to "extract" $x$ from this equation. First multiply both sides by $100$. On the left, you get simply $(100+x)(80)$. On the right, you get $(100)(100)$, which is $10000$. So our new equation is $$(100+x)(80)=10000.$$ So something, namely $100+x$, multiplied by $80$, is $10000$. What is the something? The idea is to divide both sides by $80$. On the left, you get $100+x$. On the right, you get $10000/80$. By calculator or by hand division, you get that $10000/80=125$. So our new equation is $$100+x=125.$$ Now subtract $100$ from both sides. We get $$x=25.$$ This tells you that you must apply a $25$ percent markup so that a $20$ percent discount will leave the price unchanged. This may look long, but that is only because I have done the calculations in great detail. Note Since we are dealing with percentages, the answer is independent of the actual initial price, which for simplicity we took to be $100$. Let's use the same reasoning to solve a different and harder problem. You want the final discount to be say $17$ percent. Let us ask what percent markup there should be so that at the end, after the discount, you end up selling the originally $100$ dollar item for $105$ dollars. The process will be almost exactly the same, except that the numbers will be a lot uglier, so you will have to use a calculator. Let the desired markup be $x$ percent. Then the price is $100+x$. You want to apply a $17$ percent discount to that. So the new price would be $83$ percent of $100+x$. You want this new price to be $105$. So $83$ percent of $100+x$ is $105$. That gives you the equation $$\frac{(100+x)(83)}{100}=105.$$ You want to extract $x$ from this equation. The procedure is in outline much the same as before. First multiply both sides by $100$. So our new equation is $$(100+x)(83)=10500.$$ Now divide both sides by $83$. We will not get a simple integer on the right, so I will round off, and sloppily still write "$=$" when I mean almost equal. If you are following this with a calculator, you should get something like $$100+x=126.506.$$ Now, like before, subtract $100$ from both sides. We get $$x=26.506.$$ Of course this is absurd precision. For all practical purposes, the markup should be $26.5$ percent. I hope there is enough detail in the above calculations to enable you to solve problems of the same general kind with not much difficulty. Share edited May 31, 2011 at 21:21 answered May 31, 2011 at 20:29 André NicolasAndré Nicolas 515k4747 gold badges584584 silver badges1k1k bronze badges $\endgroup$ 1 $\begingroup$ thanks for writing this all out, it was easy to follow. Also completely different from Jonas approach, but with the same answer. Reviewing his equation, his is the same as yours except = 1 and not 100. $\endgroup$ Chuck Savage – Chuck Savage 2011-05-31 20:52:24 +00:00 Commented May 31, 2011 at 20:52 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 algebra-precalculus 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 Related What are the steps to solve this simple algebraic equation? 1 function for inverting 3 How to write a function which depends on another function which in turn depends on the first function... 1 $\frac {(x-y)}{x}$ Can it be simplified? 1 Create a set of system of linear equations to answer the following. 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8779	https://www.wyzant.com/resources/answers/933397/understanding-a-piecewise-probability-density-function-and-deriving-its-cdf	Understanding a Piecewise Probability Density Function and Deriving Its CDF \| Wyzant Ask An Expert Log inSign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us All Questions Search for a Question Find an Online Tutor Now Ask a Question for Free Login WYZANT TUTORING Log in Sign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us Subject ZIP Search SearchFind an Online Tutor NowAsk Ask a Question For Free Login StatisticsProbability Bim B. asked • 09/28/23 Understanding a Piecewise Probability Density Function and Deriving Its CDF I have a piecewise Probability Density Function (PDF) defined as follows: h(x) = { 0 if x < 1 1.5 (x - 1)^2 if 1 <= x < 2 1.5 (3 - x)^2 if 2 <= x < 3 0 if x >= 3 } What is the CDF Follow •2 Add comment More Report 2 Answers By Expert Tutors Best Newest Oldest By: Kevin S.answered • 09/28/23 Tutor 5.0(379) Experienced Statistics Tutor and Researcher with 25+ Years Experience See tutors like this See tutors like this Let's denote the CDF as H(x). For x<1: H(x) = 0 For 1<=x<2: We integrate 1.5∗(x−1)^2 from 1 to x: H(x) = 0.5 (x - 1)^3 + 0 For 2<=x<3: We integrate 1.5∗(3−x)^2 from 2 to x: H(x) = -0.5 (3 - x)^3 + C , where C is the constant of integration. To find C, we know that H(2)=0.5 (from the previous step). So, C=0.5+0.5∗(3−2)^3=0.5+0.5=1, H(x) = - 0.5 (3 - x)^3 For x>=3x>=3: H(x) = 1 Putting it all together, the CDF H(x)H(x) is: H(x) = { 0 if x < 1 0.5 (x - 1)^3 if 1 <= x < 2 0.5 (3 - x)^3 if 2 <= x < 3 1 if x >= 3 } Upvote • 1Downvote Comment •1 More Report Patrick F. tutor I think there is a problem here. Please have a look at my video. Report 09/28/23 Patrick F.answered • 09/28/23 Tutor 5(14) Inspiring Math Teacher and former Actuary About this tutor› About this tutor› Upvote • 0Downvote Add comment More Report Still looking for help? Get the right answer, fast. Ask a question for free Get a free answer to a quick problem. Most questions answered within 4 hours. 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8780	https://math.umd.edu/~mboyle/courses/411sp13/n.pdf	SCALAR PRODUCTS, NORMS AND METRIC SPACES 1. Definitions Below, “real vector space” means a vector space V whose ﬁeld of scalars is R, the real numbers. The main example for MATH 411 is V = Rn. Also, keep in mind that “0” is a many splendored symbol, with meaning depending on context. It could for example mean the number zero, or the zero vector in a vector space. Deﬁnition 1.1. A scalar product is a function which associates to each pair of vectors x, y from a real vector space V a real number, < x, y >, such that the following hold for all x, y, z in V and α in R: (1) < x, x > ≥0, and < x, x > = 0 if and only if x = 0. (2) < x, y > = < y, x >. (3) < x + y, z > = < x, z > + < y, z >. (4) < αx, y > = α < x, y >. The dot product is deﬁned for vectors in Rn as x·y = x1y1 +· · ·+xnyn. The dot product is an example of a scalar product (and this is the only scalar product we will need in MATH 411). Deﬁnition 1.2. A norm on a real vector space V is a function which associates to every vector x in V a real number, \|\|x\|\|, such that the following hold for every x in V and every α in R: (1) \|\|x\|\| ≥0, and \|\|x\|\| = 0 if and only if x = 0. (2) \|\|αx\|\| = \|α\|\|\|x\|\|. (3) (Triangle Inequality for norm) \|\|x + y\|\| ≤\|\|x\|\| + \|\|y\|\|. The standard Euclidean norm on Rn is deﬁned by \|\|x\|\| = √x · x. There are other useful norms, as we’ll see. Deﬁnition 1.3. A metric space is a set X together with a function d which associates to each pair of points x, y from X a real number, d(x, y), such that the following hold for all x, y, z in X: (1) d(x, y) ≥0, and d(x, y) = 0 if and only if x = y. (2) d(x, y) = d(y, x). (3) (Triangle Inequality) d(x, z) ≤d(x, y) + d(y, z). Above, that function d is the distance function, also called the metric. An example of a distance function is the usual distance between vectors x, y in Rn: d(x, y) = \|\|x −y\|\| = √< x −y, x −y > . 2. Relations What we see in the familiar examples above is perfectly general: (1) Given a scalar product < ·, · > on a real vector space V , the deﬁnition \|\|x\|\| = √< x, x > deﬁne a norm on V . 1 2 SCALAR PRODUCTS, NORMS AND METRIC SPACES (2) Given a norm \|\| · \|\| on V , the deﬁnition d(x, y) = \|\|x −y\|\| deﬁnes a distance function making V a metric space. These two claims are checked simply by verifying that the deﬁning properties hold. For example, to show property (1) of a metric space is true for the function d(x, y) = \|\|x −y\|\|, you note that it easily follows from property (1) for the norm. All the other properties are likewise easy to check, except for checking that the triangle inequality for the norm holds when \|\|x\|\| is deﬁned by √< x, x >. For this, you prove the Cauchy Schwarz Inequality ﬁrst, then make the argument. The proof of the Cauchy Schwarz Inequality only uses the properties listed in the deﬁnition of the scalar product. 3. Important norms on Rn To consider more than one norm, we use a little more notation. For x in Rn, deﬁne the following three norms: \|\|x\|\|1 = \|x1\| + · · · + \|xn\| \|\|x\|\|2 = p (x1)2 + · · · + (xn)2 \|\|x\|\|∞= max{\|x1\|, \|x2\|, . . . , \|xn\|} . The norm \|\| · \|\|2 is also called the standard norm, or the Euclidean norm. The norm \|\| · \|\|∞\| is also called the supremum norm, or the sup norm. Associated to each of these norms is the distance function it deﬁnes. Let us call these respectively d1, d2, d∞. The distance d2 is the Euclidean distance – it is the standard idea of distance on Rn, known to Euclid (when n ≤3). Exercise. Draw the “unit circle” for each of these three metrics for the case R2, where “unit circle” means the set of points which are distance 1 from the origin. 4. Convergence For a sequence x1, x2, . . . of points in a metric space X, convergence of the sequence to a point x in X is denoted limk→∞xk = x. We deﬁne this to be true if and only if lim k→∞dist(xk, x) = 0 . Going back to MATH 410: this last condition means that for any ϵ > 0, there is a number M such that k > M = ⇒dist(xk, x) < ϵ . We will see below that the distances d1, d2, d∞deﬁne the same notion of con-vergence on Rn: if limk→∞dist(xk, x) = 0 if d is any one of these three, then limk→∞dist(xk, x) = 0 if it is one of the others as well. 5. Some inequalities For real numbers x1, . . . , xn: considering cross terms we see \|x1\|2 + · · · + \|xn\|2 ≤(\|x1\| + · · · + \|xn\|)2 and therefore, taking square roots, p \|x1\|2 + · · · + \|xn\|2 ≤\|x1\| + · · · + \|xn\| . SCALAR PRODUCTS, NORMS AND METRIC SPACES 3 We can add couple of easily veriﬁed inequalities to get: max i \|xi\| ≤ p \|x1\|2 + · · · + \|xn\|2 ≤\|x1\| + · · · + \|xn\| ≤n max i \|xi\| . In the notation of norms, we can write the last line as \|\|x\|\|∞≤\|\|x\|\|2 ≤\|\|x\|\|1 ≤n\|\|x\|\|∞. We see that given any two of these norms, \|\| · \|\|i and \|\| · \|\|j say, that there is a positive constant C such that \|\|x\|\|i ≤C\|\|x\|\|j for all x. Consequently, whenever lim k dj(xk, x) = 0 it must also be true that lim k di(xk, x) = 0 which means that convergence of a sequence to x in the dj metric implies conver-gence of the sequence to x in the di metric. So! If we want to check convergence of a sequence of points in Rn, we can use any of the three criteria, at our convenience A closely related equivalent criterion for convergence of a sequence xk to a point x in Rn is componentwise convergence, as discussed in the text of Fitzpatrick. 6. Some remarks Deﬁnition 6.1. Two norms (call them \|\| · \|\|a and \|\| · \|\|b) on a real vector space V are equivalent if there exist positive numbers C1, C2 such that for all x in V , \|\|x\|\|a ≤C1\|\|x\|\|b , and \|\|x\|\|b ≤C2\|\|x\|\|a . Here is a fact. Any two norms on Rn are equivalent! So, all norms on Rn determine the same notion of convergence – this is not a special property of the particular three norms we looked at above. Proving that would be an interesting exercise within your powers after we ﬁnish the early part of the course on metric space topology. The deﬁnitions of scalar product and norm generalize to inﬁnite dimensional vector spaces, and especially to spaces of functions. That is one reason for spelling out the chain of logic here. There are many metrics which do not arise from norms. For example, if we take a subset of a metric space, with the same deﬁnition for distance, it is again a metric space. For example, the circle in R2 with the standard metric is a metric space. Also, in contrast to the situation for norms, there are (vastly) many metrics on Rn which are not equivalent.
8781	https://www.investopedia.com/terms/e/embezzlement.asp	Skip to content Top Stories Inflation Just Ticked Up Again—Protect Your Savings Fast Are You Ready for the Fed's Rate Decision Next Week? Is a Common Medicare Mistake Draining Your Savings? How Much Cash Americans Really Have in the Bank Table of Contents Table of Contents What Is Embezzlement? How Embezzlement Works Common Method of Embezzlement Preventing Embezzlement FAQs The Bottom Line What Is Embezzlement, and How Does It Happen? 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 August 29, 2025 Reviewed by Caitlin Clarke Fact checked by Yarilet Perez Fact checked by Yarilet Perez Full Bio Yarilet Perez is an experienced multimedia journalist and fact-checker with a Master of Science in Journalism. She has worked in multiple cities covering breaking news, politics, education, and more. Her expertise is in personal finance and investing, and real estate. Learn about our editorial policies What Is Embezzlement? Embezzlement, a type of white-collar crime, involves the unlawful misappropriation of assets by an individual or entity entrusted with them. The embezzler gains lawful possession of these assets but subsequently breaches their fiduciary duty for personal gain. Examples range from false expense claims to complex Ponzi schemes. At its core, embezzlement signifies a misuse of trust, with civil and criminal liabilities for those involved. Key Takeaways Embezzlement involves the misappropriation of assets by a person with lawful access for unauthorized personal use. It is a breach of fiduciary duty and can lead to both civil and criminal penalties, including fines and imprisonment. Embezzlement schemes can range from small acts like skimming cash to large-scale frauds involving millions. Businesses can prevent embezzlement by vetting employees, conducting audits, and implementing strong internal controls. Ponzi schemes are a notable example of embezzlement where funds are deceptively used for personal gain instead of their intended purpose. How Embezzlement Works and Its Implications Individuals who are entrusted with access to an organization’s funds are expected to safeguard those assets for their intended use. It is illegal to intentionally access that money and convert it to personal use. Such activities can include diverting funds to accounts that appear to be authorized to receive payments or transfers. However, the account is a front that allows the individual, or a third party they are collaborating with, to take the funding. For instance, an embezzler might create bills and receipts for business activities that never took place or services that were never rendered to disguise the transfer of funds as a legitimate transaction. An embezzler might collaborate with a partner who is listed as a consultant or contractor who issues invoices and receives payment, yet never actually performs the duties they are charging for. The nature of embezzlement can be both small and large. Embezzlement can be as minor as a clerk stealing a few dollars from the cash register. On a larger scale, executives may falsely expense millions and transfer funds to personal accounts. Depending on the scale of the crime, embezzlement may be punishable by large fines and time in jail. Take the Next Step to Invest 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. Common Methods Used in Embezzlement Schemes Embezzlement occurs when someone steals or misappropriates what they were entrusted to manage or safeguard. The property or asset need not be of substantial value for embezzlement to occur. Although closely related, it differs from fraud in that the embezzler had authorization to use or oversee the property or funds.1 Embezzlement is sometimes combined with other frauds, like Ponzi schemes. In such cases, the embezzler scams investors to entrust them with their assets to invest on their behalf but instead uses the money for personal gain and enrichment. Maintaining the fraud often includes seeking out new investors to bring in more money to appease prior investors. 150 years The number of years Bernie Madoff was sentenced to for orchestrating the largest Ponzi scheme in history.2 Embezzlers may also take assets other than money. An embezzler might claim the real estate, company vehicles, smartphones, and other hardware such as laptops that belong to an organization for personal use. Government employees may also embezzle by taking local, state, or national funds for themselves. Such instances may occur when funding is disbursed to fulfill contracts or to support projects, and a member of the staff skims some of the money that was earmarked. People who embezzle can be charged with a criminal offense and/or held civilly responsible for their crimes. Punishment can range from paying monetary damages and restitution to victims to incarceration. White-collar offenses don't prevent offenders from being prescribed lengthy prison sentences, ones traditionally handed to violent offenders. Strategies for Preventing Embezzlement in the Workplace Theft and embezzlement cost businesses about $400 billion annually and cause over 50% of business failures.3 However, employers can develop strategies to combat these white-collar crimes. Embezzlement starts with the breach of trust of a person endowed with the authority to care for the property or money of another. Reasonably, one of the first steps an employer can take is to carefully vet prospective employees. In addition to conducting thorough background checks, assessing character traits via personality tests could reveal undesirable behaviors. A security and monitoring program could also deter corporate crime, especially when performed by a dedicated risk management team or an independent, third party. These risk managers can create internal controls that monitor behaviors and allow for the anonymous reporting of suspicious activities, as well as conduct periodic audits that expose misappropriations. Early detection helps to mitigate losses and protect the company's reputation and the people it serves. Employers should make clear that they have a no-tolerance policy regarding illegal acts such as embezzling and communicate the consequences of such violations. Every company should promote a culture of honesty and fairness, encouraging its employees to remain vigilant and report instances of wrongdoing. How Does One Legally Prove Embezzlement? To legally prove embezzlement, the claimant must prove that the perpetrator had a fiduciary responsibility to the victim and that the embezzled asset was acquired through that relationship and conveyed to the accused intentionally.4 The key to spotting embezzlement is that it involves a betrayal of trust or duty. While this looks different in every state, generally, these four factors must be present:5 There must be a fiduciary relationship between the two parties. That is, there must be a reliance by one party on the other. The defendant must have acquired the property through that relationship, The defendant's actions must have been intentional, and not the result of an error; The defendant must have taken ownership of the property (at least temporarily), transferred the property to someone else, or destroyed or hidden the property. What Is the Punishment for Embezzlement? A person can be held civilly and criminally responsible for embezzling. Punishments range from monetary fines and restitution to imprisonment. What Is a White Collar Crime? A white-collar crime is a non-violent crime committed by a business professional who breached trust for economic gain. White-collar crimes include fraud, theft, counterfeiting, embezzlement, money laundering, and other fraudulent schemes.6 The Bottom Line Embezzlement is a serious white-collar crime where individuals misuse assets they are entrusted to manage. It involves a breach of fiduciary duty and can lead to severe legal consequences, including fines and imprisonment. Deterring embezzlement starts with rigorous hiring practices and implementing strong internal controls. These measures, coupled with fostering a culture of honesty and transparency, are essential in protecting organizations from significant financial losses and reputational harm. 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. Nolo. "What Is Embezzlement?" U.S. Department of Justice, Southern District of New York. "United States V. Bernard L. Madoff And Related Cases." Association of Certified Fraud Examiners. "Report to the Nation," Page 2. Cornell University, Legal Information Institute. "Embezzlement." U.S. Department of Justice. "Embezzlement." Federal Bureau of Investigations. "White-Collar Crime." 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. 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8782	https://mindyourdecisions.com/blog/2023/10/31/3-puzzles-about-shaded-areas-and-circles/	Skip to content Mind Your Decisions Math Videos, Math Puzzles, Game Theory. By Presh Talwalkar 3 Puzzles About Shaded Areas And Circles If you buy from a link in this post, I may earn a commission. This does not affect the price you pay. As an Amazon Associate I earn from qualifying purchases. Learn more. Posted October 31, 2023 By Presh Talwalkar. Read about me, or email me. Problem 1 I read this was a PSAT question, which is a standardized test given to US students aged about 15 or 16 who most likely have not taken a course in trigonometry. If each of the 3 semicircles below has an area equal to 72, and is the midpoint of , what is the area of the shaded region? Problem 2 This is adapted from the 2020 AMC 10B, problem 14. A regular hexagon with side length 2 has semicircles constructed in its interior of each side. What is the shaded area inside the hexagon not covered by the semicircles? Problem 3 Thanks to Andrei for the suggestion! A square has a side length of 7. On each corner, a quarter circle with radius 3 is constructed with the corner as its center. And on the midpoint of each side, a semicircle with radius 2 is constructed with the midpoint as its center. What is the difference between the orange shaded areas and the blue shaded areas? As usual, watch the video for solutions. 3 Puzzles About Shaded Areas And Circles Or keep reading. . . "All will be well if you use your mind for your decisions, and mind only your decisions." It costs thousands of dollars to run a website and your support matters. If you like the posts and videos, please consider a monthly pledge on Patreon. You may also consider a one-time donation to support my work. . . . . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To PSAT Question Three Semicircles (Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks). Thanks Erik for alerting me of typos! Problem 1 There’s a remarkably easy way to calculate the area! Since Q is the midpoint of the diameter PR, it is the center of the left semicircle. Since the three semicircles have the same area, all three are congruent. The two semicircles on the right pass through Q, so their intersections with semicircle Q will be congruent and have the same area. Thus the two green regions are equal in area and will cancel out in the calculation. Thus the area of the claw is equal to the area of a 20 degree sector. Since each semicircle has 180 degrees, and an area of 72, the answer is: area claw= area 20 degree sector= 72(20/180)= 8 That’s it! Problem 2 Divide the hexagon into a grid of equilateral triangles of side length 1, as shown. The shaded area is then 6 times the area of a single blue slice. But each blue slice is equal to the area of 2 equilateral triangles minus a circular sector with a central angle equal to 1. Both the triangles have a side of 1 and the circle has a radius of 1. A single blue slice has area equal to: 2(s2√3)/4 – πr2(60/360) = 2(√3)/4 – π/6 = (3√3 – π)/6 This area times 6 gives the shaded area as 3√3 – π ≈ 2.05. Problem 3 At first I had tried to calculate the area of the overlapping region. But this is a far too challenging computation. So I thought there must be another method. Let each of the orange areas be equal to a, the overlapping areas be b, and the blue areas be c. We want to calculate: 4a – 4c But we do know: a + 2b = area quarter circle c + 2b = area semicircle Multiplying each equation by 4 and subtracting gives: 4a – 4c = 4(area quarter circle) – 4(area semicircle) = 4(π32/4) – 4(π22/2) = π And like magic we have found the answer! I received a nice email from Atanas who has developed code in 3 languages if you want to simulate this result yourself! Check it out on Github: Problem 1Math StackExchange Problem 2AoPS Published by PRESH TALWALKAR I run the MindYourDecisions channel on YouTube, which has over 1 million subscribers and 200 million views. I am also the author of The Joy of Game Theory: An Introduction to Strategic Thinking, and several other books which are available on Amazon. (As you might expect, the links for my books go to their listings on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.) By way of history, I started the Mind Your Decisions blog back in 2007 to share a bit of math, personal finance, personal thoughts, and game theory. It's been quite a journey! I thank everyone that has shared my work, and I am very grateful for coverage in the press, including the Shorty Awards, The Telegraph, Freakonomics, and many other popular outlets. 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8783	https://pmc.ncbi.nlm.nih.gov/articles/PMC4289612/	Use of non-benzodiazepine sedative hypnotics and risk of falls in older men - 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. 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 J Gerontol Geriatr Res . Author manuscript; available in PMC: 2015 Jan 11. Published in final edited form as: J Gerontol Geriatr Res. 2014 May 26;3(3):158. doi: 10.4172/2167-7182.1000158 Search in PMC Search in PubMed View in NLM Catalog Add to search Use of non-benzodiazepine sedative hypnotics and risk of falls in older men Susan J Diem Susan J Diem, MD, MPH 1 Department of Medicine, University of Minnesota, Minneapolis, MN 2 Division of Epidemiology & Community Health, University of Minnesota, Minneapolis, MN Find articles by Susan J Diem 1,2, Susan K Ewing Susan K Ewing, MS 3 California Pacific Medical Center Research Institute, San Francisco, CA Find articles by Susan K Ewing 3, Katie L Stone Katie L Stone, PhD 3 California Pacific Medical Center Research Institute, San Francisco, CA Find articles by Katie L Stone 3, Sonia Ancoli-Israel Sonia Ancoli-Israel, PhD 4 Departments of Psychiatry and Family & Preventive Medicine, University of California – San Diego, La Jolla, CA Find articles by Sonia Ancoli-Israel 4, Susan Redline Susan Redline, MD, MPH 5 Harvard Medical School, Boston, MA Find articles by Susan Redline 5, Kristine E Ensrud Kristine E Ensrud, MD, MPH 1 Department of Medicine, University of Minnesota, Minneapolis, MN 2 Division of Epidemiology & Community Health, University of Minnesota, Minneapolis, MN 6 Center for Chronic Disease Outcomes Research, Minneapolis VA Health Care System, Minneapolis, MN Find articles by Kristine E Ensrud 1,2,6; the Osteoporotic Fractures in Men (MrOS) Study Group Author information Article notes Copyright and License information 1 Department of Medicine, University of Minnesota, Minneapolis, MN 2 Division of Epidemiology & Community Health, University of Minnesota, Minneapolis, MN 3 California Pacific Medical Center Research Institute, San Francisco, CA 4 Departments of Psychiatry and Family & Preventive Medicine, University of California – San Diego, La Jolla, CA 5 Harvard Medical School, Boston, MA 6 Center for Chronic Disease Outcomes Research, Minneapolis VA Health Care System, Minneapolis, MN ✉ Corresponding Author: Susan J. Diem, MD, MPH, Epidemiology Clinical Research Center, University of Minnesota, 1100 Washington Avenue S., Suite 201, Minneapolis, MN 55415, Phone: (612) 626-9199; Fax: (612) 626-9505, sdiem@umn.edu Issue date 2014 Jul 1. PMC Copyright notice PMCID: PMC4289612 NIHMSID: NIHMS648697 PMID: 25587493 The publisher's version of this article is available at J Gerontol Geriatr Res Abstract Background To ascertain whether use of non-benzodiazepine sedative-hypnotics is associated with risk of falls and compare this to risk of falls associated with use of benzodiazepines. Methods Among 4450 community-dwelling men, aged 71 years and older, enrolled in the population-based prospective cohort study, Osteoporotic Fractures in Men (MrOS), use of nonbenzodiazepine sedative-hypnotics and benzodiazepines was assessed by interview and verified from medication containers at the third annual visit of the MrOS study. Falls in the subsequent one-year period were ascertained by tri-annual questionnaires and a computerized dictionary used to categorize type of medication. Results In age-adjusted models, non-benzodiazepine sedative hypnotic use was associated with an increased risk of any falls (one or more falls) (RR 1.44, 95% CI 1.15, 1.81) and recurrent falls (2 or more falls) (RR 1.51, 95% CI 1.07, 2.14). Use of benzodiazepines was associated with a similar increase in age-adjusted risk of falling. Depressive symptoms, inability to stand from a chair, and instrumental activities of daily living (IADL) impairment modestly attenuated these associations. The association between non-benzodiazepine sedative-hypnotic use and falls was most pronounced among men without a history of falls in the previous year: in a multivariable model controlling for multiple potential confounders, the RR of any falls was 1.74 (95% CI 1.13, 2.68) in this subgroup. Conclusions Use of non-benzodiazepine sedative-hypnotics is associated with an increased risk of falls. Non-pharmacologic approaches to sleep disturbances may represent the safest approach to sleep difficulties in older adults. Keywords: falls, sedative hypnotics, benzodiazepines, men, zolpidem, adverse drug effects INTRODUCTION Nonbenzodiazepine sedative-hypnotics, such as zolpidem, zaleplon, and eszopiclone, are often advocated as safer alternatives to benzodiazepines for the treatment of sleep disturbances due to their short half-life and preservation of normal sleep architecture.[1-5] However, limited data are available about their safety in older patients, in particular regarding postural instability, falls, and fractures. These so-called “Z-drugs” affect the same receptor as benzodiazepines, suggesting that their risks may be similar. Clinical trials of zolpidem in healthy younger adults have demonstrated central nervous system side effects, including impaired cognitive and motor function, particularly in the first few hours after use.[6-8] In addition, observational data have suggested an association between non-benzodiazepine sedative-hypnotics and risk of fracture.[9,10] However, these studies have relied on administrative data bases and thus had limited or no information on important potential confounders, such as baseline physical function, cognitive function, and comorbidities. To determine whether use of non-benzodiazepine sedative-hypnotics is associated with an increased risk of falls, and to compare this to the risk of falls observed with benzodiazepine use, we ascertained use of non-benzodiazepine sedative-hypnotics and benzodiazepines in a cohort of 4450 men aged 71 years and older enrolled in the Osteoporotic Fractures in Men study (MrOS) and followed them prospectively for incident falls during one year of follow-up. METHODS Participants From March 2000 through April 2002, 5994 men who were at least 65 years of age were recruited for participation in the baseline examination of the prospective Osteoporotic Fractures in Men (MrOS) study. Men were recruited from population based listings in Birmingham, AL; Minneapolis, MN; Palo Alto, CA; Pittsburgh, PA; Portland, OR; and San Diego, CA. Men with a history of bilateral hip replacement and men who were unable to walk without the assistance of another person were excluded.[11,12] To be included in the present analysis, men must have attended a 3 rd clinic examination between March 2007 and March 2009, completed a medication inventory at the 3 rd exam, and returned at least two follow-up questionnaires regarding falls in the subsequent one year period. Of the original cohort, 1043 men had died prior to the 3 rd exam and 168 had terminated participation in the study; 4682 of the original cohort, (98% of survivors) attended the 3 rd exam (baseline for this analysis). 4588 men returned at least two follow-up questionnaires in the subsequent one year period; of these, 4471 completed a medication inventory. 19 men who reported use of both a non-benzodiazepine sedative hypnotic and a benzodiazepine were excluded, as were two men reporting use of ramelteon, a melatonin receptor agonist. In a secondary analysis, we restricted the cohort to those 2722 men participating in an ancillary study evaluating sleep disorders, titled the MrOS Sleep Study, who had attended an earlier exam and completed a measure of subjective sleep quality (average 3.4 ± 0.5 years between ancillary sleep exam and 3 rd exam). The Institutional Review Board (IRB) at each center approved the study protocol and written informed consent was obtained from all subjects. Medication Use Participants attending the clinic examination were asked to bring all current (any use within the last 30 days) prescription and nonprescription medications with them to clinic. Interviewers completed a medication history for each participant, including name of medication and frequency of use. All medications recorded by the clinics were stored in an electronic medications inventory database (San Francisco Coordinating Center, San Francisco, CA). Each medication was matched to its ingredient(s) based on the Iowa Drug Information Service (IDIS) Drug Vocabulary (College of Pharmacy, University of Iowa, Iowa City, IA). Zolpidem, zaleplon, and eszopiclone were categorized as non-benzodiazepine sedative-hypnotics; benzodiazepines included lorazepam, clonazepam, alprazolam, temazepam, diazepam, triazolam, oxazepam, clorazepate, and chlordiazepoxide. Participants were categorized as users of a medication if they reported any use in the last 30 days. Ascertainment of Falls At 4-month intervals, participants were queried by mailed questionnaire about the number of times they had fallen during the interval. Participants who fell in the previous 4 months were asked how many times (1, 2, 3, 4, or ≥5). Participants who did not initially return a tri-annual questionnaire or did not adequately complete the questionnaire were followed up with a telephone call. The present study includes fall reports during the one year period following the subject's Visit 3 examination. Other Measurements Participants completed questionnaires and were interviewed at the examination by trained staff, who asked about race/ethnicity, educational achievement, health status, smoking status, alcohol use, medical history, and falls in previous year. Men were asked whether they had difficulty performing any of 5 independent activities of daily living (preparing meals, shopping, heavy housework, walking 2-3 blocks, and climbing up 10 steps). A composite functional impairment score expressed the total number of activities ranging from 0 to 5 that a participant reported difficulty performing. Physical performance measures included chair stand time and walking speed. For the chair stand, participants were asked to stand from a chair without using their arms; the time required to complete five chair stands without using arms was recorded. Cognitive function was assessed with the Teng Modified Mini-Mental State Examination (mMMSE) (maximum score 100). Depressive symptoms were evaluated using the Geriatric Depression Scale short form (GDS). The total number of selected comorbid conditions reported by participants (stroke, diabetes mellitus, Parkinson's disease, chronic obstructive pulmonary disease, myocardial infarction, angina, and congestive heart failure) was summed for each participant. Body weight was measured by using a balance beam scale at all sites, except at one site (Portland) which used a digital scale. Height was measured by using a wall-mounted Harpenden stadiometer (Holtain, UK). Body Mass Index (BMI) was calculated as weight in kilograms divided by the square of height in meters. Physical activity was assessed using the Physical Activity Scale for the Elderly (PASE). Among participants in the MrOS Sleep Study, sleep complaints were measured using the Pittsburgh Sleep Quality Index (PSQI), a 19-item self-report questionnaire that assesses sleep quality over a one month time period (range 0-21), with higher scores reflecting poorer sleep quality.[19,20] Statistical Analysis Characteristics of the men at the 3 rd exam were compared by medication use category (non-user, benzodiazepine user, non-benzodiazepine sedative-hypnotic user) using either analyses of variance or Kruskal-Wallis tests (continuous variables) or chi-squared tests of homogeneity (categorical variables). For the primary analysis examining the association between medication use and risk of falls, log binomial and Poisson models were used to estimate the relative risk (RR) and 95% confidence interval (CI) of any falls (1 or more falls vs. 0) and recurrent falls (2 or more falls vs. 1 or none) by medication use category, with nonusers as the referent group. Models were initially adjusted for age and then further adjusted for potential confounders one at a time. In addition, covariates known to be risk factors for falls and characteristics related to use of sedativehypnotics were examined for inclusion in a multivariable model, with those factors related to falls, non-benzodiazepine sedative-hypnotic use, or benzodiazepine use at p <0.1 included. Interactions between medication and history of falls were explored. We also conducted secondary analyses in which the analysis was limited to the 2722 men who had data about self-reported sleep quality available from the earlier sleep exam. All analyses were performed using SAS software (Version 9.1, SAS Institute Inc., Cary, NC). RESULTS Of the 4450 men included in this analysis, 94 men reported use of zolpidem (n=81), zaleplon (n=3), or eszopiclone (n=10), and 177 reported use of a benzodiazepine. Baseline characteristics of the cohort overall and by use status appear in Table 1. Sedative-hypnotic users were more likely to report fair or poor health and more IADL impairments than non-users. They were more likely to have difficulty standing from a chair without using their arms and were more likely to have fallen in the previous year. Benzodiazepine users and non-benzodiazepine sedative-hypnotic users were similar in many characteristics, although non-benzodiazepine sedative-hypnotic users were more likely to be college-educated, had higher mean MMSE, and reported more alcohol consumption. Table 1. Baseline Characteristics \| Characteristic \| Overall \| Non-users \| Benzodiazepine Users \| Non-benzodiazepine sedative hypnotic users \| P-value \| :---: :---: :---: \| \| \| (N=4450) \| (N= 4179) \| (N= 177) \| (N= 94) \| \| \| Age, years, mean ± SD \| 79.5 ± 5.3 \| 79.5 ± 5.3 \| 79.1 ± 5.3 \| 80.4 ± 5.7 \| 0.18 \| \| Caucasian, n (%) \| 3991 (89.7) \| 3744 (89.6) \| 162 (91.5) \| 85 (90.4) \| 0.69 \| \| College education or greater, n (%) \| 2476 (55.6) \| 2314 (55.4) \| 92 (52.0) \| 70 (74.5) \| <0.001 \| \| Fair, poor, very poor self-reported health, n (%) \| 710 (16.0) \| 639 (15.4) \| 48 (27.3) \| 23 (24.5) \| <0.001 \| \| Number of IADL impairments (range 0-5), mean ± SD \| 0.49 ± 1.0 \| 0.47 ± 1.02 \| 0.83 ± 1.19 \| 0.76 ± 1.26 \| <0.001 \| \| Walking speed, m/sec, mean ± SD \| 1.11 ± 0.2 \| 1.12 ± 0.24 \| 1.04 ± 0.28 \| 1.10 ± 0.25 \| <0.001 \| \| Inability to stand up without using arms, n (%) \| 3860 (10.0) \| 347 (9.6) \| 23 (16.2) \| 16 (19.5) \| <0.001 \| \| Any fall in previous year, n (%) \| 1371 (30.8) \| 1249 (29.9) \| 79 (44.6) \| 43 (45.7) \| <0.001 \| \| Number of co-morbidities (range 0-7), n (%) \| \| \| \| \| 0.003 \| \| None \| 2462 (55.4) \| 2332 (55.9) \| 87 (49.2) \| 43 (45.7) \| \| \| 1 co-morbidity \| 1257 (28.3) \| 1183 (28.3) \| 48 (27.1) \| 26 (27.7) \| \| \| 2+ co-morbidities \| 727 (16.4) \| 660 (15.8) \| 42 (23.7) \| 25 (26.6) \| \| \| MMSE score (range 0-100), mean ± SD \| 92.0 ± 7.23 \| 91.9 ± 7.3 \| 91.7 ± 6.6 \| 94.1 ± 5.3 \| 0.002 \| \| GDS score (range 0-15), mean ± SD \| 2.1 ± 2.4 \| 2.0 ± 2.3 \| 3.5 ± 3.4 \| 2.8 ± 2.9 \| <0.001 \| \| PASE score, mean ± SD \| 126.7 ± 70.5 \| 128.2 ± 70.5 \| 104.4 ± 68.9 \| 101.2 ± 58.9 \| <0.001 \| \| Smoking status, n (%) \| \| \| \| \| 0.59 \| \| Never \| 1721 (38.9) \| 1625 (39.1) \| 59 (33.5) \| 37 (39.4) \| \| \| Past \| 2616 (59.1) \| 2446 (58.9) \| 114 (64.8) \| 56 (59.6) \| \| \| Current \| 87 (2.0) \| 83 (2.0) \| 3 (1.7) \| 1 (11) \| \| \| ≥12 alcoholic drinks in past 12 months, n (%) \| 2697 (61.2) \| 2532 (61.2) \| 99 (56.6) \| 66 (70.2) \| 0.09 \| \| BMI, kg/m 2, mean ± SD \| 27.1 ± 3.9 \| 27.1 ± 3.9 \| 27.0 ± 4.0 \| 27.3 ± 4.6 \| 0.81 \| \| PSQI score (range 0-21), mean ± SD \| 5.5 ± 3.2 \| 5.3 ± 3.0 \| 7.9 ± 4.3 \| 8.9 ± 3.4 \| <0.001 \| \| Fell at least once in 1 st year post V3, n (%) \| 1353 (30.4) \| 1241 (29.7) \| 71 (40.1) \| 41 (43.6) \| <0.001 \| \| Fell at least twice in 1 st year post V3, n (%) \| 741 (16.7) \| 677 (16.2) \| 40 (22.6) \| 24 (25.5) \| 0.005 \| Open in a new tab P values are from ANOVA for normally distributed continuous variables and Kruskal-Wallis for skewed continuous variables. P values for categorical data are from a χ 2 test for homogeneity. Abbreviations: IADL, Instrumental Activities of Daily Living; MMSE, Mini-Mental State Examination; GDS, Geriatric Depression Scale; PASE, Physical Activity Scale for the Elderly; BMI, Body Mass Index; PSQI, Pittsburgh Sleep Quality Index Non-benzodiazepine sedative-hypnotic use and falls In age-adjusted models, non-benzodiazepine sedative-hypnotic use was associated with an increased risk of any falls (RR 1.44, 95% CI 1.15, 1.81) and recurrent falls (RR 1.5l, 95% CI 1.07, 2.14). The addition of potential confounders one at a time modestly attenuated the associations, with GDS having the greatest impact on the association between nonbenzodiazepine sedative-hypnotic use and risk of any falls (RR 1.30; 95% CI 1.07, 1.58). IADL impairment, inability to stand from a chair, and GDS all attenuated the risk of recurrent falls for non-benzodiazepine sedative-hypnotic users (Figure 1a). Figure 1a-b. Association between Non-benzodiazepine Sedative-hypnotic use, Benzodiazepine Use and Risk of Falls. Open in a new tab model adjusted for age, site, GDS score, educational status, BMI, comorbidity index, ability to stand from a chair, PASE score, self-reported health status, IADL impairments, alcohol use, and MMSE. Abbreviations: IADL, Instrumental Activities of Daily Living; MMSE, Mini-Mental State Examination; GDS, Geriatric Depression Scale; BMI, body mass index; PASE, Physical Activity Scale for the Elderly In multivariable models that simultaneously adjusted for age, site, GDS score, educational status, BMI, comorbidity index, ability to stand from a chair, PASE score, self-reported health status, IADL impairments, alcohol use, and MMSE, use of non-benzodiazepine sedative-hypnotics appeared to be associated with modest increases in the risks of any fall and recurrent falls [(any falls: RR 1.37; 95% CI 1.09, 1.71; p=0.01); (recurrent falls: RR 1.44; 95% CI 0.99, 2.09); p=0.06]. In the subset of participants who attended the Sleep visit (n=2722), results were similar (any falls: RR 1.42; 95% CI 1.08, 1.86); (recurrent falls: RR 1.44; 95% CI .95, 2.16). The addition of the PSQI to the models did not significantly change the results. There was a significant interaction between use of non-benzodiazepine sedativehypnotics and history of falls in the age-adjusted model for any falls (p=0.01); for recurrent falls the interaction was not significant (p=0.24), possibly due to limited power. We subsequently stratified on history of falls in the previous year. The risk of falling with non-benzodiazepine sedative hypnotic use appeared to be most pronounced among men without a falling history. For men without a history of previous falls, non-benzodiazepine sedative hypnotic use was associated with an increased risk of any falls in age-adjusted models (RR 1.83, 95% CI 1.23, 2.70) (Figure 2). The point estimate for the age-adjusted relative risk of recurrent falls (RR 1.77, 95% CI 0.83, 3.79) was similar to that for the risk of any falls but did not reach significance, likely due to low power. In the multivariable model, non-benzodiazepine sedative-hypnotic use was significantly associated with an increased risk of any falls [RR 1.74 (95% CI: 1.13, 2.68)]; for recurrent falls, the relative risk was 1.61 (95% CI: 0.62, 4.13). For the subset of men who had previously fallen, there was no evidence of an association between use of nonbenzodiazepine sedative-hypnotics and risk of subsequent falls in age-adjusted or multivariable models. Figure 2. Association between Non-benzodiazepine Sedative-hypnotic Use and Risk of Falls among men without a history of falls. Open in a new tab model adjusted for age, site, GDS score, educational status, BMI, comorbidity index, ability to stand from a chair, PASE score, self-reported health status, IADL impairments, alcohol use, and MMSE. Abbreviations: IADL, Instrumental Activities of Daily Living; MMSE, Mini-Mental State Examination; GDS, Geriatric Depression Scale; BMI, body mass index; PASE, Physical Activity Scale for the Elderly Benzodiazepine use and falls In age-adjusted models, benzodiazepine use was associated with an increased risk of any falls (RR 1.35; 95% CI 1.12, 1.61) and recurrent falls (RR 1.40; 95% CI 1.06, 1.85). The addition of IADL impairments, ability to stand from a chair, and GDS score each attenuated the association between benzodiazepine use and risk of any and recurrent falls (Figure 1b). Use of benzodiazepines was not associated with risk of any falls [RR 1.09 (95% CI: 0.90, 1.33)] or risk of recurrent falls [RR 1.08 (95% CI: 0.79, 1.47)] in the full multivariable model. Addition of the PSQI to the multivariable model in the subset of men with this measurement did not significantly alter the results. There was no evidence of an interaction between benzodiazepine use and history of falls (p=0.41 for any falls and p=0.70 for recurrent falls). DISCUSSION In this cohort of older men, use of non-benzodiazepine sedative-hypnotics and use of benzodiazepines were each associated with an increase in the age-adjusted risk of falling. The association between benzodiazepine use and falls appeared to be largely explained by greater disability, poorer physical performance and greater level of depressive symptoms among men taking benzodiazepines. In contrast, the association between non-benzodiazepine sedative-hypnotic use and falls was only modestly attenuated by consideration of potential confounders. Furthermore, this association depended on fall history and was most pronounced among men without a fall history. These findings are consistent with the limited existing epidemiologic data. In a case-control study of patients aged 65 years and older, investigators found a significantly increased risk of hip fracture in users of zolpidem (adjusted OR 1.95; CI 1.09-3.51), a rate similar for benzodiazepines. A recent retrospective cohort study found that the risk of nonvertebral fractures and dislocations was higher in the 90 days following an initial prescription for zolpidem than the risk prior to the prescription in members 65 years of age and older; this increased risk for zolpidem was similar to or higher than that associated with several benzodiazepines. Because these medications act via the benzodiazepine receptor-GABA complex, a similar side effect profile for non-benzodiazepine sedative-hypnotics and benzodiazepines is plausible. Indeed, clinical trials have reported comparable rates of CNS side effects, such as drowsiness, fatigue, impaired cognitive and motor function, postural sway, and ataxia.[6-8,21-24] Due to its observational nature, our analysis has several limitations, most notably confounding by indication. These medications are typically prescribed for sleep disturbances. Because sleep disturbances have been linked to an increased risk of falls, the observed association between use of non-benzodiazepine sedative-hypnotics and risk of falls may be due to the underlying sleep disturbances that prompted their initiation. To address this issue, we controlled for the PSQI in multivariable models and did not observe any significant change in our results. However, the PSQI was not available at Visit 3 in the cohort and thus we utilized a PSQI measurement from an earlier visit in a subset of our original analytic cohort. Another potential source of confounding is the possibility of channeling bias – i.e., if these drugs were preferentially prescribed to patients with a higher risk of falls due to the perception that these drugs have a safer side effect profile than benzodiazepines. We attempted to address these issues by controlling for multiple potential confounders, as well as by excluding subjects who had reported falls in the year prior to the study visit. Nevertheless, the potential for unmeasured confounding remains a limitation of this analysis. Our analysis is also limited by a lack of detailed information about the reported falls. Information regarding time of day of the falls or the mechanism of the falls is not available in the cohort. We also do not have information about dose of medication or frequency of use; it is possible than subjects were not taking these medications at the time of a fall. We found a weaker association between benzodiazepine use and risk of falls than previously reported. This finding may be due to temporal trends in benzodiazepine prescription patterns, due to growing recognition of the adverse effects of benzodiazepines in the elderly. We also found that the association between use of non-benzodiazepine sedative hypnotics and risk of falls was more pronounced for men who had not fallen in the previous year. This finding may be due to the strong association between previous falls and risk of subsequent falls: the additional risk of a sedative hypnotic may be negligible for men already falling, whereas for men not previously falling, the addition of one of these medications may “tip the balance.” Despite these limitations, this work adds to the growing body of literature suggesting that use of the non-benzodiazepine sedative-hypnotics may be associated with adverse effects such as falls. Future randomized clinical trials of this class of medications should include falls as a safety outcome. In addition, non-pharmacologic approaches to sleep disturbances, such as cognitive behavioral therapy, may represent the safest approach to this common problem in older adults. ACKNOWLEDGMENTS Results were presented at the Gerontological Society of America annual meeting on November 15, 2012 in San Diego, CA. FUNDING The Osteoporotic Fractures in Men (MrOS) Study is supported by National Institutes of Health funding. The following institutes provide support: the National Institute of Arthritis and Musculoskeletal and Skin Diseases (NIAMS), the National Institute on Aging (NIA), the National Center for Research Resources (NCRR), and NIH Roadmap for Medical Research under the following grant numbers: U01 AR45580, U01 AR45614, U01 AR45632, U01 AR45647, U01 AR45654, U01 AR45583, U01 AG18197, U01 AG027810, and UL1 RR024140. 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8784	https://www.osmosis.org/learn/Antibiotics_-_Beta_lactam_and_beta_lactamase_inhibitor_combinations:_Nursing_pharmacology	SALE EXTENDED! Act fast, these savings won't last much longer. Lowest prices only available until September 2 at 11:59 PM PT. Unlock savings Skip to the video Antibiotics - Beta lactam and beta lactamase inhibitor combinations: Nursing pharmacology Nursing School CoursesNursing pharmacologyAnti-infective medications No notes for this video yet Try adding a note below Assessments Review questions 0 / 8 complete Want to see more? Log in or sign up to access the full notes and additional features. Notes \| \| \| \| --- \| ANTIBIOTICS: BETA LACTAM/BETA LACTAMASE INHIBITOR COMBINATIONS \| \| \| \| DRUG NAME \| Penicillins: ampicillin/sulbactam (Unasyn), amoxicillin/clavulanate (Augmentin), piperacillin/tazobactam (Zosyn) Cephalosporins: ceftolozane/tazobactam (Zerbaxa), ceftazidime/avibactam (Avycaz) \| \| \| CLASS \| Beta-lactam/beta-lactamase inhibitor combinations \| \| \| MECHANISM of ACTION \| Beta-lactamase inhibitors bind to beta-lactamase enzymes in the bacteria and prevent the breakdown of the beta-lactam ring in the antibiotic molecule \| \| \| INDICATIONS \| Infections caused by penicillin-resistant bacteria like Haemophilus influenzae, methicillin-sensitiveStaphylococcus aureus, Bacteroides fragilis,Proteus spp., Escherichia coli, Klebsiella spp.,and Acinetobacter spp. Cephalosporin/beta-lactamase inhibitor combination: Pseudomonas aeruginosa \| \| \| ROUTE(S) of ADMIN \| Amoxicillin-clavulanate, ceftolozane-tazobactam: PO Ampicillin-sulbactam: IV, IM Piperacillin-tazobactam, ceftazidime-avibactam: IV \| \| \| SIDE EFFECTS \| Allergic reactions (e.g., urticaria, pruritus, anaphylaxis) Superinfections (e.g., vaginitis, candidiasis, Clostridioides difficile infection) Nausea, gastrointestinal upset, diarrhea, taste disturbance Clavulanate: increased gastrointestinal side effects; cholestatic hepatitis \| \| \| CONTRAINDICATIONS & CAUTIONS \| Hypersensitivity to other beta-lactam antibiotics Hepatic or renal disease Pregnancy and breastfeeding Drug interactions: Aminoglycoside antibiotics Oral contraceptive medications \| \| \| \| \| \| --- \| NURSING CONSIDERATIONS for ANTIBIOTICS: BETA LACTAM/BETA LACTAMASE INHIBITOR COMBINATIONS \| \| \| \| ASSESSMENT & MONITORING \| Assessment Current signs of infection: unusual fussiness, decreased appetite, rubbing or tugging on the ear Weight Vital signs Presence of drainage in ear canal Laboratory test results: renal and liver function, CBC Monitoring Side effects Evaluate the therapeutic effect: resolution of the ear infection and absence of symptoms \| \| \| CLIENT EDUCATION \| About otitis media Purpose of medication: to treat the ear infection and solve the child’s symptoms Administration + Take with food at evenly spaced intervals + Shake oral solution well + Use oral solution dispenser + May take with juice, milk, formula + Complete entire course of antibiotic Side effects + Nausea, vomiting, diarrhea - Increase fluid intake, frequent diaper changes + Notify emergency services - Allergic reaction: itchy rash, swelling of the face or throat, difficulty breathing + Notify healthcare provider - Superinfection: white patches on the inner cheeks, tongue, roof of the mouth, severe or bloody diarrhea - Liver impairment Fatigue, anorexia, dark urine, yellowing of the skin or eyes \| \| Transcript Watch video only Content Reviewers Antonella Melani, MD, Lisa Miklush, PhD, RNC, CNS, Ashley Mauldin MSN, APRN, FNP-BC, Gabrielle Proper, RN, BScN, MN Contributors Kaylee Neff, Aileen Lin, MScBMC Beta-lactams are a group of antibiotics that contain a beta-lactam ring in their structure that is essential for the antibacterial activity of antibiotics like penicillins and cephalosporins. However, over time, due to their widespread use, some bacteria have acquired resistance by developing enzymes called beta-lactamases. For this reason, some beta-lactam antibiotics are often combined with a class of medications known as beta-lactamase inhibitors in order to treat infections caused by beta-lactam-resistant bacteria like Haemophilus influenzae, methicillin-sensitive Staphylococcus aureus, Bacteroides fragilis, Proteus spp., Escherichia coli, Klebsiella spp., and Acinetobacter spp. Additionally, cephalosporin-containing combinations are effective against Pseudomonas aeruginosa. Now, beta-lactam/beta-lactamase inhibitor combinations typically contain either a penicillin or a cephalosporin combined with a beta-lactamase inhibitor. Commonly used penicillin beta-lactamase inhibitor combinations include amoxicillin-clavulanate, which is given orally; ampicillin-sulbactam, which is given intramuscularly and intravenously; and piperacillin-tazobactam, which is given intravenously. Commonly used cephalosporin beta-lactamase inhibitor combinations include ceftazidime-avibactam, which is given intravenously; and ceftolozane-tazobactam, which is given orally. Once administered, beta-lactam antibiotics act by inhibiting bacterial cell wall synthesis, which kills the bacteria. However, in resistant bacteria, the beta-lactamase enzyme binds to the beta-lactam ring within the antibiotic and breaks it down, thus inactivating the antibiotic. This is where the beta-lactamase inhibitors come into play, by binding to the beta-lactamase enzymes in the bacteria. As a result, beta-lactamase inhibitors prevent the beta-lactamase enzyme from inactivating the beta-lactam antibiotic. As a result, the antibiotics are left free to go about their job of killing the bacteria. The beta-lactam/beta-lactamase inhibitor combinations are generally well tolerated, and the few side effects they produce are usually the results of the antibiotic, or beta-lactam, counterpart. Allergy to beta-lactams is quite common, and clients typically present with urticaria, pruritus, and some can even progress to anaphylaxis. Other common side effects include headaches and gastrointestinal disturbances, such as nausea, vomiting, and diarrhea. Beta-lactams can also affect the healthy bacterial flora, making clients more susceptible to superinfections by more resistant pathogens, including vaginitis, candidiasis, and more rarely, Clostridioides difficile infection or CDI for short. In particular, clavulanate increases the gastrointestinal side effects of amoxicillin, causing antibiotic-associated diarrhea, which can range from frequent loose stools to more severe colitis. Additionally, clients on amoxicillin-clavulanate can also develop cholestatic hepatitis. Now, beta-lactam/beta-lactamase inhibitor combinations are contraindicated in clients allergic to any beta-lactam antibiotics due to the risk of cross-reactivity between the different classes. They should also be used with caution in clients with hepatic or renal disease, as well as during pregnancy and breastfeeding. With respect to drug interactions, the beta-lactam/beta-lactamase inhibitor combinations should not be given along with aminoglycoside antibiotics, since this combination can result in reduced antibacterial effects of both medications. Finally, penicillins increase the effects of methotrexate and warfarin and decrease the effects of oral contraceptives. Okay, when caring for a pediatric client with otitis media who has been prescribed a beta-lactam/beta-lactamase inhibitor combination like amoxicillin-clavulanate, first perform a baseline assessment by asking the parents or caregivers about current signs of infection, such as unusual fussiness, decreased appetite, as well as rubbing or tugging on the ear. Then, assess weight, vital signs, especially temperature; and the ear canal, making note of the presence of any drainage. Lastly, review recent laboratory test results including renal and liver function tests, as well as CBC. Summary Beta-lactam antibiotics, named after their beta-lactam ring in their chemical structure, are a type of antibiotic that kills bacteria. Beta-lactamase is an enzyme that these bacteria produce to disable beta-lactam antibiotics. Beta-lactamase inhibitors are another type of antibiotic that are co-administered with beta-lactam antibiotics, to prevent bacteria from disabling these antibiotics using their enzymes. The combination of a beta-lactam and a beta-lactamase inhibitor is called a "beta-lactam/beta-lactamase inhibitor" (BL/BLI) combination. BL/BLI combinations are used to treat many types of infections, including lung infections, skin infections, urinary tract infections, and stomach infections. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. For more information, see ourCookie Policy Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. 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8785	https://courses.lumenlearning.com/calculus2/chapter/geometric-calculations-of-parametric-curves/	Geometric Calculations of Parametric Curves Learning Outcomes Find the area under a parametric curve Use the equation for arc length of a parametric curve Apply the formula for surface area to a volume generated by a parametric curve Integrals Involving Parametric Equations Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations x(t)=t−sint,y(t)=1−costx(t)=t−sint,y(t)=1−cost. Suppose we want to find the area of the shaded region in the following graph. To derive a formula for the area under the curve defined by the functions x=x(t),y=y(t),a≤t≤bx=x(t),y=y(t),a≤t≤b, we assume that x(t)x(t) is differentiable and start with an equal partition of the interval a≤t≤ba≤t≤b. Suppose t0=a<t1<t2<⋯<tn=bt0=a<t1<t2<⋯<tn=b and consider the following graph. We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is y(x(¯ti))y(x(¯ti)) for some value ¯ti¯ti in the ith subinterval, and the width can be calculated as x(ti)−x(ti−1)x(ti)−x(ti−1). Thus the area of the ith rectangle is given by Ai=y(x(¯ti))(x(ti)−x(ti−1))Ai=y(x(¯ti))(x(ti)−x(ti−1)). Then a Riemann sum for the area is An=n∑i=1y(x(¯ti))(x(ti)−x(ti−1))An=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)). Multiplying and dividing each area by ti−ti−1ti−ti−1 gives An=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)ti−ti−1)(ti−ti−1)=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)Δt)ΔtAn=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)ti−ti−1)(ti−ti−1)=n∑i=1y(x(¯ti))(x(ti)−x(ti−1)Δt)Δt. Taking the limit as nn approaches infinity gives A=limn→∞An=∫bay(t)x′(t)dtA=limn→∞An=∫bay(t)x′(t)dt. Note that as the value of nn approaches ∞∞, the change in xx over smaller and smaller time intervals can be rewritten as the instantaneous rate of change of xx with respect to tt at some value within the sub-interval of tt. Recall that this fact is known as the Mean Value Theorem, and we briefly review it to clarify the proofs within this section. Recall: Mean Value Theorem Let ff be continuous over the closed interval [a,b][a,b] and differentiable over the open interval (a,b)(a,b). Then, there exists at least one point c∈(a,b)c∈(a,b) such that f′(c)=f(b)−f(a)b−af′(c)=f(b)−f(a)b−a The preceding result leads to the following theorem. theorem: Area under a Parametric Curve Consider the non-self-intersecting plane curve defined by the parametric equations x=x(t),y=y(t),a≤t≤bx=x(t),y=y(t),a≤t≤b and assume that x(t)x(t) is differentiable. The area under this curve is given by A=∫bay(t)x′(t)dtA=∫bay(t)x′(t)dt. Example: Finding the Area under a Parametric Curve Find the area under the curve of the cycloid defined by the equations x(t)=t−sint,y(t)=1−cost,0≤t≤2πx(t)=t−sint,y(t)=1−cost,0≤t≤2π. Show Solution Using the theorem, we have A=∫bay(t)x′(t)dt=∫2π0(1−cost)(1−cost)dt=∫2π0(1−2cost+cos2t)dt=∫2π0(1−2cost+1+cos2t2)dt=∫2π0(32−2cost+cos2t2)dt=3t2−2sint+sin2t4\|2π0=3π.A=∫bay(t)x′(t)dt=∫2π0(1−cost)(1−cost)dt=∫2π0(1−2cost+cos2t)dt=∫2π0(1−2cost+1+cos2t2)dt=∫2π0(32−2cost+cos2t2)dt=3t2−2sint+sin2t4\|2π0=3π. Watch the following video to see the worked solution to Example: Finding the Area under a Parametric Curve. For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window). try it Find the area under the curve of the hypocycloid defined by the equations x(t)=3cost+cos3t,y(t)=3sint−sin3t,0≤t≤πx(t)=3cost+cos3t,y(t)=3sint−sin3t,0≤t≤π. Hint Use the theorem, along with the identities sinαsinβ=12[cos(α−β)−cos(α+β)]sinαsinβ=12[cos(α−β)−cos(α+β)] and sin2t=1−cos2t2sin2t=1−cos2t2. Show Solution A=3πA=3π (Note that the integral formula actually yields a negative answer. This is due to the fact that x(t)x(t) is a decreasing function over the interval [0,2π][0,2π]; that is, the curve is traced from right to left.) Try It Arc Length of a Parametric Curve In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph. Given a plane curve defined by the functions x=x(t),y=y(t),a≤t≤bx=x(t),y=y(t),a≤t≤b, we start by partitioning the interval [a,b][a,b] into n equal subintervals: t0=a<t1<t2<⋯<tn=bt0=a<t1<t2<⋯<tn=b. The width of each subinterval is given by Δt=(b−a)nΔt=(b−a)n. We can calculate the length of each line segment: d1=√(x(t1)−x(t0))2+(y(t1)−y(t0))2d2=√(x(t2)−x(t1))2+(y(t2)−y(t1))2etc.d1=√(x(t1)−x(t0))2+(y(t1)−y(t0))2d2=√(x(t2)−x(t1))2+(y(t2)−y(t1))2etc. Then add these up. We let s denote the exact arc length and snsn denote the approximation by n line segments: s≈n∑k=1sk=n∑k=1√(x(tk)−x(tk−1))2+(y(tk)−y(tk−1))2s≈n∑k=1sk=n∑k=1√(x(tk)−x(tk−1))2+(y(tk)−y(tk−1))2. If we assume that x(t)x(t) and y(t)y(t) are differentiable functions of t, then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval [tk−1,tk][tk−1,tk] there exist ˆtk^tk and ˜tk~tk such that x(tk)−x(tk−1)=x′(ˆtk)(tk−tk−1)=x′(ˆtk)Δty(tk)−y(tk−1)=y′(˜tk)(tk−tk−1)=y′(˜tk)Δt.x(tk)−x(tk−1)=x′(^tk)(tk−tk−1)=x′(^tk)Δty(tk)−y(tk−1)=y′(~tk)(tk−tk−1)=y′(~tk)Δt. Therefore ss becomes s≈n∑k=1sk=n∑k=1√(x′(ˆtk)Δt)2+(y′(˜tk)Δt)2=n∑k=1√(x′(ˆtk))2(Δt)2+(y′(˜tk))2(Δt)2=(n∑k=1√(x′(ˆtk))2+(y′(˜tk))2)Δt.s≈n∑k=1sk=n∑k=1√(x′(^tk)Δt)2+(y′(~tk)Δt)2=n∑k=1√(x′(^tk))2(Δt)2+(y′(~tk))2(Δt)2=(n∑k=1√(x′(^tk))2+(y′(~tk))2)Δt. This is a Riemann sum that approximates the arc length over a partition of the interval [a,b][a,b]. If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives s=limn→∞n∑k=1sk=limn→∞(n∑k=1√(x′(ˆtk))2+(y′(˜tk))2)Δt=∫ba√(x′(t))2+(y′(t))2dt.s=limn→∞n∑k=1sk=limn→∞(n∑k=1√(x′(^tk))2+(y′(~tk))2)Δt=∫ba√(x′(t))2+(y′(t))2dt. When taking the limit, the values of ˆtk^tk and ˜tk~tk are both contained within the same ever-shrinking interval of width ΔtΔt, so they must converge to the same value. We can summarize this method in the following theorem. theorem: Arc Length of a Parametric Curve Consider the plane curve defined by the parametric equations x=x(t),y=y(t),t1≤t≤t2x=x(t),y=y(t),t1≤t≤t2 and assume that x(t)x(t) and y(t)y(t) are differentiable functions of t. Then the arc length of this curve is given by s=∫t2t1√(dxdt)2+(dydt)2dts=∫t2t1√(dxdt)2+(dydt)2dt. At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function y=F(x)y=F(x). Then y(t)=F(x(t))y(t)=F(x(t)) and the Chain Rule gives y′(t)=F′(x(t))x′(t)y′(t)=F′(x(t))x′(t). Substituting this into the theorem gives s=∫t2t1√(dxdt)2+(dydt)2dt=∫t2t1√(dxdt)2+(F′(x)dxdt)2dt=∫t2t1√(dxdt)2(1+(F′(x))2)dt=∫t2t1x′(t)√1+(dydx)2dt.s=∫t2t1√(dxdt)2+(dydt)2dt=∫t2t1√(dxdt)2+(F′(x)dxdt)2dt=∫t2t1√(dxdt)2(1+(F′(x))2)dt=∫t2t1x′(t)√1+(dydx)2dt. Here we have assumed that x′(t)>0x′(t)>0, which is a reasonable assumption. The Chain Rule gives dx=x′(t)dtdx=x′(t)dt, and letting a=x(t1)a=x(t1) and b=x(t2)b=x(t2) we obtain the formula s=∫ba√1+(dydx)2dxs=∫ba√1+(dydx)2dx, which is the formula for arc length obtained in the Introduction to the Applications of Integration. Example: Finding the Arc Length of a Parametric Curve Find the arc length of the semicircle defined by the equations x(t)=3cost,y(t)=3sint,0≤t≤πx(t)=3cost,y(t)=3sint,0≤t≤π. Show Solution The values t=0t=0 to t=πt=π trace out the red curve in Figure 9. To determine its length, use the theorem: s=∫t2t1√(dxdt)2+(dydt)2dt=∫π0√(−3sint)2+(3cost)2dt=∫π0√9sin2t+9cos2tdt=∫π0√9(sin2t+cos2t)dt=∫π03dt=3t\|π0=3π.s=∫t2t1√(dxdt)2+(dydt)2dt=∫π0√(−3sint)2+(3cost)2dt=∫π0√9sin2t+9cos2tdt=∫π0√9(sin2t+cos2t)dt=∫π03dt=3t\|π0=3π. Note that the formula for the arc length of a semicircle is πrπr and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity. Watch the following video to see the worked solution to Example: Finding the Arc Length of a Parametric Curve. For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window). try it Find the arc length of the curve defined by the equations x(t)=3t2,y(t)=2t3,1≤t≤3x(t)=3t2,y(t)=2t3,1≤t≤3. Hint Use the theorem. Show Solution s=2(1032−232)≈57.589s=2(1032−232)≈57.589 Try It We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive x-axis, the parametric equations for this curve can be written as x(t)=140t,y(t)=−16t2+2tx(t)=140t,y(t)=−16t2+2t where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions x(t)x(t) and y(t)y(t) using v as an independent variable, so as to eliminate any confusion with the parameter t: x(v)=140v,y(v)=−16v2+2vx(v)=140v,y(v)=−16v2+2v. Then we write the arc length formula as follows: s(t)=∫t0√(dxdv)2+(dydv)2dv=∫t0√1402+(−32v+2)2dv.s(t)=∫t0√(dxdv)2+(dydv)2dv=∫t0√1402+(−32v+2)2dv. The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression we can use a formula from Appendix A, ∫√a2+u2du=u2√a2+u2+a22ln\|u+√a2+u2\|+C∫√a2+u2du=u2√a2+u2+a22ln\|u+√a2+u2\|+C. We set a=140a=140 and u=−32v+2u=−32v+2. This gives du=−32dvdu=−32dv, so dv=−132dudv=−132du. Therefore ∫√1402+(−32v+2)2dv=−132∫√a2+u2du=−132[(−32v+2)2√1402+(−32v+2)2+14022ln\|(−32v+2)+√1402+(−32v+2)2\|]+C∫√1402+(−32v+2)2dv=−132∫√a2+u2du=−132⎡⎢ ⎢⎣(−32v+2)2√1402+(−32v+2)2+14022ln\|(−32v+2)+√1402+(−32v+2)2\|⎤⎥ ⎥⎦+C and s(t)=−132[(−32t+2)2√1402+(−32t+2)2+14022ln\|(−32t+2)+√1402+(−32t+2)2\|]+132[√1402+22+14022ln\|2+√1402+22\|]=(t2−132)√1024t2−128t+19604−12254ln\|(−32t+2)+√1024t2−128t+19604\|+√1960432+12254ln(2+√19604).s(t)=−132[(−32t+2)2√1402+(−32t+2)2+14022ln\|(−32t+2)+√1402+(−32t+2)2\|]+132[√1402+22+14022ln\|2+√1402+22\|]=(t2−132)√1024t2−128t+19604−12254ln\|(−32t+2)+√1024t2−128t+19604\|+√1960432+12254ln(2+√19604). This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus: ddx∫xaf(u)du=f(x)ddx∫xaf(u)du=f(x). Therefore s′(t)=ddt[s(t)]=ddt[∫t0√1402+(−32v+2)2dv]=√1402+(−32t+2)2=√1024t2−128t+19604=2√256t2−32t+4901.s′(t)=ddt[s(t)]=ddt[∫t0√1402+(−32v+2)2dv]=√1402+(−32t+2)2=√1024t2−128t+19604=2√256t2−32t+4901. One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to s(13)=(132−132)√1024(13)2−128(13)+19604−12254ln\|(−32(13)+2)+√1024(13)2−128(13)+19604\|+√1960432+12254ln(2+√19604)≈46.69feet.s(13)=(132−132)√1024(13)2−128(13)+19604−12254ln\|(−32(13)+2)+√1024(13)2−128(13)+19604\|+√1960432+12254ln(2+√19604)≈46.69feet. This value is just over three quarters of the way to home plate. The speed of the ball is s′(13)=2√256(13)2−16(13)+4901≈140.34ft/ss′(13)=2√256(13)2−16(13)+4901≈140.34ft/s. This speed translates to approximately 95 mph—a major-league fastball. Surface Area Generated by a Parametric Curve Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function y=f(x)y=f(x) from x=ax=a to x=bx=b, revolved around the x-axis: S=2π∫baf(x)√1+(f′(x))2dxS=2π∫baf(x)√1+(f′(x))2dx. We now consider a volume of revolution generated by revolving a parametrically defined curve x=x(t),y=y(t),a≤t≤bx=x(t),y=y(t),a≤t≤b around the x-axis as shown in the following figure. The analogous formula for a parametrically defined curve is S=2π∫bay(t)√(x′(t))2+(y′(t))2dtS=2π∫bay(t)√(x′(t))2+(y′(t))2dt provided that y(t)y(t) is not negative on [a,b][a,b]. Example: Finding Surface Area Find the surface area of a sphere of radius rr centered at the origin. Show Solution We start with the curve defined by the equations x(t)=rcost,y(t)=rsint,0≤t≤πx(t)=rcost,y(t)=rsint,0≤t≤π. This generates an upper semicircle of radius r centered at the origin as shown in the following graph. When this curve is revolved around the xx-axis, it generates a sphere of radius rr. To calculate the surface area of the sphere, we use the above equation: S=2π∫bay(t)√(x′(t))2+(y′(t))2dt=2π∫π0rsint√(−rsint)2+(rcost)2dt=2π∫π0rsint√r2sin2t+r2cos2tdt=2π∫π0rsint√r2(sin2t+cos2t)dt=2π∫π0r2sintdt=2πr2(−cost\|π0)=2πr2(−cosπ+cos0)=4πr2.S=2π∫bay(t)√(x′(t))2+(y′(t))2dt=2π∫π0rsint√(−rsint)2+(rcost)2dt=2π∫π0rsint√r2sin2t+r2cos2tdt=2π∫π0rsint√r2(sin2t+cos2t)dt=2π∫π0r2sintdt=2πr2(−cost\|π0)=2πr2(−cosπ+cos0)=4πr2. This is, in fact, the formula for the surface area of a sphere. Watch the following video to see the worked solution to Example: Finding Surface Area. For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end. You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window). try it Find the surface area generated when the plane curve defined by the equations x(t)=t3,y(t)=t2,0≤t≤1x(t)=t3,y(t)=t2,0≤t≤1 is revolved around the x-axis. Hint Use the above equation. When evaluating the integral, use a u-substitution. Show Solution A=π(494√13+128)1215A=π(494√13+128)1215 Candela Citations CC licensed content, Original 7.2 Calculus of Parametric Curves. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 2. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at Licenses and Attributions CC licensed content, Original 7.2 Calculus of Parametric Curves. Authored by: Ryan Melton. License: CC BY: Attribution CC licensed content, Shared previously Calculus Volume 2. Authored by: Gilbert Strang, Edwin (Jed) Herman. Provided by: OpenStax. Located at: License: CC BY-NC-SA: Attribution-NonCommercial-ShareAlike. License Terms: Access for free at
8786	https://math.stackexchange.com/questions/563771/prove-that-if-a-function-f-has-a-jump-at-an-interior-point-of-the-interval-a	calculus - Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function. - 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 Prove that if a function f f has a jump at an interior point of the interval [a,b][a,b] then it cannot be the derivative of any function. Ask Question Asked 11 years, 10 months ago Modified7 years, 4 months ago Viewed 6k times This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. Prove that if a function f f has a jump at an interior point of the interval [a,b][a,b] then it cannot be the derivative of any function. I know that for f f is differentiable in (a,b)(a,b) and that it has one-sided derivative f′+(a)≠f′−(b)f+′(a)≠f−′(b) at the endpoints. If C C is a real number between f′+(a)f+′(a) and f′−(b)f−′(b), then there exists c∈(a,b)c∈(a,b) such that f′(c)=C f′(c)=C. How can I use this to prove the above? calculus proof-writing Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 12, 2013 at 14:30 Diane VanderwaifDiane Vanderwaif asked Nov 12, 2013 at 12:15 Diane VanderwaifDiane Vanderwaif 4,480 8 8 gold badges 53 53 silver badges 112 112 bronze badges 10 4 See en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis).lhf –lhf 2013-11-12 12:16:38 +00:00 Commented Nov 12, 2013 at 12:16 What do you mean by "can not be derivative at any function"? As it is written it doesn't make sense. Is " of any function " instead?Sergio Parreiras –Sergio Parreiras 2013-11-12 12:29:05 +00:00 Commented Nov 12, 2013 at 12:29 3 You should at least look at the link of lhf's comment above. It has the answer to your question in it.coffeemath –coffeemath 2013-11-16 14:31:36 +00:00 Commented Nov 16, 2013 at 14:31 1 @DianeVanderwaif, why did you started TWO bounties for the same question in the same day? For those who interested the same question is hereNorbert –Norbert 2013-11-22 10:32:14 +00:00 Commented Nov 22, 2013 at 10:32 2 To copper.hat: This is very strange. What you mean to say is that any function which satisfies intermediate value theorem can't have jump discontinuity. This is plain wrong. However if we add the constraint of monotonicity then it is OK. Thus a monotone function satisfying IVT can't have jump discontinuity. The proof that "derivatives don't have jump discontinuity" is based on mean value theorem as shown in my answer to this problem.Paramanand Singh –Paramanand Singh♦ 2013-11-23 09:21:48 +00:00 Commented Nov 23, 2013 at 9:21 \|Show 5 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 14 Save this answer. +25 This answer has been awarded bounties worth 25 reputation by Community Show activity on this post. This a standard result that derivatives don't have jump discontinuity. Let c∈(a,b)c∈(a,b) then f′(c)=lim x→c f(x)−f(c)x−c f′(c)=lim x→c f(x)−f(c)x−c exists. Let us assume that limits lim x→c+f′(x)=A,lim x→c−f′(x)=B lim x→c+f′(x)=A,lim x→c−f′(x)=B exist. Now let's handle the case for x→c+x→c+ first. Clearly then x>c x>c and we have f(x)−f(c)x−c=f′(d)f(x)−f(c)x−c=f′(d) for some d∈(c,x)d∈(c,x). As x→c+x→c+, d→c+d→c+ and we get f′(c)=lim x→c+f(x)−f(c)x−c=lim d→c+f′(d)=A f′(c)=lim x→c+f(x)−f(c)x−c=lim d→c+f′(d)=A Similarly by considering x→c−x→c− we can show that B=f′(c)B=f′(c) so that A=B A=B and f′(x)f′(x) is continuous at c c and therefore does not have jump discontinuity. It may happen however that one or both of the limits A,B A,B don't exist or are ±∞±∞. Update: I am bit surprised to see that in the comments to the question people have linked this result with IVT (intermediate value theorem) for derivatives. These two properties of derivatives (IVT and no jump discontinuity) are not derivable from each other. Rather they are both derived from Mean Value Theorem in completely different ways. ~~Further Update: I had a look at the Wikipedia article dealing with Darboux theorem (IVT for derivatives). Even the wikipedia makes a mistake that any function satisfying IVT can't have jump discontinuity. This is totally unexpected from wikipedia and I don't know whom to complain for this.~~ ~~ Here is a very simple example to prove my point. Let f(0)=0,f(1)=1 f(0)=0,f(1)=1 and f(x)=1−x f(x)=1−x for x∈(0,1)x∈(0,1). This function satisfies IVT on [0,1][0,1] and is yet having jumps at the end-points. What is true is the following: ~~ ~~A function f f which is monotone and satisfies IVT on [a,b][a,b] does not have jump discontinuity and is therefore continuous in [a,b][a,b]~~ Even more update: Due to the paraphrasing of the comment by copper.hat in the question I misinterpreted the Wikipedia article. According to copper.hat comment if g(x)g(x) takes all values in interval [g(a),g(b)][g(a),g(b)] as x x varies in [a,b][a,b] then g(x)g(x) can't have jumps in [a,b][a,b]. This statement is wrong. Wikipedia however has a different definition. It says that a function is Darboux function if it satisfied intermediate value property. The intermediate value property is defined as follows: let f f be defined on interval I I. If for any [a,b]⊆I[a,b]⊆I the function f f takes all values between f(a)f(a) and f(b)f(b) for some value of x∈(a,b)x∈(a,b) then it is said to have intermediate value property on I I. I missed the part of any [a,b]⊆I[a,b]⊆I and thought that intermediate value property of f f on an interval [a,b][a,b] is supposed to mean that f f must take all values between f(a)f(a) and f(b)f(b) for some x∈(a,b)x∈(a,b). Note the subtle difference in Wikipedia version and my interpretation. Wikipedia prescribes a very strong condition where we have to check every subinterval [a,b][a,b] of the domain of defintion I I of function f f whereas in my interpretation we only need to check this for I I and not any subintervals of I I. To put formally let Wikipedia version of IVT be called WIVT and my version be called PIVT. Then a function f f satisfies WIVT if it satisfies PIVT on every sub-interval of I I. A function satisfying WIVT does not have jumps whereas a function satisfying PIVT may have jumps. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 23, 2013 at 12:21 answered Nov 23, 2013 at 9:12 Paramanand Singh♦Paramanand Singh 92.6k 15 15 gold badges 159 159 silver badges 349 349 bronze badges 9 This proof is false. The relevant result is the Darboux theorem which is a bit deeper.Selim Ghazouani –Selim Ghazouani 2013-11-23 09:57:02 +00:00 Commented Nov 23, 2013 at 9:57 1 can you explain what is the fault in my proof? I have mentioned that wikipedia article is wrong so don't give wiki references. And Darboux theorem has nothing to do with the question here.Paramanand Singh –Paramanand Singh♦ 2013-11-23 09:58:09 +00:00 Commented Nov 23, 2013 at 9:58 and if you are unable to find fault here please cancel your downvote. I am not desperate for reputation. But I definitely don't want to be penalized without any reason.Paramanand Singh –Paramanand Singh♦ 2013-11-23 09:59:37 +00:00 Commented Nov 23, 2013 at 9:59 2 Thanks Selim for response. But I proved that if A A and B B exist then they must be equal. I have also mentioned in my answer that A,B A,B may not exist. Also when x→c+x→c+ then since c<d<x c<d<x we must have d→c+d→c+. Existence of the limit lim x→c+f′(x)lim x→c+f′(x) is an assumption. What I want to show is that if this limit exists then it must equal f′(c)f′(c).Paramanand Singh –Paramanand Singh♦ 2013-11-23 11:36:30 +00:00 Commented Nov 23, 2013 at 11:36 1 I have not proved that f′(x)f′(x) is guaranteed to be continuous. But I proved that if both the limits lim x→c+f′(x)lim x→c+f′(x) and lim x→c+f′(x)lim x→c+f′(x) exist then they must be equal to f′(c)f′(c). Note the word "if" in bold in last sentence.Paramanand Singh –Paramanand Singh♦ 2013-11-23 11:38:53 +00:00 Commented Nov 23, 2013 at 11:38 \|Show 4 more comments This answer is useful 3 Save this answer. Show activity on this post. Just trying to add some clarity to the tricky part of the proofs presented above. Let f:[a,b]→R f:[a,b]→R be differentiable in (a,b)(a,b) and c c and interior point of the interval, c∈(a,b)c∈(a,b). We'll prove that lim x→c+f′(x)=A lim x→c−f′(x)=B⎫⎭⎬⎪⎪⇒A=B=f′(c)lim x→c+f′(x)=A lim x→c−f′(x)=B}⇒A=B=f′(c) For each x∈(c,b)x∈(c,b), f f is continuous in [c,x][c,x] and differentiable in (c,x)(c,x) so we can apply the mean value theorem to f f on the interval [c,x][c,x]. Thus, for each interval [c,x][c,x], there exists a d(x)∈(c,x)d(x)∈(c,x) such that f′(d(x))=f(x)−f(c)x−c f′(d(x))=f(x)−f(c)x−c Since f f is differentiable at c c, making x→c+x→c+ on the RHS of the expression gives: lim x→c+f(x)−f(c)x−c=f′(c)lim x→c+f(x)−f(c)x−c=f′(c) The LHS of the equality is a bit trickier, we would like to say that lim x→c+f′(d(x))=lim x→c+f′(x)=A lim x→c+f′(d(x))=lim x→c+f′(x)=A but that is not obvious, since we can't assume f′f′ to be continuous (on the right) at c c. However we know that lim x→c+f′(x)=A lim x→c+f′(x)=A, we'll use ϵ−δ ϵ−δ definition of limit. Let ϵ>0 ϵ>0, ∃δ>0:if 0<x−c<δ then\|f′(x)−A\|<ϵ∃δ>0:if 0<x−c<δ then\|f′(x)−A\|<ϵ (notice that we write 0<x−c<δ 0<x−c<δ instead of 0<\|x−c\|<δ 0<\|x−c\|<δ because we're taking the right hand limit, that is x>c x>c). The MVT guarantees that d(x)∈(c,x)d(x)∈(c,x), that is c<d(x)<x⇒0<d(x)−c<x−c c<d(x)<x⇒0<d(x)−c<x−c But this means 0<x−c<δ⇒0<d(x)−c<δ⇒\|f′(d(x))−A\|<ϵ 0<x−c<δ⇒0<d(x)−c<δ⇒\|f′(d(x))−A\|<ϵ thus lim x→c+f′(d(x))=A lim x→c+f′(d(x))=A as desired. We've seen f′(d(x))=f(x)−f(c)x−c−→−−x→c+A=f′(c)f′(d(x))=f(x)−f(c)x−c→x→c+A=f′(c) Similarly, we can prove B=f′(c)B=f′(c) and we're done. The answer to your question follows directly from this result. Let f f be differentiable in (a,c)∪(c,b)(a,c)∪(c,b). Assume f′f′ has a jump discontinuity at c c: lim x→c+f′(x)=A≠B=lim x→c−f′(x)lim x→c+f′(x)=A≠B=lim x→c−f′(x) (to say we have a jump we require the limits to \textbf{exist} and to be different!). By the result, if f f was also differentiable at c c, we'd have A=B A=B, that contradicts the hypothesis of having a jump: A≠B A≠B. Therefore f f can't be differentiable at c c. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered May 26, 2018 at 8:02 Gerard GraciaGerard Gracia 341 1 1 silver badge 7 7 bronze badges 1 Good proof. One comment, though. I think you should clarify that, for example, lim x→c+f(x)−f(c)x−c=f′(c)lim x→c+f(x)−f(c)x−c=f′(c) , is not technically accurate. Rather, you have that lim x→c+f(x)−f(c)x−c=f′+(c)lim x→c+f(x)−f(c)x−c=f′+(c) i.e. this is the definition of the right-side derivative. Similarly, lim x→c−f(c)−f(x)c−x=f′(c)lim x→c−f(c)−f(x)c−x=f′(c) , is incorrect and should instead be written as: lim x→c−f(c)−f(x)c−x=f′−(c)lim x→c−f(c)−f(x)c−x=f′−(c) NOW, by assumption, we know that f′(c)f′(c) exists. Therefore, the left and right hand derivatives must be equal. Then your argument proceeds.S.C. –S.C. 2022-09-21 18:43:57 +00:00 Commented Sep 21, 2022 at 18:43 Add a comment\| You must log in to answer this question. 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Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in Calculus Volume 3 4.2 Limits and Continuity Calculus Volume 34.2 Limits and Continuity Contents Contents Highlights Table of contents Preface 1 Parametric Equations and Polar Coordinates 2 Vectors in Space 3 Vector-Valued Functions 4 Differentiation of Functions of Several Variables Introduction 4.1 Functions of Several Variables 4.2 Limits and Continuity 4.3 Partial Derivatives 4.4 Tangent Planes and Linear Approximations 4.5 The Chain Rule 4.6 Directional Derivatives and the Gradient 4.7 Maxima/Minima Problems 4.8 Lagrange Multipliers Chapter Review 5 Multiple Integration 6 Vector Calculus 7 Second-Order Differential Equations A \| Table of Integrals B \| Table of Derivatives C \| Review of Pre-Calculus Answer Key Index Search for key terms or text. Close Learning Objectives 4.2.1 Calculate the limit of a function of two variables. 4.2.2 Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach. 4.2.3 State the conditions for continuity of a function of two variables. 4.2.4 Verify the continuity of a function of two variables at a point. 4.2.5 Calculate the limit of a function of three or more variables and verify the continuity of the function at a point. We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. It turns out these concepts have aspects that just don’t occur with functions of one variable. Limit of a Function of Two Variables Recall from The Limit of a Function the definition of a limit of a function of one variable: Let f(x)f(x)f(x) be defined for all x≠a x≠a x≠a in an open interval containing a.a.a. Let L L L be a real number. Then lim x→a f(x)=L lim x→a f(x)=L lim x→a f(x)=L if for every ε>0,ε>0,ε>0, there exists a δ>0,δ>0,δ>0, such that if 0<\|x−a\|<δ 0<\|x−a\|<δ 0<\|x−a\|<δ for all x x x in the domain of f,f,f, then \|f(x)−L\|<ε.\|f(x)−L\|<ε.\|f(x)−L\|<ε. Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables. Definition Consider a point (a,b)∈R 2.(a,b)∈ℝ 2.(a,b)∈ℝ 2. A δ δ δ disk centered at point (a,b)(a,b)(a,b) is defined to be an open disk of radius δ δ δ centered at point (a,b)(a,b)(a,b)—that is, {(x,y)∈R 2∣∣(x−a)2+(y−b)2<δ 2}{(x,y)∈ℝ 2\|(x−a)2+(y−b)2<δ 2}{(x,y)∈ℝ 2\|(x−a)2+(y−b)2<δ 2} as shown in the following graph. Figure 4.14 A δ δ δ disk centered around the point (2,1).(2,1).(2,1). The idea of a δ δ δ disk appears in the definition of the limit of a function of two variables. If δ δ δ is small, then all the points (x,y)(x,y)(x,y) in the δ δ δ disk are close to (a,b).(a,b).(a,b). This is completely analogous to x x x being close to a a a in the definition of a limit of a function of one variable. In one dimension, we express this restriction as a−δ<x<a+δ.a−δ<x<a+δ.a−δ<x<a+δ. In more than one dimension, we use a δ δ δ disk. Definition Let f f f be a function of two variables, x x x and y.y.y. The limit of f(x,y)f(x,y)f(x,y) as (x,y)(x,y)(x,y) approaches (a,b)(a,b)(a,b) is L,L,L, written lim(x,y)→(a,b)f(x,y)=L lim(x,y)→(a,b)f(x,y)=L lim(x,y)→(a,b)f(x,y)=L if for each ε>0 ε>0 ε>0 there exists a small enough δ>0 δ>0 δ>0 such that for all points (x,y)(x,y)(x,y) in a δ δ δ disk around (a,b),(a,b),(a,b), except possibly for (a,b)(a,b)(a,b) itself, the value of f(x,y)f(x,y)f(x,y) is no more than ε ε ε away from L L L (Figure 4.15). Using symbols, we write the following: For any ε>0,ε>0,ε>0, there exists a number δ>0 δ>0 δ>0 such that \|f(x,y)−L\|<ε whenever 0<(x−a)2+(y−b)2−−−−−−−−−−−−−−−√<δ.\|f(x,y)−L\|<ε whenever 0<(x−a)2+(y−b)2<δ.\|f(x,y)−L\|<ε whenever 0<(x−a)2+(y−b)2<δ. Figure 4.15 The limit of a function involving two variables requires that f(x,y)f(x,y)f(x,y) be within ε ε ε of L L L whenever (x,y)(x,y)(x,y) is within δ δ δ of (a,b).(a,b).(a,b). The smaller the value of ε,ε,ε, the smaller the value of δ.δ.δ. Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in The Limit Laws. Theorem 4.1 Limit laws for functions of two variables Let f(x,y)f(x,y)f(x,y) and g(x,y)g(x,y)g(x,y) be defined for all (x,y)≠(a,b)(x,y)≠(a,b)(x,y)≠(a,b) in a neighborhood around (a,b),(a,b),(a,b), and assume the neighborhood is contained completely inside the domain of f.f.f. Assume that L L L and M M M are real numbers such that lim(x,y)→(a,b)f(x,y)=L lim(x,y)→(a,b)f(x,y)=L lim(x,y)→(a,b)f(x,y)=L and lim(x,y)→(a,b)g(x,y)=M,lim(x,y)→(a,b)g(x,y)=M,lim(x,y)→(a,b)g(x,y)=M, and let c c c be a constant. Then each of the following statements holds: Constant Law: lim(x,y)→(a,b)c=c lim(x,y)→(a,b)c=c lim(x,y)→(a,b)c=c (4.2) Identity Laws: lim(x,y)→(a,b)x=a lim(x,y)→(a,b)x=a lim(x,y)→(a,b)x=a (4.3) lim(x,y)→(a,b)y=b lim(x,y)→(a,b)y=b lim(x,y)→(a,b)y=b (4.4) Sum Law: lim(x,y)→(a,b)(f(x,y)+g(x,y))=L+M lim(x,y)→(a,b)(f(x,y)+g(x,y))=L+M lim(x,y)→(a,b)(f(x,y)+g(x,y))=L+M (4.5) Difference Law: lim(x,y)→(a,b)(f(x,y)−g(x,y))=L−M lim(x,y)→(a,b)(f(x,y)−g(x,y))=L−M lim(x,y)→(a,b)(f(x,y)−g(x,y))=L−M (4.6) Constant Multiple Law: lim(x,y)→(a,b)(c f(x,y))=c L lim(x,y)→(a,b)(c f(x,y))=c L lim(x,y)→(a,b)(c f(x,y))=c L (4.7) Product Law: lim(x,y)→(a,b)(f(x,y)g(x,y))=L M lim(x,y)→(a,b)(f(x,y)g(x,y))=L M lim(x,y)→(a,b)(f(x,y)g(x,y))=L M (4.8) Quotient Law: lim(x,y)→(a,b)f(x,y)g(x,y)=L M for M≠0 lim(x,y)→(a,b)f(x,y)g(x,y)=L M for M≠0 lim(x,y)→(a,b)f(x,y)g(x,y)=L M for M≠0 (4.9) Power Law: lim(x,y)→(a,b)(f(x,y))n=L n lim(x,y)→(a,b)(f(x,y))n=L n lim(x,y)→(a,b)(f(x,y))n=L n (4.10) for any positive integer n.n.n. Root Law: lim(x,y)→(a,b)f(x,y)−−−−−−√n=L−−√n lim(x,y)→(a,b)f(x,y)n=L n lim(x,y)→(a,b)f(x,y)n=L n (4.11) for all L L L if n n n is odd and positive, and for L≥0 L≥0 L≥0 if n n n is even and positive provided that f(x,y)≥0 f(x,y)≥0 f(x,y)≥0 for all (x,y)≠(a,b)(x,y)≠(a,b)(x,y)≠(a,b) in neighborhood of (a,b)(a,b)(a,b). The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions. Example 4.8 Finding the Limit of a Function of Two Variables Find each of the following limits: lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6) lim(x,y)→(2,−1)2 x+3 y 4 x−3 y lim(x,y)→(2,−1)2 x+3 y 4 x−3 y lim(x,y)→(2,−1)2 x+3 y 4 x−3 y Solution First use the sum and difference laws to separate the terms: lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)=(lim(x,y)→(2,−1)x 2)−(lim(x,y)→(2,−1)2 x y)+(lim(x,y)→(2,−1)3 y 2)−(lim(x,y)→(2,−1)4 x)+(lim(x,y)→(2,−1)3 y)−(lim(x,y)→(2,−1)6).lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)=(lim(x,y)→(2,−1)x 2)−(lim(x,y)→(2,−1)2 x y)+(lim(x,y)→(2,−1)3 y 2)−(lim(x,y)→(2,−1)4 x)+(lim(x,y)→(2,−1)3 y)−(lim(x,y)→(2,−1)6).lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)=(lim(x,y)→(2,−1)x 2)−(lim(x,y)→(2,−1)2 x y)+(lim(x,y)→(2,−1)3 y 2)−(lim(x,y)→(2,−1)4 x)+(lim(x,y)→(2,−1)3 y)−(lim(x,y)→(2,−1)6). Next, use the constant multiple law on the second, third, fourth, and fifth limits: =(lim(x,y)→(2,−1)x 2)−2(lim(x,y)→(2,−1)x y)+3(lim(x,y)→(2,−1)y 2)−4(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)−lim(x,y)→(2,−1)6.=(lim(x,y)→(2,−1)x 2)−2(lim(x,y)→(2,−1)x y)+3(lim(x,y)→(2,−1)y 2)−4(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)−lim(x,y)→(2,−1)6.=(lim(x,y)→(2,−1)x 2)−2(lim(x,y)→(2,−1)x y)+3(lim(x,y)→(2,−1)y 2)−4(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)−lim(x,y)→(2,−1)6. Now, use the power law on the first and third limits, and the product law on the second limit: =(lim(x,y)→(2,−1)x)2−2(lim(x,y)→(2,−1)x)(lim(x,y)→(2,−1)y)+3(lim(x,y)→(2,−1)y)2−4(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)−lim(x,y)→(2,−1)6.=(lim(x,y)→(2,−1)x)2−2(lim(x,y)→(2,−1)x)(lim(x,y)→(2,−1)y)+3(lim(x,y)→(2,−1)y)2−4(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)−lim(x,y)→(2,−1)6.=(lim(x,y)→(2,−1)x)2−2(lim(x,y)→(2,−1)x)(lim(x,y)→(2,−1)y)+3(lim(x,y)→(2,−1)y)2−4(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)−lim(x,y)→(2,−1)6. Last, use the identity laws on the first six limits and the constant law on the last limit: lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)=(2)2−2(2)(−1)+3(−1)2−4(2)+3(−1)−6=−6.lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)=(2)2−2(2)(−1)+3(−1)2−4(2)+3(−1)−6=−6.lim(x,y)→(2,−1)(x 2−2 x y+3 y 2−4 x+3 y−6)=(2)2−2(2)(−1)+3(−1)2−4(2)+3(−1)−6=−6. 2. Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law, lim(x,y)→(2,−1)(4 x−3 y)=lim(x,y)→(2,−1)4 x−lim(x,y)→(2,−1)3 y=4(lim(x,y)→(2,−1)x)−3(lim(x,y)→(2,−1)y)=4(2)−3(−1)=11.lim(x,y)→(2,−1)(4 x−3 y)=lim(x,y)→(2,−1)4 x−lim(x,y)→(2,−1)3 y=4(lim(x,y)→(2,−1)x)−3(lim(x,y)→(2,−1)y)=4(2)−3(−1)=11.lim(x,y)→(2,−1)(4 x−3 y)=lim(x,y)→(2,−1)4 x−lim(x,y)→(2,−1)3 y=4(lim(x,y)→(2,−1)x)−3(lim(x,y)→(2,−1)y)=4(2)−3(−1)=11. Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law: lim(x,y)→(2,−1)(2 x+3 y)=lim(x,y)→(2,−1)2 x+lim(x,y)→(2,−1)3 y=2(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)=2(2)+3(−1)=1.lim(x,y)→(2,−1)(2 x+3 y)=lim(x,y)→(2,−1)2 x+lim(x,y)→(2,−1)3 y=2(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)=2(2)+3(−1)=1.lim(x,y)→(2,−1)(2 x+3 y)=lim(x,y)→(2,−1)2 x+lim(x,y)→(2,−1)3 y=2(lim(x,y)→(2,−1)x)+3(lim(x,y)→(2,−1)y)=2(2)+3(−1)=1. Therefore, according to the quotient law we have lim(x,y)→(2,−1)2 x+3 y 4 x−3 y=lim(x,y)→(2,−1)(2 x+3 y)lim(x,y)→(2,−1)(4 x−3 y)=1 11.lim(x,y)→(2,−1)2 x+3 y 4 x−3 y=lim(x,y)→(2,−1)(2 x+3 y)lim(x,y)→(2,−1)(4 x−3 y)=1 11.lim(x,y)→(2,−1)2 x+3 y 4 x−3 y=lim(x,y)→(2,−1)(2 x+3 y)lim(x,y)→(2,−1)(4 x−3 y)=1 11. Checkpoint 4.6 Evaluate the following limit: lim(x,y)→(5,−2)x 2−y y 2+x−1−−−−−−−−−√3.lim(x,y)→(5,−2)x 2−y y 2+x−1 3.lim(x,y)→(5,−2)x 2−y y 2+x−1 3. Since we are taking the limit of a function of two variables, the point (a,b)(a,b)(a,b) is in R 2,ℝ 2,ℝ 2, and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward (a,b).(a,b).(a,b). If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken. Example 4.9 Limits That Fail to Exist Show that neither of the following limits exist: lim(x,y)→(0,0)2 x y 3 x 2+y 2 lim(x,y)→(0,0)2 x y 3 x 2+y 2 lim(x,y)→(0,0)2 x y 3 x 2+y 2 lim(x,y)→(0,0)4 x y 2 x 2+3 y 4 lim(x,y)→(0,0)4 x y 2 x 2+3 y 4 lim(x,y)→(0,0)4 x y 2 x 2+3 y 4 Solution The domain of the function f(x,y)=2 x y 3 x 2+y 2 f(x,y)=2 x y 3 x 2+y 2 f(x,y)=2 x y 3 x 2+y 2 consists of all points in the x y-plane x y-plane x y-plane except for the point (0,0)(0,0)(0,0) (Figure 4.16). To show that the limit does not exist as (x,y)(x,y)(x,y) approaches (0,0),(0,0),(0,0), we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point (0,0).(0,0).(0,0). First, consider the line y=0 y=0 y=0 in the x y-plane.x y-plane.x y-plane. Substituting y=0 y=0 y=0 into f(x,y)f(x,y)f(x,y) gives f(x,0)=2 x(0)3 x 2+0 2=0 f(x,0)=2 x(0)3 x 2+0 2=0 f(x,0)=2 x(0)3 x 2+0 2=0 for any value of x.x.x. Therefore the value of f f f remains constant for any point on the x-axis,x-axis,x-axis, and as y y y approaches zero, the function remains fixed at zero. Next, consider the line y=x.y=x.y=x. Substituting y=x y=x y=x into f(x,y)f(x,y)f(x,y) gives f(x,x)=2 x(x)3 x 2+x 2=2 x 2 4 x 2=1 2.f(x,x)=2 x(x)3 x 2+x 2=2 x 2 4 x 2=1 2.f(x,x)=2 x(x)3 x 2+x 2=2 x 2 4 x 2=1 2. This is true for any point on the line y=x.y=x.y=x. If we let x x x approach zero while staying on this line, the value of the function remains fixed at 1 2,1 2,1 2, regardless of how small x x x is. Choose a value for ε ε ε that is less than 1/2 1/2 1/2—say, 1/4.1/4.1/4. Then, no matter how small a δ δ δ disk we draw around (0,0),(0,0),(0,0), the values of f(x,y)f(x,y)f(x,y) for points inside that δ δ δ disk will include both 0 0 0 and 1 2.1 2.1 2. Therefore, the definition of limit at a point is never satisfied and the limit fails to exist. Figure 4.16 Graph of the function f(x,y)=(2 x y)/(3 x 2+y 2).f(x,y)=(2 x y)/(3 x 2+y 2).f(x,y)=(2 x y)/(3 x 2+y 2). Along the line y=0,y=0,y=0, the function is equal to zero; along the line y=x,y=x,y=x, the function is equal to 1 2.1 2.1 2. In a similar fashion to a., we can approach the origin along any straight line passing through the origin. If we try the x-axis x-axis x-axis (i.e., y=0),y=0),y=0), then the function remains fixed at zero. The same is true for the y-axis.y-axis.y-axis. Suppose we approach the origin along a straight line of slope k.k.k. The equation of this line is y=k x.y=k x.y=k x. Then the limit becomes lim(x,y)→(0,0)4 x y 2 x 2+3 y 4=lim(x,y)→(0,0)4 x(k x)2 x 2+3(k x)4=lim(x,y)→(0,0)4 k 2 x 3 x 2+3 k 4 x 4=lim(x,y)→(0,0)4 k 2 x 1+3 k 4 x 2=lim(x,y)→(0,0)(4 k 2 x)lim(x,y)→(0,0)(1+3 k 4 x 2)=0 lim(x,y)→(0,0)4 x y 2 x 2+3 y 4=lim(x,y)→(0,0)4 x(k x)2 x 2+3(k x)4=lim(x,y)→(0,0)4 k 2 x 3 x 2+3 k 4 x 4=lim(x,y)→(0,0)4 k 2 x 1+3 k 4 x 2=lim(x,y)→(0,0)(4 k 2 x)lim(x,y)→(0,0)(1+3 k 4 x 2)=0 lim(x,y)→(0,0)4 x y 2 x 2+3 y 4=lim(x,y)→(0,0)4 x(k x)2 x 2+3(k x)4=lim(x,y)→(0,0)4 k 2 x 3 x 2+3 k 4 x 4=lim(x,y)→(0,0)4 k 2 x 1+3 k 4 x 2=lim(x,y)→(0,0)(4 k 2 x)lim(x,y)→(0,0)(1+3 k 4 x 2)=0 regardless of the value of k.k.k. It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation x=y 2.x=y 2.x=y 2. Substituting y 2 y 2 y 2 in place of x x x in f(x,y)f(x,y)f(x,y) gives lim(x,y)→(0,0)4 x y 2 x 2+3 y 4=lim(x,y)→(0,0)4(y 2)y 2(y 2)2+3 y 4=lim(x,y)→(0,0)4 y 4 y 4+3 y 4=lim(x,y)→(0,0)1=1.lim(x,y)→(0,0)4 x y 2 x 2+3 y 4=lim(x,y)→(0,0)4(y 2)y 2(y 2)2+3 y 4=lim(x,y)→(0,0)4 y 4 y 4+3 y 4=lim(x,y)→(0,0)1=1.lim(x,y)→(0,0)4 x y 2 x 2+3 y 4=lim(x,y)→(0,0)4(y 2)y 2(y 2)2+3 y 4=lim(x,y)→(0,0)4 y 4 y 4+3 y 4=lim(x,y)→(0,0)1=1. By the same logic in a., it is impossible to find a δ δ δ disk around the origin that satisfies the definition of the limit for any value of ε<1.ε<1.ε<1. Therefore, lim(x,y)→(0,0)4 x y 2 x 2+3 y 4 lim(x,y)→(0,0)4 x y 2 x 2+3 y 4 lim(x,y)→(0,0)4 x y 2 x 2+3 y 4 does not exist. Checkpoint 4.7 Show that lim(x,y)→(2,1)(x−2)(y−1)(x−2)2+(y−1)2 lim(x,y)→(2,1)(x−2)(y−1)(x−2)2+(y−1)2 lim(x,y)→(2,1)(x−2)(y−1)(x−2)2+(y−1)2 does not exist. Interior Points and Boundary Points To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology. Definition Let S be a subset of R 2 ℝ 2 ℝ 2 (Figure 4.17). A point P 0 P 0 P 0 is called an interior point of S S S if there is a δ δ δ disk centered around P 0 P 0 P 0 contained completely in S.S.S. A point P 0 P 0 P 0 is called a boundary point of S S S if every δ δ δ disk centered around P 0 P 0 P 0 contains points both inside and outside S.S.S. Figure 4.17 In the set S S S shown, (−1,1)(−1,1)(−1,1) is an interior point and (2,3)(2,3)(2,3) is a boundary point. Definition Let S be a subset of R 2 ℝ 2 ℝ 2 (Figure 4.17). S S S is called an open set if every point of S S S is an interior point. S S S is called a closed set if it contains all its boundary points. An example of an open set is a δ δ δ disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a δ δ δ disk but not the other half, then the set is neither open nor closed. Definition Let S be a subset of R 2 ℝ 2 ℝ 2 (Figure 4.17). An open set S S S is a connected set if it cannot be represented as the union of two or more disjoint, nonempty open subsets. A set S S S is a region if it is open, connected, and nonempty. The definition of a limit of a function of two variables requires the δ δ δ disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the δ disk δ disk δ disk is not contained inside the domain. By definition, some of the points of the δ disk δ disk δ disk are inside the domain and some are outside. Therefore, we need only consider points that are inside both the δ δ δ disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point. Definition Let f f f be a function of two variables, x x x and y,y,y, and suppose (a,b)(a,b)(a,b) is on the boundary of the domain of f.f.f. Then, the limit of f(x,y)f(x,y)f(x,y) as (x,y)(x,y)(x,y) approaches (a,b)(a,b)(a,b) is L,L,L, written lim(x,y)→(a,b)f(x,y)=L,lim(x,y)→(a,b)f(x,y)=L,lim(x,y)→(a,b)f(x,y)=L, if for any ε>0,ε>0,ε>0, there exists a number δ>0 δ>0 δ>0 such that for any point (x,y)(x,y)(x,y) inside the domain of f f f and within a suitably small distance positive δ δ δ of (a,b),(a,b),(a,b), the value of f(x,y)f(x,y)f(x,y) is no more than ε ε ε away from L L L (Figure 4.15). Using symbols, we can write: For any ε>0,ε>0,ε>0, there exists a number δ>0 δ>0 δ>0 such that \|f(x,y)−L\|<ε whenever 0<(x−a)2+(y−b)2−−−−−−−−−−−−−−−√<δ.\|f(x,y)−L\|<ε whenever 0<(x−a)2+(y−b)2<δ.\|f(x,y)−L\|<ε whenever 0<(x−a)2+(y−b)2<δ. Example 4.10 Limit of a Function at a Boundary Point Prove lim(x,y)→(4,3)25−x 2−y 2−−−−−−−−−−√=0.lim(x,y)→(4,3)25−x 2−y 2=0.lim(x,y)→(4,3)25−x 2−y 2=0. Solution The domain of the function f(x,y)=25−x 2−y 2−−−−−−−−−−√f(x,y)=25−x 2−y 2 f(x,y)=25−x 2−y 2 is {(x,y)∈R 2∣∣x 2+y 2≤25},{(x,y)∈ℝ 2\|x 2+y 2≤25},{(x,y)∈ℝ 2\|x 2+y 2≤25}, which is a circle of radius 5 5 5 centered at the origin, along with its interior as shown in the following graph. Figure 4.18 Domain of the function f(x,y)=25−x 2−y 2−−−−−−−−−−√.f(x,y)=25−x 2−y 2.f(x,y)=25−x 2−y 2. We can use the limit laws, which apply to limits at the boundary of domains as well as interior points: lim(x,y)→(4,3)25−x 2−y 2−−−−−−−−−−√=lim(x,y)→(4,3)(25−x 2−y 2)−−−−−−−−−−−−−−−−−−√=lim(x,y)→(4,3)25−lim(x,y)→(4,3)x 2−lim(x,y)→(4,3)y 2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=25−4 2−3 2−−−−−−−−−−√=0.lim(x,y)→(4,3)25−x 2−y 2=lim(x,y)→(4,3)(25−x 2−y 2)=lim(x,y)→(4,3)25−lim(x,y)→(4,3)x 2−lim(x,y)→(4,3)y 2=25−4 2−3 2=0.lim(x,y)→(4,3)25−x 2−y 2=lim(x,y)→(4,3)(25−x 2−y 2)=lim(x,y)→(4,3)25−lim(x,y)→(4,3)x 2−lim(x,y)→(4,3)y 2=25−4 2−3 2=0. See the following graph. Figure 4.19 Graph of the function f(x,y)=25−x 2−y 2−−−−−−−−−−√.f(x,y)=25−x 2−y 2.f(x,y)=25−x 2−y 2. Checkpoint 4.8 Evaluate the following limit: lim(x,y)→(5,−2)29−x 2−y 2−−−−−−−−−−√.lim(x,y)→(5,−2)29−x 2−y 2.lim(x,y)→(5,−2)29−x 2−y 2. Continuity of Functions of Two Variables In Continuity, we defined the continuity of a function of one variable and saw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for f(x)f(x)f(x) to be continuous at point x=a:x=a:x=a: f(a)f(a)f(a) exists. lim x→a f(x)lim x→a f(x)lim x→a f(x) exists. lim x→a f(x)=f(a).lim x→a f(x)=f(a).lim x→a f(x)=f(a). These three conditions are necessary for continuity of a function of two variables as well. Definition A function f(x,y)f(x,y)f(x,y) is continuous at a point (a,b)(a,b)(a,b) in its domain if the following conditions are satisfied: f(a,b)f(a,b)f(a,b) exists. lim(x,y)→(a,b)f(x,y)lim(x,y)→(a,b)f(x,y)lim(x,y)→(a,b)f(x,y) exists. lim(x,y)→(a,b)f(x,y)=f(a,b).lim(x,y)→(a,b)f(x,y)=f(a,b).lim(x,y)→(a,b)f(x,y)=f(a,b). Example 4.11 Demonstrating Continuity for a Function of Two Variables Show that the function f(x,y)=3 x+2 y x+y+1 f(x,y)=3 x+2 y x+y+1 f(x,y)=3 x+2 y x+y+1 is continuous at point (5,−3).(5,−3).(5,−3). Solution There are three conditions to be satisfied, per the definition of continuity. In this example, a=5 a=5 a=5 and b=−3.b=−3.b=−3. f(a,b)f(a,b)f(a,b) exists. This is true because the domain of the function f f f consists of those ordered pairs for which the denominator is nonzero (i.e., x+y+1≠0).x+y+1≠0).x+y+1≠0). Point (5,−3)(5,−3)(5,−3) satisfies this condition. Furthermore, f(a,b)=f(5,−3)=3(5)+2(−3)5+(−3)+1=15−6 2+1=3.f(a,b)=f(5,−3)=3(5)+2(−3)5+(−3)+1=15−6 2+1=3.f(a,b)=f(5,−3)=3(5)+2(−3)5+(−3)+1=15−6 2+1=3. 2. lim(x,y)→(a,b)f(x,y)lim(x,y)→(a,b)f(x,y)lim(x,y)→(a,b)f(x,y) exists. This is also true: lim(x,y)→(a,b)f(x,y)=lim(x,y)→(5,−3)3 x+2 y x+y+1=lim(x,y)→(5,−3)(3 x+2 y)lim(x,y)→(5,−3)(x+y+1)=15−6 5−3+1=3.lim(x,y)→(a,b)f(x,y)=lim(x,y)→(5,−3)3 x+2 y x+y+1=lim(x,y)→(5,−3)(3 x+2 y)lim(x,y)→(5,−3)(x+y+1)=15−6 5−3+1=3.lim(x,y)→(a,b)f(x,y)=lim(x,y)→(5,−3)3 x+2 y x+y+1=lim(x,y)→(5,−3)(3 x+2 y)lim(x,y)→(5,−3)(x+y+1)=15−6 5−3+1=3. 3. lim(x,y)→(a,b)f(x,y)=f(a,b).lim(x,y)→(a,b)f(x,y)=f(a,b).lim(x,y)→(a,b)f(x,y)=f(a,b). This is true because we have just shown that both sides of this equation equal three. Checkpoint 4.9 Show that the function f(x,y)=26−2 x 2−y 2−−−−−−−−−−−√f(x,y)=26−2 x 2−y 2 f(x,y)=26−2 x 2−y 2 is continuous at point (2,−3).(2,−3).(2,−3). Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of two variables is continuous at a point (x 0,y 0)(x 0,y 0)(x 0,y 0) in its domain if for every ε>0 ε>0 ε>0 there exists a δ>0 δ>0 δ>0 such that, whenever (x−x 0)2+(y−y 0)2−−−−−−−−−−−−−−−−√<δ(x−x 0)2+(y−y 0)2<δ(x−x 0)2+(y−y 0)2<δ it is true, \|f(x,y)−f(a,b)\|<ε.\|f(x,y)−f(a,b)\|<ε.\|f(x,y)−f(a,b)\|<ε. This definition can be combined with the formal definition (that is, the epsilon–delta definition) of continuity of a function of one variable to prove the following theorems: Theorem 4.2 The Sum of Continuous Functions Is Continuous If f(x,y)f(x,y)f(x,y) is continuous at (x 0,y 0),(x 0,y 0),(x 0,y 0), and g(x,y)g(x,y)g(x,y) is continuous at (x 0,y 0),(x 0,y 0),(x 0,y 0), then f(x,y)+g(x,y)f(x,y)+g(x,y)f(x,y)+g(x,y) is continuous at (x 0,y 0).(x 0,y 0).(x 0,y 0). Theorem 4.3 The Product of Continuous Functions Is Continuous If g(x)g(x)g(x) is continuous at x 0 x 0 x 0 and h(y)h(y)h(y) is continuous at y 0,y 0,y 0, then f(x,y)=g(x)h(y)f(x,y)=g(x)h(y)f(x,y)=g(x)h(y) is continuous at (x 0,y 0).(x 0,y 0).(x 0,y 0). Theorem 4.4 The Composition of Continuous Functions Is Continuous Let g g g be a function of two variables from a domain D⊆R 2 D⊆ℝ 2 D⊆ℝ 2 to a range R⊆R.R⊆ℝ.R⊆ℝ. Suppose g g g is continuous at some point (x 0,y 0)∈D(x 0,y 0)∈D(x 0,y 0)∈D and define z 0=g(x 0,y 0).z 0=g(x 0,y 0).z 0=g(x 0,y 0). Let f f f be a function that maps R ℝ ℝ to R ℝ ℝ such that z 0 z 0 z 0 is in the domain of f.f.f. Last, assume f f f is continuous at z 0.z 0.z 0. Then f∘g f∘g f∘g is continuous at (x 0,y 0)(x 0,y 0)(x 0,y 0) as shown in the following figure. Figure 4.20 The composition of two continuous functions is continuous. Let’s now use the previous theorems to show continuity of functions in the following examples. Example 4.12 More Examples of Continuity of a Function of Two Variables Show that the functions f(x,y)=4 x 3 y 2 f(x,y)=4 x 3 y 2 f(x,y)=4 x 3 y 2 and g(x,y)=cos(4 x 3 y 2)g(x,y)=cos(4 x 3 y 2)g(x,y)=cos(4 x 3 y 2) are continuous everywhere. Solution The polynomials g(x)=4 x 3 g(x)=4 x 3 g(x)=4 x 3 and h(y)=y 2 h(y)=y 2 h(y)=y 2 are continuous at every real number, and therefore by the product of continuous functions theorem, f(x,y)=4 x 3 y 2 f(x,y)=4 x 3 y 2 f(x,y)=4 x 3 y 2 is continuous at every point (x,y)(x,y)(x,y) in the x y-plane.x y-plane.x y-plane. Since f(x,y)=4 x 3 y 2 f(x,y)=4 x 3 y 2 f(x,y)=4 x 3 y 2 is continuous at every point (x,y)(x,y)(x,y) in the x y-plane x y-plane x y-plane and g(x)=cos x g(x)=cos x g(x)=cos x is continuous at every real number x,x,x, the continuity of the composition of functions tells us that g(x,y)=cos(4 x 3 y 2)g(x,y)=cos(4 x 3 y 2)g(x,y)=cos(4 x 3 y 2) is continuous at every point (x,y)(x,y)(x,y) in the x y-plane.x y-plane.x y-plane. Checkpoint 4.10 Show that the functions f(x,y)=2 x 2 y 3+3 f(x,y)=2 x 2 y 3+3 f(x,y)=2 x 2 y 3+3 and g(x,y)=(2 x 2 y 3+3)4 g(x,y)=(2 x 2 y 3+3)4 g(x,y)=(2 x 2 y 3+3)4 are continuous everywhere. Functions of Three or More Variables The limit of a function of three or more variables occurs readily in applications. For example, suppose we have a function f(x,y,z)f(x,y,z)f(x,y,z) that gives the temperature at a physical location (x,y,z)(x,y,z)(x,y,z) in three dimensions. Or perhaps a function g(x,y,z,t)g(x,y,z,t)g(x,y,z,t) can indicate air pressure at a location (x,y,z)(x,y,z)(x,y,z) at time t.t.t. How can we take a limit at a point in R 3?ℝ 3?ℝ 3? What does it mean to be continuous at a point in four dimensions? The answers to these questions rely on extending the concept of a δ δ δ disk into more than two dimensions. Then, the ideas of the limit of a function of three or more variables and the continuity of a function of three or more variables are very similar to the definitions given earlier for a function of two variables. Definition Let (x 0,y 0,z 0)(x 0,y 0,z 0)(x 0,y 0,z 0) be a point in R 3.ℝ 3.ℝ 3. Then, a δ δ δ ball in three dimensions consists of all points in R 3 ℝ 3 ℝ 3 lying at a distance of less than δ δ δ from (x 0,y 0,z 0)(x 0,y 0,z 0)(x 0,y 0,z 0)—that is, {(x,y,z)∈R 3∣∣(x−x 0)2+(y−y 0)2+(z−z 0)2−−−−−−−−−−−−−−−−−−−−−−−−−√<δ}.{(x,y,z)∈ℝ 3\|(x−x 0)2+(y−y 0)2+(z−z 0)2<δ}.{(x,y,z)∈ℝ 3\|(x−x 0)2+(y−y 0)2+(z−z 0)2<δ}. To define a δ δ δ ball in higher dimensions, add additional terms under the radical to correspond to each additional dimension. For example, given a point P=(w 0,x 0,y 0,z 0)P=(w 0,x 0,y 0,z 0)P=(w 0,x 0,y 0,z 0) in R 4,ℝ 4,ℝ 4, a δ δ δ ball around P P P can be described by {(w,x,y,z)∈R 4∣∣(w−w 0)2+(x−x 0)2+(y−y 0)2+(z−z 0)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√<δ}.{(w,x,y,z)∈ℝ 4\|(w−w 0)2+(x−x 0)2+(y−y 0)2+(z−z 0)2<δ}.{(w,x,y,z)∈ℝ 4\|(w−w 0)2+(x−x 0)2+(y−y 0)2+(z−z 0)2<δ}. To show that a limit of a function of three variables exists at a point (x 0,y 0,z 0),(x 0,y 0,z 0),(x 0,y 0,z 0), it suffices to show that for any point in a δ δ δ ball centered at (x 0,y 0,z 0),(x 0,y 0,z 0),(x 0,y 0,z 0), the value of the function at that point is arbitrarily close to a fixed value (the limit value). All the limit laws for functions of two variables hold for functions of more than two variables as well. Example 4.13 Finding the Limit of a Function of Three Variables Find lim(x,y,z)→(4,1,−3)x 2 y−3 z 2 x+5 y−z.lim(x,y,z)→(4,1,−3)x 2 y−3 z 2 x+5 y−z.lim(x,y,z)→(4,1,−3)x 2 y−3 z 2 x+5 y−z. Solution Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, the identity law, and the constant law, lim(x,y,z)→(4,1,−3)(2 x+5 y−z)=2(lim(x,y,z)→(4,1,−3)x)+5(lim(x,y,z)→(4,1,−3)y)−(lim(x,y,z)→(4,1,−3)z)=2(4)+5(1)−(−3)=16.lim(x,y,z)→(4,1,−3)(2 x+5 y−z)=2(lim(x,y,z)→(4,1,−3)x)+5(lim(x,y,z)→(4,1,−3)y)−(lim(x,y,z)→(4,1,−3)z)=2(4)+5(1)−(−3)=16.lim(x,y,z)→(4,1,−3)(2 x+5 y−z)=2(lim(x,y,z)→(4,1,−3)x)+5(lim(x,y,z)→(4,1,−3)y)−(lim(x,y,z)→(4,1,−3)z)=2(4)+5(1)−(−3)=16. Since this is nonzero, we next find the limit of the numerator. Using the product law, difference law, constant multiple law, and identity law, lim(x,y,z)→(4,1,−3)(x 2 y−3 z)=(lim(x,y,z)→(4,1,−3)x)2(lim(x,y,z)→(4,1,−3)y)−3 lim(x,y,z)→(4,1,−3)z=(4 2)(1)−3(−3)=16+9=25.lim(x,y,z)→(4,1,−3)(x 2 y−3 z)=(lim(x,y,z)→(4,1,−3)x)2(lim(x,y,z)→(4,1,−3)y)−3 lim(x,y,z)→(4,1,−3)z=(4 2)(1)−3(−3)=16+9=25.lim(x,y,z)→(4,1,−3)(x 2 y−3 z)=(lim(x,y,z)→(4,1,−3)x)2(lim(x,y,z)→(4,1,−3)y)−3 lim(x,y,z)→(4,1,−3)z=(4 2)(1)−3(−3)=16+9=25. Last, applying the quotient law: lim(x,y,z)→(4,1,−3)x 2 y−3 z 2 x+5 y−z=lim(x,y,z)→(4,1,−3)(x 2 y−3 z)lim(x,y,z)→(4,1,−3)(2 x+5 y−z)=25 16.lim(x,y,z)→(4,1,−3)x 2 y−3 z 2 x+5 y−z=lim(x,y,z)→(4,1,−3)(x 2 y−3 z)lim(x,y,z)→(4,1,−3)(2 x+5 y−z)=25 16.lim(x,y,z)→(4,1,−3)x 2 y−3 z 2 x+5 y−z=lim(x,y,z)→(4,1,−3)(x 2 y−3 z)lim(x,y,z)→(4,1,−3)(2 x+5 y−z)=25 16. Checkpoint 4.11 Find lim(x,y,z)→(4,−1,3)13−x 2−2 y 2+z 2−−−−−−−−−−−−−−−√.lim(x,y,z)→(4,−1,3)13−x 2−2 y 2+z 2.lim(x,y,z)→(4,−1,3)13−x 2−2 y 2+z 2. Section 4.2 Exercises For the following exercises, find the limit of the function. lim(x,y)→(1,2)x lim(x,y)→(1,2)x lim(x,y)→(1,2)x 61. lim(x,y)→(1,2)5 x 2 y x 2+y 2 lim(x,y)→(1,2)5 x 2 y x 2+y 2 lim(x,y)→(1,2)5 x 2 y x 2+y 2 Show that the limit lim(x,y)→(0,0)5 x 2 y x 2+y 2 lim(x,y)→(0,0)5 x 2 y x 2+y 2 lim(x,y)→(0,0)5 x 2 y x 2+y 2 exists and is the same along the paths: y-axis y-axis y-axis and x-axis,x-axis,x-axis, and along y=x.y=x.y=x. For the following exercises, evaluate the limits at the indicated values of x and y.x and y.x and y. If the limit does not exist, state this and explain why the limit does not exist. 63. lim(x,y)→(0,0)4 x 2+10 y 2+4 4 x 2−10 y 2+6 lim(x,y)→(0,0)4 x 2+10 y 2+4 4 x 2−10 y 2+6 lim(x,y)→(0,0)4 x 2+10 y 2+4 4 x 2−10 y 2+6 lim(x,y)→(11,13)1 x y−−√lim(x,y)→(11,13)1 x y lim(x,y)→(11,13)1 x y 65. lim(x,y)→(0,1)y 2 sin x x lim(x,y)→(0,1)y 2 sin x x lim(x,y)→(0,1)y 2 sin x x lim(x,y)→(0,0)sin(x 8+y 7 x−y+10)lim(x,y)→(0,0)sin(x 8+y 7 x−y+10)lim(x,y)→(0,0)sin(x 8+y 7 x−y+10) 67. lim(x,y)→(π/4,1)y tan x y+1 lim(x,y)→(π/4,1)y tan x y+1 lim(x,y)→(π/4,1)y tan x y+1 lim(x,y)→(0,π/4)sec x+2 3 x−tan y lim(x,y)→(0,π/4)sec x+2 3 x−tan y lim(x,y)→(0,π/4)sec x+2 3 x−tan y 69. lim(x,y)→(2,5)(1 x−5 y)lim(x,y)→(2,5)(1 x−5 y)lim(x,y)→(2,5)(1 x−5 y) lim(x,y)→(4,4)x ln y lim(x,y)→(4,4)x ln y lim(x,y)→(4,4)x ln y 71. lim(x,y)→(4,4)e−x 2−y 2 lim(x,y)→(4,4)e−x 2−y 2 lim(x,y)→(4,4)e−x 2−y 2 lim(x,y)→(0,0)9−x 2−y 2−−−−−−−−−√lim(x,y)→(0,0)9−x 2−y 2 lim(x,y)→(0,0)9−x 2−y 2 73. lim(x,y)→(1,2)(x 2 y 3−x 3 y 2+3 x+2 y)lim(x,y)→(1,2)(x 2 y 3−x 3 y 2+3 x+2 y)lim(x,y)→(1,2)(x 2 y 3−x 3 y 2+3 x+2 y) lim(x,y)→(π,π)x sin(x+y 4)lim(x,y)→(π,π)x sin(x+y 4)lim(x,y)→(π,π)x sin(x+y 4) 75. lim(x,y)→(0,0)x y+1 x 2+y 2+1 lim(x,y)→(0,0)x y+1 x 2+y 2+1 lim(x,y)→(0,0)x y+1 x 2+y 2+1 lim(x,y)→(0,0)x 2+y 2 x 2+y 2+1√−1 lim(x,y)→(0,0)x 2+y 2 x 2+y 2+1−1 lim(x,y)→(0,0)x 2+y 2 x 2+y 2+1−1 77. lim(x,y)→(0,0)ln(x 2+y 2)lim(x,y)→(0,0)ln(x 2+y 2)lim(x,y)→(0,0)ln(x 2+y 2) For the following exercises, complete the statement. A point (x 0,y 0)(x 0,y 0)(x 0,y 0) in a plane region R R R is an interior point of R R R if _____. 79. A point (x 0,y 0)(x 0,y 0)(x 0,y 0) in a plane region R R R is called a boundary point of R R R if _____. For the following exercises, use algebraic techniques to evaluate the limit. lim(x,y)→(2,1)x−y−1 x−y√−1 lim(x,y)→(2,1)x−y−1 x−y−1 lim(x,y)→(2,1)x−y−1 x−y−1 81. lim(x,y)→(0,0)x 4−4 y 4 x 2+2 y 2 lim(x,y)→(0,0)x 4−4 y 4 x 2+2 y 2 lim(x,y)→(0,0)x 4−4 y 4 x 2+2 y 2 lim(x,y)→(0,0)x 3−y 3 x−y lim(x,y)→(0,0)x 3−y 3 x−y lim(x,y)→(0,0)x 3−y 3 x−y 83. lim(x,y)→(0,0)x 2−x y x√−y√lim(x,y)→(0,0)x 2−x y x−y lim(x,y)→(0,0)x 2−x y x−y For the following exercises, evaluate the limits of the functions of three variables. lim(x,y,z)→(1,2,3)x z 2−y 2 z x y z−1 lim(x,y,z)→(1,2,3)x z 2−y 2 z x y z−1 lim(x,y,z)→(1,2,3)x z 2−y 2 z x y z−1 85. lim(x,y,z)→(0,0,0)x 2−y 2−z 2 x 2+y 2−z 2 lim(x,y,z)→(0,0,0)x 2−y 2−z 2 x 2+y 2−z 2 lim(x,y,z)→(0,0,0)x 2−y 2−z 2 x 2+y 2−z 2 For the following exercises, evaluate the limit of the function by determining the value the function approaches along the indicated paths. If the limit does not exist, explain why not. lim(x,y)→(0,0)x y+y 3 x 2+y 2 lim(x,y)→(0,0)x y+y 3 x 2+y 2 lim(x,y)→(0,0)x y+y 3 x 2+y 2 Along the x-axis x-axis x-axis(y=0)(y=0)(y=0) Along the y-axis y-axis y-axis(x=0)(x=0)(x=0) Along the path y=2 x y=2 x y=2 x 87. Evaluate lim(x,y)→(0,0)x y+y 3 x 2+y 2 lim(x,y)→(0,0)x y+y 3 x 2+y 2 lim(x,y)→(0,0)x y+y 3 x 2+y 2 using the results of previous problem. lim(x,y)→(0,0)x 2 y x 4+y 2 lim(x,y)→(0,0)x 2 y x 4+y 2 lim(x,y)→(0,0)x 2 y x 4+y 2 Along the x-axis (y=0)(y=0)(y=0) Along the y-axis (x=0)(x=0)(x=0) Along the path y=x 2 y=x 2 y=x 2 89. Evaluate lim(x,y)→(0,0)x 2 y x 4+y 2 lim(x,y)→(0,0)x 2 y x 4+y 2 lim(x,y)→(0,0)x 2 y x 4+y 2 using the results of previous problem. Discuss the continuity of the following functions. Find the largest region in the x y-plane x y-plane x y-plane in which the following functions are continuous. f(x,y)=sin(x y)f(x,y)=sin(x y)f(x,y)=sin(x y) 91. f(x,y)=ln(x+y)f(x,y)=ln(x+y)f(x,y)=ln(x+y) f(x,y)=e 3 x y f(x,y)=e 3 x y f(x,y)=e 3 x y 93. f(x,y)=1 x y f(x,y)=1 x y f(x,y)=1 x y For the following exercises, determine the region in which the function is continuous. Explain your answer. f(x,y)=x 2 y x 2+y 2 f(x,y)=x 2 y x 2+y 2 f(x,y)=x 2 y x 2+y 2 95. f(x,y)={x 2 y x 2+y 2 0 if(x,y)≠(0,0)if(x,y)=(0,0)}f(x,y)={x 2 y x 2+y 2 if(x,y)≠(0,0)0 if(x,y)=(0,0)}f(x,y)={x 2 y x 2+y 2 if(x,y)≠(0,0)0 if(x,y)=(0,0)} (Hint: Show that the function approaches different values along two different paths.) f(x,y)=sin(x 2+y 2)x 2+y 2 f(x,y)=sin(x 2+y 2)x 2+y 2 f(x,y)=sin(x 2+y 2)x 2+y 2 97. Determine whether g(x,y)=x 2−y 2 x 2+y 2 g(x,y)=x 2−y 2 x 2+y 2 g(x,y)=x 2−y 2 x 2+y 2 is continuous at (0,0).(0,0).(0,0). Create a plot using graphing software to determine where the limit does not exist. Find where in the coordinate plane f(x,y)=1 x 2−y f(x,y)=1 x 2−y f(x,y)=1 x 2−y is continuous. 99. Determine the region of the x y-plane x y-plane x y-plane in which the function g(x,y)=arctan(x y 2 x+y)g(x,y)=arctan(x y 2 x+y)g(x,y)=arctan(x y 2 x+y) is continuous. Use technology to support your conclusion. Determine the region of the x y-plane x y-plane x y-plane in which f(x,y)=ln(x 2+y 2−1)f(x,y)=ln(x 2+y 2−1)f(x,y)=ln(x 2+y 2−1) is continuous. Use technology to support your conclusion. (Hint: Choose the range of values for x and y x and y x and y carefully!) 101. At what points in space is g(x,y,z)=x 2+y 2−2 z 2 g(x,y,z)=x 2+y 2−2 z 2 g(x,y,z)=x 2+y 2−2 z 2 continuous? At what points in space is g(x,y,z)=1 x 2+z 2−1 g(x,y,z)=1 x 2+z 2−1 g(x,y,z)=1 x 2+z 2−1 continuous? 103. Show that lim(x,y)→(0,0)1 x 2+y 2 lim(x,y)→(0,0)1 x 2+y 2 lim(x,y)→(0,0)1 x 2+y 2 does not exist at (0,0)(0,0)(0,0) by plotting the graph of the function. [T] Evaluate lim(x,y)→(0,0)−x y 2 x 2+y 4 lim(x,y)→(0,0)−x y 2 x 2+y 4 lim(x,y)→(0,0)−x y 2 x 2+y 4 by plotting the function using a CAS. Determine analytically the limit along the path x=y 2.x=y 2.x=y 2. 105. [T] Use a CAS to draw a contour map of z=9−x 2−y 2−−−−−−−−−√.z=9−x 2−y 2.z=9−x 2−y 2. What is the name of the geometric shape of the level curves? Give the general equation of the level curves. What is the maximum value of z?z?z? What is the domain of the function? What is the range of the function? True or False: If we evaluate lim(x,y)→(0,0)f(x)lim(x,y)→(0,0)f(x)lim(x,y)→(0,0)f(x) along several paths and each time the limit is 1,1,1, we can conclude that lim(x,y)→(0,0)f(x)=1.lim(x,y)→(0,0)f(x)=1.lim(x,y)→(0,0)f(x)=1. 107. Use polar coordinates to find lim(x,y)→(0,0)sin x 2+y 2√x 2+y 2√.lim(x,y)→(0,0)sin x 2+y 2 x 2+y 2.lim(x,y)→(0,0)sin x 2+y 2 x 2+y 2. You can also find the limit using L’Hôpital’s rule. Use polar coordinates to find lim(x,y)→(0,0)cos(x 2+y 2).lim(x,y)→(0,0)cos(x 2+y 2).lim(x,y)→(0,0)cos(x 2+y 2). 109. Discuss the continuity of f(g(x,y))f(g(x,y))f(g(x,y)) where f(t)=1/t f(t)=1/t f(t)=1/t and g(x,y)=2 x−5 y.g(x,y)=2 x−5 y.g(x,y)=2 x−5 y. Given f(x,y)=x 2−4 y,f(x,y)=x 2−4 y,f(x,y)=x 2−4 y, find lim h→0 f(x+h,y)−f(x,y)h.lim h→0 f(x+h,y)−f(x,y)h.lim h→0 f(x+h,y)−f(x,y)h. 111. Given f(x,y)=x 2−4 y,f(x,y)=x 2−4 y,f(x,y)=x 2−4 y, find lim h→0 f(1+h,y)−f(1,y)h.lim h→0 f(1+h,y)−f(1,y)h.lim h→0 f(1+h,y)−f(1,y)h. PreviousNext 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-NonCommercial-ShareAlike 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: Gilbert Strang, Edwin “Jed” Herman Publisher/website: OpenStax Book title: Calculus Volume 3 Publication date: Mar 30, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 24, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . 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8788	https://www.facebook.com/merriamwebster/posts/synecdoche-using-a-part-of-something-to-stand-in-for-the-whole-thingmetonymy-the/10156849625943308/	'Synecdoche': using a part of... - Merriam-Webster Dictionary \| Facebook Log In Log In Forgot Account? Merriam-Webster Dictionary's Post Merriam-Webster Dictionary November 5, 2018 · 'Synecdoche': using a part of something to stand in for the whole thing 'Metonymy': the use of the name of one thing to represent something related to it merriam-webster.com On Synecdoche and Metonymy When you're left to your own rhetorical devices. All reactions: 174 6 comments 55 shares Like Comment Most relevant Patricia Di Paola "The pen is mightier than the sword," from Edward Bulwer Lytton's play Richelieu. This sentence has two metonyms: "Pen" stands for "the written word." 6y 4 Patricia Di Paola The phrase "hired hands" can be used to refer to workers. This is an example of a synecdoche. 6y 3 Top fan Phillip Cano Metonymy is v meta 6y Nelson Daniel Stella Jummai Uchechi Nnodi 6y Nelson Daniel Mbasiti Julius Jesse 6y View 1 reply
8789	https://www.intmath.com/differentiation-transcendental/table-derivatives.php	Table of Derivatives Skip to main content Interactive Mathematics Home Tutoring Features Reviews Pricing FAQ Problem Solver More Lessons Forum Interactives Blog Contact About SearchSearch IntMath Search for: Close Login Create Free Account Search IntMath Search: Close Home Tutoring Features Reviews Pricing FAQ Problem Solver More Lessons Forum Interactives Blog Contact About Login Create Free Account Want Better Math Grades? ✅ Unlimited Solutions ✅ Step-by-Step Answers ✅ Available 24/7 ➕Free Bonuses ($1085 value!) Invalid email address Thank you for booking, we will follow up with available time slots and course plans. Home Differentiation of Transcendental Functions Table of Derivatives On this page Differentiation of Transcendental Functions 1. Derivatives of Sin, Cos and Tan Functions 2. Derivatives of Csc, Sec and Cot Functions Differentiation interactive applet - trigonometric functions 3. Derivatives of Inverse Trigonometric Functions 4. Applications: Derivatives of Trigonometric Functions 5. Derivative of the Logarithmic Function 6. Derivative of the Exponential Function 7. Applications: Derivatives of Logarithmic and Exponential Functions Table of Derivatives Related Sections Math Tutoring Need help? Chat with a tutor anytime, 24/7. Chat now Online Calculus Solver Solve your calculus problem step by step! Online Calculus Solver IntMath Forum Get help with your math queries: See Forum Table of Derivatives Following are the derivatives we met in previous chapters: Introduction to Differentiation Applications of Differentiation; and this chapter, Differentiation of Transcendental Functions. 1. Powers of x General formula d d x u n\displaystyle\frac{d}{{\left.{d}{x}\right.}}{u}^{n}d x du n=n u n−1 d u d x\displaystyle={n}{u}^{{{n}-{1}}}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}=n u n−1 d x d u, where u\displaystyle{u}u is a function of x\displaystyle{x}x. Particular cases and examples d d x c\displaystyle\frac{d}{{\left.{d}{x}\right.}}{c}d x dc=0\displaystyle={0}=0 d d x x\displaystyle\frac{d}{{\left.{d}{x}\right.}}{x}d x dx=1\displaystyle={1}=1 d d x x n\displaystyle\frac{d}{{\left.{d}{x}\right.}}{x}^{n}d x dx n=n x n−1\displaystyle={n}{x}^{{{n}-{1}}}=n x n−1 d d x x 7\displaystyle\frac{d}{{\left.{d}{x}\right.}}{x}^{7}d x dx 7=7 x 6\displaystyle={7}{x}^{6}=7 x 6 2. Trigonometric Functions Trigonometry General formulas (a) d d x sin⁡u=(cos⁡u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \sin{{u}}={\left( \cos{{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dsin u=(cos u)d x d u d d x cos⁡u=−(sin⁡u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \cos{{u}}=-{\left( \sin{{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dcos u=−(sin u)d x d u d d x tan⁡u=(sec⁡2 u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \tan{{u}}={\left({{\sec}^{2}{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dtan u=(sec 2 u)d x d u Particular cases and examples d d x sin⁡3 x=3 cos⁡3 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \sin{{3}}{x}={3} \cos{{3}}{x}d x dsin 3 x=3 cos 3 x d d x sin⁡x 2=2 x cos⁡x 2\displaystyle\frac{d}{{\left.{d}{x}\right.}}{ \sin{{x}}^{2}=}\ {2}{x}\ { \cos{{x}}^{2}}d x dsin x 2=2 x cos x 2 d d x sin⁡x=cos⁡x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \sin{{x}}= \cos{{x}}d x dsin x=cos x d d x cos⁡x=−sin⁡x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \cos{{x}}=- \sin{{x}}d x dcos x=−sin x d d x cos⁡3 x=−3 sin⁡2 x\displaystyle\frac{d}{{\left.{d}{x}\right.}}{{\cos}^{3}{x}}=-{3}\ {{\sin}^{2}{x}}d x dcos 3 x=−3 sin 2 x d d x tan⁡x=sec⁡2 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \tan{{x}}={{\sec}^{2}{x}}d x dtan x=sec 2 x d d x 5 tan⁡7 x=3 5 sec⁡2 7 x\displaystyle\frac{d}{{\left.{d}{x}\right.}}{5} \tan{{7}}{x}={35}\ {{\sec}^{2}{7}}{x}d x d5 tan 7 x=3 5 sec 2 7 x Trigonometry General formulas (b) - reciprocals d d x csc⁡u=(−csc⁡u cot⁡u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \csc{{u}}={\left(- \csc{{u}} \cot{{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dcsc u=(−csc u cot u)d x d u d d x sec⁡u=(sec⁡u tan⁡u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \sec{{u}}={\left( \sec{{u}} \tan{{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dsec u=(sec u tan u)d x d u d d x cot⁡u=(−csc⁡2 u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \cot{{u}}={\left(-{{\csc}^{2}{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dcot u=(−csc 2 u)d x d u Particular cases and examples d d x csc⁡x=−csc⁡x cot⁡x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \csc{{x}}=- \csc{{x}} \cot{{x}}d x dcsc x=−csc x cot x d d x sec⁡x=sec⁡x tan⁡x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \sec{{x}}= \sec{{x}} \tan{{x}}d x dsec x=sec x tan x d d x cot⁡x=−csc⁡2 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \cot{{x}}=-{{\csc}^{2}{x}}d x dcot x=−csc 2 x Exponential and Logarithmic Functions General formulas d d x e u=(e u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}{e}^{u}={\left({e}^{u}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x de u=(e u)d x d u d d x b u=(b u ln⁡(b))d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}{b}^{u}={\left({b}^{u} \ln{{\left({b}\right)}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x db u=(b u ln(b))d x d u d d x ln⁡(u)=(1 u)d u d x=u′u\displaystyle\frac{d}{{\left.{d}{x}\right.}} \ln{{\left({u}\right)}}={\left(\frac{1}{{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}=\frac{{{u}'}}{{u}}d x dln(u)=(u 1)d x d u=u u′ Particular cases and examples d d x e x=e x\displaystyle\frac{d}{{\left.{d}{x}\right.}}{e}^{x}={e}^{x}d x de x=e x d d x 3 x=3 x ln⁡(3)=1.0 9 8 6×3 x\displaystyle\frac{d}{{\left.{d}{x}\right.}}{3}^{x}={3}^{x} \ln{{\left({3}\right)}}={1.0986}\times{3}^{x}d x d3 x=3 x ln(3)=1.0 9 8 6×3 x d d x ln⁡(x)=1 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \ln{{\left({x}\right)}}=\frac{1}{{x}}d x dln(x)=x 1 d d x ln⁡(x 4)=4 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \ln{{\left({x}^{4}\right)}}=\frac{4}{{x}}d x dln(x 4)=x 4 d d x ln⁡(5 x)=1 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \ln{{\left({5}{x}\right)}}=\frac{1}{{x}}d x dln(5 x)=x 1 Inverse Trigonometric Functions General formulas d d x arcsin⁡u=(1 1−u 2)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \arcsin{{u}}={\left(\frac{1}{\sqrt{{{1}-{u}^{2}}}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x darcsin u=(1−u 21)d x d u d d x arccsc u=(−1∣u∣u 2−1)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{arccsc}\ {u}={\left(-\frac{1}{{{\left\|{u}\right\|}\sqrt{{{u}^{2}-{1}}}}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x darccsc u=(−∣u∣u 2−11)d x d u d d x arccos⁡u=(−1 1−u 2)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \arccos{{u}}={\left(-\frac{1}{\sqrt{{{1}-{u}^{2}}}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x darccos u=(−1−u 21)d x d u d d x arcsec u=(1∣u∣u 2−1)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{arcsec}{u}={\left(\frac{1}{{{\left\|{u}\right\|}\sqrt{{{u}^{2}-{1}}}}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x darcsec u=(∣u∣u 2−11)d x d u d d x arctan⁡u=(1 1+u 2)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \arctan{{u}}={\left(\frac{1}{{{1}+{u}^{2}}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x darctan u=(1+u 2 1)d x d u d d x arccot u=(−1 1+u 2)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{arccot}\ {u}={\left(-\frac{1}{{{1}+{u}^{2}}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x darccot u=(−1+u 2 1)d x d u Particular cases d d x arcsin⁡x=1 1−x 2\displaystyle\frac{d}{{\left.{d}{x}\right.}} \arcsin{{x}}=\frac{1}{\sqrt{{{1}-{x}^{2}}}}d x darcsin x=1−x 21 d d x arccsc x=−1∣x∣x 2−1\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{arccsc}\ {x}=-\frac{1}{{{\left\|{x}\right\|}\sqrt{{{x}^{2}-{1}}}}}d x darccsc x=−∣x∣x 2−11 d d x arccos⁡x=−1 1−x 2\displaystyle\frac{d}{{\left.{d}{x}\right.}} \arccos{{x}}=-\frac{1}{\sqrt{{{1}-{x}^{2}}}}d x darccos x=−1−x 21 d d x arcsec x=1∣x∣x 2−1\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{arcsec}{x}=\frac{1}{{{\left\|{x}\right\|}\sqrt{{{x}^{2}-{1}}}}}d x darcsec x=∣x∣x 2−11 d d x arctan⁡x=1 1+x 2\displaystyle\frac{d}{{\left.{d}{x}\right.}} \arctan{{x}}=\frac{1}{{{1}+{x}^{2}}}d x darctan x=1+x 2 1 d d x arccot x=−1 1+x 2\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{arccot}\ {x}=-\frac{1}{{{1}+{x}^{2}}}d x darccot x=−1+x 2 1 Hyperbolic Functions The hyperbolic functions are defined as follows: sinh⁡x=e x−e−x 2\displaystyle \sinh{{x}}=\frac{{{e}^{x}-{e}^{{-{x}}}}}{{2}}sinh x=2 e x−e−x cosh⁡x=e x+e−x 2\displaystyle \cosh{{x}}=\frac{{{e}^{x}+{e}^{{-{x}}}}}{{2}}cosh x=2 e x+e−x tanh⁡x=sinh⁡x cosh⁡x=e x−e−x e x+e−x\displaystyle \tanh{{x}}=\frac{{ \sinh{{x}}}}{{ \cosh{{x}}}}=\frac{{{e}^{x}-{e}^{{-{x}}}}}{{{e}^{x}+{e}^{{-{x}}}}}tanh x=cosh x sinh x=e x+e−x e x−e−x csch x=1 sinh⁡x\displaystyle\text{csch}\ {x}=\frac{1}{{ \sinh{{x}}}}csch x=sinh x 1 sech x=1 cosh⁡x\displaystyle\text{sech}\ {x}=\frac{1}{{ \cosh{{x}}}}sech x=cosh x 1 coth⁡x=1 tanh⁡x\displaystyle \coth{{x}}=\frac{1}{{ \tanh{{x}}}}coth x=tanh x 1 General formulas d d x sinh⁡u=(cosh⁡u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \sinh{{u}}={\left( \cosh{{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dsinh u=(cosh u)d x d u d d x csch u=(−coth⁡u csch u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{csch}{u}={\left(- \coth{{u}}\text{csch}{u}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dcsch u=(−coth u csch u)d x d u d d x cosh⁡u=(sinh⁡u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \cosh{{u}}={\left( \sinh{{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dcosh u=(sinh u)d x d u d d x sech u=(−tanh⁡u sech u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{sech}{u}={\left(- \tanh{{u}}\text{sech}{u}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dsech u=(−tanh u sech u)d x d u d d x tanh u=(1−tanh⁡2 u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{tanh}{u}={\left({1}-{{\tanh}^{2}{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dtanh u=(1−tanh 2 u)d x d u d d x coth⁡u=(1−coth⁡2 u)d u d x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \coth{{u}}={\left({1}-{{\coth}^{2}{u}}\right)}\frac{{{d}{u}}}{{\left.{d}{x}\right.}}d x dcoth u=(1−coth 2 u)d x d u Particular cases d d x sinh⁡x=cosh⁡x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \sinh{{x}}= \cosh{{x}}d x dsinh x=cosh x d d x csch x=−coth⁡x csch x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{csch}\ {x}=- \coth{{x}}\text{csch}\ {x}d x dcsch x=−coth x csch x d d x cosh⁡x=sinh⁡x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \cosh{{x}}= \sinh{{x}}d x dcosh x=sinh x d d x sech x=−tanh⁡x sech x\displaystyle\frac{d}{{\left.{d}{x}\right.}}\text{sech}\ {x}=- \tanh{{x}}\text{sech}\ {x}d x dsech x=−tanh x sech x d d x tanh⁡x=1−tanh⁡2 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \tanh{{x}}={1}- \tanh{{2}}{x}d x dtanh x=1−tanh 2 x d d x coth⁡x=1−coth⁡2 x\displaystyle\frac{d}{{\left.{d}{x}\right.}} \coth{{x}}={1}- \coth{{2}}{x}d x dcoth x=1−coth 2 x 7. 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8790	https://web.infn.it/VirgoRoma/images/tesi_g23/intini_thesis.pdf	Metodologie innovative per scremare l’insieme di potenziali candidati di onde gravitazionali continue nell’era dei rivelatori avanzati LIGO-Virgo Facoltà di Scienze Matematiche, Fisiche e Naturali Corso di Laurea Magistrale in Fisica Candidato Giuseppe Intini Matricola 1227552 Relatore Dr.ssa Paola Leaci Correlatore Prof. Sergio Frasca Anno Accademico 2015/2016 Metodologie innovative per scremare l’insieme di potenziali candidati di onde gravitazionali continue nell’era dei rivelatori avanzati LIGO-Virgo Tesi di Laurea Magistrale. Sapienza – Università di Roma © 2016 Giuseppe Intini. Tutti i diritti riservati Questa tesi è stata composta con L AT EX e la classe Sapthesis. Email dell’autore: intinigiuseppe@gmail.com 3 Sommario Il progetto di collaborazione internazionale LIGO-Virgo ha come obiettivo la ricerca di segnali gravitazionali e la comprensione della struttura dell’Universo derivante dalla loro rivelazione. In quest’ambito si pone il problema di riuscire a trovare tali segnali e ricostruirne i parametri che caratterizzano le relative sorgenti. In particolare la ricerca all-sky di segnali continui, vale a dire la ricerca di quelle sorgenti di cui non si conosce a priori l’esistenza e che sono oggetto di interesse della presente tesi, richiede uno studio molto dettagliato. Dovendo, infatti, ricostruire un segnale senza aver alcuna informazione sulla sorgente che lo emette, questa procedura risulta piuttosto onerosa dal punto di vista computazionale, dato che è necessario esplorare tutto lo spazio quadridimensionale dei parametri che caratterizzano queste sorgenti (coordinate eclittiche β e λ, frequenza di rotazione e sue derivate temporali). Si devono inoltre identiﬁcare deboli segnali in dati estremamente rumorosi, cercando di ricostruire la posizione delle relative sorgenti in modo quanto più accurato possibile, in maniera da poter fare in un secondo tempo ricerche più mirate nella direzione stimata. Parte della pipeline che si occupa della ricerca di questo tipo di sorgenti, sviluppata ed utilizzata nel gruppo Virgo dell’università "La Sapienza" di Roma, è stata migliorata per quanto riguarda i tempi di calcolo relativi alla determinazione dei parametri delle sorgenti simulate e rivelate. I dati di partenza sono stati quelli del secondo e del quarto run scientiﬁco di VIRGO (VSR2 e VSR4), mentre il terzo (VSR3) è stato escluso per via della scarsa qualità dei dati e la poca sensibilità. VSR2 è cominciato il 7 Luglio 2009 ed è terminato l’8 Gennaio 2010, con un duty factor dell’80,4% per un totale di ∼159 giorni di science mode data. VSR4, invece, è cominciato il 3 Giugno 2011 ed è terminato il 5 Settembre dello stesso anno, con un duty factor dell’81% per un totale di 76 giorni di dati . A questi dati sono stati aggiunti segnali continui artiﬁciali (Software Injections) allo scopo di veriﬁcare l’abilità dell’algoritmo di ricerca (in seguito presentato) di identiﬁcarli e determinare la precisione nella stima dei parametri ad essa associati. Lo studio si è quindi mosso nella direzione di riconoscere ed analizzare pattern di candidati prodotti da una certa sorgente simulata rispetto al rumore di fondo presente nei dati considerati. In questo senso si è sviluppato un nuovo algoritmo che punta a ridurre il grande numero di candidati (dell’ordine di 108), ricercando proprio all’interno dello spazio a quattro dimensioni sopra introdotto e analizzando quali siano quelli con alta statistica di rivelazione e che esibiscono pattern ben precisi. Inﬁne, studiando le proprietà della stessa statistica di rivelazione si sono ricostruiti i parametri delle sorgenti. Sebbene l’algoritmo possa godere di ulteriori miglioramenti, i risultati ottenuti sono abbastanza promettenti: l’algoritmo di scrematura dei candidati infatti è riuscito in 2 ore a rimuovere il 99,65% dei candidati di partenza senza perdere alcun candidato proveniente da una sorgente; questo ridurrebbe di molto i tempi di calcolo della pipeline del gruppo. Dall’altro lato l’algoritmo ha potuto rivelare con buona precisione 161 sorgenti in 2,5 ore (per un totale di 4,5 ore, includendo anche l’algoritmo di pulizia) contro le 240 rivelate utilizzando l’algoritmo precedente in più di 5 giorni. In questa dissertazione verrà brevemente presentato il contesto in cui si inserisce questo lavoro; si descriverà quindi il tema delle onde gravitazionali e della loro rivela-zione e si presenterà l’algoritmo di riconoscimento dei segnali usato dal gruppo Virgo di Roma, su cui questo lavoro si basa. Inﬁne si descriverà dettagliatamente il nuovo 4 algoritmo sviluppato, esponendone i dettagli ed i miglioramenti che hanno portato a ridurre drasticamente i tempi di calcolo nonché i risultati ottenuti e i possibili sviluppi futuri. 5 Indice 1 Introduzione 7 1.1 Tipi di segnali gravitazionali . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.1.1 Segnali continui . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.2 Strumentazione . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.2.1 Rivelazione interferometrica di segnali gravitazionali . . . . . . . . 15 1.2.2 Cenni sulla Rivelazione Risonante . . . . . . . . . . . . . . . . . . . . 19 1.2.3 Rete dei rivelatori interferometrici terrestri . . . . . . . . . . . . . . . 19 1.2.4 Virgo+ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2 Procedure 29 2.1 Tipi di analisi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.2 Pipeline del gruppo Virgo di Roma . . . . . . . . . . . . . . . . . . . . . . . . 30 2.2.1 Rimozione dei glitches nel dominio del tempo . . . . . . . . . . . . . 31 2.2.2 Studio tempo-frequenza . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.2.3 Trasformata di Hough . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.3 Clustering e coincidenze . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3 Nuovi algoritmi per scremare grandi moli di candidati 39 3.1 Dal segnale rivelato al pattern di candidati . . . . . . . . . . . . . . . . . . . 43 3.1.1 Identiﬁcazione dei pattern nel set di candidati ottenuti . . . . . . . 44 3.2 Pulizia dei candidati . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.2.1 Dimensione delle celle da analizzare . . . . . . . . . . . . . . . . . . 48 3.2.2 Margini di miglioramento dell’algoritmo . . . . . . . . . . . . . . . . 49 3.3 Clustering dei candidati . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.4 Analisi dei Cluster e stima dei parametri delle sorgenti recuperate . . . . . 51 3.5 Risultati . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 4 Conclusioni 55 A Script 57 A.1 Pulizia dei dati . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 A.2 Creazione dei cluster . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 A.3 Recupero dei parametri delle sorgenti . . . . . . . . . . . . . . . . . . . . . . 65 B Tabelle Dati 77 Bibliograﬁa 107 7 Capitolo 1 Introduzione In questo capitolo si tratteranno cenni dell’emissione di onde gravitazionali e della relativa rivelazione. Le onde gravitazionali, teorizzate da Einstein nel 1916 , sono increspature dello spazio tempo che si propagano alla velocità della luce e sono fondamentalmente prodotte da accelerazioni di masse asimmetriche. Nel tentativo di rendere compatibile la gravità con la teoria relativistica, Einstein parte dal principio di equivalenza1 e giunge al concetto secondo cui la gravità non è una forza basata sull’azione a distanza, bensì è l’effetto della curvatura dello spazio-tempo dovuta alla presenza di una massa. Tale principio è formalizzato nelle equazioni di Einstein Gµν = Rµν −1 2gµνR = χTµν, χ = 8πG c4 , (1.1) dove Rµν = g αβRµναβ è la contrazione del tensore di Riemann, R = g µνRµν è lo scalare di curvatura, gµν è il tensore della metrica, Tµν è il tensore energia-impulso, G è la costante di gravitazione universale e c la velocità della luce. Queste equazioni non lineari contengono già al loro interno la propagazione del campo gravitazionale sotto forma di onda. Si pone però il problema dell’impossibilità di trovare una soluzione esatta a causa della forte non linearità. Si possono, tuttavia, trovare soluzioni in casi approssimati in cui la variazione del tensore metrico sia piccola rispetto al tensore stesso. L’esempio classico consiste nel considerare un spazio Minkowskiano2 con piccole perturbazioni variabili nel tempo (gµν = ηµν +hµν). Si ottiene facilmente (vedi ) che tali piccole perturbazioni 1Il principio di Equivalenza Forte afferma che in un campo gravitazionale, per ogni punto dello spazio tempo, è sempre possibile trovare un Sistema di Riferimento Localmente Inerziale, ovvero un intorno sufﬁcientemente piccolo in cui tutte le leggi della Fisica si comportano come in assenza di gravità, ovvero come predetto dalla Relatività Speciale (vedi ad esempio [10, 19]). 2Vale a dire uno spazio piatto, con metrica ηµν =     −1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1    . (1.2) 8 1. Introduzione rispettano le equazioni delle onde □hµν = −16πG c4 Tµν (1.3a) ∂h µ λ ∂xµ = 0, (1.3b) dove hµν ≡hµν + 1 2ηµνhλλ. Notare che la parte destra dell’equazione (1.3a) si annulla nel vuoto, ovvero dove supponiamo di studiare tale onda. Dalla soluzione delle equazioni (1.3a) e (1.3b) si ottiene che, all’interno della sorgente, h µ0 = 0 µ = 0,1,2,3 (1.4a) hik(t,r) = 2G c4,r · · d2 dt2 1 c2 Z V T00(r −ct,xn)xi xkd3x ¸ i,k,n = 1,2,3. (1.4b) La grandezza da derivare due volte nell’equazione (1.4b) è l’integrale del primo termine del tensore energia impulso (vale a dire la densità di massa-energia) moltiplicato per le due coordinate xi e xk. Come è noto questo termine è il momento di quadrupolo. Data la simmetria del sistema (hi j = h ji) abbiamo 6 gradi di libertà, ma possiamo ancora imporre 4 condizioni di gauge (detta Transverse-Traceless) che riducono a 2 i gradi di libertà del problema. Il risultato sarà quello tipico di un’onda che si propaga come ci saremmo aspettati (ovvero proporzionale a cos(ωt −κx)). Considerando, allora, che questa onda si propaghi lungo l’asse x, la perturbazione prende la forma: hT T µν =      0 0 0 0 0 0 0 0 0 0 h+ h× 0 0 h× −h+     ·cos(ωt −κx), (1.5) con ω e κ pulsazione e modulo del (tri-)vettore d’onda3. Dalla matrice in equazione (1.5) appare immediato il fatto che i due gradi di libertà corrispondono a due polarizzazioni dell’onda gravitazionale che modiﬁcano la distanza tra due punti in maniera differente. Ricordando che la quadri-distanza (ovvero la distanza invariante della Relatività Speciale) è deﬁnita da ds2 = g µνdxµdxν, (1.6) si può dedurre l’effetto dell’onda gravitazionale sulle distanze. Osserviamo quindi in ﬁgu-ra 1.1 cosa succede ad un anello di punti soggetti ad un’onda gravitazionale polarizzata + o ×4. L’onda polarizzata + tenderà a modiﬁcare la coordinata di un fattore proporzionale a sé stessa, come si può vedere trovando la componente y e z del vettore posizione controvariante x2 = g 2νxν = (1+h+)x2 e x3 = g 3νxν = (1−h+)x3. L’onda polarizzata × tenderà, invece, a modiﬁcare la coordinata di un fattore proporzionale alla coordinata ortogonale: x2 = g 2νxν = x2 +h×x3 e x3 = g 3νxν = x3 +h×x2. Quanto detto ﬁnora spiega perché le onde gravitazionali siano molto più difﬁcili da rivelare rispetto alle onde elettromagnetiche: 3Si ricava inoltre dall’equazione (1.3b) che, dato κµ = (ω,⃗ κ), kµkµ = 0. 4Notare che l’onda gravitazionale non ha effetto sul singolo punto se non in relazione ad un altro punto. La metrica modiﬁca le distanze, ma non ha un effetto diretto sulle coordinate. 9 (a) Effetto onda polarizzata + (b) Effetto onda polarizzata × Figura 1.1. Effetto delle onde gravitazionali su un anello di particelle. • come è noto, almeno su scala macroscopica, il campo gravitazionale è intrinseca-mente più debole; • in secondo luogo lo sviluppo in multipoli ci mostra che il potenziale di quadrupolo scala con una potenza in meno della distanza da una certa sorgente (∼r −3) rispetto al momento necessario per la generazione di onde elettromagnetiche (∼r −2). La prima prova indiretta dell’esistenza delle onde gravitazionali è stata attestata nel 1981 , dopo che nel 1974 Hulse e Taylor scoprirono la prima radio-pulsar5 in un sistema binario di stelle di neutroni (PSR B1913+16 ). Essi notarono una diminuzione del periodo orbitale di tale pulsar attorno alla propria compagna in perfetto accordo con quanto predetto della teoria della Relatività Generale di Einstein. Negli anni successivi, quanto trovato da Hulse e Taylor è stato confermato da ulteriori studi, condotti sia sullo stesso sistema che su altri analoghi. Merita menzione l’unico sistema oggi noto composto da due pulsar, PSR J0737-3039. La prova diretta dell’esistenza delle onde gravitazionali si è avuta il 14 Settembre 2015, quando sono state rivelate per la prima volta onde gravitazionali provenienti dalla coalescenza di due buchi neri. 5Una pulsar è una stella di neutroni che emette radiazione elettromegnetica in coni ristretti che si osservano come impulsi emessi ad intervalli estremamente regolari. La radiazione emessa è dovuta alla rapida rotazione lungo un asse non parallelo al suo elevato momento magnetico. 10 1. Introduzione Figura 1.2. Forma d’onda emessa da sorgenti continue. 1.1 Tipi di segnali gravitazionali La teoria della Relatività Generale è stata sviluppata nel tempo cercando di prevedere quali segnali gravitazionali potessero plausibilmente essere presenti nell’Universo. Ad oggi è possibile individuare 4 classi di sorgenti di onde gravitazionali, due di tipo continuo e due di tipo transiente : • Onde gravitazionali continue da stelle di neutroni rotanti • Fondo Stocastico di onde gravitazionali • Burst • Coalescenza di sistemi binari compatti. Onde gravitazionali continue da stelle di neutroni rotanti Le onde gravitazionali di ti-po continuo (CW dal termine inglese Continuous Wave) sono onde quasi mono-cromatiche (ovvero con forme d’onda del tipo in ﬁgura 1.2). Questo tipo di onde è prodotto da stelle di neutroni rapidamente rotanti caratterizzate da una simmetria non assiale6. Questa asimmetria può essere, per esempio, dovuta a stress magnetici interni. Com’è stato detto nella sezione precedente, essendo fenomeni di debole intensità, si ricercano solo sorgenti presenti nella nostra galassia. Oggetti di questo tipo sono, ad esempio, Crab e Vela. Le frequenze di questi segnali sono legate a quelle di rotazione della stella ed al tipo di rottura di simmetria che le caratterizza. Pur essendo prodotte da sorgenti di scarsa intensità, la continuità che le caratterizza le rende ottimi candidati per studiare l’evoluzione delle sorgenti. È inoltre da evidenziare che, a differenza dei fenomeni impulsivi, questi invece possono essere studiati retroattivamente, andando a cercare nei dati già ottenuti la presenza o meno del segnale e garantendo quindi un’ulteriore conferma della validità del risultato. 6Si noti che oggetti rotanti a simmetria assiale non producono onde gravitazionali in quanto caratterizzare da variazione temporale nulla del momento di quadrupolo. 1.1 Tipi di segnali gravitazionali 11 Figura 1.3. Esempio di forma d’onda che potrebbe avere il rumore stocastico di fondo. Figura 1.4. Esempio di forma d’onda di un segnale di tipo burst. Fondo stocastico di onde gravitazionali Questo tipo di rumore stocastico di fondo ha una duplice natura: cosmologica ed astroﬁsica. Tale segnale è infatti composto sia da una radiazione residua prodotta nei primissimi istanti di vita dell’Universo, sia da una sovrapposizione delle emissioni gravitazionali di migliaia di sorgenti troppo lontane per essere distinte singolarmente. Questo è un fenomeno continuo e non transiente; a differenza del precedente però, il fondo stocastico di onde gravitazionali è molto difﬁcile da rivelare, ma riuscire ad identiﬁcarlo potrebbe permetterci di conoscere più approfonditamente i primissimi istanti dell’Universo dopo il Big Bang. In ﬁgura 1.3 si nota un esempio di come potrebbe presentarsi il rumore stocastico di fondo se venisse rivelato. Burst Si tratta di fenomeni di brevissima durata (anche inferiore al millisecondo) prodot-ti da esplosioni di supernovae o da Gamma Ray Burst. Essendo fenomeni transienti estremamente energetici, si presuppone siano caratterizzati da una forte emissione gravitazionale. In ﬁgura 1.4 è possibile osservare una forma d’onda attesa per un segnale di questo tipo. Come è possibile notare, ci si attende che il fenomeno sia molto più energetico di un fenomeno che produce segnali continui anche se, appunto, di durata estremamente limitata. 12 1. Introduzione Figura 1.5. Forma d’onda prevista dalla coalescenza di un sistema binario di due oggetti compatti. Coalescenza di sistemi binari compatti Una delle principali sorgenti di onde gravitazio-nali sono i sistemi binari di oggetti compatti formati da stelle di neutroni e/o buchi neri. È possibile distinguere tre fasi nel processo ﬁsico di coalescenza che genera il segnale: lo spiraleggiamento, il merger ed il ringdown. Durante lo spiraleggiamento i due oggetti compatti emettono onde gravitazionali quasi continue a frequenze multiple del loro periodo7 e non possono quindi essere rivelate dai rivelatori interferometrici terrestri come Virgo e LIGO (che sono sen-sibili a frequenze al di sopra dei 10 Hz mentre l’orbita di due oggetti lontani dalla coalescenza è, solitamente, un fenomeno caratterizzato da frequenze inferiori ad 1 Hz), ma dovrebbero essere rivelabili da antenne spaziali (che si concentrano su frequenze molto minori di 1 Hz; si veda, ad esempio, il progetto eLISA ). La seconda e la terza fase sono gli effetti che hanno luogo quando un sistema binario è vicino al collasso: il segnale gravitazionale continuo diventa transiente attraversando una rapida fase di merger in cui i due corpi si avvicinano. Il segnale atteso da questa fase ha una forma d’onda di tipo chirp che raggiunge alte fre-quenze (vedi ﬁgura 1.5). Questa fase si conclude quando i due corpi si stabilizzano formando un nuovo oggetto (ringdown). Il segnale rivelato il 14 Settembre 2015 è stato emesso da un sistema di due buchi neri e ci ha permesso di osservare non solo per la prima volta le onde gravitazionali in maniera diretta, ma anche l’esistenza di buchi neri di massa intermedia8 e la loro coalescenza. In ﬁgura 1.6 vediamo il segnale effettivamente rivelato dai due rivelatori LIGO. Si può distin-guere facilmente la fase di merger tra i ∼0.32 ed i ∼0.43 secondi, mentre un po’ meno evidente è ringdown ﬁnale tra ∼0.43 secondi e la ﬁne. I corpi in coalescenza erano una coppia di buchi neri binari di masse 36M⊙e 29M⊙che, collassando in un buco nero di 62M⊙, hanno emesso 3M⊙in onde gravitazionali (che, data la teoria moderna, sono l’unica forma di energia che un buco nero può perdere). 7Un’unica frequenza pari al doppio di quella di rivoluzione in caso di orbite circolari, mentre una forma più complicata caratterizzata dalla sovrapposizione dei multipli della frequenza di rivoluzione in caso di orbite ellittiche. 8Prima della scoperta eravamo a conoscenza dell’esistenza di buchi neri piccoli (ﬁno a 10M⊙), che si osservano in fase di nascita o di buchi neri supermassivi (di masse ≳106M⊙), di cui si osservano gli effetti gravitazionali. 1.1 Tipi di segnali gravitazionali 13 Figura 1.6. In ﬁgura il segnale effettivamente rivelato dai due rivelatori LIGO Hanford (a sinistra, in rosso) e Livingston (a destra, in blu): nella parte superiore del graﬁco vediamo il segnale effettivamente rivelato dall’interferometro in funzione del tempo; il graﬁco centrale presenta la curva ottenuta dalla simulazione numerica paragonata con quella ricostruita in base al segnale letto; inﬁne si può notare l’andamento della frequenza nel tempo. 14 1. Introduzione 1.1.1 Segnali continui Poiché, come verrà descritto in sezione 2.1, lo studio speciﬁco si è focalizzato sul-l’individuazione di segnali emessi da stelle di neutroni rotanti e dotate di una certa asimmetria9, mostriamo brevemente qual è la forma d’onda emessa da tale tipo di sor-gente. Partiamo dalla forma generica di un segnale gravitazionale espresso nella base delle sue polarizzazioni: hµν(t) = h+(t)eµν + (t)+h×(t)eµν × (t), (1.7) dove h+ e h× sono le forme d’onda nelle due polarizzazioni + e ×, mentre eµν +,× rappresen-tano la base del tensore polarizzazione rispettivo. Si nota che le forme d’onda del segnale sono del tipo h+(t) = h0 µ1+cos2ι 2 ¶ cos(Φ(t)) h+(t) = h0 cosι sin(Φ(t)), (1.8) dove t è il tempo nel sistema di riferimento del rivelatore, ι è l’angolo di inclinazione dell’asse di rotazione della stella rispetto alla linea di vista; Φ(t) è la fase del segnale e h0 è l’ampiezza del segnale gravitazionale espressa da h0 = 4π2G c4 Izzϵf 2 d ; (1.9) G è la costante gravitazionale e c la velocità della luce; Izz è il momento d’inerzia principale della stella (supponendo che sia allineato con l’asse di rotazione), mentre ϵ = Iyy −Ixx Izz (1.10) è la sua ellitticità equatoriale (essendo Iyy e Ixx gli altri due assi di inerzia) e f la frequenza del segnale (che si veriﬁca essere pari a 2 volte la frequenza di rotazione della stella, escludendo cioè deformazioni più complesse) [8, 13, 19]. 1.2 Strumentazione Come detto all’inizio del capitolo 1, l’effetto di un’onda gravitazionale che attraversa lo spazio è quello di modiﬁcare le distanze, secondo quanto mostrato in ﬁgura 1.1. Per misurare questa variazione non possiamo però usare uno strumento che sia esso stesso soggetto alla medesima deformazione, come ad esempio uno standard metro campione. Pertanto si ricorre all’interferometria laser e ad uno dei postulati della Relatività Ristretta, ossia l’invarianza della velocità della luce. Un rivelatore interferometrico di onde gra-vitazionali sfrutta le proprietà di interferenza della luce per osservare la differenza di lunghezza dei bracci indotta dal passaggio dell’onda. Un interferometro di Michelson a bracci ortogonali è uno strumento che fa uso dell’interferometria per misurare le distanze attraverso le differenze di cammino ottico 9Altri tipi di emissioni gravitazionali per una stella di neutroni sono: la precessione libera della stella che emette con una frequenza pari a quella di rotazione; i cosiddetti r-modes, modi toroidali che agiscono sulla superﬁcie della stella, causando un’emissione attesa ∼3/4frotazione . 1.2 Strumentazione 15 Figura 1.7. Schema di un interferometro tipo LIGO-Virgo. di due fasci di fotoni. Tale strumento è costituito da un laser che, puntato su un beam splitter, divide il suo fascio in due parti; ciascuno di questi fasci, attraversando un braccio di lunghezza L, impatta contro uno specchio posto alla ﬁne del braccio stesso e torna al beam splitter. I due fasci si ricombinano sul beam splitter e interferiscono, permettendo così di misurare le distanze di cammino ottico percorse. Come si è accennato all’inizio del capitolo, si mostrerà in sezione 1.2.1 che la differenza di cammino ottico (2∆L) indotta da un’onda gravitazionale è proporzionale al cammino ottico stesso (2L) e all’ampiezza h dell’onda gravitazionale, ossia ∆L ∝Lh. L’aggiunta di cavità Fabry-Peròt lungo i bracci permette di aumentare il cammino ottico (lungo ciascun braccio), ampliﬁcando così la variazione ∆L dovuta al passaggio dell’onda gravitazionale. In ﬁgura 1.7 possiamo vedere un esempio di interferometro di Michelson a braccia ortogonali con cavità Fabry-Peròt, sistema analogo agli interferometri Virgo e LIGO. 1.2.1 Rivelazione interferometrica di segnali gravitazionali Esaminiamo come l’onda gravitazionale generica (nella TT gauge) modiﬁchi i per-corsi dei fasci nell’interferometro e, di conseguenza, come sia possibile la rivelazione di un segnale gravitazionale. Grazie ad un sistema di sospensioni mediante il quale gli specchi sono isolati dalle vibrazioni terrestri , possiamo considerare gli specchi come se fossero posti in un sistema di riferimento localmente inerziale e quindi studiare come (in TT gauge) vengono modiﬁcate le distanze in seguito al passaggio di un’onda gravitazionale. Consideriamo un’onda che si propaghi lungo l’asse z e polarizzata + (per semplicità di conto). Nel piano z = cost. l’onda gravitazionale ha ampiezza h+(t) = h0 cos(ωGWt), (1.11) con ωGW frequenza dell’onda stessa. Tale onda modiﬁca la distanza propria di Minkowsky in modo che ds2 = −c2dt2 +(1+h+(t))dx2 +(1−h+(t))dy2 +dz2. (1.12) I fotoni, essendo particelle di tipo luce, si muovono lungo le geodetiche di norma nulla (ossia ds2 = 0), quindi per fotoni che viaggiano nel braccio dell’interferometro parallelo 16 1. Introduzione all’asse x varrà ds2 = −c2dt2 +(1+h+(t))dx2 = 0 (1.13) dx = ± s 1 1+h+(t)cdt ≃± µ 1−1 2h+(t) ¶ cdt, (1.14) dove il segno dipende dal percorso che sta facendo il fascio di luce (diciamo "+" per il tratto dal beam splitter allo specchio e "-" per il ritorno). Consideriamo allora un fotone che parte dal beam splitter al tempo iniziale t0, attraversa il braccio di lunghezza Lx, colpisce lo specchio al tempo t1 e torna al beam splitter al tempo t2. Il percorso del fotone sarà 2Lx = t1 Z t0 µ 1−1 2h+(t) ¶ cdt + t2 Z t1 µ 1−1 2h+(t) ¶ cdt = c(t2 −t0)−c 2 t2 Z t0 h+(t)dt. (1.15) Inserendo la deﬁnizione di h+ dell’equazione (1.11) si può ottenere facilmente che 2Lx = c(t2 −t0)−c 2 t2 Z t0 h0 cos(ωGWt)dt = = c(t2 −t0)−ch0 2ωGW (sin(ωGWt2)−sin(ωGWt0)). (1.16) Dato che studiamo un’approssimazione al prim’ordine in h0, possiamo invertire l’equa-zione (1.16) per poter sostituire t2 = t0 + 2L c +O(h0) all’interno della funzione seno (che poiché moltiplica h0 è già di ordine 1). Infatti: ch0 2ωGW sin(ωGWt2) = ch0 2ωGW sin µ ωGW 2L c ¶ +O(h2 0) ∼ch0 2ωGW sin µ ωGW 2L c ¶ . (1.17) Sostituendolo quindi nella stessa equazione (1.16) otteniamo: 2Lx = c(t2 −t0)−ch0 2ωGW (sin(ωGW(t0 + Lx c ))−sin(ωGWt0)) = = c(t2 −t0)−ch0 2ωGW 2sin µ ωGW Lx c ¶ cos µ ωGW µ t0 + Lx 2c ¶¶ = = c(t2 −t0)−h0Lx sin ³ ωGW Lx c ´ ωGW Lx c cos µ ωGW µ t0 + Lx 2c ¶¶ = = c(t2 −t0)−h0Lx cos µ ωGW µ t0 + Lx 2c ¶¶ sinc µ ωGW Lx c ¶ . (1.18) Notiamo ora che il tempo iniziale è contenuto solo nel coseno e quindi rappresenta una sorta di fase del segnale e permette di riscrivere il tutto come 2Lx = c(t2 −t0)−h+ µ t0 + Lx 2c ¶ Lxsinc µ ωGW Lx c ¶ . (1.19) 1.2 Strumentazione 17 Considerando che il percorso in assenza di onde gravitazionali è 2Lx = c(t2 −t0), lo sfasamento del segnale in un singolo braccio indotto dal passaggio dell’onda è quindi ∆φx = ωL c ·∆Lx = LxωLh+ ³ t0 + Lx 2c ´ c sinc µ ωGW Lx c ¶ . (1.20) In maniera del tutto analoga (osservando che, come accennato, l’onda gravitazionale agisce in maniera opposta sulle direzioni ortogonali), si nota che lo sfasamento sull’altro braccio è uguale ed opposto a quello mostrato sul primo, raddoppiando così la differenza dei cammini ottici e quindi lo sfasamento totale sul segnale rivelato. Cavità Fabry-Peròt Come accennato all’inizio di questa sezione e mostrato in ﬁgura 1.7, i rivelatori interferometrici Virgo e LIGO migliorano la rivelazione attraverso cavità Fabry-Peròt, che consentono di aumentare la lunghezza effettiva dei bracci. Infatti, se la lunghezza di un braccio è uguale alla lunghezza d’onda del laser (o ad un suo multiplo) l’interferenza di tipo Fabry-Peròt è costruttiva ed i segnali si sommano in intensità. Così, detta R la riﬂettività di uno specchio che costituisca la cavità Fabry-Peròt e considerando 1 la riﬂettività dello specchio all’estremità del braccio, la probabilità che un fotone esca dalla cavità verso il rivelatore dopo n riﬂessioni all’interno del braccio è p(n) = Rn−1 ∞ P n=1 Rn−1 = Rn−1(1−R). (1.21) Risulta quindi utile valutare il numero medio di salti effettuati da un fotone < n >= ∞ X n=1 np(n) = ··· = 1 1−R , (1.22) che porta a valutare uno storage time della cavità τs = 2L c < n >. Questa grandezza indica quanto tempo in media un fotone rimane all’interno della cavità, permettendoci di valutare il cammino ottico effettivo: < L >= cτs. Queste grandezze possono essere riscritte in funzione della ﬁnesse di una cavità Fabry-Pérot, deﬁnita come il rapporto tra il free spectral range10 (δωL) e l’ampiezza a mezza altezza delle curve di interferenza ∆ωL11: F = δωL ∆ωL = π 4 pR1R2 1−pR1R2 , (1.23) con R1 ed R2 riﬂettività dei due specchi. Da qui si vede che lo storage time può essere scrit-to, nelle approssimazioni di specchio terminale del Michelson perfettamente riﬂettente e 10Il free spectral range rappresenta la differenza di lunghezza d’onda tra due picchi di trasmissione dell’interferometro. 11 Il modulo quadro del segnale trasmesso da un interferometro Fabry-Peròt è \|Et \|2 = E2 o t2 1 t2 2 1+(r1r2 cos(2kLL)), con r 2 i = Ri e t2 i = Ti (con i = 1,2 rispettivamente riﬂettività e trasmettività dell’i-esimo specchio), kL vettore d’onda del laser, E0 intensità iniziale del segnale e Et intensità del segnale trasmesso. 18 1. Introduzione 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 frequenza [pf] Figura 1.8. Differenza tra la rispo-sta di interferometro di Michel-son normale (in in verde) e con cavità Fabry-Perot (in rosso). La frequenza è in unità di fp. specchio di uscita del Fabry-Peròt con R ≃1, τs ≃L c F π . (1.24) L’effetto del passaggio dell’onda gravitazionale in una cavità Fabry-Peròt merita ulteriore attenzione. Se infatti in assenza di onda gravitazionale l’interferenza allo specchio di uscita della cavità è solo costruttiva, l’onda gravitazionale crea uno sfasamento relativo tra i fasci all’uscita della cavità che dipende da quanti salti un determinato fotone ha compiuto. Si vede facilmente che ogni salto crea uno sfasamento di ∆φ ≃4F π kL∆L = 2F π kLLh0 cos(ωGWt) = 2F π kLLh+, (1.25) con kL vettore d’onda del fascio. Tralasciando semplici calcoli, si ottiene che, deﬁnita la pole frequency fp ≡ 1 4πτs , lo sfasamento in valore assoluto in un interferometro Fabry-Peròt è \|∆φFP\| ≃h0 4F π kLL 1 q 1+(fGW/fp)2 = = h0TFP(fGW), (1.26) dove fGW è la frequenza dell’onda gravitazionale e TFP(fGW) la funzione di trasferimento del segnale. In ﬁgura 1.8 si può vedere la differenza tra l’interferometro di Michelson normale e quello con cavità Fabry-Pérot. I risultati che sono stati presentati ﬁnora sono fatti per un caso particolare: onda che si propaghi lungo l’asse z con polarizzazione +. Il caso generico si ottiene attraverso calcoli analoghi: per una trattazione più completa si veda M. Maggiore: Gravitational Waves: Theory and Experiments, Capitolo 9. In generale si mostra che nelle equazioni (1.20) e (1.25) per avere i risultati in caso di polarizzazione e direzione di propagazione generici è sufﬁciente sostituire il termine h+ con il termine 1 2hxx −hyy = 1 2h+(1+cos2 θ)cos2φ+h× cosθsin2φ, (1.27) dove (90−θ) e φ sono gli angoli corrispondenti rispettivamente a latitudine e longitudine in coordinate polari. 1.2 Strumentazione 19 Come si può notare dall’equazione (1.27), l’interferometro ha degli angoli ciechi per le due polarizzazioni, angoli cioè dove la differenza di lunghezza sui due bracci indotta dal passaggio dell’onda è la stessa e quindi l’effetto dell’onda è nullo. Questo è evidente annullando i contributi dell’equazione (1.27) per le singole polarizzazioni. 1.2.2 Cenni sulla Rivelazione Risonante Sebbene i sistemi di rivelazione moderna si siano mostrati efﬁcaci nella ricerca di onde gravitazionali, è interessante ricordare il primo esperimento condotto da Weber utilizzando uno strumento completamente diverso da quello di cui si è ﬁnora parlato: il rivelatore risonante. La rivelazione risonante si proponeva di rivelare le onde gravitazionali misurando la messa in moto delle particelle al passaggio dell’onda. In tal senso, essendo la forza indotta dall’onda periodica come l’onda stessa, si è pensato che potesse essere efﬁcace utilizzare un oggetto con frequenza di risonanza pari a quella dell’onda. L’onda gravitazionale, incidendo sul rivelatore risonante, avrebbe dovuto mettere in moto l’antenna e dei rivelatori piezoelettrici avrebbero dovuto misurarne il passaggio. Tale strumento portò il ﬁsico Joseph Weber ad affermare di aver rivelato per la prima volta onde gravitazionali nel 1969 . Questi risultati sono stati considerati molto controversi ed il fatto che molti rivelatori di questo genere siano stati costruiti, ma nessuno abbia rivelato altri risultati consistenti con quelli di Weber, ha fatto sì che tale rivelazione fosse considerata sbagliata. Il problema di tale strumento era la scarsa sensibilità, nonostante le barre risonanti fossero molto massive per incrementarne la risposta e i rivelatori piezoelettrici molto sensibili. La debolezza del segnale gravitazionale non consentiva tuttavia di osservare segnali forti che provenissero dal di fuori della nostra galassia. Un’ulteriore problematica di tali rivelatori era dovuta al fatto che la frequenza dell’onda gravitazionale doveva essere esattamente pari alla risonanza dello strumento perché vi fosse rivelazione . Il passaggio storico dai rivelatori risonanti a quelli interferometrici ha segnato il passaggio dagli esperimenti di piccola scala (una dozzina di ricercatori) a quelli di collaborazione internazionale che abbiamo oggi (più di 1000 ricercatori). 1.2.3 Rete dei rivelatori interferometrici terrestri Sulla base di un accordo internazionale si è deciso di installare una rete di rivelatori che permettesse non solo di rivelare segnali gravitazionali, ma anche di conoscere con buona precisione la loro provenienza. Tale rete (oggi in fase di sviluppo) è costituita da rivelatori cosiddetti di seconda generazione: Virgo nella sua versione Advanced. Si tratta di un rivelatore interferometrico con lun-ghezza dei bracci pari a 3 km, situato a Cascina (Pisa). È oggi in fase di aggiorna-mento. LIGO è la coppia di interferometri americani: i due apparati sono situati a Hanford (Wa-shington) ed a Livingston (Luisiana) e sono già attivi nella loro versione Advanced. I due interferometri hanno bracci di 4 km. GEO600 situato a sud di Hannover, in Germania, è un piccolo interferometro con bracci di 600 m e viene essenzialmente usato come laboratorio per lo studio delle compo-20 1. Introduzione nenti che poi saranno usate sui rivelatori più grandi. Rimane oggi un interferometro di prima generazione. Kagra è un interferometro in costruzione in Giappone. Il progetto prevede bracci di 3 km. Ligo-India (o INDIGO) è un progetto mirato alla costruzione di un interferometro in India. Questo è motivato dalla necessità di avere un interferometro nell’emisfero australe per migliorare la precisione nella stima della direzione di provenienza di un segnale gravitazionale. Oltre a questi interferometri è interessante accennare a due progetti futuri: da un lato gli interferometri spaziali che, essendo privi di rumore sismico, saranno in grado di rivelare segnali a frequenze molto più basse (nonostante altri tipi di rumori caratteristici impediscano di rivelare quelli a frequenze più alte). Il progetto eLISA dovrebbe prevedere un interferometro triangolare (quindi non più a bracci ortogonali) in grado di osservare le frequenze ﬁno ad 1 Hz . Un altro progetto su cui si sta lavorando sono gli interfero-metri di terza generazione, sotterranei, con bracci pari o superiori a 10 km e specchi a temperature vicine ai 20 K . 1.2.4 Virgo+ Essendo stato svolto il lavoro sui dati sperimentali di Virgo+, speciﬁcatamente sui run scientiﬁci 2 e 4, presentiamo una breve descrizione di tale apparato. Il paragrafo si concluderà con la descrizione delle migliorie apportate nel frattempo a Virgo+, trasfor-mato nell’apparato Advanced, che dovrebbe effettuare la prima presa dati nel secondo semestre di quest’anno. D’ora in avanti si indicherà per semplicità di notazione "Virgo" intendendo però la versione + dell’interferometro. Tipi di dissipazione in Virgo Presentiamo i vari tipi di rumore che inﬂuenzano le prese dati di Virgo . Rumore sismico L’insieme delle oscillazioni del terreno viene classiﬁcato come rumore sismico, che è rilevante alle basse frequenze. Tale tipo di rumore impedisce di rivelare l’emissione di onde gravitazionali di sistemi binari di corpi compatti coa-lescenti in fase a spirale (vedere 1.1). Questo limite può essere superato solo da interferometri spaziali come quello del progetto eLISA . Shot noise Si tratta delle ﬂuttuazioni in lettura del laser ed è possibile mostrare che tale rumore si riduce all’aumentare della potenza del laser e che il rapporto segnale/ru-more è massimo quando in uscita si osserva la frangia scura. Pressione di radiazione La luce emessa dal laser esercita una pressione di radiazione sugli specchi dell’interferometro, causandone uno spostamento. Tale effetto au-menta all’aumentare della potenza del laser, e deve quindi essere minimizzato insieme allo shot noise. 1.2 Strumentazione 21 Figura 1.9. Curva di sensibilità di Virgo. Rumore termico Si tratta delle ﬂuttuazioni indotte dal moto microscopico delle parti-celle di specchi e sospensioni ed è quindi strettamente legato alla temperatura e alla geometria del sistema. In prossimità delle frequenze di risonanza dei modi normali delle varie componenti si hanno alcuni tra i picchi più elevati. In ﬁgura 1.9 osserviamo la sensibilità di Virgo. È evidente che per fare uno studio del segnale alcune frequenze caratterizzate da alti picchi di rumore vadano esclude dai dati perché troppo disturbate. Cenni sull’apparato sperimentale Virgo è un rivelatore interferometrico la cui struttura, mostrata in ﬁgura 1.10, è quella di un Michelson con bracci ortogonali con un cammino ottico di 267 km indotto da bracci lunghi 3 km ciascuno con cavità Fabry-Peròt di ﬁnesse 14012. Ciascuna cavità è costituita da uno specchio detto input più vicino alla sorgente laser e uno specchio detto end posto all’estremità di ciascun braccio dell’interferometro. Una caratteristica peculiare dei due specchi end è quella di essere concavi con raggio di curvatura Rc = 3450 m: in questo modo è possibile collimare ulteriormente il fascio laser che altrimenti, poiché non perfettamente coerente, avrebbe uno spot via via crescente a ogni riﬂessione all’interno della cavità. Virgo è inoltre munito di un sistema di power recycling in ingresso al beam splitter. Tale sistema consente di far rientrare la luce riﬂessa all’ingresso del Michelson attraverso uno specchio aggiuntivo, senza il quale, tornando verso il laser, provocherebbe dei disturbi nel funzionamento del laser stesso. Con questa tecnica è possibile aumentare l’intensità del fascio laser riducendo, tra l’altro, lo shot noise. Inﬁne è presente un apparato mode cleaner 12Il cammino ottico si ricava facendo Leff = L∗< n >, con < n > dato dall’equazione (1.22). 22 1. Introduzione Figura 1.10. Apparato ottico di Virgo. posto in entrata al power recycling con lo scopo di ripulire il laser dalle sue componenti non gaussiane . Gli specchi che formano la cavità Fabry-Peròt, il beam splitter e lo specchio utilizzato per ottenere il power recicling sono stati pensati di dimensioni e masse tali da ridurre il più possibile l’effetto della pressione di radiazione che il laser imprime sulle loro superﬁci. Ciascuno specchio è sostenuto da un sistema di sospensioni con cui forma il superat-tenuatore (ﬁgura 1.11). Questo sistema è stato costruito con lo scopo di ridurre il rumore sismico, che è la principale fonte di rumore a basse frequenze. Il superattenuatore è co-stituito da un pendolo invertito, sei sospensioni e un payload. Quest’ultimo è formato da uno specchio collegato alle sospensioni tramite una marionetta e dalla massa di reazione, come indicato in ﬁgura 1.12. Ogni elemento del superattenuatore è sospeso al precedente tramite un cavo (ﬁgura 1.11). L’ultimo di questi elementi (denominato Filtro 7) è costituito da quattro gambe a cui sono attaccate altrettante bobine e sorregge la marionetta tramite cavo. Come pos-siamo vedere in ﬁgura 1.12, questa presenta quattro bracci con agli estremi dei magneti: facendo passare corrente nelle bobine è possibile quindi controllare i gradi di libertà della marionetta riducendo ulteriormente i moti residui delle sospensioni ﬁno a raggiungere il punto di lavoro dell’interferometro (meno di 1 µrad per i gradi di libertà angolari, meno di 1 µm per gli spostamenti lungo la direzione del fascio laser). Lo specchio, in silice fusa, pende dalla marionetta tramite quattro ﬁbre di lunghezza 0.70 m e diametro (285±10) µm composte del suo stesso materiale. Queste sono saldate in punti diametralmente opposti sullo specchio chiamati orecchie: tale sistema specchio-ﬁbre è detto sospensione monolitica. La scelta dello stesso materiale piuttosto che di materiali differenti per ﬁbre e specchio è stata fatta per ridurre il rumore termico. Come si vede in ﬁgura 1.12, sullo specchio sono presenti quattro magneti utilizzati per controllare i suoi gradi di libertà tramite altrettante bobine: queste sono sorrette dalla massa di reazione la cui forma ad anello protegge la superﬁcie laterale dello specchio. La 1.2 Strumentazione 23 Figura 1.11. Sistema di sospensione in Virgo. 24 1. Introduzione Figura 1.12. Struttura del payload con marionetta, massa di reazione e sospensione monolitica. massa di reazione presenta delle aperture per far passare le ﬁbre dello specchio ed anche questa pende dalla marionetta tramite quattro cavi d’acciaio. In entrata al Michelson di Virgo è presente un laser Nd:YAG con λ = 1064 nm che produce una potenza di 17 W in grado di ridurre lo shot noise rispetto alla precedente conﬁgurazione di Virgo . Tenendo conto delle caratteristiche tecniche descritte e dei tipi di dissipazione, mini-mizzati tramite le scelte che caratterizzano l’apparato sperimentale di Virgo, l’interfero-metro risulta sensibile ad un intervallo di frequenza tra ∼10 Hz e 10 kHz. Miglioramenti in Advanced Virgo Advanced Virgo (AdV) è un progetto di miglioramento del rivelatore Virgo che lo ha portato ad essere un interferometro di seconda generazione. Il progetto è quello di arrivare entro il 2020 circa ad aumentare la sensibilità di un ordine di grandezza rispetto al suo predecessore , che corrisponde ad aumentare il tasso di rivelazione di 3 ordini di grandezza. In ﬁgura 1.13 è possibile vedere la sensibilità teorica di AdV mentre in ﬁgura 1.14 si osserva la differenza tra AdV e Virgo+. Rispetto a Virgo+ la maggior parte dei sottosistemi di rivelazione hanno avuto signiﬁcativi miglioramenti in modo da essere 1.2 Strumentazione 25 Figura 1.13. Curva di sensibilità teorica di AdV nelle frequenze di acquisizione dati; si possono notare anche gli andamenti delle diverse componenti di rumore. compatibili con la sensibilità voluta. Il progetto AdV è stato ﬁnanziato nel dicembre del 2009 ed è oggi in avanzato stato di installazione. Di seguito esporremo brevemente i principali miglioramenti previsti nella progettazione di AdV, mentre in ﬁgura 1.15 è possibile osservare la nuova conﬁgurazione ottica . Conﬁgurazione ottica dell’interferometro AdV sarà un interferometro dual-recycled. Oltre al già citato power recycling, infatti, sarà inserita anche una cavità signal-recycling (SR). Quest’ultima componente ottica, opportunamente regolata, permet-terà di modiﬁcare la forma della curva di sensibilità ottimizzando il rivelatore per differenti sorgenti astroﬁsiche. In più, per ridurre l’impatto del rumore termico del rivestimento degli specchi nell’intervallo delle medie frequenze, saranno aumenta-te le dimensioni dello spot del laser sulle masse test. Pertanto risulta necessario spostare il beam waist al centro della cavità Fabry-Peròt (circa ad 1,5 km dallo spechio input). La ﬁnesse della cavità stessa sarà maggiore rispetto a quella di Virgo: è stato scelto un valore di riferimento di 433. Avendo un fascio più largo è stato necessario installare un nuovo telescopio mode matching all’entrara/uscita dell’interferometro. Aumento della potenza del laser Per aumentare la sensibilità ad alte frequenze è neces-sario un interferometro con maggior potenza. La sensibilità di riferimento per AdV è stata calcolata assumendo 125 W all’entrata dell’interferometro, dopo l’Input Mode Cleaner. Considerando perciò le perdite del sistema di iniezione, il laser 26 1. Introduzione Figura 1.14. Confronto tra la curva di sensibilità teorica di AdV e quella misurata in Virgo+. Figura 1.15. Apparato ottico di AdV. 1.2 Strumentazione 27 deve produrre una potenza di almeno 175W. Dopo il primo anno di misurazioni a potenza minore (60W), si progetta quindi di installare in seguito un laser da 200 W. Altri miglioramenti Le masse degli specchi sono state raddoppiate (ﬁno a 42 kg) per ge-stire la nuova pressione di radiazione, nonché per ridurre ancora il rumore termico degli specchi stessi. Anche i payload sono stati ridisegnati per aiutare a ridurre il rumore sismico, mentre i superattenuatori di Virgo erano già sufﬁcientemente avanzati da gestire le migliorie previste per AdV (anche se sono in progetto ulteriori miglioramenti per il futuro). 29 Capitolo 2 Procedure Una volta ottenuti i dati dall’interferometro, devono essere analizzati: esistono vari tipi di analisi differenti, in base al segnale ed al tipo di studio che si vuole effettuare. Qui punteremo la nostra attenzione solo su segnali di onde gravitazionali continue prodotti da stelle di neutroni rapidamente rotanti e caratterizzate da una certa asimmetria. Dopo una sintetica spiegazione dei tipi di analisi possibili nel contesto della rivelazione di segnali di CW, descriveremo la pipeline utilizzata dal gruppo Virgo dell’università di Roma "La Sapienza" per la ricerca su tutto il cielo di tali segnali, dalla quale si partirà per lo studio in esame. 2.1 Tipi di analisi Il modo in cui si effettuano ricerche di segnali CW dipende da quante informazioni (grazie alle osservazioni di emissione di radiazione elettromagnetica) si hanno sulla sorgente. Esistono quindi: Ricerche mirate per stelle di neutroni note, per le quali si conoscono accuratamente posizione nel cielo ed evoluzione in frequenza (es. Crab e Vela). Ricerche dirette a stelle di neutroni per le quali si conosce solo la posizione nel cielo (es. Centro galattico, Scorpius X1, Cassiopeia A). Ricerche Narrow Band metodologia intermedia tra quelle appena descritte. Le ricerche Narrow Band sono effettuate per segnali di cui si conosce con una certa precisione sia posizione nel cielo che evoluzione ma, supponendo che l’emissione delle onde possa avvenire per fenomeni ad una frequenza leggermente diversa da quella attesa1, si vuole effettuare una ricerca un po’ più ampia sulla frequenza dell’onda gravitazionale2 e sulla sua derivata. Ricerche su tutto il cielo (o cieche) per le quali non si ha alcuna informazione sul segna-le ricercato, ma si procede studiando un ampio spazio dei parametri e cercando di riconoscere e ricostruire i parametri di ipotetiche sorgenti. 1Per esempio, se il nucleo della stella ruotasse in maniera diversa dalla crosta. 2Si ricorda che per stelle di neutroni con asimmetria assiale rapidamente rotanti, la frequenza di emissione di onde gravitazionali è fGW = 2frot con frot frequenza di rotazione della stella. 30 2. Procedure Figura 2.1. Diagramma di ﬂusso con i tratti salienti della pipeline del gruppo Virgo della Sapienza. La strategia utilizzata per estrarre i deboli segnali CW dai dati rumorosi dei rivelatori si basa sul cosiddetto ﬁltro adattato matched ﬁltering. La ben nota tecnica di matched ﬁltering consiste nel correlare i dati con un template (ﬁltro) che descrive l’ampiezza e l’evoluzione della fase del segnale al passare del tempo. Pertanto, una ricerca di segnali di CW consiste nell’utilizzare il metodo matched ﬁltering su un insieme di template, i cui parametri associati alla fase devono essere scelti per mappare una certa regione di interesse. Per una ricerca di sorgenti non note di CW, si tratta di scandagliare l’intero cielo, un grande intervallo di frequenze e di derivate temporali della frequenza (spindown). Purtroppo, il numero di template aumenta all’aumentare della durata temporale dei dati da analizzare3. Quindi, il matched ﬁltering diventa troppo oneroso dal punto di vista computazionale per analizzare lunghe serie di dati e ampie regioni dello spazio dei parametri. Pertanto, per ridurre i tempi di calcolo è necessario ricorrere a metodi gerarchici. In questo tipo di strategie i dati vengono divisi in segmenti più corti. Ogni segmento è analizzato in maniera coerente per poi ricombinare in l’informazione ottenu-ta dall’analisi dei differenti segmenti attraverso un’analisi incoerente (ciò signiﬁca che viene persa l’informazione sulla fase). La sensibilità in ampiezza cresce nel migliore dei casi come la radice quarta del numero di segmenti . 2.2 Pipeline del gruppo Virgo di Roma La pipeline di seguito descritta è stata sviluppata dal gruppo Virgo dell’università "La Sapienza" di Roma nell’ambito della ricerca su tutto il cielo di onde gravitazionali continue. In ﬁgura 2.1 si nota il diagramma di ﬂusso che rappresenta le fasi salienti di tale pipeline . 3Periodi d’integrazione dell’ordine di pochi mesi o anni sono necessari per consentire una rivelazione adeguata. 2.2 Pipeline del gruppo Virgo di Roma 31 È stata successivamente veriﬁcata l’efﬁcienza di tale pipeline partendo dai dati dei run scientiﬁci di Virgo, VSR2 e VSR4, ed iniettando in essi dei segnali continui simulati (vedi tabella B.1), noti come software injections. Questo viene fatto per testare l’abilità di un algoritmo (che analizza tali dati) ad identiﬁcare i medesimi segnali e recuperarne i parametri delle relative sorgenti. La distribuzione dei parametri delle sorgenti simulate è uniforme in frequenza, spindown, longitudine eclittica λ e coseno della latitudine eclittica cosβ4. 2.2.1 Rimozione dei glitches nel dominio del tempo Prima di essere analizzati, i dati ottenuti dal rivelatore vengono ripuliti di quei disturbi noti come glitches nel dominio del tempo5. La procedura per rimuovere questi effetti è suddivisa in due parti: l’identiﬁcazione di tali eventi e la loro rimozione . Identiﬁcazione degli eventi A causa dei rumori non stazionari la sensibilità del rivela-tore varia nel tempo. Questo implica che, se vogliamo imporre una soglia per identiﬁcare gli eventi di rumore, questa deve essere adattiva, vale a dire deve cambiare nel tempo. Chiamiamo xi il nostro campione di candidati. Il background è stimato dalla media auto-regressiva del valore assoluto e del quadrato di xi: yi = xi + w yi−1 (2.1a) qi = x2 i + wqi−1, (2.1b) con y0 = 0, q0 = 0 e w = e−δt τ , (2.2) dove δt è il tempo di campionamento e τ è il termine di memoria (in questo caso aventi dimensioni di un tempo) che è stato usato per la media auto-regressiva. Essendo w un termine che agisce come peso, il fattore di normalizzazione Zi è costruito da w in base alla formula Zi = (1−wZi−1), con Z0 = 0. La media e la deviazione standard sono quindi 4Tale scelta è stata fatta data la nota equivalenza tra la superﬁce di una sfera e quella di un cilindro dimostrata da Archimede. 5All’interno dei dati sono presenti alcuni disturbi nel dominio del tempo e della frequenza che, se non propriamente rimossi, riducono in maniera signiﬁcativa la sensibilità di un certo algoritmo di ricerca, ﬁno addirittura a rendere ciechi in certe frequenze o a certi tempi. L’effetto varia notevolmente in base alla natura o all’ampiezza del disturbo. Questi disturbi si classiﬁcano essenzialmente in due classi, che discutiamo brevemente di seguito. Glitch nel dominio del tempo Disturbi istantanei che possono essere approssimati con una delta di Dirac nel dominio, appunto, del tempo. Queste linee aumentano il livello di rumore del rivelatore in un’ampia banda in frequenza (come ci si aspetta facendo la trasformata in frequenza di una delta). Linee spettrali di frequenza costante Nella maggior parte dei casi si tratta di disturbi la cui origine è nota (vedi sezione 1.2.4), come le linee di calibrazione o linee la cui origine è stata scoperta studiando il comportamento del rivelatore e dell’ambiente circostante. Questi prendono la forma di picchi molto stretti nel dominio della frequenza e possono essere eliminati rimuovendo i dati corrispondenti ad un intorno delle frequenze affette. 32 2. Procedure calcolati come µi = yi Zi (2.3a) σi = r qi Zi −µ2. (2.3b) La soglia utile ad identiﬁcare i time glitches viene quindi imposta alla statistica di rivela-zione detta Critical Ratio (CR), deﬁnita come CR = xi −µi σi . (2.4) Il valore del tempo di memoria τ dipende dall’apparato. Per l’analisi di VSR2 e di VSR4 (come per tutte le analisi riguardanti Virgo+ fatte dal gruppo di Virgo della Sapienza) è stato ﬁssato tale tempo a 600 s, mentre la soglia CRthr è stata ﬁssata a 6. Deﬁnita quindi la soglia adattiva, la procedura fa ciò che segue: • quando un segnale xi va sopra soglia (CR > CRthr), si considera avvenuto un evento; • un evento termina dopo che il segnale è rimasto sotto soglia per un tempo maggiore del tempo morto, un valore ﬁssato ad 1 s; • l’evento è descritto da alcuni parametri. Qui utilizzeremo il suo tempo d’inizio e la sua durata, deﬁnita come il tempo dall’inizio dell’evento al suo termine sottratto del tempo morto6. Rimozione degli eventi Si deﬁnisce ora il valore detto bordo (che in questo caso è stato scelto di 0,1s). Questo valore indica quanti secondi prima e dopo l’evento sono utilizzati nella pulizia dei dati. A questo punto i dati tra l’inizio dell’evento e la sua ﬁne (inizio+durata) vengono messi a 0; i tempi compresi tra (inizio-bordo) e l’inizio vengono linearmente mandati a 0; i tempi compresi tra la ﬁne e (ﬁne+bordo) vengono linearmente mandati da 0 al valore misurato del segnale. 2.2.2 Studio tempo-frequenza Una volta eliminati i disturbi nel dominio del tempo, il segnale subisce una serie di studi nel dominio tempo-frequenza in modo da riuscire a ricostruire il più possibile il segnale in funzione del tempo e della posizione celeste. Questa parte si divide in tre pro-cessi diversi: la creazione del cosiddetto SFDB (Short Fourier Database), la creazione della peakmap e la correzione dell’effetto Doppler dovuto al moto di rotazione e rivoluzione della Terra. SFDB La creazione dello Short Fourier Database (SFDB) è descritta diffusamente in . Brevemente l’SFDB consiste in una collezione di Fast Fourier Transform (FFT) ottenute sui dati ripuliti come spiegato nella sezione precedente. Come è noto, una FFT induce 6Gli altri parametri, che trascuriamo in questa trattazione, sono istante della massima ampiezza e massima ampiezza dell’evento. 2.2 Pipeline del gruppo Virgo di Roma 33 una discretizzazione in frequenza pari all’inverso della sua durata (tempo di coerenza). Si sceglieranno quindi FFT sufﬁcientemente corte da garantire che la frequenza di un segnale non si sposti mai più di un passo di discretizzazione a causa dello spindown della sorgente e dell’effetto Doppler dovuto ai moti della Terra. Inoltre, per ridurre la dispersione di potenza dovuta alla lunghezza ﬁnita di ciascuna FFT, queste sono interallacciate a metà della loro durata. Il tempo di coerenza per i dati usati è di TFFT = 8192 s, che corrisponde ad una risoluzione in frequenza di δf = 1 TFFT = 1,22x10−4 Hz [2, 8]. Peakmap Dall’SFDB si costruisce quindi la cosiddetta peakmap: una mappa nel dominio tempo-frequenza, creata scegliendo tra i dati delle FFT i massimi locali (o picchi) più signiﬁcativi. Sono considerati più signiﬁcativi quei picchi per i quali un dato estimatore R superi un valore di soglia. Mostriamo come viene stabilito l’estimatore R. Cominciamo creando una media auto-regressiva dello spettro di potenza che rappre-senti una sorta di valore atteso per una data frequenza. Per fare questo prendiamo uno spettro di potenza con risoluzione in frequenza δf . Sia allora xi il valore assoluto di tale spettro per ogni campione i = 1,...,N, essendo N la lunghezza di FFT [2, 6, 8]. Deﬁniamo le seguenti grandezze: w = e−δf τ (2.5a) yi = xi + w yi−1 (2.5b) Zi = (1−wZi−1) con Z0 = 0. (2.5c) Queste grandezze sono molto simili a quelle deﬁnite nella sezione 2.2.1 per rimuovere i disturbi nel dominio del tempo. Da notare comunque che questa volta il termine di memoria è dimensionato come una frequenza. A questo punto la media auto-regressiva sarà: µi = yi Zi , (2.6) mentre l’estimatore che cercavamo sarà semplicemente il rapporto R = xi µi . (2.7) Il concetto che si cela dietro questa procedura è quello di valutare un termine medio dello spettro di potenza per poter identiﬁcare quei picchi che ne escono fuori. A questo punto, come detto prima, sono selezionati quei picchi il cui R superi una soglia, stabilita empiricamente e che risulta pari a Rthr = p2,5. Una volta costruita la peakmap, questa viene quindi ripulita da quei picchi dovuti ai disturbi nel dominio delle frequenze. Correzione dell’effetto Doppler A questo punto si effettua la correzione dell’effetto Doppler. Questa procedura è con-cettualmente semplice, sebbene sia complicata a livello pratico (poiché non è semplice calcolare con la precisione necessaria la velocità della Terra). Mostreremo di seguito brevemente alcune considerazioni a riguardo. 34 2. Procedure Ciò che stiamo cercando di fare è di spostare le frequenze dei picchi di un ammontare corrispondente alla variazione di frequenza che subirebbe un segnale proveniente da ognuna delle direzioni di cielo considerate. Ricordiamo che l’effetto Doppler, ipotizzando di poter trascurare il moto proprio della sorgente (che è vero se la distanza della sorgente è ≳10pc) , è del tipo f = f0 Ã 1+ ⃗ VT · ˆ n c ! , (2.8) dove VT è la somma delle velocità di rotazione e rivoluzione terrestre. Mettiamoci ora in coordinate eclittiche geocentriche e trascuriamo la rotazione terrestre. Sotto queste condizioni la velocità della Terra dipende solo dalla longitudine eclittica del Sole λS, in quanto la sua latitudine eclittica βS ≡0 7. Supponiamo allora di avere una sorgente con coordinate eclittiche λ e β. Facendo il prodotto scalare tra il versore posizione nel cielo8 ed il versore velocità della Terra9 si ottiene f = f0 Ã 1+ \| ⃗ VT \| c cos(β)sin(λ−λS) ! . (2.9) A questo punto conviene considerare una griglia nel cielo per effettuare queste correzioni. Inizialmente si farà un’analisi su una griglia lasca, per poi rafﬁnarla in seguito (vedi il successivo paragrafo 2.2.3) [2, 8]. Per costruire la griglia consideriamo l’equazione (2.9) e, considerando il moto della Terra circolare, inseriamo per comodità ¯ ¯⃗ VT ¯ ¯ = RΩORB, dove R è il raggio dell’orbita della Terra e ΩORB è la velocità angolare della Terra. Si richiede che un passo in una delle due direzioni del cielo corrisponda ad un passo nella frequenza corretta dall’effetto Doppler. Quindi: δf = 1 TFFT = df dλδλ = df dβδβ ⇒ δλ = ¯ ¯ ¯ ¯ ¯ δf df dλ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ δf f0 ΩORBR c cos(β)cos(λ−λS) ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ cδf f0ΩORBR cos(β)cos(λ−λS) ¯ ¯ ¯ ¯, (2.10a) δβ = ¯ ¯ ¯ ¯ ¯ ¯ δf df dβ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ δf f0 −ΩORBR c sin(β)sin(λ−λS) ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ cδf f0ΩORBR sin(β)sin(λ−λS) ¯ ¯ ¯ ¯. (2.10b) L’unico limite di questo discorso è che λS varia nel tempo (data la rotazione della Terra), mentre noi siamo interessati ad una discretizzazione che resti costante per tutti i tempi, in quanto vorremo poi spostarci in uno spazio indipendente dal tempo tramite la Tra-sformata di Hough (vedi 2.2.3). Per rimediare a questo massimizzeremo la parte della funzione che contiene λS: 7In coordinate eclittiche geocentriche il moto relativo del Sole rispetto alla Terra avviene lungo l’eclittica. Poiché la latitudine eclittica βS misura la distanza angolare di un corpo celeste dal piano dell’eclittica, per il Sole tale valore è zero. 8 ˆ n = (cos(β)cos(λ),cos(β)sin(λ),sin(β)). 9Se consideriamo il versore Terra-Sole ˆ v = (cos(λS),sin(λS),0), si ottiene facilmente il versore velocità della Terra. Poiché la Terra si muove in senso antiorario, la derivata del versore equivale, in approssimazione di orbita circolare, ad aggiungere π/2 a λS. 2.2 Pipeline del gruppo Virgo di Roma 35 δλ = max λS∈[−π:π] ¯ ¯ ¯ ¯ cδf f0ΩORBR cos(β)cos(λ−λS) ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 2 ND cosβ ¯ ¯ ¯ ¯, (2.11a) δβ = max λS∈[−π:π] ¯ ¯ ¯ ¯ cδf f0ΩORBR sin(β)sin(λ−λS) ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 2 ND sinβ ¯ ¯ ¯ ¯, (2.11b) con ND = 2f0ΩORBR cδf . A questo punto si può creare la griglia sul cielo, ricordando che ND10, e quindi la griglia, dipendono dalla frequenza in studio. Poiché, come è stato mostrato, l’effetto Doppler dipende tanto dalla posizione del segnale quando dalla frequenza a cui questo è emesso, per mantenere il segnale all’in-terno di un singolo passo di frequenza per tutta la durata della FFT è necessario che il numero di punti nel cielo (Nsky: sky patches) cresca con il quadrato della frequenza (infatti Nsky ≃4πN 2 D). In particolare abbiamo quindi 2492 punti a 20 Hz e 81244 punti a 128 Hz. Per via del ragionamento di cui sopra è più comodo costruire la griglia in coordinate eclittiche . 2.2.3 Trasformata di Hough Ogni peakmap corretta è l’input della Frequency-Hough Transform. Questo strumento si basa sull’idea di mappare un punto con coordinate (x, y) in una retta su un piano ortogonale (m,q) secondo l’equazione y = mx + q. (2.12) In questo modo, se abbiamo dei pattern lineari, è abbastanza chiaro che il punto di intersezione tra le rette nel piano ortogonale fornirà esattamente quei valori di m e di q che regolano l’andamento della retta. Nel caso speciﬁco, si trasforma il piano (tempo, frequenza) nel piano ortogonale f / ˙ f , ossia (frequenza, spindown) secondo la relazione f = f0 + ˙ f (t −t0), (2.13) dove f0 e ˙ f sono frequenza e spindown della sorgente al tempo di osservazione iniziale t0, mentre t è il tempo di inizio associato ad una peakmap. Per fare questo si costruisce una griglia nel piano f / ˙ f . Poiché la trasformazione dalla peakmap al piano di Hou-gh non è computazionalmente dipendente dal valore dell’intervallo di frequenza (che impatta invece sulla dimensione del piano di Hough), riduciamo il passo di discretiz-zazione ad 1/10 del valore ottenuto dalla FFT, in modo da ridurre le perdite dovute agli effetti di digitalizzazione. Per questo motivo la risoluzione ﬁnale in frequenza è δfH = 1,22x10−5 Hz. Per quanto riguarda i valori dello spindown da ricercare, è opportuno sceglierli in maniera tale che da un lato si includano valori di spindown della maggior parte delle pulsar note e dall’altro che non aumenti troppo il tempo di calcolo computazionale. La 10ND è in sostanza il numero di passi di ciascuna delle coordinate celesti inﬂuenzati dall’effetto Doppler. 36 2. Procedure scelta fatta è stata [−1,0×10−10,+1,5×10−11] Hz s−1. L’asimmetricità del range è data dal fatto che, sebbene siano noti effetti che fanno aumentare la frequenza di rotazione di una stella di neutroni, solitamente la perdita di energia tende a farla diminuire, rendendo più comuni gli spindown negativi. È immediato osservare che lo step in spindown debba scalare con il reciproco del tempo di osservazione, ossia δ ˙ f = δf /Tobs. Evidenziamo che i dati usati per questo studio, (VSR2 e VSR4) avendo tempi di osservazione diversi, hanno quindi un δ ˙ f diverso: per VSR2 δ ˙ f = 7,63×10−12 Hz s−1, che corrisponde ad avere Nsd = 16 valori di spindown diversi, mentre per VSR4, δ ˙ f = 1,5×10−11 Hz s−1, e Nsd = 9 . Per un dato punto nel cielo, il risultato della Frequency-Hough Transform è un isto-gramma nel piano f / ˙ f chiamato Mappa di Hough. A questo punto vengono scelti i candidati più signiﬁcativi (ovvero le celle con ampiezza maggiore nella mappa di Hough) e si esegue una ricerca più rafﬁnata intorno ai relativi parametri ottenendo i candidati rafﬁnati di primo livello. Questa ricerca si svolge con la stessa procedura già descritta, usando una sovra-risoluzione in β, λ e ˙ f in modo da ridurre maggiormente le perdite indotte dagli effetti di digitalizzazione . La scelta del parametro di sovra-risoluzione è stata fatta cercando di bilanciare la riduzione della perdita per effetti di digitalizzazione e l’aumento dei tempi di calcolo. Essendo, invece, la griglia in frequenza già sovra-risoluta non si sono fatti ulteriori aumenti di risoluzione. Per lo spindown è stata applicata una sovra-risoluzione di un fattore K ˙ f = 6, ovvero l’intervallo della griglia lasca tra un valore di spindown ed il successivo (da entrambi i lati) è stato diviso ancora in 6 parti. Lo studio nella scala rafﬁnata include i 12 valori minori e gli 11 maggiori rispetto allo spindown del candidato scelto nella griglia lasca. Si è fatta questa scelta poiché, facendo in parallelo la ricerca rafﬁnata anche per la posizione, per via della correlazione dei parametri lo spindown può essere trovato anche fuori dal suo originario intervallo lasco. La griglia in posizione viene rafﬁnata con un fattore di sovra-risoluzione pari a ˆ Ksky = 5 per ognuna delle due coordinate, dividendo effettivamente la posizione di cielo lasca in 25 nuovi punti e selezionando tutti i punti ﬁni all’interno della griglia lasca . 2.3 Clustering e coincidenze I dati ottenuti da questa analisi vengono messi in matrici n × 9 che contengono ciascuna 10 Hz di dati; n è il numero di candidati per ogni matrice ed è di ordine 108. Poiché il numero di candidati scala come 1/f per via della correzione Doppler , ma aumenta come f 2 per via dell’aumento dei punti nel cielo, il totale dei candidati aumenta proporzionalmente alla frequenza. Le 9 componenti di questo vettore sono in ordine: la frequenza f , le coordinate eclittiche β e λ, lo spindown ˙ f , l’ampiezza di segnale rivelato A, il Critical Ratio CR, gli errori sulle coordinate eclittiche δβ e δλ, l’ampiezza di segnale emesso dall’onda gravitazionale h (aggiunta solo dopo la calibrazione). Si osserva innanzitutto che la probabilità che il j-esimo valore in frequenza dell’i-esima FFT abbia una statistica R compresa tra x ed x +dx si vede essere pari a e−x . È immediato, allora, rendersi conto che la probabilità che questo valore sia un massimo locale è pari alla probabilità che i due elementi primi vicini in frequenza siano minori, con probabilità 1−e−x ciascuno. La probabilità p0 che un picco venga scelto (e sia quindi sopra soglia) è 2.3 Clustering e coincidenze 37 p0 = Zinf Rthr e−x ¡ 1−e−x¢2 dx = e−Rthr −e−2Rthr + 1 3e−3Rthr. (2.14) Detto n il numero di conteggi nella mappa di Hough, scriviamo quindi il CR come CR = n −Np0 p Np0(1−p0) , (2.15) dove N è il numero delle FFT. Questo parametro, che dipende dalla soglia imposta su R, può essere utilizzato come mostrato in per valutare quale sia il valore di soglia migliore. Inoltre è un ottimo parametro da usare per valutare se un candidato sia associabile ad una sorgente o meno. Si vede che il CR così deﬁnito segue una distribuzione binomiale . I passi successivi sono molto importanti in quanto lo studio su cui verte la presente dissertazione propone miglioramenti ad una parte delle procedure usate ﬁnora per quanto riguarda il tempo di calcolo. Il concetto di base che guida i passaggi successivi è l’assunzione che una sorgente generi un candidato con alta statistica di rivelazione (alto CR quindi) su ogni rivelatore diverso (o su ogni run dello stesso rivelatore, come nel caso in esame), mentre mai (o quasi mai) questo succeda a causa del rumore. Di conseguenza, se una distanza nello spazio dei parametri (che dobbiamo deﬁnire) tra due candidati di run diversi con alta statistica di rivelazione è minore o uguale ad una certa soglia (tale da considerarsi quasi lo stesso punto), questi candidati si considerano essere generati dalla stessa sorgente e quindi avranno parametri molto vicini a quelli della sorgente stessa. I candidati associati a queste sorgenti verranno usati successivamente per analisi più approfondite. La procedura di ricerca di questo tipo di candidati tra quelli prodotti dalle diverse prese dati dei rivelatori è detta tecnica delle coincidenze. Prima di cercare le coincidenze, in ogni caso, è necessario raggruppare i candidati in cluster, in modo tale da ridurre il costo computazionale della procedura. Un cluster è una collezione di candidati tali che la distanza d nello spazio dei parametri a quattro dimensioni tra ogni coppia di punti di questa collezione sia minore di una soglia dclust. È quindi necessario, innanzi tutto, deﬁnire una sorta di metrica tra parametri che hanno diverse dimensioni ﬁsiche. Deﬁniamo quindi una differenza adimensionale tra due parametri dello stesso tipo: ∆(a)x ≡x2 −x1 δx con δx ≡δx1 +δx2 2 , (2.16) con x1 e x2 generiche coordinate (f , λ, β, ˙ f ) nello spazio dei parametri e δx1, δx2 i relativi passi11. A questo punto si può deﬁnire una distanza d d ≡ q¡ ∆(a) f ¢2 + ¡ ∆(a) ˙ f ¢2 + ¡ ∆(a)λ ¢2 + ¡ ∆(a)β ¢2 (2.17) ed una soglia dclust che è stata ﬁssata empiricamente a 2 . Dopo aver trovato i cluster per i due set di dati, le frequenze dei candidati vengono ricondotte ad uno stesso tempo per essere rese compatibili, sottraendo a ciascuna di esse lo spindown relativo moltiplicato 11Ricordiamo che per λ e β le risoluzioni δλ e δβ non sono costanti ma dipendono dalla frequenza e da β -come si evince dalle equazioni (2.11) 38 2. Procedure per la differenza tra i tempi di inizio di VSR4 e VSR2. Questo problema si pone perché VSR2 e VSR4 sono dati presi ad intervalli temporali molto lontani. Dopo di ciò, per ogni cluster del primo set di dati (ad esempio VSR2) viene effettuato un controllo con i cluster del secondo set (ad esempio VSR4) per vedere quali di questi si sovrappongano con il primo: se i parametri sono compatibili i cluster vengono considerati potenzialmente coincidenti. A questo punto, per ogni coppia di cluster potenzialmente coincidenti, ogni candidato del primo cluster viene confrontato con tutti i candidati del secondo cluster valutandone la distanza: se almeno una distanza è minore di dcoin = 2, i due cluster vengono considerati coincidenti. Ottenuti i cluster coincidenti si cerca la coppia dei candidati, presi ciascuno da uno dei cluster, la cui distanza tra loro sia la minima e quei candidati vengono considerati coincidenti. La scelta di dcoin è stata basata su uno studio di segnali software-injection in modo da ridurre la probabilità di falso allarme, ma allo stesso tempo da recuperare il numero più alto possibile di segnali . Una volta trovati i segnali si procede con una procedura di ranking per ridurre il numero di candidati in coincidenza e si conclude la procedura con il cosiddetto follow up. L’idea alla base di tale procedura è quella di ripetere l’analisi ﬁnora fatta aumentando la sensibilità. Questo viene fatto utilizzando i parametri ottenuti nell’analisi precedente per fare una ricerca più accurata solo nell’intorno del segnale ricostruito in una sorta di ricerca targeted. Maggiori dettagli sulla teoria delle coincidenze può essere trovata in appendice di . 39 Capitolo 3 Nuovi algoritmi per scremare grandi moli di candidati In questo capitolo si descriverà il lavoro fatto, riportando i risultati più importanti e presentando brevemente ulteriori percorsi di ricerca che sono emersi durante il lavoro e che saranno oggetto di studio in una prossima fase. Nel lavoro svolto si è andati ad analizzare la procedura di coincidenze descritta in sezione 2.3 al ﬁne di migliorarla riducendo l’insieme di potenziali candidati di onde continue ad un sottoinsieme più signiﬁcativo. Dovendo infatti il vecchio algoritmo confrontare grosse moli di dati, risultava piuttosto lento. Il problema fondamentale che si è riscontrato durante il presente studio è che tra i candidati ottenuti attraverso le procedure precedenti, molti sono dei falsi allarmi, ossia dovuti a disturbi strumentali non ancora rimossi. Inoltre non è sufﬁciente cercare i massimi relativi di CR all’interno dei dati, in quanto da un lato tale grandezza tende ad oscillare parecchio anche a causa di ﬂuttuazioni di rumore e dall’altro il massimo valore di CR, anche qualora sia effettivamente attribuibile ad un segnale continuo, non garantisce sempre di recuperare con esattezza i parametri che caratterizzano la relativa sorgente. La soluzione migliore è quella di cercare di identiﬁcare i candidati generati effetti-vamente da una data sorgente e discriminare così quelli dovuti a disturbi. Il metodo di porre una soglia ﬁssa sul CR in modo da eliminare i candidati con CR inferiore risulta inefﬁcace, in quanto il fondo di rumore in alcune frequenze supera di molto i candidati associabili ai segnali più deboli situati in altre frequenze. Punto chiave della soluzione, allora, è stato notare come i candidati verosimilmente generati da una stessa sorgente (e quindi con alta statistica di rivelazione) si raggruppino lungo curve molto precise (pattern) nello spazio quadridimensionale dei parametri (f , λ, β e ˙ f ). Identiﬁcare i candidati che giacciono su queste curve permetterebbe allora di ripu-lire efﬁcacemente i candidati e quindi di ricostruire molto più facilmente e/o velocemente i parametri della sorgente che li ha generati. Nonostante questo sia stato il risultato più importante dello studio, sono state svolte altre ricerche per proporre metodi alternativi da applicare sui dati scremati per recuperare correttamente i parametri di una certa sorgente. Evidenziamo comunque come la scrematura non levi i candidati più importanti, lascian-do la possibilità di utilizzare il vecchio metodo su questo insieme di candidati molto ridot-to e velocizzando notevolmente i tempi di calcolo. Nelle ﬁgure da 3.1 a 3.6 sono rafﬁgurati 40 3. Nuovi algoritmi per scremare grandi moli di candidati Figura 3.1. Pattern di candidati proiettati nel piano f /λ; la colour bar verticale indica il valore di CR associato ad ogni candidato. i candidati a frequenza compresa tra 68.885 e 69.0025 Hz corrispondenti alla software injection di parametri £ f = 69,01 Hz, λ = 239,97◦, β = −41,003◦, −8,279110−11 Hz s−1¤ . Nelle ﬁgure 3.1, 3.2, 3.3 e 3.4 sono molto visibili i pattern prodotti dai candidati generati da una sorgente; nella sezione 3.1 si spiegherà quali siano tali pattern e per quale motivo si formino. Nelle ﬁgure 3.5 e 3.6 non risulta evidente alcun pattern; come vedremo questo fatto non solo è facilmente spiegabile, ma è anche utile perché permetterà di studiare un pattern separatamente dall’altro (vedi sezione 3.2). 41 Figura 3.2. Pattern di candidati proiettati nel piano f /β; la colour bar verticale indica il valore di CR associato ad ogni candidato. Figura 3.3. Pattern di candidati proiettati nel piano f / ˙ f ; la colour bar verticale indica il valore di CR associato ad ogni candidato. 42 3. Nuovi algoritmi per scremare grandi moli di candidati Figura 3.4. Pattern di candidati proiettati nel piano λ/β; la colour bar verticale indica il valore di CR associato ad ogni candidato. Figura 3.5. Candidati proiettati nel piano λ/ ˙ f ; non è visibile alcun pattern. La colour bar verticale indica il valore di CR associato ad ogni candidato. 3.1 Dal segnale rivelato al pattern di candidati 43 Figura 3.6. Candidati proiettati nel piano β/ ˙ f ; non è visibile alcun pattern. La colour bar verticale indica il valore di CR associato ad ogni candidato. 3.1 Dal segnale rivelato al pattern di candidati Il passo fondamentale è stato quello di scoprire come e perché i candidati generati da una stessa sorgente esibiscono i pattern mostrati nelle ﬁgure 3.1, 3.2, 3.3 e 3.4. In altre parole cerchiamo di capire come un segnale, generato da una data sorgente con determinati parametri, generi un set diverso di candidati. Mettiamoci per semplicità nel caso di un rivelatore completamente privo di ogni tipo di rumore. Supponiamo di rivelare un unico segnale proveniente da una speciﬁca sorgente. Tale sorgente avrà parametri fsour, λsour, βsour e ˙ fsour = 0 (per semplicità). Nel giungere al rivelatore la frequenza del segnale sarà modiﬁcata per via dell’effetto Doppler (dovuto ai moti di rotazione e di rivoluzione della Terra), con una regola che è quella dell’equazione (2.9) con λS dipendente dal tempo. Chiamiamo questa nuova frequenza fD(t) = fD(λS(t)). Ricordiamo che la durata di ogni FFT è scelta in modo tale che la variazione di frequenza dovuta all’effetto Doppler non sia più grande di δf = 1 TFFT (vedi 2.2.2). Quindi, per ogni tempo di inizio associato ad una FFT, si ha una frequenza fD ricostruita. Una volta selezionati i picchi e costruita la peakmap (secondo quanto detto nella sezione 2.2.2), per ogni posizione di cielo considerata, il segnale viene corretto in frequen-za per rimuovere l’effetto Doppler che sarebbe indotto se il segnale provenisse da tale posizione. Considerando trascurabile la rotazione della Terra, troveremo la frequenza 44 3. Nuovi algoritmi per scremare grandi moli di candidati corretta attraverso l’equazione fcorr(ti,λj,βk) = fD 1 1+ ΩorbRorb c cos(βk)sin(λj −λ0) ∼fD µ 1−ΩorbRorb c cos(βk)sin(λj −λ0) ¶ , (3.1) con ti i-esimo tempo di FFT (con i = 1...N, essendo N numero totale delle FFT) e λj e βk rispettivamente j-esimo e k-esimo valore delle coordinate eclittiche (il cui numero massimo dipende dalla discretizzazione della griglia descritta in sezione 2.2.2); λ0 è una fase che dipende dall’istante di inizio e dall’istante di ﬁne della presa dati e che quindi va valutato per ogni diverso run. Si noti che l’ultimo termine della equazione (3.1) è un’espansione di Taylor al primo ordine che trascura i valori di ordine f 10−8, la qual cosa cambierà lievemente il valore della statistica ﬁnale senza alterare quanto è stato descritto. Ci si aspetta che se le coordinate eclittiche considerate sono quelle da cui proviene realmente il segnale (quindi λj ≃λsour e βj ≃βsour) le correzioni applicate agli elementi della peakmap selezionati, uno per ciascuno dei tempi di inizio delle FFT, corrispondano tutte alla stessa frequenza, cioè quella del segnale reale (in quanto abbiamo ipotizzato spindown nullo). Analogamente nelle porzioni di cielo immediatamente vicine a quella di provenienza del segnale, essendo la frequenza discreta, la correzione dei segnali fornirà il più delle volte la stessa frequenza. Viceversa, più lontane sono le coordinate considerate rispetto a quelle della sorgente, più la correzione di frequenza fornirà valori diversi ed apparentemente casuali. A questo punto interviene la mappa di Hough grazie alla quale si ottiene un conteggio di tutti i picchi nella stessa posizione di cielo con un certo spindown ˙ f e con una frequenza iniziale fD(ti) che evolve come fD(ti) = fD(t0)−˙ f (ti −t0). (3.2) In questo modo, se vado a studiare i candidati in base alla statistica di rivelazione, vedrò che i candidati con maggior CR saranno molto vicini e giaceranno lungo le curve di livello dell’equazione: f = f0 Ã 1+ \| ⃗ VT \| c cos(β)sin(λ−λ0) ! + ˙ f τ, (3.3) dove τ è un parametro di tempo che corrisponde alla durata temporale del run. 3.1.1 Identiﬁcazione dei pattern nel set di candidati ottenuti Sebbene in teoria il pattern descritto dall’equazione 3.3 sia facile da identiﬁcare, il procedimento non è semplice come può apparire. In assenza di rumore il punto con statistica di rivelazione più alta è quello con i parametri più vicini a quelli reali. Il rumore, tuttavia, inﬂuenza in maniera signiﬁcativa la statistica di rivelazione rendendo impossibile identiﬁcare i parametri del segnale solo basandosi sul massimo valore di CR; un suo alto valore resta comunque un ottimo modo per stimare i parametri del segnale. D’altro canto appare evidente che l’andamento del CR in funzione della variazione di frequenza (e quindi degli altri parametri) segue come una distribuzione binomiale . 3.2 Pulizia dei candidati 45 Una complicazione nell’utilizzo dei pattern per stimare i parametri di una sorgente corrisponde al fatto che questi giacciano su ipersuperﬁci tridimensionali dello spazio quadridimensionale dei parametri (infatti sono regolati dall’unica equazione (3.3)). An-che se è possibile fare alcune approssimazioni per studiare il punto di intersezione, ci sono troppi parametri liberi per poter ricostruire i segnali da semplici intersezioni. 3.2 Pulizia dei candidati Piuttosto che ricercare le sorgenti attraverso lo studio dei pattern, si è notato che l’e-quazione (3.3), opportunamente impiegata, permette di ridurre il numero di candidati in modo da applicare successivamente (in maniera più veloce) la tecnica delle coincidenze. Su questo si basa l’algoritmo sviluppato. Per capire l’idea di base, supponiamo di selezionare un punto nel quadrispazio dei parametri. Come è stato detto precedentemente, se esiste una sorgente con parametri abbastanza simili a quelli del punto selezionato, nell’intorno di quel punto i candidati ad alta statistica giaceranno sulla curva descritta dall’equazione (3.3). Se invece consideria-mo un punto con parametri molto diversi da qualsiasi sorgente (iniettata nei dati), quello che vedremo sarà solo rumore distribuito in maniera (almeno apparentemente) casuale. Possiamo allora valutare quale sia la statistica di rivelazione dei candidati distribuiti lungo la curva (in un intorno del punto selezionato) rispetto a quelli distribuiti lungo un’altra curva che intersechi la prima nel punto in esame (o meglio, in un intorno del punto stesso). Se riveleremo molti candidati con maggiore statistica di rivelazione sulla prima curva rispetto alla seconda, saremo in presenza di una sorgente. Viceversa, se i candidati con alta statistica sono ugualmente distribuiti lungo la prima e la seconda curva, saremo in un punto di solo rumore. Spieghiamo ora nel dettaglio come si è implementato quanto descritto. Innanzitutto è necessario ridurre i gradi di libertà del problema che altrimenti rimarrebbe computa-zionalmente costoso. Una ispezione dei dati (inizialmente visiva e poi seguita da ﬁt) ci ha permesso di veriﬁcare che i candidati sulla proiezione del quadrispazio f / ˙ f giacciono sempre lungo pattern a forma di rette abbastanza precise (avremo in realtà forme allarga-te dovute da un lato agli errori sperimentali e dall’altro alla proiezione in 2 dimensioni di uno spazio quadridimensionale). Il fatto che, come si può vedere anche dalle ﬁgure 3.5 e 3.6, lo spindown non sia direttamente correlato con le posizioni nel cielo garantisce che si possano analizzare separatamente (con i dovuti accorgimenti) il pattern dovuto all’effetto Doppler e quello dovuto allo spindown; infatti per due parametri scorrelati la variazione di uno non incide sul pattern dell’altro. D’altronde, osservando invece i pattern indotti da λ e β sulla frequenza (ﬁgure 3.1 e 3.2) si può notare che il pattern, per quanto preciso, è inﬂuenzato da un parametro nascosto (ovvero l’altra coordinata celeste). Il pattern indotto dallo spindown, invece, essendo strettamente bidimensionale, si può analizzare con facilità. Per quanto detto sopra, applichiamo l’algoritmo sopra descritto usando un pattern ridotto rispetto a quello quadridimensionale. Tale pattern sarà del tipo f = f0 + ˙ f τ, (3.4) con τ parametro con le dimensioni di un tempo. 46 3. Nuovi algoritmi per scremare grandi moli di candidati (a) Disposizione dei candidati corrispon-denti ad una sorgente nel piano fre-quenza/spindown prima e dopo aver applicato l’algoritmo di pulizia qui descritto (b) Disposizione dei candidati di puro rumo-re nel piano frequenza/spindown pri-ma e dopo aver applicato l’algoritmo di pulizia qui descritto. Figura 3.7. Pattern ed efﬁcacia dell’algoritmo di pulizia. 3.2 Pulizia dei candidati 47 Figura 3.8. Esempio di selezione delle rette da valutare. Le dimensioni non sono quelle reali, ma l’immagine è a titolo esempliﬁcativo. L’idea sopra citata si implementa quindi secondo quanto segue: si valutano dapprima i candidati all’interno di un’area rettangolare con i lati lunghi paralleli alla retta descritta dall’equazione (3.4) e centrata sul punto scelto. Si confrontano quindi con quelli che giacciono in un rettangolo con i lati lunghi ortogonali alla primo, sempre centrato sul punto scelto. Si dovrebbe quindi valutare il coefﬁciente angolare della retta descritta dall’equazione (3.4), ma è abbastanza evidente che non sia signiﬁcativo: sia il passo, sia i possibili valori di frequenza e spindown sono molto diversi. Si valuta, invece, quale porzione di spazio frequenza e spindown attraversino: il secondo tende ad attraversare tutti i valori possibili, mentre la frequenza varia di un inﬁnitesimo rispetto al range di valori che può assumere (come si può vedere bene anche in ﬁgura 3.7). Questo ci spinge a riscalare le grandezze in modo tale che, con le grandezze riscalate, la retta risulti avente un coefﬁciente angolare τ quasi nullo, che esprima cioè variazione piccola o nulla di frequenza per una grande variazione di spindown. Il problema della scala sarà affrontato alla ﬁne di questo paragrafo, in quanto la decisione deve essere basata sul tipo di algoritmo che vogliamo implementare. Ricapitolando: abbiamo rettangoli con i lati lunghi a frequenza costante (che chia-meremo verticali), nei quali troviamo tutti i candidati ad alta statistica nel caso in cui questi corrispondano a segnali, e rettangoli con i lati lunghi a spindown costante (che chiameremo orizzontali) che ci servono come controprova. Scegliamo un valore nel nostro spazio riscalato, ovvero una cella corrispondente a uno qualunque dei valori discreti di frequenza f j e spindown e ˙ fk (in ﬁgura 3.8 rappresentata dalla cella azzurra). Prendiamo un rettangolo verticale (in ﬁgura 3.8 in giallo) centrato sulla retta di equazione f = f j ed un rettangolo orizzontale (in ﬁgura 3.8 in rosso) centrato sulla retta di equazione ˙ f = ˙ fk che contengano il punto scelto. In ciascuno di questi 48 3. Nuovi algoritmi per scremare grandi moli di candidati rettangoli valutiamo la media dei CR dei candidati al loro interno, che chiamiamo CRV (per la verticale) e CRO (per l’orizzontale). Vogliamo fare in modo che quanto più CRV è alto rispetto a CRO, tanti più candidati nella cella azzurra si vogliono tenere. Viceversa, se CRV ≲CRO, si ipotizza che i candidati presenti siano di solo rumore e quindi si vogliono eliminare completamente. Deﬁniamo quindi una soglia sul CR CRthr = m CRO CRV , (3.5) dove m è il massimo valore di CR nella cella selezionata. Scarteremo quindi tutti i candidati nella cella selezionata che abbiano CR minore della soglia e terremo gli altri. In questa maniera si tende ad eliminare tutti (o quasi tutti) i candidati nel caso in cui CRm O ≳ CRm V , vale a dire quando non c’è una prevalenza di candidati con CR maggiore lungo la verticale. Nel caso in cui, invece, CRm V ≫CRm O , vale a dire quando consideriamo di avere trovato i candidati generati da una sorgente, si terranno solo quelli più signiﬁcativi. Questo procedimento viene iterato per tutti i valori di j e k dettati dalla discretizza-zione dello spazio f / ˙ f . Questo metodo tende a ridurre notevolmente il numero di candidati in esame, pas-sando da circa 108 (rispettivamente 187855700 e 187117000 candidati per VSR2 e VSR4) a circa 105 (rispettivamente 753109 e 564131 candidati), con una percentuale di dati esclusi del 99,65%. Nelle ﬁgure 3.7 si vede l’effetto della scrematura su un set di candidati associabili ad una sorgente e su un altro contenente candidati associabili a solo rumore. 3.2.1 Dimensione delle celle da analizzare Ci poniamo ora il problema della scelta della dimensione delle celle da usare per l’analisi descritta. L’idea di utilizzare la discretizzazione scelta nel capitolo 2 funziona solo per lo spindown, mentre tale scelta non si può applicare per la frequenza: è necessario infatti che le celle siano abbastanza larghe da contenere tutti i candidati generati da un segnale, qualora questo sia presente. Empiricamente si vede che la scelta migliore è quella di usare come larghezza me-tà della variazione Doppler massima. Infatti i candidati con maggior CR sono situati nello spazio in coordinate eclittiche con posizione molto vicina a quella vera, inoltre la variazione di frequenza dovuta allo spindown è tale da far sì che la larghezza scelta sia appropriata. Tale larghezza risulta essere pari a ∆f = f j10−4, con f j frequenza in esame (centro della cella azzurra in ﬁgura 3.8). Ricordando che i dati sono raccolti in matrici cia-scuna da 10 Hz, come detto in sezione 2.3, per semplicità di applicazione dell’algoritmo, la discretizzazione per ogni matrice è calcolata attraverso la formula ∆f = fini10−4, dove fini è la frequenza minima dell’intervallo, vale a dire a passo costante per ogni intervallo di 10 Hz. È poi necessario scegliere la lunghezza dei rettangoli orizzontale e verticale (rispetti-vamente rosso e giallo in ﬁgura 3.8). Visto che si vuole ridurre il più possibile numero di candidati, si pone un segmento sufﬁcientemente piccolo da non venire inﬂuenzato da picchi "lontani", ma sufﬁcientemente grande per fare statistica. Si è scelto di considerare per entrambi i rettangoli 11 celle centrate su quella in analisi (vale a dire 5 celle da un lato ed altrettante dall’altro, come mostrato in ﬁgura 3.9a). 3.2 Pulizia dei candidati 49 Questa suddivisione, da sola, non è sufﬁciente. Si pone infatti il rischio che i candidati con alto CR, capitando sul bordo di una cella, non siano completamente inclusi in un rettangolo verticale pregiudicando così il funzionamento dell’algoritmo; in questi casi (come si può osservare in ﬁgura 3.9a) il rettangolo verticale, pur spostandosi, conterrà ogni volta solo una parte della banda ad alto CR. Per eliminare questo problema di bordo, si deve evitare di muoversi lungo l’asse della frequenza per passi interi; avendo deﬁni-to correttamente la dimensione dei rettangoli e mantenendola comunque come sopra descritto, muoversi per passi frazionati permetterà di trovare sempre una condizione in cui la sorgente sia interamente contenuta nell’area di analisi. Il passo verrà ridotto di un fattore 3 per una questione di simmetria, triplicando di conseguenza il numero j dei passi in frequenza; questo ha a tutti gli effetti creato delle sotto-celle con larghezza in frequenza ∆f /3 e dimensione in spindown invariata. In ognuno dei passi della screma-tura dei candidati, completata l’analisi e la determinazione della soglia come descritto dall’equazione (3.5), anziché selezionare i candidati nell’intera cella di analisi, lo si farà solo all’interno della sua sotto-cella centrale (vedi ﬁgure 3.9b). 3.2.2 Margini di miglioramento dell’algoritmo L’algoritmo appena descritto, sebbene si sia mostrato molto efﬁcace, ha anche fatto trasparire un limite in alcuni casi particolari. Quando una serie di candidati di rumore con alta statistica di rivelazione sono abbastanza vicini a quelli prodotti da una sorgente c’è il rischio che non vengano esclusi. Questo avviene perché il segmento verticale in analisi include nel suo bordo alcuni candidati relativi alla sorgente che, aumentando la media verticale rispetto a quella orizzontale, diminuiscono il valore di soglia. Sebbene non sia un limite per l’efﬁcacia dell’algoritmo, riuscire a risolvere questo problema potrebbe ridurre ulteriormente il tempo di calcolo e le probabilità di falso allarme delle analisi successive. Si investigheranno quindi metodi per mitigare questo effetto. Un altro limite dell’algoritmo presentato è dovuto ad un effetto di bordo. Poiché i dati sono stati immagazzinati in matrici contenenti 10 Hz ciascuna, se si fa l’analisi solo sulle singole matrici, i candidati più al bordo avranno una media fatta su un rettangolo orizzontale più corto rispetto agli altri. Un metodo per risolvere questo problema è quello di aggiungere ad ogni matrice un certo numero di candidati della matrice precedente e di quella successiva. Un modo molto conservativo per aumentare le dimensioni di queste matrici eliminando il problema è quello di aggiungere a destra ed a sinistra dati inclusi nella lunghezza di un rettangolo orizzontale. Ad esempio: alla matrice 60−70 Hz aggiungeremo ∆f 11 = 0,03 Hz a sinistra di 60 Hz ed a destra di 70 Hz. Chiaramente questo non risolve il problema nei bordi in frequenza del set di dati. 50 3. Nuovi algoritmi per scremare grandi moli di candidati (a) Senza sovrarisoluzione: la cella su cui si applica la selezione è quella centrale. Si può osservare come i candidati ad alto CR, capitando sul bordo dell’intervallo di analizzare, non diano contributo massimo sulla media e, quindi, sulla selezione dei candidati della retta stessa. (b) Con sovrarisoluzione: la cella su cui si applica la selezione è quella centrale. Si può osservare come quando si applica selezione sulla retta, tutta la retta venga inclusa nella media, risolvendo il problema di bordo. Figura 3.9. Scelta delle celle per l’algoritmo di selezione. 3.3 Clustering dei candidati 51 3.3 Clustering dei candidati Il clustering che viene eseguito per lo studio successivo dei candidati è molto diverso da quello deﬁnito in precedenza per la pipeline (vedi sezione 2.3). Si cerca infatti di raggruppare tutti i candidati compatibili con una singola sorgente, piuttosto che raggrup-pare quelli molto vicini. In questo senso, la strategia usata (per ognuno dei due set di dati) è la seguente: • Si seleziona il candidato con massimo CR tra quelli del set di dati in studio. • Si seleziona un intorno in frequenza di tale candidato che contenga tutti i candidati compatibili con una sorgente in quel punto. Il semi-intervallo scelto è quello della correzione Doppler ∆f = f010−4, con f0 frequenza del punto di massimo. • Si genera un cluster che raccoglie i candidati così selezionati, che vengono quindi rimossi dal set dei candidati in esame. • Si itera il procedimento sul set dei candidati ﬁno a che sono stati tutti classiﬁcati nei cluster. Questa procedura raggruppa i candidati ottenuti dalla pulizia in 12327 cluster per il VSR2 e 11476 per il VSR4; di tale cluster vengono immediatamente eliminati i cluster che non contengano almeno 3 candidati (ritenendoli non interessanti per lo studio). Resteranno quindi il 95% dei cluster di partenza. Si procede inﬁne con un’ulteriore scrematura dei cluster. L’idea è che i cluster che contengono candidati compatibili con una certa sorgente devono avere il CR massimo più alto dei cluster contenenti residui di rumori. Pertanto, si ordinano i cluster in frequenza e, per ognuno di essi, si seleziona un set di cluster contenente esso stesso ed i 4 intorno a lui (2 prima e 2 dopo). Il cluster sarà scartato qualora il massimo valore di CR qui trovato sia minore o uguale della mediana dei massimi di CR dei cluster nel set. Questa procedura riduce il numero di cluster del 60 %. 3.4 Analisi dei Cluster e stima dei parametri delle sorgenti recu-perate A questo punto i cluster che abbiamo ottenuto verranno utilizzati per studiare i pa-rametri delle sorgenti. Ricordiamo che prima di effettuare questa procedura di stima, poiché stiamo lavorando con prese dati molto distanti nel tempo, è necessario ricorreg-gere la frequenza di ciascun candidato di VSR2 aggiungendo alla sua frequenza quella variazione indotta dal suo spindown; questa variazione sarà ˙ f ∆tVSR4-VSR2. Vogliamo quindi veriﬁcare quali cluster nei due set di dati siano compatibili tra loro: deﬁniamo allora una distanza minima in frequenza dfr per cui due cluster si considerino in coincidenza. Perché due cluster siano in coincidenza, dovranno sovrapporsi a meno di un fattore dfr. Pertanto si impone la condizione secondo cui il massimo dei minimi di frequenza dei due cluster (sottratto dell’eventuale dfr) sia minore del minimo dei massimi di frequenza dei due cluster. Una volta ottenuta la matrice delle possibili coincidenze tra i cluster di VSR2 e VSR4, scegliamo quali di queste coppie siano effettivamente compatibili con potenziali sorgenti. 52 3. Nuovi algoritmi per scremare grandi moli di candidati Ci si aspetta infatti che ogni cluster sia generato al più da una sorgente, quindi ogni cluster deve avere al più un "compagno". Per ogni cluster di ogni run, allora, scegliamo tra le coincidenze possibili quella il cui compagno abbia CR maggiore. Questo procedimento riduce il numero di coincidenze da 7363 a 2777. Deﬁnita allora la distanza descritta dall’equazione (2.17), per ogni candidato del primo cluster della coppia, si valuta la distanza con ogni candidato della seconda coppia. La minore di queste distanze ci indicherà quali candidati siano in coincidenza. Consi-derando che un alto CR indica un’alta probabilità che i parametri siano quelli corretti, ricaviamo i parametri della sorgente calcolandoli mediante la media dei parametri dei candidati in coincidenza pesata mediante il loro CR; detto xi generico parametro dell’i-esimo candidato coincidente (con i = 1,2) e CRi il rispettivo Critical Ratio, il parametro ricavato sarà: xdet = x1CR1 + x2CR2 CR1 +CR2 . (3.6) A questo punto è utile deﬁnire uno o più parametri stimatori (una sorta di pesi) che valutino quanto sia verosimile che una sorgente trovata non sia un artefatto di qualche ge-nere. Per valutare questo peso dobbiamo però fare una considerazione. La distribuzione attesa del CR sull’ipersuperﬁcie del quadrispazio dei parametri deve necessariamente es-sere una Bernoulliana. Questo deriva da come è stato ricavato il CR. Ci aspettiamo quindi che un segnale proveniente da una sorgente produrrà una sorta di campana con un picco più o meno alto mentre un rumore residuo dovrebbe essere distribuito in maniera più casuale. Risulta pertanto ragionevole creare un peso stimatore con i parametri della curva (altezza e varianza). Il problema è che le curve con pochi candidati vengono di solito valutate meglio delle curve con più candidati, cosa che pregiudica la bontà dello stimatore, quindi è necessario inserire nello stimatore anche il numero di candidati. Si è visto empiricamente che dei buoni stimatori sono: P1 = −σ2 1σ2 2K1K2 (3.7) P2 = P1 M1M2 , (3.8) con σ2 l scarto quadratico medio, Kl numero di candidati e Ml massimo CR del l-esimo cluster. Ottenuti questi due stimatori facciamo un’ulteriore scrematura. Una supposizione che potremmo fare, ad esempio, è che non ci aspettiamo di trovare due sorgenti con una frequenza uguale a meno di un fattore minore di 0,05 Hz1; per questo, trovata una sorgente, escludiamo tutte le le altre che abbiano frequenza entro 0,05 Hz da quella trovata e che abbiano P2 minore di quello della sorgente selezionata. Inﬁne mettiamo una soglia sui pesi ed eliminiamo le sorgenti con peso inferiore. Si veriﬁca in maniera empirica che delle soglie ragionevoli sono 108 e 106 rispettivamente su P1 e P2. 1Questa scelta è stata fatta supponendo che non ci sia mai più di una sorgente per ogni intervallo di 0,1 Hz; sebbene per il nostro caso questa scelta sia valida (data la conoscenza a priori delle sorgenti iniettate), nel caso generale invece, per deﬁnire tale soglia è opportuno valutare la sensibilità dello strumento speciﬁco in uso per la rivelazione. 3.5 Risultati 53 3.5 Risultati Un primo risultato importante è che, durante il primo algoritmo di pulizia descritto in sezione 3.2, è stata preservata la totalità dei candidati associabili alle sorgenti. Per veriﬁcarlo è stata effettuata un’analisi dei canditati ad alto CR residui dopo l’algoritmo di pulizia confrontandoli con i parametri delle sorgenti iniettate e cercandone le compati-bilità. Questa analisi è stata utile per veriﬁcare che il metodo di rimozione del rumore non rimuovesse anche candidati signiﬁcativi. Questo risultato è già importante di per sé perché dà un’indicazione del fatto che l’algoritmo lavora nella direzione giusta. Infatti, anche indipendentemente dai risultati della seconda parte dell’algoritmo (clustering e recupero dei parametri), riuscire a eseguire una prima scrematura così signiﬁcativa dei dati permetterà di rendere molto più veloce l’algoritmo di clustering e coincidenze già in uso dal gruppo. La seconda fase dell’algoritmo descritto in questa dissertazione è quella di clustering dei candidati e recupero dei parametri. Ricordiamo che è necessario che l’algoritmo trovi con alta precisione la frequenza e la posizione nel cielo, in modo da riuscire ad individuare la sorgente con il successivo follow up che utilizzerà in modo più importante le coordinate eclittiche. Dalla tabella in appendice B.2 si può subito notare che i dati recuperati con frequenza compatibile (entro 0,015 Hz) con quella delle sorgenti sono 226 contro 272, vale a dire l’83%. Una parte di questi, tuttavia, nell’analisi dei parametri ha rivelato una posizione nel cielo non coerente con quella della sorgente, mentre 161 sono state individuate con buona precisione. Abbiamo ottenuto quindi in maniera piuttosto precisa che il 60% delle sorgenti sono state comunque recuperate. Notare che in questa analisi si sono usate solamente quelle sorgenti generalmente considerate rivelabili, ovvero quelle con SNR pari o superiore a 3. Per fare un paragone, sugli stessi dati, neppure per l’algoritmo originale è stato possi-bile individuare alcune sorgenti (32 delle 272), che sono poi state recuperate utilizzando una variante dello stesso decisamente più pesante a livello computazionale. Il vantaggio notevole dell’algoritmo qui presentato è che i risultati sono stati ottenuti in 4 ore e mezza (15973 secondi) di elaborazione rispetto ai tempi dell’algoritmo prece-dente che ha impiegato più di 5 giorni. Pertanto si è ottenuta una riduzione nel tempo di calcolo del 96%. L’elaborazione di entrambi gli algoritmi è avvenuta sulla medesima macchina con le stesse condizioni di CPU. 55 Capitolo 4 Conclusioni Come si è visto in sezione 3.5, la parte di scrematura dei candidati non solo si è mostra-ta molto efﬁciente (eliminando il 99,65% dei candidati senza rimuovere alcuna sorgente), ma anche molto più veloce del clustering proposto in precedenza. Implementare quindi questa tecnica a monte del clustering potrebbe portare ad un deciso risparmio di tempo oltre che, probabilmente, ad un rischio molto più basso di falsi allarmi. Oltre a questi risultati ottenuti, l’algoritmo presenta comunque anche possibilità di evoluzione e di miglioramento, su cui è però necessaria un’indagine e una valutazione più approfondita. La ricerca per la scrematura, infatti, si è svolta su un pattern ridotto (proiettato cioè da uno spazio quadridimensionale su uno bidimensionale) e basato sullo spindown. Questo è stato possibile perché l’effetto Doppler e lo spindown che creano il pattern sono scorrelati. Può valere la pena, in un eventuale studio successivo, di cercare il modo di applicare una seconda soglia per la selezione dei candidati analoga a quella qui descritta, ma per il pattern legato all’effetto Doppler. Questo potrebbe ridurre ulteriormente il numero di candidati. Per poter implementare, però, questa nuova soglia è necessario studiare il parametro λ0 che compare nell’equazione 3.1 e che dovrebbe essere un termine dipendente dal tempo di inizio e dal tempo di ﬁne della presa dati. L’analisi effettuata con il presente studio si è focalizzata sul pattern dovuto allo spindown in quanto appariva il più efﬁciente ed immediato (la valutazione è stata fatta sullo studio empirico dei pattern). Per quanto riguarda la procedura di clustering e recupero di parametri, il discorso è più ampio. Si è voluta cercare una strada alternativa a quella già esistente che, per recuperare tutte le sorgenti, richiedeva un costo computazionale estremamente elevato. Il fatto che l’algoritmo proposto abbia recuperato il 60 % dei segnali con buona precisione in un tempo signiﬁcativamente inferiore, è un’indicazione che la direzione intrapresa è valida e promettente. Ricordiamo che l’utilizzo dell’algoritmo qui presentato ha permesso ridurre i tempi di calcolo del 96%rispetto all’impiego dell’algoritmo originale . Il numero di candidati persi non è tuttavia irrilevante, indicando quindi che bisogna approfondire ulteriormente questo approccio per poter ottenere un risultato più vicino a quanto sperato. Gli elementi su cui lavorare sono legati al fatto che presumibilmente esiste ancora una grande quantità di rumore presente dopo la scrematura. Si ritiene che si sarebbero potuti ottenere risultati migliori dopo una pulizia ancora più efﬁcace di quella ottenuta con il nuovo algoritmo. Infatti è possibile che, per esempio, per cercare la minima distanza si debba prima ridurre il numero dei candidati al solo picco di CR. 56 4. Conclusioni Per questo motivo, dopo la costruzione dei cluster e la ricerca di quelli in coincidenza, vale la pena studiare come scremare candidati nei cluster coincidenti per poi valutare la distanza tra i punti come descritto in sezione 3.4. Il risultato nel complesso è comunque positivo perché mette a disposizione un si-stema che è in grado di discernere una quota signiﬁcativa dei segnali, permettendo di compiere successivamente analisi più approfondite su un sottoinsieme ridotto delle frequenze, riducendone quindi drasticamente l’impatto computazionale. 57 Appendice A Script A.1 Pulizia dei dati function cleaned=data_cleaning(cand,sddnat,varargin) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Questa funzione si occupa di studiare le matrici dei candidati % (studiata su matrici del tipo VSR2_1_060070_ref) al fine di % rimuovere molti dei candidati non associabili a sorgenti. Al % momento è stato testato sui VSR2 e VSR4 con le sorgenti % iniettate da Sabrina. % %%%%%%%%ARGOMENTI IN INPUT%%%%%%%%%% % %%%%%%%% OBBLIGATORI %%%%%%%%%%%%%%% % % La funzione necessita di una matrice cand(Nx6) di N candidati % del tipo VSR#1######_ref composta da (freq, #, #, sd, #, CR) % gli altri parametri della matrice non sono utilizzati (i.e. % potrebbe essere maggiore di 6). % % La funzione necessita anche dello step minimo in spindown, % sddnat. % % % % % Gli argomenti opzionali sono sempre in coppie del tipo: % ..., ’tipo di argomento’, valore, ... % % Gli argomenti opzionali sono: 58 A. Script % ’save’,filename-> salva i risultati su file; % filename e’ il nome del file.mat incluso di destinazione. % ’dstep’, valore-> larghezza dello step lasco in frequenza. % Di standard è fr/10^4. (Vedi documentazione). %’sdlenght’, valore -> lunghezza del rettangolo a frequenza % costante. %’frlenght’, valore -> lunghezza del rettangolo a spindown % costante. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Di Giuseppe Intini - intinigiuseppe@gmail.com -Versione Maggio 2016 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %verifica l’esistenza degli argomenti opzionali e li elabora. %Inizializza gli argomenti iniziali. if ~exist(’varargin’,’var’) varargin=[]; end if mod(length(varargin),2)==1 display(’Attenzione, vargin dispari!!! \n La seguente variabile non verrà... ...considerata:’) display(varargin{length(varargin)}) end fr=min(cand(:,1)); frout=max(cand(:,1)); span=frout-frin; saving=0; fsteps=floor(3span10^4/fr); sdlenght=5; frlenght=5; A.1 Pulizia dei dati 59 V=floor(length(varargin)/2); for i=1:V selected=varargin{2i}; switch varargin{2i-1} case ’save’ saving=1; filename=selected; case ’fsteps’ fstep=floor(3span/selected); case ’sdlenght’ sdlenght=floor(selected); case ’frlenght’ frlenght=floor(selected3); otherwise display(’Unknown varargin argument:’) display(varargin{2i-1}) display(’Argument excluded’) end end % Inizializza le funzioni di calcolo delle posizioni nella % griglia. Alcune sono inutili nella funzione di per sé ma % sono controlli in caso si 60 A. Script %decida di intervenire a livello debug. % Per la frequenza info(1)=length(cand); fr_sd.fr=@(index)[fr+(index-1)span/fsteps, fr+indexspan/fsteps]; fr_sd.frin=@(freq)max(floor((freq-fr).fsteps/(span)(1+10^-12))+1,1); % Per lo spindown s=floor(min(cand(:,4))/sddnat)sddnat; fr_sd.sd=@(index) [s+(index-1)sddnat, s+indexsddnat]; fr_sd.sdin=@(sd) floor((sd-s)(1+10^-24)/sddnat)+1; Sstep=fr_sd.sdin(max(cand(:,4))); %numero di passi in spindown %Inizializza la griglia. for i=1:(fsteps+1) for j=1:Sstep fr_sd.table(i,j).cand=[]; % "Casella" i-esima in frequenza % e j-esima in spindown end end N=length(cand); A.1 Pulizia dei dati 61 % Organizza i candidati in tabelle. Questo passaggio è un po’ pesante % perché allarga le matrici. for n=1:N i=fr_sd.frin(cand(n,1)); j=fr_sd.sdin(cand(n,4)); fr_sd.table(i,j).cand=[fr_sd.table(i,j).cand; cand(n,:)]; end cleaned=[]; F=floor((frlenght-1)/2); S=floor((sdlenght-1)/2); % Applicazione del metodo vero e proprio. NB: per minimizzare la % complessità c’è un unico ciclo che fa tutte le caselle in spindown % per poi avanzare di 1 in frequenza e ricominciare. for n=1:((fsteps+1)Sstep) i=1+floor((n-1)/Sstep); %calcolo il numero della cella in frequenza. j=mod(n-1,Sstep)+1; %calcolo il numero della cella in spindown. t=fr_sd.table(i,j).cand; if isempty(t) continue end cellf=max(i+(-1:+1),1); %Larghezza cella in frequenza. La cella di % partenza è sovrarisoluta. cellf=min(cellf,(fsteps+1)); %problemi di bordo cellf=unique(cellf); %problemi di bordo Tf=max(i+(-F:+F),1); %Lunghezza del rettangolo a spindown costante Tf=min(Tf,(fsteps+1)); Tf=unique(Tf); Tsd=max(j+(-S:+S),1); %Lunghezza del rettangolo a frequenza costante Tsd=min(Tsd,Sstep); Tsd=unique(Tsd); %Media del rettangolo a spindown costante c=[]; for l=Tf if ~isempty(fr_sd.table(l,j).cand) c=[c;fr_sd.table(l,j).cand(:,6)]; end end Mo=mean(c); 62 A. Script %Media del rettangolo a frequenza costante c=[]; for l=cellf for h=Tsd if ~isempty(fr_sd.table(l,h).cand) c=[c;fr_sd.table(l,h).cand(:,6)]; end end end Mv=mean(c); m=max(t(:,6)); ii=find(t(:,6)>(mMo/Mv)); cleaned=[cleaned;t(ii,:)]; end if saving==1 save(filename,’cleaned’) end end A.2 Creazione dei cluster function clust=clustercreation(cand) % Questa funzione si occupa di raggruppare i candidati formando % cluster in frequenza. % L’algoritmo cerca un picco, ne sceglie un intorno e ne crea un % cluster. % %%%%%%%%%%%%%% ARGOMENTI IN INPUT %%%%%%%%%%%%% % %%%%%%%%%%%%%%%%%% OBBLIGATORI %%%%%%%%%%%%%%%%%%%% % % La funzione necessita di una matrice cand(Nx6) di N candidati del tipo VSR#1###### % gli altri parametri della matrice non sono utilizzati (i.e. potrebbe essere maggiore % % % % % Gli argomenti opzionali sono sempre in coppie del tipo: ...,’tipo di argomento’, val % % Gli argomenti opzionali sono: A.2 Creazione dei cluster 63 % ’selection’, ’fr’ % -> impone una soglia sulla frequenza intorno al massimo di CR. % ’selection’, ’CR’ % -> impone una soglia sul CR rispetto al picco. % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Di Giuseppe Intini - intinigiuseppe@gmail.com - Versione Maggio 2016 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %verifica l’esistenza degli argomenti opzionali e li elabora. %Inizializza gli argomenti iniziali. if ~exist(’varargin’,’var’) varargin=[]; end if mod(length(varargin),2)==1 display(’Attenzione, vargin dispari!!! \n... ... La seguente variabile non verrà considerata:’) display(varargin{length(varargin)}) end soglia=0; V=floor(length(varargin)/2); for i=1:V selected=varargin{2i}; switch varargin{2i-1} case ’selection’ if selected==’fr’ 64 A. Script soglia=1; elseif selected==’CR’ soglia==2; end otherwise display(’Unknown varargin argument:’) display(varargin{2i-1}) display(’Argument excluded’) end end exam=cand; clust=[]; while ~isempty(exam) [~,ind]=max(exam(:,6)); f0=exam(ind(1),1); kk=find(abs(exam(:,1)-exam(ind(1),1))<f010^-4); if soglia==0; ii=kk; elseif soglia==1 ii=find(abs(exam(:,1)-exam(ind(1),1))exam(ind(1),6)0.36); A.3 Recupero dei parametri delle sorgenti 65 end clust(length(clust)+1).cand=exam(ii,:); clust(length(clust)).fr=exam(ind(1),1); clust(length(clust)).lam=exam(ind(1),2); clust(length(clust)).bet=exam(ind(1),3); clust(length(clust)).sd=exam(ind(1),4); exam(kk,:)=[]; end end A.3 Recupero dei parametri delle sorgenti function [R,info]=source_recovery(fr,varargin) if ~exist(’varargin’,’var’) varargin=[]; end if mod(length(varargin),2)==1 display(’Attenzione, vargin dispari!!! \n... ...La seguente variabile non verrà considerata:’) display(varargin{length(varargin)}) end folder=[]; 66 A. Script saving=0; dist=0.01; V=floor(length(varargin)/2); for i=1:V selected=varargin{2i}; switch varargin{2i-1} case ’folder’ folder=selected; if folder(length(folder))~=’/’ folder(length(folder)+1)=’/’; end case ’save’ saving=1; filename=selected; case ’dist’ dist=selected; otherwise display(’Unknown varargin argument:’) display(varargin{2i-1}) display(’Argument excluded’) end end A.3 Recupero dei parametri delle sorgenti 67 frdnat(2)=1.2207e-04; frdnat(4)=1.2207e-04; sddnat(2)=7.6270e-12; sddnat(4)=1.5016e-11; for RUN=2:2:4 string=sprintf(’%sVSR%d_1_%03d%03d_ref’,folder,RUN,fr,frout); load(string) string=sprintf(’VSR%d_1_%03d%03d_ref.cand;’,RUN,fr,frout); eval([’cand=’ string]) info(RUN/2).cand(1)=length(cand); cleaned=data_cleaning(cand,sddnat(RUN)) eval(sprintf(’cand%d=cand;’,RUN)) info(RUN/2).cand(2)=length(cleaned); end run2=cand2; run4=cand4; 68 A. Script clust=clustercreation(run2); info(1).clust(2)=length(clust); clust=clustercreation(run4); info(2).clust(2)=length(clust2); ii=find(abs([clust(:).n])<3); clust(ii)=[]; ii=find(abs([clust2(:).n])<3); clust2(ii)=[]; info(1).clust(3)=length(clust); info(2).clust(3)=length(clust); info(1).cand(3)=sum([clust(:).n]); info(2).cand(3)=sum([clust2(:).n]); %Procedura di pulizia dei cluster. [~,s]=sort([clust(:).fr]); c=clust(s); [~,s]=sort([clust2(:).fr]); c2=clust2(s); clust=[]; A.3 Recupero dei parametri delle sorgenti 69 clust2=[]; for i=1:length(c) z=i+(-2:2); z=min(z,length(c)); z=max(z,1); z=unique(z); if c(i).M>median([c(z).M]) clust=[clust,c(i)]; end end for i=1:length(c2) z=i+(-2:2); z=min(z,length(c2)); z=max(z,1); z=unique(z); if c2(i).M>median([c2(z).M]) clust2=[clust2,c2(i)]; end end info(1).clust(4)=length(clust); info(2).clust(4)=length(clust); info(1).cand(4)=sum([clust(:).n]); 70 A. Script info(2).cand(4)=sum([clust2(:).n]); match=[]; for j=1:length(clust2) for i=1:length(clust) M=min(clust(i).frmax, clust2(j).frmax); m=max(clust(i).frmin, clust2(j).frmin); if M>m-dist match=[match; i,j,clust(i).M,clust2(j).M]; end end %j100/length(clust2) end info(3).match(1)=length(match); old=match; match=[]; for j=1:length(clust) ii=find(old(:,1)==j); if isempty(ii) continue end [~,k]=max(old(ii,4)); match=[match; old(ii(k),:)]; A.3 Recupero dei parametri delle sorgenti 71 end old=match; match=[]; for j=1:length(clust2) ii=find(old(:,2)==j); if isempty(ii) continue end [~,k]=max(old(ii,3)); match=[match; old(ii(k),:)]; end info(3).match(2)=length(match); for i=1:length(clust) clust(i).normcand=clust(i).cand; clust(i).normcand(:,1)=clust(i).normcand(:,1)/frdnat(RUN); clust(i).normcand(:,2)=... ... clust(i).normcand(:,2)./clust(i).normcand(:,7); clust(i).normcand(:,3)=... ...clust(i).normcand(:,3)./clust(i).normcand(:,8); clust(i).normcand(:,4)=clust(i).normcand(:,4)/sddnat(RUN); end for i=1:length(clust2) 72 A. Script clust2(i).normcand=clust2(i).cand; clust2(i).normcand(:,1)=clust2(i).normcand(:,1)/frdnat(RUN); clust2(i).normcand(:,2)=... ...clust2(i).normcand(:,2)./clust2(i).normcand(:,7); clust2(i).normcand(:,3)=... ...clust2(i).normcand(:,3)./clust2(i).normcand(:,8); clust2(i).normcand(:,4)=clust2(i).normcand(:,4)/sddnat(RUN); end x=[[clust(:).variance]; [clust(:).n]; [clust(:).fr];... ... [clust(:).M]; [clust(:).lam];[clust(:).bet];[clust(:).sd]]’; y=[[clust2(:).variance]; [clust2(:).n]; [clust2(:).fr];... ... [clust2(:).M]; [clust2(:).lam];[clust2(:).bet];[clust2(:).sd]]’; result=[x(match(:,1),:), y(match(:,2),:)]; [P,s]=sort(-result(:,4).result(:,11).result(:,2).result(:,9)... ..../(result(:,1).result(:,8))); result=result(s,:); result=result(:,[3,10,5,12,6,13,7,14,1,8,2,9,4,11]); index=match(s,:); ii=1:(i-1); R=sort(result,1); source=[]; for i=1:length(result); a=index(i,1); A.3 Recupero dei parametri delle sorgenti 73 b=index(i,2); cand=clust(a).normcand; cand2=clust2(b).normcand; table=zeros(length(cand(:,1)),length(cand2(:,1))); for j=1:length(cand(:,1)) table(j,:)=(... (cand(j,1)-cand2(:,1)).^2+... (cand(j,2)-cand2(:,2)).^2+... (cand(j,3)-cand2(:,3)).^2+... (cand(j,4)-cand2(:,4)).^2 ... ).^0.5; end [d,j,k]=minmin(table); cand=clust(a).cand; cand2=clust2(b).cand; source(i,:)=[cand(j,[1:4,6]),cand2(k,[1:4,6]),d]; end S=[]; for i=1:4 S(:,i)=(source(:,i).source(:,5)+source(:,5+i).source(:,10))... ..../(source(:,5)+source(:,10)); end S(:,5)=source(:,5).source(:,10); 74 A. Script R=S; info(3).S(1)=length(S); R=[R,-P]; for i=length(R):-1:1 ii=1:(i-1); jj=find(abs(R(i,1)-R(ii,1))<0.05); if ~isempty(jj) R(i,:)=[]; end end info(3).S(2)=length(R); ii=find(R(:,6)<10^8); R(ii,:)=[]; ii=find(R(:,6)./R(:,5)<10^6); R(ii,:)=[]; info(3).S(3)=length(R); if saving==1 save(filename,’R’,’info’) A.3 Recupero dei parametri delle sorgenti 75 end end 77 Appendice B Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 20,042 −53,99 −13,594 -6,52E-11 10,884 20,256 −112,16 17,586 -7,38E-11 15,149 20,472 66,859 −37,871 -1,79E-11 12,36 20,689 −176,39 44,063 -6,55E-11 8,6386 20,904 −59,9 44,746 -8,11E-12 4,1542 1 21,119 −136,15 16,637 -7,45E-12 1,6861 21,335 7,272 −45,704 -1,65E-11 3,1428 2 21,551 −71,295 −74,982 -1,13E-11 30,264 21,767 119,66 −60,886 -9,03E-11 14,742 21,983 −38,277 8,988 -5,85E-11 72,692 3 22,199 105,33 −21,903 -7,94E-11 1,2568 22,413 −7,1333 −17,046 9,28E-12 0,24837 22,63 −116,9 4,913 -2,36E-11 1,3023 22,844 24,557 −55,425 -3,64E-11 0,78806 23,062 9,652 60,354 1,06E-12 0,63724 23,277 174,59 14,963 -7,09E-11 0,43998 23,492 57,121 −22,487 -1,00E-10 0,6815 23,709 119,74 −8,1215 -3,29E-11 2,2749 23,925 141,71 −60,536 -5,86E-11 2,6815 24,141 31,633 −46,006 -6,28E-11 1,3166 78 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 24,356 57,246 16,545 -5,49E-11 1,6363 24,571 56,881 −3,5948 -8,26E-11 51,307 4 24,788 30,482 75,666 -6,80E-12 2,7647 25,003 −12,983 1,0751 -1,95E-11 61,869 5 25,219 −178,42 36,457 -1,81E-11 0,41338 25,433 81,189 −16,832 -1,55E-11 6,9427 7 25,651 −137,33 −25,604 -6,67E-11 22,987 8 25,866 −62,999 −8,634 -9,94E-11 33,201 26,083 −78,668 −63,606 6,18E-12 3,1222 26,297 51,368 −21,43 -9,15E-11 5,6734 26,513 −170,03 −16,135 -1,38E-11 6,84 9 26,728 161,16 −38,908 1,13E-11 0,23824 26,945 140,7 −30,736 -2,83E-12 16,414 10 27,16 50,056 49,352 -5,76E-11 2,6095 27,376 −2,4195 30,503 2,52E-12 16,419 27,591 166,64 −45,13 -5,41E-12 16,096 11 27,807 132,66 −11,705 -5,27E-11 1,0725 28,023 −64,745 −33,779 -5,24E-12 3,6551 28,238 −68,309 55,606 -1,21E-11 36,856 12 28,454 −61,221 −43,227 -8,16E-11 0,68229 28,671 152,78 27,13 1,32E-11 0,35829 28,887 −175,67 57,392 -5,44E-11 23,875 13 29,101 22,641 26,814 1,28E-11 1,1039 29,317 −147,37 −60,443 -3,10E-11 54,214 14 29,532 55,944 39,211 -3,73E-11 1,8154 29,749 −56,681 −7,6167 -7,21E-11 0,4304 29,965 103,54 −4,0192 -1,59E-11 0,97871 30,18 115,08 −45,51 -8,84E-11 39,543 15 30,395 47,843 29,041 -7,65E-11 2,6658 30,611 −45,687 −2,0687 -9,52E-11 64,467 16 79 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 30,827 −168,4 −9,8653 -3,31E-11 0,27898 31,043 −126,76 −43,379 -4,27E-11 0,59556 31,26 66,737 51,823 -1,01E-11 1,2528 31,474 87,964 82,573 -9,29E-11 0,26571 31,691 51,813 −22,52 2,13E-12 49,208 31,906 −109,45 −41,756 -3,70E-11 3,7437 17 32,122 −32,466 −29,813 3,45E-12 92,899 18 32,338 5,4797 −65,521 -8,16E-11 78,056 19 32,553 −29,18 −55,871 -3,08E-11 3,1065 20 32,768 1,3703 37,998 -4,77E-11 53,204 21 32,984 0,087909 −35,799 -3,90E-11 0,22865 33,201 50,635 −9,4928 -7,38E-11 30,411 22 33,416 159,23 −52,668 -6,64E-11 1,3082 33,631 −125,88 54,18 -7,48E-11 34,147 23 33,847 128,62 32,186 -7,81E-11 0,28993 34,064 133,23 18,727 -6,01E-11 1,6597 34,278 100,02 −41,343 -6,98E-11 2,819 34,494 117,84 −67,905 -3,37E-11 3,8826 24 34,709 135,49 −26,175 -7,07E-11 50,198 25 34,926 17,839 −14,589 -1,48E-11 0,72723 35,142 −112,6 −47,348 -7,08E-11 21,827 26 35,358 −34,925 −11,36 -2,96E-11 1,1435 35,574 130,4 44,462 -1,66E-11 77,964 27 35,788 −84,201 9,4739 1,32E-12 0,94796 36,004 64,372 49,648 -6,13E-11 11,156 28 36,22 −12,616 61,759 -1,38E-11 4,0697 36,437 −104,71 6,2461 -8,58E-11 0,65347 36,651 −152,23 −11,598 -4,42E-11 4,3996 29 36,867 −50,527 −4,6661 -1,34E-11 3,9145 30 37,083 165,66 −34,239 1,24E-11 0,28219 80 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 37,299 114,73 36,915 -4,88E-11 28,468 31 37,514 −53,509 71,2 -9,56E-11 3,3491 32 37,731 −75,519 −66,131 -5,78E-11 0,65931 37,945 −71,61 −15,716 -4,82E-12 50,025 33 38,162 72,291 −14,538 -5,28E-11 1,2833 38,377 −154,87 −25,34 -4,21E-11 0,20374 38,592 99,601 21,624 3,54E-13 0,27642 38,81 94,136 5,3345 -4,38E-11 4,0261 34 39,026 101,47 15,991 -4,01E-11 1,9223 39,241 20,748 −71,437 1,18E-11 0,62508 39,457 −86,291 60,137 -2,48E-11 2,2808 39,671 −165,88 45,552 -1,93E-11 0,20274 39,888 −20,228 −52,201 -5,04E-12 3,2149 40,103 112,98 −34,362 -1,84E-11 0,232 40,319 −109,38 47,231 -2,37E-11 11,556 40,534 −31,478 18,708 -7,11E-11 11,74 36 40,75 82,96 −38,433 -1,97E-12 1,6037 40,966 −133,58 −41,684 -5,47E-11 1,4039 41,181 −152,59 21,516 -6,12E-11 4,7849 37 41,397 89,932 −22,992 -3,98E-11 3,7375 38 41,614 −122,37 15,69 -5,77E-11 10,558 39 41,829 7,1357 −54,267 5,51E-12 1,546 42,045 103,18 45,339 -7,89E-12 0,40836 42,26 145,15 −23,435 7,74E-12 9,2754 40 42,477 −31,613 34,551 1,91E-12 29,945 41 42,693 −128,55 −18,616 1,22E-11 76,817 42 42,907 140,07 −34,06 -1,24E-12 11,023 43 43,123 65,62 15,581 -8,10E-11 0,31221 43,34 122,96 13,896 -4,30E-12 9,5238 44 43,554 −105,51 −33,331 -3,67E-11 0,94051 81 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 43,769 139,86 85,683 -3,13E-11 0,2522 43,987 −13,343 −40,766 -1,19E-11 0,23183 44,202 0,13456 −55,616 -7,60E-12 35,797 44,417 −137,65 −33,454 -2,90E-12 1,7992 45 44,634 −53,108 4,6767 -1,65E-11 25,315 46 44,85 26,841 32,799 -4,70E-14 1,075 45,066 129,66 48,677 -6,93E-12 11,624 47 45,281 35,184 −25,802 8,86E-12 1,179 45,496 171,17 −3,6545 -9,97E-11 19,063 48 45,712 85,146 29,446 -8,03E-11 29,941 49 45,927 −128,74 −50,894 -3,32E-11 51,949 46,144 57,403 −23,961 -7,24E-11 0,97891 46,359 125,17 28,07 -8,73E-11 0,44522 46,575 6,4423 70,974 -2,06E-11 8,5649 51 46,79 57,872 25,835 -4,36E-11 22,549 52 47,007 76,723 −11,369 3,66E-13 19,263 53 47,223 −49,683 14,201 -3,05E-11 1,0595 47,439 156,26 −13,999 -7,62E-11 2,4529 56 47,654 75,925 −39,475 -3,55E-11 1,214 47,869 173,82 19,867 -9,87E-11 0,47034 48,086 20,053 −5,0628 -5,81E-11 14,45 57 48,301 −172,64 43,095 -6,44E-12 50,985 58 48,517 −107,21 22,997 -4,94E-11 0,30996 48,731 −66,681 71,768 -5,64E-11 18,088 48,948 −35,526 −72,276 1,10E-11 0,34711 49,164 −15,303 −6,9376 -5,01E-11 39,124 59 49,38 110,46 46,327 -5,22E-12 1,0715 49,595 171,66 −2,6237 1,05E-11 1,0243 49,812 −142,49 −21,144 -8,07E-11 0,66599 50,026 80,502 −24,789 -6,25E-11 5,5437 60 82 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 50,243 155,46 −51,944 -9,48E-11 0,27827 50,458 −57,799 6,1533 -6,36E-11 3,3951 50,674 26,12 37,183 -2,40E-11 17,26 50,889 107,63 −34,067 -8,82E-11 0,32439 51,105 −177,03 −2,6474 -7,08E-11 70,33 61 51,32 93,367 −66,944 -7,06E-11 0,42016 51,536 −176,99 −6,4205 9,26E-12 2,6862 62 51,752 −142,88 −17,403 -5,96E-11 0,5155 51,968 65,556 23,867 -5,37E-11 60,599 63 52,184 −10,593 −58,351 7,73E-12 0,98982 52,399 −134,46 −1,3221 -6,59E-11 1,4676 52,614 −107,39 9,5011 -9,43E-11 0,25505 52,831 69,881 9,0432 1,23E-11 1,8541 64 53,046 122,46 −15,711 -3,54E-11 0,58869 53,261 −92,327 22,657 -2,62E-12 0,24219 53,477 23,938 −8,6407 -7,60E-11 10,927 65 53,693 31,074 42,062 -7,05E-11 7,314 66 53,91 124,38 51,641 -8,98E-11 0,65512 54,124 144,3 −26,974 -8,68E-11 1,014 54,341 −149,73 7,5472 -8,66E-11 0,36829 54,557 130,28 12,303 -4,96E-12 0,41029 54,772 −27,704 −53,158 -3,79E-11 0,4639 54,989 −15,302 −58,637 -9,81E-11 36,484 67 55,203 −130,27 6,0523 -4,69E-11 7,9372 68 55,42 173,16 27,43 -2,89E-11 80,603 69 55,636 −100,87 50,811 5,01E-12 11,073 55,852 −89,552 −56,545 -9,55E-11 4,4443 70 56,067 −72,241 53,725 -8,09E-11 12,669 71 56,282 82,405 18,427 -1,33E-11 0,33526 56,498 113,4 9,2384 -6,66E-11 18,871 72 83 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 56,714 −15,675 −50,144 -2,58E-11 1,3795 56,928 86,266 49,628 -1,08E-11 1,8641 57,145 −32,963 13,14 -5,83E-11 0,26901 57,36 24,556 14,441 -7,24E-11 15,675 73 57,576 119,95 22,882 -2,91E-11 3,1489 74 57,791 115,79 43,932 -4,01E-11 20,069 75 58,008 −123,06 46,851 -3,71E-11 61,128 76 58,224 −162,84 −19,508 -7,93E-11 1,5827 78 58,44 −43,759 44,054 -9,94E-11 2,806 58,655 −152,08 28,287 -1,73E-12 0,77438 58,87 −152,81 52,508 -7,68E-11 7,1712 79 59,087 −94,698 15,748 -8,64E-12 39,451 80 59,302 −148,19 5,7289 5,40E-12 72,005 81 59,518 28,787 −81,671 -5,15E-11 43,948 82 59,735 144,26 −1,7001 -6,97E-11 43,702 83 59,951 −127,53 −2,9505 -9,20E-11 1,684 60,165 127,2 28,436 -4,23E-11 3,1578 60,38 85,998 −32,024 -8,96E-11 8,7281 85 60,596 −178,96 3,4587 1,24E-11 1,4853 60,812 −31,095 −14,48 -7,19E-11 11,054 86 61,028 2,7365 −55,674 -4,01E-11 29,52 87 61,243 −115,96 −23,517 -8,82E-11 0,67522 88 61,458 87,318 −7,9825 -3,07E-11 10,789 90 61,675 17,521 −10,409 -3,23E-11 0,87968 61,892 58,065 −16,873 4,86E-12 0,35912 62,107 −150,87 −11,681 -5,97E-11 6,7587 92 62,321 80,275 7,398 -9,46E-11 0,45967 62,538 −141,46 10,247 -7,14E-11 0,22013 62,753 76,956 56,046 7,53E-12 0,40945 62,969 −132,44 −60,982 -3,05E-11 25,845 93 84 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 63,185 −47,27 80,781 -2,80E-11 7,3444 94 63,402 150,11 −29,76 -1,55E-11 19,285 95 63,616 −49,734 −14,416 -1,33E-11 7,1951 96 63,832 24,22 −33,467 -5,06E-11 13,222 97 64,048 112,69 −15,491 -1,71E-11 2,3511 64,264 104,85 50,075 -1,18E-11 17,27 98 64,481 176,3 39,809 9,81E-12 24,194 99 64,695 36,139 24,972 -7,25E-11 52,734 64,911 −100,79 43,491 -5,25E-11 51,188 100 65,126 −21,356 21,272 -6,43E-11 0,981 65,342 43,108 37,515 1,02E-11 1,8996 65,559 −61,818 −8,9556 -5,35E-11 54,023 101 65,774 −61,904 8,5844 -5,71E-11 0,33703 65,99 56,862 34,615 -3,08E-11 0,58342 66,205 −119,25 72,805 -6,68E-12 0,98699 66,421 161,62 20,29 -6,44E-12 0,42167 66,636 102,05 −37,381 -6,32E-11 1,0613 66,853 115,77 13,607 -8,51E-11 0,39136 67,068 −27,778 43,415 3,62E-12 18,115 102 67,284 35,039 65,605 -4,94E-11 17,341 67,5 −59,277 −35,139 -2,65E-11 1,2974 103 67,715 137,82 −53,565 -4,09E-11 0,63968 105 67,931 34,958 −48,584 -6,22E-11 11,066 107 68,148 −22,701 −0,57597 -4,61E-11 41,775 109 68,363 73,187 −68,835 -7,30E-11 0,40641 68,578 −11,378 −21,18 -8,62E-11 0,40387 68,795 77,888 −0,87476 -3,92E-11 22,155 110 69,01 −120,03 −41,003 -8,28E-11 70,014 111 69,225 44,764 55,667 4,17E-12 41,573 113 69,44 −9,6865 9,703 -2,43E-11 0,56814 85 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 69,656 −34,348 −21,202 -4,51E-11 0,58696 69,872 −45,02 −38,027 -5,90E-12 1,8672 70,089 23,847 −33,464 -1,59E-11 14,114 116 70,304 −135,66 14,014 -3,89E-11 3,7107 70,519 −1,8238 −25,922 -9,02E-11 0,72594 70,734 48,182 34,945 -9,58E-11 6,8299 118 70,952 100,09 −61,315 1,41E-11 34,952 121 71,166 −173,32 51,689 -1,15E-11 28,234 122 71,382 114,4 −25,912 2,67E-12 12,636 123 71,598 17,538 −2,248 -3,67E-11 1,8058 71,815 −155,25 29,592 -2,08E-11 64,505 124 72,03 76,073 29,791 -9,95E-11 81,352 125 72,245 124,05 −25,867 -1,34E-11 0,33623 72,461 49,516 8,8118 -1,56E-11 22,565 127 72,676 −160,45 35 7,44E-12 4,1211 72,892 176,76 61,015 -9,54E-11 13,839 129 73,109 68,159 −10,952 1,23E-11 1,8394 73,324 −65,846 41,775 6,63E-12 42,529 130 73,54 −28,054 40,912 -9,53E-11 28,634 131 73,756 −163,13 48,53 -7,63E-11 62,455 132 73,972 −152,84 −33,383 -7,66E-11 2,5691 133 74,187 −126,23 26,769 -2,84E-11 3,9892 74,402 47,178 7,9324 -3,94E-11 75,328 134 74,617 −112,65 −28,72 -8,44E-11 0,46107 74,835 118,47 −67,092 -4,69E-11 3,8258 75,051 23,674 14,692 -2,63E-11 6,6003 75,266 −9,6751 −65,252 -7,41E-11 0,95492 137 75,48 −95,158 −3,656 -3,37E-11 1,1231 75,696 93,051 66,928 6,92E-12 7,7441 141 75,913 168,63 −28,974 1,40E-12 22,916 144 86 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 76,129 61,489 −3,5841 -7,34E-11 1,5769 76,344 −51,932 −54,687 -8,62E-11 77,889 145 76,56 −48,896 −11,117 -7,50E-11 3,451 146 76,776 139,75 −22,549 -3,30E-11 31,658 147 76,991 156,82 14,481 -6,77E-11 16,771 148 77,208 100,83 83,388 -3,94E-11 2,2627 77,422 169,96 −31,814 -3,36E-11 0,2334 77,638 72,927 36,05 -4,20E-11 0,58991 77,855 −25,419 −8,4039 -7,53E-11 2,9369 78,07 −132,84 −14,712 -8,80E-11 2,6106 78,287 −135,2 22,064 -7,40E-11 0,4348 78,502 171,59 71,742 -9,42E-11 44,613 78,717 99,496 23,939 -8,56E-11 14,057 150 78,933 138,19 −54,611 -4,78E-11 4,1012 79,148 33,385 26,257 -3,84E-11 76,054 151 79,363 102,02 −48,416 3,89E-12 40,969 79,581 −65,214 7,8586 -9,34E-11 0,26251 79,795 −16,843 −12,197 -5,07E-11 24,433 152 80,011 100,25 67,533 -2,59E-12 7,5078 153 80,227 5,6276 −36,906 -9,30E-11 25,53 156 80,443 −154,71 −26,799 -3,15E-11 3,9577 157 80,658 −58,4 −75,765 1,07E-12 94,318 158 80,875 −177,15 −61,227 -5,88E-11 0,38211 81,091 −136,47 29,978 -2,02E-11 0,4844 81,305 141,18 −20,581 -4,75E-11 0,59995 81,522 −12,149 −11,659 -3,78E-11 3,8669 159 81,737 −76,58 −1,1049 -5,06E-12 68,351 160 81,953 58,446 −6,4261 -1,13E-11 34,355 161 82,169 83,833 −24,69 -3,79E-11 3,956 162 82,385 −24,865 −79,726 -7,71E-11 76,051 163 87 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 82,599 80,073 −2,7572 -8,41E-11 0,28935 82,815 −72,506 0,062056 -3,12E-11 56,696 164 83,03 97,674 24,543 -5,77E-11 23,644 165 83,246 −106,18 44,449 -6,77E-11 8,1375 167 83,463 −177,1 −37,614 -3,92E-11 6,0624 169 83,679 −65,071 16,85 -7,11E-12 24,269 170 83,894 −32,305 −21,806 -4,51E-11 13,489 171 84,111 96,712 −36,052 -7,07E-12 2,2838 84,325 −176,08 5,0921 -5,36E-12 4,267 172 84,542 177,47 33,397 -7,02E-11 0,22448 84,757 −24,716 −24,344 -7,05E-11 8,8507 173 84,972 26,381 −25,559 -4,96E-11 28,684 174 85,188 −71,306 −3,3049 -1,25E-11 0,37903 85,405 −20,986 42,024 -2,27E-11 5,1778 175 85,62 −101,24 −19,462 5,52E-12 0,31803 85,835 73,181 35,035 -5,35E-11 0,6983 86,052 146,35 4,9567 -2,84E-13 2,2546 86,268 −68,751 −29,98 -1,10E-12 4,198 177 86,483 −109,93 −9,3523 -1,99E-11 6,1687 180 86,699 −116,02 −4,1463 -9,05E-11 0,67767 86,914 63,469 26,818 -9,95E-11 0,62203 87,131 115,1 −50,22 -2,89E-11 10,427 182 87,347 157,82 57,383 -2,06E-11 15,705 87,562 −25,949 60,099 -4,15E-11 4,5513 183 87,778 −162,83 26,951 1,34E-11 1,5081 87,993 −101,54 −3,68 -1,81E-11 0,85275 88,208 −174,47 −53,049 -7,50E-12 46,175 184 88,424 48,55 47,637 -8,88E-11 0,51322 88,64 −163,6 28,507 -3,24E-11 9,3855 185 88,857 93,412 −31,811 -6,08E-11 0,9539 88 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 89,071 −11,641 13,385 -6,52E-11 0,46844 89,287 −47,926 −20,305 9,57E-12 1,3071 89,503 107,29 −19,136 -7,41E-11 10,714 186 89,718 −76,489 −18,305 -2,60E-11 9,7571 187 89,935 158,51 −13,993 -7,54E-11 1,4643 90,151 −69,727 17,756 -2,17E-11 4,0602 90,367 −161,63 20,797 9,45E-12 2,8021 90,583 86,823 −26,575 -4,39E-11 0,44247 90,798 101,94 5,3815 -5,07E-11 70,23 189 91,012 −177,06 −18,906 -1,64E-11 0,38608 91,228 −126,24 31,927 -9,78E-11 0,39365 91,445 −131,99 −3,6332 -5,51E-11 1,2571 91,661 −53,043 −26,555 -4,01E-11 17,184 190 91,876 176,84 −26,956 -1,19E-11 0,24132 92,093 −52,925 −0,22144 -5,20E-11 0,44125 92,308 −19,919 −66,564 -8,44E-11 5,9999 191 92,524 168,3 49,546 -8,99E-11 0,38673 92,738 −19,431 1,6989 -1,58E-11 0,76538 92,954 101,11 −36,634 -1,70E-11 11,234 192 93,17 92,536 15,669 -3,89E-11 4,5315 193 93,385 −20,098 49,057 -4,23E-11 0,38757 93,601 −146,74 72,121 -3,06E-11 31,525 194 93,819 −9,4632 −13,074 -8,38E-11 41,167 195 94,034 96,936 −49,266 -2,53E-11 16,045 196 94,249 −144,02 −3,3334 -7,33E-12 35,559 198 94,464 −53,849 29,686 -7,29E-11 1,2177 200 94,68 −86,328 −71,418 -1,13E-11 61,316 201 94,897 72,932 −34,548 1,38E-11 37,347 95,112 −13,236 22,547 -9,41E-12 49,47 202 95,329 40,89 20,835 -4,34E-11 7,1302 203 89 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 95,542 −143,74 −44,454 1,47E-11 89,814 204 95,758 −13,609 −9,8748 -6,99E-11 3,1188 206 95,974 −131,96 −10,808 -7,71E-11 96,95 209 96,19 −17,203 72,433 -2,19E-11 1,1551 96,407 −152,18 −27,72 7,85E-12 6,8482 210 96,623 42,442 −22,444 -4,35E-11 5,9551 211 96,837 164,94 22,677 -4,89E-11 30,2 212 97,054 77,573 23,967 -7,26E-11 0,63576 97,268 124,63 24,263 -9,01E-12 1,4984 97,485 −14,115 −27,062 7,23E-12 29,07 213 97,7 −87,265 6,8008 -3,94E-11 30,124 214 97,916 143,95 −35,077 -4,92E-12 58,512 215 98,131 −70,847 −46,971 -3,25E-11 6,5646 216 98,347 61,275 −36,472 -7,21E-11 0,84094 98,564 163,95 −79,37 1,24E-11 1,538 98,778 30,837 51,175 -5,44E-11 42,798 217 98,996 126,27 8,0499 -4,83E-11 0,2077 99,21 171,29 58,962 -5,08E-11 36,109 219 99,425 −75,226 −29,536 4,08E-13 73,992 220 99,642 96,846 −27,308 -9,98E-11 1,7865 99,858 −26,352 8,3867 1,48E-11 81,792 221 100,07 −141,46 15,765 5,23E-13 0,42647 223 100,29 −36,212 30,15 -4,55E-11 1,7852 100,5 108,75 −38,515 -7,54E-12 4,5626 224 100,72 100,04 33,603 -9,85E-11 15,733 225 100,94 84,115 52,734 1,26E-11 1,6755 101,15 145,39 −49,283 -5,77E-11 9,559 226 101,37 −6,2342 −2,27 -4,88E-11 0,54233 101,58 −84,178 22,451 -9,35E-11 14,471 227 101,8 −92,432 −8,2166 -8,03E-11 0,31778 229 90 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 102,02 −35,464 −20,62 1,00E-11 1,2128 102,23 −40,82 −19,575 -8,71E-11 33,069 232 102,45 −10,344 58,329 -7,99E-12 66,757 233 102,66 −57,102 −57,889 -3,10E-11 1,4575 102,88 −98,768 −29,236 -7,46E-11 0,7633 103,09 −142,46 66,296 -4,41E-11 3,035 234 103,31 −38,331 −14,349 -7,22E-11 0,46665 103,52 −59,362 −10,99 -1,40E-11 27,611 235 103,74 140,62 84,367 -6,84E-11 0,48766 237 103,96 101,87 −50,653 -2,32E-11 4,47 239 104,17 93,375 3,8093 6,01E-12 2,2551 240 104,39 21,219 −49,122 8,00E-12 10,67 241 104,6 44,787 58,143 -6,28E-11 39,517 242 104,82 −101,25 58,291 -2,15E-11 7,5268 243 105,04 12,514 −28,169 -8,92E-11 8,1677 244 105,25 38,673 54,12 -9,91E-11 29,463 246 105,47 65,703 10,7 -7,64E-11 43,95 249 105,68 79,847 −28,536 -9,85E-12 27,627 251 105,9 −15,324 15,2 -8,12E-11 0,22296 106,11 −114,79 3,5585 -1,55E-11 1,5053 106,33 77,519 69,033 -6,99E-11 2,8045 106,54 −129,25 58,499 -2,70E-11 1,541 106,76 32,831 −31,691 -6,48E-11 0,3849 106,98 −132,82 75,706 -6,56E-11 7,7263 252 107,19 −14,424 −50,334 -9,98E-11 3,5933 253 107,41 2,9156 −17,134 -1,11E-11 0,23208 107,62 −32,316 −47,472 -1,96E-11 11,274 254 107,84 130,02 −8,1821 -5,20E-11 0,39586 108,05 121,26 18,376 -9,36E-11 0,32414 108,27 −112,93 10,256 -1,23E-11 0,22553 91 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 108,49 135,06 50,174 -6,20E-11 4,9405 255 108,7 −131,3 −15,157 -4,38E-11 0,29901 108,92 144,28 70,591 6,33E-12 0,21998 109,13 29,078 −43,356 -4,98E-11 2,4731 109,35 −143,45 −30,114 -6,34E-11 2,5952 109,56 −86,086 −35,782 -9,71E-11 64,337 257 109,78 −166,37 17,793 -4,47E-11 0,42831 110 −134,11 2,9419 -4,42E-12 76,53 110,21 140,86 54,906 -6,57E-11 1,604 110,43 140,8 −54,392 -1,99E-11 4,025 259 110,64 −163,18 13,634 -6,13E-11 1,2343 110,86 162,32 −44,639 -2,82E-11 0,23399 111,08 136,09 −13,366 -5,79E-11 47,287 260 111,29 −167,19 −2,5763 -4,85E-11 3,2046 262 111,51 154,55 38,525 -7,74E-11 0,79436 264 111,72 55,133 15,61 -7,25E-11 3,9056 267 111,94 −13,867 16,039 8,91E-12 0,7034 269 112,15 138,29 16,745 -2,08E-11 0,56082 112,37 169 63,146 1,34E-11 17,772 270 112,58 −97,844 −36,8 -4,16E-11 1,4644 112,8 −168,21 40,617 9,89E-12 3,9942 113,02 122,37 −64,805 -2,40E-11 36,989 271 113,23 138,44 75,181 8,88E-12 1,8397 272 113,45 −103,9 −28,063 -2,81E-11 0,39636 274 113,66 −41,861 57,366 -8,84E-11 33,656 275 113,88 106,59 4,4653 -8,51E-11 0,5749 277 114,1 −33,972 −52,186 -8,13E-11 0,49871 114,31 −115,53 −21,247 -2,62E-11 51,894 279 114,53 −173,8 39,37 -7,12E-11 0,37227 114,74 3,9869 42,073 -3,76E-11 0,50211 92 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 114,96 −150,38 20,832 -3,15E-12 2,1018 115,17 25,497 −0,56317 -6,01E-11 0,27105 115,39 −146,37 75,945 -1,69E-11 82,704 280 115,61 −164,26 50,72 -6,47E-11 10,441 281 115,82 −22,552 −57,305 8,88E-12 20,37 282 116,04 157 −48,499 -1,59E-11 31,05 283 116,25 −173,48 14,197 -7,25E-11 4,3069 285 116,47 −66,638 −51,555 -1,94E-11 16,296 287 116,68 99,684 −8,7381 -5,26E-11 0,7368 116,9 −5,8739 44,505 -5,88E-11 80,57 290 117,11 89,828 −25,255 -3,62E-12 4,9022 291 117,33 159,81 −46,159 -4,10E-11 67,247 117,55 −148,7 21,193 -6,47E-11 0,30899 117,76 −89,164 −67,356 -2,79E-11 47,567 293 117,98 116,19 −44,192 -7,18E-11 4,7424 296 118,19 49,765 72,694 -6,80E-11 0,95531 118,41 −137,82 −59,709 -7,27E-11 2,052 118,62 −33,653 −33,125 -2,06E-11 0,96242 118,84 61,513 −38,595 -7,37E-11 1,2802 119,06 −148,88 45,901 -6,98E-11 4,0898 297 119,27 −8,4923 27,346 -5,16E-11 0,32264 299 119,49 167,23 29,472 -6,64E-11 18,767 302 119,7 −163,43 −7,2724 -4,00E-11 61,927 305 119,92 −117,71 −21,946 -5,53E-11 5,9987 307 120,13 −62,159 44,341 -6,57E-12 3,612 120,35 106,56 −62,037 -4,40E-11 15,272 309 120,57 161,26 18,086 -4,94E-11 2,8454 120,78 −40,632 −70,148 7,28E-12 83,924 121 −176,23 −33,483 -1,72E-11 4,2193 121,21 54,058 −27,101 -4,12E-11 27,322 310 93 Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 121,43 −168,47 39,456 -1,14E-11 1,1672 121,65 −19,615 −2,2747 1,11E-11 23,683 311 121,86 27,818 34,334 -1,15E-11 1,4234 122,08 87,881 −67,371 -8,90E-11 31,432 122,29 2,821 −22,077 1,32E-12 12,924 312 122,51 21,438 −43,051 -9,65E-11 23,974 314 122,72 −57,971 −65,794 -7,16E-11 24,437 315 122,94 43,936 32,53 -5,82E-11 0,21009 123,16 10,997 −27,569 -1,94E-11 3,2185 316 123,37 −105,66 −10,959 -6,50E-11 17,789 317 123,59 79,465 −70,775 -6,42E-11 2,9013 123,8 −100,78 42,758 -2,15E-11 5,0732 318 124,02 −125,22 18,284 -6,89E-11 0,69886 124,23 164,04 −24,82 -6,51E-11 0,66094 124,45 80,543 −17,51 -1,42E-11 72,547 319 124,67 173,33 44,784 -1,09E-11 0,70919 124,88 57,438 27,583 -1,67E-11 2,61 320 125,1 −33,813 6,6681 -1,55E-11 3,8096 125,31 159,8 −53,546 4,43E-14 10,868 322 125,53 140,59 −38,94 -2,18E-11 3,8036 125,74 177,39 −34,266 -8,04E-11 24,5 326 125,96 −150,24 79,195 -2,42E-11 71,86 126,18 5,5331 57,731 -9,73E-11 7,9505 327 126,39 115,06 −50,525 -4,56E-11 18,89 328 126,61 −7,0553 −12,112 -9,14E-12 1,2204 126,82 51,82 −29,126 -3,31E-11 23,557 329 127,04 17,295 −16,779 -8,69E-11 1,7326 331 127,25 147,88 −7,8076 -9,62E-11 7,1703 332 127,47 140,56 −0,40879 -1,07E-11 15,949 333 127,68 −144,26 50,559 -2,44E-11 0,82354 94 B. Tabelle Dati Tabella B.1. Sorgenti iniettate. Gli indici nell’ultima colonna corrispondono a quelli dei segnali ricostruiti più vicini. Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] SNR Indice 95 Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 1 20,905 317,74 23,595 2,01E-11 2 21,323 23,227 −14,279 -9,69E-11 3 21,982 311,7 −36,786 -2,68E-11 4 24,569 94,049 53,4 -3,33E-12 5 25,004 335,14 −5,9228 2,21E-11 6 25,155 148,29 47,336 -2,64E-11 7 25,433 78,374 27,787 1,42E-11 8 25,65 230,23 22,035 -1,90E-11 9 26,513 193,58 −27,97 2,32E-11 10 26,945 142,37 34,057 1,04E-13 11 27,591 163,48 −50,843 1,41E-11 12 28,238 284,85 −53,038 2,55E-12 13 28,886 195,67 58,65 2,12E-13 14 29,318 226,47 −59,715 3,01E-11 15 30,178 135,7 −42,396 -1,27E-11 16 30,606 308,53 26,154 -1,31E-11 17 31,905 255,65 −37,114 9,54E-12 18 32,122 327,7 −30,73 -3,29E-12 19 32,337 352,02 58,285 7,23E-12 20 32,552 336 −52,121 -7,52E-12 21 32,768 346,64 47,093 1,79E-11 22 33,2 40,375 −34,507 2,04E-11 23 33,63 241,11 −44,056 1,13E-11 24 34,494 126 73,269 1,20E-11 25 34,705 141,42 26,542 -3,26E-11 26 35,141 258,94 −36,34 1,19E-11 27 35,574 131,52 −47,916 6,32E-12 28 36,004 73,65 63,904 -1,58E-11 29 36,649 203,3 −27,742 -1,50E-11 30 36,867 310,27 −4,1768 4,85E-13 96 B. Tabelle Dati Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 31 37,299 111,17 46,925 3,30E-11 32 37,513 321,58 56,305 2,53E-11 33 37,945 287,82 −12,398 -3,99E-12 34 38,81 108,79 25,092 2,16E-11 35 39,987 8,1775 −23,984 -1,07E-10 36 40,533 325,44 −16,674 2,07E-11 37 41,179 212 31,418 1,31E-11 38 41,397 87,001 −29,72 1,06E-11 39 41,613 243,36 17,204 -3,85E-12 40 42,26 144,62 −22,818 5,05E-13 41 42,477 329,97 −33,341 7,12E-12 42 42,693 229,82 −19,224 -7,66E-12 43 42,907 141,86 −32,532 -1,32E-11 44 43,34 123,4 13,047 -1,13E-11 45 44,417 222,33 −29,867 -1,52E-12 46 44,634 303,21 −7,4299 -1,50E-11 47 45,066 130,72 49,937 3,38E-12 48 45,494 172,82 21,6 -2,39E-11 49 45,709 99,085 42,136 -2,40E-11 50 46,027 80,332 70,435 1,28E-11 51 46,576 346,24 −74,145 1,12E-11 52 46,789 53,284 −23,875 7,82E-12 53 47,007 77,238 12,065 -5,01E-12 54 47,073 76,29 −12,746 -6,59E-11 55 47,157 91,644 −71,973 -3,46E-11 56 47,437 160,6 −11,118 1,20E-11 57 48,085 14,334 −5,2345 2,07E-11 58 48,3 186,78 41,883 -1,55E-11 59 49,165 337,34 5,1571 2,37E-11 60 50,028 153,79 −64,016 -5,08E-11 97 Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 61 51,104 185,39 24,694 3,32E-11 62 51,525 164,21 31,837 -1,10E-10 63 51,965 104,46 −40,317 -6,28E-11 64 52,831 71,941 1,6542 -1,73E-11 65 53,476 20,321 −6,8892 -1,68E-11 66 53,692 23,663 41,252 1,98E-11 67 54,987 340,16 −53,078 2,78E-11 68 55,203 234,06 4,3686 1,85E-11 69 55,42 176,74 −29,327 2,38E-11 70 55,848 236,29 40,451 -1,79E-11 71 56,062 276,18 46,457 -2,51E-11 72 56,497 111,38 −21,907 1,72E-11 73 57,359 18,775 −11,65 2,65E-11 74 57,575 119,25 28,386 1,82E-11 75 57,79 116,56 45,925 -2,77E-11 76 58,008 242,04 −44,484 3,05E-11 77 58,09 292,82 −76,668 1,82E-11 78 58,222 231,21 −38,499 -3,46E-11 79 58,867 203,9 49,41 2,02E-12 80 59,088 266,13 −12,809 1,10E-11 81 59,302 211,37 −1,6168 -2,99E-12 82 59,518 323,51 −84,994 1,06E-12 83 59,733 142,12 18,93 1,81E-11 84 59,797 314,73 −70,293 -1,02E-11 85 60,377 81,716 38,164 1,69E-12 86 60,81 326,83 −17,472 1,01E-11 87 61,027 0,53586 51,163 2,02E-11 88 61,211 340,27 −76,306 -2,32E-11 89 61,369 333,5 −65,598 2,03E-11 90 61,458 85,894 18,232 1,09E-11 98 B. Tabelle Dati Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 91 61,614 231,31 69,797 -5,20E-13 92 62,103 191,99 −66,546 -8,09E-12 93 62,969 232,73 −59,124 2,51E-11 94 63,184 312,3 79,328 1,77E-11 95 63,402 148,82 −33,631 1,60E-11 96 63,616 310,48 −8,5954 1,70E-11 97 63,831 20,954 −27,422 2,19E-11 98 64,264 104,04 51,146 6,27E-12 99 64,48 175,08 39,137 -2,26E-12 100 64,91 262,21 39,157 1,61E-11 101 65,554 303,97 −7,1753 -3,79E-11 102 67,068 332,31 43,524 1,87E-12 103 67,49 349,47 48,872 -4,77E-11 104 67,61 329,27 −53,464 -8,00E-11 105 67,68 332,86 48,085 -3,29E-11 106 67,85 336,39 43,11 -8,33E-11 107 67,933 325,99 −73,863 -2,99E-11 108 67,99 336,16 −29,101 -7,15E-11 109 68,147 335,17 −5,4845 1,76E-11 110 68,795 76,105 −15,889 1,60E-11 111 69,007 238,86 −32,94 9,67E-12 112 69,099 261,05 −69,75 -9,72E-11 113 69,225 47,156 57,798 -2,07E-11 114 69,282 242,03 41,905 -1,21E-11 115 69,335 239,81 −54,9 -1,46E-11 116 70,09 22,036 −32,811 1,66E-11 117 70,446 32,726 45,083 -2,98E-11 118 70,731 52,511 55,551 -2,15E-11 119 70,813 18,992 −36,664 -5,08E-11 120 70,875 33,149 −40,853 -8,65E-11 99 Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 121 70,952 106,58 −60,928 -1,64E-11 122 71,166 187,46 51,945 6,20E-13 123 71,382 114,31 −25,839 3,91E-12 124 71,815 206,83 30,241 2,00E-11 125 72,028 73,251 38,037 9,49E-12 126 72,193 78,693 58,185 -5,70E-11 127 72,459 52,779 14,126 -3,77E-11 128 72,539 85,743 −49,367 -5,17E-11 129 72,887 159,9 65,93 -1,40E-11 130 73,324 293,31 −42,647 -7,58E-12 131 73,536 336,58 33,558 -4,42E-13 132 73,754 198,75 45,286 5,66E-12 133 73,969 205,69 −37,615 -1,57E-11 134 74,402 44,138 11,645 2,93E-11 135 75,14 32,167 −21,454 -5,75E-11 136 75,198 25,075 17,532 -6,27E-11 137 75,292 20,443 −11,262 -3,47E-11 138 75,398 18,96 −6,9341 -2,53E-11 139 75,553 34,573 50,015 -2,03E-11 140 75,622 33,393 25,013 2,04E-11 141 75,695 96,545 −65,981 -3,47E-11 142 75,747 27,06 −33,668 -4,00E-13 143 75,848 68,738 56,291 -5,46E-11 144 75,913 168,79 −28,407 -3,19E-12 145 76,34 322,85 −49,911 7,22E-12 146 76,556 310,54 −8,7847 -1,68E-11 147 76,776 139,72 −26,047 1,79E-11 148 76,989 157,64 21,227 8,07E-12 149 78,552 221,19 −37,241 -4,05E-11 150 78,715 96,343 30,459 2,10E-11 100 B. Tabelle Dati Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 151 79,148 30,101 −34,507 2,85E-11 152 79,794 340,55 13,948 1,79E-11 153 80,011 98,505 67,444 3,58E-12 154 80,076 101,12 66,644 1,96E-11 155 80,142 95,896 66,127 -5,07E-11 156 80,224 8,9185 −8,4738 7,23E-12 157 80,441 199,7 −38,421 -3,22E-11 158 80,658 296,83 −76,023 -1,42E-11 159 81,522 345,88 −10,792 -3,01E-12 160 81,732 288,9 −21,469 -7,85E-11 161 81,952 57,946 −9,2633 1,04E-12 162 82,168 82,331 −28,887 2,11E-11 163 82,38 308,26 −77,25 -1,41E-11 164 82,815 293,76 0,61341 2,29E-11 165 83,03 96,242 −30,294 1,38E-11 166 83,119 99,739 39,198 -1,05E-10 167 83,245 257,72 43,7 9,11E-12 168 83,339 104,57 29,63 -3,61E-11 169 83,463 185,06 −37,465 2,09E-12 170 83,679 295,27 15,152 3,86E-12 171 83,891 329,88 21,753 -3,28E-11 172 84,325 184,32 9,9469 9,02E-12 173 84,757 334,26 −19,166 1,88E-11 174 84,973 21,804 −29,189 3,81E-11 175 85,405 337,95 39,369 2,31E-11 176 86,118 146,3 2,4756 -4,46E-11 177 86,268 291,12 −29,686 9,19E-13 178 86,318 147,44 −10,831 -6,54E-11 179 86,369 144,65 −3,3902 -7,87E-11 180 86,483 250,38 −1,0324 -9,99E-12 101 Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 181 86,549 147,25 −3,2076 -2,20E-11 182 87,131 114,79 −51,964 -3,21E-12 183 87,561 333,11 58,403 1,40E-11 184 88,208 185,84 −53,036 -2,96E-12 185 88,638 195,69 28,291 -1,57E-11 186 89,502 106,31 27,117 1,85E-11 187 89,718 284,9 −15,411 1,84E-11 188 90,007 180 89,816 0 189 90,797 100,61 16,244 1,94E-11 190 91,66 306,92 −22,728 1,60E-11 191 92,304 340,24 −64,498 -2,81E-11 192 92,954 101 −37,57 -1,52E-12 193 93,169 91,296 20,736 1,17E-11 194 93,601 216,53 71,208 -9,01E-12 195 93,817 348,28 −4,1615 3,49E-11 196 94,034 94,699 −50,672 2,50E-11 197 94,175 92,245 46,903 -4,36E-11 198 94,25 216,72 12 1,87E-11 199 94,38 88,901 52,266 -3,09E-11 200 94,438 82,596 −44,79 -3,54E-11 201 94,68 275,77 −70,636 1,88E-11 202 95,112 346,64 22,867 -1,07E-11 203 95,328 38,596 19,981 1,64E-11 204 95,592 352,65 4,2319 -2,82E-11 205 95,663 347,36 55,209 -1,33E-11 206 95,754 342,06 9,0694 -5,18E-13 207 95,815 346,68 6,8149 -5,59E-11 208 95,877 356,12 11,789 -1,45E-11 209 95,971 229,7 7,5019 5,73E-12 210 96,407 207,14 −27,579 -6,16E-12 102 B. Tabelle Dati Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 211 96,622 41,197 −25,094 -1,12E-11 212 96,836 165,43 26,514 8,21E-12 213 97,485 346,22 −28,449 -6,30E-12 214 97,699 275,5 1,3399 2,04E-11 215 97,917 144,32 −35,243 1,08E-12 216 98,131 289,99 −44,71 2,31E-11 217 98,777 26,931 51,258 1,97E-11 218 98,847 298,42 −53,157 -2,58E-11 219 99,209 172,78 59,652 2,18E-11 220 99,425 283,64 −29,34 -1,39E-11 221 99,858 333,95 11,697 -3,33E-12 222 100 16,5 89,234 1,81E-11 223 100,06 147,04 −85,302 -6,56E-11 224 100,5 109,34 −39,009 -5,44E-12 225 100,72 98,312 38,312 -8,04E-13 226 101,15 142,72 −52,071 -9,99E-12 227 101,58 277,66 14,118 2,86E-12 228 101,69 153,55 −31,337 -9,63E-12 229 101,79 161,74 −42,661 -6,90E-11 230 101,91 135,98 −51,78 1,55E-12 231 101,96 145,52 −46,036 9,91E-12 232 102,23 320,13 10,438 -3,01E-12 233 102,45 348,85 58,611 -3,82E-12 234 103,09 220,08 65,489 -1,65E-11 235 103,52 302,25 −2,4018 1,76E-11 236 103,66 209,71 57,962 -5,09E-12 237 103,72 216,93 79,934 -1,39E-11 238 103,8 219,66 65,697 9,18E-12 239 103,96 100,35 −51,567 6,15E-12 240 104,17 95,508 8,1333 -3,03E-11 103 Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 241 104,39 21,942 −50,277 -1,58E-11 242 104,6 42,211 60,875 -1,41E-11 243 104,82 261,88 58,037 2,70E-11 244 105,03 10,143 −29,854 -3,93E-11 245 105,12 12,025 −23,438 8,04E-12 246 105,25 37,61 57,33 -1,41E-11 247 105,3 3,0616 −27,595 -8,29E-12 248 105,37 9,5767 −29,468 -9,97E-11 249 105,47 64,985 16,757 -5,78E-11 250 105,52 15,278 27,054 -9,57E-11 251 105,68 78,729 −28,203 1,43E-11 252 106,98 229,8 75,068 3,62E-12 253 107,19 342,81 −49,231 -2,39E-11 254 107,62 327,08 −46,719 2,07E-14 255 108,49 132,93 52,768 7,17E-12 256 108,86 193,26 −31,842 6,49E-12 257 109,56 282,1 −32,493 2,02E-11 258 109,89 23,998 −34,383 -3,56E-11 259 110,43 141,11 −55,591 1,39E-11 260 111,07 136,28 −20,629 1,94E-11 261 111,19 147,55 −1,1481 -2,68E-11 262 111,28 131,92 −30,494 -6,20E-11 263 111,44 129,68 5,49 -7,12E-11 264 111,49 134,58 −12,934 -8,14E-12 265 111,6 127,23 −6,7692 2,01E-12 266 111,66 142,96 3,3174 -1,19E-11 267 111,72 99,422 32,046 -8,48E-11 268 111,84 140,12 −13,549 3,32E-12 269 111,96 137,1 −19,496 7,87E-12 270 112,37 169 62,723 -8,32E-12 104 B. Tabelle Dati Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 271 113,02 121,41 −66,09 1,53E-11 272 113,19 241,72 79,362 -4,08E-11 273 113,33 127,05 66,712 -3,05E-11 274 113,44 127,17 63,429 -1,05E-10 275 113,66 319,1 53,968 1,60E-11 276 113,79 69,923 −65,564 2,23E-11 277 113,89 107,2 −81,074 -1,12E-10 278 113,98 103,16 68,931 6,34E-12 279 114,31 246,22 21,335 2,38E-11 280 115,39 217,19 76,277 1,56E-11 281 115,6 196,9 52,062 1,66E-11 282 115,82 337,45 −57,626 -1,95E-12 283 116,04 157,77 −49,694 1,57E-11 284 116,13 157,1 51,948 1,11E-11 285 116,25 187,18 15,717 2,59E-11 286 116,32 151,14 −47,658 -1,05E-10 287 116,47 294,5 −50,51 2,16E-11 288 116,54 162,66 −47,833 -4,13E-11 289 116,62 157,89 −43,413 -8,26E-11 290 116,9 353,3 42,664 5,07E-12 291 117,11 90,43 −25,958 -1,26E-11 292 117,64 90,734 −27,038 -5,34E-11 293 117,76 288,66 −69,893 -5,15E-11 294 117,85 91,771 −17,024 -1,27E-11 295 117,91 90,921 −28,349 -2,73E-11 296 117,98 116,73 −47,206 1,44E-11 297 119,05 213,94 45,315 3,38E-12 298 119,12 217,38 34,848 -4,43E-11 299 119,23 211,06 33,694 -6,82E-11 300 119,29 219,03 52,468 -1,76E-11 105 Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 301 119,34 212,72 51,643 -1,06E-11 302 119,49 167,76 −32,288 1,08E-11 303 119,54 215,32 44,47 8,61E-12 304 119,61 199,14 57,889 -6,15E-11 305 119,7 197,88 30,62 -1,63E-13 306 119,82 207,78 54,942 -2,02E-11 307 119,92 243,4 −18,004 -2,21E-11 308 120,02 4,5206 −75,01 -5,27E-11 309 120,35 104,5 −63,524 2,29E-11 310 121,21 52,589 −28,712 1,09E-11 311 121,64 341,38 −13,569 -4,75E-11 312 122,29 3,745 19,793 3,62E-12 313 122,41 67,075 65,221 -2,84E-11 314 122,5 20,254 −43,272 4,21E-12 315 122,72 303,97 −63,834 4,62E-12 316 123,16 10,484 −26,643 6,01E-12 317 123,37 255,03 −13,741 -2,86E-11 318 123,8 260,11 42,592 -3,53E-12 319 124,45 81,033 −18,483 -3,14E-11 320 124,88 56,167 26,8 2,23E-11 321 125,23 331,81 −10,882 -2,76E-12 322 125,31 159,78 −53,559 1,03E-13 323 125,37 323,35 18,115 -6,47E-11 324 125,43 328,07 0,64572 -8,64E-11 325 125,61 323,8 13,313 -2,49E-11 326 125,74 178,67 −33,286 -5,95E-11 327 126,17 4,4971 55,307 1,80E-11 328 126,39 114,41 −52,553 2,66E-11 329 126,82 49,992 −27,787 2,56E-11 330 126,95 343,8 65,532 -3,34E-11 106 B. Tabelle Dati Tabella B.2. Parametri recuperati. indice Frequenza [Hz] λ [rad] β [rad] ˙ f [Hzs−1] 331 127,03 17,753 −16,775 -6,84E-11 332 127,25 148,86 −12,04 -4,77E-12 333 127,47 142,29 1,1835 1,32E-11 107 Bibliograﬁa J. Aasi et al. «Einstein@Home all-sky search for periodic gravitational waves in LIGO S5 data». In: Phys. Rev. D 87 (4 feb. 2013), p. 042001. DOI: 10.1103/PhysRevD.87. 042001. URL: J. Aasi et al. «First low frequency all-sky search for continuous gravitational wave signals». In: Phys. Rev. D 93 (4 feb. 2016), p. 042007. DOI: 10.1103/PhysRevD.93. 042007. URL: B. P . Abbott et al. «Observation of Gravitational Waves from a Binary Black Hole Mer-ger». In: Phys. Rev. Lett. 116 (6 feb. 2016), p. 061102. DOI: 10.1103/PhysRevLett. 116.061102. URL: 061102. B. P . Abbott et al. «Prospects for Observing and Localizing Gravitational-Wave Transients with Advanced LIGO and Advanced Virgo». Ver. v3. In: Living Reviews in Relativity (2016). DOI: 10.1007/lrr-2016-1. F . Acernese et al. «Advanced Virgo: a second-generation interferometric gravitatio-nal wave detector». In: Classical and Quantum Gravity 32.2 (2015), p. 024001. URL: P . Astone, S. Frasca e C. Palomba. «The short FFT database and the peak map for the hierarchical search of periodic sources». In: Classical and Quantum Gravity 22.18 (2005), S1197. URL: P . Astone, S. Frasca e C. Palomba. «The short FFT database and the peak map for the hierarchical search of periodic sources». In: Classical and Quantum Gravity 22.18 (2005), S1197. URL: P . Astone, A. Colla, S. D’Antonio, S. Frasca e C. Palomba. «Method for all-sky sear-ches of continuous gravitational wave signals using the frequency-Hough tran-sform». In: Phys. Rev. D 90 (4 ago. 2014), p. 042002. DOI: 10.1103/PhysRevD.90. 042002. URL: C. Caprini et al. «Science with the space-based interferometer eLISA. II: gravita-tional waves from cosmological phase transitions». In: Journal of Cosmology and Astroparticle Physics 2016.04 (2016), p. 001. URL: A. Einstein. Sitzungsber. K. Preuss. Akad. Wiss. 1. 688, 1916. R. Hulse e J. Taylor. «Discovery of a Pulsar in a Close Binary System.» In: Bulletin of the American Astronomical Society 6 (1974). 108 Bibliograﬁa M. Kramer e N. Wex. «The double pulsar system: a unique laboratory for gravity». In: Classical and Quantum Gravity 26.7 (2009), p. 073001. URL: iop.org/0264-9381/26/i=7/a=073001. P . Leaci, the LIGO Scientiﬁc Collaboration e the Virgo Collaboration. «Searching for continuous gravitational wave signals using LIGO and Virgo detectors». In: Journal of Physics: Conference Series 354.1 (2012), p. 012010. URL: org/1742-6596/354/i=1/a=012010. M. Maggiore. Gravitational Waves: Theory and Experiments. A cura di O. Oxford. Volume 1 voll. 2007. ISBN: 0198570740. D. Pascucci. «Interferometric measurement of the Quality factor for test masses of Virgo». Laurea Magistrale in Astroﬁsica e Scienze dello Spazio. Università degli Studi di Napoli "Federico II", 2012/2013. M. Punturo et al. «The third generation of gravitational wave observatories and their science reach». In: Classical and Quantum Gravity 27.8 (2010), p. 084007. URL: Schutz. «Gravitational wave sources and their detectability». In: Classical and Quantum Gravity 6.12 (1989), pp. 1761–1780. J. Taylor e J. Weisberg. «A new test of general relativity. Gravitational radiation and the binary pulsar PSR 1913+16». In: Astrophysucal Journal (253 1982), pp. 908–920. DOI: 10.1086/159690. V.Ferrari e L. Gualtieri. «General Relativity». 2010. J. Weber. «Evidence for Discovery of Gravitational Radiation». In: Phys. Rev. Lett. 22 (24 giu. 1969), pp. 1320–1324. DOI: 10.1103/PhysRevLett.22.1320. URL:
8791	https://prepp.in/question/two-cards-are-drawn-successively-without-replaceme-64490e85cb8aedb68af5393c	Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is Hello, Guest Login / Register All Categories+ Test SeriesQuizzesPrevious Year PapersLive TestsLive QuizzesCurrent AffairsVideosNewsContact Us Get App All Exams Test series for 1 year @ ₹349 onlyEnroll Now ×× Home Mathematics Probability of Random Experiments two cards are drawn successively without replaceme Question Mathematics Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is 1/26 1/221 4/223 1/13 Solution The correct answer is 1/221 Calculating Probability of Drawing Two Aces This problem asks for the probability of drawing two aces in a row from a standard deck of 52 cards, with the important condition that the first card drawn is not replaced. Drawing cards without replacement means that the total number of cards, and potentially the number of specific cards, changes after each draw. This makes the events dependent. Understanding Probability Without Replacement When events are dependent, the outcome of the first event affects the probability of the second event. To find the probability of two dependent events A and B both happening, we use the formula: P(A and B)=P(A)×P(B∣A) P(A \text{ and } B) = P(A) \times P(B\|A) P(A and B)=P(A)×P(B∣A) Where P(B∣A) P(B\|A) P(B∣A) is the probability of event B happening given that event A has already happened. Step-by-Step Probability Calculation Let's apply this to our problem of drawing two aces successively: Event A: Drawing an ace on the first draw. Event B: Drawing an ace on the second draw, given that the first card drawn was an ace and was not replaced. Probability of Drawing the First Ace A standard deck has 52 cards. There are 4 aces in the deck. The probability of drawing an ace on the first draw (Event A) is: P(First card is Ace)=Number of Aces Total Number of Cards P(\text{First card is Ace}) = \frac{\text{Number of Aces}}{\text{Total Number of Cards}} P(First card is Ace)=Total Number of Cards Number of Aces P(First card is Ace)=4 52=1 13 P(\text{First card is Ace}) = \frac{4}{52} = \frac{1}{13} P(First card is Ace)=52 4=13 1 Probability of Drawing the Second Ace (Given the First was an Ace) After drawing one ace and not replacing it, the deck now has: Total cards remaining: 52−1=51 52 - 1 = 51 52−1=51 Aces remaining: 4−1=3 4 - 1 = 3 4−1=3 The probability of drawing another ace from this modified deck (Event B given A) is: P(Second card is Ace\|First card was Ace)=Number of remaining Aces Total remaining Cards P(\text{Second card is Ace \| First card was Ace}) = \frac{\text{Number of remaining Aces}}{\text{Total remaining Cards}} P(Second card is Ace\|First card was Ace)=Total remaining Cards Number of remaining Aces P(Second card is Ace\|First card was Ace)=3 51=1 17 P(\text{Second card is Ace \| First card was Ace}) = \frac{3}{51} = \frac{1}{17} P(Second card is Ace\|First card was Ace)=51 3=17 1 Combined Probability of Drawing Two Successive Aces Now, we multiply the probabilities of the two dependent events to find the probability of drawing two aces successively: P(Drawing Two Aces)=P(First card is Ace)×P(Second card is Ace\|First card was Ace) P(\text{Drawing Two Aces}) = P(\text{First card is Ace}) \times P(\text{Second card is Ace \| First card was Ace}) P(Drawing Two Aces)=P(First card is Ace)×P(Second card is Ace\|First card was Ace) P(Drawing Two Aces)=1 13×1 17 P(\text{Drawing Two Aces}) = \frac{1}{13} \times \frac{1}{17} P(Drawing Two Aces)=13 1×17 1 P(Drawing Two Aces)=1×1 13×17 P(\text{Drawing Two Aces}) = \frac{1 \times 1}{13 \times 17} P(Drawing Two Aces)=13×17 1×1 P(Drawing Two Aces)=1 221 P(\text{Drawing Two Aces}) = \frac{1}{221} P(Drawing Two Aces)=221 1 Thus, the probability of drawing two aces successively without replacement from a well-shuffled pack of 52 cards is 1 221 \frac{1}{221} 221 1. Revision Table: Key Probability Concepts \| Concept \| Description \| How it Applies Here \| --- \| Probability \| The chance of a specific event occurring. \| Used to quantify the likelihood of drawing aces. \| \| Without Replacement \| The drawn item is not returned to the sample space. \| Crucially changes the probabilities for the second draw. \| \| Dependent Events \| Events where the outcome of one affects the probability of the other. \| Drawing the second card's outcome depends on the first card drawn and not replaced. \| \| Conditional Probability \| The probability of an event occurring given that another event has already occurred. \| Calculated for drawing the second ace after the first was drawn. \| \| Multiplication Rule (Dependent) \| P(A∩B)=P(A)×P(B∣A) P(A \cap B) = P(A) \times P(B\|A) P(A∩B)=P(A)×P(B∣A) \| Formula used to find the probability of both aces being drawn. \| Additional Information: Deck Composition and Probability Understanding the composition of a standard 52-card deck is essential for solving card probability problems. Total Cards: 52 Suits: 4 (Hearts, Diamonds, Clubs, Spades) Cards per Suit: 13 (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K) Total Aces: 4 (one in each suit) Total Face Cards (J, Q, K): 12 (3 in each suit) Total Number Cards (2-10): 36 (9 in each suit) When solving probability problems, always identify if the events are independent (with replacement) or dependent (without replacement), as this significantly changes the calculation method for successive events. Download PDF Was this answer helpful? 0 0 Important Questions from Probability of Random Experiments A problem is given to three students A, B and C, whose probabilities of solving the problem independently are 1 2\rm \frac{1}{2}2 1,3 4\rm \frac{3}{4}4 3and p respectively. If the probability that the problem can be solved is 29 32\rm \frac{29}{32}32 29, then what is the value of p? Mathematics View Answer If three dice are rolled under the condition that no two dice show the same face, then what is the probability that one of the faces is having the number 6? Mathematics View Answer Three fair dice are tossed once. What is the probability that they show different numbers that are in AP? Mathematics View Answer What is the probability of getting a composite number in the list of natural numbers from 1 to 50? Mathematics View Answer The completion of a construction job may be delayed due to strike. The probability of strike is 0.6. The probability that the construction job gets completed on time if there is no strike is 0.85 and the probability that the construction job gets completed on time if there is a strike is 0.35. What is the probability that the construction job will not be completed on time ? Mathematics View Answer Need Expert Advice? 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8792	https://artofproblemsolving.com/wiki/index.php/Common_factorizations?srsltid=AfmBOoqNnSYHDfM23Atquua5p7EfKvcn4up2lsBmdSrke-7zWAVFMO20	Art of Problem Solving Common factorizations - 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 Common factorizations Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Common factorizations These are common factorizations. Contents [hide] 1 Basic Factorizations 2 Vieta's/Newton Factorizations 3 Circulant Identities 4 Other Resources Basic Factorizations Vieta's/Newton Factorizations These factorizations are useful for problem that could otherwise be solved by Newton's sums or problems that give a polynomial, and ask a question about the roots. Combined with Vieta's formulas, these are excellent, useful factorizations. Circulant Identities The matrices above are called circulant matrices. In general, the determinant of a circulant matrix will be a multiple of the sum of the entries in any of its rows/columns. Other Resources More Factorizations Sum and difference of powers Retrieved from " Category: Algebra 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.
8793	https://jingyan.baidu.com/article/e75aca850e5e4f552edac6a8.html	圆锥侧面积的计算公式-百度经验分享到一键分享 QQ空间新浪微博百度云收藏人人网腾讯微博百度相册开心网腾讯朋友百度贴吧豆瓣网搜狐微博百度新首页 QQ好友和讯微博更多... 百度分享新闻网页贴吧知道经验音乐图片视频地图百科文库发布经验百度首页登录写经验领红包首页分类美食/营养游戏/数码手工/爱好生活/家居健康/养生运动/户外职场/理财情感/交际母婴/教育时尚/美容认证悬赏令回享商城视频经验知道百度经验>生活/家居>生活常识圆锥侧面积的计算公式浏览：12124 \| 更新：2022-11-26 08:10 设圆锥的底面半径为r,高为h,母线长为l 圆锥侧面展开图是一个扇形,半径为l,弧长为2πr ∴圆锥侧面积=(1/2)(2πr)l=πrl 拓展资料圆锥的侧面积将圆锥的侧面沿母线展开，是一个扇形,这个扇形的弧长等于圆锥底面的周长,而扇形的半径等于圆锥的母线的长. 圆锥的侧面积就是弧长为圆锥底面的周长×母线/2；没展开时是一个曲面。圆锥的侧面积公式是怎么来的 ① S = π R L 圆锥侧面积=n/360×π×R2=1/2LR （n指扇形顶角度数,R是圆锥底面半径,L指母线）圆锥的侧面积推导,需要把圆锥展开； ② 数学上规定,圆锥的顶点到该圆锥底面圆周上任意一点的连线叫圆锥的母线； ③ 沿圆锥的任意一条母线剪开展开成平面图形即为一个扇形； ④ 展开后的扇形的半径就是圆锥的母线, 展开后的扇形的弧长就是圆锥底面周长； ⑤ 通过展开,就把求立体图形的侧面积转化为了求平面图形的面积. 设圆锥的母线长为 L ,设圆锥的底面半径为 R , 则展开后的扇形半径为 L ,弧长为圆锥底面周长（2πR）扇形的面积公式为：S = （1/2）× 扇形半径 × 扇形弧长. = （1/2）× L × （2πR） = π R L 即圆锥的侧面积为：圆锥底面半径与圆锥母线长的乘积的π倍. 经验内容仅供参考，如果您需解决具体问题(尤其法律、医学等领域)，建议您详细咨询相关领域专业人士。展开阅读全部【word版】圆锥的侧面积练习题专项练习_即下即用圆锥的侧面积练习题完整版下载，海量试题试卷，全科目覆盖，随下随用，简单方便，即刻下载，试卷解析，强化学习，尽在百度教育百度教育广告 2025完整版圆锥面积-含完整资料-在线下载 360文库广告查看更多圆锥体表面积计算的公式-点击查看百度教育广告查看更多换一批相关经验 πrl是什么公式是什么2022.07.25 韦达定理公式是什么2022.08.10 圆锥侧面积πrl怎么推导2023.03.19 相对原子质量怎么求2018.12.22 凸透镜成像规律表格是什么2022.08.10 今日支出元写经验有钱赚 >> 商光豪作者的经验雪碧能放冰箱里冻成冰能爆吗初学者如何使用excel 车子大灯好像就一个灯泡，近光不亮远... 流量开关和压力开关是不是得先接一个... 单片机中的AJMP LJMP SJMP JMP有什么... 相关推荐圆锥的表面积计算公式-点击查看涵盖全册名校课堂答案，圆锥的表面积计算公式精准解析，海量知识点汇总，助力学习更上一层楼!找圆锥的表面积计算公式?点这里，快速提升学习成绩! 百度教育圆锥面积计算公式-点击查看涵盖全册名校课堂答案，圆锥面积计算公式精准解析，海量知识点汇总，助力学习更上一层楼!找圆锥面积计算公式?点这里，快速提升学习成绩! 百度教育圆锥表面积公式-点击查看涵盖全册名校课堂答案，圆锥表面积公式精准解析，海量知识点汇总，助力学习更上一层楼!找圆锥表面积公式?点这里，快速提升学习成绩! 百度教育广告如要投诉，请到百度经验投诉中心，如要提出意见、建议，请到百度经验管理吧反馈。此内容有帮助? 7 我的财富值去登录我的现金去登录帮助意见反馈投诉举报 ©2025Baidu使用百度前必读百度经验协议作者创作作品协议企业推广京ICP证030173号-1 京网文【2023】1034-029号分享收藏返回顶部新浪微博QQ 空间 ◆ 请扫描分享到朋友圈辅助模式
8794	https://sites.ualberta.ca/~vbouchar/MAPH464/section-symmetric-group.html	(\DeclareMathOperator{\Tr}{Tr} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} ) MA PH 464 - Group Theory in Physics: Lecture Notes Vincent Bouchard Section 1.8 The symmetric group (S_n) ¶ Objectives You should be able to: Sketch the proof of Cayley's theorem. Use standard notations (permutations, cycles) to describe elements of the symmetry group (S_n\text{.}) Calculate the parity of a permutation. Describe conjugacy classes in (S_n) in terms of cycle structures and integer partitions. Many of the concepts that we just studied abstractly become alive when we study the symmetric group (S_n\text{,}) which is the group of permutations of (n) elements. Everything becomes clear and explicit, since we kind of all known intuitively how permutations work. Subsection 1.8.1 Cayley's theorem ¶ But in fact studying the symmetric group (S_n) is fundamental, because of what is known as Cayley's theorem, which basically states that any finite group (G) of order (n) is isomorphic (that is, identical as abstract group) to a subgroup of the symmetric group (S_n\text{.}) More precisely, the statement of Cayley's theorem is that (G) is a subgroup of the group of permutations of the elements of (G\text{,}) which is (S_n) since (G) has order (n\text{.}) So from this point of view, studying symmetric groups and their subgroups basically means studying all finite groups. Cayley's theorem is quite deep, since it says that we can realize any abstract finite group as a subgroup of a permutation group. In other words, it connects abstract group theory to the very concrete world of permutations. Isn't it cool? As we will see, Cayley's theorem gives an example of a nice representation which exists for all finite groups, and is called the regular representation. Why is Cayley's theorem true? Without going through a formal proof, let us see how it goes. Consider the multiplication table of a finite group (G = {g_1, \ldots, g_n }\text{.}) The (i)'th row of the table is given by the elements ({ g_i g_1, \ldots, g_i g_n }\text{.}) By the Rearrangement Theorem Theorem 1.4.2, this is the same set of elements as (G\text{,}) but in a different order. Thus we can assign to the (i)'th row an element of the permutation group of (n) elements, that is an element of (S_n\text{,}) which acts as ({ g_1, \ldots, g_n } \mapsto {g_i g_1, \ldots, g_i g_n}\text{.}) As a result, to every element (g_i \in G) we can assign a permutation (s_i \in S_n\text{;}) we have constructed a map from (G) to a subgroup of (S_n\text{,}) and one can show that this map is a group isomorphism. This is the essence of Cayley's theorem. Example 1.8.1. (\mathbb{Z}_3) as a subgroup of (S_3). As an example, consider the group of three elements (G = \mathbb{Z}_3) with multiplication table Table 1.4.5. We see that the map assigns to the element (e \in G) the identity permutation (e \in S_3\text{;}) to (a \in G) it assigns the cyclic permutation (1 \mapsto 2, 2 \mapsto 3, 3 \mapsto 1) in (S_3\text{;}) and to (b \in G) it assigns the cyclic permutation (1 \mapsto 3, 2 \mapsto 1, 3 \mapsto 2\text{.}) Thus we have identified (G = \mathbb{Z}_3) with the alternating subgroup (A_3 \subset S_3) consisting of even permutations. Subsection 1.8.2 Notation Now let us study the symmetric group (S_n) in more detail. First we introduce standard notation. An element of (S_n) corresponds to a permutation of (n) elements. It is customary to denote it by its action on the (n) elements ({1, \ldots, n }\text{.}) For instance, the cyclic rotation of the (n) elements, which takes 1 to 2, 2 to 3, etc., would be denoted by \begin{equation} \pi = \begin{pmatrix} 1 \amp 2 \amp \cdots \amp n-1 \amp n \ 2 \amp 3 \amp \cdots \amp n \amp 1 \end{pmatrix}. \end{equation} Generically, a permutation (\pi) that takes (1) to (\pi(1)\text{,}) (2) to (\pi(2)\text{,}) and so on, would be denoted by \begin{equation} \pi = \begin{pmatrix} 1 \amp 2 \amp \cdots \amp n-1 \amp n \ \pi(1) \amp \pi(2) \amp \cdots \amp \pi(n-1) \amp \pi(n) \end{pmatrix}. \end{equation} Example 1.8.2. The symmetric group (S_3). As an example, consider (S_3\text{.}) It has (3! = 6) elements; the identity permutation ((\pi_1)), three permutations corresponding to a single exchange of two elements ((\pi_2,\pi_3,\pi_4)), and two permutations that permute the three elements ((\pi_5, \pi_6)). In the notation introduced above, the six elements are \begin{align} e = \pi_1 = \begin{pmatrix} 1 \amp 2 \amp 3\1 \amp 2 \amp 3 \end{pmatrix}, \qquad \pi_2 = \begin{pmatrix} 1 \amp 2 \amp 3\2 \amp 1 \amp 3 \end{pmatrix},\qquad \pi_3 = \begin{pmatrix} 1 \amp 2 \amp 3\3 \amp 2 \amp 1 \end{pmatrix},\ \pi_4 = \begin{pmatrix} 1 \amp 2 \amp 3\1 \amp 3 \amp 2 \end{pmatrix}, \qquad \pi_5 = \begin{pmatrix} 1 \amp 2 \amp 3\3 \amp 1 \amp 2 \end{pmatrix}, \qquad \pi_6 = \begin{pmatrix} 1 \amp 2 \amp 3\2 \amp 3 \amp 1 \end{pmatrix}. \end{align} To construct the multiplication table for (S_3\text{,}) we need to multiply elements in (S_3\text{,}) where by multiplication here we mean composition of permutations. For instance, the composition (\pi_4 \circ \pi_5) can be constructed as follows. Start with (\pi_5\text{;}) it takes (1 \mapsto 3, 2 \mapsto 1, 3 \mapsto 2\text{.}) Then, combining with the action of (\pi_4\text{,}) we get (1 \mapsto 3 \mapsto 2\text{,}) (2 \mapsto 1 \mapsto 1\text{,}) and (3 \mapsto 2 \mapsto 3\text{,}) which is (\pi_2\text{.}) Therefore, (\pi_4 \circ \pi_5 = \pi_2\text{.}) In fact, we see explicitly that (S_3) is non-Abelian. Consider (\pi_5 \circ \pi_4\text{.}) We obtain (1 \mapsto 1 \mapsto 3\text{,}) (2 \mapsto 3 \mapsto 2) and (3 \mapsto 2 \mapsto 1\text{,}) which is (\pi_3\text{.}) Therefore (\pi_5 \circ \pi_4 = \pi_3\text{,}) hence (\pi_4 \circ \pi_5 \neq \pi_5 \circ \pi_4\text{.}) Subsection 1.8.3 Cycles Since we are dealing with permutations of set of (n) elements, it is clear that after applying a given permutation a certain number of times, we will come back to the initial elements. In other words, consider for instance the element (1\text{.}) A given permutation (\pi) will take (1 \to \pi(1)\text{.}) Applying it again, we will get (1 \to \pi(1) \to \pi(\pi(1))\text{.}) After a number of application, say (r\text{,}) we will necessarily get back (\pi^r(1) = 1\text{.}) The numbers (1, \pi(1), \pi^2(1), \ldots, \pi^{r-1}(1)\text{,}) that are reached from (1) by (\pi) form what we call a “cycle”. More precisely: Definition 1.8.3. Cycle. Let (\pi \in S_n\text{,}) (i \in {1, \ldots, n}) and let (r) be the smallest positive integer such that (\pi^r(i) = i\text{.}) Then the set of (r) distinct elements ({\pi^k(i) }_{k=0}^{r-1}) is called a cycle of (\pi) of length (r\text{,}) or an (r)-cycle generated by (i). We denote a given cycle by ((i~ \pi(i)~ \ldots~ \pi^{r-1}(i) )\text{.}) It is clear that a given permutation (\pi) breaks up the set of (n) elements ({1,\ldots,n}) into disjoint cycles. We can then denote the permutation (\pi) by the cycle decomposition of ({1,\ldots,n}) that it implies. Example 1.8.4. The cycle structure of permutations in (S_3). Consider the permutation (\pi_5) for (S_3) as in the previous example. Start with (1\text{.}) Then (\pi_5(1) = 3\text{,}) (\pi_5^2(1) = \pi_5(3) = 2\text{,}) and (\pi_5^3(1) = \pi_5(2) = 1\text{.}) Thus (\pi_5) decomposes ({1,2,3}) into a single cycle of length (3\text{.}) In cycle notation, we write (\pi_5 = (132)\text{.}) We could have written (\pi_5 = (321)) or (\pi_5 = (213)) as well; those are the same permutations. In this notation the numbers defining the cycles can be moved around cyclically without changing anything. Similarly, (\pi_6\text{,}) which is the other permutation that exchanges the three elements, has cycle decomposition (\pi_6 = (123)\text{.}) Consider instead the permutation (\pi_2\text{.}) Start with (1\text{.}) Then (\pi_2(1) = 2\text{,}) and (\pi_2^2(1) = \pi_2(2) = 1\text{.}) Thus (1) generates a (2)-cycle given by ((12)\text{.}) There is another cycle however. Start with (3\text{.}) Then (\pi_2(3)=3\text{,}) hence (3) generates a (1)-cycle given by ((3)\text{.}) Therefore, the set of three elements is broken up into two cycles by (\pi_2\text{,}) that is (\pi_2 = (12)(3)\text{.}) Similarly, we get (\pi_3= (13)(2)\text{,}) (\pi_4 = (1)(23)\text{,}) and (\pi_1=(1)(2)(3)\text{.}) With the notion of cycles we can define permutations that are “cyclic”: Definition 1.8.5. Cyclic permutations. If (\pi \in S_n) has a cycle of length (r) and all other cycles of (\pi) have only one element, then (\pi) is called a cyclic permutation of length (r). Example 1.8.6. Cyclic permutations in (S_3). In the example of (S_3\text{,}) we then have that (\pi_5) and (\pi_6) are cyclic permutations of length (3\text{,}) while (\pi_2,\pi_3,\pi_4) are cyclic permutations of length (2\text{.}) Thus in (S_3) all permutations are cyclic. But for instance in (S_4\text{,}) we have the permutation (\pi = (12)(34)\text{,}) which is not cyclic. A particularly useful type of cyclic permutation is of length two: Definition 1.8.7. Transposition. A cyclic permutation of length (2) (i.e. a permutation that only exchanges two elements) is called a transposition. Example 1.8.8. Transpositions in (S_3). To come back to (S_3\text{,}) there are three transpositions, namely (\pi_2, \pi_3) and (\pi_4\text{.}) Now that we understand cycles, we can easily compute composition of permutations. Given two permutations (\pi_1) and (\pi_2\text{,}) the composition (\pi_1 \circ \pi_2) can be computed easily by “multiplying cycles”. This is easier understood in terms of an example. Consider the product of cycles ((15)(234)(253)) in (S_5\text{.}) To find the corresponding permutation, we consider each element of ({1,\ldots,5 }\text{,}) and, starting from the far right of the product of cycles, we keep track of what the element becomes under the action of the cycles. In our case, consider first (1\text{.}) The first two cycles on the right leave it invariant, while the last cycle sends its to (5\text{.}) Thus (1 \to 5\text{.}) Consider (2\text{.}) The first cycle sends it to (5\text{,}) the second cycle then leaves (5) invariant, and then the last cycle sends it to (1\text{.}) Thus (2 \to 1\text{.}) Consider (3\text{.}) Then we get (3 \to 2 \to 3 \to 3\text{.}) Consider (4\text{.}) We get (4 \to 4 \to 2 \to 2\text{.}) Finally, for (5) we get (5 \to 3 \to 4 \to 4\text{.}) The corresponding permutation is then: \begin{equation} \pi = \begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \amp 5\ 5 \amp 1 \amp 3 \amp 2 \amp 4 \end{pmatrix}. \end{equation} (A consistency check here is that each number (1) to (5) only appear once in the final result of the permutation, which must be the case.) Equivalently, we can write the resulting permutation in terms of its cycle structure, which is (\pi=(1542)(3)\text{.}) Hence we get that \begin{equation} (15)(234)(253)=(1542)(3). \end{equation} This is how we multiply cycles. Example 1.8.9. Multiplication of cycles vs composition of permutations. Let us just show in an example that multiplication of cycles and composition of permutations are equivalent. Consider (\pi_1 = (123)(45)) and (\pi_2 = (1)(2)(345)) in (S_5\text{.}) Suppose that we want to compute the composition (\pi_1 \circ \pi_2\text{.}) The resulting permutation is easily calculated to be \begin{equation} \pi_1 \circ \pi_2=\begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \amp 5\ 2 \amp3 \amp 5 \amp 4 \amp 1 \end{pmatrix}. \end{equation} We claim that this is equivalent to the product of cycles ((123)(45)(1)(2)(345)\text{.}) When we consider products of cycles, we can remove cycles of length (1\text{,}) since they do not do anything in the product. So we want to compute ((123)(45)(345)\text{.}) Under this product, we have (1 \to 1 \to 1 \to 2\text{,}) (2 \to 2 \to 2 \to 3\text{,}) (3 \to 4 \to 5 \to 5\text{,}) (4 \to 5 \to 4 \to 4\text{,}) (5 \to 3 \to 3 \to 1\text{,}) which is indeed the permutation above. In fact, a few minutes of thought will make it clear that the way we defined products of cycles is precisely equivalent to composition of the corresponding permutations. A very interesting result for the symmetric group is that any permutation can in fact be written as the product of a number of transpositions. This is just saying that every permutation can be done in steps, where in each step we only exchange two objects. Let's see how it goes. We start by studying how a cycle can be written as a product of transpositions: Proposition 1.8.10. Decomposition of cycles into products of transpositions. An (r)-cycle ((i_1 i_2 \ldots i_r)) can be decomposed into the product of (r-1) transpositions: \begin{equation} (i_1 i_2 \ldots i_r) = (i_1 i_r)(i_1 i_{r-1}) \cdots (i_1 i_3) (i_1 i_2). \end{equation} Note however that the decomposition is not unique; in particular, since the square of any transposition is the identity, we could insert the square of any transposition in the product on the RHS without changing the end result. Proof. The proof simply involves multiplying the transpositions on the RHS. Start with (i_1\text{.}) Under the product of transpositions, we get (i_1 \to i_2 \to \ldots \to i_2\text{.}) Consider (i_2\text{:}) we get (i_2 \to i_1 \to i_3 \to \ldots \to i_3\text{.}) Iterating, we get that (i_k \to i_{k+1}) for all (k=1,\ldots,r-1\text{,}) and (i_r \to i_1\text{.}) Thus the resulting permutation is the cyclic permutation of length (r) that has cycle structure ((i_1 i_2 \ldots i_r)\text{.}) Definition 1.8.11. Parity. We define the parity of the cycle as being even if the number of transpositions in the decomposition is even, and odd if the number if transpositions is odd. Note that even though the decomposition is not unique, the parity of the decomposition is uniquely defined. We can then apply these results to permutations themselves, not just to individual cycles. We obtain: Proposition 1.8.12. Decomposition of permutations into products of transpositions. Any permutation can be decomposed as a product of transpositions (and length (1) cycles). The parity (even or odd, according to whether the decomposition has an even or odd number of transpositions) of the permutation is unique. In fact, the parity of a permutation can be determined directly from its cycle structure, without having to work out its decomposition into transpositions, since each (r)-cycle can always be decomposed into the product of (r-1) transpositions. Example 1.8.13. Decomposition of a permutation in (S_6). As an example, consider the permutation (\pi=(145)(23)(6)) in (S_6\text{.}) Looking at the cycle structure, the (3)-cycle is even (can be decomposed into two transpositions), the (2)-cycle is of course odd (it is a transposition itself), and the (1)-cycle does not contribute to the parity. Thus (\pi) is odd, since (2+1=3\text{.}) In fact, a decomposition is easy to find following Proposition 1.8.10. We get: \begin{equation} \pi = (15)(14)(23)(6). \end{equation} Of course, this decomposition is not unique (for instance ((15)(14)(24)(24)(23)(6)) would also work), but its parity is uniquely defined. Definition 1.8.14. The alternating group. The set of even permutations of (n) elements is denoted by (A_n) and is called the alternating group. Clearly, (A_n) is a subgroup of (S_n\text{,}) since the product of two even permutations is even, it contains the identity, and the inverse of an even permutation is also even. Subsection 1.8.4 Conjugacy classes Now let us study conjugacy classes of (S_n\text{.}) Recall that conjugacy classes are defined as being subsets of elements that are conjugate to each other, where two elements (x,y \in G) are conjugate if (y = g x g^{-1}) for some (g \in G\text{.}) The main result for (S_n) is the following theorem: Theorem 1.8.15. Conjugacy classes and cycle structure. Two permutations in (S_n) are conjugate if and only if they have the same cycle structure, meaning that they have the same number of cycles of equal length. To clarify the notation, if we go back to (S_3\text{,}) then the three permuations ((12)(3)\text{,}) ((13)(2)) and ((1)(23)) have the same cycle structure, hence are conjugate. The two permutations ((123)) and ((132)) are then also conjugate, and the identity permutation ((1)(2)(3)) is of course only conjugate to itself. Thus there are three conjugacy classes in (S_3\text{.}) Proof. We will only show that conjugate permutations have the same cycle structure, and leave the other direction as an exercise. Consider two permutations (\pi, \sigma \in S_n\text{.}) We want to check whether (\sigma \circ \pi \circ \sigma^{-1}) has the same cycle structure as (\pi\text{.}) If (\pi) and (\sigma) are given by the permutations: \begin{equation} \pi = \begin{pmatrix} 1 \amp 2 \amp \ldots \amp n \ \pi(1) \amp \pi(2) \amp \ldots \amp \pi(n) \end{pmatrix}, \qquad \sigma = \begin{pmatrix} 1 \amp 2 \amp \ldots \amp n \ \sigma(1) \amp \sigma(2) \amp \ldots \amp \sigma(n) \end{pmatrix}, \end{equation} then it follows that \begin{equation} \sigma \circ \pi \circ \sigma^{-1} = \begin{pmatrix} \sigma(1) \amp \sigma(2) \amp \ldots \amp \sigma(n) \ \sigma(\pi(1)) \amp \sigma(\pi(2)) \amp \ldots \amp \sigma(\pi(n)), \end{pmatrix}, \end{equation} since (\sigma^{-1}) takes the element (\sigma(i)) and sends it to (i\text{.}) Thus we can see the permutation (\sigma \circ \pi \circ \sigma^{-1}) as applying (\sigma) to the symbols in the cycle decomposition of (\pi\text{.}) The result will generally be different from (\pi\text{,}) but its cycle structure will necessarily be the same. The argument may become clearer with a simple example. Consider (\pi=(123)(4) \in S_4\text{,}) and (\sigma = (12)(34) \in S_4\text{.}) The corresponding permutations are \begin{equation} \pi = \begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \ 2 \amp 3 \amp 1 \amp 4 \end{pmatrix}, \qquad \sigma = \begin{pmatrix} 1 \amp 2 \amp 3 \amp 4 \ 2 \amp 1 \amp 4 \amp 3 \end{pmatrix}. \end{equation} The conjugate permutation is then \begin{equation} \sigma \circ \pi \circ \sigma^{-1} = \begin{pmatrix} 2 \amp 1 \amp 4 \amp 3 \ 1 \amp 4 \amp 2 \amp 3 \end{pmatrix}, \end{equation} which has cycle structure (\sigma \circ \pi \circ \sigma^{-1} = (2 1 4)(3)\text{.}) It is indeed the same cycle structure as (\pi\text{.}) In fact, it could have been obtained directly by simply applying (\sigma) to the symbols in the cycle decomposition (\pi = (123)(4)\text{,}) which gives directly (\sigma \circ \pi \circ \sigma^{-1} = (2 1 4)(3)\text{.}) It follows from Theorem 1.8.15 that all one has to do to find the conjugacy classes in (S_n) is to construct the possible cycle structures. This in turn is equivalent to partitioning the set ({1,\ldots,n }) into disjoint subsets of various lengths. What we now show is that this problem can be reformulated into the problem of finding integer partitions of the positive integer (n\text{.}) Consider a permutation (\pi \in S_n) and its cycle decomposition. Let (\nu_k) be the number of (k)-cycles in its decomposition. We introduce a new notation which gives the cycle structure of the permutation (i.e. the number of cycles of each length): we write ((1^{\nu_1}, 2^{\nu_2}, \ldots, n^{\nu_n})) to denote that the cycle decomposition of (\pi) has (\nu_1) cycles of length (1\text{,}) (\nu_2) cycles of length (2\text{,}) and so on. The total number of elements in the cycle decomposition must of course be (n\text{.}) This means that the sum (\sum_{k=1}^n k \nu_k = n\text{.}) We can rewrite this sum as: \begin{equation} (\nu_1 + \nu_2 + \ldots + \nu_n) + (\nu_2 + \nu_3 + \ldots + \nu_n) + (\nu_3 + \ldots + \nu_n ) + \ldots + (\nu_n) = n. \end{equation} We then introduce the notation (\lambda_j := \sum_{k=j}^n \nu_k\text{,}) so that the sum above becomes \begin{equation} \lambda_1 + \lambda_2 + \lambda_3 + \ldots + \lambda_n = n. \end{equation} It is also clear that (\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n \geq 0\text{,}) since (\lambda_1 - \lambda_2 = \nu_1\text{,}) (\lambda_2 - \lambda_3 = \nu_2\text{,}) and so on, and the (\nu_i) are by definition non-negative integers. The result of this combinatorial analysis is that the cycle structure of permutations in (S_n) are in one-to-one correspondence with decreasing sequences of non-negative integers (\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n) such that (\sum_{k=1}^n \lambda_k = n\text{.}) Such a decreasing sequence of non-negative integers is called a partition of (n) (we usually omit the (\lambda_k)'s that are zero). To be precise, we only showed how the (\lambda_k)'s can be constructed out of the cycle structure, that is out of the (\nu_i)'s. But it works the other way around as well. Given a partition of (n\text{,}) that is a sequence of (\lambda_k\text{,}) then we define (\nu_k = \lambda_k - \lambda_{k+1}\text{,}) with (\nu_n = \lambda_n\text{.}) (\nu_k) then gives us the number of cycles of length (k) in the cycle structure associated to the partition. Example 1.8.16. Conjugacy classes in (S_4) and partitions. This construction will become clearer with an example. Let us construct the conjugacy classes in (S_4\text{,}) corresponding to the different cycle structures in (S_4\text{.}) This corresponds to different partitions of (4\text{.}) The five distinct partitions of (4) are easily found to be (4\text{,}) (3+1\text{,}) (2+2\text{,}) (2+1+1) and (1+1+1+1\text{.}) What this tells us directly is that there are five distinct conjugacy classes in (S_4\text{.}) For fun, let us find out what kind of cycle structures these partitions correspond to. Consider first the partition (4\text{,}) which means (\lambda_1 = 4\text{,}) (\lambda_2 = \lambda_3 = \lambda_4 = 0\text{.}) To get the cycle structure, we compute (\nu_1 = \lambda_1 - \lambda_2 = 4\text{,}) (\nu_2 = \nu_3 = \nu_4 = 0\text{.}) Thus the corresponding cycle structure consists in (4) cycles of length (1\text{.}) There is in fact a unique permutation of (S_4) with this cycle structure, which is the identity permutation (\pi = (1)(2)(3)(4)\text{.}) Consider the partition (3+1\text{.}) We get (\nu_1 = 2\text{,}) (\nu_2 = 1\text{,}) (\nu_3=\nu_4=0\text{.}) So the cycle structure consists in (2) cycles of length (1\text{,}) and one cycle of length (2\text{.}) The corresponding permutations are transpositions, and there are (6) of them in (S_4\text{.}) Consider the partition (2+2\text{.}) We get (\nu_1 = 0\text{,}) (\nu_2 = 2\text{,}) (\nu_3=\nu_4=0\text{.}) The cycle structure is two cycles of length (2\text{.}) There are (3) such permutations in (S_4\text{.}) Consider (2+1+1\text{.}) We get (\nu_1=1\text{,}) (\nu_2=0\text{,}) (\nu_3=1) and (\nu_4=0\text{.}) The cycle structure is one cycle of length (3) and one cycle of length (1\text{,}) which corresponds to cyclic permutations of length (3\text{.}) There are (8) such permutations in (S_4\text{.}) Finally, (1+1+1+1\text{,}) which gives (\nu_1 = \nu_2 = \nu_3 = 0\text{,}) (\nu_4=1\text{.}) The cycle structure is a single cycle of length (4\text{,}) corresponding to a cyclic permutation of length (4\text{.}) There are (6) such permutations in (S_4\text{.}) The result is that there are five conjugacy classes in (S_4\text{,}) corresponding to five possible cycle structures, corresponding to the five partitions of (4\text{.}) Counting the number of elements in each conjugacy classes, we end up with (1 + 6 + 3 + 8 + 6 = 24 = 4!\text{,}) which is indeed the order of (S_4\text{.}) For those of you who like combinatorics, it is also fun to count in general the number of permutations of (S_n) of a given cycle structure. We will leave that as an exercise, but the result is the following. Consider a cycle structure ((1^{\nu_1}, 2^{\nu_2}, \ldots, n^{\nu_n})) with (\sum_k k \nu_k = n\text{.}) The number of permutation in (S_n) with this cycle structure is precisely: \begin{equation} \frac{n!}{\prod_{j=1}^n j^{\nu_j} \nu_j !}. \end{equation} Let us check that this formula is consistent with what we wrote above in (S_4\text{.}) Consider for instance the cycle structure given by one cycle of length (3) and one cycle of length (1\text{.}) The number of permutations with that cycle structure is (\frac{4!}{3 \cdot 1! \cdot 1 \cdot 1!} = 8\text{,}) as written above. Similarly, the number of permutations with one cycle of length (2) and two cycles of length (1) is (\frac{4!}{2 \cdot 1! \cdot 1^2 \cdot 2!} = 6\text{,}) and so on and so forth. Remark 1.8.17. Note that partitions of an integer (n) can be encoded diagrammatically into Young diagrams. A Young diagram is a finite collection of boxes arranged in left-justified rows, with the rows “weakly decreasing” (i.e., such that each row has the same or shorter length than the row right above it). It is then clear that Young diagrams with (n) boxes contain precisely the same information as integer partitions of (n\text{,}) with the integers in the partition given by the length of the rows of the Young diagram. Young diagrams are useful for studying representations of the symmetric group, in which case they are upgraded to Young tableaux.
8795	https://www.gauthmath.com/solution/1837832182958113/Identify-the-constant-difference-for-a-hyperbola-with-foci-3-0-and-3-0-and-a-poi	Solved: Identify the constant difference for a hyperbola with foci (-3,0) and (3,0) and a point on [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Questions Question Identify the constant difference for a hyperbola with foci (-3,0) and (3,0) and a point on the hyperbola (7,0). 14 10 2 Show transcript Expert Verified Solution 100%(3 rated) Answer The answer is Option 4: 6 Explanation Option 1: 14 This option suggests the constant difference is 14. Option 2: 10 This option suggests the constant difference is 10. Option 3: 2 This option suggests the constant difference is 2. Option 4: 6 This option suggests the constant difference is 6. Recall the definition of a hyperbola A hyperbola is defined as the set of all points such that the absolute difference of the distances to two fixed points (the foci) is constant. Calculate the distances from the point (7,0) to the foci (-3,0) and (3,0) Distance from (7,0) to (-3,0) is $$\|7 - (-3)\| = \|7 + 3\| = 10$$∣7−(−3)∣=∣7+3∣=10 Distance from (7,0) to (3,0) is $$\|7 - 3\| = 4$$∣7−3∣=4 Find the absolute difference of these distances The absolute difference is $$\|10 - 4\| = 6$$∣10−4∣=6 So Option 4 is correct. Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related Identify the constant difference for a hyperbola with foci -6,0 and 6,0 and a point on the hyperbola 4,0. 8 14 12 10 100% (2 rated) Identify the constant difference for a hyperbola with foci hyperbola 4,0. -6,0 and 6,0 and a point on the 10 12 8 14 94% (106 rated) Identify the constant difference for a hyperbola with foci -6,0 and 6,0 and a point on the hyperbola 4, 0. 12 8 10 14 98% (286 rated) Identify the constant difference for a hyperbola with foci -7,0 and 7,0 and a point on the hyperbola 10,0 17 14 7 3 100% (2 rated) Identify the constant difference for a hyperbola with foci 0,-5 and 0,5 and a point on the hyperbola 0,3. 8 6 2 4 93% (533 rated) Write an equation for the hyperbola with foci 12,0 and -12,0 and constant difference of 16. 100% (3 rated) What is an equation for the hyperbola with foci 9,0 and -9,0 and constant difference of 14? 93% (589 rated) Write an equation for the hyperbola. foci at 10,0 and -10,0 and a constant difference of 12. 99% (239 rated) Write an equation for the hyperbola. foci at 0,26 and 0,-26 and a constant difference of 20. 100% (3 rated) Write an equation for the hyperbola. foci at 50,0 and -50,0 and a constant difference of 28. Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
8796	https://math.stackexchange.com/questions/3365243/sum-of-squares-of-3-consecutive-numbers-is-not-a-perfect-square	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 Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Sum of squares of 3 consecutive numbers is not a perfect square Ask Question Asked Modified 4 years, 2 months ago Viewed 2k times 2 $\begingroup$ I'm trying to show that $(n-1)^2+n^2+(n+1)^2=a^2$ does not have a solution for $n,a\in \Bbb N$. I've written $(n-1)^2+n^2+(n+1)^2=3n^2+2$, so what I need to show is that $3n^2+2$ cannot be a perfect square. From here and here I understand that there is some relationship between perfect squares and modulo, but I fail to see which is it or how should apply it in my case. I do understand, however, that, in the case of 5 consecutive numbers, $n^2+2$ needs to be a multiple of 5. modular-arithmetic square-numbers Share edited May 9, 2020 at 3:47 Skyfall 4511 silver badge77 bronze badges asked Sep 22, 2019 at 6:25 PatricioPatricio 1,65411 gold badge1010 silver badges2020 bronze badges $\endgroup$ 1 2 $\begingroup$ Your title says something completely different from your actual question. Of course the sum of 3 consecutive natural numbers can be a perfect square, e.g. 2+3+4=9. $\endgroup$ bof – bof 2019-09-22 06:28:13 +00:00 Commented Sep 22, 2019 at 6:28 Add a comment \| 3 Answers 3 Reset to default 5 $\begingroup$ You are nearly there. Now just use the fact that $a^2 \ne 2 \pmod 3$. The proof is here, which relies on the fact that you can just check $n=0,1,2$ (or even better, $n=-1,0,1$) for a result that holds for all natural $n$. Share edited Sep 22, 2019 at 6:35 answered Sep 22, 2019 at 6:27 Toby MakToby Mak 17.1k44 gold badges3131 silver badges4646 bronze badges $\endgroup$ 2 $\begingroup$ Why "even better" $n=-1,0,1$? $\endgroup$ Patricio – Patricio 2019-09-23 11:05:14 +00:00 Commented Sep 23, 2019 at 11:05 $\begingroup$ Out of any three consecutive numbers, there will always be one that is $0 \pmod 3, 1 \pmod 3$ and $2 \pmod 3$. $n=-1,0,1$ is a better choice because $(-1)^2 = 1^2 = 1 $ and $0^2 = 0$ which are even smaller numbers than $2^2$. $\endgroup$ Toby Mak – Toby Mak 2019-09-24 02:02:47 +00:00 Commented Sep 24, 2019 at 2:02 Add a comment \| 1 $\begingroup$ The formulas for the sum of consecutive squares up to $n$ is $\frac{n(n+1)(2n+1)}{6}$. So, the sum of $k$ consecutive squares starting with $n$ is $\displaystyle \frac{ (n+k-1)(n+k)(2n + 2k -1) }{6} - \frac{ (n-1)n(2n-1)}{6}$ (I had an off-by-one error so I solved for 4 through 7 instead of 3 through 6.) For $k = 4$ we get $2(2n^2 + 6n + 7 $. For this expression to be a square, the right hand side must be of the form $2x^2$. But $2n^2 + 6n + 7$is always odd. For $k = 5$ we get$5( n^2 + 4n + 6 )$. Can the right hand-side be divisible by 5? Working modulo 5, the right-hand side gives $n^2 - n + 1$ which doesnt have a root. (You can verify that by checking $n= 0$through $n=4$.) So the original quadratic term cant be divisible by 5, and the sum of squares is not a square. For $k = 6$ we get $6n^2 + 30n + 55$. Theres no obvious modulus to try here, but 4 is usually a convenient one, because squares mod 4 have values 0 or 1. The polynomial is always equal to 3 mod 4, so it cant be a square. For $k = 7$ we get $7(n^2 + 6n + 13)$. The same trick as $k = 5$ works here. Plug in the values $n = 0$ through $n = 6$ into the polynomial, computing mod 7. The polynomial never has value 0, so it cant be divisible by 7. I hope this answer your questions and feel free to reach out of you have any doubt. thanks. Share answered Jul 25, 2021 at 14:24 user868051user868051 6977 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ Toby Mak's hint settles the question but I would like to rephrase the answer without using congruences to make it comprehensible to those not-so-much into math, as this result comes up on the first page in Google search. In order to show that $$ (n - 1)^2 + n^2 + (n + 1)^2 \neq a^2 $$ $$ n, a \in \mathbb N $$ We will try to perform reductio ad absurdum by assuming that there is $a \in \mathbb N$ that satisfies the equation: $$ (n - 1)^2 + n^2 + (n + 1)^2 = a^2 $$ We expand the left-hand side of the equation and thus get: $$ 3n^2 + 2 = a^2 $$ Notice that the left-hand side of the equation divided by $3$ gives a remainder of $2$. That implies a question - what is the remainder of the right-hand side when divided by $3$? We've got three cases to consider $$ a = 3k + 0 \lor a = 3k + 1 \lor a = 3k + 2, k \in \mathbb N $$ It's clear that if $a = 3k$ then the remainder of $a^2$ is $0$. If $a = 3k + 1$ , then $a^2 = 3(3k^2 + 2k) + 1$ and thus the remainder is $1$. If $a = 3k + 2$ , then $a^2 = 3(3k^2 + 4k + 1) + 1$, so the remainder is again $1$. Hence, we arrived to a contradiction as it is imposible for a number which divided by 3 gives the remainder of $2$ to be equal to a number which divided by $3$ gives a remainder of $0$ or $1$. Q.E.D. Share edited May 9, 2020 at 4:25 answered May 9, 2020 at 4:11 SkyfallSkyfall 4511 silver badge77 bronze badges $\endgroup$ 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 modular-arithmetic square-numbers See similar questions with these tags. 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8797	https://jn.nutrition.org/article/S0022-3166(22)02441-5/fulltext	Effects of Eating Frequency, Snacking, and Breakfast Skipping on Energy Regulation: Symposium Overview1,2 - The Journal of Nutrition Skip to Main ContentSkip to Main Menu Login to your account Email/Username Your email address is a required field. E.g., j.smith@mail.com Password Show Your password is a required field. Forgot password? [x] Remember me Don’t have an account? Create a Free Account If you don't remember your password, you can reset it by entering your email address and clicking the Reset Password button. 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Ok Full length articleVolume 141, Issue 1p144-147 January 2011 Open Archive Download Full Issue Download started Ok Effects of Eating Frequency, Snacking, and Breakfast Skipping on Energy Regulation: Symposium Overview1,2 Megan A.McCrory Megan A.McCrory Correspondence To whom correspondence should be addressed. E-mail: mmccror@purdue.edu mmccror@purdue.edu Affiliations Department of Foods and Nutrition, Purdue University, West Lafayette IN 47907 Department of Psychological Sciences, Purdue University, West Lafayette IN 47907 Ingestive Behavior Research Center, Purdue University, West Lafayette IN 47907 Search for articles by this author 3,4,5mmccror@purdue.edu ∙ Wayne W.Campbell Wayne W.Campbell Affiliations Department of Foods and Nutrition, Purdue University, West Lafayette IN 47907 Ingestive Behavior Research Center, Purdue University, West Lafayette IN 47907 Search for articles by this author 3,5 Affiliations & Notes Article Info 3 Department of Foods and Nutrition, Purdue University, West Lafayette IN 47907 4 Department of Psychological Sciences, Purdue University, West Lafayette IN 47907 5 Ingestive Behavior Research Center, Purdue University, West Lafayette IN 47907 Footnotes: 1 Published as a supplement to The Journal of Nutrition. Presented as part of the symposium entitled “Eating Patterns and Energy Balance: A Look at Eating Frequency, Snacking, and Breakfast Omission” given at the Experimental Biology 2009 meeting, April 19, 2009, in New Orleans, LA. This symposium was sponsored by the American Society for Nutrition Energy and Macronutrient Metabolism RIS. The symposium was chaired by Megan A.McCrory and Wayne W. Campbell. Guest Editor for this symposium publication was Anna Maria Siega-Riz. Guest Editor disclosure: No conflicts to disclose. 2 Author disclosures: M. A. McCrory and W. W. Campbell, no conflicts of interest. DOI: 10.3945/jn.109.114918 External LinkAlso available on ScienceDirect External Link Copyright: © 2011 American Society for Nutrition. Published by Elsevier Inc. User License: Elsevier user license \| Elsevier's open access license policy Download PDF Download PDF Outline Outline Abstract Eating frequency Snacking Breakfast skipping Integration of findings to date and suggestions for future research Acknowledgments Literature Cited Article metrics Related Articles Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley bluesky Add to my reading list More More Download PDF Download PDF Cite Share Share Share on Email X Facebook LinkedIn Sina Weibo Add to Mendeley Bluesky Add to my reading list Set Alert Get Rights Reprints Download Full Issue Download started Ok Previous articleNext article Show Outline Hide Outline Abstract Eating frequency Snacking Breakfast skipping Integration of findings to date and suggestions for future research Acknowledgments Literature Cited Article metrics Related Articles Abstract The ASN hosted a symposium entitled “Eating Patterns and Energy Balance: A Look at Eating Frequency, Snacking, and Breakfast Omission” at the Experimental Biology 2009 annual meeting on April 19, 2009, in New Orleans, LA. The symposium was chaired by Megan McCrory and co-chaired by Wayne Campbell, both from Purdue University. The goal of the symposium was to bring together experts to provide an overview of research on the potential role of eating patterns in the development of overweight and obesity. Studies on eating frequency, snacking, and breakfast skipping were highlighted. In particular, evidence both for and against their roles were discussed, methodological issues that underlie controversies were addressed, and suggested future directions for research were outlined. Appetite regulation and hormonal effects were also reviewed. Megan McCrory introduced the session then discussed studies on eating frequency and energy regulation in free-living adults consuming self-selected diets. Heather Leidy summarized the state of the research on eating frequency and energy regulation in adults from controlled feedings studies. Didier Chapelot discussed various usages of “snack” and argued for a physiological basis to distinguish snacks from meals. Mark Pereira presented information on the effects of breakfast skipping and the macronutrient composition of breakfast in energy regulation and mood. A panel discussion/question and answer session ended the symposium. The symposium was videotaped and can be viewed at www.nutrition.org. Excess body weight is a major public health problem affecting over 1.5 billion people worldwide (1 1. James, WP The epidemiology of obesity: the size of the problem. J Intern Med. 2008; 263:336-352 Crossref Scopus (555) PubMed Google Scholar ). Fundamentally, excess body weight results from long-term positive energy balance, i.e. when energy intake persistently exceeds energy expenditure. Multifarious factors impact a person's ability to successfully achieve and maintain energy balance consistent with a healthy body weight. Despite many scientific advances over the years about the role of dietary intake in energy balance, many questions remain about how people should eat to prevent weight gain or to permanently maintain weight loss. For example, controversy exists regarding the relative importance of the timing of eating events for weight control. The complexity of ingestive behavior makes it extremely difficult to isolate and determine the contribution of a given behavioral aspect of eating to the problem of overeating and obesity. However, we think critical evaluations and discussions of defined ingestive behavior-related topics are important to advance our understanding and approaches used to helping people successfully control their body weight. For this symposium, we chose to focus on research related to the timing and regularity of food intake and their potential roles in human energy balance. Experts involved in studying eating frequency, snacking, and breakfast skipping reviewed the literature to date and highlighted important barriers to progress in the field such as methodological shortcomings or incongruence among studies in definitions of terms. It was our hope that this symposium would bring together interested parties to interact and provide a stimulus for new collaborations and directions in research on eating patterns and energy regulation. Herein, we provide a brief overview of the topics covered by each speaker in the symposium. Please consult the 4 accompanying papers for expanded lists of bibliographic citations. Eating frequency Megan McCrory started the symposium by providing an overview of studies on eating frequency and energy regulation in free-living adults consuming self-selected diets (2 2. McCrory, MA ∙ Howarth, NC ∙ Roberts, SB ... Eating frequency and energy regulation in free-living subjects consuming self-selected diets. J Nutr. 2011; 141:148-153 Full Text Full Text (PDF) Scopus (67) PubMed Google Scholar ). As early as the 1960s, Fabry et al. (3 3. Fabry, P ∙ Fodor, J ∙ Hejl, Z ... The frequency of meals: its relation to overweight, hypercholesterolaemia, and decreased glucose-tolerance. Lancet. 1964; 2:614-615 Crossref Scopus (202) PubMed Google Scholar ) observed an inverse relationship between eating frequency, defined as the number of eating occasions per day, and the prevalence of excess adiposity and increasing risk factors for chronic disease such as poor glycemic control and high blood cholesterol. Since then, the majority of cross-sectional studies on eating frequency and adiposity have shown similar results. However, a key problem contributing to most of those studies is that many individuals underreport their energy intake relative to their actual energy intake (or energy need for weight maintenance) (4 4. Schoeller, DA Limitations in the assessment of dietary energy intake by self-report. Metabolism. 1995; 44:18-22 Abstract Full Text (PDF) Scopus (382) PubMed Google Scholar ). First shown by Bellisle et al. (5 5. Bellisle, F ∙ McDevitt, R ∙ Prentice, AM Meal frequency and energy balance. Br J Nutr. 1997; 77:S57-S70 Crossref PubMed Google Scholar ) in 1997, the greater the magnitude of underreporting of energy intake, the less frequently individuals also report eating. It follows that the apparent inverse relation between eating frequency and adiposity may be an artifact and in large part can be attributed to the underreporting of eating frequency concomitant with the underreporting of energy intake. The impact of energy intake underreporting on observed relationships between eating frequency and BMI is illustrated further by showing that when implausible energy intake reporters are identified objectively with strict standards (6 6. Huang, TT ∙ Roberts, SB ∙ Howarth, NC ... Effect of screening out implausible energy intake reports on relationships between diet and BMI. Obes Res. 2005; 13:1205-1217 Crossref Scopus (237) PubMed Google Scholar , 7 7. McCrory, MA ∙ Hajduk, CL ∙ Roberts, SB Procedures for screening out inaccurate reports of dietary energy intake. Public Health Nutr. 2002; 5:873-882 Crossref Scopus (172) PubMed Google Scholar ) and excluded from the analysis, the relation between eating frequency and BMI becomes positive (6 6. Huang, TT ∙ Roberts, SB ∙ Howarth, NC ... Effect of screening out implausible energy intake reports on relationships between diet and BMI. Obes Res. 2005; 13:1205-1217 Crossref Scopus (237) PubMed Google Scholar ,8 8. Huang, TT ∙ Howarth, NC ∙ Lin, BH ... Energy intake and meal portions: associations with BMI percentile in U.S. children. Obes Res. 2004; 12:1875-1885 Crossref Scopus (171) PubMed Google Scholar ). It is important to note that besides eating frequency, underreporting of other aspects of a person's diet might affect total energy intake. Huang et al. (6 6. Huang, TT ∙ Roberts, SB ∙ Howarth, NC ... Effect of screening out implausible energy intake reports on relationships between diet and BMI. Obes Res. 2005; 13:1205-1217 Crossref Scopus (237) PubMed Google Scholar ,8 8. Huang, TT ∙ Howarth, NC ∙ Lin, BH ... Energy intake and meal portions: associations with BMI percentile in U.S. children. Obes Res. 2004; 12:1875-1885 Crossref Scopus (171) PubMed Google Scholar ) provided evidence that eating frequency, portion consumed, and energy density were each underreported. Randomized intervention studies with individuals who consumed self-selected diets using high vs. low eating frequencies generally do not support the notion that eating frequency affects body weight. This conclusion should be viewed with caution, because the self-reporting of dietary intake contributes to uncertainty about adherence to the dietary goals of studies in which different eating frequencies are behaviorally prescribed. Overall, based on studies in free-living individuals, there is a lack of evidence to support a role for eating frequency in energy balance primarily due to self-reporting bias of energy intake and weaknesses in study design. Next, Heather Leidy discussed eating frequency and energy regulation in controlled feeding studies (9 9. Leidy, HJ ∙ Harris, CT ∙ Campbell, WW The effect of eating frequency on appetite control and food intake: brief synopsis of controlled feeding studies. J Nutr. 2011; 141:154-157 Full Text Full Text (PDF) Scopus (111) PubMed Google Scholar ). She examined studies conducted during the past 15 y in which food was provided either in a laboratory environment or for consumption at home and reviewed outcomes of energy expenditure, appetite control, and hormonal profiles relating to the regulation of food intake. Those controlled studies showed no effect of eating frequency on energy expenditure, but skipping a meal reduced fat oxidation. Studies examining nibbling (small, frequent meals) compared to gorging (large, infrequent meals) under isoenergetic conditions over a range of meal frequencies from 2 to 12 meals/d provided conflicting evidence, but over a narrower range suggest there may be some tendency for a 6-meals/d pattern to improve appetite control relative to a 3-meals/d pattern. Furthermore, reducing eating frequency to <3 meals/d may impair appetite control. It is, however, important to consider that any practical recommendation toward a higher eating frequency for better weight control stemming from these studies also needs to include some advice about limiting the energy intake at each eating occasion; otherwise, the advice to simply increase eating frequency could lead to excess energy intake and weight gain. As Leidy et al. (9 9. Leidy, HJ ∙ Harris, CT ∙ Campbell, WW The effect of eating frequency on appetite control and food intake: brief synopsis of controlled feeding studies. J Nutr. 2011; 141:154-157 Full Text Full Text (PDF) Scopus (111) PubMed Google Scholar ) point out, it is probably premature to make any recommendations about a particular eating frequency for weight control at this point, due to several factors, including the short-term nature of most of the studies conducted to date, their relatively small sample sizes, and the many different design approaches that have been used, which preclude definitive conclusions from being made. Overall, based on controlled feeding studies, there may be a small benefit to appetite of eating 6 meals/d relative to eating 3 meals/d, and eating <3 meals/d may negatively affect appetite. However, it is premature to make specific recommendations on eating frequency due to several methodological weaknesses. Snacking The effects of snacking on energy regulation were reviewed by Didier Chapelot (10 10. Chapelot, D. The role of snacking in energy balance: a biobehavioral approach. J Nutr. 2011; 2011:158-162 Full Text Full Text (PDF) Scopus (94) Google Scholar ). At the heart of his presentation was the fundamental problem of what defines a snack. Culturally, a snack is often defined as something consumed between meals, i.e. breakfast, lunch, and dinner. Certain types of foods, particularly those perceived as unhealthy due to low nutritional quality (e.g. micronutrient poor; high in fat and/or added sugars; or low in fiber, such as cookies and potato chips), are also referred to as snacks. But confusion lies in the fact that micronutrient-dense, high-fiber foods usually viewed in a more positive light, such as apples, can also be eaten as snacks. Well-controlled studies conducted in the laboratory (11–14 11. Chapelot, D ∙ Marmonier, C ∙ Aubert, R ... Consequence of omitting or adding a meal in man on body composition, food intake, and metabolism. Obesity (Silver Spring). 2006; 14:215-227 Crossref Scopus (56) PubMed Google Scholar 12. Marmonier, C ∙ Chapelot, D ∙ Louis-Sylvestre, J Metabolic and behavioral consequences of a snack consumed in a satiety state. Am J Clin Nutr. 1999; 70:854-866 Full Text Full Text (PDF) Scopus (50) PubMed Google Scholar 13. Marmonier, C ∙ Chapelot, D ∙ Louis-Sylvestre, J Effects of macronutrient content and energy density of snacks consumed in a satiety state on the onset of the next meal. Appetite. 2000; 34:161-168 Crossref Scopus (132) PubMed Google Scholar 14. Marmonier, C ∙ Chapelot, D ∙ Fantino, M ... Snacks consumed in a nonhungry state have poor satiating efficiency: influence of snack composition on substrate utilization and hunger. Am J Clin Nutr. 2002; 76:518-528 Full Text Full Text (PDF) Scopus (83) PubMed Google Scholar ) and those outside of the laboratory that account for implausible energy intake reporting (15 15. Howarth, NC ∙ Huang, TTK ∙ Roberts, SB ... Eating patterns and dietary composition in relation to BMI in younger and older adults. Int J Obes (Lond). 2007; 31:675-684 Crossref Scopus (187) PubMed Google Scholar ) provide evidence to suggest that regardless of snack macronutrient composition, consuming snacks can lead to overeating and, potentially, weight gain. The primary reason is a lack of compensation in later energy intake; i.e. less is not consumed at other eating times such as the next meal. But Chapelot's group has also shown that eating between what are usually considered the 3 main meals of the day may not always have that effect. Some individuals in French society eat what can be thought of as a small 4th meal consumed between traditional lunch and dinner times; it is called a goûter. Experiments in the laboratory showed that, in contrast to non-goûter eaters, goûter eaters did in fact compensate at the next meal after eating a goûter not only by eating later but also by eating less. Also, their blood glucose decreased just before they chose to eat the goûter. Non-goûter eaters who were provided with the same food, but were not hungry when they chose to eat the food, did not compensate at the next meal. Following the goûter, fatty acid oxidation was suppressed in the non-goûter eaters. Chapelot proposes a biologically based definition of a snack, which is eating during a period of satiety, rather than simply eating between meals. This biological definition supports the idea that eating at times other than breakfast, lunch, and dinner may not contribute to overeating and obesity provided one is hungry prior to eating and the eating episode results in a metabolic state that does not reduce fatty acid oxidation. However, in societies where obesity is prevalent, this type of eating between meals may be rare. Eating between meals in a non-hungry state is more common and based on preliminary experimental evidence should be discouraged, because it is likely to promote excess weight gain. Consistent with that idea are the findings of Howarth et al. (15 15. Howarth, NC ∙ Huang, TTK ∙ Roberts, SB ... Eating patterns and dietary composition in relation to BMI in younger and older adults. Int J Obes (Lond). 2007; 31:675-684 Crossref Scopus (187) PubMed Google Scholar ) from U.S. national survey data showing that eating 3 times/d was associated with a mean BMI value in the normal weight range, whereas eating >3 times/d was associated with mean BMI values in the overweight range. Taken together, both controlled feeding studies and free-living studies provide tentative evidence that snacking, or eating between main meals, especially in a non-hungry state, is detrimental to energy regulation and likely leads to weight gain. However, additional research is needed to confirm this suggestion. Breakfast skipping Lastly, Mark Pereira (16 16. Pereira, M ∙ Erickson, E ∙ McKee, P ... Breakfast frequency and quality may affect glycemia and appetite in adults and children. J Nutr. 2011; 141:163-168 Full Text Full Text (PDF) Scopus (132) PubMed Google Scholar ) reviewed studies on breakfast skipping in both adults and children. Data from cross-sectional studies show consistently that skipping breakfast is associated with excess body weight and markers of insulin resistance. However, as discussed by McCrory (2 2. McCrory, MA ∙ Howarth, NC ∙ Roberts, SB ... Eating frequency and energy regulation in free-living subjects consuming self-selected diets. J Nutr. 2011; 141:148-153 Full Text Full Text (PDF) Scopus (67) PubMed Google Scholar ) in this symposium, it is possible that the inverse association could be due to underreporting of energy intake, because there is some evidence to suggest that both breakfast and snacks are underreported. Therefore, prospective and experimental studies on the role of breakfast skipping in obesity and chronic disease risk are needed. In addition, the nutritional quality of breakfast macronutrient composition needs to be considered. Breakfast meals higher in fiber and whole grains may assist with daily appetite control and prevent weight gain. In children and adults, pilot experimental studies conducted by Pereira's group on breakfast frequency and quality showed that breakfasts higher in whole grains and fiber had positive influences on appetite control, insulin resistance, and mood relative to breakfasts containing primarily refined grains or skipping breakfast (water condition). This research is currently being expanded on a larger scale and provides important preliminary evidence that a nutritious breakfast meal is a key factor in promoting healthy body weight, chronic disease risk reduction, and positive mental health. Integration of findings to date and suggestions for future research It is clear that examining any one aspect of eating patterns, e.g. eating frequency, snacking, or breakfast skipping, is difficult to do in isolation, because in the real world they are much intertwined. However, it is important to try to disentangle any separate effects on energy regulation, so that more specific dietary advice for weight management on when, how often, how much, and what to eat at each eating occasion can be given. Cross-sectional studies accounting for reporting plausibility as presented by McCrory et al. (2 2. McCrory, MA ∙ Howarth, NC ∙ Roberts, SB ... Eating frequency and energy regulation in free-living subjects consuming self-selected diets. J Nutr. 2011; 141:148-153 Full Text Full Text (PDF) Scopus (67) PubMed Google Scholar ) together with several study design results discussed by Chapelot (10 10. Chapelot, D. The role of snacking in energy balance: a biobehavioral approach. J Nutr. 2011; 2011:158-162 Full Text Full Text (PDF) Scopus (94) Google Scholar ) and Pereira et al. (16 16. Pereira, M ∙ Erickson, E ∙ McKee, P ... Breakfast frequency and quality may affect glycemia and appetite in adults and children. J Nutr. 2011; 141:163-168 Full Text Full Text (PDF) Scopus (132) PubMed Google Scholar ) tentatively support that snacks and breakfasts may be key eating occasions important for energy regulation. As explained by Leidy et al. (9 9. Leidy, HJ ∙ Harris, CT ∙ Campbell, WW The effect of eating frequency on appetite control and food intake: brief synopsis of controlled feeding studies. J Nutr. 2011; 141:154-157 Full Text Full Text (PDF) Scopus (111) PubMed Google Scholar ), it is difficult to make definitive conclusions about the relative importance of eating frequency per se to energy regulation and obesity based on the controlled feeding trials conducted to date. However, there is some suggestion that eating somewhere between 3 and 6 times/d may be preferred for energy regulation and weight control, as well as most practical to sustain. Based on the totality of evidence presented in this symposium, we propose a theoretical model (Fig. 1) that integrates the findings from studies on eating frequency, snacking, and breakfast skipping and includes skipping of any meal to take into account eating frequencies of less than the traditional 3 meals/d. The model also helps explain the apparent contrast in results from studies differing in design, from feeding trials in which energy intake is controlled (in the real world, “planned”), to observational studies that do or do not for plausibility of energy intake reporting (in the real world, “unplanned”), to longitudinal and experimental studies in which diets are self-selected and therefore energy intake is not controlled and may or may not exceed energy requirements (also “unplanned” in the real world). This model depicts that an eating frequency in the range of 3–6 times/d is preferred provided that energy intake at each eating occasion is monitored so that total daily energy intake does not exceed energy requirements. It also suggests that when energy intake at each eating occasion is not monitored, there is a risk for appetite dysregulation and hence weight gain, particularly at eating frequencies above 6 times/d. Figure viewer FIGURE 1 Theoretical model of the interactions of eating frequency and appetite regulation consumed under controlled and ad libitum feeding conditions. Based on studies to date, an eating frequency in the range of 3–6 times/d is associated with successful appetite regulation provided that energy intake at each eating occasion is monitored so that total daily energy intake does not exceed energy requirements. When energy intake at each eating occasion is not monitored, there is a risk for appetite dysregulation and hence weight gain, particularly at eating frequencies above 6 times/d. The model also describes how snacking, the French goûter, and meal skipping align with different eating frequencies. More carefully controlled, laboratory-based studies need to be conducted, and from those results the most promising eating patterns for weight control should be further tested for effectiveness in behavioral interventions. Potential interactions with factors such as dietary composition, food form, nutritional quality, and portion served also must be tested. Importantly, questions on the standardization of definitions of eating patterns, such as what is considered breakfast skipping, and whether snacking on all types of food (e.g. an apple vs. a candy bar) affects energy intake, still remain. Additionally, it is unknown whether different eating patterns have different effects during attempted weight loss compared to weight maintenance or in different groups of individuals such as dietary restrained (weight conscious) compared to unrestrained individuals. Finally, compliance with dietary recommendations in behavioral interventions or dietary manipulations in feeding studies must be carefully monitored to have the confidence in the individual's fidelity to the eating patterns being tested and hence the results of the study. Without these, outcomes of dietary studies will likely be uncertain. Acknowledgments We thank Didier Chapelot, Heather Leidy, and Mark Pereira for comments on an earlier draft of this manuscript. M.A.M. wrote the manuscript and conceptualized the figure. W.W.C. commented on the manuscript and contributed to the conceptualization of the figure. Both authors read and approved the final version of the paper. Literature Cited 1. James, WP The epidemiology of obesity: the size of the problem. J Intern Med. 2008; 263:336-352 Crossref Scopus (555) PubMed Google Scholar 2. McCrory, MA ∙ Howarth, NC ∙ Roberts, SB ... Eating frequency and energy regulation in free-living subjects consuming self-selected diets. J Nutr. 2011; 141:148-153 Full Text Full Text (PDF) Scopus (67) PubMed Google Scholar 3. Fabry, P ∙ Fodor, J ∙ Hejl, Z ... The frequency of meals: its relation to overweight, hypercholesterolaemia, and decreased glucose-tolerance. Lancet. 1964; 2:614-615 Crossref Scopus (202) PubMed Google Scholar 4. Schoeller, DA Limitations in the assessment of dietary energy intake by self-report. Metabolism. 1995; 44:18-22 Abstract Full Text (PDF) Scopus (382) PubMed Google Scholar 5. Bellisle, F ∙ McDevitt, R ∙ Prentice, AM Meal frequency and energy balance. Br J Nutr. 1997; 77:S57-S70 Crossref PubMed Google Scholar 6. Huang, TT ∙ Roberts, SB ∙ Howarth, NC ... Effect of screening out implausible energy intake reports on relationships between diet and BMI. Obes Res. 2005; 13:1205-1217 Crossref Scopus (237) PubMed Google Scholar 7. McCrory, MA ∙ Hajduk, CL ∙ Roberts, SB Procedures for screening out inaccurate reports of dietary energy intake. Public Health Nutr. 2002; 5:873-882 Crossref Scopus (172) PubMed Google Scholar 8. Huang, TT ∙ Howarth, NC ∙ Lin, BH ... Energy intake and meal portions: associations with BMI percentile in U.S. children. Obes Res. 2004; 12:1875-1885 Crossref Scopus (171) PubMed Google Scholar 9. Leidy, HJ ∙ Harris, CT ∙ Campbell, WW The effect of eating frequency on appetite control and food intake: brief synopsis of controlled feeding studies. J Nutr. 2011; 141:154-157 Full Text Full Text (PDF) Scopus (111) PubMed Google Scholar 10. Chapelot, D. The role of snacking in energy balance: a biobehavioral approach. J Nutr. 2011; 2011:158-162 Full Text Full Text (PDF) Scopus (94) Google Scholar 11. Chapelot, D ∙ Marmonier, C ∙ Aubert, R ... Consequence of omitting or adding a meal in man on body composition, food intake, and metabolism. Obesity (Silver Spring). 2006; 14:215-227 Crossref Scopus (56) PubMed Google Scholar 12. Marmonier, C ∙ Chapelot, D ∙ Louis-Sylvestre, J Metabolic and behavioral consequences of a snack consumed in a satiety state. Am J Clin Nutr. 1999; 70:854-866 Full Text Full Text (PDF) Scopus (50) PubMed Google Scholar 13. Marmonier, C ∙ Chapelot, D ∙ Louis-Sylvestre, J Effects of macronutrient content and energy density of snacks consumed in a satiety state on the onset of the next meal. Appetite. 2000; 34:161-168 Crossref Scopus (132) PubMed Google Scholar 14. Marmonier, C ∙ Chapelot, D ∙ Fantino, M ... Snacks consumed in a nonhungry state have poor satiating efficiency: influence of snack composition on substrate utilization and hunger. Am J Clin Nutr. 2002; 76:518-528 Full Text Full Text (PDF) Scopus (83) PubMed Google Scholar 15. Howarth, NC ∙ Huang, TTK ∙ Roberts, SB ... Eating patterns and dietary composition in relation to BMI in younger and older adults. Int J Obes (Lond). 2007; 31:675-684 Crossref Scopus (187) PubMed Google Scholar 16. Pereira, M ∙ Erickson, E ∙ McKee, P ... Breakfast frequency and quality may affect glycemia and appetite in adults and children. J Nutr. 2011; 141:163-168 Full Text Full Text (PDF) Scopus (132) PubMed Google Scholar Figures (1)Figure Viewer Article metrics Related Articles Open in viewer Effects of Eating Frequency, Snacking, and Breakfast Skipping on Energy Regulation: Symposium Overview1,2 Hide CaptionDownloadSee figure in Article Toggle Thumbstrip FIGURE 1 Download .PPT Go to Go to Show all references Expand All Collapse Expand Table Authors Info & Affiliations Home Access for Developing Countries Articles & Issues Articles In Press Current Issue List of Issues Editor's Choice Archive Journal Information Advertising Information Aims & Scope Editorial Team Contact Information For Authors Researcher Academy Rights & Permissions Submit a Manuscript Society Information American Society for Nutrition ASN Journals Advances in Nutrition The American Journal of Clinical Nutrition Current Developments in Nutrition The content on this site is intended for healthcare professionals. 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8798	https://www.omnicalculator.com/statistics/residual	Residual Calculator We have prepared this residual calculator for you to calculate the residuals for the linear regression analysis. Residual is one of the most important metrics used to assess the accuracy of your linear regression analysis. It tells you the performance of the linear regression and how accurate it is. If everything is still unclear for you, scroll down now to understand what linear regression and residuals are, how to find residuals, and how to calculate the sum of squares residuals in statistics. Furthermore, you will find some practical examples to help you understand the concept better. We will also explain the application of the residual graph. What is linear regression? Linear regression is a statistical approach that attempts to explain the relationship between 2 variables. It can be shown as: y = a × x + b y = a × x + b where y is the dependent variable, whereas x is the independent variable. Linear regression aims to explain the relationship between y and x. Specifically, it models the change in y for any changes in x. y x y x y x We define a as the slope. It controls the change in y per unit change in x. The second parameter b is the intercept and it is the value of y when x equals zero. a y x b y x Linear regression is a very powerful tool as it can help you to predict the "future". For example, we can use linear regression to predict future stock prices. Let's say we model the stock price of Company Alpha using the following model: stock price stock price = 1.5 × GDP growth + 20 stock price = 1.5 × GDP growth + 20 If the expected GDP growth of the following year is 10%, stock price of Company Alpha is: 10% stock price 1.5 × 10 + 20 = $35 1.5 × 10 + 20 = $35 However, it is important that you understand not all relationships are linear. If your data can't be explained by using just a straight line, you might want to try out other regression methods. Please visit our quadratic regression calculatorand exponential regression calculator. What is residual? – The residual definition Let's say you have now modeled a linear relationship between y and x using linear regression. The next vital step to take is to estimate the accuracy of your linear model. And this is where the calculation of the residual comes in. So, how to find the residual? y x The residual definition is the difference between the observed value and the predicted value of a certain point in the model. If the observed value is larger than the predicted value, the residual is positive. If the predicted value is larger than the observed value, the residual is negative. The further away the residual is from zero, the less accurate the model is in predicting that particular point. However, to assess the performance of the whole linear model, we need to sum all the residuals up. This is when we need to calculate the sum of squared residuals to prevent the positive value from being offset by the negative residuals. Theory aside, let's dive into how to calculate the residuals in statistics to help you understand the process now. How to calculate residual in statistics? – The residual formula As we mentioned previously, residual is the difference between the observed value and the predicted value at one point. We can calculate the residual as: e = y − ŷ where: For instance, say we have a linear model of y = 2 × x + 2. One of the actual data points we have is (2, 7), which means that when x equals 2, the observed value is 7. However, according to the model, the ŷ, the predicted value, is 2 × 2 + 2 = 6. Hence, according to the equation above, the residual, e, is 7 − 6 = 1. To assess the whole linear model, determining the residual of a single data point is not enough since you will probably have many data points. So, now we need to sum up all the individual residuals. And to capture both the positive and negative deviations, we will need to take the sum of e² instead of e. A square e² will turn all the negative residuals into positive ones. The sum of squares residuals calculation can be done using the following equation: Σ(e²) = e₁² + e₂² + e₃² + … + en² So, if the model of y = 2 × x + 2 has 3 data points of (1, 4), (2, 7) and (3, 5); the predicted values of each point will be: And the individual residuals will be: So, we can calculate the sum of squares residuals as: Σ(e²) = 0² + 1² + (-3)² = 0 + 1 + 9 = 10 How to use the residual plot or residual graph? Now, let's take some time to talk about what a residual plot is after we have discussed the residual meaning and the residual formula. A residual graph is a plot of the residuals calculated against the predicted value, i.e., the residuals will be on the y-axis, and the predicted value will be the x-axis. So, why do we need to plot the residual graph? The primary usage of the residual plot is to assess if a linear model is a good model for the data. By definition, the residuals in the linear model should be random. So, if the residuals in the residual plot look totally random, you have got yourself a good model. On the other hand, if the residuals on the plot seem to follow a certain pattern, it might mean that a linear model is not suitable for your data, and you should consider other models, such as the quadratic model instead. FAQ What is the sum of squares residuals? The sum of squares residuals is one of the metrics used to analyze the accuracy of your linear model. The larger the sum of squares residuals, the less accurate your model is. Why do you need to use sum of squares residuals? The main reason we need to use the sum of squares residuals instead of the sum of residuals is that the negative residuals and positive residuals might offset each other. This would make the linear model more accurate than it is. Can we explain every relationship with linear regression? Mathematically speaking, yes, you can. However, it might not be wise to do that. Some relationship is not linear, and fitting a linear model to it might lead to poor results. What is a residual plot? A residual plot is a graph plotted with residuals on the y-axis and predicted value on the x-axis. It allows you to assess if the linear model is a good fit. Bonferroni correction Christmas tree Outlier Significant figures
8799	https://ag-weitze-schmithusen.github.io/ModularGroup/doc/manual.pdf	The ModularGroup Package Finite-index subgroups of (P)SL2(Z) Version 2.0.0 14 July 2022 Luca L. Junk Gabriela Weitze-Schmithüsen Luca L. Junk Email: junk@math.uni-sb.de Homepage: Address: Saarland University Department of Mathematics Postfach 15 11 50 66041 Saarbrücken Germany Gabriela Weitze-Schmithüsen Email: weitze@math.uni-sb.de Homepage: Address: AG Weitze-Schmithüsen FR 6.1 Mathematik Universität des Saarlandes D-66041 Saarbrücken The ModularGroup Package 2 Copyright © 2022 by Luca L. Junk and Gabriela Weitze-Schmithüsen Acknowledgements Supported by Project I.8 of SFB-TRR 195 ’Symbolic Tools in Mathematics and their Application’ of the German Research Foundation (DFG). Contents 1 Introduction 4 1.1 General aims of the ModularGroup package . . . . . . . . . . . . . . . . . . . . . 4 1.2 Technicalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2 Subgroups of SL2(Z) 5 2.1 Construction of modular subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.2 Computing with modular subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3 Subgroups of PSL2(Z) 17 3.1 Construction of projective modular subgroups . . . . . . . . . . . . . . . . . . . . . 17 3.2 Computing with projective modular subgroups . . . . . . . . . . . . . . . . . . . . . 18 References 25 Index 26 3 Chapter 1 Introduction 1.1 General aims of the ModularGroup package This GAP package provides methods for computing with finite-index subgroups of the modular groups SL2(Z) and PSL2(Z). This includes, but is not limited to, computation of the generalized level, index or cusp widths. It also implements algorithms described in [Hsu96] and [HL14] for testing if a given group is a congruence subgroup. Hence it differs from the Congruence package [DJKV18], which can be used - among other things - to construct canonical congruence subgroups of SL2(Z). 1.2 Technicalities A convenient way to represent finite-index subgroups of SL2(Z) is by specifying the action of genera-tor matrices of SL2(Z) on the right cosets by right multiplication. For example, one could choose the generators S = 0 −1 1 0 T = 1 1 0 1 and represent a subgroup as a tuple of transitive permutations (σS,σT) describing the action of S and T. This is exactly the way this package internally treats such subgroups. We use the convention that 1 corresponds to the coset of the identity matrix. Note that such a representation as a tuple of permutations is only unique up to relabelling of the cosets, i.e. up to simultaneous conjugation (fixing the 1 coset by our convention). 4 Chapter 2 Subgroups of SL2(Z) For representing finite-index subgroups of SL2(Z), this package introduces the new object ModularSubgroup. As stated in the introduction, a ModularSubgroup essentially consists of the two permutations σS and σT describing the coset graph with respect to the generators S and T (with the convention that 1 corresponds to the identity coset). So explicitly specifying these permutations is the canonical way to construct a ModularSubgroup. Though you might not always have a coset graph of your subgroup at hand, but rather a list of gener-ating matrices. Therefore we implement multiple constructors for ModularSubgroup: three that take as input two permutations describing the coset graph with respect to different pairs of generators of SL2(Z), and one that takes a list of SL2(Z) matrices as generators. 2.1 Construction of modular subgroups 2.1.1 Constructors ▷ModularSubgroup(s, t) (operation) Returns: A modular subgroup. Constructs a ModularSubgroup object corresponding to the finite-index subgroup of SL2(Z) de-scribed by the permutations s and t. This constructor tests if the given permutations actually describe the coset action of the matrices S = 0 −1 1 0 , T = 1 1 0 1 by checking that they act transitively and satisfy the relations s4 = (s3t)3 = s2ts−2t−1 = 1 Upon creation, the cosets are renamed in a standardized way to make the internal interaction with existing GAP methods easier. (The fact that 1 corresponds to the identity coset is not changed by this) Example gap> G := ModularSubgroup( > (1,2)(3,4)(5,6)(7,8)(9,10), > (1,4)(2,5,9,10,8)(3,7,6)); 5 The ModularGroup Package 6 ▷ModularSubgroupST(s, t) (operation) Returns: A modular subgroup. Synonymous for ModularSubgroup (see above). ▷ModularSubgroupRT(r, t) (operation) Returns: A modular subgroup. Constructs a ModularSubgroup object corresponding to the finite-index subgroup of SL2(Z) de-termined by the permutations r and t which describe the action of the matrices R = 1 0 1 1 T = 1 1 0 1 on the right cosets. A check is performed if the permutations actually describe such an action on the cosets of some subgroup. Upon creation, the cosets are renamed in a standardized way to make the internal interaction with existing GAP methods easier. (The fact that 1 corresponds to the identity coset is not changed by this) Example gap> G := ModularSubgroupRT( > (1,9,8,10,7)(2,6)(3,4,5), > (1,4)(2,5,9,10,8)(3,7,6)); ▷ModularSubgrouSJ(s, j) (operation) Returns: A modular subgroup. Constructs a ModularSubgroup object corresponding to the finite-index subgroup of SL2(Z) de-termined by the permutations s and j which describe the action of the matrices S = 0 −1 1 0 J = 0 1 −1 1 on the right cosets. A check is performed if the permutations actually describe such an action on the cosets of some subgroup. Upon creation, the cosets are renamed in a standardized way to make the internal interaction with existing GAP methods easier. (The fact that 1 corresponds to the identity coset is not changed by this) Example gap> G := ModularSubgroupSJ( > (1,2)(3,6)(4,7)(5,9)(8,10), > (1,5,6)(2,3,7)(4,9,10)); The ModularGroup Package 7 ▷ModularSubgroup(gens) (operation) Returns: A modular subgroup. Constructs a ModularSubgroup object corresponding to the finite-index subgroup of SL2(Z) gen-erated by the matrices in gens. No test is performed to check if the generated subgroup actually has finite index! This constructor implicitly computes a coset table of the subgroup. Hence it might be slow for very large index subgroups. Example gap> G := ModularSubgroup([ > , > , > > ]); 2.1.2 Getters for the coset action ▷SAction(G) (operation) Returns: A permutation. Returns the permutation σS describing the action of the matrix S on the cosets of G. ▷TAction(G) (operation) Returns: A permutation. Returns the permutation σT describing the action of the matrix T on the cosets of G. ▷RAction(G) (operation) Returns: A permutation. Returns the permutation σR describing the action of the matrix R on the cosets of G. ▷JAction(G) (operation) Returns: A permutation. Returns the permutation σJ describing the action of the matrix J on the cosets of G. ▷CosetActionOf(A, G) (operation) Returns: A permutation. Returns the permutation σA describing the action of the matrix A ∈SL2(Z) on the cosets of G. 2.2 Computing with modular subgroups 2.2.1 Index (IndexSL2Z) ▷Index(G) (attribute) Returns: A natural number. For a given modular subgroup G this method returns its index in SL2(Z). As G is internally stored as permutations (s,t) this is just LargestMovedPoint(s,t) (or 1 if the permutations are trivial). The ModularGroup Package 8 Example gap> G := ModularSubgroup((1,2)(3,5)(4,6), (1,3)(2,4)(5,6)); gap> Index(G); 6 2.2.2 GeneralizedLevel (GeneralizedLevelSL2Z) ▷GeneralizedLevel(G) (attribute) Returns: A natural number. This method calculates the general Wohlfahrt level (i.e. the lowest common multiple of all cusp widths) of G as defined in [Woh64]. Example gap> G := ModularSubgroup((1,2)(3,5)(4,6), (1,3)(2,4)(5,6)); gap> GeneralizedLevel(G); 2 2.2.3 RightCosetRepresentatives (RightCosetRepresentativesSL2Z) ▷RightCosetRepresentatives(G) (attribute) Returns: A list of words. This function returns a list of representatives of the (right) cosets of G as words in S and T. Example gap> G := ModularSubgroup((1,2),(2,3)); gap> RightCosetRepresentatives(G); [ , S, ST ] 2.2.4 GeneratorsOfGroup (GeneratorsOfGroupSL2Z) ▷GeneratorsOfGroup(G) (attribute) Returns: A list of words. Calculates a list of generators (as words in S and T) of G. This list might include redundant generators (or even duplicates). Example gap> G := ModularSubgroup((1,2)(3,5)(4,6), (1,3)(2,4)(5,6)); gap> GeneratorsOfGroup(G); [ S^-2, T^-2, ST^-2S^-1 ] The ModularGroup Package 9 2.2.5 MatrixGeneratorsOfGroup ▷MatrixGeneratorsOfGroup(G) (attribute) Returns: A list of matrices. Calculates a list of generator matrices of G. This list might include redundant generators (or even duplicates). Example gap> G := ModularSubgroup((1,2)(3,5)(4,6), (1,3)(2,4)(5,6)); gap> MatrixGeneratorsOfGroup(G); [ [ [ -1, 0 ], [ 0, -1 ] ], [ [ 1, -2 ], [ 0, 1 ] ], [ [ 1, 0 ], [ 2, 1 ] ] ] 2.2.6 IsCongruence (IsCongruenceSL2Z) ▷IsCongruence(G) (attribute) Returns: True or false. This method test whether a given modular subgroup G is a congruence subgroup. It is essentially an implementation of an algorithm described in [HL14]. Example gap> G := ModularSubgroup([ > , > > ]); gap> IsCongruence(G); true 2.2.7 Cusps (CuspsSL2Z) ▷Cusps(G) (attribute) Returns: A list of rational numbers and infinity. This method computes a list of inequivalent cusp representatives with respect to G. Example gap> G := ModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> Cusps(G); [ infinity, 0, 1, 2, 3/2, 5/3 ] The ModularGroup Package 10 2.2.8 CuspWidth (CuspWidthSL2Z) ▷CuspWidth(c, G) (operation) Returns: A natural number. This method takes as input a cusp c (a rational number or infinity) and a modular group G and calculates the width of this cusp with respect to G. Example gap> G := ModularSubgroup( > (1,2,6,3)(4,11,15,12)(5,13,16,14)(7,17,9,18)(8,19,10,20)(21,24,22,23), > (1,4,5)(2,7,8)(3,9,10)(6,15,16)(11,20,21)(12,19,22)(13,23,17)(14,24,18) > ); gap> CuspWidth(-1, G); 3 gap> CuspWidth(infinity, G); 3 2.2.9 CuspsEquivalent (CuspsEquivalentSL2Z) ▷CuspsEquivalent(p, q, G) (operation) Returns: True or false. Takes two cusps p and q and a modular subgroup G and checks if they are equivalent modulo G, i.e. if there exists a matrix A ∈G with Ap = q. Example gap> G := ModularSubgroup( > (1,2,6,3)(4,11,15,12)(5,13,16,14)(7,17,9,18)(8,19,10,20)(21,24,22,23), > (1,4,5)(2,7,8)(3,9,10)(6,15,16)(11,20,21)(12,19,22)(13,23,17)(14,24,18) > ); gap> CuspsEquivalent(infinity, 1, G); false gap> CuspsEquivalent(-1, 1/2, G); true 2.2.10 CosetRepresentativeOfCusp (CosetRepresentativeOfCuspSL2Z) ▷CosetRepresentativeOfCusp(c, G) (operation) Returns: A word in S and T. For a cusp c this function returns a right coset representative A of G such that A∞and c are equivalent with respect to G. Example gap> G := ModularSubgroup( > (1,2,6,3)(4,11,15,12)(5,13,16,14)(7,17,9,18)(8,19,10,20)(21,24,22,23), > (1,4,5)(2,7,8)(3,9,10)(6,15,16)(11,20,21)(12,19,22)(13,23,17)(14,24,18) > ); The ModularGroup Package 11 gap> CosetRepresentativeOfCusp(4, G); TS 2.2.11 IndexModN (IndexModNSL2Z) ▷IndexModN(G, N) (operation) Returns: A natural number. For a modular subgroup G and a natural number N this method calculates the index of the projec-tion ¯ G of G in SL2(Z/NZ). Example gap> G := ModularSubgroup((1,2)(3,5)(4,6), (1,3)(2,4)(5,6)); gap> IndexModN(G, 2); 6 2.2.12 Deficiency (DeficiencySL2Z) ▷Deficiency(G, N) (operation) Returns: A natural number. For a modular subgroup G and a natural number N this method calculates the so-called deficiency of G from being a congruence subgroup of level N. The deficiency of a finite-index subgroup Γ of SL2(Z) was introduced in [WS15]. It is defined as the index [Γ(N):Γ(N)∩Γ] where Γ(N) is the principal congruence subgroup of level N. Example gap> G := ModularSubgroup([ > , > > ]); gap> Deficiency(G, 2); 2 gap> Deficiency(G, 4); 1 2.2.13 Deficiency (DeficiencySL2ZAttr) ▷Deficiency(G) (attribute) Returns: A natural number. Shorthand for Deficiency(G, GeneralizedLevel(G)). Example gap> G := ModularSubgroup([ The ModularGroup Package 12 > , > > ]); gap> Deficiency(G); 2 gap> Deficiency(G, GeneralizedLevel(G)); 2 2.2.14 Projection ▷Projection(G) (operation) Returns: A projective modular subgroup. For a given modular subgroup G this function calculates its image ¯ G under the projection π:SL2(Z) →PSL2(Z). Example gap> G := ModularSubgroup([ > , > > ]); gap> Projection(G); 2.2.15 Conjugate (ConjugateSL2Z) ▷Conjugate(G, A) (operation) Returns: A ModularSubgroup. Conjugates the group G by A and returns the group A−1 ∗G∗A. 2.2.16 NormalCore (NormalCoreSL2Z) ▷NormalCore(G) (attribute) Returns: A modular subgroup. Calculates the normal core of G in SL2(Z), i.e. the maximal subgroup of G that is normal in SL2(Z). Example gap> G := ModularSubgroup([ > , > > ]); gap> NormalCore(G); The ModularGroup Package 13 2.2.17 QuotientByNormalCore (QuotientByNormalCoreSL2Z) ▷QuotientByNormalCore(G) (attribute) Returns: A finite group. Calculates the quotient of SL2(Z) by the normal core of G. Example gap> G := ModularSubgroup([ > , > > ]); gap> QuotientByNormalCore(G); 2.2.18 AssociatedCharacterTable (AssociatedCharacterTableSL2Z) ▷AssociatedCharacterTable(G) (attribute) Returns: A character table. Returns the character table of SL2(Z)/N where N is the normal core of G. Example gap> G := ModularSubgroup([ > , > > ]); gap> AssociatedCharacterTable(G); CharacterTable( ) 2.2.19 IsElementOf (IsElementOfSL2Z) ▷IsElementOf(A, G) (operation) Returns: True or false. This function checks if a given matrix A is an element of the modular subgroup G. Example gap> G := ModularSubgroup([ > , > > ]); gap> IsElementOf(, G); false gap> IsElementOf(, G); true The ModularGroup Package 14 2.2.20 Genus (GenusSL2Z) ▷Genus(G) (attribute) Returns: A non-negative integer. Computes the genus of the quotient G\H via an algorithm described in [Sch04]. Example gap> G := ModularSubgroup((1,2),(2,3)); gap> Genus(G); 0 2.3 Miscellaneous The following functions are mostly helper functions used internally and are only documented for sake of completeness. 2.3.1 DefinesCosetActionST ▷DefinesCosetActionST(s, t) (operation) Returns: True or false. Checks if two given permutations s and t describe the action of the generator matrices S and T on the cosets of some subgroup. This is the case if they satisfy the relations s4 = (s3t)3 = s2ts−2t−1 = 1 and act transitively. Example gap> s := (1,2)(3,4)(5,6)(7,8)(9,10);; gap> t := (1,4)(2,5,9,10,8)(3,7,6);; gap> DefinesCosetActionST(s,t); true 2.3.2 DefinesCosetActionRT ▷DefinesCosetActionRT(r, t) (operation) Returns: True or false. Checks if two given permutations r and t describe the action of the generator matrices R and T on the cosets of some subgroup. This is the case if they satisfy the relations (rt−1r)4 = ((rt−1r)3t)3 = (rt−1r)2t(rt−1r)−2t−1 = 1 and act transitively. The ModularGroup Package 15 Example gap> r := (1,9,8,10,7)(2,6)(3,4,5);; gap> t := (1,4)(2,5,9,10,8)(3,7,6);; gap> DefinesCosetActionRT(r,t); true 2.3.3 DefinesCosetActionSJ ▷DefinesCosetActionSJ(s, j) (operation) Returns: True or false. Checks if two given permutations s and j describe the action of the generator matrices S and J on the cosets of some subgroup. This is the case if they satisfy the relations s4 = (s3 j−1s−1)3 = s2 j−1s−2 j = 1 and act transitively. Example gap> s := (1,2)(3,4)(5,6)(7,8)(9,10);; gap> j := (1,5,6)(2,3,7)(4,9,10);; gap> DefinesCosetActionSJ(s,j); true 2.3.4 CosetActionFromGenerators ▷CosetActionFromGenerators(gens) (operation) Returns: A tuple of permutations. Takes a list of generator matrices and calculates the coset graph (as two permutations σS and σT) of the generated subgroup of SL2(Z). Example gap> CosetActionFromGenerators([ > , > > ]); [ (1,2,5,3)(4,8,10,9)(6,11,7,12), (1,4)(2,6)(3,7)(5,10)(8,12,9,11) ] 2.3.5 STDecomposition ▷STDecomposition(A) (operation) Returns: A word in S and T. Takes a matrix A ∈SL2(Z) and decomposes it into a word in the generator matrices S and T. Example gap> M := [ [ 4, 3 ], [ -3, -2 ] ];; The ModularGroup Package 16 gap> STDecomposition(M); S^2T^-1S^-1T^2S^-1T^-1S^-1 2.3.6 RTDecomposition ▷RTDecomposition(A) (operation) Returns: A word in R and T. Takes a matrix A ∈SL2(Z) and decomposes it into a word in the generator matrices R and T. Example gap> M := [ [ 4, 3 ], [ -3, -2 ] ];; gap> RTDecomposition(M); (RT^-1R)^2T^-1R^-1(TR^-1T)^2R^-1T^-1R^-1TR^-1 2.3.7 SJDecomposition ▷SJDecomposition(A) (operation) Returns: A word in S and J. Takes a matrix A ∈SL2(Z) and decomposes it into a word in the generator matrices S and J. Example gap> M := [ [ 4, 3 ], [ -3, -2 ] ];; gap> SJDecomposition(M); S^3J(S^-1J^-1)^2S^-1JS^-1 2.3.8 STDecompositionAsList ▷STDecompositionAsList(A) (operation) Returns: A list representing a word in S and T. Takes a matrix A ∈SL2(Z) and decomposes it into a word in the generator matrices S and T. The word is represented as a list in the format [[generator, exponent], ... ] Example gap> M := [ [ 4, 3 ], [ -3, -2 ] ];; gap> STDecompositionAsList(M); [ [ "S", 2 ], [ "T", -1 ], [ "S", -1 ], [ "T", 2 ], [ "S", -1 ], [ "T", -1 ], [ "S", -1 ], [ "T", 0 ] ] Chapter 3 Subgroups of PSL2(Z) Analogous to finite-index subgroups of SL2(Z), we define a new type ProjectiveModularSubgroup for representing subgroups of PSL2(Z). It consists essentially of two permutations σS and σT describ-ing the action of S and T on the cosets of the given subgroup, where S and T are the images of S and T in PSL2(Z). The methods implemented for PSL2(Z) subgroups are mostly the same as for SL2(Z) subgroups and behave more or less identically. Nevertheless we list them here. 3.1 Construction of projective modular subgroups 3.1.1 Constructors ▷ProjectiveModularSubgroup(s, t) (operation) Returns: A projective modular subgroup. Constructs a ProjectiveModularSubgroup object corresponding to the finite-index subgroup of PSL2(Z) described by the permutations s and t. This constructor tests if the given permutations actually describe the coset action of S and T on some subgroup by checking that they act transitively and satisfy the relations s2 = (st)3 = 1 Upon creation, the cosets are renamed in a standardized way to make the internal interaction with extisting GAP methods easier. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,4)(5,6)(7,8)(9,10), > (1,4)(2,5,9,10,8)(3,7,6)); If you want to construct a ProjectiveModularSubgroup from a list of generators, you can lift each generator to a matrix in SL2(Z), construct from these a ModularSubgroup, and then project it to PSL2(Z) via Projection. 17 The ModularGroup Package 18 3.1.2 Getters for the coset action ▷SAction(G) (operation) Returns: A permutation. Returns the permutation σS describing the action of S on the cosets of G. ▷TAction(G) (operation) Returns: A permutation. Returns the permutation σT describing the action of T on the cosets of G. 3.2 Computing with projective modular subgroups 3.2.1 Index (IndexPSL2Z) ▷Index(G) (attribute) Returns: A natural number. For a given projective modular subgroup G this method returns its index in PSL2(Z). As G is internally stored as permutations (s,t) this is just LargestMovedPoint(s,t) (or 1 if the permutations are trivial). Example gap> G := ProjectiveModularSubgroup((1,2),(2,3)); gap> Index(G); 3 3.2.2 GeneralizedLevel (GeneralizedLevelPSL2Z) ▷GeneralizedLevel(G) (attribute) Returns: A natural number. This method calculates the general Wohlfahrt level (i.e. the lowest common multiple of all cusp widths) of G as defined in [Woh64]. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> GeneralizedLevel(G); 12 The ModularGroup Package 19 3.2.3 RightCosetRepresentatives (RightCosetRepresentativesPSL2Z) ▷RightCosetRepresentatives(G) (attribute) Returns: A list of words. This function returns a list of representatives of the (right) cosets of G as words in S and T. Example gap> G := ProjectiveModularSubgroup((1,2),(2,3)); gap> RightCosetRepresentatives(G); [ , S, ST ] 3.2.4 GeneratorsOfGroup (GeneratorsOfGroupPSL2Z) ▷GeneratorsOfGroup(G) (attribute) Returns: A list of words. Calculates a list of generators (as words in S and T) of G. This list might include redundant generators. Example gap> G := ProjectiveModularSubgroup((1,2)(3,5)(4,6), (1,3)(2,4)(5,6)); gap> GeneratorsOfGroup(G); [ T^-2, ST^-2S^-1 ] 3.2.5 IsCongruence (IsCongruencePSL2Z) ▷IsCongruence(G) (attribute) Returns: True or false. This method test whether a given modular subgroup G is a congruence subgroup. It is essentially an implementation of an algorithm described in [Hsu96]. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,5)(4,6), > (1,3)(2,4)(5,6) > ); gap> IsCongruence(G); true 3.2.6 Cusps (CuspsPSL2Z) ▷Cusps(G) (attribute) Returns: A list of rational numbers and infinity. This method computes a list of inequivalent cusp representatives with respect to G. The ModularGroup Package 20 Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> Cusps(G); [ infinity, 0, 1, 2, 3/2, 5/3 ] 3.2.7 CuspWidth (CuspWidthPSL2Z) ▷CuspWidth(c, G) (operation) Returns: A natural number. This method takes as input a cusp c (a rational number or infinity) and a modular group G and calculates the width of this cusp with respect to G. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,7)(4,8)(5,9)(6,10)(11,12), > (1,3,4)(2,5,6)(7,10,11)(8,12,9) > ); gap> CuspWidth(-1, G); 3 gap> CuspWidth(infinity, G); 3 3.2.8 CuspsEquivalent (CuspsEquivalentPSL2Z) ▷CuspsEquivalent(p, q, G) (operation) Returns: True or false. Takes two cusps p and q and a projective modular subgroup G and checks if they are equivalent modulo G, i.e. if there exists A ∈G with Ap = q. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,7)(4,8)(5,9)(6,10)(11,12), > (1,3,4)(2,5,6)(7,10,11)(8,12,9) > ); gap> CuspsEquivalent(infinity, 1, G); false gap> CuspsEquivalent(-1, 1/2, G); true The ModularGroup Package 21 3.2.9 CosetRepresentativeOfCusp (CosetRepresentativeOfCuspPSL2Z) ▷CosetRepresentativeOfCusp(c, G) (operation) Returns: A word in S and T. For a cusp c this function returns a right coset representative A of G such that A∞and c are equivalent with respect to G. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,7)(4,8)(5,9)(6,10)(11,12), > (1,3,4)(2,5,6)(7,10,11)(8,12,9) > ); gap> CosetRepresentativeOfCusp(4, G); TS 3.2.10 LiftToSL2ZEven ▷LiftToSL2ZEven(G) (operation) Returns: A modular subgroup. Lifts a given subgroup G of PSL2(Z) to an even subgroup of SL2(Z), i.e. a group that contains −1 and whose projection to PSL2(Z) is G. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,7)(4,8)(5,9)(6,10)(11,12), > (1,3,4)(2,5,6)(7,10,11)(8,12,9) > ); gap> LiftToSL2ZOdd(G); 3.2.11 LiftToSL2ZOdd ▷LiftToSL2ZOdd(G) (operation) Returns: A modular subgroup. Lifts a given subgroup G of PSL2(Z) to an odd subgroup of SL2(Z), i.e. a group that does not contain −1 and whose projection to PSL2(Z) is G. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,7)(4,8)(5,9)(6,10)(11,12), > (1,3,4)(2,5,6)(7,10,11)(8,12,9) > ); gap> LiftToSL2ZOdd(G); The ModularGroup Package 22 3.2.12 IndexModN (IndexModNPSL2Z) ▷IndexModN(G, N) (operation) Returns: A natural number. For a projective modular subgroup G and a natural number N this method calculates the index of the projection ¯ G of G in PSL2(Z/NZ). Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> IndexModN(G, 2); 6 3.2.13 Deficiency (DeficiencyPSL2Z) ▷Deficiency(G, N) (operation) Returns: A natural number. For a projective modular subgroup G and a natural number N this method calculates the so-called deficiency of G from being a congruence subgroup of level N. The deficiency of a finite-index subgroup Γ of PSL2(Z) was introduced in [WS15]. It is defined as the index [Γ(N):Γ(N)∩Γ] where Γ(N) is the principal congruence subgroup of level N. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> Deficiency(G, 4); 4 3.2.14 Deficiency (DeficiencyPSL2ZAttr) ▷Deficiency(G) (attribute) Returns: A natural number. Shorthand for Deficiency(G, GeneralizedLevel(G)). Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> Deficiency(G); 4 The ModularGroup Package 23 gap> Deficiency(G, GeneralizedLevel(G)); 4 3.2.15 Conjugate (ConjugatePSL2Z) ▷Conjugate(G, A) (operation) Returns: A ProjectiveModularSubgroup. Conjugates the group G by (the redidue class in PSL2(Z) of) A and returns the group A−1 ∗G∗A. 3.2.16 NormalCore (NormalCorePSL2Z) ▷NormalCore(G) (attribute) Returns: A projective modular subgroup. Calculates the normal core of G in PSL2(Z), i.e. the maximal subgroup of G that is normal in PSL2(Z). Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> NormalCore(G); 3.2.17 QuotientByNormalCore (QuotientByNormalCorePSL2Z) ▷QuotientByNormalCore(G) (attribute) Returns: A finite group. Calculates the quotient of PSL2(Z) by the normal core of G. Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> QuotientByNormalCore(G); 3.2.18 AssociatedCharacterTable (AssociatedCharacterTablePSL2Z) ▷AssociatedCharacterTable(G) (attribute) Returns: A character table. Returns the character table of PSL2(Z)/N where N is the normal core of G. The ModularGroup Package 24 Example gap> G := ProjectiveModularSubgroup( > (1,2)(3,6)(4,8)(5,9)(7,11)(10,13)(12,15)(14,17)(16,19)(18,21)(20,23)(22,24), > (1,3,7,4)(2,5)(6,9,8,12,14,10)(11,13,16,20,18,15)(17,21,22,19)(23,24) > ); gap> AssociatedCharacterTable(G); CharacterTable( ) 3.2.19 IsElementOf (IsElementOfPSL2Z) ▷IsElementOf(A, G) (operation) Returns: True or false. This function checks if the image of a given matrix A in PSL2(Z) is an element of the group G. Example gap> G := ProjectiveModularSubgroup((1,2),(2,3)); gap> IsElementOf(, G); true gap> IsElementOf(, G); false 3.2.20 Genus (GenusPSL2Z) ▷Genus(G) (attribute) Returns: A non-negative integer. Computes the genus of the quotient G\H via an algorithm described in [Sch04]. Example gap> G := ProjectiveModularSubgroup((1,2),(2,3)); gap> Genus(G); 0 References [DJKV18] Ann Dooms, Eric Jespers, Alexander Konovalov, and Helena Verrill. Congruence, con-gruence subgroups of SL(2, Z), Version 1.2.2. Congruence, Feb 2018. GAP package. 4 [HL14] Thomas Hamilton and David Loeffler. Congruence testing for odd subgroups of the mod-ular group. LMS Journal of Computation and Mathematics, 17(1):206–208, 2014. 4, 9 [Hsu96] Tim Hsu. Identifying congruence subgroups of the modular group. Proceedings of the American Mathematical Society, 124(5), April 1996. 4, 19 [Sch04] Gabriela Schmithuesen. An algorithm for finding the Veech group of an origami. Experi-mental Mathematics, 13(4):459–472, 2004. 14, 24 [Woh64] Klaus Wohlfahrt. An extension of F. Klein’s level concept. Illinois J. Math., 8(3):529–535, 09 1964. 8, 18 [WS15] Gabriela Weitze-Schmithuesen. The deficiency of being a congruence group for Veech groups of origamis. International Mathematics Research Notices, 2015(6):1613–1637, 2015. 11, 22 25 Index AssociatedCharacterTable AssociatedCharacterTablePSL2Z, 23 AssociatedCharacterTableSL2Z, 13 Conjugate ConjugatePSL2Z, 23 ConjugateSL2Z, 12 CosetActionFromGenerators, 15 CosetActionOf, 7 CosetRepresentativeOfCusp CosetRepresentativeOfCuspPSL2Z, 21 CosetRepresentativeOfCuspSL2Z, 10 Cusps CuspsPSL2Z, 19 CuspsSL2Z, 9 CuspsEquivalent CuspsEquivalentPSL2Z, 20 CuspsEquivalentSL2Z, 10 CuspWidth CuspWidthPSL2Z, 20 CuspWidthSL2Z, 10 Deficiency DeficiencyPSL2Z, 22 DeficiencyPSL2ZAttr, 22 DeficiencySL2Z, 11 DeficiencySL2ZAttr, 11 DefinesCosetActionRT, 14 DefinesCosetActionSJ, 15 DefinesCosetActionST, 14 GeneralizedLevel GeneralizedLevelPSL2Z, 18 GeneralizedLevelSL2Z, 8 GeneratorsOfGroup GeneratorsOfGroupPSL2Z, 19 GeneratorsOfGroupSL2Z, 8 Genus GenusPSL2Z, 24 GenusSL2Z, 14 Index IndexPSL2Z, 18 IndexSL2Z, 7 IndexModN IndexModNPSL2Z, 22 IndexModNSL2Z, 11 IsCongruence IsCongruencePSL2Z, 19 IsCongruenceSL2Z, 9 IsElementOf IsElementOfPSL2Z, 24 IsElementOfSL2Z, 13 JAction, 7 LiftToSL2ZEven, 21 LiftToSL2ZOdd, 21 MatrixGeneratorsOfGroup, 9 ModularSubgroup, 5 ModularSubgroupGens, 7 ModularSubgroupRT, 6 ModularSubgroupST, 6 ModularSubgrouSJ, 6 NormalCore NormalCorePSL2Z, 23 NormalCoreSL2Z, 12 Projection, 12 ProjectiveModularSubgroup, 17 QuotientByNormalCore QuotientByNormalCorePSL2Z, 23 QuotientByNormalCoreSL2Z, 13 RAction, 7 RightCosetRepresentatives RightCosetRepresentativesPSL2Z, 19 RightCosetRepresentativesSL2Z, 8 RTDecomposition, 16 26 The ModularGroup Package 27 SAction SActionPSL2Z, 18 SActionSL2Z, 7 SJDecomposition, 16 STDecomposition, 15 STDecompositionAsList, 16 TAction TActionPSL2Z, 18 TActionSL2Z, 7

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