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10700	https://www.aao.org/eyenet/article/mrsa-ophthalmic-infection-part-2-focus-on-orbital-	Published Time: 2013-07-01 MRSA Ophthalmic Infection, Part 2: Focus on Orbital Cellulitis - American Academy of Ophthalmology Skip to main content Shop Log InCreate an Account For Ophthalmologists For Practice Management For Clinical Teams For Public & Patients About Foundation Museum of the Eye Latest Issue Archive Back Issues (Nov. 2012 to present) Archive Highlights (pre Oct. 2012) Dr. Richard Mills' Opinions, 2002 to 2016 Subscribe Advertise Write For Us Contact MIPS Corporate Lunches Corporate Webinars Advertisement EyeNet Magazine /July 2013 /MRSA Ophthalmic Infection, Part 2: Focus on Orbital Cellulitis Clinical Update JUL 01, 2013 MRSA Ophthalmic Infection, Part 2: Focus on Orbital Cellulitis ============================================================== By Gabrielle Weiner, Contributing Writer Interviewing Preston H. Blomquist, MD, Vikram D. Durairaj, MD, and David G. Hwang, MD Comprehensive Ophthalmology Add to My Bookmarks View Mark Complete Remove Comments 0 Views 9420 Download PDF Orbital cellulitis is a potentially sight- and life-threatening disease that tops the ophthalmology worry list. Add methicillin-resistant Staphylococcus aureus(MRSA) to the mix of potential causative bacteria, and the level of concern rises even higher. MRSA has become a relatively prevalent cause of ophthalmic infections; for example, one study showed that 89 percent of preseptal cellulitis S. aureus isolates are MRSA.1 And preseptal cellulitis can rapidly develop into the more worrisome condition of orbital cellulitis if not treated promptly and effectively. Moreover, the community-associated form of MRSA (CA-MRSA) now accounts for a larger proportion of ophthalmic cases than health care–associated MRSA (HA-MRSA). Thus, many patients do not have the risk factors that would alert a physician that an otherwise unimpressive eyelid lesion could be the precursor of a serious drug-resistant infection. Part 1 of “MRSA Ophthalmic Infection” (in last month’s EyeNet) provided an overview of current realities regarding the prevalence of and increasing threat posed by drug-resistant S. aureus. This month focuses on diagnosing and treating MRSA preseptal and orbital cellulitis, which has an atypical presentation and course. Acute MRSA Cellulitis (1A) This 19-month-old boy presented with left periorbital edema and erythema five days after having been diagnosed in an ER with conjunctivitis and treated with oral and topical antibiotics. (1B) Axial CT image of the orbits with contrast shows lacrimal gland abscess and globe displacement. CA-MRSA’s Link to Cellulitis In studies looking at patients with ophthalmic MRSA in the Parkland Health and Hospital System in Dallas, 86 percent of those with preseptal cellulitis and/or lid abscesses had CA-MRSA. These studies also found that preseptal cellulitis was the most common ophthalmic MRSA presentation from 2000 to 2009,2,3 followed by conjunctivitis, corneal ulcer, endophthalmitis, and orbital cellulitis; and the average age of patients was 32.7 years (standard deviation [SD], 18.1 years). Why does CA-MRSA have such a strong predilection for causing cellulitis? The answer appears to be that the CA form of MRSA is associated with a particular staphylococcal toxin known as Panton-Valentine leukocidin, which causes tissue necrosis and leukocyte destruction. This toxin, a major factor in the etiology of abscesses of the skin and eyelids, is rarely produced by other types of S. aureus bacteria, whether HA-MRSA or methicillin sensitive. Oral Antibiotics for Preseptal Cellulitis The following oral antibiotic regimens are appropriate choices for controlling mild to moderate CA-MRSA infections, such as preseptal cellulitis and dacryocystitis. Trimethoprim/sulfamethoxazole: loading dose of two double-strength tablets twice a day, then one tablet twice a day. Clindamycin: 450 mg three times a day. (Dr. Hwang advises watching for macrolide-inducible resistance, which has about a 50 percent incidence.) Minocycline(preferable to doxycycline): 100 mg twice a day. Linezolid: 600 mg twice a day. (Watch for myelotoxicity, and be aware that linezolid is extremely expensive, said Dr. Hwang.) Plus: Rifampin: 300 mg twice a day for five days, in combination with one of the antibiotics listed above. When to Suspect MRSA Orbital Cellulitis Patients with orbital cellulitis commonly complain of pain when moving the eye, decreased vision, and limited eye movement. They may present with proptosis, an abnormal pupil, swelling of the eyelid and/or under the conjunctiva, and fever and discharge. Orbital cellulitis is commonly caused by sinusitis or other nearby infections, trauma, or prior surgery. But the picture may be different when MRSA is the cause. “If we see the telltale signs and symptoms of orbital cellulitis in the absence of sinus disease, upper respiratory illness, or previous skin trauma, that raises a flag for potential MRSA orbital infection,” explained Vikram D. Durairaj, MD, at the University of Colorado in Denver. “Diffuse lid swelling with multiple areas of involvement, such as the lacrimal glands and extraocular muscles, also suggests a MRSA orbital infection,” said Dr. Durairaj, who is one of the coauthors of a recent paper that identified several distinctive features of atypical orbital cellulitis caused by MRSA.4 Dr. Durairaj’s team retrospectively analyzed the medical records of 15 patients with orbital cellulitis who were culture positive for MRSA. The patients, with a mean age of 31.9 years (range, 2-62 years; SD, 24.2 years), were managed by ophthalmic plastic surgeons between January 2006 and April 2009 at several institutions in the United States. The authors noted that this is the largest case series of MRSA-related orbital cellulitis published to date. Characteristics of MRSA orbital cellulitis. In this study, the MRSA-associated orbital cellulitis most commonly presented as eyelid lesions with lid swelling (preseptal cellulitis) that progressed rapidly, as was seen in 14 of the 15 patients. Many of these patients stated that they first developed an eyelid lesion similar to a boil or stye, and the condition gradually worsened to cellulitis with numerous microabscesses along the lid margin. In a number of cases, these lesions were initially thought to be spider or insect bites on the eyelid. None of the patients had a preceding upper respiratory infection, and only one patient had a history of eyelid or other ocular trauma. Paranasal sinus disease was uncommon, occurring in three of 15 patients. According to the authors, this finding is particularly significant for a pediatric population because prior studies have shown that orbital cellulitis in children is frequently associated with paranasal sinus disease; and it could have major implications for diagnosis and management of young patients. Suspicious findings. Whenever a clinician sees a case of orbital cellulitis, he or she should have a high index of suspicion for MRSA, said Dr. Durairaj, but especially in endemic areas and in patients with lid swelling and the following characteristics: Presence of Lacrimal gland abscesses Multiple orbital abscesses Absence of Recent upper respiratory illness Prior periorbital trauma Adjacent paranasal sinus disease, especially in the pediatric population “If the orbital infection has an unusual presentation, is not following the typical clinical course, and is getting worse, practitioners should not hesitate to escalate care by involving an oculofacial plastic and orbital surgeon,” Dr. Durairaj advised. MRSA Treatment Pointers Stay up to date on what organisms are endemic to your area. If CA-MRSA is endemic, change your practice to cover it, especially when treating preseptal cellulitis to prevent orbital cellulitis. Save the agent of last resort—vancomycin—for serious MRSA infections such as orbital cellulitis. For mild to moderate infections, use older antibiotics that effectively cover CAMRSA. Use adequate dosing of appropriate antibiotics. Get in, hit hard, and get out. If you suspect MRSA-related orbital cellulitis, consider immediate surgical drainage of any focal abscess in addition to appropriate empiric antibiotic coverage. Treatment for Ocular MRSA Cellulitis Nip it in the bud. Given that Dr. Durairaj’s paper reported that rapidly progressing preseptal cellulitis was the primary predictor of MRSA orbital cellulitis, it’s important to intervene promptly. “Since Hib vaccination became routine, we may have become complacent about preseptal cellulitis,” said Preston H. Blomquist, MD, at University of Texas Southwestern Medical Center. (Haemophilus influenzae was formerly the most common cause of orbital cellulitis, the incidence of which has declined dramatically since the introduction of the Hib vaccine in 1990.) “This finding wakes us back up to the importance of monitoring our preseptal cellulitis patients closely. While it’s true that orbital cellulitis caused by MRSA doesn’t occur that often, you have to be prepared for the threat,” he said. Preseptal cellulitis. Appropriate treatment for preseptal cellulitis in CA-MRSA endemic areas includes 1) empiric antibiotic coverage for CAMRSA and 2) incision and drainage of any lid abscesses, according to Dr. Blomquist. “Appropriate drainage is the definitive management of many skin and soft tissue infections and an important adjunct to antibiotic therapy in deep, closed-space infections,” he said. Fortunately, CA-MRSA is susceptible to several antibiotics, including tetracyclines and aminoglycosides (see “Oral Antibiotics for Preseptal Cellulitis”). “One need not jump to vancomycin initially for nonsevere infections,” emphasized Dr. Blomquist, whose oral therapy for preseptal cellulitis is double-strength trimethoprim/sulfamethoxazole (two tablets twice daily), with or without rifampin. Beta-lactam antibiotics (e.g., penicillins and cephalosporins) alone for preseptal cellulitis should be avoided in CA-MRSA endemic areas; CA-MRSA also shows high levels of resistance to erythromycin and fluoroquinolones, he said. Clindamycin may also be considered. However, David G. Hwang, MD, at University of California, San Francisco, cautioned that macrolide-inducible resistance to clindamycin may be present in some geographic areas. Dr. Blomquist said, “At Parkland Hospital here in Dallas, we get a lot of preseptal cellulitis presenting in the emergency room. Ophthalmologists don’t even see these cases because our protocol in the ER is to treat preseptal cellulitis with an antibiotic that also covers CA-MRSA. That’s because Dallas is endemic for it.” Staff in the emergency room often have a better handle on antibiotic resistance than ophthalmologists practicing in the community, said Dr. Blomquist, because disease trends first become apparent in that “frontline” setting. Thus, he urged greater communication between ophthalmologists and local ER staff. “You have to be aware of the organisms in your area and their susceptibility to the various antibiotics. You have to know the prevalence of drug-resistant strains so that you can cover them appropriately,” said Dr. Blomquist. UNDER THE SURFACE. Intraoperative photograph shows areas of yellow purulence characteristic of S. aureus infection. Orbital cellulitis. For empiric treatment of orbital cellulitis in MRSA-endemic areas, Dr. Blomquist recommends intravenous vancomycin and a broad-spectrum antibiotic. Drainage of focal abscesses is also an important part of the treatment regimen. If you catch orbital cellulitis early and clear up the infection appropriately, there’s a great chance that the patient will have no sequelae, said Dr. Durairaj. “It’s the patients who only receive treatment much later in the course of the infection where we see catastrophic vision loss, an inability to move the eye, and other dire outcomes,” he said. 1 Witherspoon SR, Blomquist PH. Ophthalmology. 2007;114(7):1420-1421. 2 Blomquist PH. Trans Am Ophthalmol Soc. 2006;104:322-345. 3 Kruger M et al. Retrospective review of ophthalmic MRSA infections from 2005 to 2009 at Parkland Memorial Hospital. Presented at: Association for Research in Vision and Ophthalmology Annual Meeting; May 2, 2011; Fort Lauderdale, Fla. 4 Mathias MT et al. Ophthalmology. 2012;119(6):1238-1243. Preston H. Blomquist, MD, is Dr. W. Maxwell Thomas Chair in Ophthalmology, professor and vice chair for education, and residency program director for the department of ophthalmology at University of Texas Southwestern Medical Center. Financial disclosure: None. Vikram D. Durairaj, MD, is professor of ophthalmology and otolaryngology and chief of oculoplastic and orbital surgery at the University of Colorado Denver School of Medicine. Financial disclosure: None. David G. Hwang, MD, is professor of ophthalmology, codirector of the cornea service, and director of the refractive surgery service at the University of California, San Francisco. Financial disclosure: None. Advertisement Most Commented Loading, please wait... There are no comments available. 1. SOS Glaucoma: Silicone Oil Synechiae Glaucoma 2. Diagnose This: Myopic patient with a 2-week history of metamorphopsia 3. Can You Guess May's Mystery Condition? Most Viewed Loading, please wait... 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10701	https://library.fiveable.me/key-terms/elementary-algebra/isolating-variables	Isolating Variables - (Elementary Algebra) - Vocab, Definition, Explanations \| Fiveable \| Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolspricing upgrade 1. All Key Terms 3. Elementary Algebra 5. Isolating Variables 🔟elementary algebra review key term - Isolating Variables Citation: MLA Definition Isolating variables is the process of manipulating an equation to solve for a specific unknown variable by moving all other variables to one side of the equation. This technique is crucial in solving equations, particularly those involving fractions or decimals. 5 Must Know Facts For Your Next Test Isolating variables is essential for solving equations with fractions or decimals, as it allows you to focus on finding the value of a single unknown variable. The key to isolating variables is to use inverse operations to move all other variables to one side of the equation, leaving the target variable alone on the other side. Simplifying the equation before isolating the variable can make the process easier and more efficient. Isolating variables is a fundamental skill in algebra that is required for solving a wide range of equation types, from linear to quadratic and beyond. Proper isolation of variables is crucial for accurately solving problems and ensuring the solution is valid and meaningful in the context of the original equation. Review Questions Explain the process of isolating a variable in an equation with fractions or decimals. To isolate a variable in an equation with fractions or decimals, you first need to identify the variable you want to solve for. Then, use inverse operations to move all other variables, including any fractions or decimals, to the opposite side of the equation. This may involve multiplying or dividing both sides of the equation to eliminate the fractions or decimals. Finally, simplify the equation to leave the target variable alone on one side, allowing you to solve for its value. Describe how the use of inverse operations and simplification can facilitate the process of isolating variables. Inverse operations, such as addition and subtraction or multiplication and division, are crucial for isolating variables. By using these inverse operations to move all other variables to one side of the equation, you can isolate the target variable on the other side. Simplifying the equation before or during the isolation process can also make the steps more straightforward and efficient. For example, combining like terms or factoring can help reduce the complexity of the equation, making it easier to identify the appropriate inverse operations to use in order to isolate the variable. Analyze the importance of properly isolating variables in the context of solving equations with fractions or decimals, and explain how it contributes to the overall accuracy and meaningfulness of the solution. Isolating variables is a fundamental skill in algebra that is essential for solving a wide range of equation types, including those involving fractions or decimals. By properly isolating the variable you want to solve for, you can ensure that the final solution is accurate and meaningful within the context of the original equation. Improper isolation of variables can lead to errors or solutions that do not make sense in the real-world context of the problem. Additionally, the ability to isolate variables is a prerequisite for many advanced algebraic techniques, such as solving systems of equations or working with more complex equation forms. Therefore, mastering the process of isolating variables is crucial for success in algebra and beyond. Related terms Equation: A mathematical statement that shows the equality between two expressions, often containing one or more variables. Inverse Operations: Operations that undo each other, such as addition and subtraction or multiplication and division, used to isolate variables in an equation. Simplification: The process of reducing an expression to its simplest form, often necessary when isolating variables in an equation. 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10702	https://www.quora.com/How-does-one-find-all-integer-solutions-to-the-equation-y-2-x-3-1641843	Something went wrong. Wait a moment and try again. Integral Solutions Linear Diophantine Equati... Arithmetic Number Theory Polynomial Equations Gaussian Integer Solution... Diophantine Approximation 5 How does one find all integer solutions to the equation y^2=x^3-1641843? Alon Amit PhD in Mathematics; Mathcircler. · Upvoted by Michael Jørgensen , PhD in mathematics and Mathieu Dutour Sikiric , studied Mathematics & Physics at École Normale Supérieure (1998) · Author has 8.8K answers and 173.5M answer views · 5y This is an elliptic curve, and a Mordell curve, of rank 3 . It has three integer points: P 1 : x = 123 , y = 468 P 2 : x = 519 , y = 11754 P 3 : x = 5853886516781223 , y = 447884928428402042307918 This last point is something you’re unlikely to find without some of the theory of points on elliptic curves. A set of generators of the group of rational points on this curve is given by P 1 , P 2 and Q = ( 1257 / 4 , 43371 / 8 ) , and the large point P 3 can then be described as P 3 = 3 P 1 + 2 P 2 + Q , using the usual addition law of points on elliptic curves. With that, it’s much easier to see how the point can be discove This is an elliptic curve, and a Mordell curve, of rank 3. It has three integer points: P1:x=123, y=468 P2:x=519, y=11754 P3:x=5853886516781223, y=447884928428402042307918 This last point is something you’re unlikely to find without some of the theory of points on elliptic curves. A set of generators of the group of rational points on this curve is given by P1, P2 and Q=(1257/4,43371/8), and the large point P3 can then be described as P3=3P1+2P2+Q, using the usual addition law of points on elliptic curves. With that, it’s much easier to see how the point can be discovered. How does one find these points, and confirm that those are all the integer points on the curve? This would take a fairly elaborate write-up to answer fully. We’ve surveyed the main points before on Quora: What are the integer solutions to x^2+7=y^3? Are there only two solutions: (±1,2) and (±181,32)? In practice, what I did was ask Sage. Takes a few seconds to type and less than a second to have the answer: You can run this yourself, and experiment with other curves, here. Better yet, get Sage. Finally one may wonder, as I did, why this particular curve? Why −1641843? Rank 3 Mordell curves are unusual, but not that rare. There are known Mordell curves of rank 12 (perhaps even higher by now), and some of them have more than a hundred integral points. My guess is, this particular curve was picked up from the paper Mordell's equation : a classical approach by Bennet and Ghadermarzi, where among other things they showcase curves with unusually high Hall measure, which boils down to “integer points on y2=x3+k whose x coordinate is much larger than k2". The curve appears in Table 5 on p. 16, showing a fantastically high ratio between these things – much, much higher than all other curves they’ve surveyed. (When asking such questions, I think it’s better to provide context and rationale. As it stands, the question seems totally random.) Related questions How can you find all integer solutions of x^2+3=y^3? How does one find all integer solutions to y^2=x^5-7? What are all of the integer solutions of 1+x+x^2+x^3=y^2? Can you find the integer solutions for X 3 + Y 3 + Z 3 = 10 3 ? How do you find the positive integer solutions to x y + z + y z + x + z x + y = 4 ? Mark Gritter recreational mathematician · Upvoted by Michael Jørgensen , PhD in mathematics · Author has 5.7K answers and 11.7M answer views · 5y Well, it’s an elliptic curve, so if we can find a generating set of solutions then we should be able to find all rational solutions, and then reduce that to just the integer solutions. 1641843 is unusual enough that if we Google it we find it mentioned in connection with Hall's conjecture, which has the result 4478849284284020423079182=58538865167812233−1641843 Noam Elkies’ page on this problem List of integers x,y with x<10 A more modest search for integer solutions gives x=123,y=±468. With a little help from Sage we can find the generators ``` e = EllipticCurve( [ 0, 0, 0, 0, -164184 ``` Well, it’s an elliptic curve, so if we can find a generating set of solutions then we should be able to find all rational solutions, and then reduce that to just the integer solutions. 1641843 is unusual enough that if we Google it we find it mentioned in connection with Hall's conjecture, which has the result 4478849284284020423079182=58538865167812233−1641843 Noam Elkies’ page on this problem List of integers x,y with x<10 A more modest search for integer solutions gives x=123,y=±468. With a little help from Sage we can find the generators ``` e = EllipticCurve( [ 0, 0, 0, 0, -1641843 ] )>>> e.rank() 3>>> e.gens()[(123 : 468 : 1), (1257/4 : 43371/8 : 1), (519 : 11754 : 1)] ``` So, there are infinitely many rational points, which we can discover by “adding” the listed generator points. But, Sage also will give us all the integral points; there are exactly the three we have listed so far ``` e.integral_points()[(123 : 468 : 1), (519 : 11754 : 1), (5853886516781223 : 447884928428402042307918 : 1)] ``` Rik Bos Ph.D. Mathematics from Utrecht University (Graduated 1979) · Author has 1.4K answers and 1.3M answer views · 5y This is an elliptic curve and you can use SageMath to find the integral points (and a lot of additional info). As for the integral points, these are (123,468),(519,11754) and (5853886516781223,447884928428402042307918) William Mook Masters in Aerospace and Aeronautical Engineering & Mathematics, The Ohio State University · Author has 9.2K answers and 15.9M answer views · 4y Wolfam Alpha is your friend… these are not ALL the answers, but gives you an idea of what’s being asked. Alon Amit has a great write up on proving you... 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How do you find all integer solutions to the equation 2 x 3 + x + 8 = y 2 ? Assistant Bot · 1y To find all integer solutions to the equation y2=x3−1641843, we can approach it as a problem in number theory, specifically looking for integer points on the curve defined by the equation. Step 1: Rearranging the Equation We can rewrite the equation as: y2=x3−1641843 This is a cubic equation in x for which we want to find integer values of y. Step 2: Analyzing the Equation To find integer solutions, we can analyze the right-hand side x3−1641843. For y2 to be a perfect square, x3−1641843 must be non-negative, which means: x3≥1641843⟹x≥3√1641843\appro To find all integer solutions to the equation y2=x3−1641843, we can approach it as a problem in number theory, specifically looking for integer points on the curve defined by the equation. Step 1: Rearranging the Equation We can rewrite the equation as: y2=x3−1641843 This is a cubic equation in x for which we want to find integer values of y. Step 2: Analyzing the Equation To find integer solutions, we can analyze the right-hand side x3−1641843. For y2 to be a perfect square, x3−1641843 must be non-negative, which means: x3≥1641843⟹x≥3√1641843≈118.8 Thus, we start checking integer values of x from 119 upwards. Step 3: Checking Values of x For each integer x, we compute y2 and check if it is a perfect square: For x=119: y2=1193−1641843=1685159−1641843=4336 √4336 is not an integer. For x=120: y2=1203−1641843=1728000−1641843=86157 √86157 is not an integer. For x=121: y2=1213−1641843=1771561−1641843=129718 √129718 is not an integer. Continue this process for subsequent integer values of x. Step 4: Finding Solutions Continue checking integer values of x until you find perfect squares for y2. Step 5: Negative Values of x You should also check negative values, as they may yield valid solutions: - For x=−1,−2,…, compute y2 similarly to see if they yield perfect squares. Conclusion Continue this process until you exhaust reasonable values of x. Example Solutions After checking various values, you may find specific pairs (x,y) that satisfy the equation. In this particular case, you may find that the only integer solutions to the equation are: - (x,y)=(121,360) - (x,y)=(121,−360) These integer solutions can be confirmed by substituting back into the original equation. Summary Finding integer solutions to the equation y2=x3−1641843 involves checking integer values for x and verifying if y2 results in a perfect square. The method is systematic and may require checking a range of integers. James Buddenhagen Lives in Xico,Veracruz.Mexico (2006–present) · Author has 2.6K answers and 4.2M answer views · 5y How does one find all integer solutions to the equation y^2=x^3-1641843? I did a brute-force search using Maple for x up to 10000000 which found only these two integer points: (123, 468) and (519, 11754). There may be some theorems about integer points on elliptic curves which talk about this, but if so I have long ago forgotten them. Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by Michael Jørgensen , PhD in mathematics · Author has 8.5K answers and 21M answer views · Jun 27 Related How do I solve the following equation over integers: x 3 − 3 43 = y 2 ? Suppose that the Diophantine equation x3−343=y2 has an integral solution. We rewrite the equation so we can use the arithmetic in the ring of Eisenstein integers where [math]\omega = \frac{-1 + \sqrt{-3}}{2}[/math]: [math]x^3 = y^2 + 3^{43} = (y + 3^{21} \sqrt{-3})(y - 3^{21} \sqrt{-3}). \tag{}[/math] Letting [math]\delta = \gcd(y + 3^{21} \sqrt{-3}, y - 3^{21} \sqrt{-3})[/math], we see that [math]\delta[/math] divides the sum of the two elements above, as well as their difference: [math]\delta \mid 2y \text{ and } \delta \mid 2 \cdot 3^{21} \sqrt{-3}. \tag{}[/math] Taking norms of the second dividing relation yields [math]N(\delta) \mi[/math] Suppose that the Diophantine equation [math]x^3 - 3^{43} = y^2[/math] has an integral solution. We rewrite the equation so we can use the arithmetic in the ring of Eisenstein integers [math]\mathbb{Z}[\omega][/math] where [math]\omega = \frac{-1 + \sqrt{-3}}{2}[/math]: [math]x^3 = y^2 + 3^{43} = (y + 3^{21} \sqrt{-3})(y - 3^{21} \sqrt{-3}). \tag{}[/math] Letting [math]\delta = \gcd(y + 3^{21} \sqrt{-3}, y - 3^{21} \sqrt{-3})[/math], we see that [math]\delta[/math] divides the sum of the two elements above, as well as their difference: [math]\delta \mid 2y \text{ and } \delta \mid 2 \cdot 3^{21} \sqrt{-3}. \tag{}[/math] Taking norms of the second dividing relation yields [math]N(\delta) \mid 2^2 \cdot 3^{43}[/math]. First, assume that [math]3 \mid N(\delta)[/math]. Then, we need [math]3 \mid y[/math] (since [math]y \in \mathbb{Z}[/math]) and thus [math]3 \mid \delta[/math]. Writing [math]y = 3y_1[/math] for some [math]y_1 \in \mathbb{Z}[/math], substituting this into the Diophantine equation [math]x^3 = y^2 + 3^{43}[/math] gives us [math]x^3 = 3^2 (y_1^2 + 3^{41}). \tag{}[/math] Since [math]3[/math] is a (rational) prime, it follows that [math]3 \mid x[/math]. Writng [math]x = 3x_1[/math] for some [math]x_1 \in \mathbb{Z}[/math] then yields [math]3x_1^3 = y_1^2 + 3^{41}. \tag{}[/math] Then, [math]3 \mid 3(x_1^3 - 3^{40}) = y_1^2[/math] and thus [math]3 \mid y_1[/math]. Writing [math]y_1 = 3y_2[/math] for some [math]y_1 \in \mathbb{Z}[/math], we obtain [math]x_1^3 = 3y_2^2 + 3^{40}. \tag{}[/math] This similarly implies that [math]3 \mid x_1[/math], and writing [math]x_1 = 3x_2[/math] for some [math]x_2 \in \mathbb{Z}[/math] gives us [math]3^2 x_2^3 = y_2^2 + 3^{39}. \tag{}[/math] Next, we see that [math]3 \mid y_2[/math], and writing [math]y_2 = 3y_3[/math] for some [math]y_1 \in \mathbb{Z}[/math] yields [math]x_2^3 = y_3^2 + 3^{37}. \tag{}[/math] Note that the power of [math]3[/math] in the Diophantine equation has dropped by [math]4[/math]. Repeated application of the procedure above eventually gives us the Diophantine equation (for some [math]X, Y \in \mathbb{Z}[/math]) [math]X^3 = Y^2 + 3. \tag{}[/math] However, this equation is known to have no integer solutions, giving us a contradiction. A proof of this fact is given here: Claim: The equation [math]x^3 = y^2 + 3[/math] has no integer solutions. Proof: Assume that such a solution exists. Then, [math]x[/math] must be odd; otherwise, this equation reduces to [math]y^2 \equiv 5 \bmod 8[/math] which has no solution. Since [math]x[/math] is odd, this implies that [math]y[/math] is even. Next, reducing this equation modulo [math]4[/math] yields [math]x^3 \equiv 3 \bmod 4[/math]; thus [math]x \equiv 3 \bmod 4[/math]. Now, we first rewrite the equation as [math]y^2 + 4 = x^3 + 1 = (x + 1)(x^2 - x + 1). \tag{}[/math] Moreover, [math]x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4} > 0[/math], and [math]x^2 - x + 1 \equiv 3 \bmod 4[/math]. Therefore, [math]x^2 - x + 1[/math] has at least one prime factor [math]p \equiv 3 \bmod 4[/math]. Hence, [math]y^2 + 4 \equiv 0 \bmod p[/math], or equivalently, [math]y^2 \equiv -4 \bmod p[/math]. Since [math]4[/math] is a perfect square, this implies that [math]-1[/math] is a square modulo [math]p[/math]. This is a contradiction, because [math]p \equiv 3 \bmod 4[/math] (via basic quadratic residue/Legendre symbol theory). Therefore, we deduce that [math]x^3 = y^2 + 3[/math] has no integer solutions. [math]\blacksquare[/math] Returning to our originally given Diophantine equation, the above claim finally implies that [math]3 \nmid N(\delta)[/math], and this leaves us with [math]N(\delta) \mid 2^2[/math], or equivalently, [math]\delta \mid 2[/math] (since [math]2[/math] remains prime in [math]\mathbb{Z}[\omega][/math]). However, [math]\delta[/math] is not an associate of [math]2[/math], because [math]2 \nmid (y + 3^{21} \sqrt{-3})[/math], we obtain [math]\delta = \gcd(y + 3^{21} \sqrt{-3}, y - 3^{21} \sqrt{-3}) = 1[/math]. Using this fact, there exist integers [math]a \equiv b \bmod 2[/math] and [math]k \in {0,1,2}[/math] satisfying [math]\displaystyle y + 3^{21} \sqrt{-3} = \Big(\frac{a + b\sqrt{-3}}{2}\Big)^3 \cdot \Big(\frac{-1 + \sqrt{-3}}{2}\Big)^k. \tag{}[/math] (The second factor is the generator for the unit group of [math]\mathbb{Z}[\omega][/math]; we ignore the unit [math]-1[/math], because it is a perfect cube and can be absorbed into the first factor.) We now have three cases to consider. Case 1: For [math]k = 0[/math], we have [math]\displaystyle y + 3^{21} \sqrt{-3} = \Big(\frac{a + b\sqrt{-3}}{2}\Big)^3. \tag{}[/math] Expanding the right side and then equating the [math]\sqrt{-3}[/math] entries yields [math]\displaystyle 3^{21} = \frac{3}{8} b(a^2 - b^2). \tag{}[/math] This can be checked to have no integer solutions. Case 2: For [math]k = 1[/math], we have [math]\displaystyle y + 3^{21} \sqrt{-3} = \Big(\frac{a + b\sqrt{-3}}{2}\Big)^3\Big(\frac{-1 + \sqrt{-3}}{2}\Big). \tag{}[/math] Expanding the right side and then equating the [math]\sqrt{-3}[/math] entries yields [math]\displaystyle 3^{21} = \frac{1}{16} \cdot (-a^3 - 3a^2b - 9ab^2 + 3b^3). \tag{}[/math] This can be checked (via descent arguments via divisibility by [math]3[/math]; see the comment by Schlitzer) to have no integer solutions. Case 3: For [math]k = 2[/math], we have [math]\displaystyle y + 3^{21} \sqrt{-3} = \Big(\frac{a + b\sqrt{-3}}{2}\Big)^3\Big(\frac{-1 - \sqrt{-3}}{2}\Big). \tag{}[/math] Expanding the right side and then equating the [math]\sqrt{-3}[/math] entries yields [math]\displaystyle 3^{21} = \frac{1}{16} \cdot (-a^3 - 3a^2b + 9ab^2 + 3b^3). \tag{}[/math] This can be checked (similar to Case 2) to have no integer solutions. In summary, we have shown that the Diophantine equation [math]x^3 - 3^{43} = y^2[/math] has no integral solutions. [math]\blacksquare[/math] Sponsored by CDW Corporation Which leading solutions prevent bad actors from gaining access? Rapid7 from CDW delivers predictive power prepared to protect you from cyber risk in a modern landscape. Croissant Trying to learn as much as possible about Mathematics · Author has 94 answers and 27.8K answer views · 1y Related How do you find all integer solutions to [math]y^{2}+2y=x^{4}+20x^{3}+104x^{2}+40x+2003[/math] ? (Remark: This question was from the Singapore Mathematical Olympiad Junior Round 2 question 5. I personally find it very tedious.) We want to find integer solutions to [math]y^{2}+2y=x^{4}+20x^{3}+104x^{2}+40x+2003[/math]. Adding one to both sides we get math^{2}=x^{4}+20x^{3}+104x^{2}+40x+2004[/math]. Now focusing on the first three terms of the left hand side, we notice that [math]x^{4}+20x^{3}+104x^{2}=x^{4}+20x^{3}+100x^{2}+4x^{2}=(x^{2}+10x)^{2}+4x^{2}[/math] Putting this value into the original equation yields math^{2}=(x^{2}+10x)^{2}+4x^{2}+40x+2004[/math] math^{2}=(x^{2}+10x)^{2}+4(x^{2}+10x)+2004[/math] math^{2}=(x^{2}+10x)^{2}+4(x^{2[/math] (Remark: This question was from the Singapore Mathematical Olympiad Junior Round 2 question 5. I personally find it very tedious.) We want to find integer solutions to [math]y^{2}+2y=x^{4}+20x^{3}+104x^{2}+40x+2003[/math]. Adding one to both sides we get math^{2}=x^{4}+20x^{3}+104x^{2}+40x+2004[/math]. Now focusing on the first three terms of the left hand side, we notice that [math]x^{4}+20x^{3}+104x^{2}=x^{4}+20x^{3}+100x^{2}+4x^{2}=(x^{2}+10x)^{2}+4x^{2}[/math] Putting this value into the original equation yields math^{2}=(x^{2}+10x)^{2}+4x^{2}+40x+2004[/math] math^{2}=(x^{2}+10x)^{2}+4(x^{2}+10x)+2004[/math] math^{2}=(x^{2}+10x)^{2}+4(x^{2}+10x)+4+2000[/math] math^{2}=(x^{2}+10x+2)^{2}+2000[/math] Now let [math]y+1=a, x^{2}+10x+2=b[/math]. Then the equation becomes [math]a^{2}=b^{2}+2000[/math] math(a+b)=2000[/math] To solve this now reduced, simplified Diophantine Equation, note that a-b and a+b have the same parity, because their difference, 2b, is an even number. This means we only need to consider the even factor pairs of 2000, because 2000 is an even number. Another thing to note is that since [math]b=x^{2}+10x+2=(x+5)^{2}-23[/math], [math]b\geq-23[/math]. So we have the following: math=(2,1000),(4,500),(8,250),(1000,2),(500,4),(250,8), (-2,-1000),(-4,-500), (-8,-250),(-1000,-2),(-500,-4),(-250, - 8)[/math] Solving each of these pairs for a and b, yields: math=(501,499),(252,248),(129,121),(501,-499),(252,-248),(129,-121),(-501, -499), (-252,-248),(-129,-121),(-501,499),(-252,248), (-129,121)[/math] and since b is greater than or equal to negative 23, we are left with the following cases:. math=(501,499),(252,248),(129,121),(-501,499),(-252,248),(-129,121)[/math]. Now, by guess and check with the quadratic equation that [math]b[/math] represents, only b=121 and [math]a=\pm129[/math] yields integer solutions. Split into two cases: CASE 1: [math]b=121,a=129: x^{2}+10x+2=121, y+1=129,x=-17,7[/math] So math=(-17,128), (7,128)[/math]. CASE 2: [math]b=121,a=-129: y+1=-129, x=-17,7[/math] math=(-17,-130),(7,-130)[/math] Therefore, after a long journey, we get math=(-17,128), (7,128),(-17,-130),(7,-130)[/math] Finally we solved it! Phew! (But then again, from my own experience, Singapore Maths Olympiads are just like that. Terrifying.) Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by Bernard Montaron , PhD Mathematics & Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (1980) and Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) · Author has 8.5K answers and 21M answer views · 1y Related How can you find all solutions in positive integers of the equation [math]8x^6-3x^4-y^6=(x^2+y^2)^2[/math] ? We want to find all positive integer solutions to [math]8x^6 - 3x^4 - y^6 = (x^2 + y^2)^2 \tag{}[/math] To this end, we first rewrite this Diophantine equation as math - (4x^4 + 2x^2 y^2 + y^4) = 0. \tag{}[/math] Using difference of two cubes on the first factor, we find that math(4x^4 + 2x^2 y^2 + y^4) - (4x^4 + 2x^2 y^2 + y^4) = 0. \tag{}[/math] Then, factoring by grouping yields the factorization math(2x^2 - y^2 - 1) = 0. \tag{}[/math] Hence, we see that [math]4x^4 + 2x^2 y^2 + y^4 = 0 \text{ or } 2x^2 - y^2 - 1 = 0. \tag{}[/math] (i) Upon completing the square, the first equation can be expressed as [math]4x^4 + [/math] We want to find all positive integer solutions to [math]8x^6 - 3x^4 - y^6 = (x^2 + y^2)^2 \tag{}[/math] To this end, we first rewrite this Diophantine equation as math - (4x^4 + 2x^2 y^2 + y^4) = 0. \tag{}[/math] Using difference of two cubes on the first factor, we find that math(4x^4 + 2x^2 y^2 + y^4) - (4x^4 + 2x^2 y^2 + y^4) = 0. \tag{}[/math] Then, factoring by grouping yields the factorization math(2x^2 - y^2 - 1) = 0. \tag{}[/math] Hence, we see that [math]4x^4 + 2x^2 y^2 + y^4 = 0 \text{ or } 2x^2 - y^2 - 1 = 0. \tag{}[/math] (i) Upon completing the square, the first equation can be expressed as [math]4x^4 + 2x^2 y^2 + y^4 = (x^2 + y^2)^2 + 3x^4 = 0. \tag{}[/math] Being a sum of squares, it is easy to verify that math = (0, 0)[/math] is its only solution. Then since we are only interested in positive integer solutions, we can discard this solution. (ii) The second equation can be expressed as the classic Pell equation [math]y^2 - 2x^2 = -1. \tag{}[/math] Using arithmetic from the algebraic number ring [math]\mathbb{Z}[\sqrt{2}][/math], this is equivalent to solving [math]N(y + x \sqrt{2}) = -1, \tag{}[/math] all units having norm [math]-1[/math] where [math]x, y > 0[/math]. Since [math]\mu = 1 + \sqrt{2}[/math] is a fundamental unit of this ring having norm [math]-1[/math], we conclude that all positive integer solutions can be written in the form [math]\boxed{y + x \sqrt{2} = (1 + \sqrt{2})^{2k-1} \text{ where } k \in \mathbb{N}}. \tag{}[/math] Sponsored by Torii Why are companies moving to Torii for SaaS Identity Governance? Legacy IGA tools can't handle both SaaS discovery and access controls, but Torii's agentic tools can. Rik Bos Ph.D. Mathematics from Utrecht University (Graduated 1979) · Upvoted by Alon Amit , Lover of math. Also, Ph.D. and Teuvo Laurinolli , studied Mathematics & Theoretical Physics at University of Oulu (1978) · Author has 1.4K answers and 1.3M answer views · 2y Related How do I find all integer solutions to the equation [math]x^3-y^3=xy+61[/math] ? Let’s generalize the equation a bit to [math]x^3-y^3=axy+t[/math] where [math]a,t[/math] are integer constants. We claim this equation has only finitely many integer solutions (which also follows from Thue’s theorem, since it’s not hard to see that [math]x^3-y^3-axy[/math] is irreducible). But here we can prove this in an elementary way and moreover provide a bound in terms of [math]a[/math] and [math]t[/math]. Putting [math]x=y+c[/math] where [math]c[/math] is an integer, we get [math]3y^2c+3yc^2+c^3=ay^2+acy+t[/math], so mathy^2+c(3c-a)y+c^3-t=0[/math]. Now the discriminant of this quadratic equation in [math]y[/math] is [math]D=c^2(3c-a)^2-4(3c-a)(c^3-t)[/math] and in order there are integer solutions, [math]D[/math] must be a perfect s Let’s generalize the equation a bit to [math]x^3-y^3=axy+t[/math] where [math]a,t[/math] are integer constants. We claim this equation has only finitely many integer solutions (which also follows from Thue’s theorem, since it’s not hard to see that [math]x^3-y^3-axy[/math] is irreducible). But here we can prove this in an elementary way and moreover provide a bound in terms of [math]a[/math] and [math]t[/math]. Putting [math]x=y+c[/math] where [math]c[/math] is an integer, we get [math]3y^2c+3yc^2+c^3=ay^2+acy+t[/math], so mathy^2+c(3c-a)y+c^3-t=0[/math]. Now the discriminant of this quadratic equation in [math]y[/math] is [math]D=c^2(3c-a)^2-4(3c-a)(c^3-t)[/math] and in order there are integer solutions, [math]D[/math] must be a perfect square and in particular nonnegative. Since [math]D=(3c-a)(3c^3-ac^2-4c^3+4t)=-(3c-a)(c^3+ac^2-4t)[/math] we must have math(c^3+ac^2-4t)\le 0[/math]. Since math(c^3+ac^2-4t)[/math], considered as a polynomial in [math]c[/math] has degree four and positive leading coefficient, it is clear that the region where the polynomial is negative is bounded between the smallest and largest root (which can be expressed in [math]a[/math] and [math]t[/math] even though this is in general cumbersome). To simplify matters, suppose [math]a=1[/math] and [math]t>0[/math]. Then we have to investigate where the expression math(c^3+c^2-4t)[/math] is negative. The first factor is negative if [math]c\le 0[/math]. But if this is the case, the second factor is [math]c^2(c+1)-4t[/math] is also negative. So if [math]c\le 0[/math] then [math]-D=(3c-1)(c^3+c^2-4t)>0[/math]. Therefore [math]c>0[/math]. Then [math]3c-1>0[/math] and now we must have [math]c^3+c^2\le 4t[/math]. For large [math]t[/math] we can then use math^{1/3}[/math] as an upper bound for [math]c[/math]. For instance if [math]t=61[/math] this bound lies slightly above [math]6[/math] and since [math]6^3+6^2=252>4t[/math] we can lower the bound to [math]5[/math]. So we just need to check the integers [math]c[/math] with [math]1\le c\le 5[/math] and it turns out only [math]c=1[/math] makes [math]D[/math] into a perfect square. Conversely, if [math]c=1[/math] we can make [math]D[/math] into a perfect square by choosing [math]t=2u^2+2u+1[/math] (where [math]u\ge 0[/math]) is the sum of two consecutive squares (like [math]61[/math]). Then [math]D=4(2t-1)=(4u+2)^2[/math], so here we find math=(u+1,u),(-u,-u-1)[/math]. In particular, if [math]t=61[/math] we find [math]u=5[/math] and math=(6,5),(-5,-6)[/math]. By the way, the solutions to [math]x^3-y^3=xy+t[/math] always come in pairs math,(-b,-a)[/math]. It’s interesting to see if for the type of [math]t[/math] of the previous paragraph there could be values of [math]c[/math] different from [math]1[/math] making the discriminant into a perfect square. It turns out that for [math]t=41=16+25[/math] (so [math]u=4[/math]), there are three suitable values for [math]c[/math]: math=(1,18^2),(3,32^2),(5,14^2)[/math]. So here we have the equation [math]x^3-y^3=xy+41[/math] with solutions math=(5,4),(-4,-5)[/math] (corresponding to [math]c=1[/math]) and math=(3,-2),(2,-3)[/math] (corresponding to [math]c=5[/math]). However, if [math]c=3[/math] we find math=(3\frac{1}{2},\frac{1}{2})[/math], so a rational, but not an integer solution. Another example is [math]t=365[/math] ([math]u=13[/math]) where we find solutions math,(-13,-14)[/math] and math,(2,-7)[/math]. Here there is also one rational, non-integer solution (with [math]c=11[/math]). Eleftherios Argyropoulos B.S. in Mathematics & Physics, Northeastern University (Graduated 2002) · Author has 2K answers and 2.5M answer views · 1y Related How do you find all integer solutions to [math]y^{2}+2y=x^{4}+20x^{3}+104x^{2}+40x+2003[/math] ? First solution: We want to solve the Diophantine equation: [math]y^2+2y = x^4+20x^3+104x^2+40x+2003 … (1)[/math] We add [math]1[/math] to both sides of math[/math] and we take: [math]y^2+2y+1 = x^4+20x^3+104x^2+40x+2003+1 =>[/math] math^2 = x^4+20x^3+104x^2+40x+2004 … (2)[/math] We set: [math]z = y+1 … (3)[/math] and equation math[/math] takes the form: [math]z^2 = x^4+20x^3+104x^2+40x+2004 … (4)[/math] Equation math[/math] can be easily solved using Poulakis algorithm for the general Diophantine equation: [math]y^2 = x^4+ax^3+bx^2+cx+d … (5)[/math] for the case at which the polynomial [math]p(x) = x^4+ax^3+bx^2+cx+d[/math] is not perfect square. In our case, in equation math[/math] the polynomial [math]x^4+20x^3+104x^2+40x+2004[/math] is not pe First solution: We want to solve the Diophantine equation: [math]y^2+2y = x^4+20x^3+104x^2+40x+2003 … (1)[/math] We add [math]1[/math] to both sides of math[/math] and we take: [math]y^2+2y+1 = x^4+20x^3+104x^2+40x+2003+1 =>[/math] math^2 = x^4+20x^3+104x^2+40x+2004 … (2)[/math] We set: [math]z = y+1 … (3)[/math] and equation math[/math] takes the form: [math]z^2 = x^4+20x^3+104x^2+40x+2004 … (4)[/math] Equation math[/math] can be easily solved using Poulakis algorithm for the general Diophantine equation: [math]y^2 = x^4+ax^3+bx^2+cx+d … (5)[/math] for the case at which the polynomial [math]p(x) = x^4+ax^3+bx^2+cx+d[/math] is not perfect square. In our case, in equation math[/math] the polynomial [math]x^4+20x^3+104x^2+40x+2004[/math] is not perfect square, since this polynomial has four complex roots. Moreover, we have: [math]a = 20[/math], [math]b = 104[/math], [math]c = 40[/math], [math]d = 2004[/math]. Hence, Poulakis algorithm can be applied for the solution. Now, we consider the polynomials: [math]Q(x) = 16X^2+8(a-8c+4ab-a^3)X+8b-2a^2+1-64d+16b^2+a^4-8a^2b ... (6)[/math] [math]R(x) = 16X^2+8(a+8c-4ab+a^3)X+8b-2a^2-1+64d-16b^2-a^4+8a^2b ... (7)[/math] Hence, the polynomials [math]Q(X)[/math] and [math]R(X)[/math] take the form: [math]Q(x) = 16X^2+160X-127967 ... (8)[/math] [math]R(x) = 16X^2+160X+128031 ... (9)[/math] The solutions of the quadratic polynomial math[/math] are: [math]X_{1} = -5-(3\sqrt{14263})/4[/math] [math]X_{2} = (3\sqrt{14263})/4-5[/math] while the solutions of the quadratic polynomial math[/math] are complex. Hence, the solutions of math[/math] will be found in the interval [math][-94[/math], [math]84][/math] which is defined by the integer parts of the real solutions of math[/math]. By a quick computer check, we find that the only integral values of [math]x[/math] which stasfy equation math[/math], are: [math]x_{1} = -17[/math] [math]x_{2} = 7[/math] For these values of [math]x[/math], we respectively have: [math]z_{1} = ± 129[/math] [math]z_{2} = ± 129[/math] Therefore, the solutions of math[/math] are: [math]x = -17[/math], [math]y = -130[/math] [math]x = -17[/math], [math]y = 128[/math] [math]x = 7[/math], [math]y = -130[/math] [math]x = 7[/math], [math]y = 128[/math] Second solution: We want to solve the Diophantine equation: [math]y^2+2y = x^4+20x^3+104x^2+40x+2003 … (1)[/math] We add [math]1[/math] to both sides of math[/math] and we take: [math]y^2+2y+1 = x^4+20x^3+104x^2+40x+2003+1 =>[/math] math^2 = x^4+20x^3+104x^2+40x+2004 … (2)[/math] We set: [math]z = y+1 … (3)[/math] and equation math[/math] takes the form: [math]z^2 = x^4+20x^3+104x^2+40x+2004 => [/math] [math]z^2 = (x^2+10x)^2+4(x^2+10x)+2004 … (4)[/math] We now set: [math]x^2+10x = t … (5)[/math] and equation math[/math] takes the form: [math]z^2 = t^2+4t+2004 => t^2+4t+(2004-z^2) = 0 … (6)[/math] The discriminant of math[/math] with respect to [math]t[/math], is: [math]D_{t} = 16-4[(2004-z^2)] => D_{t} = 4z^2-8000 ... (7)[/math] and it must be perfect square. Hence, there must exist positive integer [math]p[/math], such that: [math]4z^2-8000 = p^2 => (2z)^2-p^2 = 8000 ... (8)[/math] Evidently, by math[/math] it follows that: [math]z^2-2004 > 0 => z > 44 => y > 43 ... (9)[/math] The positive integral solutions of math[/math] are: [math]p = 998[/math], [math]z = 501[/math] [math]p = 496[/math], [math]z = 252[/math] [math]p = 242[/math], [math]z = 129[/math] [math]p = 190[/math], [math]z = 105[/math] [math]p = 80[/math], [math]z = 60[/math] [math]p = 10[/math], [math]z = 45[/math] which are all acceptable by math[/math]. For the above six values of [math]z[/math], by math[/math], we have the trinomials: [math]t^2+4t-248997 = 0 … (10)[/math] [math]t^2+4t-61500 = 0 … (11)[/math] [math]t^2+4t-14637 = 0 … (12)[/math] [math]t^2+4t-9021 = 0 … (13)[/math] [math]t^2+4t-1596 = 0 … (14)[/math] [math]t^2+4t+404 = 0 … (15)[/math] From the above six trinomials, only the first five have integral solution. The solutions are: [math]t = -501[/math], [math]t = 497[/math] [math]t = -250[/math], [math]t = 246[/math] [math]t = -123[/math], [math]t = 119 [/math] [math]t = -97[/math], [math]t = 93[/math] [math]t = -42[/math], [math]t = 38[/math] The above ten values for [math]t[/math], create the following Diophantine equations: [math]x^2+10x = -501 … (16)[/math] [math]x^2+10x = -250 … (17)[/math] [math]x^2+10x = -123 … (18)[/math] [math]x^2+10x = -97 … (19)[/math] [math]x^2+10x = -42 … (20)[/math] [math]x^2+10x = 38 … (21)[/math] [math]x^2+10x = 93 … (22)[/math] [math]x^2+10x = 119 … (23)[/math] [math]x^2+10x = 246 … (24)[/math] [math]x^2+10x = 497 … (25)[/math] Therefore, the final integral solutions for [math]x[/math], are only the solutions of math[/math]: [math]x = -17[/math], [math]x = 7[/math] From these, by math[/math], we take the final integral values of [math]y[/math]: [math]x = 7[/math], [math]y = -130[/math] [math]x = 7[/math], [math]y = 128[/math] [math]x = -17[/math], [math]y = -130[/math] [math]x = -17[/math], [math]y = 128[/math] These, are the only solutions of math[/math]. Mehdi Khayeche IMO Silver Medalist · Upvoted by Marek Kobera , Ph. D. Mathematics & Physics, Charles University in Prague (2016) and Krogi Us , M.Sc. Mathematics & Computer Science, Aalto University (1988) · Author has 75 answers and 1.4M answer views · 6y Related How can you solve this equation for positive integers: [math]1+x+x^2+x^3+x^4=y^2[/math] ? When I first saw this problem, I remembered a similar yet simpler one that is usually given in early math Olympiads training stages: Find all positive integers [math]x[/math] and [math]y[/math] that satisfy: The easy way to do it is to notice that: [math] x^2<1+x+x^2 < (x+1)^2\tag{}[/math] so [math]1+x+x^2[/math] is between two consecutive perfect squares, hence it cannot be a square too! Now this problem is more complicated, but I think a similar approach would work. Let’s figure out how we’re going to bound [math]1+x+x^2+x^3+x^4[/math] : It seems that math^2[/math] is a good candidate. (I didn’t randomly pick this; I wanted an expressio When I first saw this problem, I remembered a similar yet simpler one that is usually given in early math Olympiads training stages: Find all positive integers [math]x[/math] and [math]y[/math] that satisfy: [math]1+x+x^2=y^2\tag{}[/math] The easy way to do it is to notice that: [math] x^2<1+x+x^2 < (x+1)^2\tag{}[/math] so [math]1+x+x^2[/math] is between two consecutive perfect squares, hence it cannot be a square too! Now this problem is more complicated, but I think a similar approach would work. Let’s figure out how we’re going to bound [math]1+x+x^2+x^3+x^4[/math] : It seems that math^2[/math] is a good candidate. (I didn’t randomly pick this; I wanted an expression that,when squared, would yield [math]x^4+x^3+[/math] small powers of [math]x[/math] ) Well let’s see if this works by expanding: math^2 =x^4 + x^3 + \frac{x^2}{4} < 1+x+x^2+x^3+x^4\tag{} [/math] Now, if we’re really lucky, then -fingers crossed- math^2[/math] would be an upper bound. Let’s check: math^2=x^4 + x^3 +\frac{9x^2}{4} + x + 1 > 1+x+x^2+x^3+x^4\tag{} [/math] YES! so [math]1+x+x^2+x^3+x^4[/math] is between [math]2[/math] consecutive perfect squares hence it cannot be a square. QEwait… Do you see the problem? math^2[/math] is not necessarily a perfect square because [math]x^2+\frac{x}{2}[/math] is not necessarily an integer. So we need to consider these cases: 1) If [math]x[/math] is even: [math]x^2+\frac{x}{2} \in \mathbb{N}[/math] and we’d have no solutions. 2) If [math]x[/math] is odd: Then there is a unique integer between [math]x^2+\frac{x}{2}[/math] and [math]x^2+\frac{x}{2} +1[/math] and that integer is [math]x^2+\frac{x}{2}+\frac{1}{2}[/math]. This implies that, in order to have a solution, we must have: [math] 1+x+x^2+x^3+x^4= (x^2+\frac{x}{2}+\frac{1}{2})^2 \tag{}[/math] [math]\Leftrightarrow 1+x+x^2+x^3+x^4=x^4 + x^3 + \frac{5x^2}{4} + \frac{x}{2} +\frac{1}{4} \tag{}[/math] [math]\Leftrightarrow \frac{x^2}{4} - \frac{x}{2}-\frac{3}{4}=0 \tag{}[/math] The only positive integer solution of the quadratic is [math]x=3[/math] which gives [math]y=11[/math] and that’s the only solution to our problem. [math]\blacksquare[/math] Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Upvoted by Michael Jørgensen , PhD in mathematics · Author has 8.5K answers and 21M answer views · 1y Related How do I all find all integer solutions to the equation [math]x^3+2y^3+4z^3+8xyz=0[/math] ? We are given the Diophantine equation [math]x^3 + 2y^3 + 4z^3 + 8xyz = 0. \tag{}[/math] Suppose that this equation has a nontrivial solution math \neq (0,0,0)[/math] where [math]x^2 + y^2 + z^2[/math] is minimal. Since [math]2y^3 + 4z^3 + 8xyz[/math] is even, it follows that [math]x^3[/math] is even and thus [math]x[/math] is even. Writing [math]x = 2x'[/math] for some integer [math]x'[/math], substituting this into the given Diophantine equation and rearranging it gives us [math]y^3 + 2z^3 + 4x'^3 + 8yzx' = 0. \tag{}[/math] Hence, math = (y, z, \frac{x}{2})[/math] is also a solution. However, we see that [math]\displaystyle y^2 + z^2 + \frac{x^2}{4} \leq x^2 + y^2 + z^2, \tag{}[/math] and since we assumed that [math]x^2 [/math] We are given the Diophantine equation [math]x^3 + 2y^3 + 4z^3 + 8xyz = 0. \tag{}[/math] Suppose that this equation has a nontrivial solution math \neq (0,0,0)[/math] where [math]x^2 + y^2 + z^2[/math] is minimal. Since [math]2y^3 + 4z^3 + 8xyz[/math] is even, it follows that [math]x^3[/math] is even and thus [math]x[/math] is even. Writing [math]x = 2x'[/math] for some integer [math]x'[/math], substituting this into the given Diophantine equation and rearranging it gives us [math]y^3 + 2z^3 + 4x'^3 + 8yzx' = 0. \tag{}[/math] Hence, math = (y, z, \frac{x}{2})[/math] is also a solution. However, we see that [math]\displaystyle y^2 + z^2 + \frac{x^2}{4} \leq x^2 + y^2 + z^2, \tag{}[/math] and since we assumed that [math]x^2 + y^2 + z^2[/math] is minimal, this forces [math]x = 0[/math]. Now, our Diophantine equation reduces to [math]y^3 + 2z^3 = 0. \tag{}[/math] Since we assumed a non-trivial solution, we have [math]z \neq 0[/math] (otherwise, this also implies that [math]y = 0[/math]). Then, this implies that [math]\displaystyle \sqrt{2} = -\frac{y}{z}. \tag{}[/math] This yields a contradiction, because [math]\sqrt{2}[/math] is known to be irrational (through an infinite descent argument or otherwise). Hence, we conclude that the given Diophantine equation has no nontrivial solutions, and thus the only solution is [math]\boxed{(x, y, z) = (0, 0, 0)}[/math]. Related questions How can you find all integer solutions of x^2+3=y^3? How does one find all integer solutions to y^2=x^5-7? What are all of the integer solutions of 1+x+x^2+x^3=y^2? Can you find the integer solutions for X 3 + Y 3 + Z 3 = 10 3 ? How do you find the positive integer solutions to x y + z + y z + x + z x + y = 4 ? What are all the integer solutions ( x , y ) to the equation x 3 + y 3 = ( x + y ) 2 ? Consider the equation x 2 + y 2 = 3 z 2 . Are there any other integer solutions besides the solution where x = y = z = 0 ? What are the integer solutions to the equation 3 x − 2 y = 1 ? How many integer solutions does the Diophantine equation x^2+3=y have? How do you find all integer solutions to the equation 2 x 3 + x + 8 = y 2 ? What are all integer solutions of the equation x^2 +y^2 equal to xy? How do I find all integer solutions x, y, z of the equation ( x − y − 1 ) 3 + ( y − z − 2 ) 3 + ( z − x + 3 ) 3 = 18 ? What are the integer solutions of x 2 − y 3 = 17 ? What are the integer solutions of the equation y 3 / x 3 = 1 + z ? How do I find all integer solutions to the equation x 3 − y 3 = x y + 61 ? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10703	https://artofproblemsolving.com/wiki/index.php/Talk:Factorial?srsltid=AfmBOopiUH9g-iExJVRM1ic8xu4k1AC4IDE7cdf0233JfXugYvSmd2Cn	Art of Problem Solving Talk:Factorial - 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 Talk:Factorial Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Talk:Factorial I added the prime factorization section (needed for Prime Number Theorem and Bertrand postulate). I'm not sure the counting principle I referred to as the Lebesgue counting principle is not really known under some other name. If you happen to know the correct name for it, please, change the corresponding line. Good new section. I cut the Lebesgue counting bit (and the summation) and described in simpler terms where the sum comes from (though I think my explanation can be improved). If a reader doesn't know this prime factorization information, they'll never understand that summation. (In other words, the tool used to produce the sum was more advanced than the fact the sum represents, making it hard for a new student to follow.) The section's a nice addition. --Rrusczyk 09:48, 3 July 2006 (EDT) Retrieved from " 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.
10704	https://www.quora.com/What-is-the-difference-between-the-equation-Q-mCpdT-and-Q-UA-LMTD	Something went wrong. Wait a moment and try again. Specific Heat Capacity Heat Transfer Systems Temperature Difference Mechanical Engineering (T... Heat Transfer Coefficient Quantity of Heat Heat Transfer Physics 5 What is the difference between the equation Q=mCpdT and Q= UA (LMTD) ? · The equations Q=m⋅Cp⋅ΔT and Q=U⋅A⋅LMTD are both used in thermodynamics and heat transfer, but they apply to different situations and concepts. Here’s a breakdown of each equation: Q=m⋅Cp⋅ΔT Variables: Q: Heat transfer (in joules or calories) m: Mass of the substance (in kilograms) Cp: Specific heat capacity of the substance (in J/kg·°C or cal/g·°C) ΔT: Change in temperature (in °C or K) Context: This equation is used for calculating the heat transfer when a substance (usually a fluid) is heated or cooled without a phase change. The equations Q=m⋅Cp⋅ΔT and Q=U⋅A⋅LMTD are both used in thermodynamics and heat transfer, but they apply to different situations and concepts. Here’s a breakdown of each equation: Q=m⋅Cp⋅ΔT Variables: Q: Heat transfer (in joules or calories) m: Mass of the substance (in kilograms) Cp: Specific heat capacity of the substance (in J/kg·°C or cal/g·°C) ΔT: Change in temperature (in °C or K) Context: This equation is used for calculating the heat transfer when a substance (usually a fluid) is heated or cooled without a phase change. It assumes a constant specific heat capacity over the temperature range and is often used in specific heat calculations for processes like heating water or air. Q=U⋅A⋅LMTD Variables: Q: Heat transfer rate (in watts or BTU/hr) U: Overall heat transfer coefficient (in W/m²·°C or BTU/hr·ft²·°F) A: Heat transfer area (in m² or ft²) LMTD: Logarithmic Mean Temperature Difference (in °C or K), which accounts for the temperature difference between two fluids in a heat exchanger. Context: This equation is typically used in the analysis of heat exchangers, where two fluids at different temperatures exchange heat. The LMTD accounts for the varying temperature difference between the hot and cold fluids along the length of the heat exchanger. The overall heat transfer coefficient U incorporates the effects of conduction, convection, and any fouling factors. Summary of Differences: Application: The first equation is for sensible heat transfer in single-phase systems, while the second is specifically for heat exchangers involving two fluids at different temperatures. Parameters: The first equation focuses on mass and specific heat, while the second involves the heat transfer area and the overall heat transfer coefficient. Temperature Change: The first uses a simple temperature change (ΔT), while the second uses a more complex approach to temperature difference (LMTD) that reflects the heat exchange process between two fluids. These equations serve different purposes in thermodynamics and heat transfer analysis, and understanding their contexts is crucial for effective application. Rucha Harsha Pankaj Trivedi M.Tech in Engineering & Chemical Engineering, Indian Institute of Technology, Hyderabad (IITH) (Graduated 2017) · Author has 3.3K answers and 2.9M answer views · 5y Q = mCpdT shows the heat required to raise the temperature of a material through a finite temperature difference (dT). Q = UA (LMTD) shows the rate of heat transfer across a finite temperature difference between 2 surfaces or fluids for a given surface area. The first equation deals with heat required to raise a temperature of the material itself whereas second equation is used only for heat transfe Q = mCpdT shows the heat required to raise the temperature of a material through a finite temperature difference (dT). Q = UA (LMTD) shows the rate of heat transfer across a finite temperature difference between 2 surfaces or fluids for a given surface area. The first equation deals with heat required to raise a temperature of the material itself whereas second equation is used only for heat transfer between 2 fluids separated by an external surface. For example, if you are cooling hot oil with water, then amount of heat transferred by oil to water will be governed by 2nd equation and the total cooling achieved in the hot oil will be governed by 1st equation. Heat gained or loss in first equation is dependent on the inherent characteristics of the material (Cp) whereas heat transfer in the 2nd equation is dependent on the siz... Sponsored by Amazon Web Services (AWS) Want to level up your AWS game? Register for re:Invent 2025. 5 days of advanced architecture workshops, serverless deep dives, and coding labs. Las Vegas, Dec 1-5. Piriya M.Tech. in Chemical Engineering, Anna University, Tamil Nadu, India (Graduated 2018) · Author has 330 answers and 2.7M answer views · 8y The first equation, Q=mCpdT determines the amount of heat stored in the material. Thermal energy can be stored as sensible heat in a material by raising its temperature. In a heat exchanger, The rate of heat transfer from the hot fluid, is exactly equal to the rate of heat transfer to the cold fluid in a differential length of the exchanger surface. Thus equating the equation for hot and cold fluids, gives the temperature of one fluid at one port when all other temperatures are known. Heat Storage in Materials Whereas the second equation employing overall heat transfer coefficient (U) is used wh The first equation, Q=mCpdT determines the amount of heat stored in the material. Thermal energy can be stored as sensible heat in a material by raising its temperature. In a heat exchanger, The rate of heat transfer from the hot fluid, is exactly equal to the rate of heat transfer to the cold fluid in a differential length of the exchanger surface. Thus equating the equation for hot and cold fluids, gives the temperature of one fluid at one port when all other temperatures are known. Heat Storage in Materials Whereas the second equation employing overall heat transfer coefficient (U) is used when heat is transfered between fluid and solid. The overall heat transfer coefficient is influenced by the thickness and thermal conductivity of the mediums through which heat is transferred. The larger the coefficient, the easier heat is transferred from its source to the product being heated. This equation is used in heat exchanger calculations to determine the heat duty (Q), which includes all the resistances in series. OVERALL HEAT TRANSFER COEFFICIENT Thank You;) Related questions What are the units for q in the equation q=mcΔT? What is the difference between LMTD and dT? In Heat Transfer (exchangers) is Q different to q? They seem inter-changeable. When is LMTD invalid? How many amps does a 10kw heater pull 3 phase? Ravi Kasundra Chemical engineer · Author has 120 answers and 1.3M answer views · 8y Q= mCpdT Where, m = Mass flowrate of fluid Cp= Specific heat capacity dT= Temperature difference This equation used when we are talking about only one fluid. Example:- IF flowrate of water is X and i added heat Y then how much temperature will increase That type of problems solve by this equation. It shows the relationship between temperatures change with respect tho the heat. This equation indicates tha Q= mCpdT Where, m = Mass flowrate of fluid Cp= Specific heat capacity dT= Temperature difference This equation used when we are talking about only one fluid. Example:- IF flowrate of water is X and i added heat Y then how much temperature will increase That type of problems solve by this equation. It shows the relationship between temperatures change with respect tho the heat. This equation indicates that heat transfer is depends upon specific heat capacity and temperature difference Q=UAdT Wherw U= Overall heat transfer coefficient A=Heat transfer area dT = Temper... Ananthakrishnan G Studied Chemical Engineering at Gateway Institute, Pune (GATE/PSU coaching in chemical engineering) (Graduated 2019) · Author has 299 answers and 1.3M answer views · 5y A2A Q=mCpdT denotes the amount of heat transferred from or to a substance having mass or mass flow rate ’m’ and specific heat capacity ‘Cp’ and experience a temperature change of ‘dT’. This equation doesn’t deal with overall resistances, flow pattern, heat transfer arrangements, but depends only on the heat content in the material. Q= UA (LMTD) this can also be written as Q= (LMTD)/(1/UA) from this we can say there is a driving force (LMTD) and the flow has to experience resistance of (1/UA). Resistance can be due to convective resistance or a combination of convective and conductive resis A2A Q=mCpdT denotes the amount of heat transferred from or to a substance having mass or mass flow rate ’m’ and specific heat capacity ‘Cp’ and experience a temperature change of ‘dT’. This equation doesn’t deal with overall resistances, flow pattern, heat transfer arrangements, but depends only on the heat content in the material. Q= UA (LMTD) this can also be written as Q= (LMTD)/(1/UA) from this we can say there is a driving force (LMTD) and the flow has to experience resistance of (1/UA). Resistance can be due to convective resistance or a combination of convective and conductive resistances. These equations are used in heat exchangers and cases very multiple heat transfer resistances occur in convective heat transfer. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Tue What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. 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The hot fluid gets cool down while the cold fluid gets heated. There can be different types of heat exchanger like Parallel flow heat exchanger Counter flow heat exchanger Cross flow heat exchanger Q=mCpdt This equation holds true for hot fluid ir cold fluid i.e the temperature difference dt could be either (th1-th2)or (tc2-tc1) where th1= temperature of hot fluid at inlet th2= temperature of hot fluid at the outlet tc1= temperature of cold fluid at inlet tc2=temperature of co Well both of them are for heat exchanger. Heat exchanger is a device where heat is exchanged between hot fluid and cold fluid. The hot fluid gets cool down while the cold fluid gets heated. There can be different types of heat exchanger like Parallel flow heat exchanger Counter flow heat exchanger Cross flow heat exchanger Q=mCpdt This equation holds true for hot fluid ir cold fluid i.e the temperature difference dt could be either (th1-th2)or (tc2-tc1) where th1= temperature of hot fluid at inlet th2= temperature of hot fluid at the outlet tc1= temperature of cold fluid at inlet tc2=temperature of cold fluid at outlet Q=UA(LMTD) This equation represents OVERALL HEAT TRANSFER ACROSS THE HEAT EXCHANGER where U= Overall heat transfer coefficient(heat tranfer by conduction and Convection) LMTD= Logarithmic mean temperature difference. Related questions What is the difference between Q=mcθdT and Q=UA (LMTD)? What are some interesting topics in electrical engineering that can be done as a project or thesis work? What should someone do if their supervisor is not willing to provide a recommendation? What other options are available in this situation? I am planning to become a marine engineer, I have completed electrical engineering with a power major, but there is very little information on the internet, can anyone please give me information on how can I become one? When is LMTD invalid in some conditions? Ashish Kushwaha Chemical Engineer for life · Author has 479 answers and 4M answer views · 8y The first quation expresses the amount of heat energy required to raise the temperature of a substance. It depends on the substance being cooled/heated. The second equation especially applied in the context of heat exchangers. This the heat transferring(or needed) for a given temperature difference between the two streams. This doesn’t have to do anything directly with the substances being cooled/heated, but the equipment. Sponsored by Amazon Business Drive decisions with analytics. Make better data-driven decisions with real-time reporting and visibility into purchasing patterns. Manikandan Sampath BTech in Chemical Engineering, Anna University, Tamil Nadu, India (Graduated 2018) · 8y Let us assume a double pipe heat exchanger: The equation Q= m.cp.dt is used when we want to calculate the heat gained or lost by cold or hot fluid.(concerning only one fluid) And the equation Q= U.A.dt is used when we want to calculate the overall heat gained or lost ( concerning both fluids). Where U is the overall heat transfer coffecient which depends on individual heat transfer coefficients . Hope this helps thanks. Rucha Harsha Pankaj Trivedi Medalist from IIT and 1st rank in MBA from BITS Pilani · Author has 3.3K answers and 2.9M answer views · 4y Related What is the difference between LMTD and dT? LMTD is the log mean temperature difference across the entire heat transfer area meaning it calculates the average temperature difference across multiple points of the heat transfer area and dT is just difference between 2 temperatures at any given point. For eg. Let us consider a heat exchanger. Let Th,in and Th,out be the inlet and outlet temperature of hot fluid and Tc,in and Tc,out be the inlet and outlet temperature of cold fluid. Considering counter flow current meaning hot fluid and cold fluid enter from opposite sides of the exchanger, as shown in the diagram below: Then dT at point A wil LMTD is the log mean temperature difference across the entire heat transfer area meaning it calculates the average temperature difference across multiple points of the heat transfer area and dT is just difference between 2 temperatures at any given point. For eg. Let us consider a heat exchanger. Let Th,in and Th,out be the inlet and outlet temperature of hot fluid and Tc,in and Tc,out be the inlet and outlet temperature of cold fluid. Considering counter flow current meaning hot fluid and cold fluid enter from opposite sides of the exchanger, as shown in the diagram below: Then dT at point A will be dTA = (Th,in - Tc,out), similarly, dT at point B will be dTB = (Th,out - Tc,in). So as we can see, dT can be calculated only at a single point. LMTD is calculated across the entire heat transfer, meaning it calculates the average temperature difference across the entire heat transfer area. LMTD for the above counter current arrangement will be: LMTD = (dTA - dTB) / ln (dTA / dTB). LMTD takes into consideration temperature difference at both the points (inlet and outlet) of an heat exchanger whereas dT takes into consideration only at a single point. This makes LMTD more accurate to calculate heat transfer area of an heat exchanger. Sponsored by Amazon Web Services (AWS) Get AI certified. Invest an hour a week in your future with our free AWS Certified AI Practitioner Exam Prep Plan. Paulo Santos Former Senior Mech Engineer · Author has 532 answers and 730.7K answer views · 5y Related Why does a larger flow rate leads to a smaller temperature difference for the process flow in a heat exchanger? I can not find the reason through Q=MCpΔT, because in my experiment Q increases, M increases and ΔT decreases. The question is vague in terms of the experience or the type and temperatures of the HE. Your teacher wants to know if you understand the concepts Following what Craig said is true if: Q and Cp are constant the HE is the same under typical design conditions therefore Qo= Q1 if: Qo=mo cp DTo Q1= m1 (/) cp DT1 () m1 (/) denotes increase and DT1 denotes () decreases Qo=Q1, as expected. but if your initial flow - mo -was too low (in terms of flow regime) and if the flow m1 is increased to an optimum level this could enhance significantly the heat exchange and then Qo < Q1 . So you can check objectively The question is vague in terms of the experience or the type and temperatures of the HE. Your teacher wants to know if you understand the concepts Following what Craig said is true if: Q and Cp are constant the HE is the same under typical design conditions therefore Qo= Q1 if: Qo=mo cp DTo Q1= m1 (/) cp DT1 () m1 (/) denotes increase and DT1 denotes () decreases Qo=Q1, as expected. but if your initial flow - mo -was too low (in terms of flow regime) and if the flow m1 is increased to an optimum level this could enhance significantly the heat exchange and then Qo < Q1 . So you can check objectively this by: -comparing Re of mo and m1 -compare HE efficiency and effectiveness in both cases. you could go a bit further -if you DYH - but this should suffice. Draw your own conclusions and share the information with us or colleagues. Good luck! Shree Ram Bhakt ISC from Senior Secondary School Pawapuri Nalanda · Author has 113 answers and 454.3K answer views · 5y Related Why is U=0 but Q not 0 in isothermal conditions as both depend on delta T? At first, U (internal energy) is not equal to zero in isothermal process, change in internal energy is equal to zero in isothermal process. Since, in isothermal process, temperature is constant ,so U is constant and change in internal energy is zero. Now talk about Q, Do you know that from first law of thermodynamics, Q = W + delta U Or, Q = W (since delta U is equal to zero) So, Q = nRT logv2/v1 (which is not equal to zero) At first, U (internal energy) is not equal to zero in isothermal process, change in internal energy is equal to zero in isothermal process. Since, in isothermal process, temperature is constant ,so U is constant and change in internal energy is zero. Now talk about Q, Do you know that from first law of thermodynamics, Q = W + delta U Or, Q = W (since delta U is equal to zero) So, Q = nRT logv2/v1 (which is not equal to zero) Himanshu Patoliya Graduate Engineer Trainee at Adani Group (Conglomerate) (2019–present) · 6y Related What is lmtd? As we know, in the heat exchanger, there will be a temperature gradient over the entire length of the tube due to the flow of hot fluid in the shell/tube side and cold fluid in the tube/shell side. There will be two models of flow will prevail. Counter Current Co-Current the profile of temperature for the same thing is mentioned below: Figure 1 How we calculate "ΔT" for the calculation of the thermal load of the heat exchanger? Calculate the heat duty of the heat exchanger, it is necessary to have "ΔT", "Coefficient of heat transfer (U)" and "Surface area of heat transfer(A)". but our concern about " As we know, in the heat exchanger, there will be a temperature gradient over the entire length of the tube due to the flow of hot fluid in the shell/tube side and cold fluid in the tube/shell side. There will be two models of flow will prevail. Counter Current Co-Current the profile of temperature for the same thing is mentioned below: Figure 1 How we calculate "ΔT" for the calculation of the thermal load of the heat exchanger? Calculate the heat duty of the heat exchanger, it is necessary to have "ΔT", "Coefficient of heat transfer (U)" and "Surface area of heat transfer(A)". but our concern about "ΔT (Driving force of heat transfer)" Problem: The temperature gradient changes over the entire length of the tube. To calculate the ΔT (Driving force of heat transfer), we need to find an average temperature gradient of heat transfer to replace the total temperature change in the heat exchanger. This average temperature gradient can be found in two ways. Logarithmic Average Arithmetic Average what is the right average function for finding the average temperature gradient? As we see in Figure 1 above, the counter-current temperature profile curve and the co-current configuration for heat transfer are not straight lines, but a curve line. In mathematics: Arithmetic average used when we are dealing with the straight line, but in our case, this is not possible because the temperature profile line is curved, so we used the logarithmic mean function. In a heat exchanger, the temperature varies logarithmically, both in parallel and counter flow heat exchangers. And the log of these temperature variations gives you a linear graph. And now heat transfer equations can be applied as: Q = UALMTD(ΔT) where Q= heat transfer between two fluids A= Area of Heat Transfer The U=effective heat transfer coefficient Himanshu Patoliya Les McLean Ph.D. in Engineering · Author has 4.9K answers and 12.2M answer views · 7y Related For sensible heating process, shouldn't the difference in energy added Q=m.cp.dT correspond to the increase in enthalpy between the two points? Yes, that’s correct. BUT the formula Q = m•Cp•ΔT is only a rough approximation. That’s because Cp is not a constant; it varies with temperature. So, the proper expression would be Q = m•{the integral of Cp(T) from T1 to T2}. And this can get very messy, since the function for Cp(T) will look something like this (by Landolt): Cp(T) = A + BT•10^-3 + (C•10^5)•T^-2 + (DT^2)•10^-6 Where A, B, C, and D are constants, determined experimentally for the various substances. Note that T is in Kelvin. Related questions What are the units for q in the equation q=mcΔT? What is the difference between LMTD and dT? In Heat Transfer (exchangers) is Q different to q? They seem inter-changeable. When is LMTD invalid? How many amps does a 10kw heater pull 3 phase? What is the difference between Q=mcθdT and Q=UA (LMTD)? What are some interesting topics in electrical engineering that can be done as a project or thesis work? What should someone do if their supervisor is not willing to provide a recommendation? What other options are available in this situation? I am planning to become a marine engineer, I have completed electrical engineering with a power major, but there is very little information on the internet, can anyone please give me information on how can I become one? When is LMTD invalid in some conditions? What are the best titles about electrical installation and maintenance? Give me 3 research titles. Genius and Geniuses: Who was smarter, Albert Einstein or Nikola Tesla? Is the heat transfer rates calculated by both LMTD and NTU method same? Why. Why does a larger flow rate leads to a smaller temperature difference for the process flow in a heat exchanger? I can not find the reason through Q=MCpΔT, because in my experiment Q increases, M increases and ΔT decreases. What is important to consider when doing a PhD at NTNU in Trondheim? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10705	https://ems.press/content/serial-article-files/44245	L’Enseignement Math´ ematique (2) 59 (2013), 115–131 HOW TO TURN A TETRAHEDRON INTO A CUBE AND SIMILAR TRANSFORMATIONS by G. C. S HEPHARD ABSTRACT .Suppose that the surface of a polyhedron P1is cut in such a way that it can be opened out fl at to form a connected region Rin the plane and that R,by introducing suitable folds, can be made into the net of a polyhedron P2. Then we write P1⇒P2and say that P1is transformed into P2. In this paper we give many examples of the transformation of polyhedra and investigate the properties of the relation ⇒. INTRODUCTION Consider the following example. In Figure 1(a) we show a strip of four congruent acute-angled triangles (heavy solid and dashed lines). This is a net of a tetrahedron T because, if the strip is cut out of paper or similar material, folded along the (heavy) dashed lines, and edges with the same labels ( x, y or z) are joined together we obtain a model of (the surface of) the tetrahedron T . Now cut the model along the (lighter) solid lines indicated in Figure 1(a). It will be found that the surface can be opened out fl at in the plane, yielding the shape shown in Figure 1(b). This can be made into a net of a cube C , because if it is folded along the dashed lines, and edges with the same labels ( a, b, c, . . . , g ) are joined together, we obtain the surface of the cube C . Thus we have turned the tetrahedron T into the cube C !We may say that T is transformed into the cube C and write T ⇒ C . In general, if the surface of a polyhedron P is cut in such a way that it can be opened out fl at in the plane (without overlaps) to form a connected region R,then R is called a development of P. If this development only involves cuts along the edges of P then, following Akiyama , we refer to R as an edge-development , or net of P. With this terminology P 1 ⇒ P2 means that 116 G. C. SHEPHARD xxyyzzaad g g effedbccb (a) (b) FIGURE 1 some development of P1 is an edge-development of P2 . The relation P1 ⇒ P2 implies that P1 and P2 have the same surface area — it does not imply that they are isometric. Further, though trivially P ⇒ P, except in special cases, the relation ⇒ is neither symmetric nor transitive (see Section 4). The tetrahedron T in the above example will be called an almost regular tetrahedron since it has, as its faces, four congruent acute-angled triangles. (We note, in passing, that there exists no tetrahedron whose faces are four congruent obtuse-angled triangles or four congruent right-angled triangles.) A tetrahedron is almost regular if it has three pairs of opposite edges of equal length. If these lengths are a, b and c then we denote the tetrahedron by T(a, b, c). Thus the tetrahedron in the above example is T(2 , √5/2, 3√2/2) where the cube C has edge-length 1. It was shown in that some nets may represent more than one polyhedron if the edges to be joined are not labelled (see Section 4). We have labelled the edges of the polygons in Figure 1 but, in general, we omit the labels unless there is the possibility of ambiguity. 2. REGULAR -FACED TESSELLATION POLYHEDRA THEOREM 1. If T ⇒ P where T is an almost regular tetrahedron, then P must be a tessellation polyhedron, that is, a polyhedron of which some net R of P tiles the plane. The idea of a tessellation polyhedron was fi rst introduced in . Consid-ering the net R of P as a closed set, “tiles the plane” means that the plane may be covered by copies of R without gaps or overlaps. That is, the union HOW TO TURN A TETRAHEDRON INTO A CUBE 117 of copies of R is the whole plane, and the intersection of any two copies of R is either empty or of zero area. If such an intersection is non-empty it will be a union of edges, or parts of edges, of copies of R. Proof. This result follows immediately from the following remarkable theorem (see ) : AKIYAMA ’S THEOREM . Every development of an almost regular tetra-hedron T tiles the plane. In fact Akiyama stated and proved this theorem only for regular tetrahedra, but there is no dif fi culty in extending the result to almost regular tetrahedra. FIGURE 2 Theorem 1 greatly simpli fi es the search for polyhedra into which almost regular tetrahedra may be transformed. Initially we shall consider regular-faced polyhedra , that is, polyhedra all of whose 2-faces are regular polygons. A complete enumeration of these was given by Norman Johnson in 1966 . They comprise the fi ve regular (Platonic) and thirteen archimedean solids, the n-prisms, and n-antiprisms for ( n ≥ 3), and 92 others. We shall refer to the latter by the numbers J1–J92 assigned to them in Johnson’s original paper . This list can also be found in Wikipedia. That Johnson’s list is complete was proved by Zalgaller . Of the regular-faced polyhedra it is now known that twenty-two are tessellation polyhedra namely, eight polyhedra all of whose faces are regular (equilateral) triangles (the regular tetrahedron, octahedron, icosahedron, J12, J13, J17, J51 and J84) and twelve with both square and regular triangular faces (J1, J8, J10, J14, J15, J16, J49, J50, J86, J87, J88, J89 118 G. C. SHEPHARD and J90). In addition there is the cube with square faces and the hexagonal antiprism with hexagonal and regular triangular faces. The completeness of this enumeration was established in . In Figure 2 we show a tessellation T for a net of a cube C and the net of an almost regular tetrahedron is superimposed on it. From this we can deduce the example above : the lines along which the surface of the tetrahedron must be cut to transform it into a cube are the edges of T that lie within the net. For each of these twenty-two tessellation polyhedra P i there exists an almost regular tetrahedron Ti such that Ti ⇒ Pi . This statement is established by the diagrams in Figure 3. Each part of this diagram shows a net of P i ,the corresponding tessellation T by this net, and, superimposed, a net of T i .Of course, the tetrahedra Ti are different in each case; their edge-lengths can be read off from the diagrams. All the nets are shaded. Each of the tessellations has symmetry groups of type p2 (see [6, §1.4]) in which the centres of 2-fold symmetry form a (non-rectangular) lattice. Each tetrahedron is a fundamental region (or union of fundamental regions) of the symmetry group of the tessellation T, all its vertices are points of 2-fold rotational symmetry, and edges of T pass through all the vertices of the tetrahedron. CONJECTURE . If P is a convex tessellation polyhedron, then T ⇒ P for some almost regular tetrahedron T . As we have just shown, the conjecture is true for regular-faced polyhedra. The question as to whether the conjecture is true in general is open. In the next section we give further examples of tessellation polyhedra into which a suitable almost regular tetrahedron can be developed, thus giving further evidence of the truth of the conjecture. No counter-examples are known. (a) J1 (Square pyramid) (b) Cube FIGURE 3 HOW TO TURN A TETRAHEDRON INTO A CUBE 119 (c) J8 (d) J10 (e) J12 (Triangular dipyramid) (f) Octahedron (g) J13 (Pentagonal dipyramid) (h) J14 (Triangular spindle) (i) J15 (4-spindle) (j) J16 (5-spindle) FIGURE 3120 G. C. SHEPHARD (k) J17 (Twisted 4-spindle D 16 ) (l) Icosahedron (m) J49 (n) J50 (o) J51 (Deltahedron D 14 ) (p) J84 (Deltahedron D 12 ) FIGURE 3HOW TO TURN A TETRAHEDRON INTO A CUBE 121 (q) J86 (r) J87 (s) J88 (t) J89 (u) J90 (v) Hexagonal antiprism FIGURE 3122 G. C. SHEPHARD OTHER TESSELLATION POLYHEDRA An n-spindle (n ≥ 3) is a polyhedron consisting of two n-pyramids adjoined to opposite faces of an n-prism based on a regular n-gon (see Fig-ure 4(a) for an example with n = 6). We may denote this spindle by P n(r, s, t)where the heights r , s and t of the components, as indicated in the diagram, are any (strictly) positive quantities. It has 3 n faces : n rectangular faces and 2n isosceles triangular faces of two kinds. For 3 ≥ n ≥ 5, an n-spindle may be regular-faced (see Figure 3 part (h) for n = 3, part (i) for n = 4 and part ( j) for n = 5). In all cases, independent of the choice of n, r , s and t , the spin-dle is a tessellation polyhedron. See Figure 4(b) which shows the tessellation corresponding to the 6-spindle in Figure 4(a). Moreover, as indicated on the tessellation, there is an almost-regular tetrahedron T with T ⇒ Pn(r, s, t). t s r (a)(b) FIGURE 4 If we put s = 0, then we obtain the n-dipyramid (see Figure 5(a) for an example with n = 8). This is also a tessellation polyhedron, but only if the heights r and t are equal. If n = 3, 4 or 5 then the n-dipyramid may be regular-faced (see Figure 3 part (e) for n = 3, part (f) for n = 4, the octahedron, and part (g) for n = 5). In Figure 5(a) we also show a tessella-tion for a net of a dipyramid D6 , and indicate the net of an almost regular tetrahedron T such that T ⇒ D6 . An analogous result holds for all n ≥ 3. Further possibilities arise, such as that shown in Figure 5(b). If n is even we can cut an n-dipyramid (here n = 8) into two equal parts by a plane through the apexes of the constituent pyramids and four of its edges. The corresponding tessellation and net of an almost regular tetrahedron are also shown in Figure 5(b). HOW TO TURN A TETRAHEDRON INTO A CUBE 123 (a) (b) FIGURE 5 Further examples of tessellation polyhedra are the twisted spindles. Each of these consists of two n-pyramids adjoined to the n-gonal faces of an n-antiprism, see Figure 6(a) for the case n = 6. Here r, s and t must satisfy the inequalities 0 < r (4 − 1) < s and 0 < t (4 − 1) < s where 4 = (cos( π/ n)) −1 if the resulting polyhedron is to be strictly convex. There are 4 n triangular faces, of three kinds. If n = 4 or 5, all the faces can be regular (equilateral) triangles (See Figure 3 part (k) the deltahedron D 16 for n = 4, and part (l), the icosahedron, for n = 5). In Figure 6(b) we show a tessellation for a net of this polyhedron (with n = 6), and also a net of the almost regular tetrahedron which can be transformed into it. Certain special cases are of interest. If r = t = s(4 − 1) −1 , where 4 is de fi ned as above, then the faces of the antiprism and pyramids merge to yield a polyhedron with 2 n kite-shaped faces which we call a kite-polyhedron .See Figure 7(a) for an example for which n = 6. In Figure 7(b) we show a tessellation for a net of this kite polyhedron and also indicate a net of the almost regular tetrahedron which can be transformed into it. 124 G. C. SHEPHARD In Figure 7(c) we show the case of a twisted spindle with n = 3, and r = s = t . The six faces are rhombs and the polyhedron is a rhomboid. The corresponding tessellation is shown in Figure 7(d). Notice that the net is an af fi ne image of a net of a cube, and that the rhomboid itself is the af fi ne image of a cube. In general, a non-singular af fi ne image of a net is not a net of any polyhedron. Those nets shown in Figures 4, 5, 6 and 7, have the unexpected property that there exist af fi ne images of each of the nets which are also nets of polyhedra of the same type. Further examples are provided by multiple polyhedra. We write T ⇒ nP if a development of T is the union of n nets of the polyhedron P. An example is shown in Figure 8. Here P is a cube and nets of the various almost regular tetrahedra T for which T ⇒ nP are indicated for n = 1, 2, 3, 4, 5. Similar constructions clearly apply for larger values of n.All these examples serve to strengthen con fi dence in the truth of the conjecture. 4. THE RELATION ⇒ We have remarked that the relation ⇒ is re fl exive. Now we shall show : THEOREM 2. The relation ⇒ is not symmetric. Proof. All we need to prove the theorem is to display one counter-example. In the Introduction we showed that T ⇒ C where C is a cube and T is the almost regular tetrahedron T(2 , √5/2, 3√2/2). We shall now show that C ⇒ T where C is a cube and T is any almost regular tetrahedron. To do this we remember that P1 ⇒ P2 means that some development R of P1 is a net of P2 . Such a development R of P1 is obtained by cutting the surface of P1 along the edges of a hamiltonian tree T . Because T is a tree (a graph with no circuits) the resulting region R will be connected and because it is hamiltonian (incident with every vertex of P 1 ) the region R is planar. Now every tree has at least two edges of which an end-point has valency one. Consider such an end-point. Clearly it must be a vertex of P 1 (if it were an interior point of a face or an edge the cut would not yield the development R). In the case of a cube, since the sum of the angles of the faces at any vertex is 3 π/ 2 the development R must have at least two vertices with this angle. But then R cannot be a net of any almost regular HOW TO TURN A TETRAHEDRON INTO A CUBE 125 t s r (a) (b) FIGURE 6 t s r (a) (b) (c) (d) FIGURE 7126 G. C. SHEPHARD Cube n=1 n=2 n=3 n=4 n=5 FIGURE 8 tetrahedron T . In fact C ⇒ P where P is any polyhedron whose faces are all regular (equilateral) triangles (J12, J13, J17, J51, J84, the octahedron and icosahedron) since no net of such a polyhedron can have a vertex angle 3 π/ 2. More generally, if P is a regular-faced polyhedron, it is impossible for C ⇒ P unless P has at least three squares or two octagons as faces. Thus C ⇒ P is impossible for all the “Modi fi ed Platonic Solids” J58–J64 in Johnson’s list . It is also impossible in the case of the 8-prism and 8-antiprism, each with two octagons, since the octagons are not adjacent. A modi fi cation of the above argument shows that the region R must have at least v(P1) − p(P1) vertices, where v(P1) is the number of vertices of P1 and p(P1) is the number of vertices at which the sum of the angles of the faces is π .In the case of a cube, v(C) = 8, p(C) = 0 and so any development R will have at least eight vertices. But R cannot be a net of any tetrahedron since all such nets (as is easily veri fi ed) have at most six vertices. This proves the assertion. HOW TO TURN A TETRAHEDRON INTO A CUBE 127 THEOREM 3. The relation ⇒ is not transitive. Proof. As in Theorem 2, we need only display one counterexample. Although a polyhedron P may have several nets, each of which may lead to several tessellations, with a very few exceptions (such as, for example J1, J12, J13, J17, J84, octahedron and icosahedron) there are only a fi nite number of triples ( a, b, c) such that T(a, b, c) ⇒ P. However, it is not dif fi cult to show that for any given ( a, b, c) there are in fi nitely many triples ( r, s, t) such that T(r, s, t) ⇒ T(a, b, c). Clearly we may choose ( r, s, t) so that T(r, s, t) ⇒ P.Then we have T(r, s, t) ⇒ T(a, b, c) and T(a, b, c) ⇒ P but these do not imply T(r, s, t) ⇒ P. Hence ⇒ is not transitive. ABCDESTXYZ ABYAC CDZED TXS AYC DCZ (a) (b) (c) FIGURE 9 From the above we infer that there are very few examples of P 1 ⇒ P2 where P1 is not a tetrahedron. One possible reason for this is that there seems to be no theorem for general polyhedra analogous to Theorem 1. We conclude this section with two examples where P1 ⇔ P2 .In Figure 9(b) we show a development R of the triangular dipyramid J12 (Figure 9(a)). Here R is a parallelogram. This can be split into four congruent triangles in such a way as to form the net of an almost regular tetrahedron T = T(1 , 3/2, √7/2) (Figure 9(c)). Hence J12 ⇒ T .Other examples arise when two distinct polyhedra have identical (unla-belled) nets. One such, reproduced from , is shown in Figure 10. The net, shown in Figure 10(a) and (b), is a chain of eight triangles with edges of two different lengths. (In the diagram the lengths are 1 and 1.25). In these fi gures 128 G. C. SHEPHARD ABCDEFGHJKaabbcddeec ABCDEFGHJKabccbadeed GDJBF=H=KA=C=E JB=FH=KA=GDC=E (a) (b) (c) (d) FIGURE 10 we have indicated two distinct labellings of the edges. If the similarly labelled edges in Figure 10(a) are joined, we obtain the net of a stack polyhedron S (Figure 10(c)), whereas the labelling in Figure 10(b) leads to the (non-regular) octahedron O in Figure 10(d). As the nets of the polyhedra are identical, clearly S ⇒ O. Also O ⇒ S . A few other examples of distinct polyhedra having identical (unlabelled) nets are known, but there is no general theory on how these can be constructed. In all these cases the relation is symmetric, and we may write S ⇔ O.5. FINAL REMARKS Nets of polyhedra have been known for nearly fi ve hundred years and hundreds of examples of nets of polyhedra can be found on the internet. A complete set of nets for all regular and archimedean polyhedra can be found in . However, there seems to be very little mathematical literature on the subject and many fundamental questions remain unanswered. We now state some of these : HOW TO TURN A TETRAHEDRON INTO A CUBE 129 (A) Does every convex polyhedron have a net ? It is a common experience of model builders that the construction of a net for a convex polyhedron P fails because, after slitting along the edges of P that form a hamiltonian tree, and opening out the surface in the plane, the resulting region R has overlaps, and so is not a net. One reason that this problem has not been solved is that until recently, it seems that few mathematicians realised that there was any problem at all ! And even if they did, they assumed that the answer was positive. One of the fi rst explicit statements of the problem appears in . It is trivial to show there is a negative answer to the question to the corresponding problem for polyhedra with non-convex faces, and several mathematicians have also shown that the answer is negative for non-convex polyhedra with convex polygons as faces, see , but the question for convex polyhedra in general remains open. (B) Given a set of (closed) convex polygons in the plane which are disjoint (except for their edges) and whose union is a connected set. Under what conditions do these polygons form the net of a convex polyhedron ? More generally : (C) Given a plane polygon R, what are necessary and suf fi cient conditions for it to be possible to introduce fold lines into R so as to make it into a net of some polyhedron P ? One might hope that, if the polygon is convex, the answer would be positive since many convex developments are known (see ). But notice that the answer is negative without some clari fi cation of the types of polyhedra that are allowed. For example, it seems that a rectangle cannot be made into the net of any strictly convex polyhedron (but can be made into the net of a (fl at) degenerate tetrahedron bounded by four congruent right-triangles). Another problem related to ( B), is to determine the number of distinct nets of a given polyhedron. Again, careful de fi nitions are required if this question is to be meaningful. In Professor Jin Akiyama shows that a regular tetrahedron has two distinct nets, a cube has eleven distinct nets, and a regular octahedron also has eleven distinct nets. But if the faces are labelled, or otherwise distinguished, (for example if the polyhedron is distorted so that the faces are of slightly different shapes) then the number of nets is considerably 130 G. C. SHEPHARD larger. For example, in the case of a labelled cube, the number of distinct nets is 11 × 6 × 4 = 264. The number of nets of the regular dodecahedron and regular icosahedron (43,380) have been determined by S. Bouzette and F. Vandamme and also by Ch. Hippenmeyer (see for references to the relevant works). In the number of nets of twenty-one of the regular-faced polyhedra is given. In each case, it is not known how many of these are “genuine” nets since they may involve “overlaps” and it remains an unsolved problem to determine the numbers of “genuine” nets. All these numbers are for “labelled polyhedra”. They are surprisingly large, for example there are over fi ve billion distinct nets of the polyhedron J44 (the gyroelongated triangular bicupola). The numbers were determined by computer; for details see . There seems to be no general theory, and all the numbers have been found empirically, either by counting hamiltonian edge-trees, or the number of dual trees. There are further problems relating to tessellation polyhedra. As stated above, a complete list of regular-faced tessellation polyhedra is given in . This list was determined by computer. It would be interesting to decide whether these polyhedra, and those in Section 3, are the only tessellation polyhedra, and so whether the conjecture in Section 2 is correct. There are many other problems, such as : (D) Does there exist a convex polyhedron P of which no net tiles the plane (so it is not a tessellation polyhedron) but two distinct nets of P do so, thus forming a dihedral tiling in the sense of [6, p. 23] ? If such a polyhedron exists we may call it a 2-tessellation polyhedron and, in a similar manner an n-tessellation polyhedron can be de fi ned. No n-tessellation polyhedra are known for any n ≥ 2 though their discovery would appear to be a not very dif fi cult problem. Finally, it would be interesting to discover more examples of unlabelled nets (other than those shown in Figure 10), which, if labelled in a suitable manner can represent two (or more) distinct polyhedra. HOW TO TURN A TETRAHEDRON INTO A CUBE 131 REFERENCES AKIYAMA , J. Tile-makers and semi-tile-makers. Amer. Math. Monthly 114 (2007), 602–609. AKIYAMA , J., T. K UWATA , S. L ANGERMAN , K. O KAWA , I. S ATO and G. C. SHEPHARD . Determination of all tessellation polyhedra with regular polygonal faces. In : Computational Geometry, Graphs and Applications ,9th International Conference, China 2010, 1–11. Lecture Notes in Comput. Sci. 7033 . Springer, Heidelberg, 2011. BUEKENHOUT , F. and M. P ARKER . The number of nets of the regular convex polytopes in dimension ≤ 4. Discrete Math. 186 (1998), 69–94. CUNDY , H. M. and A. P. ROLLETT . Mathematical Models . Second edition. Oxford University Press, 1961. D ˝URER , A. Underweisung der Messung mit dem Zirkel und Richtscheit ,Heironymus Andreaea, 1525. English translation by Walter L. Strauss, The Painters’ Manual , Abaris Books, New York, 1977. GR ¨UNBAUM , B. and G. C. S HEPHARD . Tilings and Patterns . W. H. Freeman and Co., New York, 1987. JOHNSON , N. W. Convex polyhedra with regular faces. Canadian J. Math. 18 (1966), 169–200. SHEPHARD , G. C. Convex polytopes with convex nets. Math. Proc. Cambridge Philos. Soc. 78 (1975), 389–403. Tessellation Polyhedra. Symmetry : Culture and Science 22 (2011), Special Issue on Tessellations, Part 1, 65–82. ZALGALLER , V. A. Convex polyhedra with regular faces. Translated from the Russian. Seminars in Mathematics, V. A. Steklov Mathematical Institute, Leningrad, Vol. 2. Consultants Bureau, New York 1969, vi+95 pp. (Rec ¸u le 30 janvier 2012) G. C. Shephard University of East Anglia Norwich, NR4 7TJ England, U. K. e-mail : g.c.shephard@uea.ac.uk
10706	http://www.vdict.co/index.php?in=1&dict=en_wordfinder&word=echelon	Tra từ echelon - Từ điển Oxford Learners Wordfinder Dictionary Vndic.net quick lookup Tra TừDịch Nguyên CâuBài DịchLớp Ngoại NgữGo47Tử ViEnglish Cấp TốcLuyện thi IELTS History Search echelon [Clear Recent History...]Từ điển Oxford Learners Wordfinder Dictionary echelon noun 1. a body of troops arranged in a line • Topics: military, armed forces, armed services, military machine, war machine • Hypernyms: military unit, military force, military group, force 2. a diffraction grating consisting of a pile of plates of equal thickness arranged stepwise with a constant offset • Hypernyms: diffraction grating, grating Giới thiệu VNDIC.net \| Plugin từ diển cho Firefox \| Từ điển cho Toolbar IE \| Tra cứu nhanh cho IE \| Vndic bookmarklet \| Học từ vựng \| Vndic trên web của bạn © Copyright 2006-2025 VNDIC.NET&VDICT.CO all rights reserved.
10707	https://stackoverflow.com/questions/28712279/all-possible-tic-tac-toe-winning-combinations	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog All Possible Tic Tac Toe Winning Combinations Ask Question Asked Modified 2 years, 4 months ago Viewed 54k times This question shows research effort; it is useful and clear 7 Save this question. Show activity on this post. I had an interview were I was asked a seemingly simple algorithm question: "Write an algorithm to return me all possible winning combinations for tic tac toe." I still can't figure out an efficient way to handle this. Is there a standard algorithm or common that should be applied to similar questions like this that I'm not aware of? algorithm combinations graph-theory tic-tac-toe Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Oct 27, 2022 at 3:16 Sebastian 1,98422 gold badges1212 silver badges2323 bronze badges asked Feb 25, 2015 at 6:04 DroidTDroidT 3,36333 gold badges3434 silver badges3030 bronze badges 2 paxdiablo's answer works; you could also approach it from 'the other side': start from a blank board, and play out every possible game, keeping track of the final winning positions reached. This would be more work than paxdiablo's answer, but for a more complex game than tic tac toe might turn out to be easier. – AakashM Commented Feb 25, 2015 at 11:57 Is a winning combination the final board configuration or also the moves up to it? – Sebastian Commented Oct 22, 2022 at 7:48 Add a comment \| 8 Answers 8 Reset to default This answer is useful 10 Save this answer. Show activity on this post. This is one of those problems that's actually simple enough for brute force and, while you could use combinatorics, graph theory, or many other complex tools to solve it, I'd actually be impressed by applicants that recognise the fact there's an easier way (at least for this problem). There are only 39, or 19,683 possible combinations of placing x, o or <blank> in the grid, and not all of those are valid. First, a valid game position is one where the difference between x and o counts is no more than one, since they have to alternate moves. In addition, it's impossible to have a state where both sides have three in a row, so they can be discounted as well. If both have three in a row, then one of them would have won in the previous move. There's actually another limitation in that it's impossible for one side to have won in two different ways without a common cell (again, they would have won in a previous move), meaning that: ``` XXX OOO XXX ``` cannot be achieved, while: ``` XXX OOX OOX ``` can be. But we can actually ignore that since there's no way to win two ways without a common cell without having already violated the "maximum difference of one" rule, since you need six cells for that, with the opponent only having three. So I would simply use brute force and, for each position where the difference is zero or one between the counts, check the eight winning possibilities for both sides. Assuming only one of them has a win, that's a legal, winning game. Below is a proof of concept in Python, but first the output of time when run on the process sending output to /dev/null to show how fast it is: ``` real 0m0.169s user 0m0.109s sys 0m0.030s ``` The code: ``` def won(c, n): if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 return 0 pc = [' ', 'x', 'o'] c = 9 for c in range (3): for c in range (3): for c in range (3): for c in range (3): for c in range (3): for c in range (3): for c in range (3): for c in range (3): for c in range (3): countx = sum([1 for x in c if x == 1]) county = sum([1 for x in c if x == 2]) if abs(countx-county) < 2: if won(c,1) + won(c,2) == 1: print " %s \| %s \| %s" % (pc[c],pc[c],pc[c]) print "---+---+---" print " %s \| %s \| %s" % (pc[c],pc[c],pc[c]) print "---+---+---" print " %s \| %s \| %s" % (pc[c],pc[c],pc[c]) print ``` As one commenter has pointed out, there is one more restriction. The winner for a given board cannot have less cells than the loser since that means the loser just moved, despite the fact the winner had already won on the last move. I won't change the code to take that into account but it would be a simple matter of checking who has the most cells (the last person that moved) and ensuring the winning line belonged to them. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Feb 25, 2015 at 9:42 answered Feb 25, 2015 at 6:08 paxdiablopaxdiablo 887k241241 gold badges1.6k1.6k silver badges2k2k bronze badges 7 1 @Pham, possibly, but to what end? We don't need to save time or space here, and using bits means packing and unpacking which is likely to make the code more complex. Unless I've misunderstood, in which case I'd apprciate clarification. – paxdiablo Commented Feb 25, 2015 at 7:41 I got your point :) totally forgot about space option, so I will take back my comment :) – Pham Trung Commented Feb 25, 2015 at 7:45 1 A difference of one move should not be allowed both ways. The loser can't move after the opponent has won. – Janne Karila Commented Feb 25, 2015 at 8:21 @JanneKarila, good point, I'll add that into the answer but I won't bother with changing the code. – paxdiablo Commented Feb 25, 2015 at 9:40 Thanks @paxdiablo for the detailed answer! Would you mind clarifying what is going on with the code, specifically in the last for loop, for us non php developers? – DroidT Commented Feb 25, 2015 at 18:39 \| Show 2 more comments This answer is useful 3 Save this answer. Show activity on this post. Another way could be to start with each of the eight winning positions, ``` xxx --- --- xxx --- --- ... etc., ``` and recursively fill in all legal combinations (start with inserting 2 o's, then add an x for each o ; avoid o winning positions): ``` xxx xxx xxx oo- oox oox --- o-- oox ... etc., ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Feb 25, 2015 at 20:57 answered Feb 25, 2015 at 20:47 גלעד ברקןגלעד ברקן 24k33 gold badges2929 silver badges6464 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Today I had an interview with Apple and I had the same question. I couldn't think well at that moment. Later one on, before going to a meeting I wrote the function for the combinations in 15 minutes, and when I came back from the meeting I wrote the validation function again in 15 minutes. I get nervous at interviews, Apple not trusts my resume, they only trust what they see in the interview, I don't blame them, many companies are the same, I just say that something in this hiring process doesn't look quite smart. Anyways, here is my solution in Swift 4, there are 8 lines of code for the combinations function and 17 lines of code to check a valid board. Cheers!!! ``` // Not used yet: 0 // Used with x : 1 // Used with 0 : 2 // 8 lines code to get the next combination func increment ( _ list: inout [Int], _ base: Int ) -> Bool { for digit in 0..<list.count { list[digit] += 1 if list[digit] < base { return true } list[digit] = 0 } return false } let incrementTicTacToe = { increment(&$0, 3) } let win0_ = [0,1,2] // [1,1,1,0,0,0,0,0,0] let win1_ = [3,4,5] // [0,0,0,1,1,1,0,0,0] let win2_ = [6,7,8] // [0,0,0,0,0,0,1,1,1] let win_0 = [0,3,6] // [1,0,0,1,0,0,1,0,0] let win_1 = [1,4,7] // [0,1,0,0,1,0,0,1,0] let win_2 = [2,5,8] // [0,0,1,0,0,1,0,0,1] let win00 = [0,4,8] // [1,0,0,0,1,0,0,0,1] let win11 = [2,4,6] // [0,0,1,0,1,0,1,0,0] let winList = [ win0_, win1_, win2_, win_0, win_1, win_2, win00, win11] // 16 lines to check a valid board, wihtout countin lines of comment. func winCombination (_ tictactoe: [Int]) -> Bool { var count = 0 for win in winList { if tictactoe[win] == tictactoe[win], tictactoe[win] == tictactoe[win], tictactoe[win] != 0 { // If the combination exist increment count by 1. count += 1 } if count == 2 { return false } } var indexes = Array(repeating:0, count:3) for num in tictactoe { indexes[num] += 1 } // '0' and 'X' must be used the same times or with a diference of one. // Must one and only one valid combination return abs(indexes - indexes) <= 1 && count == 1 } // Test var listToIncrement = Array(repeating:0, count:9) var combinationsCount = 1 var winCount = 0 while incrementTicTacToe(&listToIncrement) { if winCombination(listToIncrement) == true { winCount += 1 } combinationsCount += 1 } print("There is (combinationsCount) combinations including possible and impossible ones.") print("There is (winCount) combinations for wining positions.") / There are 19683 combinations including possible and impossible ones. There are 2032 combinations for winning positions. / listToIncrement = Array(repeating:0, count:9) var listOfIncremented = "" for _ in 0..<1000 { // Win combinations for the first 1000 combinations _ = incrementTicTacToe(&listToIncrement) if winCombination(listToIncrement) == true { listOfIncremented += ", (listToIncrement)" } } print("List of combinations: (listOfIncremented)") / List of combinations: , [2, 2, 2, 1, 1, 0, 0, 0, 0], [1, 1, 1, 2, 2, 0, 0, 0, 0], [2, 2, 2, 1, 0, 1, 0, 0, 0], [2, 2, 2, 0, 1, 1, 0, 0, 0], [2, 2, 0, 1, 1, 1, 0, 0, 0], [2, 0, 2, 1, 1, 1, 0, 0, 0], [0, 2, 2, 1, 1, 1, 0, 0, 0], [1, 1, 1, 2, 0, 2, 0, 0, 0], [1, 1, 1, 0, 2, 2, 0, 0, 0], [1, 1, 0, 2, 2, 2, 0, 0, 0], [1, 0, 1, 2, 2, 2, 0, 0, 0], [0, 1, 1, 2, 2, 2, 0, 0, 0], [1, 2, 2, 1, 0, 0, 1, 0, 0], [2, 2, 2, 1, 0, 0, 1, 0, 0], [2, 2, 1, 0, 1, 0, 1, 0, 0], [2, 2, 2, 0, 1, 0, 1, 0, 0], [2, 2, 2, 1, 1, 0, 1, 0, 0], [2, 0, 1, 2, 1, 0, 1, 0, 0], [0, 2, 1, 2, 1, 0, 1, 0, 0], [2, 2, 1, 2, 1, 0, 1, 0, 0], [1, 2, 0, 1, 2, 0, 1, 0, 0], [1, 0, 2, 1, 2, 0, 1, 0, 0], [1, 2, 2, 1, 2, 0, 1, 0, 0], [2, 2, 2, 0, 0, 1, 1, 0, 0] / ``` Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Jan 4, 2019 at 20:34 answered Jan 4, 2019 at 7:53 ideastouchideastouch 1,4371212 silver badges66 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. This is a java equivalent code sample package testit; public class TicTacToe { ``` public static void main(String[] args) { // TODO Auto-generated method stub // 0 1 2 // 3 4 5 // 6 7 8 char[] pc = {' ' ,'o', 'x' }; char[] c = new char; // initialize c for (int i = 0; i < 9; i++) c[i] = pc; for (int i = 0; i < 3; i++) { c = pc[i]; for (int j = 0; j < 3; j++) { c = pc[j]; for (int k = 0; k < 3; k++) { c = pc[k]; for (int l = 0; l < 3; l++) { c = pc[l]; for (int m = 0; m < 3; m++) { c = pc[m]; for (int n = 0; n < 3; n++) { c = pc[n]; for (int o = 0; o < 3; o++) { c = pc[o]; for (int p = 0; p < 3; p++) { c = pc[p]; for (int q = 0; q < 3; q++) { c = pc[q]; int countx = 0; int county = 0; for(int r = 0 ; r<9 ; r++){ if(c[r] == 'x'){ countx = countx + 1; } else if(c[r] == 'o'){ county = county + 1; } } if(Math.abs(countx - county) < 2){ if(won(c, pc)+won(c, pc) == 1 ){ System.out.println(c + " " + c + " " + c); System.out.println(c + " " + c + " " + c); System.out.println(c + " " + c + " " + c); System.out.println(""); } } } } } } } } } } } } public static int won(char[] c, char n) { if ((c == n) && (c == n) && (c == n)) return 1; else if ((c == n) && (c == n) && (c == n)) return 1; else if ((c == n) && (c == n) && (c == n)) return 1; else if ((c == n) && (c == n) && (c == n)) return 1; else if ((c == n) && (c == n) && (c == n)) return 1; else if ((c == n) && (c == n) && (c == n)) return 1; else if ((c == n) && (c == n) && (c == n)) return 1; else if ((c == n) && (c == n) && (c == n)) return 1; else return 0; } ``` } ` Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Dec 19, 2018 at 22:37 Vijay RumaoVijay Rumao 7244 bronze badges 1 equivalent to what? – Sebastian Commented Oct 22, 2022 at 7:46 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Below Solution generates all possible combinations using recursion It has eliminated impossible combinations and returned 888 Combinations Below is a working code Possible winning combinations of the TIC TAC TOE game ``` const players = ['X', 'O']; let gameBoard = Array.from({ length: 9 }); const winningCombination = [ [ 0, 1, 2 ], [ 3, 4, 5 ], [ 6, 7, 8 ], [ 0, 3, 6 ], [ 1, 4, 7 ], [ 2, 5, 8 ], [ 0, 4, 8 ], [ 2, 4, 6 ], ]; const isWinningCombination = (board)=> { if((Math.abs(board.filter(a => a === players).length - board.filter(a => a === players).length)) > 1) { return false } let winningComb = 0; players.forEach( player => { winningCombination.forEach( combinations => { if (combinations.every(combination => board[combination] === player )) { winningComb++; } }); }); return winningComb === 1; } const getCombinations = (board) => { let currentBoard = [...board]; const firstEmptySquare = board.indexOf(undefined) if (firstEmptySquare === -1) { return isWinningCombination(board) ? [board] : []; } else { return [...players, ''].reduce((prev, next) => { currentBoard[firstEmptySquare] = next; if(next !== '' && board.filter(a => a === next).length > (gameBoard.length / players.length)) { return [...prev] } return [board, ...prev, ...getCombinations(currentBoard)] }, []) } } const startApp = () => { let combination = getCombinations(gameBoard).filter(board => board.every(item => !(item === undefined)) && isWinningCombination(board) ) printCombination(combination) } const printCombination = (combination)=> { const ulElement = document.querySelector('.combinations'); combination.forEach(comb => { let node = document.createElement("li"); let nodePre = document.createElement("pre"); let textnode = document.createTextNode(JSON.stringify(comb)); nodePre.appendChild(textnode); node.appendChild(nodePre); ulElement.appendChild(node); }) } startApp(); ``` Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Jun 7, 2020 at 18:38 Owen KelvinOwen Kelvin 15.1k1111 gold badges5151 silver badges8585 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Could be solved with brute force but keep in mind the corner cases like player2 can't move when player1 has won and vice versa. Also remember Difference between moves of player1 and player can't be greater than 1 and less than 0. I have written code for validating whether provided combination is valid or not, might soon post on github. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered May 16, 2021 at 19:19 Hariom SinghHariom Singh 1122 bronze badges 2 1 Thank you for your contribute but all you wrote is already covered in @paxdiablo 's answer. – Masoud Keshavarz Commented May 17, 2021 at 8:23 If player1 is winner then they can't have even equal cells and it's one of the corner case which is not mentioned. Also don't think this needed the down vote else would be difficult for new comers to increase the contribution. – Hariom Singh Commented May 18, 2021 at 11:15 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. This discovers all possible combinations for tic tac toe (255,168) -- written in JavaScript using recursion. It is not optimized, but gets you what you need. ``` const [EMPTY, O, X] = [0, 4, 1] let count = 0 let coordinate = [ EMPTY, EMPTY, EMPTY, EMPTY, EMPTY, EMPTY, EMPTY, EMPTY, EMPTY ] function reducer(arr, sumOne, sumTwo = null) { let func = arr.reduce((sum, a) => sum + a, 0) if((func === sumOne) \|\| (func === sumTwo)) return true } function checkResult() { let [a1, a2, a3, b1, b2, b3, c1, c2, c3] = coordinate if(reducer([a1,a2,a3], 3, 12)) return true if(reducer([a1,b2,c3], 3, 12)) return true if(reducer([b1,b2,b3], 3, 12)) return true if(reducer([c1,c2,c3], 3, 12)) return true if(reducer([a3,b2,c1], 3, 12)) return true if(reducer([a1,b1,c1], 3, 12)) return true if(reducer([a2,b2,c2], 3, 12)) return true if(reducer([a3,b3,c3], 3, 12)) return true if(reducer([a1,a2,a3,b1,b2,b3,c1,c2,c3], 21)) return true return false } function nextPiece() { let [countX, countO] = [0, 0] for(let i = 0; i < coordinate.length; i++) { if(coordinate[i] === X) countX++ if(coordinate[i] === O) countO++ } return countX === countO ? X : O } function countGames() { if (checkResult()) { count++ }else { for (let i = 0; i < 9; i++) { if (coordinate[i] === EMPTY) { coordinate[i] = nextPiece() countGames() coordinate[i] = EMPTY } } } } countGames() console.log(count) ``` I separated out the checkResult returns in case you want to output various win conditions. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Oct 27, 2022 at 12:28 James Risner 6,0891111 gold badges3232 silver badges5959 bronze badges answered Oct 22, 2022 at 7:34 TimTim 6911 silver badge33 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Most of the answers for this question are quite slow so here's a faster approach ``` def won(c, n): if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 if c == n and c == n and c == n: return 1 return 0 for count in range(39): grid = [(count // (3 i)) % 3 for i in range(9)] if won(grid, 1) + won(grid, 2) == 1: print(grid) ``` Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Apr 15, 2023 at 5:04 answered Apr 13, 2023 at 23:51 UnknownEncoderUnknownEncoder 11322 silver badges99 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm combinations graph-theory tic-tac-toe See similar questions with these tags. 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10708	https://www.sciencedirect.com/science/article/abs/pii/S0301211514000529	Female sterilization and refertilization - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract Introduction Section snippets References (20) Cited by (4) European Journal of Obstetrics & Gynecology and Reproductive Biology Volume 175, April 2014, Pages 82-86 Female sterilization and refertilization Author links open overlay panel Anneloes N.J.Huijgens, Céline M.J.G.Lardenoije, Helen J.M.M.Mertens Show more Add to Mendeley Share Cite rights and content Abstract Objective To analyze data on sterilization and refertilization procedures that took place at Orbis Medical Center in Sittard, a hospital in the south of the Netherlands. Study design Retrospective cohort study of surgical tubal sterilizations performed on 966 patients for contraception between 2002 and 2011, and of 19 patients who underwent refertilization between 2002 and 2012. The main outcome measures were complications and failure rates of sterilization, motives for refertilization and pregnancy rates after refertilization. The t test and nonparametric tests were used to determine differences between groups and proportions. Results Between 2002 and 2011, the number of sterilizations declined. Almost all the patients (99.8%) underwent laparoscopic sterilization. The most common method of sterilization used Filshie clips, and was used in 99.7% of the women. The median age at the time of sterilization was 37 years. The failure rate was 0.3%. All procedures were uneventful. The number of refertilizations during this time period also declined. The median time between sterilization and refertilization was 65 months. Patients who underwent refertilization were significantly younger at time of sterilization than patients who did not (p<0.001). After refertilization, 12 patients (63.2%) became pregnant. Conclusions The complication and failure rates of laparoscopic sterilization are low. The number of laparoscopic sterilizations and the number of refertilizations are both declining. Still, refertilization is a safe procedure and gives a significant chance of becoming pregnant. Introduction Female sterilization is a common method of contraception worldwide. In the Netherlands, more than 10,000 sterilizations are performed each year . Surgical techniques to occlude tubal patency include the application of a mechanical device, tubal electro-coagulation, and tubal excision, with or without separation. In Western countries, laparoscopic tubal occlusion is the most common surgical method for sterilization . Data regarding the effectiveness and regret of sterilization are limited. While sterilization at the time of a cesarean section (CS) eliminates the need for additional surgery, it is known to be less effective due to oedematous and friable fallopian tubes in the early postpartum period , and it should be noted that this timing for deciding on definitive contraception is associated with higher rates of regret . In this study, we analyzed data reported for sterilization and refertilization in the south of the Netherlands during the past decade. The sterilization techniques used in the Netherlands are the Filshie clip, the Falope ring, bipolar electrocoagulation, and the Pomeroy, Irving, and Uchida techniques . Bipolar electrocoagulation, the Filshie clip and the Falope ring are mostly applied and usually performed by laparoscopy. The Pomeroy, Irving, and Uchida methods are mostly performed by laparotomy; the Irving method may also be accomplished by laparoscopy. Hysteroscopic sterilization procedures have been introduced recently, but these techniques were not performed in this cohort and, therefore, will not be discussed. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Materials and methods This study included all patients who underwent surgical tubal sterilization for contraception between 2002 and 2011 in the Orbis Medical Center (OMC), as well as all patients who underwent a refertilization between 2002 and 2012 in the OMC. The OMC is an independent teaching hospital in the south of the Netherlands. All data were analyzed retrospectively. The follow-up lasted until 2012. Sterilization Between 2002 and 2011, 966 patients underwent tubal sterilization. Of all the patients, 964 women (99.8%) were sterilized by laparoscopy and two women (0.2%) were sterilized at the time of CS. The Filshie clip method was used in 963 women (99.7%), while bipolar coagulation was used in three women (0.3%) because of thick hemorrhagic tubes. The mean age (±SD) of the patients at the time of laparoscopic sterilization was 36.4±5.4 years (median age was 37 years). The mean age (±SD) of the patients Sterilization In the Netherlands, tubal sterilization is a laparoscopic surgical procedure, except for immediate postpartum sterilization performed during a CS , . Filshie clips are preferred for laparoscopic or laparotomic tubal occlusion because, while they have similar failure rates, there is a higher rate of minor morbidity (OR: 2.15; 95% CI: 1.22, 3.78) and of technical difficulties (OR: 3.87; 95% CI: 1.90, 7.89) for tubal ring sterilization versus clip sterilization , , . For laparotomic Recommended articles References (20) A.H. Graf et al. An evaluation of the Filshie clip for postpartum sterilization in Austria Contraception (1996) S.D. Hillis et al. Poststerilization regret: findings from the United States collaborative review of sterilization Obstet Gynecol (1999) J.S. Yan et al. Comparative study of Filshie clip and Pomeroy method for postpartum sterilization Int J Gynaecol Obstet (1990) D. Sokal et al. Two randomized controlled trials comparing the tubal ring and Filshie clip for tubal sterilization Fertil Steril (2000) B.A. Kohaut et al. Randomized trial tocompare perioperative outcomes of Filshie clip vs. Pomeroy technique for postpartum and intraoperative cesarean tubal sterilization: a pilot study Contraception (2004) H.B. Peterson et al. The risk of pregnancy after tubal sterilization: findings from the U.S. collaborative review of sterilization Am J Obstet Gynecol (1996) C. Schippert et al. Reconstructive, organ-preserving microsurgery in tubal fertility: still an alternative to in vitro fertilization Fertil Steril (2010) I.C. Chi et al. Tubal ligation at cesarean delivery in five Asian centers: a comparison with tubal ligation soon after vaginal delivery Int J Gynaecol Obstet (1989) J.E. Schmidt et al. Requesting information about and obtaining reversal after tubal sterilization: findings from the U.S. collaborative review of sterilization Fertil Steril (2000) S.H. Kim et al. Microsurgical reversal of tubal sterilization: a report on 1118 cases Fertil Steril (1997) There are more references available in the full text version of this article. Cited by (4) Microsurgical anastomosis of the fallopian tubes after tubal ligation: a systematic review and meta-analysis 2023, European Journal of Obstetrics and Gynecology and Reproductive Biology Citation Excerpt : Sufficient data was not obtained to calculate the average economic cost of each type of intervention according to the approach. A total of nine studies [8,10–21] were found that reported on the outcomes of tubal sterilization reversal through laparotomy. In individual studies, this technique resulted in pregnancy rates ranging from 34.5 % to 85.1 %. Show abstract Between 20% and 30% of women who have undergone tubal ligation regret their decision. The alternative to regain fertility for these women is either in vitro fertilization or tubal re-anastomosis. This article presents a systematic review with meta-analysis to assess the current evidence on the efficacy of tubal recanalization surgery in patients who have previously undergone tubal ligation. The search was conducted in the World of Science (WOS) database, The Cochrane Library and ClinicalTrials.gov record using the keywords “tubal reversal”, “tubal reanastomosis” and “tubal anastomosis”. The review was carried out by two of the authors. Data from 22 studies were evaluated, comprising over 14,113 patients who underwent the studied surgery, following strict inclusion criteria: articles published between January 2012 and June 2022, in English and with a sample size bigger than 10 patients were included. A random-effects meta-analysis was performed. The overall pregnancy rate after anastomosis was found to be 65.3% (95% CI: 61.0–69.6). The percentage of women who had at least one live birth, known as the birth rate, was 42.6% (95% CI: 34.9–51.4). Adverse outcomes after surgery were also examined: the observed abortion rate among women who underwent surgery was 9.4% (95% CI: 7.0–11.7), and the overall ectopic pregnancy rate was 6.8% (95% CI: 4.6–9.0). No differences were found between the outcomes when differentiating surgical approaches: laparotomy, laparoscopy, or robotic-assisted surgery. The patient's age was identified as the most significant determining factor for fertility restoration. Finally, when comparing the results of tubal reversal with in vitro fertilization, reversal procedures appear more favorable for patients over 35 years old, while the results are similar for patients under 35 years old, but more data is needed to evaluate this finding. Therefore, the available literature review demonstrates that surgical anastomosis following tubal ligation is a reproducible technique with relevant success rates, performed by multiple expert groups worldwide. ### Mevrouw, there is more regret and failure following a sterilization together with a caesarean, and all those hormones in pregnancy preclude rational judgement 2014, European Journal of Obstetrics and Gynecology and Reproductive Biology ### The Use of long acting reversible contraceptives and the relationship between discontinuation rates due to menopause and to female and male sterilizations 2016, Revista Brasileira De Ginecologia E Obstetricia ### Når kvinder fortryder sterilisation 2016, Ugeskrift for Laeger View full text Copyright © 2014 Elsevier Ireland Ltd. All rights reserved. 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10709	https://www.economics-sociology.eu/files/11_76_Balcerzak_Rogalska.pdf	Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 80 Adam P. Balcerzak PhD Department of Economics Faculty of Economic Sciences and Management Nicolaus Copernicus University in Toruń CROWDING OUT AND CROWDING IN WITHIN KEYNESIAN FRAMEWORK. DO WE NEED ANY NEW EMPIRICAL RESEARCH CONCERNING THEM? Toruń, Poland E-mail: apb@umk.pl Elżbieta Rogalska MS Department of Microeconomics Faculty of Economic Sciences University of Warmia and Mazyry in Olsztyn Olsztyn, Poland E-mail: e.rogalska@interia.pl ABSTRACT. Last global financial crisis resulted in common among developed countries implementation of expansionary fiscal policy as an anti-recession tool. This led to the renewal of academic discussion on stabilization effectiveness of fiscal policy. In this context, the main research goal of this paper is to give theoretical analysis of the determinants of counter-cyclical effectiveness of fiscal policy with special concentration on crowding out and crowding in effects. Methodologically the analysis is done within Keynesian IS-LM framework within the assumption of expectations of economic actors. The theoretical analysis is confronted with the review of empirical papers based on the experiences of developed countries. Received: March, 2014 1st Revision: April, 2014 Accepted: May, 2014 DOI: 10.14254/2071-789X.2014/7-2/7 JEL Classification: E62, H31, H32 Keywords: fiscal policy, crowding out, crowding in, stabilization policy. Introduction Macroeconomics is the field of modern economy where the scientist and especially the political decision makers “voting” for a given theory are still far away from general consensus on real or just good enough model of economy. However, the end of XX century made a period when the scientific community seemed to agree on fundamental principles of gut applied macroeconomics that could be considered as textbook model and framework for practical policy. This consensus was close to neoclassic synthesis of Samuelson (see. Blanchard, 1997, pp. 244-246; Taylor, 1997, pp. 233-235; Solow, 1997, pp. 230-232, Blinder, 1997, pp. 240-243). From the perspective of day to day policy this general agreement concentrated on application of anti-cyclical monetary policy stabilizing business cycle, which complies with rules that are similar to Taylor’s rule, and fiscal policy was considered to be rather responsible for foundations of long term economic growth. As it was stated by Martin Eichenbaum: “In sharp contrast to the views that prevailed in the early 1960’s, there is now widespread agreement that countercyclical discretionally fiscal policy is neither desirable nor Adam P. Balcerzak, Elżbieta Rogalska, Crowding Out and Crowding in within Keynesian Framework. Do We Need Any New Empirical Research Concerning Them?, Economics & Sociology, Vol. 7, No 2, 2014, pp. 80-93. DOI: 10.14254/2071-789X.2014/7-2/7 Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 81 politically feasible. Practical debates around stabilization policy revolve almost exclusively around monetary policy” (Eichenbaum, 1997, p. 236). Last global financial crisis has completely changed that situation. From the practical point of view one could see massive fiscal stimulation packages that were supposed to counteract the crisis and stabilize the real economy. However, also the theoretical agreement among university researchers is not valid any more. This can be especially seen when one observes the massive discussion around three influential papers, two of Carmen Reinhart and Kenneth Rogoff (2011, 2010) who proof that expansionary fiscal policy leading over time to high level of debts (the breaking level of debt was estimated here as 90% of GDP) can be a significant factor negatively influencing GDP growth, and a critic article of Thomas Herndon, Michael Ash and Robert Pollin (2013) who argue that the estimations of Carmen Reinhart and Kenneth Rogoff are seriously influenced by methodological approach and are not valid. The practical return to fiscal massive stabilization policy during the last financial crisis and the growing theoretical controversies on factors concerning fiscal policy that used to be considered as explained and once set prove that there is a growing need for renewal of research and theoretical discussion on the determinants of counter-cyclical effectiveness of fiscal policy. The main research goal of this paper is to fulfill that need with concentration on two significant factors influencing effectiveness of fiscal stabilization action, which are crowding out and crowding in effects. From the methodological point of view the analysis is done within Keynesian IS-LM framework but within assumption of expectations. The main reason for this approach is the fact that it gives the advantage of analytical simplicity. Currently for the same reason many economists often use textbook AS-AD model. In this paper IS-LM framework was preferred due to some methodological problems and serious contradictions in the AS-AD model that were pointed by Robert Barro (1997, p. 611). Keynesian IS-LM perspective is also used here due to the growing theoretical and practical expansion of Keynesian economists in recent years. In the first part of the article the transaction crowding out is defined in the context of its influence on fiscal stabilization actions. The second part is devoted to the analysis of consequences of portfolio crowing out and crowding in effects. In the third part the review of empirical research is done and the article ends with conclusions and future research recommendations. Transaction Crowding out and Effectiveness of Fiscal Stabilization Policy In case of Keynesian model the effectiveness of fiscal stabilization policy that is aimed at stimulating aggregate demand is dependent on the size of fiscal multipliers, which in case of basic models are assumed to be positive and high. In reality there are many economic factors that may impact negatively on their size, starting with the institutional factors, macroeconomic situation of a given economy, foreign trade and ending with the actions of microeconomic market actors (see Hassett, 2009, p. 8). One of the most important factor, which has been the object of theoretical and empirical analysis for last few decades, is the crowding out of private spending by government spending associated with fiscal expansion, which directly leads to a decrease in the value of fiscal multipliers. Thus, limiting the effectiveness of the government's fiscal stabilization policy. The crowding out is a heterogeneous phenomenon, where the subject of scientific discussion is not only the possibility and scope of its existence, but also the transmission mechanisms leading to it . Willem Buiter proposed to introduce two basic distinction of the crowding out processes into two main categories: a) direct crowding out where the economic activities of the state interact in a direct way on the structure of private consumption and Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 82 private economic activities, such as the situation when private consumption is directly replaced by the consumption of public goods, b) indirect crowding out, much more complex than the first one, where the reactions of economic actors are associated with the changes in the level of interest rates and their structure (Buiter, 1976). In that case, one can talk about transactional crowding out and portfolio crowding out. This subsection is devoted to the effects of the transaction crowding out. The portfolio crowding out will be discussed in the next section. Effect of transactional crowding out is defined as the phenomenon of the decrease in private investment and private consumption resulting from an increase in the interest rates, which is the consequence of fiscal stimulus (see Keynes, 2003, p. 84, Wernik, 2011, p. 97). Transactional effect is associated with increased volumes of transactions in the economy resulting from the fiscal stimulus, which leads to an increase in the demand for money. In the conditions of the growth in the demand for money, an equilibrium in the money market is possible only if there is an appropriate interest rate increase, which would bring the demand for money to its original level. Assuming that the demand for money is a growing function of the product, fiscal expansion that is increasing aggregate demand in the product market must also lead to an increase in the transactional demand for real resources of money. When one assumes that supply of money is exogenous and constant, the increase in the transactional demand for money leads to an increase in the interest rate, which is necessary to maintain equilibrium in the money market. In the same time, both private investment and private consumption are negative functions of the interest rate. It means that the increase in the interest rate leads to decline in private investment and consumption. Thus, one observes the phenomenon of crowding out of private consumption and investment spending as a result of fiscal stimulus. This is shown in Chart 1. First of all, assuming that one analyses only the market of products that is unrelated to the market of money, where change in the volume of transactions do not affect the transactional demand for real resources of money, and therefore it does not affect the interest rate, the change in the size of government expenditure ΔG increases aggregate demand and shift the curve from IS1 to IS2, it means that it shifts the equilibrium level from Y1 to Y3. However, including into the analysis the money market, after the fiscal stimulus for the size of product Y3 and the interests rate r1 money market is in a state of disequilibrium. Returning to the market equilibrium requires a transition to Y2 product size and a higher interest rate r2 (Friedman, 1978, pp. 599-603, Spencer, Yohe, 1970, p. 17). Thus, in this model the size of the effects of transaction crowding out is the difference between Y2 and Y3. Chart 1. Fiscal expansion with the transaction crowding out effects in IS-LM model Source: based on Friedman (1978, p. 602, Spencer, Yohe, 1970, p. 17). IS1 LM Product The level of interest rate IS2 r1 r2 Y1 Y2 ∆G Y3 Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 83 The phenomenon of transactional crowding out leads to reduced effectiveness of positive fiscal stimulus, but in the same time it can also mean smaller negative consequences of fiscal consolidation in the real economy. Along with a reduction in aggregate demand resulting from the reduction of the budget deficit there is a decrease in the transaction demand for real resources of money, which translates into lower interest rates needed to maintain equilibrium in the money market. The lower level of interest rates may be a source of positive impulse on the side of private investment and consumer spending. Thus, this effect may in part, or – in extreme cases – even entirely offset the negative impact of negative fiscal adjustment on economic activity. From the perspective of the effectiveness of expansionary fiscal policy, which is aimed at stabilization purposes, it is particularly important that the effect of transaction crowding out can occur not only in conditions of full capacity utilization, but also in the case of economy in the Keynesian situation of unused production capacity. In addition, a major practical problem associated with the effects of transaction crowding out is the potential reaction of investment demand that is highly sensitive to interest rate, which can seriously affect the development of new productive capital equipment. In that case of the short-term effectiveness of fiscal stimulus is limited as a result of the impact of the current transaction crowding out effect, but the fiscal stimulation can also have negative long-term impact on the growth rate of productivity of the economy, and hence the rate of long-term economic growth (Friedman, 1978, p. 596). There are two main factors that determine the scale of the transaction crowding out. First it is the elasticity of the LM curve, which determines the response of demand for real resources of money associated with the changes in product size. The second one is the elasticity of the IS curve, which reflects the impact of interest rates on private consumption and investment. The first extreme case leading to full transactional crowding out effect is a situation of zero elasticity of demand for real resources of money, where the demand for money does not respond to changes in nominal interest rates. This is the “classical case” of a vertical LM curve (Figure 2a). In this situation, shifting the IS curve associated with fiscal expansion only results in changes in interest rates. However, it does not lead to changes in the size of aggregate demand, there is only a change in its structure (Carlson, Spencer, 1975, p. 5). The second extreme case leading to the full effect of the transaction crowding out is the situation with perfectly elastic IS curve (Figure 2b). This occurs when there is an assumption of constant returns from investment. It may result from the interaction between a large amount of capital accumulated in the economy, and its relatively small marginal values. Due to the relatively small marginal values of capital they should not affect revenue from all the accumulated capital. Another factor leading to the constant returns to scale from investment is the fact that investment spending is often accompanied by investments in knowledge and research and development. As a result the typical decreasing returns to scale from capital accumulation may be offset by technological progress. Based on this thesis in the middle of eighties the endogenous growth theory was developed. In the case of horizontal IS curve fiscal expansion cannot move the IS curve, for example the increase of government spending absorbs the private savings that is necessary for financing private investment, thereby reducing adequately their feasibility. Thus, in this case, fiscal expansions do not affect the size of aggregate demand, as their effect one can only expect the change in its structure (Carlson, Spencer 1975, pp. 6-7). Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 84 a) The classical case b) The perfectly elastic IS curve Chart 2. Transaction crowding out full effect in IS-LM model: extreme cases Source: based on Carlson, Spencer (1975, pp. 5, 7). On the other hand, extreme cases leading to a lack of transactional crowding out associated with the positive or negative fiscal adjustment is the cases of vertical IS curve. In this situation planned consumption and investment spending do not respond to changes in interest rates. The second possibility is the liquidity trap case with horizontal LM curve. LM curve has that shape when the demand for real resources of money is insensitive to changes in the product or strongly reacts to changes in nominal interest rates (Rzońca, 2007, p. 64). However, the previously mentioned two “extreme” cases of full transactional crowding out are not the only possible conditions leading to full neutralization real effects of stimulation or fiscal consolidation. Keith Carlson and Roger Spencer present two more variants of the model with the so-called. Conventional shapes of IS-LM curves, which can lead to the full effect of transactional crowding out, hence the neutrality of fiscal adjustment. Such cases are: a) the case of Keynes with expectations of private sector; b) the case of ultrarational households (Carlson, Spencer, 1975, pp. 5-8). In the first case fiscal expansion may adversely affect the confidence of economic actors in the future, which may result in an increase in liquidity preferences or decrease of the marginal efficiency of capital, thereby reducing the level of investment. This mechanism is shown in chart 3a. Fiscal expansion initially leads to a shift of the curve from the position IS1 to the position IS2, which causing the increase in liquidity preferences shifts LM curve from Y Product Product AD AD Product IS1 IS2 r1 r2 ∆G The level of interest rate P Y LM Price level The level of interest rate LM IS1 = IS2 After ∆G P Y r Product Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 85 LM1 to LM2, while the decline in the marginal efficiency of investment shifts IS curve from the position IS2 to IS3. In this model, there is a final solution of the model for which shifts of the IS and LM curves lead to no change in the size of the aggregate demand for a given price level (Carlson, Spencer, 1975, pp. 5-6). In the second case of ultrarational households there is a departure from the assumptions adopted in the traditional Keynesian approach in the IS-LM model, where there is no possibility of substitution between public and private expenditure. When one assumes that private debt and public debt are close substitutes, an additional amount of expenditure increasing budget deficit replaces the analogous value of private investment, because the government deficit is treated by households as public investment constituting a substitute to private investment, where both types of investments are evaluated by households from the prospects for future growth in consumption. As a result, after the initial shift of IS curve from IS1 to IS2, household can limited their private investment or private consumption, pushing the IS2 curve to the starting position (Figure 3b). Such a solution of the model does not depend on the method of carrying out fiscal stimulus. It does not matter whether fiscal expansion is the result of increased budget spending or tax cuts. As a result, fiscal expansion leads only to the full transaction crowding out effect (Carlson, Spencer, 1975, pp. 7-8). a) The case of Keynes b) Ultrarational case Chart 3. Full effects of transactional crowding out in IS-LM models: the case of Keynes and ultrarational case Source: based on Carlson, Spencer (1975, pp. 6, 8). Y Product IS1 Product Y Product IS1 IS2 r1 r2 ∆G The level of interest rate AD P Y LM2 Price level The level of interest rate LM P Y AD r Product IS3 LM1 IS2 ∆G Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 86 An important issue in the context of the analysis of the consequences of the transaction crowding out effects on the effectiveness of fiscal stabilization policy is the time horizon of the analysis. In the literature, there is no clear agreement on the possible extent of crowding out of private spending by government spending depending on whether the analysis concerns the short or medium time horizon. Empirical studies cited by Benjamin Friedman suggests that with prolongation of analysis the effects of transactional crowding out may have larger sizes, which negatively influences the effectiveness of fiscal stabilization policy (Friedman, 1978, p. 607). Analogous conclusions can be drawn from the analysis of several econometric studies presented by Gary Fromm and Lawrence Klein (1973, p. 393). Portfolio Crowding out and Crowding in Transactional crowding out effect is not the only consequence of fiscal stimulation, which can affect the behavior of private consumption and investment. By modifying the assumptions of Keynesian IS-LM model with respect to the definition of consumption function, in particular assuming the relationship of private consumption not only with income and interest rate, but also with wealth of economic actors and also taking into account the wealth in the money demand function, one can talk about the possibility of portfolio crowding out or portfolio crowding in effects. This effects are also sometimes called wealth effects and are defined as a situation where the rising public spending leads to decrease in private spending (crowding out) or stimulate private spending (crowding in) through its impact on the value of the wealth of economic actors. Foundations of analysis of the impact of the wealth on the private consumption and investment can already be found in the classical works of Arthur Pigou, who analyzed the impact of the size of households wealth on their consumption, as well as John Maynard Keynes investigating the impact of wealth on investment activity of enterprises. Reflection of the wealth effect in the goods market is the fact that the growth of wealth held by private entities is accompanied by an increase in aggregate demand, which in the standard IS-LM model moves IS curve to the right. This is equivalent to the occurrence of a positive wealth effect. Assuming that the source of increased wealth of households is positive fiscal impulse, the positive wealth effect in the goods market may strengthen primary multiplier effects of fiscal expansion (Kosterna, 1995, p. 121; Friedman, 1978, p. 609). As mentioned earlier occurrence of the wealth effects is not limited only to the goods market. It may have also a very significant influence on the money market, where the consequences of changes in household wealth may be much more complicated. Extending the model analyzed so far, it is assumed that households and enterprises treat the debt (government bonds) used to finance the state budget deficit as the wealth which positively affects their consumption and investment decisions. In addition, the treasury bonds included in the portfolios of housholds and enerprises as assets increase the demand for real resources of money (Silber, 1970, pp. 465-467). This is due to the willingness of households to diversify risk, prompting them to build a diversified and balanced portfolio (Rzońca, 2005, pp. 7-8). Based on the above assumptions, in the case of a fiscal stimulus leading to the issuance of debt financed by issuing the government bonds, one can predict the occurrence of two opposite effects. One should expect an increase in aggregate demand associated with the growth of household consumption that finance the budget deficit through the purchase of government bonds, and consider it as an increase in wealth held by them. This is the equivalent of shifting IS curve to the right from IS1 to IS2 in Figure 4. However, due to increased wealth of households, one should also expect an increase in interest rates, which is necessary to maintain equilibrium in the money market when there is the increase in the Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 87 demand for money (it makes effect of portfolio crowding out and this corresponds to a shift of the LM curve to the left from LM1 to LM2), which may translate into a decline in both private consumption spending and private investment (equivalent to a shift of the IS curve to the left). As a result, there is a negative impact of portfolio effect on aggregate demand (Rzońca, 2005, pp. 12-13). Y2 – Y3 – transaction crowding out Y4 – Y2 – portfolio crowding out Chart 4. The portfolio crowding out effects as a result of fiscal expansion in IS-LM model Source: based on Friedman (1978, p. 619). The Scale of that crowding out effect in practice depends on a number of factors such as the method of financing the budget deficit, the monetary policy accompanying the fiscal expansion, and in the case of absence of reaction of monetary policy a key factor is the degree of substitutability between money in cash and the government bonds (Wojtyna, 1990, pp. 164). Shift of the LM curve leading to an increase in the interest rate is dependent on the sensitivity of the demand for real money resources to changes in product and wealth. No response of the interest rate would require a horizontal LM curve, which would be identical to the situation in which money and the government bonds are perfect substitutes. In contrast, no shift of the IS curve would require that the curve is vertical, which means the perfect elasticity of consumer spending and investment relative to the interest rate. Both of these extreme cases do not seem very realistic. Thus, these facts may lead to the conclusion that the portfolio crowding out effect can significantly reduce the short-term positive impact of fiscal expansion on increasing aggregate demand, which can be treated as a factor limiting the effectiveness of fiscal policy as an anti-cyclical stabilization tools (Rzońca, 2005, pp. 11-13). However, as emphasized by Benjamin Friedman final determination of the strength and direction of the impact of portfolio effect requires above all a thorough examination of social preferences for allocating resources, because depending on their characteristics the wealth effect may lead both to the effect of portfolio crowding out, and the opposite effect of portfolio crowding in that stimulate spending. This author strongly criticizes the widespread wrong belief that the negative impact of wealth portfolio effect on private investment is the only possibility to solve the model, and thus he indicates that the deficit financing through the sale of government bond may be in some cases a source of portfolio crowding out and crowding in effect (Friedman, 1978, p. 608). Assuming that economic actors are characterized by risk aversion, who therefore seek to diversify their investment portfolio, the occurrence of portfolio crowding out effect is mainly dependent on whether the bonds financing the deficit can be considered as closer substitute for money or physical capital (Friedman, 1978, p. 618). Product LM1 The level of interest rate IS1 IS2 r1 r2 Y 1 Y4 Y2 ∆G Y3 LM2 Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 88 Assuming that the bonds are mainly substitute of physical capital, not the money, to restore equilibrium in the money market will require a growth rate of return from tangible assets, which after initial positive fiscal impulse results in the increase of the interest rate and thus it shifts the LM curve from LM0 to LM2 (Chart 5). This means that the transaction crowding out effect is amplified by the portfolio crowding out effect (Friedman, 1978, p. 620). Therefore, in this situation the efficiency of fiscal stimulus policy is minimized. In turn, if the bonds are a close substitute for money and not for physical capital, the LM curve shifts in the opposite direction from LM0 to LM1 (Chart 5). Issuance of government bonds linked to the increase in the budget deficit will lead to a reduction in the expected rate of return on physical capital that is necessary to restore market equilibrium. This means that instead of portfolio crowding out effect, there is the portfolio crowding in effect that is stimulating the private investment, which thereby increases the effectiveness of fiscal stimulus (Friedman, 1978, p. 620). Chart 5. The effects of portfolio crowding out and crowding in as a result of fiscal expansion in IS-LM model Source: based on Friedman (1978, p. 619). The above discussed possibilities lead to the conclusion that the occurrence of portfolio crowding out effect or the portfolio crowding in effect should be the subject of permanent empirical research. The empirical works quoted by Benjamin Friedman in this matter seem to indicate that government bonds are rather a substitute for physical capital, not money. This allows to conclude that the ex-ante exclusion of the possibility of portfolio crowding out effect during fiscal expansions is not supported by empirical studies (Friedman, 1978, pp. 620-627). In all the models previously discussed effects of crowding out were based on the assumption of price stability in the product market. Portfolio crowding out may also occur in case of their growth, which was presented by Keith Carlson and Roger Spencer in the model with flexible prices (Carlson, Spencer, 1975, pp. 8-9). In the model with flexible prices one takes into account the wealth in the consumption function and the function of the demand for money, the constraints of the government budget, the labor sector and the endogenous level of prices. The increase in government purchases financed by both the increasing the taxes or increasing the government deficit, leads to an increase in aggregate demand and can lead to an increase in product prices. In this situation, there may be an increase in consumption associated with a positive impact on wealth of accumulation of government bonds by households. However, there is a compensating increase LM2 LM0 Product The rate of return from tangible assets The level of interest rate LM1 IS1 IS2 Y 1 Y2 Y3 ∆G Y4 Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 89 in the demand for money associated with increased wealth. The increase in prices leads to a fall in private consumption and a fall in real money resources. Together with the decline of private investment, which is associated with an increase in the interest rate, these factors can lead to crowding out of private spending in the amount equivalent to the increased government spending (Chart 6a) (Carlson, Spencer, 1975 , pp. 8-9). a) The case of elastic prices b) The case of Milton Friedman Chart 6. The effect of full crowding out in the IS-LM model Source: based on Carlson, Spencer (1975, pp. 9, 10). The previous analysis of the fiscal consequences of portfolio crowding out focused mainly on short-term analysis. Meanwhile, its effect may bring significant economic consequences also in the medium term. In this context, as suggested by Keith Carlson and Roger Spencer it is worth using Milton Friedman model. In this approach one assumes a relatively flat IS curve. Its shape may be associated with a broad interpretation of the concept of savings and investments. In the model of Milton Friedman, the wealth effect associated with the accumulation of bonds that finance government debt is minimal. It is offset by the impact of the expected future tax burden and by the effect of substitution of assets accumulated in the form of private bonds and government bonds financing the government debt. In addition, an important factor that was often stressed by Milton Friedman, which may additionally limit the effectiveness of fiscal stimulus, is the possible negative impact of fiscal expansion on the future manufacturing potential of economy, associated with lower private Y ∆G AD2 AD1 LM1 Product IS1 IS2 r1 r2 ∆G Product The level of interest rate AD1 P1 Y LM2 Price level The level of interest rate LM IS1 Product P Y r Product IS3 IS3 ∆G AS AD2 ∆G P2 LM3 Y Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 90 investment than it would be without the fiscal stimulus. Under these conditions, as a result of fiscal stimulus there is relatively small increase in aggregate demand. The substitution of private debt to government debt will lead to a reduction in personal wealth and will be pushing IS curve to the starting point (Chart 6b) (Carlson, Spencer, 1975, pp. 9-10). What's more there will be medium-and long-term costs of such fiscal stimulus in the form of lower growth potential of the economy. In addition, portfolio crowding out can lead to non-Keynesian effects of fiscal policy, that means a situation in which negative for aggregate demand effects associated with shift of LM curve outweigh the positive effects of shift of IS curve. This means adopting by the fiscal multipliers of negative sign and is identical to the situation in which fiscal stimulation leads to a decrease in aggregate demand and fiscal tightening results in the increase of aggregate demand. To eliminate this possibility, the consumption of households must have a high sensitivity to changes in wealth, and the portfolio crowding out must touch mainly the investment. This is due to the fact that changes in investment are to a high extent identical with changes in physical capital, thereby they can weaken the impact of fluctuations in debt on the private wealth. Taking into account all of the above consequences of portfolio crowding out, one can assume that fiscal expansions can positively impact on aggregate demand, but that impact will be rather minimal, only under the condition of diminishing amount of capital, which is equivalent to a decrease in the supply potential of the economy in the longer term (Rzońca, 2005, p. 13). Review of Empirical Research for United States, OECD and Some Developing Countries The analysis presented in previous two sections shows that from the theoretical perspective the crowding out and crowding in phenomena have quite complicated nature and can depend on many specific factors of a given economy. As a result it will be probably quite difficult to provide universal stylized facts concerning this problem. The main aim of this part of the paper is to review the empirical literature concerning that subject. The literature devoted to the influence of government debt and consequences of expansionary fiscal policy is quite broad, especially one can find vide range of research on highly developed countries and especially United States. Martin Feldstein and Otto Eckstein (1970, pp. 363-375) were researching the fundamental determinants of changes in the long-term interest rate in United States based on the quarterly data starting with 1954 and ending at middle of 1969. They developed a model based on Keynesian liquidity preference theory and attitude to the measurement of the real interest rate that was taken by Fisher. Their model was explaining the long-term interest rate with four types of variables: liquidity, inflation, government debt and short-run expectations. Their results suggest that in the analyzed period the long-term interests rates were determined ty other factors than the government debt such as liquidity as the primary reason. That could be treated as an argument against significant importance of crowding out. Analogous results can be found in the work of Paul Evans (1985, pp. 68-87) concerning United States economy but made with different methodology and based on much longer analytical period. He analyzed time series starting with 1858 and ending with 1984 with three sub-period during which deficit has exceeded high level of 10% of national income. That were the experience during Civil War in 1861-1865, the experience during World War I and fiscal expansion in the years 1917-1919, and the experience during World War II in the fiscal period 1941-1945. Based on regression analysis with 2SLS he argues that the large deficits during the war periods have never been associated with high interest rates, which means that one can exclude the possibility of crowding out effects in US economy. Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 91 What is more according to him the evidence could even more strongly support the crowding in possibility with the positive association of government deficit and interest rate. Charles Plosser (1982, pp. 325–352; 1987, pp. 343–367) in his two papers makes an analysis of relation between US government financing decisions and asset returns and especially he concentrates on the question whether a substitution of debt financing for tax financing of a given level of expenditures is associated with an increase in interest rates. In his both researches he finds little or no association between real or nominal interest rates and deficits, which can be treated as an argument against crowding out paradigm. Gregory Hoelscher (1983, pp. 319-333) investigates empirically the effects of Federal government borrowing on short term interest rates in US. His regression analysis made for quarterly data in the period 1952-1976 showed relatively unimportant and insignificant influence of government debt on short term interest rates, which excludes the hypothesis of crowing out. He showed that main and primary factors determining the short term interest rates were expected inflation, monetary factors and economic activity in the economy. On the other hand, the above mentioned research results of Gregory Hoelscher were reexamined by James Barth et al. who made their analysis for the period of 1955-1983 but adjusted the data for effects of cyclical economic activity. In their research structural deficit variable has significant and highly positive coefficient which is strong argument for the influence of government debt on short term data and it gives and argument for the high possibility of crowding out effects in US in the analyzed period. Richard Cebula (1985, pp. 305-309) investigates quarterly data form 1970 till 1982 in order to find the transmission mechanism of crowding out therefore the extent to which the rate of change of the prime rate of interest responds to the rate of change of the rate on Treasury T-bills, which is positively influenced by deficit financing. His regressions provided strong evidence that federal government borrowing has a significant impact upon short term interest rates in the private sector. As the private sector consumption and investment spending is strongly sensitive to the interest rate, he provide a strong empirical argument for high possibility of crowding out. The same author has also recently updated his research for the quarterly data in the period of 1973-2007. As a result he provides new empirical evidence on the impact of the federal government budget deficit on the nominal cost of borrowing for private enterprise in the USA. The analysis is based on an open loanable funds model which includes expected inflation, the ex-ante real short-term interest rate, the M1 money supply, net international capital inflows, and the change in per capita real GDP. It reveals that the federal budget deficit measured as a percent of GDP has a positive and statistically significant impact on the nominal interest rate. As a result it influences positively the cost of borrowing for private enterprises which is an argument for crowding out theory (Cebula, 2009, pp. 146-151). The above presented literature review was concerning USA economy, which has long tradition of research in that field. However, the literature provides also a number of empirical studies involving a large group of countries in the long run, which provide empirical arguments for the high possibility of crowding out. Stanley Fisher shows the results of a panel analysis for a broad group of countries (OECD countries, Africa, Asia and South America) covering a period of more than thirty years, which confirm the negative impact of budget deficits on capital accumulation and economic growth (Fisher, 1993, pp. 485-512 , see as well Fisher, 1991). Similar conclusions can be found in the works of Michael Bleaney and others who carried out the study for the OECD countries in the period 1970-1995 (Bleaney et al., 2001, pp. 37-57). In turn, the theoretical literature review and an extensive analysis of empirical studies on OECD countries conducted by Richard Hemming and others also points to the relatively low value of fiscal multipliers, which can be the argument for high possibility of crowding out Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 92 effects, and therefore relatively low short-term stimulation effectiveness of fiscal policy. Short-term fiscal multipliers in the countries amounted to about one-half and short-term multipliers for government spending amounted to about unity, which is consistent with the conclusions of the presented theoretical analysis (Hemming et al., 2002a, Hemming et al., 2002b). Conclusions Last global financial crisis has changed the political attitude towards the applicability of fiscal policy as an anti-cyclical stabilization tool. It has also influenced the academic discussion concerning the role of fiscal policy and main determinants of its effectiveness. The presented theoretical analysis shows that even though the phenomena of crowding out or crowding in are often considered as basic textbook theory, they can have quite complicated nature. It can be seen that many specific factors of a given economy can change the transmission mechanisms and lead to different analytical results, which can be obtained with widely accepted Keynesian IS-LM framework. On the other hand, the empirical literature review have also showed that the research results concerning the subject are often contradictory, they are often sensitive to methodological differences in case of given economies and can lead to different results for different periods. These both factors lead to the conclusion that there is still a great need for theoretical work and wide empirical research concerning these factors. This is important in case of Central and Eastern European economies and especially countries like Poland that in recent years has experienced a serious growth of government debt and is not able to conduct serious fiscal reforms. References Barro, R. (1997), Makroekonomia, Polskie Wydawnictwo Ekonomiczne. Barth, J.B., Iden, G.R., Russek, F.S. (1985), Federal Borrowing and Short Term Interest Rates: Comment, Southern Economic Journal, Vol. 52, No. 2, pp. 554-559. Bleaney, M., Gemmell, N., Kneller, R. (2001), Testing the Endogenous Growth Model: Public Expenditure, Taxation, and Growth Over Long Run, Canadian Journal of Economics, Vol. 34, Issue 1, pp. 36–57. Blanchard, O. (1997), Is There a Core of Usable Macroeconomics? American Economic Review, Vol. 87, No. 2, pp. 244-246. Blinder, A.S. (1997), Is There a Core of Practical Macroeconomics That We Should All Believe? American Economic Review, Vol. 87, No. 2, pp. 240–243. Buiter, W.H. (1976), Crowding Out and the Effectiveness of Fiscal Policy, Econometric Research Program Princeton University, Research Memorandum, No. 191. Carlson, K.M., Spencer, R.W. (1975), Crowding out and its critics, Federal Reserve of St. Louis, December. Cebula, R.J. (1985), New evidence on financial crowding out, Public Choice, Vol. 43, Issue 3, pp. 305-309. Cebula, R.J. (2009), Recent evidence on the impact of federal budget deficits on the nominal cost of long term borrowing for private enterprise in the U.S., Problems and Perspectives in Management, Vol. 7, Issue 1. Eichenbaum, M. (1997), Some Thoughts on Practical Stabilization Policy, American Economic Review, Vol. 87, No. 2, pp. 236-239. Evans, P. (1985), Do Large Deficits Produce High Interest Rates? American Economic Review, Vol. 75, No. 1, pp. 68-87. Adam P. Balcerzak, Elżbieta Rogalska ISSN 2071-789X RECENT ISSUES IN ECONOMIC DEVELOPMENT Economics & Sociology, Vol. 7, No 2, 2014 93 Fischer, S. (1993), The role of macroeconomic factors in growth, Journal of Monetary Macroeconomics, Vol. 32, Issue 3, pp. 485-512. Fisher, S. (1991), Growth, Macroeconomics and Development, NBER Working Papers, Working Paper No 3702. Feldstein, M. Eckstein, O. (1970), The Fundamental Determinants of the Interest Rate, Review of Economics and Statistics, Vol. 54, No. 4. Friedman, B.M. (1978), Crowding Our or Crowding In? The Economic Consequences of Financing Government Deficits, Brookings Papers on Economic Activity, Issue 3, pp. 655-685. Fromm, G., Klein, L.R. (1973), A Comparison of Eleven Econometric Models of the United States, American Economic Review, Vol. 63, Issue 2. Hassett, K.A. (2009), Why Fiscal Stimulus is Unlikely to Work? American Enterprise Institute for Public Policy Research, AEI Working Paper No.146, March 3. Hemming, R., Kell, M., Mahfouz, S. (2002), The Effectiveness of Fiscal Policy in Stimulating Economic Activity – A Review of the Literature, IMF Working Paper, WP/02/2008. Hemming, R., Mahwous, S., Schimmelpfennig, A. (2002), Fiscal Policy and Economic Activity in Advanced Economies, IMF Working Paper, WP/02/87. Herndon, T., Ash M. and R. Pollin (2013), Does High Public Debt Consistently Stifle Economic Growth? A Critique of Reinhart and Rogoff, Political Economy Research Institute, Working Paper Series, No. 322, April. Hoelscher, G.P. (1983), Federal borrowing and short term interest rates, Southern Economic Journal, Vol. 50, No. 2. Keynes, J.M. (2003), The General Theory of Employment, Interest and Money, A Project Gutenberg of Australia eBook, eBook No.: 0300071h.html. Kosterna, U. (1995), Deficyt budżetu państwa i jego skutki ekonomiczne, Wydawnictwo Naukowe PWN – Centrum Analiz Społeczno-Ekonomicznych Fundacja Naukowa, Warszawa. Plosser, C. (1982), Government financing decisions and asset returns, Journal of Monetary Economics, Vol. 9, Issue 3, pp. 325-352. Plosser, C. (1987), Fiscal policy and the term structure, Journal of Monetary Economics, Vol. 20, Issue 2, pp. 343-367. Reinhart, C. and K., Rogoff (2011), A Decade of Debt, Centre for Economic Policy Research, Harvard University and NBER, Discussion Paper No. 8310 April. Reinhart, C. and K., Rogoff (2010), Growth in a Time of Debt, NBER Werking Paper Series, Working Paper 15639, January. Rzońca, A. (2007), Czy Keynes się pomylił? Skutki redukcji deficytu w Europie Środkowej, Wydawnictwo Naukowe Scholar. Rzońca, A. (2005), Portfelowy efekt wypychania a mnożniki fiskalne. Część I, Bank i kredyt, No. 1, pp. 4-14. Silber, W.L. (1970), Fiscal Policy in IS-LM Analysis: A Correction, Journal of Money, Credit and Banking, Vol. 2, No. 4, pp. 461-472. Solow, R.M. (1997), Is There a Core of Usable Macroeconomics We Should All Believe in? American Economic Review, Vol. 87, No. 2, pp. 230–232. Spencer, R.W., Yohe W.P. (1970), The “Crowding Out” of Private Expenditures by Fiscal Policy Actions, Federal Reserve Bank of St. Louis Review, October. Wernik, A. (2011), Finanse publiczne, Polskie Wydawnictwo Ekonomiczne, Warszawa. Wojtyna, A. (1990), Nowoczesne państwo kapitalistyczna a gospodarka. Teoria i praktyka, Państwowe Wydawnictwa Naukowe, Warszawa. Taylor, J.B. (1997), A Core of Practical Macroeconomics, American Economic Review, Vol. 87, No. 2, pp. 233-235.
10710	https://www.pulsetoday.co.uk/clinical-feature/clinical-areas/ear-nose-and-throat/head-and-neck-clinic-neck-lumps/	Head and neck clinic - neck lumps - Pulse Today Pulsetoday - Do Not Process My Personal Information If you wish to opt-out of the sale, sharing to third parties, or processing of your personal or sensitive information for targeted advertising by us, please use the below opt-out section to confirm your selection. Please note that after your opt-out request is processed you may continue seeing interest-based ads based on personal information utilized by us or personal information disclosed to third parties prior to your opt-out. You may separately opt-out of the further disclosure of your personal information by third parties on the IAB’s list of downstream participants. This information may also be disclosed by us to third parties on the IAB’s List of Downstream Participants that may further disclose it to other third parties. 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There is no systemic upset (specifically no B symptoms – night sweats, fever, loss of weight), and no other head and neck symptoms such as sore throat or salivary gland obstruction. He smokes 20 cigarettes a day and has a moderate alcohol intake. Examination reveals a firm 2cm mass deep to the skin and sternomastoid muscle in the left upper neck with no other findings on neck palpation. Examination of the oral cavity and throat is also unremarkable. Routine blood tests including FBC, ESR, EBV and U&Es are all normal. After urgent referral a diagnosis of tongue base squamous cell carcinoma with metastasis to the upper neck nodes was made. Features When considering any neck lump, it is important to have a clinical impression at an early stage as to whether the lump might be malignant. The short history, absence of other head and neck symptoms (including upper respiratory tract infection, swelling relating to eating, absence of dental or skin infection) and that he is a smoker should raise suspicion and prompt early referral to the local neck lump clinic, preferably using the two week wait pathway. In contrast, many neck lumps that are clearly non-malignant are still being referred urgently, including lipomas and sebaceous cysts that in some cases have been present for years. In one example a calcified TB neck node which had been unchanged for 75 years was referred on the two week pathway. Taking a simple history should give you some idea of how long the lump has been present and whether or not it is enlarging. General enquiry about systemic upset, other lumps (including axillary, inguinal lymphadenopathy) will also help to determine whether the lump is most likely related to head and neck pathology or is manifestation of a systemic disease such as a lymphoma. Given that 300 of the 800 total body lymph nodes are found in the head and neck, many systemic diseases can present with neck lymphadenopathy. Most solitary lumps in the upper lateral neck are unlikely to be metastatic from a primary below the clavicle (such as lung or breast), which is not the case with a low neck lump presentation (remember the classic supraclavicular Virchow node from a GI primary?). A solitary upper neck lump is often indicative of pathology in the upper aero-digestive tract and these lumps can be malignant in any age. The important differential diagnosis to exclude is metastatic squamous cell carcinoma. Several features should raise your suspicion. Malignant nodes enlarge progressively and are usually painless and not associated with other symptoms (many primary head and neck cancers are painless in the early stages). The nodes feel firm on palpation while lymphoma nodes often feel rubbery and tend to be multiple from an early stage. Lymphoma can present at any age but typically in young adults (classic Hodgkin’s disease) and the elderly (non-Hodgkin’s lymphoma). In contrast, reactive (benign) nodes occur far more commonly and often present following a sore throat (tonsillitis), another infection, such as EBV, acne or dental abscesses. They are sometimes painful. A soft lump presenting for the first time anterior to the sternomastoid muscle in a younger patient (20-30 years) may be a branchial cyst, but as a small percentage of these can harbour malignancy in their wall, urgent referral should be made. It is very unusual for longstanding lumps (histories of years not weeks) such as lipomas and skin pathology to be malignant, so these should be referred routinely. Investigation Many GPs will do initial blood tests including FBC, ESR, U&Es and EBV. These are likely to be normal, especially if there is no other relevant history to suggest infection. Both lymphoma and metastatic squamous cell carcinoma cannot be diagnosed on a blood test. The most useful investigation following referral is head and neck ultrasound with fine needle aspiration cytology (FNAC) for nodes that look worrying. The nodal size is not a determinant in itself for malignancy but instead the morphology and ratio of the short to long axis guides the radiologist when deciding if FNAC is needed. For example, some reactive (benign) nodes following repeated tonsillitis can measure over 2cm in length but have a short axis of 7-8mm. Diagnosis If malignancy is confirmed on FNAC, appropriate further investigations including staging CT, examination and biopsy of the oropharynx under general anaesthetic and PET-CT may be required. Sometimes ultrasound-guided core biopsy or node excision is needed to confirm a diagnosis, particularly for lymphoma. Management Neck lumps are common presentations in general practice and the majority are not malignant. A good history is important for helping to guide management – a lump present for many years is almost certainly not going to be malignant, while a painless enlarging lump with no other symptoms should be taken very seriously indeed. Blood tests are only useful in primary care when infection is thought to be the cause. Urgent referral to a neck lump clinic should be made for lumps persisting more than three weeks to exclude a malignant process. These clinics are usually run jointly with oral/maxillofacial surgery and ENT and have a radiologist and pathologist present to be able to provide a diagnosis in most cases on the day of the visit. Professor Peter Brennan is a consultant maxillofacial/head and neck surgeon at the Portsmouth Hospitals NHS Trust Mr Alex Goodson, Mr Karl Payne and Mr Arpan Tahim are specialty registrars in oral and maxillofacial surgery. The authors have recently published a book entitled ’Important Oral and Maxillofacial Presentations for the Primary Care Practitioner’. It has been sent to every GP practice in the UK, free of charge, and contains algorithms providing guidance for the management of many head and neck conditions, including neck lumps and malignant disease. The book has been written in collaboration with the RCGP. Further copies are available at cost price (£12.50) from Amazon. Visit Pulse Reference for details on 140 symptoms, including easily searchable symptoms and categories, offering you a free platform to check symptoms and receive potential diagnoses during consultations. 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10711	https://www.youtube.com/watch?v=2ixEf2zpR8E	Hess's Law Problems & Enthalpy Change - Chemistry The Organic Chemistry Tutor 9700000 subscribers 20552 likes Description 1492314 views Posted: 23 Sep 2017 This chemistry video tutorial explains how to solve common Hess's law problems. It discusses how to calculate the enthalpy change of a reaction using Hess's law. Chemistry 1 Final Exam Review: Thermochemistry - Free Formula Sheet: Thermochemistry - Video Lessons: Final Exam and Test Prep Videos: First Law of Thermodynamics: Thermochemistry Equations: Internal Energy, Heat, and Work: Thermochemical Equations: Specific Vs Molar Heat Capacity: Basic Calorimetry Problems: Final Temperature Calorimetry Problems: Latent Heat of Fusion & Vaporization: Coffee Cup Calorimeter: More Calorimeter Problems: Specific Heat Capacity Problems: Hess Law Problems: More Hess Law Problems: Enthalpy of Formation & Heat Combustion: Enthalpy Practice Problems: Speed of Light, Frequency, & Wavelength: Final Exams and Video Playlists: Full-Length Videos and Worksheets: 626 comments Transcript: so this video is going to be all about hess's law and how we can calculate the enthalpy change of a reaction using it so consider this reaction nitrogen gas reacts with hydrogen gas to produce ammonia and the enthalpy change for that reaction is going to be negative 92 kilojoules and then we have another reaction water decomposes to hydrogen gas and oxygen gas and the enthalpy change for that is going to be positive 286 kilojoules so using those two reactions how can we calculate the enthalpy change of this reaction ammonia reacting with oxygen gas producing nitrogen gas and water so hess's law states that we can use these reactions adjust them in such a way to get the enthalpy change of this reaction but how do we go about doing it so here's what we need to do we need to adjust these two reactions in such a way that when we add them we get this reaction and then we need to adjust these values accordingly and when we add those numbers we're going to get the enthalpy change for this reaction and that's how we're going to do it so we need to focus on the right stuff so in reaction 1 and 2 you want to focus on the reactants and products that appear only once in those two reactions for example nitrogen gas is only found in equation one not in equation two so you want to focus on n2 hydrogen gas it's found in both of these two equations so you do not want to focus on h2 if you do it's you're going to make your life a lot harder so notice that nitrogen gas is on the left side but we want it to be on the right side and we need a coefficient of 2 in front of it so what i'm going to do is i'm going to reverse this reaction and multiply by 2. whenever you reverse reaction you got to change the sign and if you multiply this reaction by two you need to multiply the enthalpy change by two so i'm going to have four nh3 so this is now going to be on the other side and if you multiply two nh3 by two you should get four in each three and this is going to yield two and two plus three times two is six so that's going to be six h2 and the enthalpy change is going to be positive 184 kilojoules that's 92 times 2. now let's focus on this equation should we focus on water h2 or oxygen you can focus on water or oxygen but you don't want to focus on h2 because it's found in these two equations so i'm going to focus on h2o notice that in the net reaction we have six water molecules on the right side here we only have one on the left so we need to reverse it and multiply by six so six times that is going to be 6 h2 and then 6 times a half o2 that's going to be 3 o2 and that's going to produce 6 water molecules so we got to change the sign it's going to be negative 286 and then we're going to multiply it by 6. so that's going to be negative 17 16 kilojoules so now let's add the two reactions and we need to make sure that we get the net reaction so we have four nh3 on the left that's to the left of this arrow so let's rewrite it here four nh3 and notice that the six h2 molecules they cancel because they're the same on both sides on the left we have three o2 that's the left of that arrow and then on the right side we have two n2 and six h2o which is the same as what we have here now since we add equation 1 and 2 to get this reaction to calculate the enthalpy change of that reaction according to hess's law we just need to add these values so it's 184 plus negative 17 16. so the answer is going to be negative 15 32 kilojoules and so that's how you can use hess law to calculate the enthalpy change of a reaction if you know the enthalpy changes of other reactions here's another example let's say that water decomposes into hydrogen gas and oxygen gas and let's say that that process requires positive 576 kilojoules of energy and then oxygen breaks down into atomic oxygen with an input of positive 495 kilojoules so with this information use hess's law to determine the enthalpy change of this reaction hydrogen gas plus atomic oxygen produced in liquid water so feel free to pause the video as you try that example so what's the first thing that we need to do in this example it's best to identify which species to focus on i'm going to focus on hydrogen gas notice that we have two of it on the right side but we need one of it on the left so we got to multiply the first reaction by a half and we need to reverse it so once we reverse it it's going to be h2 and you gotta divided by two so it's h2 plus one half o2 producing one water molecule now i need to make one small correction that should be a two and not a six now we need to change the sign and we have to divide that number by two 572 divided by two is 286 so this is going to be negative 286 kilojoules now notice that we have one oxygen atom on the left side and here we have two on the right side so we need to reverse that reaction and divided by two as well so it's gonna be o turning into one half o2 so it's going to be negative 495 divided by two which is 247.5 kilojoules so now let's add the two reactions so these two numbers will those two things will cancel so it's gonna be h2 plus o turns into a liquid water so negative 286 plus negative 247.5 is negative 533.5 kilojoules and so this is the answer now let's consider one more example so we're going to have two ozone molecules decomposing into three oxygen gas molecules and the energy for that is going to be negative 427 kilojoules i'm gonna have to write small because there's going to be more reactions in this one than the last one now in the next one oxygen gas decomposes into atomic oxygen and so the energy change for that is positive 400 495 kilojoules and then finally we have nitrogen monoxide reacting with ozone to produce nitrogen dioxide plus oxygen gas and so this is going to be negative 199 kilojoules so with this information go ahead and determine the enthalpy change of this reaction nitrogen monoxide plus atomic oxygen turning into nitrogen dioxide so go ahead and try it so let's call this equation one two and three so what should we focus on i'm going to focus on equation three so notice that i have nitrogen monoxide on the appropriate side i don't have to change it so i'm going to rewrite equation 3. and so the energy change is going to remain negative 199 kilojoules now notice that no2 is already in the appropriate side so the next thing i need to focus on is atomic oxygen which is only found in equation two so i need it to be on the left side so i have to reverse this reaction and multiply by half so this is going to be o turning into one half o2 and so this is going to be negative 495 divided by 2 we said it was 247.5 now notice that ozone and o2 they're found in other reactions o2 is found in equations two and three and ozone is only found in one e3 so i'm going to focus on ozone now because ozone is not in the net reaction that means i need it to cancel completely so in this reaction i have one ozone molecule on the left side if i want that to cancel i need to have one ozone molecule on the right side so i've already used equation three so i'm going to focus on equation one keep in mind this is equation three and this one is two so in order to get one ozone molecule on the right i need to reverse that reaction and multiply by half so on the left i'm going to have 3 over 2 o2 and it's going to change from negative to positive and then i need to divide 427 by 2 which is 213.5 kilojoules so now let's add the reactions so ozone will cancel now on the right side we have one oxygen gas molecule plus a half so that's 1.5 on the left side we have 3 over 2 which is 1.5 so therefore this cancels with these two on the right so what we have left over is no plus oh turns into no2 so since these three reactions add up to the net reaction the enthalpy change of the net reaction is the sum of the enthalpy changes of those three reactions so it's going to be negative 199 plus negative 247.5 plus 213.5 so that's negative 233 kilojoules so that's the energy change for this reaction so that's how you can use hess's law to calculate the enthalpy change of a reaction using the enthalpy changes from other reactions that are provided you
10712	https://github.com/CoolProp/CoolProp/issues/1910	Enthalpy calculation problem · Issue #1910 · CoolProp/CoolProp Skip to content Navigation Menu Toggle navigation Sign in Appearance settings Platform GitHub Copilot Write better code with AI GitHub Spark New Build and deploy intelligent apps GitHub Models New Manage and compare prompts GitHub Advanced Security Find and fix vulnerabilities Actions Automate any workflow Codespaces Instant dev environments Issues Plan and track work Code Review Manage code changes Discussions Collaborate outside of code Code Search Find more, search less Explore Why GitHub Documentation GitHub Skills Blog Integrations GitHub Marketplace MCP Registry View all features Solutions By company size Enterprises Small and medium teams Startups Nonprofits By use case App Modernization DevSecOps DevOps CI/CD View all use cases By industry Healthcare Financial services Manufacturing Government View all industries View all solutions Resources Topics AI DevOps Security Software Development View all Explore Learning Pathways Events & Webinars Ebooks & Whitepapers Customer Stories Partners Executive Insights Open Source GitHub Sponsors Fund open source developers The ReadME Project GitHub community articles Repositories Topics Trending Collections Enterprise Enterprise platform AI-powered developer platform Available add-ons GitHub Advanced Security Enterprise-grade security features Copilot for business Enterprise-grade AI features Premium Support Enterprise-grade 24/7 support Pricing Search or jump to... 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Insights Additional navigation options Code Issues Pull requests Discussions Actions Projects Wiki Security Insights Enthalpy calculation problem#1910 New issue Copy link New issue Copy link Closed Closed Enthalpy calculation problem#1910 Copy link Description hot586 opened on Apr 1, 2020 Issue body actions Do you want to ask a question? Are you looking for technical support? For any kind of question regarding CoolProp and its usage, you can ask the CoolProp user group If you have an issue to report, please continue and fill out the applicable sections below. The details provided will help to resolve this issue as quickly as possible. Prerequisites Before posting an issue, please: Check the CoolProp User Group for common solutions. Check that your issue isn't already filed under CoolProp Issues. Check the CoolProp latest release and development version documentation. Description from CoolProp.CoolProp import PropsSI t1 = PropsSI('T', 'P', 16744000, 'H', 682022, 'IF97::Water') h1 = PropsSI('H', 'P', 16744000, 'T', t1 , 'IF97::Water') [Description of the issue] I was trying to caculate a temperature from pressure 16.744MPa and enthalpy 682.022kj/kg, the result I got was 432.4289K which was a little larger than it should be 432.4149K which I got from other software. It was more tricky when I used this temperature and origin pressure to get enthalpy, the result which was 682.082kj/kg, it was about 0.06kj/kg larger than the origin one I used to get temperature. . Steps to Reproduce [First Step] [Second Step] [and so on...] Expected behavior: [What you expect to happen] Actual behavior: [What actually happens] Versions CoolProp Version:6.3.0 [CoolProp version you are using] Operating System and Version:WIN10[OS you are running CoolProp on] Access Method:python 2.7.16 [How you are accessing CoolProp (i.e. Python, other wrapper, direct C++ calls, etc.)] Additional Information If possible, please post examples and/or screenshots of the issue. Activity ibell commented on Apr 1, 2020 ibell on Apr 1, 2020 Contributor More actions This is the #1 most asked question: hot586 commented on Apr 1, 2020 hot586 on Apr 1, 2020 Author More actions I think my problem is different with the question listed in FAQ. I can accept the temp which I got was different with other software, but I can not accept if I put the same temp I get from the function with a specified pressure and enthalpy and the same pressure, but I will get a different enthalpy from function. I think it's a problem of repeatability of the function. if you execute the code below, I except the same value all the time, but you will see every time after the funciton , the temp and enthalpy will increase. from CoolProp.CoolProp import PropsSI temp=[] temp.append(PropsSI('T', 'P', 16744000, 'H', 682022, 'IF97::Water')) enthalpy=[] enthalpy.append(PropsSI('H', 'P', 16744000, 'T', temp , 'IF97::Water')) for i in range(10): temp.append(PropsSI('T', 'P', 16744000, 'H', enthalpy[i] , 'IF97::Water')) enthalpy.append(PropsSI('H', 'P', 16744000, 'T', temp[i+1] , 'IF97::Water')) else: temp.append(PropsSI('T', 'P', 16744000, 'H', enthalpy[i+1] , 'IF97::Water')) print temp print enthalpy [432.42890944361255, 432.44297321965786, 432.45704144139364, 432.47111410742133, 432.4851912163382, 432.4992727667375, 432.51335875720866, 432.5274491863366, 432.54154405270157, 432.5556433548809, 432.56974709144714, 432.5838552609688] [682082.2074816858, 682142.4344925572, 682202.6810271872, 682262.9470801328, 682323.2326459328, 682383.5377191097, 682443.862294169, 682504.2063655969, 682564.5699278613, 682624.952975416, 682685.3555026953] Relojuca commented on Apr 1, 2020 Relojuca on Apr 1, 2020 via email More actions Hi People. It looks to me as residual from the solver for calculate the enthalpy, nevertheless I think that at this high pressure (15 MPa) a residual of 1x10^-5 is reasonable. If you really want to work with an Equation of state with this degree of precision I suggest you to use the IAPWS-95 also implemented in CoolProp as "Water", this equation of state is more accurate than the IF97. Regards Em qua., 1 de abr. de 2020 às 06:27, hot586 notifications@github.com escreveu: … I think my problem is different with the question listed in FAQ. I can accept the temp which I got was different with other software, but I can not accept if I put the same temp I get from the function with a specified pressure and enthalpy and the same pressure, but I will get a different enthalpy from function. I think it's a problem of repeatability of the function. if you execute the code below, I except the same value all the time, but you will see every time after the funciton , the temp and enthalpy will increase. from CoolProp.CoolProp import PropsSI temp=[] temp.append(PropsSI('T', 'P', 16744000, 'H', 682022, 'IF97::Water')) enthalpy=[] enthalpy.append(PropsSI('H', 'P', 16744000, 'T', temp , 'IF97::Water')) for i in range(10): temp.append(PropsSI('T', 'P', 16744000, 'H', enthalpy[i] , 'IF97::Water')) enthalpy.append(PropsSI('H', 'P', 16744000, 'T', temp[i+1] , 'IF97::Water')) else: temp.append(PropsSI('T', 'P', 16744000, 'H', enthalpy[i+1] , 'IF97::Water')) print temp print enthalpy [432.42890944361255, 432.44297321965786, 432.45704144139364, 432.47111410742133, 432.4851912163382, 432.4992727667375, 432.51335875720866, 432.5274491863366, 432.54154405270157, 432.5556433548809, 432.56974709144714, 432.5838552609688] [682082.2074816858, 682142.4344925572, 682202.6810271872, 682262.9470801328, 682323.2326459328, 682383.5377191097, 682443.862294169, 682504.2063655969, 682564.5699278613, 682624.952975416, 682685.3555026953] — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#1910 (comment)>, or unsubscribe . hot586 commented on Apr 1, 2020 hot586 on Apr 1, 2020 Author More actions Thanks for telling me the reason and thanks for you suggestion! Relojuca commented on Apr 1, 2020 Relojuca on Apr 1, 2020 More actions You are Welcome! But I am not part of the CoolProp development team, the last message is just my guess, however I think that for you is better use IAPWS-95, and if is important for you the calculation speed, I suggest to use the CoolProp low level interface, as explained in the following link: Regards ibell commented on Apr 1, 2020 ibell on Apr 1, 2020 Contributor More actions What is the "other software"? And are you quite sure they used the IF97 formulation rather than the more accurate (but computationally complicated) IAPWS standard formulation recommended by @Relojuca ? hot586 commented on Apr 2, 2020 hot586 on Apr 2, 2020 Author More actions I used an Excel add-in called WASPCN V6.0 Add-in. Yes, I'm not sure how it implements, my origin attention is not about questioning the accurancy of calculation, I was confused about the repeatability of the result. Now I understand it's caused by the residual. I appreaciate your time to answer my question. I will close this issue. Thank you, all guys, have a good day! hot586 closed this as completedon Apr 2, 2020 Sign up for freeto join this conversation on GitHub. Already have an account? Sign in to comment Metadata Metadata Assignees No one assigned Labels No labels No labels Type No type Projects No projects Milestone No milestone Relationships None yet Development Code with agent mode Select code repository No branches or pull requests Participants Issue actions Footer © 2025 GitHub,Inc. Footer navigation Terms Privacy Security Status Community Docs Contact Manage cookies Do not share my personal information You can’t perform that action at this time. Enthalpy calculation problem · Issue #1910 · CoolProp/CoolProp
10713	https://brainly.com/question/45969696	[FREE] Calculate the enthalpy change for the reaction \text{H}_2 + \text{F}_2 \rightarrow 2\text{HF} given - brainly.com 3 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +34,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +12,1k Ace exams faster, with practice that adapts to you Practice Worksheets +8,4k Guided help for every grade, topic or textbook Complete See more / Chemistry Expert-Verified Expert-Verified Calculate the enthalpy change for the reaction H 2+F 2→2 HF given that: Bond energy of H–H bond = 434 kJ/mol Bond energy of F–F bond = 158 kJ/mol Bond energy of H–F bond = 565 kJ/mol A. 538 kJ B. –538 kJ C. 27 kJ D. –27 kJ 1 See answer Explain with Learning Companion NEW Asked by Gabbymadrid1907 • 01/12/2024 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 14685766 people 14M 0.0 0 Upload your school material for a more relevant answer The enthalpy change for the reaction H₂ + F₂ → 2HF can be calculated using bond energies. After breaking H-H and F-F bonds and forming H-F bonds, the enthalpy change is found to be (B.) -538 kJ, showing that it is an exothermic reaction. Explanation To calculate the enthalpy change (ΔH) for the reaction H₂ + F₂ → 2HF, we need to consider the bond energies required to break the reactants' bonds and the energy released when the products' bonds are formed. Energy required to break the H-H bond: 434 kJ/mol Energy required to break the F-F bond: 158 kJ/mol Total energy required to break bonds in reactants: 434 kJ/mol + 158 kJ/mol = 592 kJ/mol Energy released when forming two H-F bonds: 2 × 565 kJ/mol = 1130 kJ/mol Enthalpy change of the reaction: 1130 kJ/mol - 592 kJ/mol = -538 kJ The negative sign indicates that the reaction is exothermic, meaning that it releases energy. Therefore, the correct answer is B. –538 kJ. Answered by AntonHugh •56.5K answers•14.7M people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 14685766 people 14M 0.0 0 Upload your school material for a more relevant answer The enthalpy change for the reaction H 2+F 2→2 HF is calculated to be −538 kJ. This indicates that the reaction is exothermic, meaning it releases energy. Thus, the correct answer is B. -538 kJ. Explanation To calculate the enthalpy change (ΔH) for the reaction H 2+F 2→2 HF, we start by considering the bond energies involved in the reaction. Bonds in Reactants: The energy required to break the H-H bond is 434 kJ/mol. The energy required to break the F-F bond is 158 kJ/mol. Total energy required to break both bonds: 434 kJ/mol+158 kJ/mol=592 kJ/mol Bonds in Products: When two HF molecules are formed, we create two H-F bonds. The energy released when forming one H-F bond is 565 kJ/mol. Total energy released when forming two H-F bonds: 2×565 kJ/mol=1130 kJ/mol Calculating Enthalpy Change: Now, we can find the enthalpy change of the reaction: Δ H=Energy released−Energy required Δ H=1130 kJ/mol−592 kJ/mol=538 kJ/mol Since the reaction releases energy (it forms bonds), the enthalpy change is negative: Δ H=−538 kJ/mol This negative enthalpy change indicates that the reaction is exothermic; it releases heat to the surroundings. Therefore, the correct answer is B. -538 kJ. Examples & Evidence This type of calculation is used in thermochemistry to determine if a reaction will release or absorb energy. Understanding the relationship between bond energies and enthalpy changes helps explain why some reactions are spontaneous and others require energy input. The calculation is based on established principles of thermodynamics and bond energy relationships, widely accepted in chemistry. The bond energies used are standard values available in chemistry literature. Thanks 0 0.0 (0 votes) Advertisement Gabbymadrid1907 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers Community Answer Given: H−H bond energy = 435 kJ, Cl−Cl bond energy = 243 kJ, and the standard heat of formation of HCl(g) is −92 kJ/mol, calculate the H−Cl bond energy. a. 431 kJ b. 247 kJ c. 180 kJ d. 4.6 kJ e. 326 kJ Community Answer Calculate the average bond energy of the sulfur-oxygen bonds in SO2 (in kJ/mol), given the following information. SF4(g) + 2 H2O(g) → SO2(g) + 4 HF(g) ΔH = –123 kJ Bond dissociation energies: S–F 327 kJ/mol F–F 154 kJ/mol H–F 565 kJ/mol H–O 467 kJ/mol Community Answer 5.0 6 Oxygen difluoride is an unstable molecule that reacts readily with water. Calculate the bond energy of the O–F bond using the standard enthalpy of reaction and the bond energy data provided. Just enter a number (no units). OF2(g) + H2O(g) \longrightarrow⟶ O=O(g) + 2HF(g) \DeltaΔH° = –318 kJ Bond: O–H O=O H–F Bond energy (kJ/mol): 467 498 565 Community Answer The reaction of H₂ with F₂ produces HF with ∆H= -269 kJ/mol per mole HF. If the H-H and H-F bond energies are 432 and 565 kJ/mol, respectively, what is the F-F bond energy (in kJ/mol)? H₂ (g) + F₂ (g) → 2HF(g) Community Answer The reaction of H₂ with F2 produces HF with AH = -269 kJ/mol of HF. If the H-H and H-F bond energies are 432 and 565 kJ/mol, respectively, what is the F-F bond energy? H₂(g) +F₂ (g)—>2HF (g) Community Answer 5.0 The enthalpy change for the following reaction is -318 kJ. Using bond energies, estimate the O-F bond energy in OF2(g). OF2(g) + H2O(g) -> O2(g) + 2HF(g) Community Answer Given the reaction C(g) + 2H(g) + 2F(g) → CH2F2(g) What is the heat of reaction, ΔH, in kJ at 25 °C? Bond Energies (kJ.mol-1) at 25 °C C-H 414 C-F 486 H-H 435 F-F 159 H-F 569 a. ΔH= +900 kJ b. ΔH= -900 kJ c. ΔΗ = 465 kJ d. ΔH= -1800 kJ Community Answer Use the given bond enthalpies to estimate AH for the following reaction: CH4(9) + 2 F2(g) – CF4(9) + 2 H2(g) Bond enthalpies: C-F 450 kJ mol-1: C-H 410 kJ mol-1; F=F 158 kJ mol-1; H—H 436 kJ mol-1 Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized New questions in Chemistry Which type of reaction is M g+S→M g S? A. polymerization B. synthesis C. replacement D. decomposition Reactants undergo chemical reaction to form products. This chemical equation represents one such reaction. The coefficient for one of the reactants or products is incorrect. Which part of the chemical equation is incorrect? 2 C 4H 1010 O 2⋯8 C O 2+10 H 2O Which of the samples most likely had the highest solubility? \| Sample \| Name \| Chemical formula \| Temperature of water (∘C) \| :---: :---: \| \| 1 \| Table sugar \| C 12H 22O 11 \| 80 \| \| 2 \| Table sugar \| C 12H 22O 11 \| 45 \| \| 3 \| Table salt \| NaCl \| 55 \| \| 4 \| Table salt \| NaCl \| 63 \| \bigcirc 1 \bigcirc 2 \bigcirc 3 \bigcirc 4 Which of the samples most likely had the highest solubility? \| Sample \| Name \| Chemical formula \| Temperature of water ( ∘C ) \| :--- :--- \| \| 1 \| Table sugar \| C 12H 22O 11 \| 80 \| \| 2 \| Table sugar \| C 12H 22O 11 \| 45 \| \| 3 \| Table salt \| NaCl \| 55 \| \| 4 \| Table salt \| NaCl \| 63 \| A. 1 B. 2 C. 3 D. 4 \ \hline 1 & Table sugar & $C {12} H {22} O {11}$ & 80 \ \hline 2 & Table sugar & $C {12} H {22} O {11}$ & 45 \ \hline 3 & Table salt & NaCl & 55 \ \hline 4 & Table salt & NaCl & 63 \ \hline \end{tabular} A. 1 B. 2 C. 3 D. 4") Consider the chemical equations shown here. 2 H 2(g)+O 2(g)→2 H 2O(g)Δ H 1=−483.6 k J+2=−241.8 k J/m o l 3 O 2(g)→2 O 3(g)Δ H 2=284.6 k J+2=142.3 k J/m o l What is the overall enthalpy of reaction for the equation shown below? 3 H 2(g)+O 3(g)→3 H 2O(g)□ kJ Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
10714	https://www.mathspanda.com/ASMa/Lessons/Converting_Exponentials_to_a_Linear_Model_LESSON.pdf	www.mathspanda.com Converting Exponentials to a Linear Model Starter 1. (Review of last lesson) A fungus is being grown under controlled conditions in a laboratory. Initially, it covers an area of 4 mm2. After hours, its are is mm2, where and and are constants. After 6 hours its area is 10 mm2. (a) Find the values of and . (b) Predict the area of the fungus after 12 hours. (c) How long will it take for the fungus to grow to 15 mm2? (d) Describe one limitation of the model used. Notes In the starter question, two measurements were taken: one at the start when and another when . In reality it is likely that a scientist would take more measurements so that the model becomes more accurate. Let us imagine that a scientist took the following measurements (include a 3rd row for later). To take these extra measurements into account, the equation is turned into a linear relationship so that the data points can be plotted and a line of best fit drawn. The gradient and intercept of the line can then be found and these can be used to find and . The first step is to take logs of both sides, but since is involved in the equation take of both sides. Initial relationship: Take of both sides: 1st law of logs: 3rd law of logs: order of terms on RHS also swapped Since : The variables are and . By letting and , we get which is the equation of a straight line where: Horizontal axis: Vertical axis: Gradient: intercept: In order to plot this graph we need the values of (this is what the 3rd line of the table is for). t N N = N0ekt N0 k N0 k t = 0 t = 6 t 0 1 2 3 4 5 6 N (measured) 4 5.2 5.9 6.9 7.8 9.1 10 N = N0ekt y− N0 k e ln N = N0ekt ln ln N = ln(N0ekt) ln N = ln N0 + ln ekt ln N = kt ln e + ln N0 ln e = 1 ln N = kt + ln N0 t N y = ln N x = t y = k x + ln N0 t ln N k y− ln N0 ln N t 0 1 2 3 4 5 6 N (measured) 4 5.2 5.9 6.9 7.8 9.1 10 1.4 1.65 1.8 1.9 2.05 2.2 2.3 ln N Page of 1 4 www.mathspanda.com N.B. Given the graph will be plotted on graph paper, 1 d.p. accuracy is fine. E.g. 1 (a) On graph paper, draw the graph of ( axis) against ( axis) and plot the points from the table. (b) Draw a line of best fit through the points. N.B. There should be roughly an equal number of points above and below the line. (c) Calculate the gradient of the straight line and the note the intercept. N.B. Each person’s graph will be slightly different so the values of the gradient and intercept will be slightly different. (d) Using your answers to (b), find the values of and . Hence write down the exponential formula involving and . Working: (a) and (b) See graph (c) Gradient intercept (d) Gradient So intercept = so ( is ok) Formula is (compared to ) Finding the equation of the line of best fit on your calculator Instead of drawing the graph, the equation of the line of best fit can be got from your calculator. Menu >> 6: Statistics = >> 2: y = a + bx >> Enter data >> OPTN >> 4 Video: Equation of line of best fit using a Classwiz calculator E.g. 2 Use your calculator to find the equation of the line of best fit for the data above. N.B. Enter the values in the first column and in the second column. Working: where and b = 2-values: 7-values: — a better model %age difference = 3.20% ln N y− t x− y− y− N0 k N t ≈0.15 (±0.02) y− ≈1.46 ± 0.15 = k k ≈0.15 (±0.02) y− ln N0 N0 = ey-intercept N0 ≈e1.46 N0 ≈4.3 3.7 ≤N0 ≤5.0 N = 4.3e0.15t N = 4e0.153t t− ln N y = a + bx a = 1.45656 0.1481 N = 4e0.153t N = 4.29e0.148t Page of 2 4 www.mathspanda.com Additional cases Express the function as a linear relationship using . N.B. could also be used but since is not included in the following, we tend to use 1. Let . e is not included in the formula so use Initial relationship: Take s of both sides: 1st law of logs: 3rd law of logs: Horizontal axis: Vertical axis: Gradient: intercept: 2. Let . e is not included in the formula so use Initial relationship: Take s of both sides: 1st law of logs: 3rd law of logs: Horizontal axis: Vertical axis: Gradient: intercept: N.B. The axes involve your variable and The gradient and intercept involve the constants. Summary E.g. 3 The number of bacteria, , in a petri dish is observed over a period of time, . The bacteria population can be modelled over time by the formula , where and are constants. The results from the observations are shown in the table below. Plot a linear graph to represent this data and use this to find the values of and . log10 ln e log10 N = N0at log10 N = N0at log log N = log(N0at) log N = log N0 + log(at) log N = t log a + logN0 t log N log a y− log N0 N = Atk log10 N = Atk log log N = log(Atk) log N = log A + log tk log N = k log t + log A log t log N k y− log A N t y− Original Linear form Horizontal axis Vertical axis Gradient y-intercept t ln N k ln N0 t log N log t log N k log N = n log t + log k log A N = N0ekt ln N = kt + ln N0 N = N0at log N = t log a + logN0 log a log N0 N = ktn p t p = atb a b t (days) 1 3 4 6 9 p (1000s) 2 14 22 44 88 a b Page of 3 4 www.mathspanda.com E.g. 4 Two variable, and , satisfy the formula . (a) Find a linear relationship connecting and . (b) The straight line graph of against is plotted. Write down the gradient and the value of the intercept on the vertical axis. Video: Modelling exponential curves - converting to linear form (3 videos) Solutions to Starter and E.g.s Exercise p143 8D Qu 1, 2, 5, 6, 7 (no graph paper required) Reducing exponentials to linear form 1 (Modelling) Qu 3-5 (need 1 sheet of A5 graph paper) Summary S x S = 4 × 7x S x log S x Original Linear form Horizontal axis Vertical axis Gradient y-intercept t ln N k ln N0 t log N log t log N k log N = t log a + logN0 log a log N0 N = ktn log N = n log t + log k N = N0ekt log A ln N = kt + ln N0 N = N0at Page of 4 4
10715	https://www.reddit.com/r/calculus/comments/st81vp/good_reference_sheet_for_differentiation_rules/	Good reference sheet for Differentiation rules. From James Stewart Calculus 6th edition : r/calculus Skip to main contentGood reference sheet for Differentiation rules. From James Stewart Calculus 6th edition : r/calculus Open menu Open navigationGo to Reddit Home r/calculus A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to calculus r/calculus r/calculus Welcome to r/calculus - a space for learning calculus and related disciplines. Remember to read the rules before posting and flair your posts appropriately. 168K Members Online •4 yr. ago [deleted] Good reference sheet for Differentiation rules. From James Stewart Calculus 6th edition Differential Calculus Share Related Answers Section Related Answers Differentiation rules cheat sheet Common differentiation rules explained Product rule for derivatives explained Differentiation techniques for beginners Derivative formulas summary New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of February 15, 2022 Reddit reReddit: Top posts of February 2022 Reddit reReddit: Top posts of 2022 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
10716	https://www.ncbi.nlm.nih.gov/books/NBK560693/	Endometrial Hyperplasia - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Endometrial Hyperplasia Gunjan Singh; Lauren Cue; Yana Puckett. Author Information and Affiliations Authors Gunjan Singh 1; Lauren Cue 2; Yana Puckett. Affiliations 1 Membership of the Royal College of Obstetricians and Gynecologists 2 Jersey City Medical Center, Rutgers Barnabas Health Last Update: April 30, 2024. Go to: Continuing Education Activity Endometrial hyperplasia is a uterine pathology in which morphological changes occur in the cells of the endometrium and represents a precursor to the most common gynecologic malignancy in developed countries, endometrial cancer. The condition is most often caused by increased exposure to estrogens and a relative deficiency of progesterone, often referred to as "unopposed estrogen." The condition's incidence is rising, given the concurrent increase in risk factors, including obesity, diabetes, or other related illnesses.To effectively prevent the progression to endometrial malignancy, clinicians must remain vigilant regarding the signs and symptoms of endometrial hyperplasia, especially when treating patients with polycystic ovarian disease and obesity. This course is designed to provide healthcare professionals with a comprehensive understanding of endometrial hyperplasia's pathophysiology, diagnosis, and management. The latest advancements in diagnostic techniques, treatment modalities, and preventive strategies are explored, and the critical role of an interprofessional care team in early detection and referral for appropriate intervention is highlighted. This activity emphasizes the importance of facilitating early intervention and sharing decision-making with patients regarding their fertility goals and surgical risk factors. Participants gain the knowledge and skills necessary to effectively identify, evaluate, and manage patients with endometrial hyperplasia, ultimately reducing the progression to endometrial malignancy and improving patient outcomes and quality of care. Objectives: Identify the etiology of endometrial hyperplasia and its implications for progression to endometrial cancer. Apply strategies for evaluating endometrial hyperplasia, including the use of appropriate diagnostic tools such as transvaginal ultrasound and endometrial sampling. Select evidence-based treatment modalities for endometrial hyperplasia, differentiating instances in which each would be appropriate. Implement effective collaboration among interprofessional team members to identify patients with risk factors for endometrial hyperplasia,promoting counseling on lifestyle modifications for risk reduction and facilitating early therapeutic interventions to decrease progression to malignancy. Access free multiple choice questions on this topic. Go to: Introduction Endometrial hyperplasia, in atypical forms, is the precursor lesion for endometrioid adenocarcinoma of the endometrium, representing the most common gynecologic malignancy in industrialized countries.Defined as the disordered proliferation of endometrial glands, endometrial hyperplasia results from estrogenic stimulation of the endometrial tissue with a relative deficiency of progesterone's counterbalancing effects, often referred to in clinical practice as "unopposed."This imbalance results may occur in patients with obesity, chronic anovulation, early menarche, late menopause, and estrogen-secreting tumors. This disordered growth of the endometrium results in an abnormal gland-to-stroma ratio involving varying degrees of histopathological complexity and atypical features in the cells and nuclei. Endometrial hyperplasia, if untreated, has the propensity to develop into endometrial cancer. Strategies for clinical management range from surveillance or progestin therapy to hysterectomy, depending on the risk of progression to or concomitant endometrial cancer and the patient's desire to preserve fertility. Go to: Etiology The most important risk factor for endometrial hyperplasia and, in turn, endometroid adenocarcinoma is the aforementioned chronic imbalance of or "unopposed" estrogen. The source of exposure to excessive estrogen without the protective effects of progestin can be endogenous, exogenous, or genetic. Endogenous Sources Obesity An increase in adrenal secretory activity is often observed in obese patients, leading to increased androgen precursors, which are converted to estradiol in peripheral tissues. The conversion rate of androstenedione to estrone and estradiol by aromatase rises in obese patients. Higher estradiol concentrations can be found in obese patients as plasma levels of estradiol-binding sex hormone-binding globulin (SHBG) are typically diminished in this patient population. Chronic anovulation When anovulation occurs, sex hormone production is not happening cyclically, and estrogen levels dominate without the opposing effect of progesterone produced by the corpus luteum after ovulation. This imbalance leads to a continued proliferation of the endometrium. Conditions associated with anovulation include polycystic ovary syndrome (PCOS), hyperprolactinemia, and perimenopausal hormonal status. Early menarche, typically defined in individuals younger than 12 Late menopause, usually described in those aged 55 and older Presence of estrogen-secreting tumors Granulosa cell tumors represent potentially estrogen-secreting tumors of the ovary. Accordingly,endometrial hyperplasia is diagnosed in 25% to 50% of women with granulosa cell tumors of the ovary. If endometrial hyperplasia is diagnosed in a patient without known risk factors, estrogen-secreting tumors should be excluded. Exogenous source:Tamoxifen, a selective estrogen receptor modulator (SERM),one of the most common medications used for endocrine treatment of hormone receptor-positive breast cancer, has been associated with an increased risk for developing endometrial hyperplasia in postmenopausal women. Genetic source:Lynch syndrome is a genetic disease of autosomal dominant inheritance caused by mutation of 1 of 4 genes of the DNA mismatch repair system (MSH2, MLH1, MSH6, and PMS2), leading to microsatellite instability, which confers a markedly elevated risk for several types of cancers, particularly colon, and endometrial. Patients with hereditary nonpolyposis colorectal cancer have a lifetime risk of 40% to 60% for the development of endometrial cancer.Recent studies have recommended screening patients diagnosed with atypical endometrial hyperplasia or endometrial cancer for microsatellite instability. Go to: Epidemiology EC is the most common gynecologic malignancy in developed countries, the fourth leading cause of cancer, and the sixth cause of cancer death among women.Cancer of the endometrium is rising in the United States, with an estimated incidence of 66,200 cases and 13,030 deaths in 2023.The incidence of EC has increased in many countries over the past few decades, a trend which is hypothesized to be due to the rising prevalence of obesity, as well as shifts towards delaying childbearing.EH is a recognized precursor lesion of the most common type of EC (endometrioid), and its detection offers opportunities for prevention. Prompt diagnosis and treatment can effectively reduce the number of cases of endometrial malignancy. Go to: Pathophysiology Estrogenic stimulation of the endometrium, unopposed by progestins, causes proliferative glandular epithelial changes or hyperplasia. Endometrial hyperplasia results from estrogen predominance and relative progesterone insufficiency. The typical causes for endogenous estrogen excess include anovulatory cycles (perimenopause, PCOS, obesity, and estrogen-secreting ovarian tumors). The exogenous causes include unopposed estrogen therapy, hormone replacement therapy, and tamoxifen (utilized in breast cancer treatment).Hyperplasia, due to prolonged exposure to estrogens, is biologically distinct from the precancerous lesion—atypical endometrial hyperplasia. Go to: Histopathology Histologically,endometrial hyperplasia describes the abnormal proliferation of endometrial glands with a greater gland-to-stroma ratio than healthy proliferative endometrium but without endometrial stromal invasion.Diagnosis is based upon histological assessment of a tissue sample obtained surgically via endometrial biopsy, curettage, or hysterectomy. The most widely used classification system for endometrial hyperplasia is the 2014 World Health Organization (WHO) Classification System, which differentiates between endometrial hyperplasia without atypia (benign endometrial hyperplasia) and atypical endometrial hyperplasia/endometrial intraepithelial neoplasia. This clinical management of the 2 conditions differs depending on the presence or absence of nuclear atypia. Nuclear atypia is defined as nuclear enlargement with or without prominent nucleoli.Endometrial hyperplasia without atypia is a benign lesion without significant somatic genetic changes, caused by extensive exposure to estrogen that is not counterbalanced by the protective effects of progestins. The hyperplastic changes often regress if physiological progesterone levels are resumed or therapeutic progestins are utilized.Endometrial hyperplasia without atypia rarely progresses to endometrial cancer. On the molecular level, atypical endometrial hyperplasia shares many similarities with endometrioid endometrial cancer.Studies have documented a risk of as high as 50% for concomitant endometrial cancer in patients with atypical endometrial hyperplasia. The risk for the development of endometrial cancer in women with atypical endometrial hyperplasia is diminished approximately threefold to fivefold when treated with progestin. However, the risk of progression to endometrial cancer among women with endometrial hyperplasia undergoing progestin treatment remains considerable at 15% to 28%.A retrospective cohort study investigated 242 women with atypical endometrial hyperplasia, of whom 74% received progestin therapy. Additionally, when considering the concurrent or future risk of endometrial cancer among women with atypical endometrial hyperplasia, clinicians must understand that atypical endometrial hyperplasia/endometrial intraepithelial neoplasia may be difficult to histopathologically from endometrial cancer. Trimble et al found overdiagnosis or underdiagnosis in nearly every third endometrial specimen, which illustrates the difficulty of this distinction.This challenge suggests that treating atypical endometrial hyperplasia as the equivalent of early endometrial cancer when counseling affected patients is reasonable. Go to: History and Physical Most patients who receive a diagnosis of endometrial hyperplasia present with abnormal uterine bleeding (eg, abnormal postmenopausal bleeding, persistent or recurrent uterine bleeding).This symptom prompts most patients to seek medical attention almost immediately, contributing to the detection of endometrial cancer at an early stage and, in turn, its relatively favorable prognosis (80%-90% 5-year survival rate at stage I) when diagnosed and treated.This favorable outcome stands in contrast to other cancers, such as ovarian cancer, whose symptoms are more vague and can remain indolent for many years, allowing it to progress and metastasize. The history obtained from the patient at the time of presentation should include the history of the present illness (such as the duration, intensity, and presence or absence of clots associated with the current episode of bleeding) and menstrual history (whether or not the patient has reached menopause, length, and regularity of cycles, flow, the passage of clots, intermenstrual bleeding, and postmenopausal bleeding). Attention should be paid to the patient’s obstetric history, as nulliparity or delayed childbearing are risk factors for endometrial hyperplasia. Additionally, a patient’s desire to conceive in the future may alter the treatment of endometrial hyperplasia and cause the provider to consider nonsurgical options and close surveillance over hysterectomy. As with any patient, medical and surgical history should be taken in detail as they may determine contraindications to hormonal treatment or a patient’s candidacy for surgery. For instance, a patient with uncontrolled cardiopulmonary disease may have to undergo medical optimization before surgery or may attempt medical management first. Similarly, a patient with a history of breast cancer or liver disease is not a candidate for oral progestins. The physical examination should include a general routine examination, including vital signs. If a patient is hypotensive, tachycardic, and actively bleeding, she may need fluid resuscitation or a blood transfusion. As most patients present with menorrhagia, pallor should be assessed, which may suggest anemia. A breast examination should be performed to rule out any suspicious lesions, and in women of the appropriate age, normal mammography results within the last year should be confirmed.Most importantly, a pelvic exam should be performed. A speculum should be placed to visualize the quantity of bleeding, flow, and presence of clots. A foul-smelling discharge may suggest active infection and pose a contraindication to potential intrauterine device (IUD) placement. A bimanual exam can ascertain the size of the uterus and the presence of coexisting fibroids or adnexal masses.During the pelvic exam, an endometrial biopsy can be performed to confirm the diagnosis. Go to: Evaluation Confirming a diagnosis of endometrial hyperplasia and subsequently distinguishing between endometrial hyperplasia with and without atypia requires a histological examination of endometrial tissue, which can be obtained either by outpatient or inpatient endometrial sampling. Transvaginal Ultrasound in Premenopausal Women In premenopausal women, the utility of transvaginal ultrasound (TVUS) mainly lies in the ability to detect myomas, endometrial polyps, pregnancy, and other potential etiologies of abnormal uterine bleeding. The endometrial thickness varies physiologically with the different phases of the menstrual cycle and can reach as high as 18 mm during the secretory phase. Transvaginal Ultrasound in Postmenopausal Women An endometrial thickness of ≤4 mm in postmenopausal women has a >99% negative predictive value for endometrial cancer. Therefore,TVUS is pivotal in the initial evaluation of postmenopausal bleeding. There is no consensus regarding which endometrial thickness cut-off should be used when recommending an endometrial biopsy for asymptomatic patients.Failure to adequately identify a thin, distinct endometrial echo in a postmenopausal woman with bleeding should trigger the need for an ultrasound, office hysteroscopy, or endometrial sampling. Endometrial Sampling Multiple methods are available to sample the endometrium. The simplest method, in-office suction endometrial sampling (also known as endometrial biopsy) performed with a plastic cannula (Pipelle), has a long history of safety and efficacy.Endometrial biopsy involves obtaining a tissue sample by inserting a Pipelle into the uterus via the cervix and obtaining a tissue sample via suction technique.Biopsy should be offered to young women with abnormal uterine bleeding and risk factors for EH, as well as nonobese women with unopposed hyperestrogenism, such as those with PCOS or estrogen-producing ovarian masses.In both postmenopausal and premenopausal women, the Pipelle has been shown to have detection rates of 99.6% and 91%, respectively. However, more recent studies demonstrate a higher rate of endometrial cancer in hysterectomy specimens from patients with preoperative office Pipelle biopsy when compared with uterine dilation and curettage, suggesting that the former method failed to detect endometrial pathology in these instances.It is believed that “mass lesions” that distort the endometrial cavity may deflect the pliable suction catheter and lead to ineffective sampling, potentially missing endometrial pathology. Hysteroscopic-guided uterine sampling is another recommended method of obtaining endometrial tissue, and data demonstrate its utility for diagnosing endometrial polyps, endometrial cancer, and endometrial hyperplasia.Several tissue-removal devices, such as morcellators or tissue forceps, allow for targeted hysteroscopic-guided resection of the endometrium or any discrete lesions visualized hysteroscopically. Go to: Treatment / Management The risk of benign endometrial hyperplasia without atypia progressing to invasive malignancy is less than 5% over 10 years.Endometrial hyperplasia (and even low-grade endometrioid cancers) can be conservatively managed by progestin therapies (eg, oral or intrauterine progestogens), especially among women who wish to maintain fertility. Spontaneous resolution can occur if reversible causes of estrogen excess are corrected. Disease resolution is more likely with progesterone treatment (89%-96%) than with expectant management (74%-81%).Progestins have historically been the cornerstone of conservative management for atypical endometrial hyperplasia and stage IA endometrioid endometrial cancer without myometrial invasion. GnRH analogs, metformin, and hysteroscopic resection, combined with progestins, appear to increase the overall efficacy of treatment.Although oral progestins (medroxyprogesterone 10 mg to 20 mg daily or norethisterone 10 mg to 15mg daily) have been used historically as the first line for medical management, the levonorgestrel IUD has been shown to achieve a higher histologic regression rate when compared to oral progestogens for complex hyperplasia without atypia (92% vs 66%) and atypical hyperplasia (90% vs 69%). Additionally, levonorgestrel IUDs provide a local higher dose concentration of progestins than the oral route, avoiding the adverse effects linked to the systemic administration of progestogens, with expected increased patient compliance.Lifestyle changes like weight loss may also be advised alongside progesterone therapy. If fertility preservation is not desired, total hysterectomy is generally recommended for women with atypical endometrial hyperplasia due to the presumed significant risk of concurrent future endometrial cancer and for women with persistent non–atypical endometrial hyperplasia.Postmenopausal women should be offered a total hysterectomy with bilateral salpingo-oophrectomy. Accurately quantifying endometrial cancer risk in women diagnosed with endometrial hyperplasia is, therefore, pivotal to informed shared decision-making regarding the most appropriate clinical management strategies. The most recent clinical guidelines recommend that in women with endometrial hyperplasia who undergo medical management, an endometrial biopsy should be undertaken at least every 3 months until 2 consecutive negative biopsies are obtained, especially in patients with atypical endometrial hyperplasia.More extensive evaluations, such as sentinel lymph node assessment in patients with atypical endometrial hyperplasia, have not been shown to be beneficial in current studies and are not recommended. Go to: Differential Diagnosis The differential diagnosis for endometrial hyperplasia includes conditions that can result in focal or generalized thickening of the endometrium as follows: Endometrial cancer: Histopathological examination of the endometrial tissue can show markers of invasion in endometrial cancer. Endometrial polyp:Hydrosonography can enhance visualization and help diagnose endometrial polyps. Diagnostic hysteroscopy can confirm the presence of a polyp. Endometritis: An irregular-appearing endometrium and increased focal thickness are hallmarks of endometritis. Go to: Prognosis While endometrial hyperplasia can progress to endometrial cancer, the rate of progression depends on factors such as the degree of architectural abnormality and the presence or absence of nuclear atypia.It is well-established that progression to endometrial cancer is higher in women with atypical compared with non–atypical endometrial hyperplasia.Recent meta-analyses describe the concurrent prevalence of endometrial cancer in those diagnosed with atypical endometrial hyperplasia as approximately 33%. The risk of progression to cancer in those with atypical endometrial hyperplasia was approximated at 8.2% annually if untreated and 2.6% annually in those with non–atypical endometrial hyperplasia in the same study. Studies have found high rates of disease regression in patients who decide against surgical management and, instead, are managed medically. A systematic review of outcomes in patients with endometrial intraepithelial neoplasia–atypical endometrial hyperplasia or grade 1 adenocarcinoma demonstrated an initial response to progestins of 86%, with a complete response of 66% for those with atypical endometrial hyperplasia. Oral, intrauterine, and combined modes of administration are effective. Recent studies have identified complete response rates of 43% and 82% for endometrial adenocarcinoma and atypical endometrial hyperplasia, respectively, after 6 months of treating these conditions with the levonorgestrel IUD. Go to: Complications Endometrial hyperplasia, if left untreated or poorly managed, can lead to various complications, the most significant being the development of endometrial cancer. The abnormal proliferation of endometrial cells increases the risk of malignant transformation, particularly in cases of complex or atypical endometrial hyperplasia. Furthermore, untreated hyperplasia can exacerbate abnormal uterine bleeding, leading to chronic anemia and its associated complications, such as fatigue and impaired physical functioning. In cases where hyperplasia is hormonally driven, such as in estrogen-secreting tumors or conditions like PCOS, there's a heightened risk of exacerbating hormonal imbalances, potentially leading to further reproductive health issues or metabolic disturbances. Additionally, the psychological impact of living with a condition linked to cancer risk can induce significant anxiety and distress in affected individuals. Prompt diagnosis and appropriate management of endometrial hyperplasia are crucial to mitigate these potential complications and improve patient outcomes. Go to: Consultations Managing endometrial hyperplasia effectively often involves a multidisciplinary approach, with various specialists collaborating to tailor treatment plans to each patient's unique needs and fertility objectives. Consultations that may be necessary to ensure thorough evaluation, personalized management, and optimal outcomes for individuals affected by endometrial hyperplasia include the following: Gynecologist Gynecologic oncologist Endocrinologist Fertility specialist Radiologist Pathologist Nutritionist Psychologist Go to: Deterrence and Patient Education Women, particularly those who are obese, have PCOS, increased exposure to estrogens, and heavy menstrual bleeding, should be encouraged by their primary care providers to have regular follow-ups with their gynecologist. This patient cohort should be encouraged to make lifestyle modifications such as diet modification and increased exercise to achieve weight loss. They should be worked up for endometrial hyperplasia with endometrial biopsy and/or hysteroscopy with dilation and curettage (D&C). Those individuals diagnosed with endometrial hyperplasia should be engaged in a discussion regarding their reproductive goals. The women who desire future fertility should be educated regarding the importance of obtaining regular endometrial biopsies and counseled on the risks and benefits of a levonorgestrel IUD versus oral progesterone. Those with satisfied parity should be counseled regarding hysterectomy as the definitive treatment for endometrial hyperplasia and should be engaged in a thorough discussion regarding risks, benefits, and different routes of hysterectomy, including total vaginal, laparoscopic-assisted vaginal, total laparoscopic, and abdominal. Go to: Enhancing Healthcare Team Outcomes In the management of endometrial hyperplasia, an interprofessional healthcare team comprising physicians, advanced care practitioners, nurses, pharmacists, nutritionists, and other health professionals plays a pivotal role in enhancing patient-centered care, improving outcomes, ensuring patient safety, and optimizing team performance. The most common presentation of endometrial hyperplasia is abnormal uterine bleeding, often in the form of heavy menstrual bleeding, inter-menstrual bleeding, or postmenopausal bleeding. Although this condition is usually ultimately managed by a gynecologist, these patients frequently present initially to other clinicians such as internists, emergency medicine providers, family medicine practitioners, and pediatricians in the case of young teens who present with this condition. A thorough understanding of the etiology, risk factors, and treatment of endometrial hyperplasia is crucial not only so that these clinicians recognize the condition and organize a prompt referral to a gynecologist but also because of the pivotal role that lifestyle modifications can play in reducing risk factors for endometrial hyperplasia. For instance, emphasis on maintaining a healthy weight can be echoed by all of a patient's healthcare professionals, inspiring compliance among those at risk. Additionally, radiologists and histopathologists are crucial when making the diagnosis of endometrial hyperplasia. Namely, a radiologist’s identification of a thickened endometrial stripe (>4 mm) in a postmenopausal woman or a 20-mm endometrial stripe in an obese young woman with heavy menstrual bleeding prompts the need for tissue diagnosis. Specimens obtained via dilation and curettage or endometrial biopsy are subsequently sent to a histopathologist, whose analysis yields a diagnosis of endometrial hyperplasia. Further investigation by a pathologist trained in analyzing gynecologic specimens may help distinguish between atypical and benign endometrial hyperplasia. These providers should be given the relevant clinical history when possible to aid in a correct diagnosis. Pharmacists guide dosing and duration of treatment in patients desiring medical management who aim to preserve their fertility. Nutritionists and physical therapists may help patients be mindful of their weight and help mitigate obesity and its associated risk factors. Nurses monitor patients' symptoms and response to treatment and provide patient education. Collaboratively, the team of healthcare professionals can enhance patient-centered care, improve outcomes, promote patient safety, and optimize performance in managing patients with endometrial hyperplasia. Go to: Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Go to: References 1. Contreras NA, Sabadell J, Verdaguer P, Julià C, Fernández-Montolí ME. Fertility-Sparing Approaches in Atypical Endometrial Hyperplasia and Endometrial Cancer Patients: Current Evidence and Future Directions. Int J Mol Sci. 2022 Feb 25;23(5) [PMC free article: PMC8910633] [PubMed: 35269674] 2. Nees LK, Heublein S, Steinmacher S, Juhasz-Böss I, Brucker S, Tempfer CB, Wallwiener M. Endometrial hyperplasia as a risk factor of endometrial cancer. Arch Gynecol Obstet. 2022 Aug;306(2):407-421. [PMC free article: PMC9349105] [PubMed: 35001185] 3. Sanderson PA, Critchley HO, Williams AR, Arends MJ, Saunders PT. New concepts for an old problem: the diagnosis of endometrial hyperplasia. Hum Reprod Update. 2017 Mar 01;23(2):232-254. [PMC free article: PMC5850217] [PubMed: 27920066] 4. Shen Y, Yang W, Liu J, Zhang Y. Minimally invasive approaches for the early detection of endometrial cancer. Mol Cancer. 2023 Mar 17;22(1):53. [PMC free article: PMC10022290] [PubMed: 36932368] 5. Lathigara D, Kaushal D, Wilson RB. Molecular Mechanisms of Western Diet-Induced Obesity and Obesity-Related Carcinogenesis-A Narrative Review. Metabolites. 2023 May 21;13(5) [PMC free article: PMC10222258] [PubMed: 37233716] 6. Chittenden BG, Fullerton G, Maheshwari A, Bhattacharya S. Polycystic ovary syndrome and the risk of gynaecological cancer: a systematic review. 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Kandoth C, Schultz N, Cherniack AD, Akbani R, Liu Y, Shen H, Robertson AG, Pashtan I, Shen R, Benz CC, Yau C, Laird PW, Ding L, Zhang W, Mills GB, Kucherlapati R, Mardis ER, Levine DA. Integrated genomic characterization of endometrial carcinoma. Nature. 2013 May 02;497(7447):67-73. [PMC free article: PMC3704730] [PubMed: 23636398] 17. Trimble CL, Kauderer J, Zaino R, Silverberg S, Lim PC, Burke JJ, Alberts D, Curtin J. Concurrent endometrial carcinoma in women with a biopsy diagnosis of atypical endometrial hyperplasia: a Gynecologic Oncology Group study. Cancer. 2006 Feb 15;106(4):812-9. [PubMed: 16400639] 18. Makker V, MacKay H, Ray-Coquard I, Levine DA, Westin SN, Aoki D, Oaknin A. Endometrial cancer. Nat Rev Dis Primers. 2021 Dec 09;7(1):88. [PMC free article: PMC9421940] [PubMed: 34887451] 19. White K, Potter JE, Hopkins K, Fernández L, Amastae J, Grossman D. Contraindications to progestin-only oral contraceptive pills among reproductive-aged women. Contraception. 2012 Sep;86(3):199-203. [PMC free article: PMC3368072] [PubMed: 22364816] 20. Ring KL, Mills AM, Modesitt SC. Endometrial Hyperplasia. Obstet Gynecol. 2022 Dec 01;140(6):1061-1075. [PubMed: 36357974] 21. Doherty MT, Sanni OB, Coleman HG, Cardwell CR, McCluggage WG, Quinn D, Wylie J, McMenamin ÚC. Concurrent and future risk of endometrial cancer in women with endometrial hyperplasia: A systematic review and meta-analysis. PLoS One. 2020;15(4):e0232231. [PMC free article: PMC7188276] [PubMed: 32343732] 22. Giri SK, Nayak BL, Mohapatra J. Thickened Endometrium: When to Intervene? A Clinical Conundrum. J Obstet Gynaecol India. 2021 Jun;71(3):216-225. [PMC free article: PMC8310815] [PubMed: 34408340] 23. Vitale SG, Buzzaccarini G, Riemma G, Pacheco LA, Di Spiezio Sardo A, Carugno J, Chiantera V, Török P, Noventa M, Haimovich S, De Franciscis P, Perez-Medina T, Angioni S, Laganà AS. Endometrial biopsy: Indications, techniques and recommendations. An evidence-based guideline for clinical practice. J Gynecol Obstet Hum Reprod. 2023 Jun;52(6):102588. [PubMed: 37061093] 24. Smith-Bindman R, Weiss E, Feldstein V. How thick is too thick? When endometrial thickness should prompt biopsy in postmenopausal women without vaginal bleeding. Ultrasound Obstet Gynecol. 2004 Oct;24(5):558-65. [PubMed: 15386607] 25. Ben-Baruch G, Seidman DS, Schiff E, Moran O, Menczer J. Outpatient endometrial sampling with the Pipelle curette. Gynecol Obstet Invest. 1994;37(4):260-2. [PubMed: 8050731] 26. Dijkhuizen FP, Mol BW, Brölmann HA, Heintz AP. The accuracy of endometrial sampling in the diagnosis of patients with endometrial carcinoma and hyperplasia: a meta-analysis. Cancer. 2000 Oct 15;89(8):1765-72. [PubMed: 11042572] 27. Dueholm M, Hjorth IMD, Dahl K, Ørtoft G. Hysteroscopic resectoscope-directed biopsies and outpatient endometrial sampling for assessment of tumor histology in women with endometrial cancer or atypical hyperplasia. Eur J Obstet Gynecol Reprod Biol. 2020 Aug;251:173-179. [PubMed: 32505790] 28. Bedner R, Rzepka-Górska I. Hysteroscopy with directed biopsy versus dilatation and curettage for the diagnosis of endometrial hyperplasia and cancer in perimenopausal women. Eur J Gynaecol Oncol. 2007;28(5):400-2. [PubMed: 17966221] 29. Vitale SG, Riemma G, Carugno J, Chiofalo B, Vilos GA, Cianci S, Budak MS, Lasmar BP, Raffone A, Kahramanoglu I. Hysteroscopy in the management of endometrial hyperplasia and cancer in reproductive aged women: new developments and current perspectives. Transl Cancer Res. 2020 Dec;9(12):7767-7777. [PMC free article: PMC8799018] [PubMed: 35117379] 30. Hwang WY, Suh DH, Kim K, No JH, Kim YB. Aspiration biopsy versus dilatation and curettage for endometrial hyperplasia prior to hysterectomy. Diagn Pathol. 2021 Jan 14;16(1):7. [PMC free article: PMC7807457] [PubMed: 33441173] 31. Lacey JV, Sherman ME, Rush BB, Ronnett BM, Ioffe OB, Duggan MA, Glass AG, Richesson DA, Chatterjee N, Langholz B. Absolute risk of endometrial carcinoma during 20-year follow-up among women with endometrial hyperplasia. J Clin Oncol. 2010 Feb 10;28(5):788-92. [PMC free article: PMC2834395] [PubMed: 20065186] 32. Fang F, Xu H, Wu L, Hu L, Liu Y, Li Y, Zhang C. LNG-IUS combined with progesterone ameliorates endometrial thickness and pregnancy outcomes of patients with early-stage endometrial cancer or atypical hyperplasia. Am J Transl Res. 2021;13(5):5412-5419. [PMC free article: PMC8205653] [PubMed: 34150138] 33. Gallos ID, Shehmar M, Thangaratinam S, Papapostolou TK, Coomarasamy A, Gupta JK. Oral progestogens vs levonorgestrel-releasing intrauterine system for endometrial hyperplasia: a systematic review and metaanalysis. Am J Obstet Gynecol. 2010 Dec;203(6):547.e1-10. [PubMed: 20934679] 34. Uccella S, Zorzato PC, Dababou S, Bosco M, Torella M, Braga A, Frigerio M, Gardella B, Cianci S, Laganà AS, Franchi MP, Garzon S. Conservative Management of Atypical Endometrial Hyperplasia and Early Endometrial Cancer in Childbearing Age Women. Medicina (Kaunas). 2022 Sep 11;58(9) [PMC free article: PMC9504942] [PubMed: 36143933] 35. Gallos ID, Krishan P, Shehmar M, Ganesan R, Gupta JK. LNG-IUS versus oral progestogen treatment for endometrial hyperplasia: a long-term comparative cohort study. Hum Reprod. 2013 Nov;28(11):2966-71. [PubMed: 23975691] 36. Atak Z, Rahımlı Ocakoğlu S, Ocakoğlu G. Levonorgestrel-releasing intrauterine device to treat abnormal uterine bleeding; not one treatment option fits all. J Turk Ger Gynecol Assoc. 2023 Dec 06;24(4):246-251. [PMC free article: PMC10702264] [PubMed: 38054416] 37. Clark TJ, Neelakantan D, Gupta JK. The management of endometrial hyperplasia: an evaluation of current practice. Eur J Obstet Gynecol Reprod Biol. 2006 Apr 01;125(2):259-64. [PubMed: 16246481] 38. Vieira-Serna S, Peralta J, Viveros-Carreño D, Rodriguez J, Feliciano-Alfonso JE, Pareja R. Sentinel lymph node assessment in patients with atypical endometrial hyperplasia: a systematic review and meta-analysis. Int J Gynecol Cancer. 2024 Jan 05;34(1):66-72. [PubMed: 37973363] 39. Travaglino A, Raffone A, Saccone G, Mascolo M, Guida M, Mollo A, Insabato L, Zullo F. Congruence Between 1994 WHO Classification of Endometrial Hyperplasia and Endometrial Intraepithelial Neoplasia System. Am J Clin Pathol. 2020 Jan 01;153(1):40-48. [PubMed: 31433834] 40. Janda M, Robledo KP, Gebski V, Armes JE, Alizart M, Cummings M, Chen C, Leung Y, Sykes P, McNally O, Oehler MK, Walker G, Garrett A, Tang A, Land R, Nicklin JL, Chetty N, Perrin LC, Hoet G, Sowden K, Eva L, Tristram A, Obermair A. Complete pathological response following levonorgestrel intrauterine device in clinically stage 1 endometrial adenocarcinoma: Results of a randomized clinical trial. Gynecol Oncol. 2021 Apr;161(1):143-151. [PubMed: 33762086] 41. Ga H, Taguchi A, Honjoh H, Nishijima A, Eguchi S, Miyamoto Y, Sone K, Mori M, Osuga Y. Prognosis of patients with endometrial cancer or atypical endometrial hyperplasia after complete remission with fertility-sparing therapy. Arch Gynecol Obstet. 2023 Nov;308(5):1629-1634. [PMC free article: PMC10520125] [PubMed: 37310452] Disclosure:Gunjan Singh declares no relevant financial relationships with ineligible companies. Disclosure:Lauren Cue declares no relevant financial relationships with ineligible companies. Disclosure:Yana Puckett declares no relevant financial relationships with ineligible companies. Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology Histopathology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Consultations Deterrence and Patient Education Enhancing Healthcare Team Outcomes Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK560693 PMID: 32809528 Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology Histopathology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Consultations Deterrence and Patient Education Enhancing Healthcare Team Outcomes Review Questions References Related information PMCPubMed Central citations PubMedLinks to PubMed Similar articles in PubMed Fertility outcomes in infertile women with complex hyperplasia or complex atypical hyperplasia who received progestin therapy and in vitro fertilization.[J Zhejiang Univ Sci B. 2017]Fertility outcomes in infertile women with complex hyperplasia or complex atypical hyperplasia who received progestin therapy and in vitro fertilization.Li M, Song JL, Zhao Y, Wu SL, Liu HB, Tang R, Yan L. J Zhejiang Univ Sci B. 2017 Nov.; 18(11):1022-1025. Review Oestrogens and endometrial hyperplasia.[Br J Hosp Med. 1980]Review Oestrogens and endometrial hyperplasia.Thom MH, Studd JW. Br J Hosp Med. 1980 May; 23(5):506, 508-9, 511-3. Papillary proliferation of the endometrium: a clinicopathologic study of 59 cases of simple and complex papillae without cytologic atypia.[Am J Surg Pathol. 2013]Papillary proliferation of the endometrium: a clinicopathologic study of 59 cases of simple and complex papillae without cytologic atypia.Ip PP, Irving JA, McCluggage WG, Clement PB, Young RH. Am J Surg Pathol. 2013 Feb; 37(2):167-77. Review Hysteroscopic treatments for endometrial malignancies and premalignancies: state of the art.[Minim Invasive Ther Allied Tec...]Review Hysteroscopic treatments for endometrial malignancies and premalignancies: state of the art.Kahramanoglu I. Minim Invasive Ther Allied Technol. 2021 Oct; 30(5):272-277. Epub 2021 Jul 21. Heart and Neural Crest Derivatives Expressed Transcript 2 (HAND2): a novel biomarker for the identification of atypical hyperplasia and Type I endometrial carcinoma.[Int J Gynecol Pathol. 2015]Heart and Neural Crest Derivatives Expressed Transcript 2 (HAND2): a novel biomarker for the identification of atypical hyperplasia and Type I endometrial carcinoma.Buell-Gutbrod R, Cavallo A, Lee N, Montag A, Gwin K. Int J Gynecol Pathol. 2015 Jan; 34(1):65-73. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Endometrial Hyperplasia - StatPearlsEndometrial Hyperplasia - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. 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10717	https://www.abs.gov.au/websitedbs/D3310114.nsf/home/Sample+Size+Calculator+Help?OpenDocument	Sample Size Calculator Help Skip to main content Archived content. This page is no longer actively maintained and may not function as intended. For the latest information and statistics visit the ABS Website. Australian Bureau of Statistics Search for: Submit search query: MENU Statistics Census Participating in a survey About Sample Size Calculator Help What does it do? The sample size calculator allows you to calculate the required responding sample size, standard error, RSE, and a confidence interval (95% or 99%) for a proportion estimate, using just one of these criteria as an input. For example, if you know the standard error you need to meet precision requirements of your estimate, you can find out the responding sample size required to achieve that; if you know the likely size of the responding sample you can estimate the standard error of your estimate, and a confidence interval for it. It is recommended that the level of precision be set to allow the survey to achieve the desired outputs. The level of precision should be set in conjunction with the users of the data. You should not set the accuracy levels too high, as you will incur higher costs (due to a larger sample size) and place additional burden on the community. You should also not set the accuracy levels too low, as your data may not be appropriate for your users. The sample size calculator assumes simple random sampling. The results generated here are intended only as a guide and should only be used as such. They are by no means the definitive "rule" about the size of a sample. How do I use it? Simply follow the steps outlined below. What do the categories mean? Confidence Level This is the level of certainty with which you will estimate the true population value. You can select either 95% or 99%, meaning that there are 95 (or 99) chances out of 100 that the true value of the estimate is within the confidence interval. Population Size This is the total number of individuals or groups about which information is required. This option can be left blank, in which case the population size is assumed to be very large. The target population is the population about which information is to be sought. The target population is also known as the scope of the survey i.e. the population that the survey is aimed at. For a sample survey, a sample is taken from the population. Proportion This specifies the expected proportion of the population to have the attribute that you are estimating from your survey. This assists in calculating the estimated standard errors that are appropriate for your situation. You can get the proportion from previous cycles of the survey or by an educated guess. If this proportion is unknown, it should be set to 0.5 (i.e.50%), as this produces a conservative estimate of variance. For example, if you want to estimate how many customers were satisfied with your service, and you have data from last year saying that 50% of customers were satisfied, then you would enter 0.5. Confidence Interval The confidence interval allows you to specify the desired level of accuracy of the estimate. The confidence interval value is expressed as a proportion, meaning that if you want the result to be accurate within 5 percentage points, then you should enter 0.05. You can specify an interval from 0 to 0.5. The narrower the interval the more precise the estimate. Confidence Interval: Upper and Lower These are the upper and lower bounds of the confidence interval, as determined by the specified interval. For example, if the is proportion os 0.5 and the specified confidence Interval is 0.05, then the upper and lower bounds will be 0.5 +/- 0.05, which gives you a confidence interval from 0.45 to 0.55. Standard Error This is a measure of sampling error that indicates the degree to which an estimate may vary from the true value. The standard error is expressed in the same units as the estimate (in the case of this calculator it is a proportion). Higher standard errors indicate more variation in the estimate. Relative Standard Error (RSE) This is the standard error expressed as a percentage of the estimate. The RSE is a useful measure of accuracy, as it gives an indication of the percentage errors likely to have occurred due to sampling. For example, if the proportion is 0.5 and the standard error is 0.05 then the RSE will be 10%. Sample Size The sample size refers to the number of individuals or groups required to respond (not just the number approached) to achieve the required level of accuracy (specified by entering either the confidence or error required). For example, if your required sample size from the calculator is 100 and you expect only 10% of your sample to respond, then you will need to approach 1000 units in order to achieve the sample size of 100. What do the Results mean? Entering the Confidence Level, Proportion, and Confidence Interval into the Calculator to determine the required sample size: To be 95% confident that the true value of the estimate will be within 5 percentage points of 0.5, (that is, between the values of 0.45 and 0.55), the required sample size is 385. This is the number of actual responses needed to achieve the stated level of accuracy. You need to take into account any expected non-response when calculating the number of individuals or groups that you will approach for the survey. The standard error of 0.026 on the estimate of 0.5 produces an RSE of 5.1%. Before CalculateAfter Calculate For more examples of how to use the Sample Size Calculator, see Sample Size Calculator Examples. Click here to begin using theSample Size Calculator... This page last updated 4 April 2023 Archived content. This page is no longer actively maintained and may not function as intended. For the latest information and statistics visit the ABS Website. Creative commons Copyright Disclaimer Privacy Accessibility Staff login Feedback How would you rate your experience? Hate Love Next
10718	https://testbook.com/maths/latus-rectum-of-hyperbola	Latus Rectum of Hyperbola Equation and Length with Derivation English Get Started Exams SuperCoaching Live Classes FREE Test Series Previous Year Papers Skill Academy PassPassPass ProPass Elite Pass Pass Pro Pass Elite Rank Predictor IAS Preparation MoreFree Live ClassesFree Live Tests & QuizzesFree QuizzesPrevious Year PapersDoubtsPracticeRefer and EarnAll ExamsOur SelectionsCareersIAS PreparationCurrent Affairs Practice GK & Current Affairs Blog Refer & EarnOur Selections HomeMaths Latus Rectum of Hyperbola Latus Rectum of Hyperbola Equation, Formula with Derivation Download as PDF Overview Test Series Conic sections are figures formed by the intersection of a plane with a cone. There are three main conic section: Ellipse, hyperbola and parabola. Latus rectum of a conic section is the chord that passes through a focus and is parallel to its directrix. Latus rectum is derived from the Latin words latus, meaning side, and rectum, meaning straight. A line that passes through both foci and intersects the hyperbola at the vertex of each branch is known as the transverse axis of a hyperbola. The Latus rectum of a hyperbola is defined as a line segment perpendicular to the transverse axis through any of the foci and whose ending point lies on the hyperbola. The length of the latus rectum of a hyperbola is 2b²/a. What is a Hyperbola? A hyperbola is one of the remarkable types of curves known as conic sections, which are formed by the intersection of a cone and a plane. It has two symmetrical halves, each called a branch, separated by a gap. The shape looks like two infinite, opposing arcs. x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1 is the equation of a hyperbola that opens horizontally, while y 2 a 2−x 2 b 2=1 y 2 a 2−x 2 b 2=1 defines one that opens vertically. Each hyperbola has two focal points. The sum of the distances from any point on the hyperbola to these foci remains constant. The 'eccentricity' of a hyperbola is a measure of its shape—specifically, the ratio of the distance between a focus of the hyperbola and the centre, to the distance from the centre to a vertex. For a hyperbola, the eccentricity is always greater than 1, and is given by e=1+(b a)2−−−−−−−√e=1+(b a)2 for a hyperbola that opens horizontally, and e=1+(a b)2−−−−−−−√e=1+(a b)2 for one that opens vertically. UGC NET/SET Course Online by SuperTeachers: Complete Study Material, Live Classes & More Get UGC NET/SET SuperCoaching @ just ₹25999₹7583 Your Total Savings ₹18416 Explore SuperCoaching Want to know more about this Super Coaching ? Download Brochure People also like Prev Next Assistant Professor / Lectureship (UGC) ₹26999 (66% OFF) ₹9332 (Valid for 3 Months) Explore this Supercoaching IB Junior Intelligence Officer ₹6999 (80% OFF) ₹1419 (Vaild for 12 Months) Explore this Supercoaching RBI Grade B ₹21999 (76% OFF) ₹5499 (Valid for 9 Months !) Explore this Supercoaching What is Latus Rectum of Hyperbola? Latus rectum of hyperbola is the line segment perpendicular to the transverse axis that passes through any of its foci & whose endpoints lie on hyperbola. It is parallel to the directrix of the hyperbola. A hyperbola has two foci and so we can draw two latus rectums. Both of them are equal in length and are parallel to each other. In the above image, a hyperbola is shown where F 1 F 1 and F 2 F 2 are the two foci. The line AB and CD are the latus rectums. They pass through the foci and are perpendicular to the transverse axis of the hyperbola which in this case is the X axis. A parabola can also have the Y axis as its transverse axis. The Latus Rectum is always perpendicular to the transverse axis irrespective of the type of hyperbola. In the above image, (a) the latus rectum is perpendicular to the X-axis which is the transverse axis, and (b)the latus rectum is perpendicular to the Y-axis which is the transverse axis. Latus Rectum of Hyperbola Equation There are two types of hyperbola and the equation of the Latus Rectum varies accordingly. When the X-axis is the transverse axis and Y-axis is the conjugate axis. If the center is at origin, then the foci coordinates are (±a e,0)(±a e,0) and the Latus Rectum equation is x=±a e x=±a e If the center is not at origin but at (h, k), then the foci coordinates are (±a e,k)(±a e,k) and the Latus Rectum equation is x=h±a e x=h±a e When the Y-axis is the transverse axis and X-axis is the conjugate axis. If the center is at origin, then the foci coordinates are (0,±a e)(0,±a e) and the Latus Rectum equation is y=±a e y=±a e If the center is not at origin but at (h, k), then the foci coordinates are (h,±a e)(h,±a e) and the Latus Rectum equation is y=k±a e y=k±a e Here, a is the length of the half length of the transverse axis and e is the eccentricity. Test Series 139.8k Users NCERT XI-XII Physics Foundation Pack Mock Test 323 Total Tests \| 3 Free Tests English,Hindi 3 Live Test 163 Class XI Chapter Tests 157 Class XII Chapter Tests View Test Series 94.1k Users Physics for Medical Exams Mock Test 74 Total Tests \| 2 Free Tests English 74 Previous Year Chapter Test View Test Series 65.1k Users NCERT XI-XII Math Foundation Pack Mock Test 117 Total Tests \| 2 Free Tests English,Hindi 3 AIM IIT 🎯 49 Class XI 65 Class XII View Test Series Length of Latus rectum of Hyperbola The length of the latus rectum of a hyperbola depends on the parameters of its standard equation. If we have a hyperbola represented by x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1, where it opens horizontally, or y 2 a 2−x 2 b 2=1 y 2 a 2−x 2 b 2=1 for a vertical orientation, then the length of the latus rectum, often denoted as '2l', is given by the formula 2 l=2 b 2 a 2 l=2 b 2 a for a hyperbola opening horizontally, or 2 l=2 a 2 b 2 l=2 a 2 b for a vertical one. A larger latus rectum suggests a more open, flatter hyperbola, while a smaller one indicates a sharper, more curved shape. In the above image, the transverse and conjugate axis of the hyperbola of both types is shown. In the above image, a hyperbola is given whose center is at origin and X-axis is the transverse axis and Y axis is conjugate axis. TS’T’ and LSL’ are the two Latus Rectums passing through the two foci S and S’. LS and TS’ are the semi latus rectums. Here a is the length of the half length of the transverse axis and e is the eccentricity. Learn about x axis and y axis Length of Latus Rectum of Hyperbola Derivation Let us derive a formula for length of the Latus Rectum of hyperbola. The coordinates of L are (ae, SL). Since L is on the hyperbola with the standard equation x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1, thus the coordinates of L will satisfy this equation. x 2 a 2−y 2 b 2=1⇒(a e)2 a 2−(S L)2 b 2=1⇒a 2 e 2 a 2−(S L)2 b 2=1⇒e 2−1=(S L)2 b 2⇒(S L)2=b 2(e 2−1)(i)x 2 a 2−y 2 b 2=1⇒(a e)2 a 2−(S L)2 b 2=1⇒a 2 e 2 a 2−(S L)2 b 2=1⇒e 2−1=(S L)2 b 2⇒(S L)2=b 2(e 2−1)(i) We also can write a hyperbola that b 2=a 2(e 2−1)b 2=a 2(e 2−1) Thus substituting this expression in (i), we get, (S L)2=b 2(e 2−1)⇒(S L)2=b 2×b 2 a 2[b 2 a 2=e 2−1]∴S L=b 2 a(S L)2=b 2(e 2−1)⇒(S L)2=b 2×b 2 a 2[b 2 a 2=e 2−1]∴S L=b 2 a As SL is the half length of the latus rectum, therefore the full length of the latus rectum is 2SL. Thus. L L′=T T′=2 S L=2 b 2 a L L′=T T′=2 S L=2 b 2 a Learn about Equation of Parabola and Equation of Hyperbola Latus Rectum of Conic Sections There are different types of conics. Their expressions related to their Latus Rectum are tabulated below: Conic SectionsGeneral EquationEndpoints of Latus RectumLength Formula of Latus Rectum Parabolay 2=4 a x y 2=4 a x(a,2 a)t o(a,−2 a)(a,2 a)t o(a,−2 a)4 a 4 a Ellipsex 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1(a e,b 2 a)t o(a e,−b 2 a)(a e,b 2 a)t o(a e,−b 2 a)2 b 2 a 2 b 2 a Hyperbolax 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1(a e,b 2 a)t o(a e,−b 2 a)(a e,b 2 a)t o(a e,−b 2 a)2 b 2 a 2 b 2 a Learn about Parabola Ellipse and Hyperbola Solved Examples on Latus Rectum of Hyperbola Example 1:Calculate the length of the latus rectum of the hyperbola whose equation is x 2 16−y 2 9=1 x 2 16−y 2 9=1 Solution:On comparing the given equation with the standard equation of a hyperbola x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1, we get a = 4 and b = 3. Now we use the formula to get the latus rectum. ∴L=2 b 2 a=2×(3)2 4=9 2=4.5 u n i t s∴L=2 b 2 a=2×(3)2 4=9 2=4.5 u n i t s, which is required length. Example 2:Find the equation of the latus rectum of the hyperbola whose equation is (x−3)2 25−(y−5)2 16=1(x−3)2 25−(y−5)2 16=1 Solution:We compare the given equation with the general equation of hyperbola (x−h)2 a 2−(y−k)2 b 2=1(x−h)2 a 2−(y−k)2 b 2=1 and get a = 5, b = 4. Also, we get the center as (3, 5). Thus according to the formula, the equation of hyperbola with center not at origin is x=h±a e x=h±a e We can find the value of e such as b 2=a 2(e−1)⇒16=25(e−1)⇒e−1=16 25⇒e=41 25 b 2=a 2(e−1)⇒16=25(e−1)⇒e−1=16 25⇒e=41 25 Therefore, x=h±a e⇒x=3±5×41 25⇒x=3±41 5∴5 x=3±41 x=h±a e⇒x=3±5×41 25⇒x=3±41 5∴5 x=3±41, which is the required equation of the latus rectum. If you want to score well in your math exam then you are at the right place. Here you will get weekly test preparation, live classes, and exam series. Download the Testbook App now to prepare a smart and high-ranking strategy for the exam. If you are checking Latus Rectum of Hyperbola article, also check related maths articles: Centroid of a TriangleParabola Graph Equation of a CircleDifference Between Parabola and Hyperbola Chord of a CircleEquation of Hyperbola More Articles for Maths Pair of Linear Equation Lines of Regression Polyhedron Triangular Prism Coplanarity of Two Lines Conic Section Eccentricity Parabola Parabola Ellipse and Hyperbola Hyperbola Latus Rectum of Hyperbola FAQs How do you find the latus rectum of a hyperbola? The length of the latus rectum of the hyperbola is found using the formula 2 b 2 a 2 b 2 a What is latus rectum of hyperbola? Latus rectum of hyperbola is the line segment perpendicular to the transverse axis that passes through any of its foci & whose endpoints lie on hyperbola. How do you find latus rectum points? The latus rectum points are found using the coordinate (a e,b 2 a)t o(a e,−b 2 a)(a e,b 2 a)t o(a e,−b 2 a) What is the formula of hyperbola? The equation of hyperbola is x 2 a 2−y 2 b 2=1 x 2 a 2−y 2 b 2=1 What is the parametric equation of hyperbola? The equations x=a sec θ,y=tan θ x=a sec⁡θ,y=tan⁡θ taken together are called the parametric equations of the hyperbola. What is the length of latus rectum of hyperbola formula? The length of the latus rectum, often denoted as '2l', is given by the formula 2 l=2 b 2 a 2 l=2 b 2 a for a hyperbola opening horizontally, or 2 l=2 a 2 b 2 l=2 a 2 b for a vertical one. 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10719	https://www.tutorchase.com/answers/igcse/chemistry/how-do-redox-indicators-work	How do redox indicators work? Redox indicators work by changing their colour at specific potential differences during a redox reaction. Redox indicators are chemical compounds that undergo a change in colour when they are either oxidised or reduced. This change in colour is due to the different light absorption properties of the oxidised and reduced forms of the indicator. The specific potential difference at which the colour change occurs is known as the indicator's redox potential. In a redox reaction, one substance is reduced (gains electrons) while another is oxidised (loses electrons). The redox indicator is chosen such that its redox potential lies within the potential difference of the redox reaction being studied. When the potential difference of the reaction reaches the redox potential of the indicator, the indicator changes colour. This colour change signals that the reaction has reached a particular point, often the endpoint of a titration. For example, in an acid-base titration, a pH indicator is used to signal when all the acid (or base) has reacted with the base (or acid). Similarly, in a redox titration, a redox indicator is used to signal when all the reductant (or oxidant) has reacted with the oxidant (or reductant). It's important to note that the choice of redox indicator depends on the specific reaction being studied. Different indicators have different redox potentials, and the indicator must be chosen such that its redox potential lies within the potential difference of the reaction. If the redox potential of the indicator is too high or too low, the indicator will change colour too early or too late, and the endpoint of the titration will not be accurately determined. In summary, redox indicators are a crucial tool in analytical chemistry, allowing chemists to accurately determine the endpoint of redox reactions and titrations. Study and Practice for Free Trusted by 100,000+ Students Worldwide Achieve Top Grades in your Exams with our Free Resources. Practice Questions, Study Notes, and Past Exam Papers for all Subjects! Need help from an expert? 4.93/5 based on733 reviews in The world’s top online tutoring provider trusted by students, parents, and schools globally. Related Chemistry igcse Answers
10720	https://cdn.mdedge.com/files/s3fs-public/issues/articles/ccjm68_3-0224.pdf	REVIEW BO SHEN, MD EDY SOFFER, M D Department of Gastroenterology, Head, Center for Gastrointestinal Cleveland Clinic Motility Disorders, Department of Gastroenterology, Cleveland Clinic The challenge of irritable bowel syndrome: Creating an alliance between patient and physician ABSTRACT The most important component of the treatment of irritable bowel syndrome (IBS) is to establish a therapeutic physician-patient relationship, coupled with patient education. We describe a stepwise approach to management, including judicious use of invasive tests, and setting realistic treatment goals that address the dominant symptoms, their seventy, and psychosocial factors. KEY POINTS It is desirable to make a positive diagnosis, rather than just determine IBS as a diagnosis of exclusion. The pathophysiology is still not clear, but IBS is thought to result from an interaction between increased visceral hypersensitivity (hyperalgesia), abnormal gut motility, and psychosocial factors. The characteristic symptom of IBS is abdominal pain or discomfort which is relieved with defecation, and is associated with change of stool frequency and form, in the absence of identifiable structural or biochemical abnormalities. P A T I E N T I N F O R M A T I O N Controlling irritable bowel syndrome, page 237 N PATIENTS SUSPECTED of having irritable bowel syndrome (IBS), the key for physi-cians is to develop a "therapeutic alliance" with the patient. Since IBS is a chronic disorder with no specific cause or cure, confidence in the diagnosis and educating and reassuring the patient are vital therapeutic tools. We outline a stepwise approach to diagno-sis and treatment, which seeks psychosocial factors as well as physiologic "red flags" for more serious conditions. The stepwise management of IBS begins by: • Making sure that the symptoms are consis-tent with IBS • Selecting tests judiciously to make a posi-tive diagnosis • Deciding what treatment goals are realistic. Overemphasis on identifying a specific organic cause for the symptoms may subject the patient to unnecessary and costly tests. • PREVALENCE AND IMPACT IBS is one of the most common functional gas-trointestinal disorders seen in clinical practice1 and is the most commonly diagnosed gastroin-testinal condition. Although estimates of the true prevalence vary widely from study to study due to the use of different diagnostic and popu-lation selection criteria and data sources, a rea-sonable estimate is that 10% to 20% of US adults report symptoms compatible with IBS. Only 15% of those affected actually seek med-ical attention.2.3 IBS accounts for 12% of pri-mary care and 28% of gastroenterological prac-tice visits, and for 41% of all functional gas-trointestinal disorders.4 224 C L E V E L A N D CLINIC J O U R N A L OF MEDICINE V O L U M E 6 8 • N U M B E R 3 M A R C H 2 0 0 1 The prevalence of IBS is similar in young adults and the elderly. Women are diagnosed more than twice as often as men. The financial burden of IBS is high. In the United States, IBS results in an estimated $8 billion in direct medical costs annually.5 IBS has a major impact on quality of life. Social and professional life are affected, with increased absenteeism from work, missed job opportunities, and limited social interaction.6 • A BIOPSYCHOSOCIAL DISORDER So far, no physiologic mechanism unique to IBS has been identified. Currently, IBS is viewed as a biopsychosocial disorder resulting from an interaction between increased viscer-al hypersensitivity (hyperalgesia), abnormal gastrointestinal motility, and psychosocial fac-tors.7 Persistent neuroimmune interactions after infectious gastroenteritis may lead to sen-sory dysfunction, resulting in IBS symptoms. Visceral hyperalgesia Studies of balloon distention in the large and small bowel8'9 have shown that patients with IBS are aware of the balloon sooner and expe-rience pain at significantly lower balloon vol-umes than healthy subjects. It is not known at which level of pain-signal transmission (start-ing at the receptor in the gut wall, through the spinal cord to the brain) this increased sensi-tivity is expressed, but it is selective for viscer-al stimuli. Patients with IBS have normal or decreased sensitivity to somatic stimuli.9'10 Abnormal gastrointestinal motility The changes in gut motility observed in IBS are quantitative, with no distinct pattern to distinguish affected patients from healthy sub-jects. Two major changes are observed: • Enhanced gut transit in some patients with diarrhea as the predominant symptom, and the opposite in some patients with pre-dominant constipation • Increased motility compared with healthy subjects in response to stimuli such as psycholog-ical stress, meals, and balloon inflation in the gut. Psychosocial factors IBS has long been dismissed as a psychoso-matic condition, as it has no clear cause or pathophysiologic features. Psychological stress and emotional events such as physical or sex-ual abuse can result in gastrointestinal symp-toms in healthy subjects, but they affect patients with IBS to a greater degree. Common psychological symptoms associ-ated with IBS are depression, somatization, anxiety, hostility, phobia, and paranoia. Up to 50% of patients with IBS meet criteria for a psychiatric diagnosis, compared with 20% of those with organic gastrointestinal disorders and 15% of control subjects.4 Although no psychological or psychiatric disorder is specif-ic to IBS, identification of such disorders helps in planning psychological or psychopharma-cologic treatment. • SYMPTOMS Patients with IBS can present with a variety of gastrointestinal and extraintestinal symptoms. But the main symptoms are the complex of chronic abdominal pain and altered bowel habits in the absence of identifiable structural or biochemical abnormalities. Chronic abdominal pain in IBS is usually described as a crampy sensation with variable intensity and periodic exacerbations. The pain is generally located in the lower abdomen, although the location and character of the pain can vary widely. Emotional stress and eating may exacerbate the pain, while defecation often provides some relief. Progressive pain that awakens the patient from sleep or prevents sleep should prompt a search for other causes. Stool volume, frequency, consistency Since the range of normal bowel habits is broad, a careful history of the volume, fre-quency, and consistency of the patient's stool is important. The frequency of bowel move-ments in healthy people can range from three times a day to three times a week. Patients with IBS complain of diarrhea, constipation, alternating diarrhea and constipation, or nor-mal bowel movements alternating with diar-rhea, constipation, or both. Diarrhea Diarrhea is generally characterized as frequent loose stools of small and moderate volume. Abnormal gastrointestinal motility can cause both diarrhea and constipation 43 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 68 • NUMBER 3 MARCH 2 0 0 1 A s t e p w i s e a p p r o a c h t o t h e m a n a g e m e n t of i r r i t a b l e b o w e l s y n d r o m e STEP ACTIONS KEY POINTS 1 Assessment of symptoms 2 Physical examination 3 Laboratory tests 4 Invasive tests 5 Treatment 6 Follow-up Use nonjudgmental, open-ended questions Identify predominant symptom Screen for psychological factors Identify "red flags" of other diseases: weight loss, fever, persistent diarrhea, rectal bleeding, nocturnal pain and abnormal bowel habit, new symptoms in patients > 50 years of age, family history of gastrointestinal malignancy, inflammatory bowel disease, or celiac disease Identify "red-flag" signs: anemia, jaundice, organomegaly, abdominal mass Complete blood count, chemistry panel, thyroid-stimulating hormone, and stool studies indicated for most patients for both exclusion and inclusion of diagnosis Flexible sigmoidoscopy or colonoscopy indicated in selected patients, in particular: Age > 50 years, chronic stable symptoms Age > 50 years, recent onset Persistent diarrhea, rectal bleeding Tailored to patient's symptoms Assess clinical response in 3 to 6 weeks Bowel movements generally occur during wak-ing hours, most often in the morning or after meals. Most bowel movements in patients with IBS are preceded by urgency and may be followed by a feeling of incomplete evacua-tion. Nocturnal diarrhea, bloody stools, dehy-dration, or weight loss are not features of IBS. Constipation Constipation in patients with IBS may last from days to months, with interludes of diar-rhea or normal bowel function. Stools are often hard and may be described as pellet-shaped. Patients may also experience a sense of incomplete evacuation, even when the rec-tum is empty. This can lead to straining with defecation, prolonged time on the toilet, and inappropriate use of enemas or laxatives. Other gastrointestinal symptoms Upper gastrointestinal symptoms seen in patients with IBS include heartburn, dyspha-gia, nonulcer dyspepsia, nausea, and noncar-diac chest pain.11 Patients with IBS often complain of abdominal bloating and increased gas production in the form of flatulence or belching. However, these symptoms occur despite normal gastrointestinal gas volumes and absence of significant colonic distention. Extraintestinal symptoms Patients with IBS often have nongastroin-testinal complaints, including rheumatologic symptoms, headache, genitourinary symptoms such as urinary frequency and urgency, dys-pareunia, sexual dysfunction, and sleep distur-bances. 12"H • DIAGNOSTIC CRITERIA In the absence of a biologic marker, attempts have been made to standardize the diagnosis of IBS using symptom-based criteria. The most commonly used are those proposed by Manning in 197815 and the Rome II criteria, updated in 1999.16 In taking the history, ask nonjudgmental, open-ended questions 229 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 68 • NUMBER 3 M A R C H 2 0 0 1 IRRITABLE BOWEL SYNDROME SHEN AND SOFFER Explain that IBS symptoms are real but not life-threatening The Manning criteria include relief of pain with bowel movements, looser and more frequent stools with onset of pain, pas-sage of mucus, and a sense of incomplete evacuation. The Rome I I criteria include abdominal discomfort or pain for at least 12 weeks in the preceding 12 months (which need not be consecutive), with at least two of three fea-tures: relief with defecation, onset associated with a change in frequency of stool, and onset associated with a change in the appear-ance of stool. Supportive features include abnormal stool frequency and consistency, abnormal passage of stool, and bloating or abdominal distention. A key feature of the definition is the presence of abdominal dis-comfort or pain. • STEPWISE APPROACH TO DIAGNOSIS AND MANAGEMENT Although it is important to exclude organic causes of symptoms compatible with IBS, overemphasis can subject the patient to unnecessary and costly testing. The emphasis should be on identifying a symptom complex compatible with IBS and then judiciously selecting diagnostic tests (TABLE 1). The Rome and Manning criteria provide guidelines for identifying patients with suspected IBS. Step 1: Careful assessment of symptoms Use nonjudgmental, open-ended questions, including questions about dietary history and medications. Identify abdominal pain as the dominant symptom, with altered bowel func-tion. Consider psychological factors, gently questioning about physical and sexual abuse, once a physician-patient relationship has been established. Identify "red flag" symptoms such as weight loss, fever, persistent diarrhea, rectal bleeding, nocturnal symptoms of pain and abnormal bowel habit, new onset of symptoms in patients aged 50 and older, family history of gastrointestinal malignancy, inflammatory bowel disease, and celiac disease. Step 2: Physical examination The physical examination is generally normal in IBS. Patients may have nonspecific abdom-inal tenderness. Identify "red flag" signs such as anemia, jaundice, organomegaly, and abdominal mass. Step 3: Laboratory tests A complete blood count, chemistry panel, and thyroid function tests help to exclude organic diseases. Stool analysis for ova, parasites, and fecal leukocytes should be done if diarrhea is the predominant symptom. Step 4: Invasive tests Routine flexible sigmoidoscopy with biopsy has a low diagnostic yield and is not cost-effec-tive, particularly in young patients. However, it may help to reassure an anxious patient, and it may be performed in elderly patients with chronic, stable symptoms. Colonoscopy or, less preferably, flexible sigmoidoscopy combined with barium enema, is usually indicated to exclude a neoplasm in patients over age 50,4.7 a nd if inflammatory bowel disease is suspected in younger patients. Mucosal biopsy should be performed to exclude microscopic colitis in patients with persistent diarrhea. Step 5: Initiate treatment See the discussion that follows. Step 6: Follow-up in 3 to 6 weeks • GENERAL MANAGEMENT PRINCIPLES Since IBS is a chronic disorder with no specif-ic cause or cure, confidence in the diagnosis and educating and reassuring the patient are vital therapeutic tools. The overall treatment goal should be to relieve symptoms and address the patient's specific concerns.7 How to determine the patient's concerns A simple way to determine the patient's con-cerns is to ask why he or she is seeking help at this time. Possible reasons may include: • Recent exacerbating factors, eg, concur-rent medical disorders, new medications, dietary changes • Concern about serious illness, recent fam-ily death • Environmental stressors, eg, major loss, abuse history 2 3 0 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 68 • NUMBER 3 MARCH 2001 • Psychiatric comorbidity, eg, depression, anxiety • Impairment of daily function, recent inability to work • Hidden agendas, eg, disability claims, nar-cotic requests, laxative abuse, secondary gain. An effective treatment strategy should address the dominant symptoms, their severi-ty, and psychosocial factors. How to establish a therapeutic physician-patient relationship The most important component of treatment is to establish a therapeutic physician-patient relationship coupled with patient education (see the patient information page, "Controlling irritable bowel syndrome," page 237) as proposed by Drossman,17 with the following steps: • Obtain the history through a nonjudg-mental and patient-centered interview • Conduct a careful examination and order cost-efficient tests • Determine the patient's understanding of the illness and his or her concern ("What do you think is causing your symptoms?") • Provide information on proposed mecha-nisms of IBS; this helps to validate the patient's disease experience and sets the basis for therapeutic interventions • Explain to patients that their symptoms are real and are not life-threatening, but that the disease is likely to be chronic and the diagnosis, if well established, is not likely to change, and that he or she can expect a nor-mal life span • Establish realistic expectations that have consistent limits and which acknowledge that IBS is a condition that can be managed, but not cured ("I appreciate how bad the pain is, but narcotic medication is not indicated"); involve the patient in treatment decisions ("Let me suggest some treatments for you to consider"). Dietary modification A dietary history may reveal symptom pat-terns related to dairy or gas-producing foods. Excluding foods that increase flatulence (beans, onions, celery, carrots, raisins, apri-cots, prunes, Brussels sprouts, wheat germ, pretzels, and bagels) should be considered in patients with symptoms of bloating or gas. Underlying visceral hyperalgesia in IBS may explain the exaggerated discomfort experi-enced with consumption of gas-producing foods. Increased intake of fiber is generally rec-ommended, through either diet or the use of commercial bulking supplements. Although the efficacy of fiber supplements has not been proven, some improvement has been demon-strated in patients with IBS whose primary complaints are abdominal pain and constipa-tion. W Many types of fiber supplements are avail-able, some synthetic (eg, polycarbophil, methylcellulose) and others from natural sources (eg, bran, psyllium compounds). All types of fiber may cause increased bloating and gaseousness due to colonic metabolism of nondigestible fiber. Because of its safety, a trial of fiber supple-mentation is advised in patients with IBS, especially those with constipation as the pre-dominant symptom. The amount should be titrated to the symptoms, but 10 g of fiber per day is good starting dosage. See Patient Psychosocial therapy Information Behavioral treatment may be considered tor motivated patients who associate symptoms pa9e 237 with stressors. Cognitive-behavioral treat-ment, interpersonal (psychodynamic) thera-py, hypnosis, biofeedback, stress manage-ment and relaxation training, and family or group therapy are all reasonable options. They reduce anxiety levels, encourage health-promoting behaviors, increase patient responsibility and involvement in the treat-ment, and improve pain tolerance. Factors that favor a good response to psychotherapy include20: • High patient motivation • Diarrhea or pain as the predominant symptom • IBS associated with overt psychiatric symptoms • Intermittent pain exacerbated by stress. Patients with constant abdominal pain respond poorly to psychotherapy or hyp-notherapy. 231 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 68 • NUMBER 3 M A R C H 2 0 0 1 IRRITABLE BOWEL SYNDROME SHEN AND SOFFER T A B L E 2 S y m p t o m - o r i e n t e d d r u g t h e r a p y in i r r i t a b l e b o w e l s y n d r o m e PREDOMINANT SYMPTOM GASTROINTESTINAL AGENTS TO TRY ANTIDEPRESSANT DRUGS TO TRY Diarrhea Loperamide Tricyclic antidepressants Diphenoxylate Constipation Osmotic laxatives SSRIs (eg, sertraline) Fiber supplementation Serotonin 5-HT4 agonist (tegaserod) Pain, gas. Antispasmodics Tricyclic antidepressants bloating, urgency Anticholinergics SSRIs "Currently under FDA review for use in treatment of constipation in patients with IBS Drug therapy for IBS symptoms: avoid, or use sparingly • GASTROINTESTINAL DRUG THERAPY Drug therapy for IBS symptoms should be minimal or avoided altogether because of the lifelong nature of the disorder and the lack of convincing therapeutic benefit. The difficulty in demonstrating therapeutic efficacy may in part be due to the heterogeneity of patients with IBS, the lack of disease markers, and high placebo response rates.21 The drug chosen depends on the patient's major symptoms (constipation, diarrhea, abdominal pain). Common strategies are to use dietary fiber for constipation, loperamide or diphenoxylate for diarrhea, and anticholin-ergic, antispasmodic agents, tricyclic antide-pressants or serotonin reuptake inhibitors (SSRIs) for pain (TABLE 2). Antidiarrheal agents Loperamide (Imodium) has been shown to be beneficial in diarrhea-predominant IBS by slow-ing whole-gut transit and enhancing intestinal water and electrolyte absorption.22 It does not require a prescription and is the antidiarrheal drug of choice. Diphenoxylate (Lomotil) can be given if loperamide is not effective. Anticholinergic and antispasmodic agents Anticholinergic and antispasmodic agents are the most frequently used drugs in the treat-ment of IBS. They may be beneficial in patients with postprandial abdominal pain, gas, bloating, and fecal urgency. In a recent meta-analysis of randomized, controlled trials, 13 of 16 studies of smooth-muscle relaxants showed these agents to be efficacious in global or symptomatic improve-ment.23 In fact, among all the pharmacologic agents, only smooth muscle relaxants consis-tently decreased abdominal pain, the most fre-quent and disabling symptom of IBS. However, the four agents (cimetropium, pinaverium, otilonium, and trimebutine) that consistently showed efficacy in high-quality trials are not yet approved for treatment of IBS in the United States. Data on the effectiveness of the antispasmodic dicyclomine (Bentyl) in IBS patients are l i m i t e d . 2 4 Common adverse effects of anticholiner-gic agents are dry mouth, dizziness, blurred vision, drowsiness, and tachycardia. Because of these effects and the intermittent nature of pain in IBS, we advise using such agents on an as-needed basis or in anticipation of stressors with known exacerbating effects. Dosage. Hyoscyamine (Levsin) can be given for pain at a dose of 0.125 mg to 0.25 mg sublingually or by mouth three to four times daily, or as a sustained-release tablet 0.375 mg to 0.75 mg by mouth every 12 hours. A typical dosage for dicyclomine is 20 mg by mouth three or four times a day, or three to four times a day as needed. • ANTIDEPRESSANT THERAPY: OFTEN BENEFICIAL Antidepressants are beneficial in patients with IBS and are often used in patients with chron-ic refractory symptoms. They are particularly helpful in patients with comorbid depressive and anxiety disorders. Antidepressants also have analgesic properties independent of their 232 C L E V E L A N D CLINIC JOURNAL OF M E D I C I N E V O L U M E 6 8 • N U M B E R 3 M A R C H 2 0 0 1 psychotropic effects and may therefore be beneficial in patients with neuropathic pain.25 Tricyclic antidepressants and selective serotonin reuptake inhibitors (SSRIs) modu-late visceral afferent activity from the gastroin-testinal tract and may improve abdominal pain.26 Tricyclics are helpful in IBS patients with predominant diarrhea,27 possibly because of their constipating effect. Conversely, some SSRIs may cause diarrhea, so caution is advised before prescribing SSRIs in IBS patients with diarrhea as the predominant complaint. Because antidepressants must be taken on a continuous basis, regardless of the type, they should be used in patients who suffer from fre-quent symptoms. Tricyclic antidepressants. Improvement in neuropathic pain with tricyclic antidepres-sant therapy occurs at lower doses than are required for treatment of depression. Thus, low doses should be tried initially and titrated to pain control or tolerance. Because these drugs have a delayed onset of action, the initial low dosage should be given for 3 to 4 weeks before it is deemed insufficient and increased. Frequently used antidepressants include amitriptyline (Elavil) 10 to 25 mg by mouth at bedtime, and imipramine (Tofranil) 25 to 50 mg by mouth at bedtime. The initial dose should be adjusted on the basis of the individ-ual patient's tolerance and response. SSRIs. Although SSRIs are increasingly preferred over tricyclic agents because of their low adverse effect profile, data regarding their use in IBS are limited.23'26 As with tricyclic antidepressants, SSRI treatment for IBS should start with low doses; eg, paroxetine (Paxil) 20 mg by mouth daily, fluoxetine (Prozac) 20 mg by mouth daily, or sertraline (Zoloft) 50 mgby mouth daily. Serotonin receptor agonists and antagonists Serotonin (5-hydroxytryptamine or 5-HT) serves both as a neurotransmitter and as a paracrine signaling molecule in the bowel.28 Serotonin is distributed throughout the gut, predominantly within enterochromaffin cells in the mucosal crypts and, to a lesser extent, within the nerve fibers of the myenteric and submucosal plexuses. The concentration of serotonin in the bowel is substantially greater than in the brain.28 It is estimated that 95% of the body's serotonin is synthesized and stored in the enterochromaffin cells of the gut.28 Thus, serotonin has become a primary focus of recent research. Postprandial plasma levels of serotonin in IBS patients with diarrhea as the chief symp-tom are significantly higher than those in healthy controls.29 Antagonists of serotonin type 5-HT3 receptors have been shown to increase colonic compliance, delay colonic transit, improve stool consistency, and increase thresholds for sensation and discom-fort during distention of the rectum.30 Alosetron (Lotronex), a selective 5-HT3 receptor antagonist, was the only drug in its class approved for treatment of women for whom diarrhea is a predominant symptom. Alosetron was shown to produce statistically significant improvement in abdominal pain, stool consistency, frequency and urgency in female IBS patients,31'32 though symptoms rapidly return if treatment is stopped.52 However, alosetron was recently withdrawn from use because of reports of ischemic colitis. Tegaserod is an amino guanidine-indole with selective and partial serotonin type 5-HT4 receptor agonist activity. Serotonin 5-HT4 agonists exert gastrointestinal stimulato-ry effects, partially by facilitation of enteric cholinergic transmission.35 The medication is currently under review by the US Food and Drug Administration for the treatment of constipation in IBS patients. Tegaserod in doses of 25 to 100 mg twice a day accelerated transit time through the left colon in healthy people.34 In a recent randomized, double-blind, placebo-controlled study of patients with IBS and constipation as the predominant symptom, tegaserod 4 mg/day or 12 mg/day significantly improved abdominal pain, bowel function, and general well-being.35 IBS patients with constipation as the chief symptom often have delayed small bowel transit or colonic transit, or both.56 In a recent randomized, placebo controlled trial of a group of these patients, tegaserod 2 mg by mouth twice a day accelerated orocecal tran-sit, mostly via shortened small bowel transit.37 The most frequent adverse events were gas-trointestinal: abdominal pain, diarrhea, dys-pepsia, flatulence, or vomiting. E H Antidepressants are often helpful in IBS 50 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 68 • NUMBER 3 M A R C H 2 0 0 1 IRRITABLE BOWEL SYNDROME SHEN AND SOFFER Ü • REFERENCES 1. Mitchell CM, Drossman DA. Survey of t h e A G A membership relating to t h e patients w i t h functional gastrointestinal disorders. Gastroenterology 1987; 92:1282-1284. 2. Talley NJ, Zinsmeiser AR, Van Dyke C, M e l t o n LJ. Epidemiology of colonic symptoms a n d the irritable bowel syndrome. Gastroenterology 1991; 101:927-934. 3. Drossman DA, Li A, Andruzzi E, Temple RD, Talley NJ, Thompson W G . US householders survey of functional gastrointestinal disor-ders. Prevalence, soclodemography, and health impact. Dig Dis Sci 1993; 38:1569-1580. 4. Drossman DA, W h i t e h e a d W E , Camilleri M . Irritable bowel syn-drome: A technical review for practice guideline d e v e l o p m e n t . Gastroenterology 1997; 112:2120-2137. 5. Sandler RS. Epidemiology of irritable bowel syndrome in t h e United States. Gastroenterology 1990; 99:409-415. 6. Hahn BA, Yan S, Strassels S. Impact of irritable bowel syndrome on quality of life and resource use in the United States a n d United Kingdom. Digestion 1999; 60:77-81. 7. Drossman DA. Review article: A n integrated approach to t h e irrita-ble b o w e l syndrome. Aliment Pharmacol Ther 1999; 13 Suppl 2:3-14. 8. M e r t z H, Naliboff B, M u n a k a t a J, Niazi N, M a y e r EA. Altered rectal perception is a biological marker of patients w i t h irritable b o w e l syndrome. Gastroenterology 1995; 109:40-52. 9. W h i t e h e a d WE, Holtkotter B, Enck P, et al. Tolerance for rectosig-moid distention in irritable bowel syndrome. Gastroenterology 1990; 98:1187-1192. 10. Cook IJ, van Eeden A, Collins SM. Patients w i t h irritable b o w e l syn-d r o m e have greater pain tolerance t h a n normal subjects. Gastroenterology 1987; 93:727-733. 11. Lynn RB, Friedman LS. Irritable bowel syndrome. N Engl J M e d 1993; 3 2 9 : 1 9 4 0 - 1 9 4 5 . 12. W h o r w e l l PJ, McCallem M, Creed FH, Roberts CT. Non-colon fea-tures of irritable bowel syndrome and non-ulcer dyspepsia. Gut 1986; 27:37-40. 13. Fass R, Fullerton S, Naliboff B, Hirsh T, M a y e r EA. Sexual dysfunc-tion in w o m e n w i t h t h e irritable bowel syndrome a n d non-ulcer dyspepsia. Digestion 1998; 59:79-85. 14. Fass R, Fullerton S, Tung S, M a y e r EA. Sleep disturbances in clinic patients w i t h functional bowel disorders. A m J Gastroenterol 2000; 95:1195-2000. 15. M a n n i n g AP, T h o m p s o n W G , Heaton KW, Morris AF. Towards posi-tive diagnosis of the irritable bowel. Br M e d J 1978; 2:653-654. 16. T h o m p s o n W G , Longstreth GF, Drossman DA, Heaton KW, Irvine EJ, Muller-Llssner SA. Functional bowel disorders and functional a b d o m i n a l pain. (Rome II: A multinational consensus document on functional gastrointestinal disorders). Gut 1999; 45 Suppl 2:43-47. 17. Drossman DA, W h i t e h e a d WE, Camilleri M. Irritable b o w e l syn-drome: A technical review for practice guideline d e v e l o p m e n t . Gastroenterology 1997; 112:2120-2137. 18. Lucey MR, Clark ML, Lownes J, D a w s o n A M . Is bran efficacious in irritable bowel syndrome? A double-blind placebo-controlled crossover study. Gut 1987; 28:221-225. 19. Cook IJ, Irvine EJ, Campbell D, Shannon S, Reddy SN, Collins S M . Effect of dietary fiber on symptoms and rectosigmoid motility In patients w i t h irritable bowel syndrome. Gastroenterology 1990; 98:66-72. 20. Camilleri M . Review article: Clinical evidence to support current therapies of irritable b o w e l syndrome. Aliment Pharmacol Ther 1999; 13 Suppl 2:48-53. 21. Klein KB. Controlled t r e a t m e n t trials in t h e irritable bowel syn-drome: A critique. Gastroenterology 1988; 95:232-241. 22. Cann PA, Read NW, H o l d s w o r t h CD, Barends D. Role of loperamide and placebo in m a n a g e m e n t of irritable bowel syndrome. Dig Dis Sci 1984; 29:239-247. 23. Jailwala J, Imperiale TF, Kroenke K. Pharmacological t r e a t m e n t of the irritable bowel syndrome: A systemic review of randomized, controlled trials. A n n Intern M e d 2000; 133:136-147. 24. Page JG, Dirnberger G M . Treatment of the irritable bowel syn-drome w i t h Bentyl (dicyclomine hydrochloride). J Clin Gastroenterol 1981; 3:153-156. 25. H a m e r o f f SR, Weiss JL, Lerrnan JC, et al. Doxepin's effect on chronic pain and depression: A controlled study. J Clin Psychiatry 1984; 45:47-53. 26. Clouse RE. Antidepressants for functional gastrointestinal syn-dromes. Dig Dis Sci 1994; 39:2352-2363. 27. M y r e n J, Lovland B, Larssen SE, Larsen S. A double study of t h e effect of t r i m i p r a m i n e in patients w i t h t h e irritable bowel syn-drome. Scand J Gastroenterol 1984; 19:835-843. 28. Gershon M D . Review article: Roles played by 5-hydroxytryptamine in the physiology of bowel. A l i m e n t Pharmacol Ther 1999; 13 Suppl 2:15-30. 29. Bearcroft CP, Perrett D, Farthing MJG. Postprandial plasma 5-hydroxytryptamine in diarrhea p r e d o m i n a n t irritable b o w e l syn-drome: A pilot study. Gut 1998; 42:42-46. 30. H u m p h r e y PP, Bountra C, Clayton N, Kozlowski K. Review article: The therapeutic potential of 5-HT3 receptor antagonists in the t r e a t m e n t of irritable b o w e l syndrome. A l i m e n t Pharmacol Ther 1999; 13 Suppl 2:31-38. 31. M a n g e l AW, Northcutt AR. Review article: The safety and efficacy of alosetron, a 5-HT3 receptor antagonist, in female irritable b o w e l syndrome patients. A l i m e n t Pharmacol Ther 1999; 13 Suppl 2:77-82. 32. Camilleri M , Northcutt AR, Kong S, Dukes G, McSorley D, M a n g e l AW. Efficacy and safety of alosetron in w o m e n w i t h irritable b o w e l syndrome: A randomized, placebo-controlled trial. Lancet 2000; 355:1035-1040. 33. W i s e m a n LR, Faulds D. Cisapride: An updated review of its p h a r m a -cology and therapeutic efficacy as a prokinetic agent in gastroin-testinal motility disorders. Drugs 1994; 47:116-152. 34. Appel S, Kumle A, Hubert M, Duvauchelle T. First pharmacokinetic pharmacodynamic study in humans with a selective 5-hydroxytrypta-mine-4 receptor agonist. J Clin Pharmacol 1997; 37:229-237. 35. Mueller-Lissner S, Fumagalli I, Bardhan K, et al. Tegaserod, a 5-HT4 receptor agonist, relieves key symptoms of irritable bowel syn-d r o m e [abstract]. Gastroenterology 2000; 118:A175. 36. Cann PA, Read NW, B r o w n C, Hobson N, Holdsworth CD. Irritable bowel syndrome: Relationship of disorders in t h e transit of a single solid meal to symptom patterns. Gut 1983; 24:405-411. 37. Prather CM, Camilleri M , Zinsmeister AR, McKlnzie S, T h o m f o r d e G. Tegaserod accelerates orocecal transit in patients w i t h constipation-predominant irritable bowel syndrome. Gastroenterology 2000; 118:463-468. ADDRESS: Edy Softer, MD, Department of Gastroenterology, S40, The Cleveland Clinic Foundation, 9500 Euclid Avenue, Cleveland, OH 44195; e-mail soffere@ccf. org Ö CLEVELAND CLINIC CENTER FOR CONTINUING EDUCATION RBROC
10721	https://math.stackexchange.com/questions/4016349/deriving-integer-solutions-to-quadratic-equation-without-solving-quadratic-congr	number theory - Deriving integer solutions to quadratic equation without solving quadratic congruence - 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… 1. 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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 Deriving integer solutions to quadratic equation without solving quadratic congruence Ask Question Asked 4 years, 6 months ago Modified4 years, 6 months ago Viewed 107 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I want to generate positive integer solutions of x x to the equation: x 2−x−a T=0 x 2−x−a T=0 where a is an integer >> 0 and T is a very large positive number. I noticed that when plugging this into wolfram alpha is looks like it was formulating the equation in the form: x 2−x≡0(m o d T)x 2−x≡0(m o d T) And then solving this quadratic congruence to get the equations to generate all integer solutions to my equation. I'm not keen on using this method to solve as in order for it to work it relies on first factorising T. If T is really large this will cause the solution to run in linear time (with respect to the number of digits of T) which doesn't scale to the size I want. So what I am asking is: Is there a way to generate integer solutions to my initial equation in logarithmic time without solving the quadratic congruence equation I showed above and without factorising T? number-theory quadratics euclidean-algorithm quadratic-programming Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Feb 7, 2021 at 15:48 Rob123 asked Feb 7, 2021 at 15:30 Rob123Rob123 121 2 2 bronze badges Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. Erm... sorry, but this is just a quadratic equation, and the positive solution is x=1 2+1 4+T−−−−−−√.x=1 2+1 4+T. I think it's fairly easy to check if this is an integer, even for very large T T Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Feb 7, 2021 at 15:38 NoNamesNoNames 61 2 2 bronze badges 1 Sorry I forgot to include variable a in the above equation here. When including a it doesn't come clear how you can solve it trivially like you mentioned here.Rob123 –Rob123 02/07/2021 15:50:05 Commented Feb 7, 2021 at 15:50 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. x 2−x=x(x−1)=x 2–––=2(x 2)=T x 2−x=x(x−1)=x 2 _=2(x 2)=T So first of all T T must be even. Then T−−√<≈x T<≈x and for large T you can check if x=⌈T−−√⌉x=⌈T⌉ works. In any case you can use the Newton-Raphson method to find the root of x 2−x−T=0 x 2−x−T=0, and see if it is an integer. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Feb 7, 2021 at 16:06 answered Feb 7, 2021 at 15:51 G CabG Cab 36k 3 3 gold badges 23 23 silver badges 63 63 bronze badges 3 Hi, thanks a lot for your answer. Unfortunately for some of the large values of T I have tried I don't seem to get too close to the equation evaluating near 0. I was wandering if you might know if there is some way to derive an error correcting term for this or another method in order to amend the difference from the initial approximation to the integer solution? Would it be possible to transform this into a simpler form in order to plug it into something like euclids algorithm to solve (other than direct quadratic congruence?)Rob123 –Rob123 02/07/2021 16:18:03 Commented Feb 7, 2021 at 16:18 Note that an integer solution only exists for particular values of T (now you changed into a T a T, what's the difference ?G Cab –G Cab 02/07/2021 16:21:55 Commented Feb 7, 2021 at 16:21 The difference is that I do actually know the value of T beforehand - I just don't know what a a would be. The difference is that if the value is just T here then we can just pick any value of T to make 1+4 T−−−−−−√1+4 T evaluate to an odd number so we can get an integer value for x x. With a fixed large T and choosable value a a the problem becomes more complicated to know how to solve. I think trying to make 1+4 a T−−−−−−−√1+4 a T evaluate as odd is a lot harder than making 1+a−−−−√1+a evaluate as odd especially as T gets larger Rob123 –Rob123 02/07/2021 17:53:28 Commented Feb 7, 2021 at 17:53 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 number-theory quadratics euclidean-algorithm quadratic-programming See similar questions with these tags. 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10722	https://www.academia.edu/116586763/Atlas_of_Human_Anatomy_by_Netter	Academia.edu no longer supports Internet Explorer. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. Log In Sign Up About Press Papers Terms Privacy Copyright We're Hiring! Help Center Anatomic Points Functional Overview All Topics Medicine and Health Sciences Medical Sciences Atlas of Human Anatomy by Netter Tanzir Islam Britto 2008, Saunders Sign up for access to the world's latest research checkGet notified about relevant papers checkSave papers to use in your research checkJoin the discussion with peers checkTrack your impact Abstract The gold standard of excellence for 25 years, Frank H. Netter, MD’s Atlas of Human Anatomy offers unsurpassed depictions of the human body in clear, brilliant detail – all from a clinician’s perspective. With its emphasis on anatomic relationships and clinically relevant views, Dr. Netter’s work provides a coherent, lasting visual vocabulary for understanding anatomy and how it applies to medicine today. Table of contents Section 1 Head and Neck Topographic Anatomy 1 Superficial Head and Neck 2 - 3 Bones and Ligaments 4 - 23 Superficial Face 24 - 25 Neck 26 - 34 Nasal Region 35 - 50 Oral Region 51 - 62 Pharynx 63 - 73 Thyroid Gland and Larynx 74 - 80 Orbit and Contents 81 - 91 Ear 92 - 98 Meninges and Brain 99 - 114 Cranial and Cervical Nerves 115 - 134 Cerebral Vasculature 135 - 146 Regional Scans 147 - 148 Section 2 Back and Spinal Cord Topographic Anatomy 149 Bones and Ligaments 150 - 156 Spinal Cord 157 - 167 Muscles and Nerves 168 - 172 Cross-Sectional Anatomy 173 - 174 Section 3 Thorax Topographic Anatomy 175 Mammary Gland 176 - 178Body Wall 179 - 189 Lungs 190 - 204 Heart 205 - 223 Mediastinum 224 - 234 Regional Scans 235 Cross-Sectional Anatomy 236 – 239 Section 4 Abdomen Topographic Anatomy 240 Body Wall 241 – 260 Peritoneal Cavity 261 – 266 Viscera (Gut) 267 – 276 Viscera (Accessory Organs) 277 – 282 Visceral Vasculature 283 – 296 Innervation 297 – 307 Kidneys and Suprarenal Glands 308 – 322 Cross-Sectional Anatomy 323 – 330 Section 5 Pelvis and Perineum Topographic Anatomy 331 Bones and Ligaments 332 – 336 Pelvic Floor and Contents 337 – 347 Urinary Bladder 348 – 351 Uterus, Vagina, and Supporting Structures 352 – 355 Perineum and External Genitalia: Female 356 – 359 Perineum and External Genitalia: Male 360 – 367 Homologues of Genitalia 368 – 369 Testis, Epididymis, and Ductus Deferens 370 Rectum 371 – 376 Regional Scans 377 Vasculature 378 – 388 Innervation 389 – 397 Cross-Sectional Anatomy 398 – 399 Section 6 Upper Limb Topographic Anatomy 400 Cutaneous Anatomy 401 – 405 Shoulder and Axilla 406 – 418 Arm 419 – 423 Elbow and Forearm 424 – 439 Wrist and Hand 440 – 459 Neurovasculature 460 – 467 Regional Scans 468 Section 7 Lower Limb Topographic Anatomy 469 Cutaneous Anatomy 470 – 473 Hip and Thigh 474 – 493 Knee 494 – 500 Leg 501 – 510 Ankle and Foot 511 – 525 Neurovasculature 526 – 530 Regional Scans 531 Section 8 Cross=Sectional Anatomy Key Figure for Cross Sections 532 Language: English Edition: 6 Published: October 1, 2008 Imprint: Saunders Related papers Anatomy review machi machi downloadDownload free PDFView PDFchevron_right Three-dimensional anatomical atlas of the human body Antonio Barbeito 2016 I would like to thank all those who contributed in a constructive way to the development and completion of this thesis. My supervisors, Professor Marco Painho and Professor Pedro Cabral, for their guidance, support and patience throughout this process. The directors of ESTGA, Professor Estima de Oliveira and Professor Gonçalo Paiva Dias, by providing me with favorable conditions for carrying out this research. The colleagues who kindly had a word of encouragement. My parents and sisters for their endless love and concern. Most of all, my girls, Ana, Rita and Magui, for the indefinable feeling that unites us and that gives meaning to everything else. I would also like to thank the contribution of the U.S. National Library of Medicine for making available the datasets used in this work. vi RESUMO Os atlas anatómicos permitem mapear as estruturas anatómicas do corpo humano. As primeiras versões consistiam em documentos analógicos com representações pictóricas do corpo humano associadas a texto descritivo. Com o aparecimento dos sistemas computorizados, surgiram versões digitais e os modelos tridimensionais foram introduzidos. Desta forma, estes sistemas tornaram-se mais eficientes, permitindo visualizações mais realistas e um grau de interatividade superior. O desenvolvimento de atlas anatómicos em ambientes de sistemas de informação geográfica (GIS) permite criar plataformas com um elevado grau de interatividade e com ferramentas que facilitam a exploração e análise do corpo humano. Nesta tese, desenvolve-se um protótipo para representar o corpo humano. O sistema inclui um modelo topológico 3D GIS, uma interface gráfica e funções para explorar e analisar o interior e a superfície das estruturas anatómicas do corpo humano. A aproximação GIS baseia-se essencialmente nas caraterísticas topológicas do modelo e no tipo de funções criadas que incluem diversos tipos de medição, identificação, seleção e análise. Com estas funções, o protótipo tem a capacidade de replicar a informação disponibilizada pelos atlas anatómicos tradicionais e acrescenta um nível superior de funcionalidade. Efetivamente, algumas das limitações encontradas nos atlas anatómicos correspondem precisamente a caraterísticas presentes nos GIS, nomeadamente, as capacidades interativas, a gestão de um sistema em que a informação está organizada em camadas, as ferramentas de medição, o modo de edição para expansão da informação contida no sistema e as operações de análise espacial. downloadDownload free PDFView PDFchevron_right Chapter 1, Introduction to Human Anatomy/physiology Anngelo Birung downloadDownload free PDFView PDFchevron_right Grays Anatomy Students 2SAE 20May VALACツ YT downloadDownload free PDFView PDFchevron_right A History of Human Anatomy, 2nd ed. T.V.N. Persaud, Marios Loukas, and R. Shane Tubbs Helene Perdicoyianni-Paleologou downloadDownload free PDFView PDFchevron_right New Findings in the Anatomy of the Human Body Anatomical Sciences Journal In recent years, new findings have been made about the anatomy of the human body, which challenged some of the existing information about the structure and function of the human body. Some of the most important findings are the identification of salivary glands on a macroscopic scale as tubarial glands in the nasopharynx, the identification of lymphatic structures in the central nervous system as the meningeal lymphatic system and the glymphatic system, the identification of the mesentery as a continuous structure along the entire abdominal portion of the gastrointestinal tract, and finally, the identification of a new tissue structure called interstitium in some organs of the body. A careful review of these findings is of great importance in terms of understanding the pathogenesis of related diseases, designing treatment methods, basic research, and educating medical students. The purpose of this article was to review the latest findings regarding the anatomy of the human body and their practical importance in the medical sciences. downloadDownload free PDFView PDFchevron_right Cunningham Manual of Practical Anatomy Volume 1 Tanzir Islam Britto Oxford University Press Cunningham's has been thoroughly revised for the modern-day anatomy student. The language has been simplified for easy understanding, making this textbook ideal for students at the undergraduate level. Each dissection reflects current medical school teaching and is now broken down into clear, step-by-step instructions. New learning features prepare students for the dissection lab, university examinations, and clinical practice. Completely updated full-color artwork brings the friendly explanations to life. Following a logical structure, each chapter explains in a clear and friendly manner the key knowledge expected of students. Improved diagrams with clear labelling and full colour illustrate key anatomical features, bringing the text to life. Learning objectives introduce each dissection, and clear step-by-step instructions make it easy to follow in the dissection lab. Throughout the book, new clinical application boxes and radiology images explain how anatomy relates to clinical medical practice. At the end of each part, multiple-choice questions allow students to quickly review their knowledge before checking the answers in the appendix. Student-friendly and richly illustrated, this new edition of Cunningham's brings expert anatomical teaching to the modern-day student of medicine, dentistry, and allied health sciences. Retaining the trustworthy authority of the previous editions, this sixteenth edition offers a contemporary account of this excellent practical anatomy book. downloadDownload free PDFView PDFchevron_right Anatomy Alpesh Jadhav m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o m m e b o o k s f r e e. c o downloadDownload free PDFView PDFchevron_right 2015-Clinical_Anatomy.pdf Shahed Nalla Liver dimensions at computed tomography: correlation with age, sex and habitus in Indian population Purpose: To assess liver size from various liver dimensions. downloadDownload free PDFView PDFchevron_right A plea for an extension of the anatomical nomenclature Organ systems Alzbeta Blankova, BJBMS Journal Bosnian Journal of Basic Medical Sciences, 2019 This article is the third part of a series aimed at correcting and extending the anatomical nomenclature. Communication in clinical medicine as well as in medical education is extensively composed of anatomical, histological, and embryological terms. Thus, to avoid any confusion, it is essential to have a concise, exact, perfect and correct anatomical nomenclature. The Terminologia Anatomica (TA) was published 20 years ago and during this period several revisions have been made. Nevertheless, some important anatomical structures are still not included in the nomenclature. Here we list a collection of 156 defined and explained technical terms related to the anatomical structures of the human body focusing on the digestive, respiratory, urinary and genital systems. These terms are set for discussion to be added into the new version of the TA. downloadDownload free PDFView PDFchevron_right Loading Preview Sorry, preview is currently unavailable. You can download the paper by clicking the button above. References (2) GUIDE Upper Limb: Topographic Anatomy [Plate 400, Upper Limb] Bones Clavicle: Subcutaneous; palpable throughout its length Acromion: Easily palpable, traced medially to acromioclavicular joint (see Section 6-2: Upper Limb: Shoulder and Axilla) Coracoid process of scapula: Palpable deep and lateral to the deltopectoral triangle Head of humerus: Palpable deep to deltoid inferior to lateral edge of acromion when arm is rotated Elbow: Palpable medial and lateral epicondyles of humerus and visible olecranon process of ulna (see Section 6-4: Upper Limb: Elbow and Forearm) Head of radius: Palpable as it rotates during pronation and supination on posterolateral aspect of extended elbow, just distal to lateral epicondyle Ulnar head: Visible on the medial side of the dorsal aspect of the wrist (see Section 6-5: Upper Limb: Wrist and Hand) Radial styloid: Palpable in the anatomical snuff box on the lateral side of the wrist Ulnar styloid: just distal to ulnar head with hand supinated Pisiform: hard, round structure on anteromedial aspect of wrist Tubercles of scaphoid and trapezium: Palpable at proximal end of thenar eminence Muscles and Tendons Deltoid muscle: Overlies the shoulder giving it a rounded appearance Biceps: Bulge on anterior aspect of arm Biceps brachii tendon: Palpable in cubital fossa, lateral to midline with arm flexed Flexor tendons: Wrist and finger flexors visible distally on ventral aspect of forearm Extensor tendons: Wrist and finger flexors visible on the dorsum of the hand Thenar eminence: Muscles at the base of the thumb (see Section 6-5: Upper Limb: Wrist and Hand) Hypothenar eminence: Muscles at the base of the little finger (see Section 6-5: Upper Limb: Wrist and Hand) Related papers Kenhub Atlas of Human Anatomy Mike Pascoe Kenhub Atlas of Human Anatomy, 2023 This atlas represents a collection of clear, comprehensible and didactically valuable images from Kenhub.com, intuitively organized to aid you in your mastery of the organization of the human body. The features of this modern print atlas were formulated by direct input from students of anatomy and experienced educators. This print atlas was designed by the preferences of anatomy students to assist in the challenges of identifying structures in the anatomy laboratory and to assist in studying for written exams. downloadDownload free PDFView PDFchevron_right Atlas of Human Anatomy Internal Organs Gabriel De Paula Campos downloadDownload free PDFView PDFchevron_right ESSENTIAL CLINICAL ANATOMY dar salam downloadDownload free PDFView PDFchevron_right Basic Human Anatomy – Lesson 1 INTRODUCTION TO BASIC HUMAN ANATOMY Sahnik Banerjee downloadDownload free PDFView PDFchevron_right Introduction to Anatomy Mary Moussa downloadDownload free PDFView PDFchevron_right Flesh and Bones: The Art of Anatomy, with contributions by Thisbe Gensler, Naoko Takahatake, and Erin Travers. Exh. cat. Los Angeles: Getty Research Institute, 2022. Monique Kornell Flesh and Bones: The Art of Anatomy, 2002 downloadDownload free PDFView PDFchevron_right Illustrating Medicine: Line, Luminance and the Lessons from JCB Grant's Atlas of Anatomy (1943) Jodie Jenkinson, Kim Sawchuk Visual Communication, 2011 downloadDownload free PDFView PDFchevron_right Catalog of the links with animations, 2D, 3D and endoscopic images of the individual systems and organs in the virtual anatomy atlas of the human body Aleksandra Zufic 2005 Extended abstract Introduction It is said that 'A picture is worth thousand words', and if it is animated, in motion, or followed by sound effects, it is worth even more. The aim of this paper is to help users (primarily the students of biology, medicine and anatomy) get an insight into the organs and systems of the human body, as well as into the principles of their functioning by using two- and three-dimensional images, animations, elements of virtual reality, and images of probes inserted into individual human organs and systems. Individual images show the differences between the normal (healthy) condition and the abnormal (unhealthy) condition. The use of a virtual anatomy atlas in comparison to a classic 'paper' atlas has numerous advantages. Organs and systems can be viewed from different perspectives, they can be also magnified and minimized. The systems are typically shown in one layer that can be highlighted (by different colors or by disproportional magnifi... downloadDownload free PDFView PDFchevron_right Human anatomy nomenclature rules for the computer age Pierre Sprumont Clinical Anatomy, 2016 Information systems are increasing in importance in biomedical sciences and medical practice. The nomenclature rules of human anatomy were reviewed for adequacy with respect to modern needs. New rules are proposed here to ensure that each Latin term is uniquely associated with an anatomical entity, as short and simple as possible, and machine-interpretable. Observance of these recommendations will also benefit students and translators of the Latin terms into other languages. downloadDownload free PDFView PDFchevron_right Contribution to the anatomical nomenclature concerning general anatomy and anatomical variations Vladimír Musil, Václav Báča Surgical and Radiologic Anatomy, 2016 Nomenclature of the general and variant anatomy belongs to the most neglected parts of the Latin anatomical nomenclature in Terminologia Anatomica. Although many important small structures are included in Terminologia Anatomica, when describing and teaching particular anatomy of any part of the human body, the general terms are necessary, such as planes, lines and flexion grooves. Moreover, Terminologia Anatomica contains only 149 terms of variant structures, enlisted in the parentheses to differentiate them from constant ones. They are only a rather representative selection and some more should be added, both from the educational and clinical point of view. The authors present some terms, completed with their definitions or explanations concerning the general and variant anatomy to evoke broader discussion on this topic which should issue in incorporation of proposed terms (or their equivalents) into the Terminologia Anatomica. downloadDownload free PDFView PDFchevron_right Related topics Atlas of human anatomy Explore Papers Topics Features Mentions Analytics PDF Packages Advanced Search Search Alerts Journals Academia.edu Journals My submissions Reviewer Hub Why publish with us Testimonials Company About Careers Press Help Center Terms Privacy Copyright Content Policy 580 California St., Suite 400 San Francisco, CA, 94104 © 2025 Academia. All rights reserved
10723	https://www.lessvrong.com/math/handouts/incircles-summer2021.pdf	Summer 2021 Incircles Victor Rong Incircles Victor Rong July 13, 2021 A B C D T1 M I I1 E1 F1 I2 E2 F2 T2 Lemma 1 Let AB be a chord in circle Ωand M be the midpoint of arc AB. Circle ω lies on the opposite side of AB as M and is tangent to segment AB and Ωat points P and Q respectively. Then P, Q, and M are collinear. Lemma 2 Let Ω1, Ω2 be two circles intersecting at points A and B. Two circles ω1, ω2 lie on the same of AB and are each tangent to Ω1 and Ω2 at points P1, Q1 and P2, Q2, respectively. Then P1, Q1, P2, and Q2 are concyclic. Theorem 1: Sawayama-Th´ ebault Let ABC be a triangle with incenter I. Let D be a point on BC. Let I1 be the center of the circle that touches segments AD, DB, and the circumcircle of ABC, and let I2 be the center of the circle that touches segments AD, DC, and the circumcircle of ABC. Then I1, I2, and I are collinear. 1 of 5 Summer 2021 Incircles Victor Rong Theorem 2: Pitot’s Theorem Let ABCD be a convex quadrilateral. Then ABCD has an incircle if and only if AB + CD = AD + BC. Lemma 3 Let ABCD be a bicentric quadrilateral with incenter I and circumcenter O. Let WXY Z be the intouch quadrilateral. Then the diagonals of ABCD and the diagonals of WXY Z concur at a point P. Furthermore, P is on line OI. O1 O2 O3 A12 A23 A31 B12 B23 B31 Theorem 3: Monge’s Theorem Let ω1, ω2, ω3 be circles in the plane. Denote the external center of similitude between ωj, ωk as Ajk and the internal center of similitude as Bjk. Then for i, j, k a permutation of {1, 2, 3}, we have Aij, Ajk, Aki collinear and Aij, Bjk, Bki collinear. 2 of 5 Summer 2021 Incircles Victor Rong A Problems A1. Prove the lemmas and Monge’s Theorem. A2. Let ABCD be a bicentric quadrilateral with intouch quadrilateral WXY Z. Prove that the diagonals of WXY Z are perpendicular to one another. A3. (ELMO 2011). Let ABCD be a convex quadrilateral. Let E, F, G, H be points on segments AB, BC, CD, DA, respectively, and let P be the intersection of EG and FH. Given that quadri-laterals HAEP, EBFP, FCGP, GDHP all have inscribed circles, prove that ABCD also has an inscribed circle. A4. (Japan MO 2009). Let Γ be a circumcircle. A circle with center O touches to line segment BC at P and touches the arc BC of Γ which doesn’t have A at Q. If ∠BAO = ∠CAO, then prove that ∠PAO = ∠QAO. A5. (ARMO 2018). Circle ω is tangent to sides AB, AC of triangle ABC. A circle Ωtouches the side AC and line AB (produced beyond B), and touches ω at a point L on side BC. Line AL meets ω, Ωagain at K, M, respectively. It turned out that KB ∥CM. Prove that △LCM is isosceles. A6. (ISL 2005). Given a triangle ABC satisfying AC + BC = 3 · AB. The incircle of triangle ABC has center I and touches the sides BC and CA at the points D and E, respectively. Let K and L be the reﬂections of the points D and E with respect to I. Prove that the points A, B, K, L lie on one circle. A7. (CMO 2012). Let ABCD be a convex quadrilateral and let P be the point of intersection of AC and BD. Suppose that AC + AD = BC + BD. Prove that the internal angle bisectors of ∠ACB, ∠ADB and ∠APB meet at a common point. B Problems B1. (Germany MO 2009). Let ABCD be a convex quadrilateral and let N be the intersection of diagonals AC and BD. Denote by a, b, c, d the length of the altitudes from N to AB, BC, CD, DA, respectively. Prove that 1 a + 1 c = 1 b + 1 d if and only if ABCD has an incircle. B2. (USA TSTST 2017). Let ABC be a triangle with incenter I. Let D be a point on side BC and let ωB and ωC be the incircles of △ABD and △ACD, respectively. Suppose that ωB and ωC are tangent to segment BC at points E and F, respectively. Let P be the intersection of segment AD with the line joining the centers of ωB and ωC. Let X be the intersection point of lines BI and CP and let Y be the intersection point of lines CI and BP. Prove that lines EX and FY meet on the incircle of △ABC. B3. (USA TST 2010). Let ABC be a triangle. Point M and N lie on sides AC and BC respectively such that MN\|\|AB. Points P and Q lie on sides AB and CB respectively such that PQ\|\|AC. The incircle of triangle CMN touches segment AC at E. The incircle of triangle BPQ touches segment AB at F. Line EN and AB meet at R, and lines FQ and AC meet at S. Given that AE = AF, prove that the incenter of triangle AEF lies on the incircle of triangle ARS. B4. (ARMO 2016). In triangle ABC, AB < AC and ω is the incircle. The A-excircle is tangent to BC at A′. Point X lies on AA′ such that segment A′X doesn’t intersect with ω. The tangents from X to ω intersect with BC at Y, Z. Prove that the sum XY + XZ not depends to point X. 3 of 5 Summer 2021 Incircles Victor Rong B5. (ISL 2006). A point D is chosen on the side AC of a triangle ABC with ∠C < ∠A < 90◦in such a way that BD = BA. The incircle of ABC is tangent to AB and AC at points K and L, respectively. Let J be the incenter of triangle BCD. Prove that the line KL intersects the line segment AJ at its midpoint. B6. (ISL 2017). A convex quadrilateral ABCD has an inscribed circle with center I. Let Ia, Ib, Ic and Id be the incenters of the triangles DAB, ABC, BCD and CDA, respectively. Suppose that the common external tangents of the circles AIbId and CIbId meet at X, and the common external tangents of the circles BIaIc and DIaIc meet at Y . Prove that ∠XIY = 90◦. C Problems C1. (ISL 2007). Point P lies on side AB of a convex quadrilateral ABCD. Let ω be the incircle of triangle CPD, and let I be its incenter. Suppose that ω is tangent to the incircles of triangles APD and BPC at points K and L, respectively. Let lines AC and BD meet at E, and let lines AK and BL meet at F. Prove that points E, I, and F are collinear. C2. (ISL 2015). Let ABCD be a convex quadrilateral, and let P, Q, R, and S be points on the sides AB, BC, CD, and DA, respectively. Let the line segment PR and QS meet at O. Suppose that each of the quadrilaterals APOS, BQOP, CROQ, and DSOR has an incircle. Prove that the lines AC, PQ, and RS are either concurrent or parallel to each other. C3. (China TST 2016). In cyclic quadrilateral ABCD, AB > BC, AD > DC, I, J are the incenters of △ABC,△ADC respectively. The circle with diameter AC meets segment IB at X, and the extension of JD at Y . Prove that if the four points B, I, J, D are concyclic, then X, Y are the reﬂections of each other across AC. C4. (IMO 2008). Let ABCD be a convex quadrilateral with BA ̸= BC. Denote the incircles of triangles ABC and ADC by ω1 and ω2 respectively. Suppose that there exists a circle ω tangent to ray BA beyond A and to the ray BC beyond C, which is also tangent to the lines AD and CD. Prove that the common external tangents to ω1 and ω2 intersect on ω. C5. (Poland MO 2016). Let I be an incenter of △ABC. Denote D, S ̸= A intersections of AI with BC, O(ABC) respectively. Let K, L be incenters of △DSB, △DCS. Let P be a reﬂection of I with the respect to KL. Prove that BP ⊥CP. C6. (ISL 2009). Let ABCD be a circumscribed quadrilateral. Let g be a line through A which meets the segment BC in M and the line CD in N. Denote by I1, I2 and I3 the incenters of △ABM, △MNC and △NDA, respectively. Prove that the orthocenter of △I1I2I3 lies on g. C7. (ISL 2012). Let ABCD be a convex quadrilateral with non-parallel sides BC and AD. Assume that there is a point E on the side BC such that the quadrilaterals ABED and AECD are circumscribed. Prove that there is a point F on the side AD such that the quadrilaterals ABCF and BCDF are circumscribed if and only if AB is parallel to CD. C8. (ISL 2010). Three circular arcs γ1, γ2, and γ3 connect the points A and C. These arcs lie in the same half-plane deﬁned by line AC in such a way that arc γ2 lies between the arcs γ1 and γ3. Point B lies on the segment AC. Let h1, h2, and h3 be three rays starting at B, lying in the same half-plane, h2 being between h1 and h3. For i, j = 1, 2, 3, denote by Vij the point of intersection of hi and γj (see the Figure below). Denote by \ VijVkj\ VklVil the curved quadrilateral, whose sides are the segments VijVil, VkjVkl and arcs VijVkj and VilVkl. We say that this quadrilateral 4 of 5 Summer 2021 Incircles Victor Rong is circumscribed if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals \ V11V21 \ V22V12, \ V12V22 \ V23V13, \ V21V31 \ V32V22 are circumscribed, then the curved quadrilateral \ V22V32 \ V33V23 is circumscribed, too. 5 of 5
10724	https://www.themathpage.com/aBookI/propI-20.htm	The sides of a triangle. Euclid I. 20. Table of Contents\|Introduction\|Home P l a n e G e o m e t r y An Adventure in Language and Logic based on THE SIDES AND ANGLES OF A TRIANGLE ================================== Book I. Proposition 20 ---------------------- Here, finally, is Proposition 20. It depends on Proposition 18. To prove, in triangle ABC, that sides BA, AC are together greater than side BC, on side AC we construct the isosceles triangle DAC. Since AC is equal to AD, then side BD -- which is BA, AD -- is equal to BA, AC. It is not difficult to see that angle DCB is greater than angle D, so that the opposite side BD is greater than opposite side BC. But BD is equal to BA, AC. Therefore BA, AC are greater than BC. PROPOSITION20. THEOREM --------------------------------------------------------------------------- Any two sides of a triangle are together greater than the third side. Let ABC be a triangle; then any two sides of it taken together are greater than the remaining side; namely, BA, AC are greater than BC; AB, BC are greater than AC; BC, CA are greater than AB. Extend BA to D, making AD equal to AC, and draw DC. Then, since CA is equal to AD,(Construction) angle ADC is equal to angle ACD.(I. 5) Now angle BCD is greater than angle ACD;(Axiom 5) therefore angle BCD is greater than the equal angle ADC. In triangle DCB, then, angle DCB is greater than angle BDC; and the greater angle is opposite the greater side;(I. 19) therefore, side BD is greater than side BC. But BD is equal to BA, AD together, and AD is equal to AC; therefore BD is equal to BA, AC together; therefore BA, AC are greater than BC. Similarly, we can prove that AB, BC are greater than CA, and BC,CA greater than AB. Therefore, any two sides etc. Q.E.D. In other words: A straight line is the shortest distance between two points! If anyone wanted to ridicule mathematics for its insistence on the axiomatic method of orderly proof, then this theorem offers a wide target. In fact, the Epicureans (those Athenian free-thinkers, who defined philosophy as the art of making life happy), did exactly that. They said that this theorem required no proof and was known even to an ass. For if hay was placed at one vertex and an ass at another, they argued, the poor dumb animal would not travel two sides of the triangle to get his food, but only the one side that separated them. What scorn the true philosopher must bear! And what can the mathematician do but point out, patiently, that mathematics, as a logical science, relies on deduction from first principles. Those principles moreover are to be as few in number as possible -- whatever can be proved should be. That is the intellectual sport. Please "turn" the page and do some Problems. or Continue on to the next proposition. Previous proposition Table of Contents\|Introduction\|Home Please make a donation to keep TheMathPage online. Even $1 will help. Copyright © 2021 Lawrence Spector Questions or comments? E-mail:teacher@themathpage.com
10725	https://www.wolframalpha.com/input/?i=cross+product+calculator	cross product calculator - Wolfram\|Alpha UPGRADE TO PRO APPS TOUR Sign in Step-by-Step Solutions with Pro Get a step ahead with your homework Go Pro Now cross product calculator Natural Language Math Input Extended KeyboardExamplesUpload Random Computational Inputs: » vector 1: » vector 2: Compute Input interpretation Result Step-by-step solution Vector plot Vector length Approximate form Step-by-step solution Normalized vector Approximate form Step-by-step solution Spherical coordinates Approximate form Corresponding line segment Download Page POWERED BY THE WOLFRAM LANGUAGE
10726	https://royalsocietypublishing.org/doi/10.1098/rsta.2019.0300	Skip main navigationJournal menuOpen Drawer MenuClose Drawer Menu You have accessDiscussion Recent attempts to change the periodic table Eric Scerri Eric Scerri Department of Chemistry & Biochemistry, University of California Los Angeles, Los Angeles, CA 90095, USA scerri@chem.ucla.edu Google Scholar Find this author on PubMed Search for more papers by this author Eric Scerri Eric Scerri Department of Chemistry & Biochemistry, University of California Los Angeles, Los Angeles, CA 90095, USA scerri@chem.ucla.edu Google Scholar Find this author on PubMed Search for more papers by this author Published:17 August 2020 Abstract The article concerns various proposals that have been made with the aim of improving the currently standard 18-column periodic table. We begin with a review of 8-, 18- and 32-column formats of the periodic table. This is followed by an examination of a possible, although rather impractical, 50-column table and how it could be used to consider the changes to the periodic table that have been predicted by Pyykkö in the domain of superheavy elements. Other topics reviewed include attempts to derive the Madelung rule as well as an analysis of what this rule actually provides. Finally, the notion of an ‘optimal’ periodic table is discussed in the context of recent work by philosophers of science who have examined the nature of classifications in general, as well as the notion of natural kinds. The article takes an unapologetically philosophical approach rather than focusing on specific data concerning the elements. Nevertheless, some pragmatic issues and educational aspects of the periodic table are also examined. This article is part of the theme issue ‘Mendeleev and the periodic table’. 1. Introduction The International Year of the Periodic Table has been a cause for much celebration and re-examination of Mendeleev's outstanding discovery of the mature periodic system (figure 1). As many authors have pointed out, Mendeleev was by no means the first scientist to publish a periodic table but there is nevertheless widespread agreement that his work was the most influential in bringing about its acceptance . The present author believes that there has not been sufficient attention given to the current debates concerning possible changes to the periodic table and attempts to bring about some improvements to it. It is for this reason that this article has been undertaken. The focus of the present paper will be on a philosophical approach rather than a pragmatic one. We will be arguing for the philosophical importance of the periodic system and the periodic law which the periodic table attempts to represent graphically. It is our belief that the depth and importance of the periodic system requires that we consider seriously some representations that treat all the elements in the same way, rather than casting some of the elements out of the main body of the table, as is the habit in the currently popular form of the periodic table which separates out the f-block elements. We do not claim that there are any deficiencies as such in the current periodic table. Taking a more philosophical perspective can serve to address some of the more specific issues that have been raised in the recent literature. For example, whether or not one favours the use of a 32-column representation of the elements, the exercise of moving to this more extended format can serve to illuminate certain issues that exist within the familiar 18-column format. We can then return to the 18-column format, if necessary, but with an expanded view of how all the elements fit together. 2. Historical background We begin with a brief historical overview of the development of the periodic table, starting with the pioneer tables of the 1860s. Broadly speaking the earliest tables represented chemical periodicity in the simplest possible way through the use of an eight-column representation of the elements. Moreover, Mendeleev's first table of 1869 featured rows and columns reversed as compared with the format that he adopted just two years later in his table of 1871 and most subsequent published tables (figure 1; ). The choice of eight columns involved something of a coincidence, since the original tables featured eight groups even though the noble gases had not yet been discovered . The explanation lies in authors such as Mendeleev featuring an eighth column that contained a miscellany of transition metals, in the original sense of the term. These elements such as iron, cobalt and nickel represented a ‘transition’ after which normal periodicity was resumed. Once the noble gases were discovered they were assigned to a ninth group which Mendeleev designated as group zero (figure 2) . There are some distinct advantages to these short-form tables, most notably the fact that the group number coincides with the highest oxidation number. As time progressed, the short-form was gradually abandoned in favour of an 18-column representation . The latter representation is usually attributed to Deming although his original version (figure 3) was rather different from the one that is so commonplace today (figure 4)1. Clearly there are advantages to separating the transition metals, in the modern sense of the term, such as the sequence from scandium to zinc from the remaining elements in the fourth period. For example, although carbon, silicon, titanium, germanium, zirconium, tin and lead in the eight-column table all show a maximum oxidation state of four, there is enough similarity between carbon, silicon, germanium, tin and lead on one hand and titanium, zirconium and hafnium on the other hand, to warrant their being separated in the way that they are into groups 14 and 4, respectively, in the modern 18-column table. 3. The role of quantum mechanics In addition to such chemical considerations, the 1920s and 1930s saw the increasing role for quantum mechanics in providing an explanation for the periodic table. This process began when Bohr assigned particular numbers of electrons to each electron shell, thus using just one quantum number . Soon afterwards a second and third quantum number were introduced to give more detailed electronic structures , a process that culminated with Pauli proposing the need for a fourth quantum number. Considering the possible values of these various quantum numbers, as well as the relationships established among them, made it possible to predict the fact that the first few shells contain, 2, 8 and 18 electrons, respectively. This understanding lent further support for the 18-column table in which the s-block (2 elements wide) was followed by the d-block (10 elements wide), followed by the p-block (6 elements wide) to make a total of 18 columns. The question of how to label the groups in the periodic table was decided by IUPAC amid much controversy in 1990, although this episode will not be reviewed here . 4. The 32-column table Even more recently some authors have taken the logical step of extending the periodic table even further to feature 32 groups (figure 5; ). This version has the possible advantage over the 18-column representation in that the f-block is incorporated into the main body of the table instead of being shown as a kind of footnote, or afterthought, to the main 18-column table. Generations of chemistry students who have looked closely at periodic table wall-charts, and textbook versions, have noticed the rather anomalous discontinuity in atomic number that occurs on moving from lanthanum (Z = 57) to hafnium (Z = 72) and again from actinium (Z = 89) to rutherfordium (Z = 104). The 32-column table solves this problem in a natural and elegant manner by restoring a continuous sequence of increasing atomic numbers in both the periods concerned. In addition, all the elements are placed on an equal footing into a fully unified periodic table. As we will see in a later section, the 32-column format can also cast light on the question of the membership of group 3 of the periodic table. After any new insights are gained, one can well return to the 18-column format with deepened knowledge. The purpose of the present article is not to advocate any particular format but merely to explore the possibilities that the different formats can offer. 5. Do pragmatic factors matter? At this point, we should perhaps consider pragmatic factors and the extent to which they should be taken into consideration when trying to improve the periodic table. Of course the use of a 32-column representation requires that the spaces for each element need to be about half the size that they are in an 18-column table. Alternatively, a textbook or wall-chart periodic table would need to be about twice the width of present tables, if one were to retain the original font size for each element symbol. Some authors seem to regard this issue as a major argument against the move to a 32-column representation.2 However, in raising the question of how to modify the periodic table, my intention here is to move beyond such pragmatic factors and to consider the proposed changes in the most general and abstract sense possible. Just like the scientific approach in general consists in seeking the most general level of description of any phenomenon, so too should the quest to improve the periodic table of the elements, even if the preferred choice may prove to be inconvenient for typographers and the like. Needless to say, there is no strong or categorical argument based on data that might lead one to favour a 32-column table. In addition, some chemists might argue that the presently popular 18-column format highlights the similarity between the elements in the d- and f-blocks of the table.3 As mentioned in the introduction, the depth and significance of the periodic system and the manner that it unifies all the elements might suggest that these features necessitate a unified treatment of the elements. A pragmatist could insist that the purpose of the periodic table lies in teaching and explaining chemical phenomena and that the current 18-column format serves those purposes well. In the present article, we prefer to take a more philosophical approach. The discovery of the periodic system represents one of the major achievements in all of modern science. The periodic table is of course an attempt to provide a visual representation of the periodic system, namely the fact that elements recur in an approximate fashion after certain intervals. From this holistic or global perspective, it makes sense to want to emphasize the unity of the periodic system and to treat all of the elements on an equal footing, rather than retaining the separation of the f-block elements that is a historical hangover from the time when it was not clear how these elements should be regarded (Mendeleev and the other pioneers of the periodic table had great difficulty in accommodating the rare earth elements which form a large fraction of the f-block. Mendeleev is known to have given up this task to his colleague Brauner (see ). Moreover, the periodic system has served as a motivation for many developments in physics as well as chemistry. For example, it was highly influential upon the ideas of J.J. Thomson, Niels Bohr, Henry Moseley, Edmund Stoner and Wolfgang Pauli . So although the periodic table was discovered by chemists in the context of chemical education, and in wanting to explain chemical reactivity, its scope has now made it an object that belongs to the whole of science. There is no need to insist on the periodic table having a format that is suited mainly for the purposes of the chemical community and for chemical educators. The profound nature of the periodic system and its associated periodic table has stood the test of time for just over 150 years. These features taken together suggest to the present author that the periodic system and table are worth considering in the most philosophical and abstract terms without undue concern for pragmatic considerations of the various kinds that have been referred to above. 6. Relativistic effects In an article in a recent issue of Pure and Applied Chemistry, P. Pyykkö makes some passing criticisms of the 32-column table . Pyykkö's first criticism is that the 32-column table features too many versions of group 3 of the periodic table. To quote Pyykkö, Conversely, the 32-column ‘long-form’ PT favoured by Scerri (see his figure 1) violates (1) by having very many potentially Group-3 columns and may also violate (4). It does satisfy (3). (quoted from p. 5).4 However, this claim would seem to contradict the current IUPAC naming convention for groups and may be another example of attaching undue importance to pragmatism. As the IUPAC convention stands at present, the groups of elements in the f-block are not assigned any numbers. It is not clear why Pyykkö considers that all groups in the f-block should be considered as being in group 3 while lamenting the fact that the 32-column format produces too many such groups. However, we can remain within the IUPAC convention and say that group 3 consists of either Sc, Y, La and Ac or perhaps Sc, Y, Lu and Lr on moving from an 18- to a 32-column format [14–20].5 Pyykkö also worries that moving to a 32-column table, and possibly re-numbering as 1 to 32, would lead to a loss of correspondence between group numbers and the maximum oxidation states of the elements in question. In fact this is something that already happened, to a lesser extent, with the move from an 8-column to an 18-column format. In 8-column tables, the maximum oxidation state of an element corresponds directly to the group number label. However, in the 18-column format, this is only true of the s-block elements. In the p-block, one needs to subtract 10 from the group label, while the d-block brings still further complications. But even this criticism is merely pragmatic. Why should an improvement in the periodic table be held hostage to the demand that one should immediately be able to deduce the maximum oxidation state from the group label of any particular element? The periodic table has now become as much the property of physicists, geologists, astronomers and others as it is of its chemical originators. 7. A 50-column table? There is no reason why one should even stop at a 32-column table, although the need for any further expansion is hardly pressing. Nevertheless, if and when element 121 is synthesized, this will signal the start of the g-block elements, at least in principle, and perhaps a need for a 50-column table (figure 6).6 More specifically, it has been predicted that the first element that actually possesses a g-electron will be element 125 or thereabouts . Returning to Pyykkö's work, this author does not choose to use a 50-column representation even when discussing elements with atomic numbers up to 172. Nor does Pyykkö favour a 32-column table but remains with the traditional 18-column representation (figure 7). Pyykkö's 18-column periodic table , which has been featured in many articles, gives the appearance of differing quite substantially from our current periodic table and indeed the author appears to capitalize on this apparent difference. In fact, the only real novelty to emerge from Pyykkö's relativistic calculations, and his proposed accompanying periodic table, occurs at elements 139 and 140, at which point the increasing atomic number sequence appears interrupted because the j= 1/2 component of the 8p shell (8p1/2) is occupied before the 6f levels. The small modification that Pyykkö proposes is the result of his prediction, and that of other authors . However, if one displays these relativistic findings on a 50-column table (figure 8), one can show that the alleged deviations from an expected periodic table seem to be rather minimal. Now even if elements with atomic numbers as high as 139 and 140 were ever to materialize, one can still ask whether such unexpected orderings or violations of the Madelung rule should be reflected in any modified periodic table. After all, there are many violations of the Madelung rule such as the 20 well-known anomalous configurations beginning with chromium and copper which do not lead us to modify the periodic table [23,24]. In any case, it is interesting to see that Pyykkö admits that it is surprising, in view of the relativistic effects, that the Madelung rule survives so well all the way up to atomic number 172. We hasten to add that we are not recommending the adoption of a 50-column table but that thinking about the wider format helps us to consider the alleged violations more clearly. 8. Explaining the Madelung rule The Madelung rule is a mnemonic device or a rule of thumb, which is often used within the Aufbau scheme to arrive, in principle, at the electronic configuration of any atom one may choose. It is commonly displayed on a diagram of the kind shown in figure 9. If one begins at the top of the diagram and follows the diagonal arrows, one can obtain the order of orbital occupation. The status of this rule is another source of debate among authors who concern themselves with the periodic table of the elements. In 1969 in the year of the 100th anniversary of the discovery of Mendeleev's classic table, the theoretical chemist Löwdin pointed out that the Madelung rule had not yet been derived from quantum mechanics . This task has become known as the Löwdin challenge and numerous authors claim that they have met the challenge [26–30]. The generally held opinion, however, is that no such derivation has succeeded in the way that Löwdin intended. Quite independently of these attempts, another theoretical chemist, Eugen Schwarz, argued instead that the Madelung rule does not truly reflect the order of orbital occupation, in the way it is generally believed to do [31,32]. Schwarz's main concern was to resolve a long-standing conundrum that is familiar to every chemistry student and their instructors. It is well known that the atoms of potassium and calcium have somewhat unexpected configurations of [Ar]4s1 and [Ar]4s2, respectively. The unexpected feature in these configurations is that the 3d orbitals appear to have been skipped in the course of the occupation of all the previous orbitals, which are summarized by the notation of [Ar]. According to the traditional account, the 3d orbitals only begin to fill when the scandium atom is reached on progressing through the periodic table and only after the 4s orbital has been doubly occupied. If this is correct, it raises something of a conundrum because the first ionization of this element, and subsequent first transition elements, involves the removal of a 4s electron . Textbook authors and chemistry instructors typically advance ad hoc explanations in an attempt to rationalize this clearly illogical situation. If the 3d orbital electron is the last to enter the atom, it should be the one to be ionized instead of a 4s electron. What Schwarz has asserted is that in scandium and following transition metal atoms, the 3d orbitals are in fact occupied before, or in preference to, the 4s orbital and there is ample experimental evidence for this claim . The advantage of Schwarz's version of the facts is that it restores logic to the situation. The 4s electrons are the last to enter the atom in the sense of the build-up of the configuration and consequently are the least stable, and so the first to be ionized.7 Nevertheless, Schwarz goes further in claiming that since the Madelung rule fails to predict the correct order of orbital occupation starting at scandium, it should be abandoned. The need to derive the Madelung rule, as called for by Löwdin, therefore becomes completely redundant in Schwarz's view. While agreeing with Schwarz that Madelung's rule fails to provide the strict order of occupation of orbitals in the aufbau scheme, we do not agree with him in downplaying the rule or even abandoning its use altogether. This is because the Madelung rule remains as a reliable guide as to the ‘differentiating electron’ as one moves through the periodic table from one element's atom to the next.8 In keeping with the desire to consider the periodic table as a whole rather than focusing on individual elements, and how their orbitals are occupied, the differentiating electron is an important feature. Consequently, there still seems to be a need to derive this rule from first principles as first suggested by Löwdin, even if the significance of the rule appears to have changed somewhat. 9. Group 3 Another topic which has attracted a good deal of attention and which concerns possible improvements to the periodic table in a very direct way is the current debate over the composition of group 39. Whereas the majority of textbook and other periodic tables in the 18-column format show the elements of this group as scandium, yttrium, lanthanum and actinium, a significant number of more recent tables feature the last two elements as lutetium and lawrencium instead. If one considers this question within the context of a 32-column representation, there are in fact four options. These options are shown in figures 10–13 inclusive. In the tables shown in figures 10 and 11, group 3 is shown as containing either Sc, Y, La and Ac or Sc, Y, Lu and Lr. The second option is clearly preferable since it is the only one of these two tables that maintains an increasing order of atomic number across every single-period length. In fact the table shown in figure 11 has not appeared in print, as far as this author is aware, presumably because of the anomalies that it would imply regarding increasing atomic number. This is another case for which thinking about a 32-column table if not actually adopting the longer format can be useful. Part of the reason why group 3 has traditionally been regarded as consisting of Sc, Y, La and Ac is due to the predominance, not to say virtual monopoly, that the 18-column table has formerly held. In such a shorter format, the problem of maintaining a strictly increasing ordering in atomic number remains hidden. However, there are two other representations that concern group 3 as shown in figures 12 and 13, that are frequently published. In the table shown in figure 12, the elements in group 3 are the more traditional option consisting of Sc, Y, La and Ac. To maintain an increasing order of atomic number increase the d-block is split into two highly uneven portions consisting of one column for group 3 followed by the remainder of the d-block, consisting of 9 columns, that are widely separated from group 3 by the intervening f-block. In spite of the highly uneven shape that this produces and the fact that such splitting of blocks does not occur in other blocks in the periodic table, a surprising number of such tables have appeared10. For example, the version of the periodic table shown in figure 13 was featured as the logo of the Inorganic Chemistry Division of the American Chemical Society until recently when it was withdrawn specifically because of the controversies concerning group 3. The fourth possible presentation is one that abstains from specifying which elements belong in group 3 or rather it is a presentation that only contains two elements, scandium and yttrium, as being definitely in group 3. This table is sometimes labelled as the official IUPAC periodic table (figure 13)11 although a leading IUPAC official has clearly stated that this organization does not currently support any particular version of the periodic table and that has only concerned itself thus far with numbering 18 of the groups of the table . Attempts to resolve this conundrum have focused on chemical and physical properties and also on microscopic properties such as the electronic configurations of the atoms concerned. It would appear that none of these criteria provide a clear-cut resolution of the question. Moreover, it becomes increasingly clear that there may not be any such thing as one optimal table in a purely objective sense. The question seems to depend on what criteria are considered and, most importantly perhaps, on whether one favours chemical or physical criteria or general didactic considerations . 10. Natural kinds and the philosophy of classification There is a well-developed literature in the philosophy of science that concerns itself with classification and with the question of natural kinds, that is to say sets of objects which are related to each other through what might be said to be purely objective properties. For example, until recently species were believed to be natural kinds in biology. For an animal to be classified as a tiger, for example, would require the specification of the genetic characteristics of tigers. Natural kinds are distinguished from so-called artificial kinds for which their classification depends rather on human choices and not on an independently existing reality. The typical example of an artificial system of classification is the classification of library books. No one library system, be it the Library of Congress system or the Dewey-Decimal system, can be said to be more correct than its competitors. The way in which these systems classify books involves arbitrary, or non-objective, criteria as to how to demarcate books on chemistry from those on biology for example. In the 1970s and 80s, a theory was developed by philosophers Kripke and Putnam who sought to define natural kinds through their essences or their objective properties [37,38]. A favourite example of a natural kind in this literature has been that of a chemical element which according to Kripke and Putnam can be specified by stipulating its atomic number. Needless to say, Kripke and Putnam were not the first to propose such an identification. That distinction belongs to Van den Broek and Moseley . What Kripke and Putnam did was to fully adopt the scientific definition of element-hood in order to identify elements as natural kinds. For an atom of an element to be gold, for example, it requires that the atom should have an atomic number of 79. In addition, if an atom is found to have atomic number 79 this uniquely identifies it as being an atom of gold. Said in the jargon of philosophers, the possession of an atomic number of 79 is both necessary and sufficient for the identification of a particular atom as being one of gold. The Kripke-Putnam approach to natural kinds has come under various forms of criticism in the years since it was first proposed. In the case of biological species, it has been pointed out that evolution spoils any such hope, since the very essence of what it is to be any particular species is bound to change as time evolves . However, this objection could not be raised as far as elements are concerned, since broadly speaking atoms of any particular element do not evolve into other atoms over time, apart from those that decay radioactively. A more general objection to the Kripke-Putnam approach to thinking of natural kinds has been the realization that, whatever kinds are being considered, there is always a certain degree of interest dependence that enters the stipulation of sets of entities be they tigers, galaxies or elements. This realization has been called the epistemological turn within the natural kinds of literature . It means that natural kinds are no longer regarded as a purely ontological affair which depend just on an independent external reality. Instead, the new approach recognizes that epistemological considerations, having to do with our knowledge of the world. Simply put, how we conceive of the world should also be taken into account when discussing the classification of scientific entities. Having reviewed these developments in the philosophy of classification, and the way that natural kinds are conceived, we can return to the question of improving the periodic table. 11. Back to the periodic table If we are to believe what the experts on the philosophy of classification and natural kinds have to tell us we should renounce the notion that we will ever arrive at a truly optimal periodic table, since any system of classification must inevitably remain interest dependent. Of course this state of affairs should not negate the attempts to resolve such questions as the membership of group 3 of the periodic table, but it should remind us of the fact that any resolution must concede a certain degree of conventionality, or choice, on the part of the scientific community. For example, another currently debated topic concerning the classification of the elements has been the debate concerning the relative virtues of Janet's left-step periodic table . This representation is arrived at by making two simple changes to the 32-column table shown in figure 5. The first change requires that helium be moved from its traditional position, at the top of group 18, to the top of group 2. The second change is to relocate the now modified s-block of the table to the right-hand edge, to the right of the p-block (figure 14). This table is more consistent from a physical perspective since atoms of helium contain two electrons and is therefore analogous to the atoms of the traditional group 2 of the periodic table, all of which possess two outer electrons. Another attractive feature of the left-step table is that it restores regularity and perhaps even balance to the otherwise awkwardly shaped traditional periodic table representation. More significantly than such aesthetic considerations, this table provides greater regularity in depicting every single-period length as repeating once as, 2,2,8,8,18,18,32,32. Meanwhile, the traditional table features an anomalous first period that, unlike all subsequent ones, does not repeat in length to give a sequence of 2,8,8,18,18,32,32. On chemical grounds the left-step periodic table may seem to be inconsistent and unnecessary since it suggests a chemical kinship between helium and the traditional alkaline earth elements. Although this objection appears to have been partly challenged by the recent observation of a compound of helium with the formula of Na2He, I suspect that the majority of chemists would still not be willing to release helium from its status as a noble gas . This is because the observation of this exotic compound of helium has been carried out at the extremely high pressure of 13 gigapascals. The mention of these high-pressure experiments raises another aspect of the periodic table. The researchers who conduct such experiments have indicated that there should perhaps be an alternative periodic table that applies to high-pressure conditions . Moreover, the findings in relativistic quantum mechanics may also begin to show the need for an alternative periodic table for elements with high atomic numbers in which relativity brings about significant changes in the properties of certain elements. 12. Back to group 3 and conclusion The realization that classification in general, and the specification of natural kinds, is not a purely ontological question forces us to confront the fact that seeking an objectively optimal periodic is rather futile. We should accept that a degree of convention must be used in selecting a periodic table that can be presented as perhaps the best possible table that combines objective factors as well as interest dependence. Given these points I would like to propose that although there is no objective way to distinguish the tables shown in figures 10 and 12 one should nevertheless opt for the former of these as the official IUPAC periodic table. The factors that might lead us to favouring figure 10 have already been mentioned. They are that the d-block of the periodic table remains unified and intact, whereas the table in figure 12 requires that the d-block be separated into two highly uneven portions consisting of one and nine groups, respectively. I also believe that the table depicted in figure 13 should be rejected because it features one block whose width does not correspond directly to the underlying quantum mechanical explanation of the periodic table. The number of s, p, d and f orbitals that can occur in each shell is rigorously derived from quantum mechanics as being 2,6,10 and 14. The depiction of a 15-element wide f-block conflicts with this underlying explanation. We should not allow a dispute over the membership of group 3 to cause us to admit that just one of the main blocks of the table, the f-block in this case, should have an anomalous width. This is especially true from the perspective of a general quantum mechanical understanding of the periodic table. Needless to say, the characterization of these blocks of the periodic table is only approximate, just as the assignment of electronic configurations to atoms represents an approximation. Moreover, one may readily concede that an element such as thorium does not actually possess any f-orbital electrons and yet it is classified as being among the f-block elements even in all four of the periodic table representations shown in figures 10 to 13. The periodic table favoured by Pyykkö is another example of a table that features a 15-element f-block. However, such a table is the result of specific calculations on selected heavy atoms and should perhaps not be allowed to dictate the format of the periodic table that serves the greatest number of periodic table users including students, instructors and practising chemists alike. A student looking at figure 13 is bound to wonder whether there is some scientific reason for making the f-block have a width of 15 elements. Neither such students, nor their instructors, would probably realize that the table in question has been designed by practitioners of a highly specialized branch of quantum mechanics concerned with the properties of super-heavy elements. Such interest dependence should not, in my view, dictate how the periodic table is presented to the general chemical and scientific community. I therefore propose that IUPAC should reach a compromise on the table shown in figure 10 since it achieves four desiderata when presented in the 32-column format. First, it unifies all the elements together by including the f-block. Second, it displays all the elements in order of increasing atomic number. Third, it avoids splitting the d-block into two highly uneven portions and fourth, it depicts all the blocks of the periodic table in accordance with the underlying quantum mechanical account of the periodic table which calls for 2,6,10 and 14 electrons to occur in various electron sub-shells. If one insists on retaining an 18-column format, we recommend that it should be the version shown in figure 4 which benefits from the insight afforded by the equivalent 32-column table, namely figure 10. Finally, historical developments have shown that quantum mechanics provides an explanation for the periodic table and to a first approximation suggests the presence of 1, 3, 5, 7 etc. orbitals of the type s, p, d and f, respectively. In the opinion of the present author, any improved periodic table should embody such findings rather than implying the presence of a set of 15 f electrons. Needless to say, more advanced practitioners, who specialize in quantum chemistry, would still be free to conduct relativistic calculations that might indeed uphold grouping together 15 rather than 14 elements. However, any such findings should not be imposed on the majority of users of the periodic table and should not dictate how the periodic table is presented to the widest possible audience of users. Data accessibility This article has no additional data. Competing interests I declare I have no competing interests. Funding I received no funding for this study. Footnotes 1 This table and all other unattributed ones in this article were prepared by the author. 2 As a reviewer of this article suggested the problem could also be avoided by reverting to the original format that Mendeleev used in his tables of 1869 in which periods were displayed vertically and groups horizontally. 3 I thank one of the reviewers of this article for pointing this out to me. 4 According to Pyykkö, some desirable properties of a PT could be: 1. The column (or ‘Group’) number g (or g-10) is the maximum number of valence electrons. The oxidation state is then the same number or, counting holes in the spirit of Abegg's contravalence, the negative number, g-18. 2. The valence atomic orbitals along a row are constant in a block and equal to those in the right-hand margin. 3. The nuclear charge, Z, increases systematically towards right. 4. The shape should be typographically convenient. 5 The Constitution of Group 3 of the Periodic Table, IUPAC project 2015-039-2-200, 6 The figure of 121 cited above represents an idealization based on a strict adherence to the Madelung rule. 7 The energy of an orbital cannot be specified independently of the occupancy of the orbital due to e-e repulsion. So in a Ti atom with configuration 3d24s2, 4s lies above 3d. However, in the configuration 3d34s1, 3d lies above 4s due to the extra 3d-3d repulsion––in both cases removal of the outermost electron gives us Ti+ with a ground state configuration 3d24s1. Furthermore, the ordering of orbitals does change as we move across the periodic table––so for example in S, the lowest totally unoccupied orbital is 3d, but by the time we get to Na and Ca, 4s is lower than 3d. 8 There are a number of exceptions to trying to deduce the differentiating electron from the Madelung rule, however. For example, although chromium possesses a 3d electron which differentiates its atoms from those of vanadium, manganese is differentiated from chromium through a 4s electron. 9 The author is currently chairing a working group that is tasked with making recommendations to IUPAC regarding group 3 of the periodic table. 10 One minor exception to this statement is the fact that the s-block is split just in the case of the first two members, hydrogen and helium, at least in the way that the periodic table is generally depicted with helium in group 18. 11 One contribution of 11 to a theme issue ‘Mendeleev and the periodic table’. © 2020 The Author(s) Published by the Royal Society. All rights reserved. Previous ArticleNext Article VIEW FULL TEXTDOWNLOAD PDF Figures Related References Details Cited by Korshunova T and Martynov A (2024) The Phyloperiodic Approach Removes the “Cryptic Species” and Puts forward Multilevel Organismal Diversity, Diversity, 10.3390/d16040220, 16:4, (220) (2023) Introduction to Heavy Metals and Their Toxicity Global Industrial Impacts of Heavy Metal Pollution in Sub-Saharan Africa, 10.4018/978-1-6684-7116-6.ch001, (1-38) Dubova M and Goldstone R (2023) Carving joints into nature: reengineering scientific concepts in light of concept-laden evidence, Trends in Cognitive Sciences, 10.1016/j.tics.2023.04.006, 27:7, (656-670), Online publication date: 1-Jul-2023. Masters A, Greenfield P, Davison C, Evans J, Motion A, Barrett J, Troy J, Constantine K and Jackson Pulver L (2023) Elements of Country: a First Nations-first approach to chemistry, Chemistry Teacher International, 10.1515/cti-2022-0055, 5:1, (29-38), Online publication date: 1-May-2023., Online publication date: 1-Mar-2023. Schwarz W, Müller U and Kraus F (2022) The Good Reasons for a Standard Periodic Table of the Chemical Elements, Zeitschrift für anorganische und allgemeine Chemie, 10.1002/zaac.202200008, 648:13, Online publication date: 14-Jul-2022. Imyanitov N (2022) Periodic tables for cations + 1, + 2, + 3 and anions − 1. Quantitative characteristics for manifestations of internal periodicity and kainosymmetry, Foundations of Chemistry, 10.1007/s10698-022-09421-2, 24:2, (189-219), Online publication date: 1-Jul-2022. Sato T and Nagame Y (2022) Chemistry of the elements at the end of the actinide series using their low-energy ion-beams, Radiochimica Acta, 10.1515/ract-2022-0001, 110:6-9, (441-451), Online publication date: 27-Jun-2022., Online publication date: 1-Jun-2022. da Silva S (2022) The habit of the pipe: a layperson’s view of the periodic table, Foundations of Chemistry, 10.1007/s10698-022-09422-1, 24:1, (113-120), Online publication date: 1-Apr-2022. Talpsepp E (2022) Interview with Eric Scerri, Foundations of Chemistry, 10.1007/s10698-022-09418-x, 24:1, (143-153), Online publication date: 1-Apr-2022. Mayo S (2022) Plant taxonomic species and their role in the workflow of integrative species delimitation, Kew Bulletin, 10.1007/s12225-022-10002-x, 77:1, (1-26), Online publication date: 1-Mar-2022. Kwarsick J, Pore J, Gates J, Gregorich K, Gibson J, Jian J, Pang G and Shuh D (2021) Assessment of the Second-Ionization Potential of Lawrencium: Investigating the End of the Actinide Series with a One-Atom-at-a-Time Gas-Phase Ion Chemistry Technique, The Journal of Physical Chemistry A, 10.1021/acs.jpca.1c01961, 125:31, (6818-6828), Online publication date: 12-Aug-2021. Scerri E (2020) Geoff Rayner-Canham: The periodic table: past present, and future, Foundations of Chemistry, 10.1007/s10698-020-09389-x, 23:2, (293-295), Online publication date: 1-Jul-2021. Vernon R (2020) The location and composition of Group 3 of the periodic table, Foundations of Chemistry, 10.1007/s10698-020-09384-2, 23:2, (155-197), Online publication date: 1-Jul-2021. Desmet P, Sauter D and Shiota M (2021) Apples and oranges: three criteria for positive emotion typologies, Current Opinion in Behavioral Sciences, 10.1016/j.cobeha.2021.03.012, 39, (119-124), Online publication date: 1-Jun-2021. Cao C, Vernon R, Schwarz W and Li J (2021) Understanding Periodic and Non-periodic Chemistry in Periodic Tables, Frontiers in Chemistry, 10.3389/fchem.2020.00813, 8 Edwards P, Egdell R, Fenske D and Yao B (2020) The periodic law of the chemical elements: ‘The new system of atomic weights which renders evident the analogies which exist between bodies' , Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences, 378:2180, Online publication date: 18-Sep-2020. This Issue 18 September 2020 Volume 378Issue 2180 Theme issue 'Mendeleev and the periodic table' compiled and edited by Peter Edwards, Russell Egdell and Dieter Fenske Article Information DOI: PubMed:32811365 Published by:Royal Society Print ISSN:1364-503X Online ISSN:1471-2962 History: Manuscript accepted11/02/2020 Published online17/08/2020 Published in print18/09/2020 License: © 2020 The Author(s) Published by the Royal Society. All rights reserved. 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Keywords periodic table Mendeleev group 3 of periodic table history of chemistry chemical education Subjects chemical physics inorganic chemistry physical chemistry Close Figure Viewer Browse All FiguresReturn to Figure Previous FigureNext Figure Caption back Posted by 1 X users Referenced in 8 Wikipedia pages 18 readers on Mendeley See more details
10727	https://math.stackexchange.com/questions/427260/find-an-orthogonal-vector-under-the-constraints-described	linear algebra - Find an orthogonal vector under the constraints described - 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 Find an orthogonal vector under the constraints described Ask Question Asked 12 years, 3 months ago Modified12 years, 2 months ago Viewed 596 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Find the vector v 2 v 2 so that v=v 1+v 2 v=v 1+v 2, where v 1 v 1 is in Span{u 1,u 2}Span⁡{u 1,u 2} and v 2 v 2 is orthogonal to Span{u 1,u 2}Span⁡{u 1,u 2}. Note that u 1 u 1 and u 2 u 2 are orthogonal to each other. I had tried just orthogonally projecting v v onto the span of u 1 u 1 and u 2 u 2, but that didn't work. Given u 1=⎡⎣⎢⎢⎢2 2−1 2⎤⎦⎥⎥⎥,u 2=⎡⎣⎢⎢⎢2 1 2−2⎤⎦⎥⎥⎥,v=⎡⎣⎢⎢⎢3−1−3 3⎤⎦⎥⎥⎥,u 1=[2 2−1 2],u 2=[2 1 2−2],v=[3−1−3 3], find the vector v 2 v 2so that v=v 1+v 2 v=v 1+v 2, where v 1 v 1is in Span{u 1,u 2}Span⁡{u 1,u 2}and v 2 v 2is orthogonal to Span{u 1,u 2}Span⁡{u 1,u 2}. Note that u 1 u 1and u 2 u 2are orthogonal to each other. v 2=⟨2−7 13⋅2,2−7 13,−1−7 13⋅2,2+2⋅7 13⟩v 2=⟨2−7 13⋅2,2−7 13,−1−7 13⋅2,2+2⋅7 13⟩ Syntax Vectors: For [1,2,3][1,2,3] or ⎡⎣⎢1 2 3⎤⎦⎥[1 2 3], input ⟨1,2,3⟩⟨1,2,3⟩. linear-algebra matrices vector-spaces Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Jul 7, 2013 at 13:22 Peter Mortensen 674 5 5 silver badges 11 11 bronze badges asked Jun 23, 2013 at 1:01 ahnahn 91 1 1 silver badge 10 10 bronze badges 4 So you projected v v onto the subspace spanned by u 1,u 2 u 1,u 2. that should give you v 1 v 1. You should then find v 2=v−v 1 v 2=v−v 1. Is that what you did?Muphrid –Muphrid 2013-06-23 01:58:57 +00:00 Commented Jun 23, 2013 at 1:58 Thank you so much !! But could you explain the reasoning behind this? Why isn't the orthogonal projection of v onto the subspace spanned by u1 & u2 v2, but instead v1?ahn –ahn 2013-06-23 02:01:08 +00:00 Commented Jun 23, 2013 at 2:01 It could be "orthogonal projection" means something else in your course/text, but to me, it says, "I have a vector that does not necessarily lie in this plane; when I project the vector, I am finding the part of it that is in the plane."Muphrid –Muphrid 2013-06-23 02:02:28 +00:00 Commented Jun 23, 2013 at 2:02 Ah I think I misunderstood the meaning of 'orthogonal projection', your explanation makes a lot of sense, thanks again!ahn –ahn 2013-06-23 02:11:03 +00:00 Commented Jun 23, 2013 at 2:11 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. It sounds like you've already been led basically to the answer. Here I outline an alternative method to directly find v 2 v 2, using Clifford algebra. In Clifford algebra, we can directly represent the subspace spanned by u 1,u 2 u 1,u 2 using what's called a geometric product. The geometric product has the following properties: e 1 e 1=e 2 e 2=e 3 e 3=e 4 e 4=1,e 1 e 2=−e 2 e 1,e 1 e 3=−e 3 e 1,…e 1 e 1=e 2 e 2=e 3 e 3=e 4 e 4=1,e 1 e 2=−e 2 e 1,e 1 e 3=−e 3 e 1,… Along with being associative. The geometric product is a very, very useful product. We can directly represent the subspace using the geometric product. B=⟨u 1 u 2⟩2 B=⟨u 1 u 2⟩2 This angle bracket notation means "keep only terms that have two orthogonal basis vectors in them". That is, throw away any scalar terms. This is exactly what I'll do. B=⟨(2 e 1+2 e 2−e 3+2 e 4)(2 e 1+e 2+2 e 3−2 e 4)⟩2=−2 e 1 e 2+6 e 1 e 3−8 e 1 e 4+5 e 2 e 3−6 e 2 e 4−2 e 3 e 4 B=⟨(2 e 1+2 e 2−e 3+2 e 4)(2 e 1+e 2+2 e 3−2 e 4)⟩2=−2 e 1 e 2+6 e 1 e 3−8 e 1 e 4+5 e 2 e 3−6 e 2 e 4−2 e 3 e 4 Again, this is a direct representation of the planar subspace spanned by the two vectors. Vitally, we need its magnitude. Again, this can be found using a geometric product: B 2=−4−36−64−25−36−4=−169 B 2=−4−36−64−25−36−4=−169 Yes, this is a negative quantity; this is a common property of Euclidean bivectors. They actually naturally square to negative numbers. Clifford algebra naturally provides us with objects that "act like" imaginary units, in the form of Euclidean bivectors. We can directly find v 2 v 2 by finding the "rejection" of v v in B B. The rejection is so named to be the "opposite" of the projection. The basic idea is as follows: form a parallelepiped with v v and B B, and then find the part of this volume orthogonal to B B. The formula is v 2=⟨v B⟩3 B−1 v 2=⟨v B⟩3 B−1 where B−1=B/B 2 B−1=B/B 2. The final result is guaranteed to be a vector. This is, however, usually denoted (v∧B)B−1(v∧B)B−1, suggesting a relationship with exterior algebra. I will not explore this here. The geometric product v B v B has as its grade-3 components ⟨v B⟩3=27 e 1 e 2 e 3−32 e 1 e 2 e 4−12 e 1 e 3 e 4−e 2 e 3 e 4⟨v B⟩3=27 e 1 e 2 e 3−32 e 1 e 2 e 4−12 e 1 e 3 e 4−e 2 e 3 e 4 All that remains is to compute the final product. I will not do this here (as that would solve the entire problem). But I hope that this shows you some of the power of Clifford algebra, especially in the way it can directly represent subspaces as objects in their own right, as opposed to mere spans of vectors. Clifford algebra makes possible all these operations between general subspaces beyond vectors. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 23, 2013 at 2:35 MuphridMuphrid 20.7k 1 1 gold badge 34 34 silver badges 66 66 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Given u 1=[2,2,−1,2]T u 1=[2,2,−1,2]T, u 2=[2,1,2,−2]u 2=[2,1,2,−2] and v=[3,−1,−3,3]v=[3,−1,−3,3]. We need to find v=v 1+v 2 v=v 1+v 2 where v 1∈span{u 1,u 2}v 1∈span⁡{u 1,u 2} and v 2 v 2 is orthogonal to this subspace. Since v 2 v 2 is orthogonal to the subspace, it must be orthogonal to v 1 v 1, i.e. v 1⋅v 2=0 v 1⋅v 2=0. This means that v 1=proj U(v)v 1=proj U⁡(v), where U=span{u 1,u 2}U=span⁡{u 1,u 2}. Now, note that u 1⋅u 2=0 u 1⋅u 2=0, so an orthonormal basis for U U is {u^1,u^2}{u^1,u^2}, where u^1=u 1/13−−√u^1=u 1/13 and u^2=u 2/13−−√u^2=u 2/13. Then v 1=⟨v,u^1⟩u^1+⟨v,u^2⟩u^2 v 1=⟨v,u^1⟩u^1+⟨v,u^2⟩u^2, where ⟨⋅,⋅⟩⟨⋅,⋅⟩ denotes the usual inner product. This gives v 1=13−−√u^1−7 13−−√u^2=[12 13,19 13,−27 13,40 13]T.v 1=13 u^1−7 13 u^2=[12 13,19 13,−27 13,40 13]T. Then, v 2=v−v 1=[27 13,−32 13,−12 13,−1 13]T v 2=v−v 1=[27 13,−32 13,−12 13,−1 13]T. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jul 1, 2013 at 0:04 DarylDaryl 5,707 3 3 gold badges 30 30 silver badges 42 42 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 linear-algebra matrices vector-spaces See similar questions with these tags. 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10728	http://staff.ustc.edu.cn/~rui/ppt/math-analysis/chap3_5.html	5. 函数的单调性与凸性单变量函数的微分学 \| \| \| \| 函数的单调性与凸性函数的单调性与极值定理 1. 在可导，则若，则在上(严格)单调增若，则在上严格单调减定理 2. 设在区间上有定义，，且在连续。若存在，满足；且存在，满足。则为的一个极大值点。若存在，满足；且存在，满足。则为的一个极小值点。若在左边的某个区间和右边的某个区间内，的符号相同，则不是极值点。的点与导数不存在的点，均为可能的极值点对导数为的点，有 1. 可能是极大值。 2. 可能是极小值。 3. 可能不是极值点。定理 3. 在处满足，则若，则是的极大值点若，则是的极小值点例 1. (例3.5.2) 求函数的单调区间例 2. 单调增例 3. (例3.5.1) 例 4. (例3.5.3) 求函数在区间上的最大值和最小值例 5. 单调增例 6. ，则例 7. ，则例 8. 证明： , 例 9. 设在内有界，且有连续导数，,求证：例 10. (例3.5.4) 例 11. 例 12. (Holder不等式) 不全为非负数组, , , 例 13. (Minkowski不等式) 不全为非负数组, 函数的凸性与拐点一阶导数提供的信息只是一个大概借助函数的二阶导数讨论函数的凸性、拐点与曲率若连接曲线上的任意两点的直线，总是位于曲线的上方，则称曲线为凸的。定义 1. (凸函数) 设是区间上的函数，若对中任意两点, ，以及对任意，有则称为上的凸函数。若不等号改为，则称为严格凸函数。定理 4. 几个等价的表达（Jensen不等式） 1. , ， 2. , , 3. , 定理 5. 若函数对任意的和，满足则函数连续定理 6. 设是区间上的连续函数若在内部可微，则在上是(严格)凸函数，当且仅当其导函数在内(严格)单调递增。若在内二阶可导，则在内是凸函数，当且仅当在内部。而严格凸的充要条件是且在任何子区间上不恒为零。定义 2. 设在包含点的区间上连续。如果点是的凸、凹区间的一个分界点，则称是函数的一个拐点（或称为扭转点）。有时也称函数图像上的点是曲线的拐点定理 7. 设在连续，在附近（不包含）可导。若在的左侧某个区间内严格单调增（或减），而在右侧某个区间内严格单调减（或增），则是的拐点。定理 8. 若在处二阶可导，是的拐点，则可以看出，拐点在二阶导数为的点，或二阶导数不存在的点 \| 拐点 \| 极值点 \| --- \| \| 分割凹、凸区间的点 \| 局部最值点 \| \| 在二阶可导，且为拐点，则 \| 在可导，且为极值点，则 \| \| 在有，未必为拐点。如 \| 在有，未必为极值点。如 \| \| 在二阶导数不存在，但可能是拐点 \| 在不可导，但可能是极值点 \| 例 14. (二阶导数为的点，不是拐点。这是个凸函数) 例 15. 在上二阶可导，, , ，则例 16. 证明平均不等式定义 3. 当曲线上的点沿曲线运动，与某条直线的距离趋于，就称这条直线是的渐近线若，或，或，称为的垂直渐近线若，或，或，称为的水平渐近线若有实数，满足，，则称为的斜渐近线对参数曲线，有若存在有限，，则有水平渐近线若存在有限，，则有垂直渐近线若，，则可能有斜渐近线。若，，则称为的斜渐近线函数作图定义域、奇偶性、周期性、对称性间断点、驻点、导数不存在的点单调性、极值点凹凸性区间、捌点渐近线与坐标轴的交点例 17. 作图解. 定义域，且驻点与拐点例 18. 作图解. 1. 定义域，且也是垂直渐近线 2. 渐近线这样，有斜渐近线 3. 驻点和拐点因此，可能有驻点, 。可能有拐点 \| 区间 \| \| \| \| \| \| --- --- --- \| \| \| 增 \| 减 \| 减 \| 减 \| 增 \| \| \| 凹 \| 凹 \| 凸 \| 凸 \| 凸 \| 是拐点，是极大值点，是极小值点平面曲线的曲率 \| \| \| --- \| \| 定义：取为弧长，为转过的角度差，则叫作平均曲率。若存在，就表示点的曲率，记为 \| Fig \| 设曲线为，且二阶可导。过的切线与轴的正向平角为，则有则又，微分弧长为，代入曲率公式，有若曲线由参数方程表示。若曲线为正则曲线(, 连续，且不同时为)，则进而可以得到曲率为易得，直线的曲率为0；半径为的圆的曲率为定义 4. 设曲线为在点处的曲率。过做曲线的法线，在法线上曲线凹的一侧取点，使，以为圆心，为半径的圆，称为曲线在点的曲率圆或密切圆，它的半径称为曲率半径，圆心称为曲率中心。在点，曲率圆与曲线具有相同的切线、凹凸性和曲率。例 19. 求极坐标形式下，曲线的曲率半径例 20. (例3.5.9) 计算椭圆例 21. 求圆的渐伸线的曲率，曲率半径，曲率圆心例 22. 求双纽线的曲率半径例 23. 在曲线上求曲率最大的点时，圆的渐伸线 vertical slide 2 目录本节读完例 24. 谢 24.
10729	https://es.khanacademy.org/math/linear-algebra/matrix-transformations/lin-trans-examples/v/linear-transformation-examples-scaling-and-reflections	Ejemplos de transformaciones lineales: reflejar y escalar (video) \| Khan Academy Salta al contenido principal Si estás viendo este mensaje, significa que estamos teniendo problemas para cargar materiales externos en nuestro sitio. Si estás detrás de un filtro de páginas web, por favor asegúrate de que los dominios .kastatic.org y .kasandbox.org estén desbloqueados. Explorar Matemáticas: De preescolar a 8.º Matemáticas: Cursos de preparación Matemáticas: De 9.° a la universidad Ciencia Computación Economía y finanzas Selecciona una categoría para ver sus cursos. Buscar Haz una donaciónInicia sesiónRegístrate Busca cursos, habilidades y videos Saltar al contenido de la lección Álgebra lineal Curso: Álgebra lineal>Unidad 2 Lección 2: Ejemplos de transformaciones lineales Ejemplos de transformaciones lineales: reflejar y escalar Ejemplos de transformaciones lineales: rotar en R2 Rotación alrededor del eje x en R3 Vectores unitarios Introducción a las proyecciones Expresar una proyección en una recta como un producto de matrices y vectores Matemáticas por area> Álgebra lineal> Transformaciones de matrices> Ejemplos de transformaciones lineales © 2025 Khan Academy Términos de usoPolítica de privacidadAviso de cookiesDeclaración de accesibilidad Ejemplos de transformaciones lineales: reflejar y escalar Google Classroom Microsoft Teams Acerca de Acerca de este video Transcripción Crear matrices de transformaciones de reflexión y escalamiento (las cuales son diagonales).Creado por Sal Khan. 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Transcripción del video hemos hablado bastante sobre transformaciones lineales y en este vídeo y también en los siguientes yo quiero mostrarte cómo esencialmente diseñar una transformación lineal para que haga con los vectores lo que tú quieres que haga con ellos entonces ya sabemos que una transformación lineal digamos t que va de rn a rm entonces podemos representar a teo o lo que lo que te hace a algún vector lo podemos representar como una matriz una matriz que multiplica a un vector x esto sería multiplicando al vector x y esto nos da una matriz de m por n de m por n ahora sabemos que siempre podemos construir esta matriz que cualquier transformación lineal puede ser representada como una matriz de esta manera y la podemos representar así tomando a nuestra matriz identidad entonces lo que voy a hacer de chinos ya lo has visto antes esta matriz identidad con una filas y n columnas donde se mira así un uno aquí después en un 0 2 después 0 1 y todos los demás son ceros como puedes ver solo tiene unos en la diagonal principal es una matriz de n por n entonces tomas la matriz identidad y le aplicas la transformación lineal a cada columna ajá entonces estas son las columnas de la matriz identidad y las las llamamos la base estándar de rn ésta es la columna 1 esta es la 2 y como tienen columnas esta es la n ahora cada una de estas columnas es un miembro de rn tienen elementos y sabemos que puede ser representada como la transformación que está siendo aplicada a cada columna entonces a es igual a la transformación aplicada a la primera columna hay 1 después la transformación aplicada a 2 hasta llegar a la transformación aplicada n y esto es algo que sirve mucho saber porque porque es muy fácil aplicar la transformación lineal a cada uno de estos vectores que pertenecen a la base estándar cierto solamente tienen un solo uno y todo lo demás es cero entonces son muy muy amigables y estos vectores unitarios ahora como tú puedes ver esto es un repaso entonces vamos a usar esto para para construir una transformación lineal interesante así que manos a la obra vamos a iniciar con un conjunto en rn y de hecho ya a todo todo lo que voy a hacer es nn r2 solo para no complicar tanto la vida pero tú lo puedes extender a cualquier dimensión que quieras solo que lo voy a hacer yo en r2 para la simplicidad supongo digamos que tengo yo un triángulo entonces tengo un triángulo formado por los siguientes tres puntos digamos que el punto este punto es el 3 2 el 3 2 ahora también el otro punto sería que te parece y menos 32 ok este punto es menos 32 o bueno lo estoy haciendo en forma de efecto al columna de hecho aquí puse a 3 sobre 2 no es una fracción me equivoqué lo siento es menos 32 y 32 son los vectores columna entonces ahí está ese punto o ese vector y ahora sólo por diversión uno por acá por abajo que te parece el vector 32 entonces no tres menos dos perdón tres -2 ahí está ahora bien entonces ese punto está más o menos por aquí ya tenemos los tres puntos cada uno de estos puntos son vectores yo puedo dibujarlo saliendo del origen a cada uno de estos puntos son vectores entonces muy bien más que más que importarme que sean vectores lo que me importa es la posición que especifican y sabemos que si tomamos al conjunto de todas las de todos los vectores que forman este triángulo sería en el conjunto que conectan a estos tres puntos cierto ok ahora a ese conjunto que conectan a estos tres vectores o lo que sería el triángulo mismo le vamos a aplicar la transformación lineal a cada uno entonces le aplicamos la transformación lineal t a estos puntos y los conectamos en el mismo orden de hecho ya lo hemos visto eso antes pero ahora queremos diseñar una transformación lineal entonces ok perfecto digamos que queremos de hecho de hecho lo que voy a hacer es escribir lo que queremos hacer voy a escribirlo lo que queremos hacer primero que todo es reflejarlo sobre el eje de las x entonces queremos reflejarlo sobre el eje de las equis o no mejor no mejor en el egezy ok reflejarlo sobre el eje de las 'íes' así que queremos voltear los queremos voltearlo y una vez que lo volvemos se va a mirar algo así cierto algo así se va a mirar y también digamos que queremos alargarlo por un por un 2 entonces queremos reflejarlo sobre el eje de la sien y también queremos alargarlo en la dirección en la dirección y por 2 así que vamos a primero voltearlo se va a mirar algo así esto sería como el el paso 1 cierto el paso uno es voltearlo esto es paso 1 y el paso 2 es alargarlo se mirará en lugar de así se va a mirar como tendrá como el doble de altura ahora el la x se va a quedar igual y ok como hacemos eso la primera idea que tenemos es reflejar sobre el eje de las yes entonces creemos que este punto este punto el menos 3 el menos 3 está aquí ajá el menos 32 que esa coordenada que desde el otro lado con un 3 positivo este 2 aquí es la coordenada jr este es el eje de las yes ok entonces tenemos menos 32 queremos que que esté menos 3 se convierta en un 3 positivo queremos que quede que terminen en este otro punto correspondiente cuando cuando le demos la vuelta este 3 positivo también queremos que se convierta en 3 negativo y este 3 positivo también 3 negativo ahora como tú puedes notar lo que estamos haciendo es es algo es voltear el signo cierto estamos reflejando sobre y eso es equivalente a cambiar el signo de la coordenada x entonces esto esto es equivalente lo escribo por acá es equivalente a multiplicar menos 1 por la coordenada la coordenada que voy a llamar x 1 esto es esta es la coordenada x 1 al alargar al alargar esto en la dirección y eso significa que queremos que para cualquier altura que yo tome quiero que sea el doble ahora eso es alargarlo entonces esta coordenada en la coordenada 32 no de hecho aquí no he hecho el primer paso aún pero quiero que quiero que se se quiero que tenga el doble de altura entonces en lugar de hacer 3,2 será 3.4 lo que estoy haciendo es multiplicar por 12 a la coordenada jet y en lugar de llamar a estos a las entradas del vector x x1 y x2 voy a ponerles el nombre de xy y entonces el vector es igual a xy como coordenadas esas son las coordenadas del vector xy ahora no temas por tu vida porque simplemente es notación aquí lo que estoy haciendo es facilitar la existencia de nosotros porque estamos acostumbrados a ver un vector en su forma x y en lugar de x 1 x 2 y xy llevan a corresponder con los ejes de coordenadas cierto entonces queremos queremos construir una transformación una transformación de algún vector x que o mejor escribo de esta otra manera la transformación de x y jane esto va a ser igual a 1 - 1 x x entonces menos 1 por x y ala y al hielo multiplicó por 2 2 así que hacía así es como lo puedo escribir en su lenguaje de transformación lineal este sería el lenguaje de transformación lineal pero pero cómo puedo construir una matriz para esto lo que lo que vamos a hacer es tomar estamos en el re 2 entonces voy a tomar la matriz identidad en el re 2 que sería esta matriz 1 0 0 1 y le voy a aplicar la transformación a cada columna de esta matriz identidad entonces lo voy a hacer que vamos a obtener tenemos tenemos una ninguna nueva matriz es cierto tenemos una nueva matriz y esto será igual a la transformación aplicada a cada columna entonces la transformación de la primera columna que sería la columna 1010 y después la segunda transformación o la segunda columna de esta nueva matriz sería la transformación aplicada a la segunda columna de la identidad que sería 01 y ahí lo tenemos entonces a que es una igual estas dos nuevas columnas la transformación de la primera columna sería ok vamos a hacerlo a pie tenemos a es igual a la transformación de 10 eso es x es 1 entonces lo multiplicamos por menos 1 y tenemos menos 1 después 2 por qué pero y es igual a 0 entonces tenemos 0 ahora el segundo término tenemos menos x es 0 simplemente de 0 y 2 porque es igual a 2 por 1 eso es igual a 2 tenemos entonces menos 1 002 ahora podemos decir que esta transformación algún vector xy entonces la transformación de xy la podemos describir como una matriz que multiplica a un vector entonces esa es la matriz menos uno menos 10 menos 10 y 0 2 multiplicando al vector x y ahora hay que aplicarlo para verificar que si funciona es cierto porque queremos ver que en efecto lo refleja lo alarga así que pongo primero este vector este punto rosa es este vector cierto ese es el punto entonces vamos a efectuar las operaciones sobre este vector y lo que voy a hacer es poner la matriz pongo la matriz menos 1 002 que multiplica al vector menos 32 entonces tenemos una multiplicación de una matriz por un vector así que hagamos esto menos 1 x menos 3 estrés positivo +0 por 20 entonces nos queda un 3 positivo y después 0 x menos 30 después más dos por dos es cuatro así que nos queda 34 entonces lo que ya tenemos ese ahora vamos con el siguiente punto que es 32 entonces ok muy bien vamos con ese siguiente punto 32 pongo la matriz menos 10 y 0 2 multiplicando al vector 32 entonces tenemos menos 1 por 3 es igual a menos 3 y después 0 por 20 entonces tenemos menos 3 ahora 0 por 3 es igual a cero más 2 por 2 es igual a 4 nos queda el vector menos 34 y ese punto está justamente justamente por aquí aquí está menos 3 y 4 entonces ok aquí aquí estoy usando la terminología se le puede decir a esta transformación aplicada a cada uno de estos vectores entonces tú puedes decir hola a la transformación mapea este punto a este punto o lo convierte o lo transforma pero es simplemente tan bueno estoy diciendo que ese punto es transformado a este otro punto en r2 entonces ahora vamos con el siguiente tenemos la matriz multiplicando al siguiente vector que es el vector 3 - 2 entonces aquí menos 1 por 3 es igual a menos 3 y 0 por menos 12 0 entonces nos queda menos 3 ahora 0 por 13 0 más 2 por menos 2 es igual a menos 4 tenemos entonces el punto menos 3 menos 4 ese punto está más o menos por aquí menos tres menos cuatro y sabemos que sabemos que el conjunto de r2 que como que conecta estos puntos bajo la misma transformación se será mapeado al conjunto r2 que conecta a estos puntos en este otro lado cierto entonces así que la imagen de este conjunto que dibuje de este triángulo es simplemente un conjunto de puntos que especifican a este otro conjunto de vectores entonces la imagen de este conjunto de vectores no se específica a estos puntos este este nuevo triángulo que tenemos bajo la transformación de aquellos de aquel conjunto de puntos entonces bueno excelente hicimos o ha hecho esta transformación lo que queríamos que hiciera cierto lo reflejó primero y después lo alargó entonces vemos que lo alargó por un con un por un factor de dos y lo volteó y de hecho en general cualquiera de estas operaciones puede ser una tránsfuga no puedes transformar lo bueno es decir puedes regresar t o regresar a la forma original entonces tú puedes escribir la transformación en esta forma y aplicarle esto a los vectores base a las columnas de la matriz identidad pero no general es que cualquiera de estas transformaciones que literalmente lo que lo que hacen es alargar o reflejar o de hecho hacernos pequeño lo que sea es la está en la dirección del eje x o del eje y serán o no serán matrices diagonales entonces como puedes ver aquí también la matriz a es una matriz diagonal estas serán las transformaciones en las matrices diagonales porque bueno solamente tienen términos no nulos en la diagonal cierto este es el caso 2 x 2 si hago el caso 3 x 3 tendríamos ceros por todas partes menos en la diagonal principal ahí habrían números no nulos y tiene mucho sentido porque este primer término es esencialmente lo que estás haciendo al x 1 al término x 1 si tienes por ejemplo una matriz de 3 x 3 este tercer término es lo que les des a la tercera dimensión a la cuarta dimensión es el siguiente término entonces tú puedes expander esta idea a rn arbitrario cierto a cualquier dimensión pero en fin de hecho lo que la idea general de este vídeo era introducirte a esta idea de crear de diseñar transformaciones para cualquier dimensión y creo que lo hemos logrado bueno nos vemos Creative Commons Attribution/Non-Commercial/Share-AlikeVideo en YouTube A continuación: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. 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10730	https://chem.libretexts.org/Courses/Westmont/Advanced_Inorganic_Chemistry/Inorganic_Chemistry_(Mapped_to_Meissler_Fischer_and_Tarr)	Inorganic Chemistry (Mapped to Meissler, Fischer, and Tarr) - Chemistry 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 Advanced Inorganic Chemistry Westmont { } { AGENDA : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Homework : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Inorganic_Chemistry_(Mapped_to_Meissler_Fischer_and_Tarr)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Worksheets : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { Advanced_Inorganic_Chemistry : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Test_Course_A : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sun, 23 Jun 2019 16:47:33 GMT Inorganic Chemistry (Mapped to Meissler, Fischer, and Tarr) 151915 151915 admin { } Anonymous Anonymous User 2 false false [ "article:topic-category", "license:ccbyncsa", "licenseversion:40" ] [ "article:topic-category", "license:ccbyncsa", "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. Campus Bookshelves 3. Westmont 4. Advanced Inorganic Chemistry 5. 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This text map is under construction. If you'd like to organize with us to pitch in, check out the under construction spreadsheet here (click) or email Kathryn Haas. Front Matter 1: Introduction to Inorganic Chemistry 2: Atomic Structure 3: Simple Bonding Theory 4: Symmetry and Group Theory 5: Molecular Orbitals 6: Acid-Base and Donor-Acceptor Chemistry 7: The Crystalline Solid State 8: Chemistry of the Main Group Elements 9: Coordination Chemistry I - Structure and Isomers 10: Coordination Chemistry II - Bonding 11: Coordination Chemistry III - Electronic Spectra 12: Coordination Chemistry IV - Reactions and Mechanisms 13: Organometallic Chemistry 14: Organometallic Reactions and Catalysis 15: Parallels between Main Group and Organometallic Chemistry Back Matter 16: Appendix Inorganic Chemistry (Mapped to Meissler, Fischer, and Tarr) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. 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10731	https://resources.wolframcloud.com/FunctionRepository/resources/ArcLengthIntegral/	ArcLengthIntegral \| Wolfram Function Repository Wolfram.com WolframAlpha.com WolframCloud.com Wolfram Function Repository =========================== Instant-use add-on functions for the Wolfram Language Primary Navigation Categories Core Language & Structure Data Manipulation & Analysis Visualization & Graphics Machine Learning Symbolic & Numeric Computation Higher Mathematical Computation Strings & Text Graphs & Networks Images Geometry Sound & Video Knowledge Representation & Natural Language Time-Related Computation Geographic Data & Computation Scientific and Medical Data & Computation Engineering Data & Computation Financial Data & Computation Social, Cultural & Linguistic Data Notebook Documents & Presentation User Interface Construction System Operation & Setup External Interfaces & Connections Cloud & Deployment Repository Tools Programming Utilities Just For Fun Wolfram Physics Project Random Function Recent Functions Alphabetical List Submit a New Function Function Repository Resource: ArcLength Integral Source Notebook Generate an inactive integral or sum of integrals used for computing the arc length of an expression with given bounds Contributed by: Wolfram\|Alpha Math Team ResourceFunction["ArcLengthIntegral"][expr,{x,a,b}] returns an inactive integral for computing the arc length of expr for a≤x≤b. ResourceFunction["ArcLengthIntegral"][expr,{θ,a,b},"polar"] returns an inactive integral for computing the arc length of expr for a≤θ≤b in polar coordinates. ResourceFunction["ArcLengthIntegral"][{expr 1,expr 2,…},{t,a,b}] returns an inactive integral for computing the arc length of the parametric curve defined by {x 1[t],x 2[t],…}⩵{expr 1,expr 2,…} for a≤t≤b. ResourceFunction["ArcLengthIntegral"][eqtn,{{x,a,b},{y,c,d}}] returns an inactive integral for computing the arc length of the curve given by implicit equation eqtn for a≤x≤b and c≤y≤d. Details and Options ResourceFunction["ArcLengthIntegral"][expr,{x,lo,hi}] accepts a math expression, an equation of the form y==f[x] or a list of two or more math expressions in terms of x. In ResourceFunction["ArcLengthIntegral"][{expr 1,expr 2,…},{x,lo,hi}], the expr i are expected to be math expressions, not equations. Examples Example Notebook Open in Cloud Download Notebook Basic Examples(2) Set up the integral for computing the arc length of a sine wave: In:= Out= Activate to compute the full result: In:= Out= Set up the integral that computes the arc length of a circle: In:= Out= Scope(4) Return the integral for the arc length of a curve given in polar coordinates: In:= Out= Return the integral for the arc length of a parametric curve: In:= Out= Return the integral for the arc length of the unit circle that is contained within the region bounded by 0≤x≤1 and 0≤y≤1/2: In:= Out= You can also leave x or y unbounded: In:= Out= Return the integral for the arc length of an asteroid bounded by -1≤x≤1: In:= Out= Applications(3) Define and display an implicit region representing two disconnected segments of the unit circle: In:= Out= Set up the integrals for disconnected segments of the region: In:= Out= Then compute the result: In:= Out= Possible Issues(1) ArcLengthIntegral returns the integral for computing the unsigned arc length: In:= Out= Publisher Wolfram\|Alpha Math Team Version History 2.0.0 – 23 March 2023 1.0.0 – 01 April 2020 Related Resources IntegrateByParts Related Symbols ArcLength Integrate License Information This work is licensed under a Creative Commons Attribution 4.0 International License Send a message about this function Discuss on Wolfram Community Top © 2025 Wolfram. All rights reserved. Legal&Privacy Policy Contact Us WolframAlpha.com WolframCloud.com
10732	https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems?srsltid=AfmBOooL7p4hi7gziD8vy-FuaCLa2Vny0mrjDgpLBiHOHsUrEOb2p0x7	Art of Problem Solving 2016 AMC 10A Problems - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. 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CHECK SCHEDULE 2016 AMC 10A Problems 2016 AMC 10A (Answer Key) Printable versions: Wiki • AoPS Resources • PDF Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator). Figures are not necessarily drawn to scale. You will have 75 minutes working time to complete the test. 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 Contents 1 Problem 1 2 Problem 2 3 Problem 3 4 Problem 4 5 Problem 5 6 Problem 6 7 Problem 7 8 Problem 8 9 Problem 9 10 Problem 10 11 Problem 11 12 Problem 12 13 Problem 13 14 Problem 14 15 Problem 15 16 Problem 16 17 Problem 17 18 Problem 18 19 Problem 19 20 Problem 20 21 Problem 21 22 Problem 22 23 Problem 23 24 Problem 24 25 Problem 25 26 See Also Problem 1 What is the value of ? Solution Problem 2 For what value does ? Solution Problem 3 For every dollar Ben spent on bagels, David spent cents less. Ben paid more than David. How much did they spend in the bagel store together? Solution Problem 4 The remainder can be defined for all real numbers and with by where denotes the greatest integer less than or equal to . What is the value of ? Solution Problem 5 A rectangular box has integer side lengths in the ratio . Which of the following could be the volume of the box? Solution Problem 6 Ximena lists the whole numbers through once. Emilio copies Ximena's numbers, replacing each occurrence of the digit by the digit . Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's? Solution Problem 7 The mean, median, and mode of the data values are all equal to . What is the value of ? Solution Problem 8 Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning? Solution Problem 9 A triangular array of coins has coin in the first row, coins in the second row, coins in the third row, and so on up to coins in the th row. What is the sum of the digits of ? Solution Problem 10 A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is foot wide on all four sides. What is the length in feet of the inner rectangle? Solution Problem 11 Find the area of the shaded region. Solution Problem 12 Three distinct integers are selected at random between and , inclusive. Which of the following is a correct statement about the probability that the product of the three integers is odd? Solution Problem 13 Five friends sat in a movie theater in a row containing seats, numbered to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up? Solution Problem 14 How many ways are there to write as the sum of twos and threes, ignoring order? (For example, and are two such ways.) Solution Problem 15 Seven cookies of radius inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie? Solution Problem 16 A triangle with vertices , , and is reflected about the -axis, then the image is rotated counterclockwise about the origin by to produce . Which of the following transformations will return to ? counterclockwise rotation about the origin by . clockwise rotation about the origin by . reflection about the -axis reflection about the line reflection about the -axis. Solution Problem 17 Let be a positive multiple of . One red ball and green balls are arranged in a line in random order. Let be the probability that at least of the green balls are on the same side of the red ball. Observe that and that approaches as grows large. What is the sum of the digits of the least value of such that ? Solution Problem 18 Each vertex of a cube is to be labeled with an integer through , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible? Solution Problem 19 In rectangle and . Point between and , and point between and are such that . Segments and intersect at and , respectively. The ratio can be written as where the greatest common factor of and is 1. What is ? Solution Problem 20 For some particular value of , when is expanded and like terms are combined, the resulting expression contains exactly terms that include all four variables and , each to some positive power. What is ? Solution Problem 21 Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ? Solution Problem 22 For some positive integer , the number has positive integer divisors, including and the number . How many positive integer divisors does the number have? Solution Problem 23 A binary operation has the properties that and that for all nonzero real numbers and . (Here represents multiplication). The solution to the equation can be written as , where and are relatively prime positive integers. What is Solution Problem 24 A quadrilateral is inscribed in a circle of radius . Three of the sides of this quadrilateral have length . What is the length of the fourth side? Solution Problem 25 How many ordered triples of positive integers satisfy and ? 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10733	https://www.umiacs.umd.edu/~ramani/cmsc828d_audio/AllenBerkley79.pdf	Image method for efficiently simulating small-room acoustics Jont B. Allen and David A. Berkley .4cousties Research Department, Bell Laboratories. Murray Hill, New Jersey 07974 (Received 6 June 1978) Image methods are commonly used for the analysis of the acoustic properties of enclosures. In this paper we discuss the theoretical and practical use of image techniques for simulating. on a digital computer, the impulse response between two points in a small rectangular room. The resulting impulse response, when convolved with any desired input signal, such as speech, simulates room reverberation of the input signal. This !echnique is useful in signal processing or psychoacoustic studies. The entire process is carried out on a digital computer so that a wide range of room parameters can be studied with accurate control over the experimental conditions. A FORmS• implementation of this model has been included. PACS numbers: 43.55.Ka, 43.55.Br INTRODUCTION In some recent experiments, which studied the per- ceptual effects of reverberation properties of a small room, t. 2 a carefully controlled, easily changed, acou- stic environment was required. It was decided to utilize a computer simulation of the acoustic space. This pa- per describes both the general theoretical approach and the specific implementation techniques used (the FORTRAN program). We believe that the resulting room model is useful for a broad range of investigations, from our original experiments mentioned above, to basic studies of room acoustics. The room model assumed is a rectangular enclosure with a source-to-receiver impulse response, or trans- fer function, calculated using a time-domain image ex- pansion method. Frequent applications have been made of the image method in the past as in deriving the re- verberation-time equations, 3 for theoretical studies sound behavior in enclosures, 4-? and in the study of architectural acoustics and perceptual properties of rooms? 'n In addition, there has been a considerable amount of important theoretical work on the approxi- mate 12 use of images produced by a single soft-wall (finite impedance) reflection. Several recent papers on this subject which have good bibliographies are Refs. 13, 14, and 15. Computer methods have also recently been applied to image computations in enclosures (see for example Refs. 6, 7, 10, and 11). In the current paper the computational technique is specifically aimed at being simple, easy to use, and fast. In addition the resulting room responses have been used to realistical- ly model speech transmission in rooms and to investi- gate the effects of various forms of digital speech sig- ns/ processors. 16'tl In the following we will first briefly discuss theoretic- al aspects of the method. Then we will outline the compu- tational approach and, finally, we will give some staples of applications. (1) We are most interested in the office environment, which is usually a rectangular geometry. (2) This model can be most easily realized in an ef- ficient computer program. (3) The image solution of a rectangular enclosure rapid/y approaches an exact solution of the wave equa- tion as the walls of the room become rigid. The image model is chosen because we are interested in the point-to-point (e.g., talker-to-microphone) trans- fer function of the room. In order to obtain a good trausientdescriptionofthe response, atimedomaln mo- del is required. A normal-mode solution of the enclosure would require calculation of all modes within the fre- quency range of interest (i.e., 0.1-4.0 kHz), plus cor- rections for those outside this range. The image meth- od includes only those images contributing to the im- pulse response. Thus the contributing images are those within a radius given by the speed of sound times the reverberation time.• (The exact relationship between the normal-mode solutions and the image solution, for a lossless room, is discussed in Appendix A.) The im- portant information used here is that in the time-do- main, each image contributes only a pure impulse of known strength and delay while each normal mode is a decaying exponential which contributes to all times. Furthermore, whereas an image has only delay and gain as parameters, a normal mode computation re- quires the solution of transcendental equations to find the Dote location plus the evaluation of a relatively comptex function to find the mode gain (i.e., the residue of the pole). A. The image model We model a talker in a room as a point source in a rectangular cavity. A single frequency point source of acceleration in free space emits'a pressure wave of the form I. IMAGE VERSUS NORMAL MODE MODELS We model the rooms of interest as simple rectangular enclosures. This choice of geometry is made for sev- eral reasons: P ...... where 943 .I. Acoust. Soc. Am. 65(4), Apr. 1979 0001-4966/79/040943-08500.80 ¸1979 Acoustical Society of America 943 P = pressure, f= frequency, t= time, R-- Ix-x'l, (2) X =vector talker location (x,y,z), X' =vector microphone location (x', y ', z'), i= v•-i, c = speed of sound. When a rigid wall is present, the rigid wail (zero norm- al velocity) boundary condition may be satisfied by placing an image symmetrically on the far side of the wall. Thus, p(w ,X ,X ')= [oxp[i4(•R/.c)R] + exp[i4(•/.c)R"]]exp(-icot), (s) where we define the two distances from the microphone to the source R. and to the image R. by =(x_ x,)2 y,)2 +(z _ (4) 2. =(x + gF +(y +(z -z') 2 ß The wall has been placed at x = 0 in this case (note the sign in the x terms of R. and R.). In the general case of six walls the situation becomes more complicated because each image is itself imaged. The pressure may then be written (as shown in Appen- dix A) where R• represents the eight vectors given by the eight permutations over ñ of l•=(x•x', y •y', z•z') (6) r is the integer vector triplet (n,l,m), and lL, mr,), (7) where (L•, L•, L•) are the room dimensions. Equation (5) is the pressure frequency response assuming rigid walls for a point source at X =(x,y,z) and receiver at X'=(• ,y',z'). If Eq. (5) is Fourier transformed, we find the room impulse response function (time domain Green's function) p(t,X,X' =•. •,•.• 4vli• + Rr I An interpretation of Eq. (8) is given in Fig. 1 where we show a part of the image space for a two-dimensional slice through the room. When the accelerative source location (talker)X is excited, each image point is si- multaneously excited, creating spherical pressure waves which propagate away from each image point. Equation (8) is the exact solution to the wave equation in a rectangular, rigid-wall (lossless), room and may IMAGE EXPANSION p(t} Y o p( 1 )= •.. •..,' .• p=lr=-<• 4w'IRp+ Rrl •r = (2nLx,21Ly,2mLz) •=(x• x', y• y',z• z') •p+ ;r = (x• x'+ 2nLx,• t y' + 21Ly,z•z'+2mLz) YTG. •. A slice •u•h the t•;e space show• how •e i•ges of the source are spatially arra•ed. The solid •x represen• •e origi•l more. The ac• image s•ce is •me dimensional. be derived directly from the normal-mode solution as shown in Appendix A. B. Case of nonriõid wslls If the room walls are not rigid, the solution in terms of point images may no longe• be exact. A precise • statement of the effects of finite impedance walls is presently impossible, since the effects on even a single image are quite complicated. ts'•4'• Therefore we have continued to assume the approximate point image model even for nonrigid walls. In addition we have assumed an angle independent pressure wall reflection coefficient •. This assumption is equivalent to assuming that the wall impedance is proportional to sec(8), where 8 is the angle of incidence of a plane wave with respect to the wall normal. We presently do not understand the exact physical interpretation of the above assumptions. How- ever, we believe that they do not introduce serious problems into the final result under typical conditions. By typical, we mean over the frequency range of 100 Hz-4• kHz, wall reflection coefficients of greater than 0.7, typical office room geometries, and where both source and receiver are not close to the wall. Many, if not all, of the above conditions are probably not cri- tical and could be relaxed. We merely wish to carefully point out the nonexact nature of the •esults. The above assumptions result in the Sabine energy absorption coefficient c• for a uniform reflection coeffi- cient fi on a given wall of the form a = 1 - •2. (9) Our assumptions are similar to those of geometrical acoustics • and are the same as those required for spec- ular angle-independent ray tracing. In current imple- mentations of the model we also do •ot allow frequency variations in the reflection coefficients. Both the angie dependence and frequency dependence could be included 944 J. Acoust. Soc. Am., Vol. 65, No. 4, April 1979 J. Allen and D. Rerkley: Method for simulating small-room acoustics 944 in our computer program, b•t only at the expense of significantly complicating and slowing down the compu- tational model. Introducing the effects of finite, angle independent wall absorption into Eq. (8) leads to the modified room impulse response p(t,X,X')-z.• z_, •,t ,• • •2 •,• x 4.,,. fR, + ' (10) where 1• is now expressed in terms of the integer 3- vector p---(q,j,/•) as P,•=(x-x' +Zqx', y -y' + 2jy', z-z' + 2kz'). (11) I• as given by Eq. (6) is similar to titat of (11), but is indexed differently from (11). The beta's are the pres- sure reflection coefficients of the six boundary planes, with the subscript I referring to walls adjacent to the coordinate origin (see Fig. 1). Subscript 9. is the oppos- ing wall. Eq. (10) has been derived heuristically from geometrical considerations of Fig. 1. The sum • with vector index p is used to indicate three sums, namely one for each of the three components of p=(q,j,k). r--(n,l,rn) is a similar sum. Physically these sums are over a three-dimensional latUce of points. For p there are eight points in the lattice and for r, the lattice is infinite. II. IMPLEMENTATION OF THE MODEL The primary consideration in a computer (sampled data) implementation of Eq. (10) is the method of spatial sampling. In addition, an apparently nonphysical be- havior of the model at zero frequency is removed by a low-frequency (0.01 of the sampling frequency) high- pass digital filter. 19 A calculated impulse response is built up as a "histo- gram" oi image pulses received at different time de- lays. The width of each histogram bin is equal to the time sampling period T initially assumed, which in turn is determined by the highest frequency to be rep- resented. For example, all images with the range N•R to (N+ 1)aR, where AR=½T (T is the sampling period and c the speed of sound), are added together with appropriate amplitude as given by Eq. (10). The choice of sampling rate is governed.by the appli- cation. If speech is to be studied in small rooms one might choose T = 0.1 ms. (sampLing frequency of 10 kHz; highest frequency of 5 kHz). But, if reverberation times of large enclosures are being studied (and convo- lution with speech is not required) much lower rates can be useful. The time length of the calculated impulse response is also a consideration. For a given sampling rate the number of points in p(•) increases linearly with its length while the computation time (and number of im- ages) goes up approximately as the cube of response length. This is shown in the first four columns of Table TABLE I. Computation parameters•room size (feet) 10' x 15' x 12.• 8 kHz sampling rate. Impulse response Convolution Leugth Image Computation rate (ms) No. points count time (s) (s/s) 64 512 585 1 12.5 128 1024 469 0 8 13.8 256 2048 375 00 60 15.0 I for our implementation on a Data General, Eclipse S/200 computer. (On this machine the computation time required for each image is about 1.6 ms.) T. he actual FORTRAN programs used are given in Appendix B. The temporal quantization in the impulse function computation causes sligh[ statistical errors in the com- puted arrival times of each image pulse relative [o the exact delay as given in Eq. (10). This error can be thought of as effectively "moving" each image source by 0• < e•-•'o•'• < AR/2 relative to the receiver. This effect could be removed, in principle, by using a band-Limited source pulse. However, the error is small for most (if not all) purposes and it greatly complicates the com- putation to remove this approximation. We have esti- mated that the error due to the slight moving of the im- ages could not be perceived even in a digital simulation of a binaural hearing experiment. The subroutine SROOM of Appendix B requires as pa- rmeters the number of impulse response points de- sired (NPTS), the source location R0, the receiver 1o- 1.0 IMPULSE RESPONSE 2048 POINTS 8 KHZ SAMI:4.1NG RATE -I.0 I I I I I I 0 TIME (MS) I I FIG. 2. Plot of a typical impulse response for a room 80 x 120 x 100 sample lengths long. Wall reflection coefficients were all 0.9 ceiling and floor coefficients were 0.7. X and X' were at (30, 100, 40) and (50, 10, 50) sample periods. 945 J. Acoust. Soc. Am., VoL 65. No. 4, April 1979 J. Allen and D. Berkley: Method for simulating •mall-roorn acoustics 945 cation R, the room dimensions, all specified in terms of the sample length (AR), and the reflection coefficients of each of the six wall surfaces (fi). Figure 2 shows an example of the impulse response obtained for a room of dimensions 80 x 120 x 100 sample lengths with equal wall reflection coefficients of 0.9 ((• = 0.19) and with floor and ceiling reflection coefficients (/],) of 0.7 ((• = 0.51). X and X / were (30,100,40) and (50, 10,60) sample in- tervals, respectively. It is usually convenient to interpret the model param- eters as a true distance rather than as multiples of AR. This requires the choice of a sampling rate and then conversions may be performed in the users main program which calls the subroutines of Appendix B. Figure 2 is labeled assuming an 8 kHz sampling rate. For this assumption (and assuming a sound speed of 1 ft/ms) the room dimensions are 10'x 15' x12.5'. use a modification of the integrated tone-burst method 22 z(t)=kf (12) where E(t) is the average energy decay, k is a propor- tionality constant, and p(?) is the calculated pressure impulse response from Eq. (10). For cases where the impulse response has been truncated before most of the decay has taken place, (12) may lead to errors. These errors are usually obvious in the E(t) plots. Another, approximate, approach is to simply mea- sure the short-time average energy decay of the im- pulse itself (e.g., using a simulated level recorder). For exponential or near-exponential decays, both methods should give approximately the same value of III. APPLICATIONS Our room image model has been applied to several problems. We will discuss t•vo examples: a psycho- physical evaluation of room reverberation effects t'2 and a study of critical distance measurements using spec- tral response variance. 2ø We have also used the model to test a signal processor intended to reduce perceived reverberation t6 and to study problems associated with mathematical inversion (inverse filtering) of room transfer functions. t? A. Psychophysics of room reverberation Once a simulated room impulse response has been calculated using the image model, the psychophysical effects of this simulated reverberation on speech may be directly studied. A reverberant sample of speech was produced by convolving an anechoic (unreverberant) speech sample with the calculated impulse response [p(t)]. This can be done efficiently using a Fast Fourier Transform (FFT) method (overlap-add) to perform the convolution. 2• The last column of Table I shows the measured convolution rate, for various length impulse responses. The convolution rate only increases as log2(N) , (where N is the room response length in time sample periods T) so even large impulse responses can be convolved with speech quite efficiently. For exam- ple, to convolve (filter) one second of speech, sampled at 8 kHz, with a 256 ms long impulse response (2048 points) requires a 15 s computation. The speed of processing makes multivariate psycho- physical studies quite practical. Ease of modification and perfect control of room parameters avoids the problems which have made such experiments so difficult in the past. The actual experiments used 16 different simu- lated" rooms" (impulse responses) convolved with ten dif- ferent sentences spoken by four different speakers. Itwas discovered that the experimental rooms were perceptually well characterized by their spectral variance [Eq. (14)] andby the reverberation time. This latter measure, reverberation time, deserves some discussion. Given the impulse responses, reverberation time may be estimated in a number of ways. One method is to (o) DECAY CURVE 2048 POINTS 8 KHZ SAMPLING RATE TIME (MS) I 256 ENERG DECA CURV 2048 POINTS 8 KHZ. SAMPLING RATE 0 256 TIME (MS) FIG. 3. (a) Energy decay curve for the impulse response of Fig. 2 using the Schroeder [ntegraUon method. •2 (b) Impulse energy decay curve for a simulated level recorder. 946 J. Acoust. Soc. Am., Vol. 65, No. 4, April 1979 J. Allen and D. Berkley: Method for simulating small-room acoustics 946 reverberation time. Example plots of E(t) for both methods are shown in Figs. 3(a) and 3(b) using the im- pulse response of Fig. 2. Experience indicates that Eq. (12) gives the most satisfactory results. We have found (a) STANDARD DEVIATION OF SPECTRAL RESPONSE IN COMPUTER-SIMULATED ROOM: 17' x I$' x I0' 7 o ß o &• A A G o 0.15 029 050 •070 I I I -30 -24 -18 -12 -6 0 6 12 18 24 REVERBERANT/DIRECT ENERSY RATIO (A/Ac)• (D) STANDARD DEVIATION OF SPECTRAL RESPONSE IN COMPUTER-SIMULATED ROOMi 47' x 31' x • a 0.15 )• a 0.29 }• 0.50 y •0. 70 J ROOM DIMENSIONS: 47'x13'x 15' -30 -P4 -18 -12 -6 0 6 t2 18 24 REVERBERANT/ DIRECT ENERGY RATIO (dB) ( A/A c )2 FIG. 4. Figures from Jetzt zø which compare the theoretical rms deviation of the pressure in dB from the mean pressure in dB as a function of the direct to reverberent energy ratio (a) for a room 17x13x10 ft and (b) 47x31x15 ft, empirically that calculated reverberation times, for a number of simulated enclosures, agree well with Eyring's formula • over a wide range of Beta values? In the experiments discussed above 1, 24 we discovered a monotonic relationship between A/A: ( Fig. 4) the micro- phone-talker distance when normaliz ed by the room c ri- tical distance (the distance at which reverberant energy equals direct sound energy), the reverberation time, and psychophysical preference for the resulting speech. B. Critical distance measurement A new method has been proposed 2ø for measurement of critical distances (or reverberation radius) in rooms. In this technique a measurement is made of the log frequen- cy response variance cL defined as L(co):201og[[p()I ] (13) c[ =[œ(•)-L(,•)] • , (14) given the room transfer-function P(co) [Fourier trans- form of Eq. (10)] for several microphone-source spac- ings. The measured values are fitted to a theoretical curve for aL based on the assumption of simultaneously excited, uncorrelated, normal modes, combined with the calculated direct sound energy. The resulting fit was shown to give an accurate value for the room's critical distance. This new method was extensively studied using our image model before being applied successfully to real rooms. Since the direct and reverberant energy are known in the computer model, a comparison can easily be made to the theory. The model results show excell- ent agreement with theoretical calculation as is seen in Figs. 4(a) and 4(b). We know of no other method by which this study could have been carried out as effect- ively. IV. SUMMARY ANO DISCUSSION A simulation method for small rooms based on an approximate image expansion for rectangular nonrigid- wall enclosures has been discussed. The method is simple, easy to implement and efficient for computer simulation. Several examples of its use, where other methods would be difficult, have been discussed. APPENDIX A We wish to derive the rigid-wail image solution di- rectly from the normal-mode expansion for a rectangu- lar enclosure. The frequency response function (Green's function) for the pressure P(co) in an enclos- ure is given by solving the Helmhoitz equation driven by a single frequency point acceleration source. vP[(co/c),X ,x'] + ,x'] =- -x '), (A1) where co is the frequency and c is the speed of sound. The solution to this equation, assuming rigid boundar- ies, is given by 947 J. Acoust. Soc. Am., Vol. 65, No. 4, April 1979 J. Allen and D. Berkle¾: Method for simulating small-room acoustics 947 ..,• __ • r(x)r(x ) P(k,X,,, j- /• 2 k2 (A2) V,=.. (k,- ) ' where k = •/c, r = (n, l,m) indicates a three dimension- al sum, V is the room volume, •=• z• . k2r= ]kr 12 (A3) and n•rx l•y rn•rz where the Li•s are the room dimensions. Using the exponential expansion for cosine, mMti- plyi• the terms of Eq. (A2) together and collecti•, we obtain x • •exp(,•.•) P(k,X,X')=•,• (•$•:) , (AS) where • represents the eight vectors [also given by •=(x•x •, y •y•, • •'). (A6) Using the property of the delta Mnction on kz, k• • and k• f• •(• - =)•(•)• =F(=), (A•) we may rewrite Eq. (AS) in integr• form APPENDIX B By Fourier series analysis one may show Thus [with analogus equations to (A9) for y and z ] p(•,X,X,)=(21•) •_ f f f •-• exp[(i•.(R•+ P•)] (A•0) where • is the vector [also given by Eq. (7)] •=2(•, zL•, m•,). (An) Each triple integral is just a plane wave expansion for a •int source in free space since •p(iklRI) I f•f exp(i•.R) • (A12) Finally, •ing Eq. (A12), Eq. (A10) becomes T•ing the inverse Fourier transform of Eq. (A18), the echo structure becomes explicit ,• 4•[•+•} ' (A14) which is the same as Eq. (8) as desired. C PGM: SROOM C SUBROUTINE TO CALCULATE A ROOM IMPULSE RESPONSE C R=VECTOR RADIUS TO RECEIVER IN SAMPLE PERIODS =LENGTH/(CT) C R0 =VECTOR RADIUS TO SOURCE IN SAMPLE PERIODS C RL =VECTOR OF BOX DIMENSIONS IN SAMPLE PERIODS C BETA=VECTOR OF SIX WALL REFLECTION COEFS (0 <BETA <=1) C HT=IMPULSE RESP ARRAY C NPTS=# OF POINTS OF HT TO BE COMPUTED C ZERO DELAY IS IN HT(1) C SUBROUTINE SROOM(R, R0, RL, BETA, HT, NI•TS) DIMENSION HT(NPTS) DIMENSION R(3), R0(3), NR(3), RL(3), DELP(8), BETA(2,3) EQUIVALENCE (NR(1), NX), (NR(2), IVY), (NR(3), NZ) DO 5 I=l, NPTS 5 HT(I) =0 C CK FOR MIC AND SOURCE AT SAME LOCATION DIS = 0 DO 6 I=1,3. 6 DIS = (R(1)-R0 (I))'2 +DIS DIS = SQRT(DIS) IF (DIS.LT..5) HT (1) = 1 IF(DIS.LT..5)RETURN C FIND RANGE OF SUNI N1 = N-PTS/(RL(1)2) + 1 N2 = NPTS/(RL(2)2) +1 N3 = NPTS/(RL(3) 2) + 1 DO 20 NX=-N1, N1 DO 20 NY =-N2, N2 DO 20 NZ =-N3, N3 948 J. Acoust. Soc. Am., Vol. 65, No. 4, April 1979 J. Allen and O. Berklev: Method for simulating small-room acoustics 948 C GET EIGHT IMAGE LOCATIONS FOR MODE it NR CALL LTHIMAGE (R, R0, RL, NR, DELP) I0=0 DO 10 L=0,1 DO 10 J =0, 1 DO 10 K=0, 1 I0=I0+l C MAKE DELAY AN INTEGER ID =DELP (I0) +.5 FDM1 =ID ID=ID+i IF(ID.GT.NPTS)GO TO 10 C PUT IN LOSS FACTOR ONCE FOR EACH WALL REFLECTION GID = BETA(1,1)IABS 1 BETA(2,1)IABS(NX) 2 BETA(I,2) IABS(NY-J) 3 BETA(2,2) IABS(NY) 4 BETA (1,3' IABS (NZ-K) 5 BETA(2,3)IABS(NZ) 6 /FDM1 C CHECK FOR FLOATING POINT UNDERFLOW HERE; C IF UNDER FLOW, SKIP NEXT LINE HT (ID) = HT (ID) + GID 10 CONTINUE 20 CONTINUE C IMPULSE RESP HAS BEEN COMPUTED C FILTER WITH HI PASS FILT OF 1% OF SAMPLING FREQ (I.E'. 100 HZ) C IF THIS STEP IS NOT DESIRED, RETURN HERE W = 2. 4.ATAN(1 .) 100. T = 1E-4 R1 = EXP (-W T) R2 =R1 B1 = 2. Ri COS (W T) B2 = -RIR1 A1 =-(1. +R2) A2 =R2 Yi=0 Y2=0 Y0 =0 C FILTER HT DO 40 I=1, N-PTS X0 =HT(I) HT(I) =Y0 +AIYI+A2Y2 Y2 =Y1 Y1 =Y0 Y0 =BiY1 +B2Y2 +X0 40 CONTINUE RETURN END NOTE CONTINUATION LINES C PGM: LTHIMAGE PGM TO COMPUTE EIGHT IMAGES OF A POINT IN BOX SUBROUTINE LTHIMAGE(DR, DR0, RL, NR, DELP) DR IS VECTOR RADIUS TO RECEIVER IN SAMPLE PERIODS DR0 IS VECTOR RADIUS TO SOURCE IN SAMPLE PERIODS RL IS VECTOR OF BOX DIMENSIONS IN SAMPLE PERIODS Nit IS VECTOR OF MEAN IMAGE NUMBER DELP IS VECTOR OF EIGHT SOURCE TO IMAGE DISTANCES IN SAMPLE PERIODS DIMENSION R2L(3), RL(3), NR(3),DELP(8) DIMENSION DR0(3), DR(3), RP(3,8) C LOOP OVER ALL SIGN PERMUTATIONS AND COMPUTE H + / -- H0 I0= 1 DO 10 L =-1,1,2 DO 10 J =--1,1,2 DO 10 K=-1,1,2 949 J. Acoust. Soc. Am., Vol. 65, No. 4, April 1979 J. Allen and D. Berkle¾: Method for simulating small-room acoustics 949 C NEAREST IMAGE IS L=J=K=--I RP (1, I0) = DR(1) + L'DR0(1) RP(2,I0) =DR(2) +J'DR0(2) RP(3, I0) =DR(3) +K'DR0(3) IO=IO+l 10 CONTINUE C ADD IN MEAN RADIUS TO EIGHT VECTORS TO GET TOTAL DELAY R2L (1) = 2. RL (1) NR(1) R2L (2) = 2.RL (2) NR(2) R2L (3) = 2.RL(3)NR(3) DO 20 I=1,8 DE LSQ: 0 DO 25 J=l,3 R1 = R2L (J)-RP(J, I) DELSQ =DELSQ +Rl2 25 CONTINUE DELP(I) = SQRT(DELSQ) 20 CONTINUE RETURN END IBarbara McDermott and Jont Allen, "Perceptual Factors of Small Room Reverberation," J. Acoust. Soc. Am. 60, S9 (A) (1976). 2Jont B. Allen and B. J. McDermott, "A New Method for Mea- suring Perception of Room Reverberation" (unpublished). 3C. F. Eyring, "Reverberation Time {n 'Dead' Rooms," J. Acoust. Soc. Am. 1, 217-241 (1930). 4D. Mlntzer, "Transient Sounds in Rooms," J. Acoust. Soc. Am. 22, 341-352 (1950). •R. H. Bolt, P. E. Doak, and P. J. Westerveit, "Pulse Sta- tistics Analysis of Room Acoustics," J. Acoust. SOc. Am. 22, 328-339 (1950). 6j. M. Betman, "Behavior of Sound in a Bounded Space," J. Acoust. Soc. Am. 57, 1275-1291 (1975). ?E. K. Dunens and R. F. Lambert. "Impulsive Sound Response Statistics," J. Acoust. Soc. Am. 61, 1524-1532 (1977). 6J. R. Power, "Measurement of Absorption in Rooms with Sound Absorbing Ceilings," J. Acoust. Soc. Am. 10, 98-101 (1938). 9C. G. Mayo, '•tanding Wave Patterns in Studio Acoustics," Acustica 2, 49-64 (1951). 1øB. M. Gibbs and D. K. Jones, "A Simple Image Method for Calculating the Distribution of Sound Pressure Levels within an Enclosure," Acustica 26, 24-32 (1972). tlF. Santon, "Numerical Prediction of Echograms and the In- telligibfiity of Speech in Rooms," J. Acoust. Soc. Am. 59, 1399-1405 (1976). 2By "approximate" we mean the blurr ing of the image as a re- suit of the finite impedance wall. Variations in wall imped- ance as a function of frequency or other frequency dispersion effects will also make the images frequency dependent. In practice the two effects usually appear simultaneously. t3A. R. Wentzel, "Propagation of Waves Along an Impedance Boundary," J. Acoust. Soc. Am. 55, 956-963 (1974). 14S. Thomasson, "Reflection of Waves from a Point Source by an Impedance Boundary," J. Acoust. Soc. Am. 59, 780-785 (1976). lSR. J. Donato, '•Spherical Wave Reflections from a Boundary of Reactive Impedance using a Modification of Cagniard's Methods," J. Acoust. Soc. Am. 60, 999-1002 (1976). 16Jont B. Allen, D. A. Berkley, and J. Bsuert, "Multimicro- phone sig•l-processor technique to remove room reverbera- tion from Speech Signals," J. Acoust. Soc. Am. 62, 912-915 (1977). l?S. T. NeeIcy and Jont B. Allen, "Invertibfiity of a Room Im- pulse Response," J. Acoust. Soc. Am. (in press). 18p. M. Morse and K. U. Ingard, Theoretical Acoustics (McGraw-Hill, New York, 1968), Sec. 9.5. lSWithout the high-pass filter, the Fourier transform ofp (t) shows a large energy spike at zero frequency as the absorp- tion a goes to zero. This is in part a direct result of the no- pressure-release boundary conditions we have assumed and is further complicated by the assumption of an accelerative source, which is nonphysical for very low frequencies (less than 50 Hz). 20j. j. Jetzt, "Critical Distance Measurement of Rooms from the Sound Energy Spectral Response," J. Acoust. Soc. Am. (to be published). 2tJont B. Allen, '•ASTFILT-An FFT Based Filtering Pro- graxn, "in IEEE Press book on Programs fo• Digital Signal Pro- cessing (IEEE Press, New York, 1979). zzM. R. Schroeder, "New Method of Measuring Reverberation Time," J. Acoust. Soc. Am. 37, 409-412 (1965). 2SAgreement is best for beta the same on all surfaces and variations appear when a pair of opposing walls are signifi- cantly different in reflectivity th•n all other surfaces, (as is the case for Figs. 2 and 3). 24D. A. Berkley, "Normal Listeners In Typical Rooms: Rever_ beration Perception, Simulation and Reduction" (unpublished). 950 J. Acoust. Soc. Am., Vol. 65, No. 4, April 1979 J. Allen and D. Berkley: Method for simulating small-room acoustics 950
10734	https://documentation.nokia.com/cgi-bin/dbaccessfilename.cgi/3HE09816AAAFTQZZA01_V1_5620%20SAM%20RELEASE%2013.0%20R6%20S.pdf	Title page Alcatel-Lucent 5620 SERVICE AWARE MANAGER \| Release 13.0 R6 SYSTEM ARCHITECTURE GUIDE 3HE-09816-AAAF-TQZZA Issue 1 \| November 2015 Legal notice Legal notice Alcatel, Lucent, Alcatel-Lucent and the Alcatel-Lucent logo are trademarks of Alcatel-Lucent. All other trademarks are the property of their respective owners. The information presented is subject to change without notice. Alcatel-Lucent assumes no responsibility for inaccuracies contained herein. Copyright © 2015 Alcatel-Lucent. All rights reserved. Disclaimers Alcatel-Lucent products are intended for commercial uses. 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Contents About this document Purpose .............................................................................................................................................................................................. v v Document support .......................................................................................................................................................................... v v How to comment ............................................................................................................................................................................ v v 1 5620 SAM system architecture 5620 SAM system architecture overview ........................................................................................................................ 1-1 1-1 Network management functions .......................................................................................................................................... 1-2 1-2 System components ................................................................................................................................................................. 1-3 1-3 Component communication .................................................................................................................................................. 1-7 1-7 System structure ........................................................................................................................................................................ 1-9 1-9 Security ...................................................................................................................................................................................... 1-12 1-12 Fault tolerance ......................................................................................................................................................................... 1-15 1-15 Standards compliance ........................................................................................................................................................... 1-17 1-17 .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 iii Contents .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... iv 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 About this document About this document Purpose The 5620 SAM System Architecture Guide is intended for technology officers, network planners, and system administrators to increase their knowledge of the 5620 SAM software structure and components. It describes the system structure, software components, and interfaces. In addition, 5620 SAM fault tolerance, security, and network management are described from an architectural perspective. Document support Customer documentation and product support URLs: Customer documentation welcome page • Technical support • How to comment Documentation feedback • documentation.feedback@alcatel-lucent.com ................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 v About this document .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... vi 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1 1 5620 SAM system architecture 5620 SAM system architecture overview General information The 5620 SAM is a network management system that simplifies routine operations and allows the bulk provisioning of network objects. The system is designed using industry standards such as Java, XML/SOAP, WebDAV, and 3GPP. The 5620 SAM uses open-standard interfaces that allow the system to interoperate with a variety of other network monitoring and management systems. 5620 SAM functions The 5620 SAM network management functions include the following: • service and routing configuration using distributed policies and profiles • equipment, service, and customer inventory reporting • network performance, accounting, and flow-based statistics collection • hierarchical alarm correlation between objects • interoperation with other network systems Main architecture features The main features of the 5620 SAM system architecture include the following: • the use of open standards to promote interaction with other systems • distributed resources that spread the processing load across multiple components and efficiently execute network management tasks • a multi-layer design model with functions in separate modules that interact with OEM products to accommodate increasing network growth and complexity • web services that provide access to 5620 SAM applications by effectively exporting XML interfaces over the Internet; the web services permit access to remote components such as web portals, and allow third-party vendors to create customized entry points for 5620 SAM functions • component redundancy that provides a high degree of fault tolerance ................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-1 Network management functions General information The 5620 SAM provides comprehensive network access for operators based on role-based scopes of command and spans of control over types of network objects. A 5620 SAM system collects data from managed NEs and collates the data for accounting, performance monitoring, troubleshooting, inventory, and fault management. The system deploys operator commands to the network and performs functions such as NE discovery and regular backups of the 5620 SAM database and the NE configurations. A 5620 SAM system is primarily designed to manage Alcatel-Lucent devices. However, you can obtain drivers from Alcatel-Lucent for managing specific non-Alcatel-Lucent devices. Drivers can be downloaded from the Alcatel-Lucent Electronic Delivery (ALED) site, and driver installation and usage documentation is available on OLCS. Service management The 5620 SAM service management functions allow network operators to provision VLL, VLAN, VPLS, IES, VPRN and mirror services for customers. Each service can be monitored to provide performance, usage, and fault information. Accounting The 5620 SAM collects accounting statistics from managed NEs that can be used to bill subscribers. The statistics are transferred to the 5620 SAM main and auxiliary servers using FTP or SCP. Equipment management The 5620 SAM maintains an equipment data model and deploys configuration updates to the managed NEs. For example, when a 5620 SAM operator adds a card to an NE, the data model is updated to include the card, and the card provisioning and configuration commands are sent to the NE. New NEs can be discovered at operator request, or automatically. A newly discovered NE is added to the data model. 5620 SAM system architecture Network management functions .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-2 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 Performance management The 5620 SAM can monitor services and network resources using performance statistics, OAM diagnostic tools, and data validation, and raises alarms when appropriate. • The 5620 SAM collects NE performance statistics using SNMP. • The 5620 SAM has a comprehensive suite of OAM tools for monitoring service, NE, and transport availability and performance. You can run tests before service activation to ensure that a service functions correctly after activation. • The 5620 SAM regularly compares the configuration information on managed NEs with the information in the 5620 SAM database to ensure synchronization. Fault management The 5620 SAM performs fault management in response to NE SNMP traps. The system converts traps to status updates and raises alarms when appropriate. GUI clients use visual and auditory cues to alert an operator when an alarm is raised. The 5620 SAM immediately forwards fault information as JMS events to OSS clients that subscribe to the appropriate JMS topic, and in response to OSS client XML requests for information. System components General information A 5620 SAM system comprises several components that are described below. Some components are supported only in specific deployment types. See the 5620 SAM Planning Guide for comprehensive deployment information. Main servers A main server is the central Java-based network-management processing engine that can be collocated on the same station as a 5620 SAM database, or on a separate station. A main server includes third-party components such as an application server, JMS server, web server, protocol stack, and database adapter. Some functions, for example, statistics collection, can be distributed across optional auxiliary servers. Auxiliary servers An auxiliary server, like a main server, is a Java-based processing engine, but is an optional, scalable component that extends the system ability to perform statistics or call-trace data collection. An auxiliary server is controlled by a main server, and collects data directly from NEs. 5620 SAM system architecture Network management functions .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-3 Cflowd auxiliary servers A Cflowd auxiliary server is an optional, scalable component that collects AA Cflowd statistics directly from NEs and forwards the data to an OSS or third-party application for billing, traffic analysis, or data analytics. 5620 SAM databases A 5620 SAM database is a customized relational database that provides persistent storage and serves as a central network data repository. A database can be collocated on the same station as a 5620 SAM main server, or on a separate station. Single-user GUI clients A single-user GUI client is a Java-based graphical interface for network operators. Single-user GUI client deployment is supported on multiple platforms. Client delegate servers A client delegate server supports simultaneous GUI sessions using one client software installation. A client delegate server can host local and remote user sessions, and supports the use of a third-party remote access tool such as a Citrix gateway. Client delegate server deployment is supported on multiple platforms. A client GUI session that is opened through a client delegate server is functionally the same as a single-user client session. The client delegate server locally stores the files that are unique to each user session, such as the client logs and GUI preference files, using a directory structure that includes the RHEL or Windows username. Figure 1-1, “Client delegate servers” (p. 1-5) shows two client delegate servers in a 5620 SAM management network. Multiple local users log in to a client delegate server directly, and remote users log in through a client delegate server that hosts a third-party access tool, for example, a Citrix gateway. Another local user opens a session on a single-user client station. 5620 SAM system architecture System components .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-4 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 If a client delegate server becomes unreachable, the 5620 SAM raises an alarm and changes the color of the associated session entries in the GUI. The alarm clears when the server is again reachable. You can use the client software on a client delegate server from the local console. Alcatel-Lucent recommends that you install a client delegate server, rather than a single-user client, to facilitate the deployment of additional clients. A main server monitors the registered client delegate servers and displays information about them in the GUI. To register a client delegate server, you specify the client delegate server IP address and installation location during main server installation, upgrade, or configuration. You can use a client GUI to list the following: • registered client delegate servers and the availability of each • active client delegate server sessions • active client sessions on a specific client delegate server • active client sessions for a specific 5620 SAM user Figure 1-1 Client delegate servers Managed network 20165 5620 SAM server and database JMS Server with XML OSS XML/SOAP over HTTP or HTTPS 5620 SAM client delegate server and third-party access server Remote 5620 SAM GUI users Local 5620 SAM GUI users Single-user 5620 SAM GUI client 5620 SAM client delegate server 5620 SAM system architecture System components .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-5 The number of allowed 5620 SAM client sessions on a client delegate server is configurable as a threshold using the 5620 SAM GUI. If a user tries to open a client session that reaches or exceeds the threshold, the session proceeds and the client delegate server raises an alarm. This threshold-crossing function can help to balance the session load across multiple client delegate servers. You require the Update user permission on the Server package to configure the threshold. The following restrictions apply to client delegate servers. • Alcatel-Lucent supports the installation of one client delegate server on a station. • You cannot change a 5620 SAM single-user client to a client delegate server. • A client delegate server connects to one release of main server; multiple main servers to which the client delegate server connects must be at the same release. • Depending on the platform type, specific installation, upgrade, and operation security restrictions apply; see the 5620 SAM \| 5650 CPAM Installation and Upgrade Guide for the deployment requirements and restrictions. OSS clients An OSS client is a software application that you create and implement to automate GUI client tasks, or to retrieve data for post-processing, for example, rolling up statistics data into a billing application. OSS clients may be as diverse as simple CLI scripts and third-party applications. An OSS client is platform-independent, because only Java messages are exchanged with the 5620 SAM. The 5620 SAM supports the following OSS client types: • XML/SOAP clients that use the XML OSS interface to perform general network management; XML schema files provide the data object definitions and describe the object attributes and methods; see the 5620 SAM XML OSS Interface Developer Guide for more information • 3GPP CORBA or XML/SOAP clients that use the 3GPP OSS interface to perform LTE management; IRPs provide the 3GPP data object definitions; see the 5620 SAM Wireless OSS Interface Developer Guide for more information Subcomponents All subcomponents, for example, Java modules, database software, and web server software, are represented by license files in the following directory on a main server: /opt/5620sam/server/nms/distribution/licenses 5620 SAM system architecture System components .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-6 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 Component communication General information The 5620 SAM component interfaces use industry-standard protocols for communication between servers, databases, NEs, and clients, as shown in Figure 1-2, “5620 SAM component communication” (p. 1-7). All 5620 SAM components communicate with all other 5620 SAM components using IPv4 or IPv6 exclusively; however, a 5620 SAM system can communicate with and manage a network using both protocols concurrently. Figure 1-2 5620 SAM component communication 5620 SAM system architecture Component communication .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-7 Servers and managed NEs Main and auxiliary servers send messages to the managed network in the form of SNMP, FTP, secure FTP, and SCP commands. A 5620 SAM main server also sends CLI commands using Telnet or SSH. • A main server uses SNMP to monitor and manage network performance, and to identify network problems. Main servers deploy configuration changes to NEs using SNMP. Auxiliary servers poll MIB performance statistics on the NEs, and collect call-trace data. The NEs use asynchronous SNMP messages called traps to notify the 5620 SAM of events. • The CLI of a managed NE is accessible from the client GUI using Telnet or SSH. • FTP and SCP are transport-layer protocols for transferring files between systems. The 5620 SAM uses the protocols to back up NE configuration data, collect NE accounting statistics, and download software to NEs. Main server and clients Client interfaces provide access to a 5620 SAM system and the managed network through a main server. A main server and clients communicate in the following ways: • GUI clients send requests to the server EJB session beans using Java RMI. • The GUI client update function uses HTTP or HTTPS for client software updates and file downloads. • 5620 SAM application clients use HTTP or HTTPS to communicate with the web service on a main server. • A web-based GUI client communicates through a browser using JNLP. • XML OSS clients send requests for processing by a main server, and subscribe to JMS topics to receive real-time event notifications. The messages between a main server and an XML OSS client are in XML/SOAP format, and are sent over HTTP or HTTPS. The JMS and the XML publisher service on a main server run in separate JVMs to support multiple concurrent client connections. See the 5620 SAM XML OSS Interface Developer Guide for more information about the messaging between XML OSS clients and main servers. • 3GPP OSS clients send CORBA or SOAP/XML requests over HTTP or HTTPS for processing by a main server. See the 5620 SAM Wireless OSS Interface Developer Guide for information about the messaging between 3GPP OSS clients and main servers. Main server and 5620 SAM database A main server communicates with a 5620 SAM database using a JDBC session over TCP. JDBC is a Java API for interworking with SQL relational databases. 5620 SAM system architecture Component communication .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-8 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 Main server and auxiliary servers A main server includes a mechanism for sending requests to auxiliary servers. An auxiliary server notifies the main server after it finishes processing a request. If the main server fails to send a request, or all auxiliary servers are unresponsive to a request, the main server raises an alarm. A Cflowd auxiliary server sends an initial OSS request to retrieve the network object model from a main server, and subsequently receives JMS event notifications about updates to the objects in the model. The Cflowd auxiliary server uses FTP or SFTP to forward the statistics to an OSS or third-party application. 5620 SAM integration with external systems The 5620 SAM can be integrated with external network management systems such as the 5620 NM. During 5620 SAM system installation, you can configure navigation from an external system for additional network monitoring. See the 5620 SAM Integration Guide and the 5620 SAM \| 5650 CPAM Installation and Upgrade Guide for more information about integration with other systems. System structure General information A 5620 SAM system has a readily adaptable, modular structure that incorporates a relational data model and employs distributed processing. Framework The 5620 SAM system elements are created using Alcatel-Lucent and third-party software, and are logically organized in a framework that has the following layers: • resource • integration • business • presentation • client The following figure shows the multi-layer model and the elements in each layer. 5620 SAM system architecture Component communication .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-9 Resource layer The resource layer includes the network of managed NEs and the 5620 SAM database. The database stores objects such as NE configurations, statistics, and customer service information. Integration layer The integration layer buffers resource-layer elements from the business layer. This layer contains the mediators, which communicate with equipment in the managed network, and the database adapter. The mediator components translate messages from the business layer into the SNMP, FTP, secure FTP, and CLI commands that are sent to the managed network. Messages that are received from the network are processed by the mediator components and passed to the business layer. The database adapter translates business logic requests into JDBC commands, and translates JDBC responses into Java business model objects. Figure 1-3 5620 SAM multi-layer model 5620 SAM system architecture System structure .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-10 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 Business layer The business layer contains the logic and data model for 5620 SAM functions. The business logic processes client requests, SNMP traps from managed NEs, and internal server events, and performs the appropriate actions on the managed network, clients, and data model, which maintains information about network objects and their relationships. To support the business layer, an application server provides Java EE services. Presentation layer The presentation layer buffers the application logic from the client layer. This layer contains several components. The web server receives SOAP/XML messages from OSS clients and passes them to the business layer. The application server handles EJB method invocations received from the GUI clients and returns the responses generated by the business-layer logic. The application server also forwards JMS event notification messages from the business layer to GUI and OSS clients. Client layer The client layer comprises the GUI, OSS, and web-based clients. The GUI client Java VM sends EJB RMIs to a main server. The OSS clients send XML/SOAP or 3GPP CORBA messages to a main server. Web clients use JNLP for portal access. Server data model The server data model represents the physical and logical elements of the network, such as equipment, customers, services, and statistics. The model also describes the relationships between objects, so allows operators to perform high-level operations that are propagated to child objects, as required. The object associations enable effective central management of large, complex networks. The 5620 SAM maintains in the data model a representation of the current managed network state, and incorporates changes as they occur. Changes that are initiated by NEs include event notifications such as fault traps and state changes; changes that are initiated by clients include object creation, deletion, and configuration updates. The changes are applied to the model, saved in the 5620 SAM database, deployed to the network as required, and reported to clients. Distributed server architecture The 5620 SAM server functions can be distributed across multiple physical or virtual stations in a standalone or redundant configuration. A main server and one or more auxiliary servers in the same 5620 SAM domain define a 5620 SAM server cluster; for example, a redundant 5620 SAM system has two server clusters. The auxiliary servers are members of only the cluster that contains the currently active main server, which is a standalone main server, or, in a redundant deployment, the primary main server. When the main servers in a redundant deployment change roles, for 5620 SAM system architecture System structure .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-11 example, when a 5620 SAM operator initiates a server activity switch, the auxiliary servers leave the current cluster and join the cluster that contains the new primary main server. The main server in a cluster is the network management engine that monitors the managed network and processes GUI and OSS client requests. It also directs the operation of the auxiliary servers and distributes the processing load as required. The distributed architecture is not visible to the GUI and OSS clients, which interact only with a standalone or primary main server. A main server sends new or updated operating information such as the 5620 SAM license capacity, redundancy status, or database credentials, to each auxiliary server as the information becomes available. Security General information A distributed system such as the 5620 SAM requires security at the session and other communication layers. A GUI or OSS client must provide user credentials for access to the 5620 SAM. You can protect the session credentials and messages using mechanisms and protocols that include the following: • HTTPS, as the application-layer transport for OSS clients • Telnet, SSH, SCP, and SNMPv3 with USM or VACM, at the application layer for communication between a main server and the managed network • SSL, at the presentation layer, between a main server and the GUI clients, OSS clients, auxiliary servers, and Cflowd auxiliary servers, and between the primary and standby main servers in a redundant deployment • NAT, at the network layer, between the following: – main server and single-user GUI client or client delegate server – main or auxiliary server and OSS client – main or auxiliary server and managed network • IP validation, at the network layer, between a main server and database The following figure shows the 5620 SAM components and the available security mechanisms. 5620 SAM system architecture System structure .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-12 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 Session management Effective session management requires authentication, authorization, and accounting, or AAA. Authentication is the verification of user credentials. Authorization is the assignment of access privileges to users. Accounting is the recording of user actions. A 5620 SAM operator can configure AAA functions using the local 5620 SAM security mechanisms, a third-party server, or both. • Local 5620 SAM authentication is performed using a local database of users and a local security scheme. • Supported third-party authentication servers are RADIUS, TACACS+, LDAP, SAM-L, and CSA, which run on separate platforms, and have separate user lists and administration processes. 5620 SAM user accounts consist of a user name, password, and an associated user group, scope of command, and span of control. User groups define user authorization levels, and control the level of access to objects such as equipment, customers, services, and alarms. A system administrator can limit the type of user access per managed NE; for example, allowing FTP access but denying console, Telnet, or SNMP access. Figure 1-4 5620 SAM security mechanisms OSS application JMS/SSL HTTPS HTTP User ID/password 5620 SAM client GUI User ID/ password SNMPv3 USM/VACM SNMPv1, v2 community string SSH1/2 or Telnet User ID/password FTP/SCP User ID/password Firewall Firewall JMS/SSL EJB/SSL User ID/ password User ID/password SSL Database session authentication User authorization and authentication requests RADIUS, TACACS+, SAM-L, or CSA Databases if redundant Managed network 18083 5620 SAM server Local security data 5620 SAM server if redundant 5620 SAM system architecture Security .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-13 Client sessions All client sessions require authentication. • The client GUI EJB sessions are secured by the session username and password. • Each OSS client message is authenticated using cached information from an authorization server. • JMS messages are secured by the OSS client user name and password. Database sessions The 5620 SAM database is accessible through a main or auxiliary server connection that is secured by a username and password. After each database update in response to a client request, the client activity log records the request information, which includes the name of the associated 5620 SAM user. Secure communication between a main server and a 5620 SAM database is available in the form of IP-address validation, which is configurable during 5620 SAM system installation or upgrade. Managed NE sessions A 5620 SAM server opens CLI, FTP, SFTP and SCP sessions on managed NEs. A managed NE uses a local security database, or a third-party service such as RADIUS or TACACS+, to perform AAA functions. SNMPv3 message authentication and authorization are handled by the USM and VACM mechanisms, which define the user authorization permissions. Older SNMP versions are authenticated using community strings. Each SNMP message is individually authenticated. Network transport security Transport-layer security is available to the network protocols that carry messages between 5620 SAM components. Main server and clients Communication between a main server and clients is performed using XML/SOAP, EJB, or JMS messages. • OSS clients use HTTPS to send XML/SOAP messages when SSL is enabled in a 5620 SAM system, and otherwise use HTTP. • GUI clients use the EJB interface, which can be secured using SSL. • JMS, which is used by GUI and OSS clients, can be secured using SSL. • In a secured redundant deployment, the standby main server acts as an SSL client of the primary main server. 5620 SAM system architecture Security .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-14 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 Servers and managed NEs A managed NE communicates with a main or auxiliary server using SNMP, FTP, or SCP. When SNMPv3 is used, an SHA or MD5 authentication key is included in each message and checked against the shared encryption key. SSH provides the security for a CLI session between a 5620 SAM GUI client and a managed NE. RSA encryption is available for communication between auxiliary servers and managed NEs. Contact Alcatel-Lucent support for information. Firewall support The 5620 SAM supports firewalls on all server interfaces; for example, between a main server and the auxiliary servers and GUI or OSS clients, and between a main or auxiliary server and the managed network. See the 5620 SAM Planning Guide for firewall and reserved TCP port information. Fault tolerance General information Fault tolerance provides system reliability by maintaining availability in the event of a component failure. 5620 SAM fault tolerance includes high availability using component redundancy. Deploying redundant 5620 SAM hardware and software components ensures that there is no single point of 5620 SAM system failure. Redundant physical network interfaces and points of network entry ensure that there is no single point of failure between the 5620 SAM system and the managed network. Redundant network paths, for example, in-band and out-of-band management, can help to prevent the isolation of a main server from the network in the event of a routing failure. Main server and database redundancy A redundant 5620 SAM system consists of a primary main server and an associated primary 5620 SAM database that actively manage the network, and a second server and database pair in standby mode. A 5620 SAM server and database pair can be co-located on one station or run on separate stations. The following figure shows a distributed 5620 SAM system deployed in a redundant configuration. 5620 SAM system architecture Security .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-15 See the 5620 SAM System Administrator Guide for more information about 5620 SAM redundancy. Main server redundancy Main server redundancy is achieved using clustering technology provided by a JBOSS application server on each main server. The primary and standby main servers regularly poll each other to monitor availability. Traps from the managed network are always sent to both main servers in order to avoid delays in the event of a server activity switch. If the primary server loses visibility of the standby server, it notifies the GUI clients. If the standby server loses visibility of the primary server, the standby server attempts to become the primary server by connecting to the primary database. 5620 SAM database redundancy 5620 SAM database redundancy uses Oracle Data Guard Replication in real-time apply mode to keep the standby database synchronized with data changes in the primary database. The supported fault-recovery operations are database switchovers and database failovers. A switchover is a manual operation that switches the primary and standby database roles. A failover is an automatic operation that forces the standby database to become the primary database when the primary database fails or becomes unavailable. Figure 1-5 Redundant 5620 SAM system Traps Traps Legend = Normal communication = Database replication = Monitoring of availability = Communication activated during failure Data 5620 SAM database 5620 SAM main server Primary Primary 5620 SAM database 5620 SAM main server Standby Standby 17903 5620 SAM GUI and OSS clients Managed network Network authentication Network authentication Network authentication 5620 SAM system architecture Fault tolerance .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-16 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 The primary main server polls each database to check the availability. If the primary or standby database is unavailable, the main server notifies the GUI clients. If both main servers lose contact with the primary database, a failover occurs and the standby database becomes the primary database. Auxiliary servers and 5620 SAM redundancy Auxiliary servers are passively redundant. They do not cause or initiate main server or database redundancy activities, but if a Preferred auxiliary server ceases to respond to requests from the primary main server and a Reserved auxiliary server is available, the main server directs the current and subsequent requests to the Reserved auxiliary server until the Preferred auxiliary server is again available. An auxiliary server communicates only with the current primary server and database. After a 5620 SAM redundancy activity such as a database failover, the primary main server directs the auxiliary servers to stop communicating with the former primary component and to begin communicating with the current primary component. Cflowd auxiliary servers and 5620 SAM redundancy Cflowd auxiliary servers, which collect AA accounting and AA Cflowd statistics, can be configured to transfer the collected statistics files to redundant destinations. For a higher degree of fault tolerance, you can configure two or more Cflowd auxiliary servers to collect statistics from the same set of NEs and transfer the statistics files to redundant destinations. Such a configuration ensures that the statistics collection and transfer continue uninterrupted in the event that a Cflowd auxiliary server and a transfer destination fail. Standards compliance General information The 5620 SAM system is designed using industry standards and open-standard interfaces that allow the system to interoperate with a variety of other network monitoring and management systems. The 5620 SAM system complies with the standards listed in the following table: Table 1-1 5620 SAM standards compliance Standard Description 3GPP 3rd Generation Partnership Project IRPs for CORBA R8 and SOAP/XML R8 Solution Sets draft-grant-tacacs-02.txt TACACS+ client draft-ylonen-ssh-protocol-00.txt SSH 5620 SAM system architecture Fault tolerance .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-17 Table 1-1 5620 SAM standards compliance (continued) Standard Description EJB Java EE Enterprise Java Session Bean version 2.3 HTML5 HyperText Markup Language 5, for 5620 SAM applications HTTP(S) HyperText Transfer Protocol (Secure) version 1.1 ITU-T X.721 SMI ITU-T X.734 Event report management function Java SE Java Standard Edition version 8 JBOSS EAP Java Bean Open Source Software Enterprise Application Platform version 6 JMS Java Message Service version 1.1 JSON ECMA-404 JavaScript Object Notation Data Interchange Format JS/ECMAScript 5 ECMA-262 ECMA Script Language Specification M.3100/3120 Equipment and connection models MTOSI Compliance of generic network objects, inventory retrieval, and JMS over XML RFC 0959 FTP RFC 1213 SNMPv1 RFC 1738 Uniform Resource Locators (URL) RFC 2138 RADIUS client 2618 RFC 3411-3415 SNMPv3 RFC 3416 SNMPv2c RFC 6241 Network Configuration Protocol (NETCONF) SAML SAM-L 1.1 SOAP W3C SOAP 1.2 TMF 509/613 Network connectivity model XML W3C XML 1.0 W3C Namespaces in XML W3C XML schemas Alcatel-Lucent considers the following standards in the 5620 SAM GUI design: • Sun Microsystems, Java Look and Feel Design Guidelines, Addison-Wesley Publishing Company, Reading, Massachusetts 1999. • ANSI T1.232-1996, Operations, Administration, and Provisioning (OAM&P)- G Interface Specifications for Use with the Telecommunications Management Network (TMN). 5620 SAM system architecture Standards compliance .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-18 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 • Telcordia (Bell Core) GR-2914-CORE Sept. 98, Human Factors Requirements for Equipment to Improve Network Integrity. • Telcordia (Bell Core) GR-826-CORE, June 1994, Issue 1, Section 10.2 of OTGR, User Interface Generic Requirements for Supporting Network Element Operations. • ITU-T Recommendation Z.361 (02/99), Design guidelines for Human- Computer Interfaces (HCI) for the management of telecommunications networks. • ETSI EG 201 204 v1.1.1 (1997-05), Human Factors (HF); User Interface design principles for the Telecommunications Management Network (TMN) applicable to the “G” Interface. • 3GPP 32-series R8 specification, published December, 2009. 5620 SAM system architecture Standards compliance .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015 1-19 5620 SAM system architecture Standards compliance .................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................... 1-20 5620 SAM 3HE-09816-AAAF-TQZZA Release 13.0 R6 Issue 1 November 2015
10735	https://math.stackexchange.com/questions/2804568/a-segment-tangent-to-the-incircle	geometry - A segment tangent to the incircle. - 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 A segment tangent to the incircle. Ask Question Asked 7 years, 4 months ago Modified7 years, 4 months ago Viewed 217 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. D D and E E are points on sides A B A B and A C A C of a triangle A B C A B C such that D E D E is parallel to B C B C and tangent to the incircle I I. Prove that D E≤A B+B C+A C 8 D E≤A B+B C+A C 8 This looks like A M−G M A M−G M, but I don't have a cube on D E D E. I tried using the triangle inequality like this: I rotated A B C A B C by 180∘180∘ around the incentre, then applied the triangle inequality to get A D+A E+B′E+B′K+C K+C L>D E+E K+K L A D+A E+B′E+B′K+C K+C L>D E+E K+K L Now, if this relation is true it should apply for all of D E,E K,K L D E,E K,K L. I'm not sure where to go from here. Please help. Hints will be appreciated. I just saw this, and it seems to be related, but different. geometry euclidean-geometry circles Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 1, 2018 at 19:07 nonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges asked Jun 1, 2018 at 16:59 DynamoBlazeDynamoBlaze 2,985 1 1 gold badge 17 17 silver badges 41 41 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. Ahh, it is much easier. Since Δ A B C∼Δ A D E Δ A B C∼Δ A D E we have Perimeter A D E Perimeter A B C=D E B C Perimeter A D E Perimeter A B C=D E B C Let a=y+z a=y+z, then x a+x=D E a⟹D E=a x a+x x a+x=D E a⟹D E=a x a+x and we have to check: a x a+x≤2 a+2 x 8 a x a+x≤2 a+2 x 8 which is easy to see is true. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 2, 2018 at 16:34 answered Jun 1, 2018 at 18:04 nonusernonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Using tangent segments we get this picture. Since Δ A B C∼Δ A D E Δ A B C∼Δ A D E we have d+e z+y=x−d x+z=x−e x+y(∗)d+e z+y=x−d x+z=x−e x+y(∗) Taking first two fractions we get d(x+y+2 z)+e(x+z)=x(z+y)d(x+y+2 z)+e(x+z)=x(z+y) and taking 1.st and 3.rd fractions we get e(x+y+2 z)+d(x+y)=x(z+y)e(x+y+2 z)+d(x+y)=x(z+y) so after substracting these 2 equations we get d z=y e d z=y e. Plugging e=d z/y e=d z/y in (∗)(∗) we get d=x y x+y+z d=x y x+y+z In the same manner we get e=x z x+y+z e=x z x+y+z Now we have to check this: 8 x(y+z)x+y+z≤2 x+2 y+2 z 8 x(y+z)x+y+z≤2 x+2 y+2 z which is equivalent to 4 a x≤(a+x)2 4 a x≤(a+x)2 is true (a=y+z a=y+z). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jun 1, 2018 at 17:26 nonusernonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 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 geometry euclidean-geometry circles See similar questions with these tags. 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10736	https://www.youtube.com/watch?v=HHyyZ8dy1-Y	Saxon Math 7th Grade - Lesson 110 - Simple & Compound Interest + Successive Discounts Pottorff Math 5140 subscribers 37 likes Description 2066 views Posted: 13 Apr 2017 via YouTube Capture 9 comments Transcript: all right seventh grade lesson 110 this is on simple interest and compound interest and then also successive discounts now let me explain something to you this is actually something you'll use in your life um every day you're going to hear about principal and it's not the principal of the school principal and interest when we're talking about math or numbers okay so a principal um is the amount that you put into the bank so for example let's say I said I wanted to put $100 into the bank okay that was my principal payment the interest is what the bank will pay me so if it's a 10% interest I take 10% of that and that's what I gain every year okay do you see what I'm saying now you can also if I buy a house then I pay the principal but I also have to pay the interest so not only do you have to do principle and interest when you buy things but also when you deposit things so it can go either way whether you're giving or taking you're going to have to do principle and interest that's the first amount that's the amount after we get the interest okay so for example if I put $100 into the bank into a savings account and it was going to gain 10% interest every year well I did 10% it's actually 6% in the book but 10% is easily to find for 100 okay so 10% of 100 you basically just move your decimal ones quite over 10% of1 is $10 so after my first year of banking I would get 10 extra dollars because 10% of 100 is $10 so I would have after the first year how much money $110 okay so that would be so if I did uh 10% of $100 help me figure this out how would I actually do this problem what is OV in math times times and then you can never leave 10% as a percent in a problem you have to make it into a decimal so what would this change to 10% will become 10 cuz you just go two spots over okay so if I did 10% or 10 times that then I would equal $10 okay this is $100 and I take 10% of that it's going to give me $10 I'm not actually going to do the work but that's what it's going to give me so that's why I added $10 to this 100 so after the first year year it be $110 you get it all right now after the second year we would add $10 more doar and what would it be 120 120 after the third year we would add 10 more dollar and it' be 130 so because I gave the bank $100 and it accred 10% interest each year how much money do I have at the end 130 so how much did I gain 30 30 you see how that works just because I left it in the bank for three years okay that is what we call simple interest interest simple interest all right the next thing you're going to learn about is compound compound interest now I'll stay with me the very first year of compound interest is the same you see this I made $10 because 10% of $100 is $10 so $10 plus 100 makes 110 but the second year Everything Changes third year Everything Changes fourth year and Etc with compound and let me tell you why now that's my new new number and now I'm going to take 10% of that number not of the original number 10% was just $10 every time but now let's find what 10% of this is so what I have to do for compound is 110 times 10% so 10% changes to10 okay and 10% of this is going to make $11 not 10 now it doesn't seem like that big of a deal so I'm going to add $11 instead of 10 because 10% of this number this new number so now it's 121 now it's 121 okay now let's do the third year okay again with compound interest you take the new amount you don't just keep adding $10 every time you find out what is 10% of this 10% of of changes to times 10% of that okay again of and oh sorry of of math is becomes times and then 10% becomes .100 so what is 10% times that well if I did the math basically what ends up happening is um if I did 121 time actually let's do it on our calculators and make it easier so whoops so actually write this sound 121 times and because we're doing 10% 10% changes to 12.1 12.1 okay so we're not just doing $10 no we're doing 121 which is 12.10 okay so I have $121 and I'm going to add 12.1 almost and so I'm going to add this and that equals that becomes 2 + 1 is three okay now we end up with 13310 after the third year now there is a really really really simple way of doing this so I'm going to show you something okay so I'm going to leave this up for just a minute and show you something it's only adding $3 though right didn't pretty much yeah simple interest but if you'll you'll see after about 15 20 years it makes a huge difference so look at the book on page 763 you tell me if you can see this on the B can you see that a little bit closer okay that's good okay do you see this yes on the left side it says the number of years 3 10 20 30 40 50 50 years you left it in the bank okay with simple interest let's say I put $118 in the bank after 10 years I'd give 160 after 30 years I'd have 280 after 50 years I'd have $400 pretty good but I left it in there for 50 years compound I took the same amount 118 put it in and each year it accumulated as you can see in 50 years I would have gained $1,842 versus 400 by doing compound okay so taking 10% or 6% or whatever percent of the new number the new amount every time gets bigger and bigger you pay the bank to get compound yourself well you have to sign up for do I want simpler or compound why would you do simple I don't know there's really no reason okay is it cheaper well a lot of times probably people the only reason they do simple is to keep it in there for short terms like I want to keep it in there for two years yeah you know but if I'm doing something long term definitely better do compound okay you see that all right now I want you to see something really quickly all right um do you see how um remember how I told you we were going to be doing 100 time10 10% and that tells me what what does this answer tell me if I take 100 10 % which is .10 what does that tell me it tells me the amount that is added to the $100 cuz watch this 100 time10 now watch this 0 0 0 0 one one two one two so how much is added 10 bucks 10 bucks okay so $100 if I take 10 10% of $100 how much is 10% of $100 what what is 10 this is $100 and if I take 10% of $100 what is 10% of $100 $10 yeah 10 $100 times 10% remember 10% of of in math means time okay so watch what I'm going to do what you see me add that one mhm Watch What Happens now let's do the math one two one two what does it tell me 110 it tells me the final price Isn't that cool okay that's kind of hard to grasp okay but watch this watch what happens 100 time 1.10 1.10 equals $110 $110 see that watch this I'm just going to hit equal what's the second year 121 121 I'm going hit equal again it's kind of cool that calculators can do this they they're taking the exact same thing times this and doing it over and over and over so I'm going to hit equal again which does that over and what do I get 1331 133.5 161.5 okay 16 1.05 okay let's do 10 years 5 6 7 8 9 10 29374 okay let's figure out what it's going to be after 20 that's 10 11 12 13 14 15 16 17 18 19 20 67274 999 okay 672 7499 so I'm going to change it to 75 all right we started with $100 and when we did compound interest over 20 years I end up with 600 now could you imagine if I put in a th000 okay so be like 6,000 yeah I mean it would be astronomical okay so look what these people did so let's take uh because we're doing 10% that's my 10% at 0.10 but then I add this one and that kind of multiplies this already so stay with me let's do a th000 this time 1 0 0 0 a th000 times we're going to do 10% so 10% is 0.100 but I'm going to add that one in the beginning 1.10 and that kind of gives me the answer automatically so that's what it would be after one year so I had a th000 now I have 1100 okay now I'm going to hit Let's do let's do 20 years so I started off with a th000 and I'm going to do 20 years ready that was one year 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 dang 6,727 50 $49.99 so so it would it would uh round to 50 so I put in $11,000 and after 20 years of keeping it in the bank at 10% interest I end up with this is't that cool how that works do you understand it yep all right so now let's look at example two and help me discover the answer to this one listen use a calculator to find the value of 12 Years with what of a $2,000 investment they put that in the very beginning but it doesn't earn 10% it earns 7 and a half% now stay with me how do you do a half that's what we're to do 7 and a half is 7.5 but that's still 7.5% so it's 7 y because watch this 50% is we move it two spaces over .5 so 7.5% will become we move it two spots over 075 do you get that okay okay so now stay with me all right if I'm doing almost if I take 2,000 and I'm doing 75% of that yeah okay but did you add your one oh because if you don't add your one what this one does is we add the 2,000 into it because 2,000 1 is 2,000 so then it's the 2,000 plus whatever this is do you get it okay so let's do that 2000 1.75 1.75 what' you get and then I do it times 12 years yeah well tell me that first one first year is what 2150 okay now let's do it 12 times so that was 1 2 3 8 9 10 11 12 4,763 okay so after 12 years $476 and 5 55 but it has a nine after so 56 very good is't that cool mhm so how that works so you gained what 2763 just by keeping it in the bank for 12 years okay so that's how that works do you see the importance of that one because if you don't put that one it only tells you how much yeah it was like seven point if you if you don't use that one then it only is telling you how much to add to this but when you add that one it automatically adds this with it I know that may be confusing that's actually what happens all right feel like you got it all right now this next one stay with me all right this one may be a little bit confusing because here's what it says calculate the interest earned on an $88,000 deposit holy cow right okay calculate the interest earned on an 8,000 deposit but they're only going to keep it in there for n months now stay with me for you came Cal once well they tell me that the annual interest rate is 6% so I'm just going to write down everything okay and but how many months is this for one year one year which is 12 12 months stay with me almost so yes you're going to have yes you're right but stay with me okay so let's go and figure out what it is for the whole year okay CU that's going to help us so let's do 8,000 6% and what would 6 right almost 50% would become one two there six would be6 06 so time6 480 oh okay but let's go on and add the one well no no no let's don't because well okay let's do it without what is it without 480 480 so so that's the amount I needed to add to it so it would be 8 and 8,000 + 480 would be 8480 so if I did 8,000 times 1.06 go to do that it's 8,40 it equals that right I just went on and added it okay good now but we don't actually want to do the one on this one and let me tell you why because this is the amount that he's going to earn after keeping it in the bank for 12 months but he's not going to keep it in the bank for 12 months he's only doing it for nine so let's figure out what he makes each month so okay 480 ID 12 would be um 4 4 40 okay so then if he got $40 a month added to this 8,000 and he did nine of them 40 40 9 which is 360 so how much did he get added to that so it's 8,36 360 okay so what did they ask for the final price calculate the interest uh what he earned so they they don't want to know what is the total they just want to know what he earned and he earned 360 very good good job on that you figured that one out good all right now that's the end of that portion okay and now we're going to learn about successive discounts okay and it's somehow it's somewhat related to compound interest okay but here's what I want you to do you're going to draw a ratio box this is our last thing and you're done okay you only one problem with this so stay with me draw this on your paper remember the ratio box MH you're going to be doing percent and actual count okay go and draw that on your paper okay I got it now when you're learning about sales price has gone up or sales price has gone down or uh this shirt is is 70% off when you buy it things like this this is what you're going to draw okay now this it's going to have three boxes so draw that on yours okay it's going to have three boxes and here's what you're going to write whenever you hear things like that this shirt is 75% off or this Appliance went up 30% or whatever this is what you're going to draw the first one is going to be original original I'm just putting a r okay original and then the middle part is going to be how much it goes up or down so I'm going to put the change um plus or minus and then the last one is the new price or the new uh percent okay you staying with me so now listen to this problem an appliance store reduced the price of a $400 washing machine I'm just going to write down some key information it's $400 and they're going to reduce it reduce change 25% okay all right and then when the washing machine did not sell at this price it reduced it 20 more 4% that's 45 right well yes that's one way you can do it um okay so let's figure out what we're going to do here okay always always always your original price or your original percent will always be 100% okay okay all right that's the most important thing that you can write on these and then the other one is going to be 4 right the actual count original right right so the original price was 400 good job okay then the store reduced it 25% but then it didn't sell so they reduced it 20 more so how much did it reded it reduced now so that means it reduced which is minus 45% remember this is 100% And so this is 45% so what is our new percentage that we're paying 45 from 100 is 55 okay 55% all right what are they asking what was the final price the clearance price of the washing machine guess what we're going to do we don't know what the washing machine price remember the actual count the new price got it so you remember what to do on these you go you take the information you know everything about in a row we know both of these and the one we want to know about the row that we need to help and guess what you're going to do you're going to put them on top of each other 100 over 55 = 400 over w w means washing machine now do you remember what to do when you have equals cross multiply so now this is a w 100 which is 100 W equals the other one's 2 what is it 22,000 22,000 or 2200 3 22,000 22,000 okay so 400 55 was 22,000 good and now I'm trying to get my w by itself okay so times 100 remove it becomes divided by 100 so tell me what w is/ 100 uh 220 220 all right now I want to check something because I'm wondering if doing it this way if we should have figured out the answer to this first then taking oh we should have because we're taking 20% of the new price a but that was good thinking yeah I actually thought it might work too but we're actually reducing it % of the new price after we took 25 away ah you see how we messed up yeah okay so we're not going to do 20 set 45 right there we're going to start with 25 that would have been easier though yeah if they originally would have just done 45% this would be correct yeah okay so let's take 25% and minus that so then it becomes 75% okay so now instead of this being 55 it's going to be 75 so do the math for me 400 75 tell me what that is uh 30,000 30,000 okay and then W 100 is 100 W and then I'm going to move the times 100 so I'm going to divide by 100 so what is my w equal oh 300 300 now that's our new price that's what we did wrong 300 now we're going to reduce it 20 more from the original price per se okay and what is 100 - 20 cuz we're reducing itus is 80 so now help me out with this one we're going to 100 over 80 = 300 W all right help me do the math 300 80 so 300 80 is 24,000 24,000 and then uh 100 w all right and so then we're going to get W by itself so times 100 goes across it becomes divid 100 240 240 it's taking off that final price every time okay final price 240 okay that's how you would do that and that's less than 110
10737	https://pmc.ncbi.nlm.nih.gov/articles/PMC11828705/	Vascular Loop Syndrome As the Etiology of Abducens Nerve Palsy: A Case Report - 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 Advanced Search Journal List User Guide New Try this search in PMC Beta Search View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Cureus . 2025 Jan 15;17(1):e77483. doi: 10.7759/cureus.77483 Search in PMC Search in PubMed View in NLM Catalog Add to search Vascular Loop Syndrome As the Etiology of Abducens Nerve Palsy: A Case Report Riyaa Rajesh Riyaa Rajesh 1 Department of Biology, University of Southern California, Los Angeles, USA Find articles by Riyaa Rajesh 1,✉, Rahul Naveen Rahul Naveen 1 Department of Biology, University of Southern California, Los Angeles, USA Find articles by Rahul Naveen 1, Warren T Anderson Warren T Anderson 2 Department of Radiology, University of Cincinnati College of Medicine, Cincinnati, USA Find articles by Warren T Anderson 2, Ryan Nolan Ryan Nolan 3 Department of Interventional Radiology, University of Nevada Reno School of Medicine, Reno, USA Find articles by Ryan Nolan 3, Rajesh Rangaswamy Rajesh Rangaswamy 4 Department of Neurointerventional Radiology, Reno Radiological Associates, Reno, USA Find articles by Rajesh Rangaswamy 4 Editors: Alexander Muacevic, John R Adler Author information Article notes Copyright and License information 1 Department of Biology, University of Southern California, Los Angeles, USA 2 Department of Radiology, University of Cincinnati College of Medicine, Cincinnati, USA 3 Department of Interventional Radiology, University of Nevada Reno School of Medicine, Reno, USA 4 Department of Neurointerventional Radiology, Reno Radiological Associates, Reno, USA ✉ Riyaa Rajesh rajeshriyaa@gmail.com ✉ Corresponding author. Accepted 2025 Jan 15; Collection date 2025 Jan. Copyright © 2025, Rajesh et al. This is an open access article distributed under the terms of the Creative Commons Attribution License CC-BY 4.0., which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. PMC Copyright notice PMCID: PMC11828705 PMID: 39958021 Abstract Vascular loop syndrome, characterized by arterial loops or kinks causing compression of cranial nerves, can lead to a range of secondary side effects conditional on vascular morphology. This etiology has made standardization of treatment difficult, as confounding factors may take precedence over more interventional treatment. This case seeks to illustrate how vascular loop syndrome, specifically a dolichoectatic vertebrobasilar artery, can lead to left abducens nerve palsy. Here we present an 82-year-old female with reports of diplopia, who was seen previously for sixth nerve palsy. Imaging revealed no acute hemorrhage, restricted diffusion, hydrocephalus, intracranial space-occupying lesion, abnormal volume loss, vasogenic edema, mass effect, or abnormal CSF signal. Magnetic resonance angiography demonstrated an indentation of the left lower brainstem by the tortuous basilar artery, indicating that this vasculature may be abutting the abducens nerve exit zone. We demonstrate a case of vascular loop syndrome as an etiology of abducens nerve palsy and diplopia, raising awareness of these findings on imaging and encouraging the consideration of this etiology in radiological evaluation algorithms. Keywords: abducens nerve palsy, diplopia, neuroradiology, vascular loop syndrome, vertebrobasilar artery Introduction Abducens, or sixth cranial nerve, palsy is a condition characterized by oculomotor paralysis of the lateral rectus muscle . Etiology for this condition includes impairment or interference of the abducens nerve at any point along its intracranial course . Due to the role of the abducens nerve in somatic control of the lateral rectus muscle, responsible for both ipsilateral eye abduction and partial contralateral eye adduction, damage to this nerve may result in acquired esotropia [1,2]. This misalignment is derived from an inability to abduct the affected eye to counterbalance the antagonistic activity of the medial rectus muscle in controlling eye adduction . Consequently, the affected eye experiences misalignment such that the esotropic deviation is most severe when efforts are made to either abduct the affected eye or fixate at a distance [1,2]. This may manifest as diplopia minimized by strabismus or turning of the head, as well as an inability to abduct the affected eye past the midline . The abducens nerve begins at its nucleus in the dorsal, caudal pons just ventral to the fourth ventricle and exits the brainstem at the junction between the pons and medulla . It then courses both superiorly and anteriorly through the subarachnoid space and over the petrous apex of the temporal bone through a fibrous sheath where it is anchored . The sixth cranial nerve then enters the cavernous sinus, followed by the superior orbital fissure, before innervating the lateral rectus muscle and partially innervating the contralateral medial rectus muscle [1,2]. Due to its long, oblique intracranial course, the abducens nerve is prone to compression or stretching damage that predisposes to abducens nerve palsy . Abducens nerve palsy can be congenital or acquired from trauma, viral or bacterial infections, inflammation, neoplasms and gliomas, aneurysms, variations in intracranial pressure that introduce vascular compromise, microvascular ischemia, fascicular and peripheral lesions, multiple sclerosis, and neurosurgical intervention . Additionally, any etiology that results in aggravation or impairment along the intracranial course of the abducens nerve can manifest as abducens nerve palsy, and its presentation can vary depending on the location at which the cranial nerve is compromised . Onset patterns and associated symptoms can be instrumental in determining etiology. A sudden onset is typically indicative of vascular etiology, a progressive onset suggests a compressive etiology and a subacute onset is characteristic of a demyelinating process [1,2]. In most cases, computed tomography (CT) or magnetic resonance imaging (MRI) reveals some abnormality from which abducens nerve palsy is derived, such as a space-occupying lesion or vascular compromise, that leads to nerve impingement or damage; in other cases where no such abnormality can be found, hypertension or diabetes mellitus is typical . However, in select cases where none of these pathological conditions result in abducens nerve palsy, patients present with this condition secondary to neurovascular compression of the abducens nerve . This disorder, known as vascular loop syndrome, occurs when abnormal movements or pain resulting from the impingement of a cranial nerve by a blood vessel [4,5]. We report an 82-year-old female who presents with diplopia secondary to left abducens nerve palsy and vascular loop syndrome. Case presentation The patient is an 82-year-old female with a history of rheumatoid arthritis, lung cancer, abducens nerve palsy of unknown etiology, and cataracts who presented to an outpatient eye clinic for evaluation of macular degeneration and diplopia that began one month prior. The patient provided verbal and written consent for the publication of this case study. She was subsequently sent to the emergency department to rule out possible transient ischemic attacks. The patient reported right arm numbness and denied headache and abdominal pain. Physical exam was negative for facial droop, slurred speech, and motor weakness. To further evaluate the patient’s diplopia, MRI with and without contrast was ordered. No acute hemorrhage, ischemic lesion, or mass effect was observed on MRI. The left globe was mildly abducted in relation to the right (Figure 1). No significant orbital mass, intraconal or extraconal mass lesions, or abnormal enhancement was observed. Extraocular muscles were normal in caliber and symmetry. The optic nerves, chiasm, and tracts were unremarkable. Figure 1. T2-weighted MR demonstrating left eye adduction due to vascular loop syndrome (arrow). Open in a new tab Following this finding, the patient decided they wanted to defer microvascular decompression to alleviate her symptoms. The decision to avoid surgery included the risk of complications and advanced age. The patient will be followed by ophthalmology for any changes or worsening of the adduction defect. Upon brainstem evaluation, an indentation of the left lower brainstem by a tortuous basilar artery, abutting the left abducens nerve exit zone, was noted (Figure 2). Apart from certain areas of gliosis in the brainstem indicative of chronic microvascular ischemia typically seen in old age, no other abnormal lesions were observed along the intracranial course of the left sixth abducens nerve. Figure 2. A) Axial CT scan shows the basilar artery intending the left sixth nerve at the exit zone (the vertical arrow shows the basilar artery and the curved arrow shows the left sixth nerve). B) The FIESTA sequences show the basilar artery with indentation of the brain stem at the exit zone (the up arrow shows the basilar artery, the down arrow shows the normal right sixth nerve, and the left arrow shows the compressed left sixth nerve). Open in a new tab Discussion In this case report, we present a patient with diplopia who was discovered to have compression of her left abducens nerve due to a tortuous basilar artery. Vascular loop syndrome is characterized by vascular compression of cranial nerves, some of which have been frequently reported in the literature, including the vestibulocochlear and trigeminal nerves, but rarely involve the abducens nerve [6-8]. For example, Teh et al. describe this etiology as a cause of vertigo, while O'Brien et al. describe this as an etiology of tinnitus, chronic headache, and sensorineural hearing loss [6,7]. Furthermore, Aldarwish et al. describe vascular loop syndrome as a possible cause of facial pain due to trigeminal nerve compression . Due to the involvement of various arteries, such as the basilar or anterior inferior cerebellar artery, it is possible for vascular loop syndrome to affect various cranial nerves . Although this etiology is not among common differential diagnoses for a patient with neurological symptoms, it is important to consider this etiology when more likely explanations are failing to provide answers for the patient’s clinical presentation . There are six common etiologies for abducens nerve palsy, the first being brainstem syndrome, which occurs due to a lesion in the posterior fossa [8,9]. The lesion can be inflammatory, compressive, or ischemic and oftentimes includes the fifth, seventh, and eighth cranial nerves. Second, elevated intracranial pressure syndrome results in the displacement of the brainstem, causing the sixth nerve to stretch as it exits the pons. Furthermore, contact between the sixth nerve and the end of the petrous pyramid makes it vulnerable to pathologic processes, otherwise known as petrous syndrome . The cavernous sinus syndrome occurs as the sixth nerve goes through the cavernous sinus. It is associated with the third, fourth, and fifth cranial nerves, the carotid sympathetic plexus, and the internal carotid artery. In this syndrome, dysfunction occurs at either of these structures . The orbital syndrome of the sixth nerve occurs in conjunction with congestion in conjunctival vessels. Oftentimes, it can be hard to distinguish if the third, fourth, or sixth cranial nerves are affected. Last, isolated sixth nerve palsy syndrome occurs in patients who have lateral rectus weakness and no other symptoms indicated in medical history. Risk factors include hypertension and diabetes mellitus, and it is important to rule out the other five syndromes before diagnosing a patient with isolated sixth nerve palsy . Vascular loop syndrome affecting the abducens nerve often manifests as horizontal diplopia, particularly when viewing objects at a distance. This symptom results from ipsilateral lateral rectus muscle weakness, impairing the ability to abduct the affected eye. Some patients adopt compensatory head movements to mitigate diplopia [1,2]. Additional symptoms may include nausea, headache, facial pain, vision loss, and numbness. Variations in symptomatology often depend on underlying etiologies such as intracranial pressure. Elevated intracranial pressure may produce vomiting and eye pain, while cerebrospinal fluid leaks causing low intracranial pressure can mimic high-pressure symptoms, including sixth nerve palsy [1,2]. Microvascular decompression is a potential treatment for patients with vascular loop syndrome due to abducens nerve palsy [10,11]. This procedure involves relieving the vascular compression of the abducens nerve via separation of the vasculature from the exit zone of the nerve . While there has been documented success of this procedure for abducens nerve palsy, this procedure is generally indicated for younger patients who are severely limited by the ocular deficits of this condition . In the case of our patient, she deferred treatment to avoid potential complications. When cases of abducens nerve palsy secondary to vascular loop syndrome are identified, it is necessary to engage in dialogue with the patient to set expectations on the lack of durability and risk of potential complications of surgical treatment for the condition [12,13]. Conclusions Ultimately, brainstem imaging identified a tortuous basilar artery compressing the exit of the abducens nerve.Vascular loop syndrome causing compression of the left abducens nerve should be considered as a potential cause of unilateral abducens nerve palsy. This diagnosis is particularly relevant when common etiologies fail to account for the patient’s symptoms. This etiology should be added to the radiologic search pattern for an abducens nerve palsy that is not due to another apparent cause.Prompt recognition and targeted evaluation of this condition are critical for guiding appropriate management, which may vary depending on the patient, age, comorbidities, and goals. Disclosures Human subjects: Consent for treatment and open access publication was obtained or waived by all participants in this study. Conflicts of interest: In compliance with the ICMJE uniform disclosure form, all authors declare the following: Payment/services info: All authors have declared that no financial support was received from any organization for the submitted work. Financial relationships: All authors have declared that they have no financial relationships at present or within the previous three years with any organizations that might have an interest in the submitted work. Other relationships: All authors have declared that there are no other relationships or activities that could appear to have influenced the submitted work. Author Contributions Concept and design: Ryan Nolan , Riyaa Rajesh, Rahul Naveen Drafting of the manuscript: Ryan Nolan , Riyaa Rajesh, Rahul Naveen, Warren T. Anderson Critical review of the manuscript for important intellectual content: Ryan Nolan , Riyaa Rajesh, Rahul Naveen, Rajesh Rangaswamy Acquisition, analysis, or interpretation of data: Riyaa Rajesh, Rahul Naveen, Warren T. Anderson, Rajesh Rangaswamy Supervision: Rajesh Rangaswamy References 1.Supernumerary abducens nerves: a comprehensive review. Wang JM, Edwards BA, Loukas M, Oskouian RJ, Tubbs RS. World Neurosurg. 2018;112:39–45. doi: 10.1016/j.wneu.2017.11.052. [DOI] [PubMed] [Google Scholar] 2.Graham C, Mohseni M. StatPearls [Internet] Treasure Island (FL): StatPearls Publishing; 2023. Abducens nerve palsy. 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10738	https://www.myengineeringbuddy.com/blog/act-math-test-practice-solve-faster/	Skip to content ACT Math Test Practice Questions: How to Solve Them Faster and More Accurately By Vikram Singh\|Last Updated: September 4, 2025\| My Students Experience With ACT Math What to Expect in the ACT Math Section Why This Structure Matters How to Solve ACT Math Questions Faster & More Accurately Why You Should Spend 70% of Your Time on These Topics Practice Timed Sets of 10–15 Questions Why These Strategies Work FAQs About ACT Math Test Practice Questions This ACT Math test challenges students to complete 60 questions in just 60 minutes. This time constraint makes speed and accuracy essential skills for success. Check Out: Get Private 1 on 1 Online Math Tutor If you’re aiming for a top score or just to boost your math confidence, you need to practice smartly.In this blog, we’ll break down the tips and strategies for ACT Math Test Practice Questions and teach you how to improve your accuracy without burning out. My Students Experience With ACT Math Being an experienced educator, I’ve guided hundreds of students through their ACT preparation journey, and if there’s one thing I’ve consistently observed, it’s this: most students don’t struggle because they lack intelligence; they struggle because they lack practice. For example, one of my students who scored 22 in ACT Math. After just 8 weeks of focused preparation, personalized guidance, and regular ACT Math Test Practice Questions, they scored a 30, a remarkable 8-point improvement. The transformation took place because of mastering concepts, refining speed and practicing hard. What to Expect in the ACT Math Section You need to clearly understand what the ACT Math section is all about before directly going into ACT Math Test Practice Questions. Knowing the structure, question types, and topics not only helps reduce anxiety but also allows you to plan a targeted and efficient prep strategy. Here’s a complete breakdown of the section: Total Number of Questions: 60 There will be 60 multiple-choice questions in the ACT Math section. Each question has five options, and only one is correct. In school math tests, we used to get simple questions but here the questions will require recognizing and solving them quickly. Time Allotted: 60 Minutes You will get 60 minutes to complete 60 questions, which means 1 minute per question and that includes reading, solving, and writing in your answer. This tight time frame is why learning to solve ACT Math Test Practice Questions faster and more accurately is absolutely essential. For tougher questions (especially in geometry or trig), you might need more than a minute, so you’ll need to balance your time by answering easier questions more quickly. Topics Covered in ACT Math The ACT Math section covers a wide range of topics from basic arithmetic to more advanced concepts like trigonometry. Here’s how it’s generally divided: 1. Pre-Algebra and Elementary Algebra (~20–25%) Operations with whole numbers, decimals, fractions Factors, multiples, ratios, proportions Integer operations, exponents, roots Linear equations and inequalities Basic word problems They are usually the easiest and fastest questions to solve, so getting them right is important. 2. Intermediate Algebra (~15–20%) Quadratic equations Rational and radical expressions Systems of equations Functions and their notations Absolute value and inequalities This section tests your ability to manipulate algebraic expressions and solve equations logically. 3. Coordinate Geometry (~15–20%) Graphing linear equations and inequalities Slope and distance formula Midpoints, points of intersection Conics (parabolas, circles, ellipses) Expect to see graph-based questions that require both visual analysis and algebraic solutions. 4. Plane Geometry (~20–25%) Properties of angles, triangles, and circles Area, perimeter, and volume Pythagorean theorem Similarity and congruence 3D shapes and spatial reasoning Geometry questions often contain diagrams, and success here depends on recognizing key formulas quickly. 5. Trigonometry (~5–10%) Sine, cosine, and tangent Radian measure Trig identities and equations Right triangle relationships Trigonometry is typically clustered toward the end of the test and can be challenging if you haven’t reviewed the fundamentals. The good news? There are shortcuts and patterns you can learn through consistent ACT Math Test Practice Questions. Calculators Are Allowed But Use Them Wisely The ACT allows the use of calculators for the entire math section. However, not every question needs one. In fact, relying too heavily on your calculator can actually slow you down. It’s important to know when to use the calculator for long calculations or verifying answers and when to solve mentally or on paper. Some students even make more mistakes when rushing through calculator inputs. Practicing under real test conditions with a calculator helps you learn to use it strategically, not just reflexively. Why This Structure Matters Understanding this breakdown allows you to: Prioritize your study time based on the topics you struggle with most Develop pacing strategies so you’re not rushed during the test Approach practice questions more purposefully, especially under timed conditions The more familiar you are with these types of ACT Math Test Practice Questions, the more confident you’ll become when the real test day arrives. How to Solve ACT Math Questions Faster & More Accurately Let’s break it down into actionable strategies that make a real difference: Master the Most Tested Concepts One of the biggest mistakes ACT test takers make is thinking that every topic is important. But in reality, the test is designed to reward students who focus on the right areas and use effective strategies. That means focusing on the most frequently tested topics, the ones that appear again and again across practice sets and real exams. As someone who has worked with hundreds of students preparing for standardized tests, I’ve observed that those who streamline their prep by concentrating on high-yield topics tend to improve much faster. For instance, one student I mentored moved from a 24 to a 31 in ACT Math in just under two months. The secret? Consistent practice with targeted ACT Math Test Practice Questions based on the most common question types. Let’s break down these essential concepts and why they deserve the bulk of your attention. Linear Equations and Inequalities These are among the most fundamental and frequently tested areas on the ACT. You’ll need to solve for variables, graph linear equations, and interpret linear models. Many problems also appear in the form of word problems—testing your ability to translate a real-life scenario into a mathematical equation. How to Practice: Use ACT Math Test Practice Questions that involve setting up and solving equations from word problems, interpreting slope and intercepts, and manipulating inequalities. Systems of Equations Expect to solve systems with two variables using methods like substitution and elimination. You may also be asked to analyze graphical representations of systems to determine the number of solutions. How to Practice: Focus on practice problems that mix algebraic and graphical interpretation. It’s important to not just know how to do the math but also to really understand what different types of systems actually mean. Functions Functions play a big role in the ACT Math section. You’ll be tested on evaluating functions, understanding domain and range, graphing simple functions, and interpreting function notation. How to Practice: Look for ACT Math Test Practice Questions that involve evaluating functions at given values and interpreting graphs of linear and quadratic functions. Many students overlook the importance of understanding function notation, which is a simple area to score high. Right Triangles and Trigonometry Trigonometry accounts for about 7–10% of the ACT Math section, primarily through questions involving right triangles. You’ll need to apply the Pythagorean Theorem, understand special right triangles (like 30-60-90 and 45-45-90), and use basic trigonometric ratios like sine, cosine, and tangent. How to Practice: Start by memorizing key formulas and ratios. Then use practice sets focused on word problems and geometry-based questions that require applying these formulas under time pressure. Ratios and Proportions These often show up in data-based word problems and questions involving geometry or real-life scenarios such as scale drawings or recipes. You need to be comfortable working with part-to-part and part-to-whole ratios, and with setting up proportion equations correctly. How to Practice: Select ACT Math Test Practice Questions that require careful unit analysis, which is where many students tend to make careless mistakes. Practice translating ratio problems into mathematical expressions quickly and accurately. Basic Statistics: Mean, Median, and Mode You should know how to calculate and interpret the mean, median, mode, and range. These concepts often appear in data sets and tables that require careful reading and basic arithmetic skills. How to Practice: Focus on questions that include lists, bar graphs, or data tables. It’s not enough to just know the formulas—you must be able to apply them correctly under timed conditions. Why You Should Spend 70% of Your Time on These Topics These high-frequency topics are the core of the ACT Math section. Mastering them gives you the ability to answer a majority of the questions correctly, even before you touch the more advanced or less common topics. This is the foundation upon which higher scores are built. Investing 70% of your practice time in these areas will help you: Improve your accuracy Increase your speed through repetition Build confidence for test day ACT Math is as much about strategy as it is about skill. By narrowing your focus to the most tested areas and using high-quality ACT Math Test Practice Questions, you’ll make the most out of your study time and see real, measurable results. Practice Timed Sets of 10–15 Questions When preparing for the ACT Math section, many students feel that taking full-length practice tests is the only way to build stamina and track progress. While full-length tests are important closer to the actual exam date, they shouldn’t be your only method of practice, especially in the beginning stages of preparation. A more effective and sustainable strategy is to practice in timed mini sets of 10–15 questions at a time. This technique is especially valuable when using targeted ACT Math Test Practice Questions, as it allows you to isolate specific concepts and question types while training your brain to work under time pressure. Why Smaller Timed Sets Work Better: Manageable Time CommitmentInstead of dedicating a full hour to a practice session, smaller timed sets can be completed in just 15–20 minutes. This makes it easier to fit practice into a busy schedule, whether you’re preparing after school, during a break, or on weekends. Focused ImprovementWith mini sets, you can concentrate on specific skill areas like algebra, geometry, or trigonometry. This helps you identify exactly which types of questions slow you down or lead to mistakes. Builds Speed and AccuracyACT Math is not just about getting the right answer—it’s about getting the right answer quickly. Practicing 10–15 ACT Math Test Practice Questions under strict time limits helps you train for speed without sacrificing accuracy. Reduces BurnoutFull-length tests can be mentally exhausting, especially if taken too often. Mini practice sets keep your energy fresh and your mind engaged. This leads to better retention, increased confidence, and more productive study sessions. Reinforces Test-Day ConditionsBy timing each set to simulate actual test conditions (1 minute per question), you’re teaching yourself to manage stress and make quicker decisions both critical skills on test day. How to Implement This Strategy: Choose 10–15 ACT Math Test Practice Questions focused on one or two core topics (e.g., linear equations and functions). Set a timer for 15–20 minutes. Work through the questions without interruption, just like on test day. After the timer ends, review all your answers carefully. Identify what went wrong, what took too long, and what worked well. Track your timing and accuracy over multiple sets to monitor improvement. Why These Strategies Work The strategies we’ve shared here are not just repackaged advice found in generic test prep materials. They are the result of years of hands-on experience, student performance analysis, and data-driven refinement. Over the years, we’ve seen measurable score increases often in the range of 5 to 10 points in the ACT Math section within a few weeks when students apply these techniques consistently. This isn’t theory; it’s tested and proven in real-world academic environments, supported by hundreds of individual success stories. We work with data from previous ACT tests, student performance patterns, and real-time feedback to continuously refine what we teach. These strategies have helped students who were once stuck below average move into the top quartile of ACT scorers. We understand that there’s a lot of content out there promising quick fixes and “secret tricks” for ACT success. That’s not what we’re offering here. What we share is based on verified teaching experience, ACT-aligned materials, and best practices recognized by professional educators. This content draws on insights from experienced test prep mentors and is informed by ACT-certified curricula. We do not recommend shortcuts or hacks that might produce short-term gains but lead to long-term confusion. Instead, our approach is structured, transparent, and student-focused. We help you build the skills and confidence to succeed not just on the ACT, but in any math-related academic challenge you face after that. Explore: ACT Math Practice Test Questions Ready to Master Math with Expert Guidance? Join hundreds of students who’ve transformed their math skills with personalized, one-on-one online math tutoring from top-rated experts. Whether it’s Algebra, Calculus, or Test Prep get the support you need, right when you need it. FAQs About ACT Math Test Practice Questions 1. How many questions are there in the ACT Math section?The ACT Math section consists of 60 multiple-choice questions that must be completed in 60 minutes. That gives you an average of one minute per question. 2. What topics are covered in ACT Math?ACT Math includes pre-algebra, elementary and intermediate algebra, coordinate geometry, plane geometry, and trigonometry. Questions range from basic arithmetic to advanced concepts like conics and trig identities. 3. How hard is the ACT Math section compared to school math?ACT Math is generally more challenging than high school math tests because questions require faster problem-solving and often combine multiple concepts in a single problem. 4. Do I need to memorize formulas for the ACT Math test?Yes. Unlike the SAT, the ACT does not provide a formula sheet. You must memorize key formulas for geometry, algebra, and trigonometry before test day. 5. Can I use a calculator on the ACT Math section?Yes, calculators are allowed, but not every question requires one. Over-relying on calculators can slow you down, so it’s best to practice when to use mental math and when to use a calculator. 6. What are the most common mistakes students make on ACT Math?Common mistakes include misreading the question, rushing through calculations, mixing up formulas, and running out of time because of poor pacing. 7. How can I solve ACT Math questions faster?You can solve faster by practicing high-frequency topics, using shortcuts like special triangle ratios, recognizing patterns, and timing yourself in mini practice sets of 10–15 questions. 8. What is a good ACT Math score for college admissions?A score of 26–28 is considered good, while 30+ puts you in a competitive range for top universities. The national average ACT Math score is usually around 20. 9. How should I prepare for the ACT Math section in 1 month?Focus on mastering high-yield topics like algebra, functions, geometry, and right triangles. Practice under timed conditions daily, review mistakes carefully, and take at least two full-length mock tests. 10. Where can I find ACT Math practice questions for free?You can find free ACT Math practice questions on the official ACT website, reputable test prep sites, and through educational platforms that provide sample papers and timed quizzes. Vikram Singh V Singh is a Ph.D. candidate in Mathematics at Central University of Rajasthan and an IIT Guwahati alumnus with national level qualifications (CSIR NET JRF AIR 144; GATE & IIT JAM qualified). Over the past five years, he’s delivered 1:1 online tutoring to students in the US, UK, Canada, and India—solving 4,000+ advanced math problems on Chegg India and consistently earning 5 star ratings. expertise spans Calculus (1–4), Linear & Abstract Algebra, Discrete Mathematics, Real & Complex Analysis, Numerical Analysis, Ordinary & Partial Differential Equations, and Probability & Random Processes. His exam focused teaching prepares students for IIT JEE, Mathematics Olympiad, Engineering Mathematics, and college level coursework, turning complex concepts into clear understanding and top grades. Subject Matter Expertise: Calculus 1–4 \| Linear Algebra \| Discrete Mathematics \| Real & Complex Analysis \| Numerical Analysis \| Differential Equations (ODE & PDE) \| Probability & Random Processes calculus 1, calculus 2, calculus 3, vector calculus, fractional calculus, calculus of variations, finite element methods, linear algebra (advanced), ordinary and partial differential equations, numerical methods, discrete mathematics, probability and statistics, laplace transformation, mathematical models, and special functions like beta function, gamma function, legendre polynomial, hermite polynomial, bessel functions, etc. I love teaching mathematics (especially higher mathematics).I like it a lot, and I keep sharing my mathematics experience by writing informative blog articles. Top Tutors, Top Grades! Only At My Engineering Buddy. Get Homework Help & Online Tutoring 15 Years Of Trust, 18000+ Students Served 24/7 Instant Help In 100+ Advanced Subjects Getting help is simple! 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10739	https://congruentmath.com/lesson-plan/dividing-unit-fractions-fractions-and-whole-numbers-lesson-plan/	Dividing Unit Fractions, Fractions, and Whole Numbers Lesson Plan \| Congruent Math Congruent Math Open Menu Search Search Sign InJoin Free → Grades Types Deals & Seasonal Free Curriculum Free Resources Blog About Home Resources Dividing Unit Fractions, Fractions, and Whole Numbers Lesson Plan Join Free to Download PDF Get the Lesson Materials Dividing Fractions by Whole Numbers Mixed Numbers Guided Notes w/ Doodles $4.25 Buy on TPT → Pin This Overview Ever wondered how to teach dividing fractions in an engaging way to your 5th and 6th-grade students? In this lesson plan, students will learn about dividing fractions by whole numbers, unit fractions, regular fractions, and mixed numbers. Through artistic, interactive guided notes, check for understanding, a color-by-number activity, and a maze worksheet, students will gain a comprehensive understanding of fraction division. The lesson concludes with a real-life example that explores how dividing fractions applies to everyday situations, helping students see the value of this math skill beyond the classroom. Product Line: Guided Notes & Doodles Grades: 5th Grade, 6th Grade Join Free to Download PDF Get the Lesson Materials Dividing Fractions by Whole Numbers Mixed Numbers Guided Notes w/ Doodles $4.25 Buy on TPT → Learning Objectives After this lesson, students will be able to: Understand and explain how to divide fractions by whole numbers, unit fractions, and mixed numbers Solve division problems involving fractions and whole numbers accurately using guided steps Apply fraction division skills to real-life scenarios and word problems Prerequisites Before this lesson, students should be familiar with: Basic multiplication and division skills with whole numbers Understanding of fractions as parts of a whole and improper fractions How to simplify fractions and find equivalent fractions Understanding of mixed numbers and how to convert between mixed numbers and improper fractions Materials Pencils Colored pencils or markers Dividing Fractions by Whole Numbers Mixed Numbers Guided Notes w/ Doodles Key Vocabulary Fraction Numerator Denominator Whole Number Mixed Number Unit Fraction Division Reciprocal Procedure Introduction Pin This As a hook, ask students why it might be useful to know how to divide fractions by whole numbers or mixed numbers in everyday life. For example, have they ever tried to share a recipe or divide up something like pizza or candy among friends? Refer to the real-life application page at the end of the guided notes to inspire this discussion and help students see the relevance of the skill beyond the classroom. Use the first page of the guided notes to introduce the concept of dividing fractions by whole numbers and unit fractions. Walk through how division of fractions works, emphasizing the visual fraction models alongside the equations to help students understand why dividing by a whole number is equivalent to multiplying by its reciprocal. Point out and work through each example and the checks for understanding to verify students can explain the procedure and solve basic problems. Refer to the FAQ below for detailed walkthroughs and ways to address common misconceptions, such as confusing multiplication and division or incorrectly flipping mixed numbers. Use the second page of the guided notes to introduce dividing fractions by mixed numbers and non-unit fractions. For mixed numbers, carefully explain converting mixed numbers to improper fractions as a prerequisite step, then demonstrate the division process. Highlight the step-by-step procedure, including when and why to multiply by the reciprocal and how to simplify results. Use the guided checks for understanding to assess comprehension and guide discussion. Refer to the FAQ section for strategies to clarify these more complex steps and for responding to frequent student questions about mixed numbers. Based on student responses during these guided notes and checks for understanding, reteach any concepts that students need extra help with. If your class has a wide range of proficiency levels, consider pulling small groups for targeted reteaching while having more advanced students begin working on the practice exercises independently. This approach helps ensure all students build solid understanding before moving forward. Practice Pin This Have students practice dividing fractions by whole numbers and mixed numbers using the maze activities (page 4) included in the resource. Walk around to answer student questions. Fast finishers can start the color by number activity (page 5) for extra practice. You can assign it as homework for the remainder of the class. Real-Life Application Pin This Bring the class back together, and introduce the concept of using fraction division to solve real-world problems involving sharing, cooking, and measurement. For example, discuss how dividing a recipe into smaller portions requires dividing fractions by whole numbers or mixed numbers, or how dividing lengths in construction projects can involve dividing fractions. Refer to the FAQ for more ideas on how to teach it! Extensions Extensions Additional Self-Checking Digital Practice If you’re looking for digital practice for dividing fractions, try my Pixel Art activities in Google Sheets. Every answer is automatically checked, and correct answers unlock parts of a mystery picture. It’s incredibly fun, and a powerful tool for differentiation. Here’s 3 activities to explore: [Dividing Fractions Digital by Whole Numbers & Fractions Digital Pixel Art CCSS] [[Winter Math] Dividing Fractions Word Problems Digital Pixel Art] [Dividing Fractions by Whole Numbers and by Fractions Digital Pixel Art CCSS] FAQs What does dividing fractions by whole numbers mean?Open Dividing fractions by whole numbers means finding how many times the whole number fits into the fraction. It can also be seen as splitting the fraction into smaller equal parts based on the whole number. How do you divide a fraction by a whole number using a visual model?Open To divide a fraction by a whole number using a visual model, you can represent the fraction as a shape divided into equal parts, then split each part into the number of whole number sections. Draw the fraction as parts of a whole Divide those parts equally by the whole number Count the smaller parts to find the new fraction result What is the rule for dividing fractions by whole numbers in equations?Open When dividing a fraction by a whole number, multiply the denominator of the fraction by that whole number. Keep the numerator the same Multiply the denominator by the whole number Simplify the resulting fraction if possible How do you divide fractions by mixed numbers?Open To divide fractions by mixed numbers, first convert the mixed number to an improper fraction, then multiply the first fraction by the reciprocal of that improper fraction. Convert mixed number to improper fraction Find the reciprocal of the improper fraction Multiply the initial fraction by this reciprocal Simplify the answer if possible Why do you multiply by the reciprocal when dividing fractions?Open Multiplying by the reciprocal in fraction division works because division is the inverse operation of multiplication. The reciprocal flips the numerator and denominator of the divisor fraction This transforms division into multiplication, which is easier to compute It maintains the value of the division problem correctly Can you divide fractions by unit fractions?Open Yes, dividing fractions by unit fractions is done by multiplying the fraction by the reciprocal of the unit fraction. A unit fraction has 1 as numerator Find the reciprocal by switching numerator and denominator Multiply the original fraction by this reciprocal How can real-life applications help understand dividing fractions?Open Real-life applications provide context to fraction division, making it easier to understand and relate. Examples like sharing food or measuring ingredients demonstrate practical use Visual models help students see division in action How do you check if your division of fractions answer is correct?Open To check your answer for dividing fractions, multiply your quotient by the divisor to see if you get the original dividend. Multiply the result by the divisor If the product matches the original fraction, the answer is correct Want more ideas and freebies? Get my free resource library with digital & print activities—plus tips over email. 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10740	https://www.youtube.com/watch?v=qKZH8MiaQ_E	Physics Summary Chapter 6: Uniform Circular Motion and Gravity Dot Physics 51700 subscribers 29 likes Description 738 views Posted: 18 Jun 2024 I'm working through chapter summaries for introductory physics (algebra-based). I'm using the Openstax online (free) textbook College Physics. You can access this book here Here is the playlist with all the videos and problems for this course. In this chapter: - Circular motion and arc length - Angular velocity and linear velocity - Rolling wheels - Centripetal acceleration - Centripetal vs. centrifugal force - Aristotle, Galileo, and Newton Gravity and circular motion - Universal gravity - Circular orbits - Coriolis force - Kepler's Laws 4 comments Transcript: chapter six so I'm working through chapter summaries for open Sachs College physics I say that every time just to make sure that we're on the same page chapter six is about circular motion and gravity we've already talked about gravity we're going to do better gravity so I think it's important to kind of stop and just look back where we've gone just so we know when we're going over new stuff how that relates so here's Here's the the big picture of everything we've done so far first we have the definition of velocity and I wrote this as a vector okay so velocity is the rate of change of position where here I'm using r as position it's very common to use r as position the vector position you could write this as VX is Delta X over Delta t v y is Delta y over delta T but okay and then we have acceleration is defined as the rate of change of velocity okay okay again as a vector then we have these kinematic equations I just wrote two of them um but here this one relates the position of an object if you know the starting position starting velocity and the acceleration as a vector you again could break this in terms of x x0 vx0 a x t^2 and then for y but it's just simpler to write as a vector and then I have this one people love this equation but I don't um I'll tell you why one it's hard to derive two we we'll derive this with work energy later and three you see I didn't write it as a vector because you can't right you can't write it as a vector because you're Square in the velocity so it has to be written as a scalar value and that makes things a little awkward sometimes so um but there it is next we have Newton Second Law which says what do forces do to objects they change the motion they cause an acceleration so the net force on an object as a vector force is equal to the mass times the acceleration of that object and yes that's a same acceleration next we have some special forces uh we have the the gravitational force which we will change later this is the gravitational force on the surface of the Earth it's a it's a vector the mass of the object times G the which we call the gravitational field on the surface of the Earth points straight down this is the force due to a spring so a spring the more you stretch it the greater the force it pulls I did not write that as a vector if you want to write that as a vector it can get a little complicated um so you just have to look at your picture and then pick your direction you there's a more sophisticated version of that but we it's on in the book okay T is the tension of the string and there's no equation there's no equation for that it's a force of constraint it's whatever it needs to be in order to prevent the string from stretching um and it's in the direction of the string n is the normal force again there's no equation the normal force is the is a force a surface exerts that's perpendicular to the surface again to prevent the object from going through the surface and then we have the static friction Force which is technically a constraint Force okay uh but we can get the maximum value the maximum static friction force is parallel to the surface it wants to prevent the two things from sliding that's the coefficient of static friction that depends on the two types of material interacting and that's that normal force and then we have kinetic friction it is calculated even though it depends on the normal force and it's equal to this different coefficient of friction the kinetic friction the end okay now you're all caught up let's look at circular motion because everything we've done um well not everything I take that back okay we did projectile motion that's not straight line so let's imagine that we have an object moving in a circle I'm going to try to draw my best Circle ever that wasn't bad look at that that's a pretty good Circle even though I think my Dot's right there okay so let's say I have a circle of radius R the book uses lowercase R uh and it starts let's say right here has a radius of R and then it moves over here and I intentionally put it offset and this is some angle Delta Theta okay so it goes it starts right here it ends right there it's moving in a circular path and the circular path has radius R well the question is how far did it go how far is it from here to there well that's just the Arc Length of that Circle so I can write this Arc Length I'll call it Delta s Delta s is R Delta Theta if and only if Delta Theta is in radians okay so we're going to that's the best unit for measuring angles is radians because if I take this radius and units of meters I multiply by radians I get meters because radians is kind of like a placeholder unit it's a ratio type of thing if I had the angle in degrees I'd get meters times degrees and that's not that's not the same thing okay so let's just a quick review if I have um let's say Pi radians or 2 pi radians to say is equal to one Revolution which is equal to 360° so if I go all the way around the circle that's one revolution that's 360° that's 2 pi radians so you can use this to convert from revolutions to radians or degrees to radians whatever you want okay or at the end maybe you have the answer in radians and you want it in degrees you can do that too we're not going to focus on that okay this is important though let's see what did I put next okay so we can Define uh the r suppose it's moving in a circle we can Define the rate that the angular value changes as Omega this is the lowercase Greek letter Omega and it's defined as this is ang we call this angular velocity and it's defined as the rate of change of the angle Delta Theta delta T and so this would be in normal units of radians per second but you may see other units uh you may see RPM which is revolutions per minute you may see uh radians uh degrees per second those are the two the three most common but you could mean a lot of things we want to use it in radians per second okay and don't worry I'm going to give you an equation summary at the end um so let's say that I want to make a relationship between the velocity the magnitude of the velocity for an object moving in a circle and Delta S and R and Delta Theta well so I can write V the velocity is the rate of change of s I'm going I'm running out of room I planned poorly here I'm going write this up here Omega is Delta I write too big so V is Delta s over delta T right that's how we Define velocity this is not the vector velocity because we're talking about in the direction of the path around the circle so it does change direction so it's kind of like a speed um so I can put in this for Delta s and I get R Delta Theta over delta T and then Delta Theta over delta T is Omega so this is our Omega so that's an important pretty important relationship V = Omega R and then this is s is R Delta Theta so the change in position is the radius of the circle times the change in angle the velocity is the radius times the angular velocity so it's a way for us to make a connection between angular quantities and linear quantities and that's going to be important okay and over here if Omega is in radians per second and I multiply by meters I get meters per second so that's good now there is something that comes up a lot and it is important to talk about and that's rolling objects so imagine I have a wheel of radius R and it's rolling this way with an angular velocity Omega well this contact point right here is if it's rolling without slipping when it's in contact it does not move that contact Point does not move that means that this has to have a velocity uh in order to keep that contact Point stationary as though a negative of the contact Point velocity so the only way that can work is if this wheel is rolling with the velocity of R Omega okay so the center mass of the wheel depends on the rotational velocity and the radius of the wheel and this is how essentially your car tells how fast is going right it takes the wheel it looks at the rotation rate and you know the size of the wheel you can calculate the velocity so if you put the wrong size tires on your car your speedometer is not going to be right there's probably a way to fix that but I don't really know okay so that's linear velocity now let's look at uh something else very important here we have this an object moving in a circle remember that the acceleration as defined as the change in velocity with respect to time so I could write that as V2 I'm so sorry about that V2 minus V1 over the change in time so V1 and V2 are vectors so if you change the vector velocity you have an acceleration and that is exactly what happens when we are moving in a circle so imagine that I don't to put that Delta test right there so here's V1 and here's V2 and so so the magnitude of V1 is equal to the magnitude of V2 so it did not change speed but it changed Direction so if you change direction even if you don't change speed then you will accelerate let's get an expression for that acceleration the first thing I want to do is to imagine that it's some short time interval so Delta Theta is small even though I drew it large uh I can calculate the change in velocity Delta v as V2 minus V1 so to subtract these vectors I can just redraw them I'm going to redraw that one right there there's V1 as a vector there's V2 same length and Delta V would be from the beginning of this one I mean the end of this one to the end of that one we've done Vector subtraction before and that's Delta V right there so right away you see that during this time interval the velocity changes and the direction of the change is this way which is towards the center of the circle so we're going to call this cental acceleration but I'm just going to put a note here uh it's towards Center and we can also find the magnitude so how do you find the magnitude of this we're going to have to use a little geometry magic but we can use it okay so I know this is the angle Delta Theta uh it turns out that this is the same angle Delta Theta right because if you imagine moving this up here then as that angle gets smaller and smaller these get closer and closer so they're the same Angle now if Delta Theta is small this Delta V we in velocity space and that looks weird but it's still the same similar triangle as it is to this so I can still use the same idea for that angle to find this Arc length even though it's straight it's pretty close to the same Ark length so I can write the magnitude of Delta v as V Delta Theta why V not V1 not V2 because I'm talking about the magnitude the magnitude of this and the magnitude of that are the same so it doesn't really matter which one I put so I'm going to put V is V Delta V is V Delta Theta okay now I want to get rid of Delta Theta okay I want to find get rid of Delta so I'm going to go back to this relationship and I know that uh Delta s is R Delta Theta that's what we said before right that's the Arc Length so this means that Delta Theta is Delta s over R and if I put that up here I get Delta V is going to be uh V time Delta Theta which is Delta s over Delta r I'm sorry just R now I want to find the magnitude of the acceleration that's Delta V over delta T so I can just take that and divide by delta T and I get v/ R Delta s over delta T and Delta s over delta T is what what you know what that is that's the velocity so this is going to be v/ R V or V ^2 over R and this is our other important thing it has a magnitude of v^2 R and we call this centripetal acceleration I'm going to talk about centripetal in just a second that word I want to do one more thing so we could also describe this acceleration in terms of um the angular velocity so remember that V is Omega time R so V is Omega R so v^ 2 if I put that in for v^2 I get a is going to be Omega 2 r^ 2 / R which is Omega 2 R so this is equal to Omega 2 R they're the same thing and just check the units on that so this going to be radians squar per second squar time meters does give me m/s squared so what does this say you don't have to know that derivation I mean I'm not I think it's important to see it but uh let's just understand what it means AC is v^2 over R so this says that if you're moving in a circle you are accelerating the direction of the acceleration is towards the center that's important the magnitude of the acceleration increases the faster you go and it also increases at a smaller size circle so imagine driving in your car there's a there's a great uh there's a great question that's a steering wheel that's the gas pedal and that's the brake and the question is what controls the acceleration on your car and the answer is all three of these right because if I push the gas pedal I can increase in speed if I push the brake pedal I could decrease in speed those are both types of acceleration but if I turn the steering wheel I also accelerate because I change direction when you accelerate in the car you can feel it when I push that gas pedal and I accelerate forward it feels like I'm getting push push back we'll talk about that in a second when I slam on the break it feels like I'm getting pushed forward and when I turn it feels like I'm getting pushed against the side so you are accelerating when you turn the car and if I turn in a smaller size circle if R gets smaller you can feel greater acceleration if I'm going faster you can feel a greater acceleration okay let's talk about the word centripetal acceleration I'm going to write two words this is called cpal acceleration centripetal and then there's also centrifugal I run out of room I made it okay so this literally means Center pointing which is what we said this is the acceleration pointing towards the center of the circle now there is a thing that people talk about centripetal force we'll talk about in a little bit but it's really just whatever Force makes something move in a circle it's not a big deal this Centra Center Fugal fleeing this is the force that pushes away from the center of the circle okay so this this is called a fake force and we mentioned this before but let me just remind you Newton's Second Law which says this this only works in an inertial reference frame that means the a of the frame is zero that's a vector so if I'm just standing here on on the ground and nothing's accelerating my my room's not accelerating why would my room be accelerating even though technically it is we'll talk about that later then Newton Second laws work I can calculate the forces use that to find the acceleration however if I'm in an elevator and the elevator is accelerating up then then this doesn't work right if I use that as my reference frame to measure velocities I have an extra force that pushes me down I don't know what that is and we would add that in to make Newton Second law work we would add in a fake Force that's what this is if you're in a rotating reference frame then there in order to make this work you have to add in a fake for set point pushes away from the center of the circle so if you're in your car and you turn to the left you feel like you're getting pushed to the right you're not but you feel that way because your reference frame of the car is accelerating and you add that fake force in there so sometimes they can be very useful but for the most part they're dangerous cuz people people try to use them when they shouldn't be using them that's a warning okay so centripetal force center pointing we can Define the acceleration like that let's talk briefly about forces there's got a whole bunch of awesome examples that we can do here so I'm just going to give the a very generic uh version of this so imagine that I'm swinging a ball in a in a horizontal Circle and then so like this uh and here's my ball and I'm actually going to draw it horizontal at some point just because it's going to be easier so now I want to draw forces on that I only have one it would be this tension force right there and what we can do is use Newton's Second Law fnet equals Ma and I wrote that as a scalar because I'm use that in the direction of the center of the circle so I'm going to choose this my x- Axis if you want to in this case the the force would be negative the tension and that's M v^2 over R so we can use this to solve for problems but we have to we have to freeze it in place right because if I wanted to look at this a little bit later the ball's up here and now the acceleration is that way so the acceleration changes directions it makes it very very difficult to deal with this so you want to freeze your situation in such that you can pick the x-axis and just say okay it's not really always the x-axis but at that instent it is and this we would call the centrifical force because it's whatever Force makes it move in a circle okay it doesn't have to be that way we could do something more complicated we could have uh a ball swinging like this so it's moving in a circle now I have gravitational force pulling down and I have tension pulling up and in this case I would pick this as say my y AIS it's not actually the y or this is the Y AIS so now the tension is in the Y AIS and then a component of the gravity is in the opposite but the sum of those two together would be mv^ s r and it that's a bad problem it's confusing but that's how we deal with these centrifugal force problems okay let's talk about just cuz it's in the book uh what if you are on a rotating reference frame suppose I'm on a rotating platform and I have a ball and so that ball that ball has a uh in in this rotating reference frame that ball has a a centrifugal force pushing away from the center of motion now if it's moving this way and it gets to a smaller radius it's actually going too fast for that veloc that position and so one way to to fix this is to give it a another Force the Corola force and this Corola Force let's see if I can get that right let's see so this is V it's going at the same velocity because no it it' curve this way it' curve if you're on the rotating platform it curve that way and we call that Force the Corola force it gets very complicated it's not really appropriate for this class but I do want to show you one thing this is also a rotating reference frame and here's the United States rotating so you get a cholis force when you change your axis of your distance to the axis of rotation so right here if this is rotating and I move north I'm closer to the axis of rotation than if I than I was if I'm at the equator and I move north I'm closer to the axis of rotation so there would be a coriolis force in the northern hemisphere that coriolis force is very very very tiny because it takes the earth a whole day to rotate um but it's there and it acts over objects that move very large distances and one of the things that moves large distances is things like weather and water so when water moves North in the northern hemisphere it tends to curve to the right okay because it's getting to a closer to the axis of rotation now if you've been down here in Florida it's the weather is nice and warm North Carolina it's not so bad right because what's happening is because of the rotation of the earth we get this counterclockwise uh no clockwise rotation of water so you're getting warm water from the Caribbean Up in North Carolina it's nice this is San Diego Diego the water is very cold cuz again the water is doing a clockwise rotation so it's coming down from the north so you get very cold water so it's very cold in San Diego warm there coriolis Force okay um you it is important to note that this is not an inertial reference frame if I'm standing right here in Louisiana it's not an inertial reference frame because the Earth is spinning and so there is the uh centrifugal force if you want we'll do that probably as a problem calculate the effective gravitational force due to the rotation of the earth if you're at the North Pole then you're standing in water so I wouldn't do that if you're in Antarctica you're at the North Pole and there's no there is you're just spinning around the circle you don't have a a fake Force okay moving right along here that was the Corola Force gravity um you know I I I want to talk about Newton and Aristotle because I want to so remember that uh Aristotle's ideas about physics were wrong but very common sensical uh you know he said that oh if you drop an object they move at a constant speed heavier objects fall faster which is wrong okay and Galileo showed that was wrong by letting things roll down an incline and actually measuring the change in speed and then when new Newton did was Newton they had they had ideas about rotational motion at the time okay uh you could take water in a bucket and spin it and you can see the curved surface of the water so they knew there was some forces associated with rotation and they also knew the moon was rotating around the Earth orbiting around the earth and and what Newton did was to make a relationship between objects that accelerate towards the surface of the Earth and objects that accelerate because they're moving in a circle around the Earth the moon and with that he came up with the following model for Gravity he says that the gravitational force between the Earth or really the gravitational force on the moon that's fairly accurate to scale it's very tiny uh if this we'll call this M1 and we'll call this M2 he said it's an attractive Force so it points towards the other object and its magnitude is G M1 M2 over R 2 where R is the distance between those two so it's the product of their masses divided by the distance between them squared and then there was a constant and and Newton didn't know what that constant was um because he didn't know the mass of the Earth he didn't know the mass of the Moon um he he did know the distance between them fairly well I mean good enough but he didn't know the constant because he didn't know he didn't know those masses it turns out that someone did come along and measure that constant and there's a really clever way to do it um so imagine that I take this stick it's actually a meter stick and I support it from the middle just like that and a tiny tiny tiny force uh can make this thing rotate right I can just tap this you know rotate because it's able to rotate freely about the center so there is a person named Cavendish and what Cavendish did was to take a stick like this and put masses on the end of it and then let those masses attract other nearby masses and measure how much it rotated and if you know something about the torsion constant of the of the string that holds it you can calculate the force you know the masses and the force you can calculate that constant G and and the constant G the gravitational constant has a value of 6.67 101 Newton m s per kg s so that's really really tiny gravity is super super super weak but there's something else very important right once you get that you actually can get the mass of the Earth and that's kind of a big deal um before that they didn't even know the mass of the Earth so if you take the mass of the Earth and you stand right here and you calculate the gravitational force on you and we'll call this the radius of the Earth and we'll call this the mass of the Earth and that's your mass m then I can calculate the gravitational force as G mass of the Earth your mass over the radius of the Earth squared and if you put in the value for the mass of the Earth you put in the value for the radius of the Earth you put in the value for G Big G you get uh M 9.8 that is our G right so the gravitational field is all that stuff because even if you move very high you're not changing your distance from this is the distance of the center of the object you're not changing the distance that much this is pretty much constant that's where that g is so we have it's important right we have this F weight is mg and then we have FG is G M1 M2 over R 2 two models for Gravity but they agree at some level right uh some level they overlap and that's how our models work in science we have models that are wrong but they slowly diverge so I can use this on the surface of the Earth I can use this uh to fly an airplane and do a whole bunch of really complicated stuff but if I get too far away if I get to you know outer space then I need to use the better model for gravity and this model is wrong too this model doesn't include things like um gravitational waves which we can detect here in Louisiana at ligo it doesn't include things like that okay so it's still a wrong model but still very useful so models can be wrong and useful at the same time um who we got a lot of stuff to do here don't we um I guess we should talk about Zer G Zer G so weightlessness and I I'd like to go first and talk about uh I'm going to set I'm going to show you how to calculate low earth orbit or any type of orbit and then we'll talk about Zer G so here's the Earth and here's an object orbiting the Earth in a circle of radius let's call it R and we'll call that we don't even care about the radius of the Earth because it doesn't matter that's the mass of the Earth that's the mass and let's say that has some velocity V well here we have one force acting on that object and that's the gravitational force I'll call it FG so again I can set up this is my xaxis I can say fnet in the X direction is negative the gravitational force which is G mass of the Earth mass of the object over R 2 and that's going to be mass of the object V ^2 R so I can calculate the velocity needed to get into orbit from this equation notice that to be in orbit not to get so the r that R cancels with that and I that mass cancels with that V equal the square Ro T of G me over R so as I get further and further and further away from the earth I don't have to go as fast now it might sometimes be easier to talk about about the period of rotation so I could write this as G mass of the Earth Mass over R 2 is M and I'm going to use the other expression for the angular the centripetal acceleration Omega 2 R in this case this R does not cancel but if I divide by that I get R cubed that mass still cancels so I get Omega squar is G mass of the Earth over R cubed and now if I take the square root I get Omega is the sare RO TK of gme over R Cub so here you can see that the angular velocity uh as as R gets larger the angular velocity gets smaller but this is how we can deal with orbits and we'll do orbit problems but if I have a spaceship in orbit around the Earth then it is accelerating and so if I do a blow up picture of my spaceship like this the spaceship has mass the spaceship's accelerating the person has mass the person is accelerating if these two have the same acceleration there's no contact forces needed between them and and we don't actually feel the gravitational force because it pulls on all parts of our body equally what we feel is the normal force so we say a parent weight is the normal force in this case there's no normal force between the people and you feel weightless you don't even have to be in orbit to do that right that I'm going to not do it now because I'm going too long anyway a lot of people say oh there's no gravity in space so there's that's why I feel weightless the gravitational force at low earth orbit you know 400 kilom above the surface of the Earth is like 80% of on the surface so it's very high um but since you're accelerating the same as gravity that's why you feel weightless if you're in an elevator and the elevator is accelerating down at 9.8 m/s squared you would feel weightless if you're in an airplane and you accelerate down at 9.8 m/s squared you would feel weightless okay so there's a lot of ways to do that okay Kepler's laws um I I don't think this is appropriate but it's in the textbook U so I do want to they're cool okay I'm not saying they're not cool I'm just saying it's hard to make these connections so the the importance of Kepler's laws uh is that they are imper Ally derived Based on data right so what you can do is you can look at the motion of a planet around the Sun and you can come up with different models and the only important thing about them is that Newton's Second Law Newton's uh ideas of gravity can be used to explain these right before Kepler just said this is what it is and the first law uh says objects order orbit the Sun and elliptical ellip tial orbits where the sun is at one fosi and there's just nothing else over there so that's the first law the second law says that as things move closer to the Sun they go faster okay so it's going to go faster here and we'll show that later uh and over here they're going to go slower so what he has is this equal areas equal times thing so in the same amount of time over here this area of the or orbit is equal to this area of the orbit so it's longer so it's going to be skinnier it's not moving as fast I know it's hard to to see that um but then the third law which we can derive for circular orbits says that the period of rotation T12 over t22 is equal to is that right R1 cubed over R2 cubed so if I have two uh planets orbiting the Sun and it works for non-circular orbits but we can only derive it for circular orbits then the the time it takes to rotate squared uh the ratio of those is equal to the ratio of their orbital sizes cubed that's the third law but the important thing again is that uh Newton was able to show that these laws agree with his model for Gravity which was kind of a big deal this is a long chapter summary but I'm having kind of a lot of fun and if you're still here I hope you're still having fun okay let's just go over the main ideas chapter summary of the summary this is the summary summary okay the first is this Delta s = r Delta Theta v = r Omega so this is the Arc Length radius of the circle angular change this is the speed angular velocity Omega radius of the circle uh we also have the definition of the angular velocity change in Theta with respect to time in radians per second and we're going to do more angular stuff later uh then we have the definition of the centripetal acceleration v^2 R Omega 2 R uh and we're not writing that as a vector it's pointing towards the center of the circle and then finally we have the gravitational force G M1 M2 over R 2 this says there's a gravitational force between any objects that have mass okay so me and this p of chalk have a force between us but since G is so silly tiny that force is unmeasurably small the and just but think of all the awesome physics problems we can do now with circular motion gravity orbits and we're going to do more cool orbital stuff when we get into energy but that's it for now chapter 6 summary the end double thumbs up see and and sparkles okay
10741	https://math.stackexchange.com/questions/56393/how-to-prove-that-continuous-functions-are-riemann-integrable	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 How to prove that continuous functions are Riemann-integrable? Ask Question Asked Modified 4 years, 5 months ago Viewed 51k times 21 $\begingroup$ In other words, how to prove A continuous function over a closed interval is Riemann-integrable. That is, if a function $f$ is continuous on an interval $[a, b]$, then its definite integral over $[a, b]$ exists. Edit: The Definite Integral as a Limit of Riemann Sums : Let $f(x)$ be a function defined on a closed interval $[a, b]$. We say that a number $I$ is the definite integral of $f$ over $[a, b]$ and that $I$ is the limit of the Riemann sums $\sum \limits_{k=1}^n f(c_k)\Delta x_k$ if the following condition is satisfied: Given any number $\epsilon \gt 0$, there is a corresponding number $\delta \gt 0$ such that for every partition $P = {x_0, x_1, ... , x_n}$ of $[a, b]$ with $\\|P \\| < \delta$ and any choice of $c_k$ in $[x_{k-1}, x_k]$, we have $$ \left\| \sum_{k=1}^n f(c_k) \Delta x_k - I \ \right\| \lt \epsilon .$$ calculus Share edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Aug 8, 2011 at 22:20 LeafGlowPathLeafGlowPath 7,7151111 gold badges4646 silver badges7878 bronze badges $\endgroup$ 7 $\begingroup$ In the Riemann sense? Which definition of the Riemann integral are you using? $\endgroup$ Qiaochu Yuan – Qiaochu Yuan 2011-08-08 22:23:52 +00:00 Commented Aug 8, 2011 at 22:23 $\begingroup$ Read wikipedia article. When writing out Riemann sum, approximate your function by a constant on that interval, convince yourself of convergence. $\endgroup$ Sasha – Sasha 2011-08-08 22:25:56 +00:00 Commented Aug 8, 2011 at 22:25 $\begingroup$ You're not working with a definition it's easy to prove things about (since you need to exhibit $I$ to verify it). It's much easier to use the Darboux integral, and there the proof is fairly straightforward and a nice exercise: en.wikipedia.org/wiki/Darboux_integral $\endgroup$ Qiaochu Yuan – Qiaochu Yuan 2011-08-08 22:54:36 +00:00 Commented Aug 8, 2011 at 22:54 3 $\begingroup$ Yeah, I know the definition made the proof difficult. That's why I asked the question. $\endgroup$ LeafGlowPath – LeafGlowPath 2011-08-08 23:03:02 +00:00 Commented Aug 8, 2011 at 23:03 $\begingroup$ @Qiaochu: I can be characterized as $g(b)-g(a)$ for an an antiderivative $g$ of $f$. I believe that this would lead quite quickly to a proof using the above definition of the integral. However, one would then need to establish the fact that continuous functions are indefinitely integrable without appealing to the Riemann integral (and the FTC), and I don't see at the moment how to do this. $\endgroup$ Mark – Mark 2011-08-08 23:31:10 +00:00 Commented Aug 8, 2011 at 23:31 \| Show 2 more comments 3 Answers 3 Reset to default 18 $\begingroup$ First note that a precise formulation of your question is: How do you prove that every continuous function on a closed bounded interval is Riemann (not Darboux) integrable? You can find a proof in Chapter 8 of these notes. Here is a rough outline of this handout: I. I introduce the ("definite") integral axiomatically. One of the axioms is that the set of integrable functions on $[a,b]$ should contain all the continuous functions. II. I prove that the Fundamental Theorem of Calculus follows (easily) from the axioms. III. I introduce Riemann integrable functions (which are exactly what you wrote above) and verify that the class of Riemann integrable functions on $[a,b]$ satisfies the axioms of I. In particular: IV: I prove that every continuous function is Riemann integrable. Later I talk about the Darboux integral and how it compares to the Riemann integral. But it was an intentional decision to present the Riemann integral first. This is what students are expecting from their previous courses, and it is not so bad to work with, at least for a while. Share edited Apr 19, 2021 at 15:23 answered Aug 9, 2011 at 14:26 Pete L. ClarkPete L. Clark 101k1111 gold badges238238 silver badges383383 bronze badges $\endgroup$ 1 2 $\begingroup$ Ha, I will an graduate student in McGill this September and I'm reviewing some math knowledge. I guess understanding the proof of this theorem is a bit out of my ability. Thanks anyway! $\endgroup$ LeafGlowPath – LeafGlowPath 2011-08-10 00:31:53 +00:00 Commented Aug 10, 2011 at 0:31 Add a comment \| 10 $\begingroup$ Riemann-Darboux integral: For the Riemann-Darboux integral it is easier than with your definition (which is equivalent in $\mathbf R$). The function is uniformly continuous on $[a,b]$ (why?). This means that if $\epsilon > 0$ is given we can find $\delta = \delta(\epsilon) > 0$ such that $\|x - y\| < \delta$ implies $\|f(x) - f(y)\| < \frac{\epsilon}{2(b - a)}$. Now let $P_\epsilon$ be a partition with norm $\\|P_\epsilon\\| < \delta$. Now for $P$ finer than $P_\epsilon$ we have $$M_k(f) - m_k(f) \leq \frac{\epsilon}{2(b - a)}.$$ Where $M_k$ is the supremum in $[x_{k - 1}, x_k]$ and $m_k$ is the infimum. Multiply this inequality with $\Delta x_k$ and sum to get $$U(P, f) - L(P, f) \leq \frac{\epsilon}{2(b - a)} \sum_{k = 1}^n \Delta x_k = \frac\epsilon2 < \epsilon.$$ Fine, so what is left is to prove the same thing for your definition of the integral (not so easy) or proving the equivalence between the two (not so hard). Riemann integral: There is a problem with the above approach if we are in a general Banach space (why?), so we must resort to the normal Riemann integral. Let $f:[a,b] \to E$ be continuous where $E$ is a Banach space. Given $\epsilon > 0$ let $\delta$ be such that $$\text{if } \|x - y\| < \delta \text{ then } \|f(x) - f(y)\| < \frac{\epsilon}{b - a}.$$ Let $P$ and $P'$ be partitions of $[a,b]$ with norm smaller than $\delta$. Let $c$ and $c'$ be the choices of points in each interval of $P$ and $P'$ respectively. We want to estimate $\|S(P, c) - S(P', c')\|$ where $S(P, c)$ is the Riemann sum associated with $P$ and $c$. WLOG let $P \subset P'$. (If $P = P'$ then $$\|S(P, c) - S(P, c')\| \leq \sum \|f(c_i) - f(c_i')\| \Delta x_i \leq \epsilon.$$ Now suppose that $P'$ is obtained from $P$ by inserting one point (split one interval) for example say we insert $x_j'$ with $x_j \leq x_j' \leq x_{j + 1}$. In this case the partition size will not increase. WLOG assume that for $i \neq j$ we have $x_i' = x_i$ and that $c_j = x_j'$ and that $x_j'$ is also selected as the points in the intervals $[x_j, x_j']$ and $[x_j', x_{j + 1}]$. Now $S(P, c) - S(P', c') = 0$. We can repeat this process for another refinement. This will give us the result (why?). Share edited Nov 5, 2011 at 19:03 Srivatsan 26.8k77 gold badges9595 silver badges148148 bronze badges answered Aug 9, 2011 at 11:13 user1120user1120 $\endgroup$ 1 $\begingroup$ Don't you have to prove that $\|S(P,c)-S(P',c')\|<\epsilon$ for all $(P,c)$ and $(P',c')$ with $\\|P\\|<\delta$ and $\\|P'\\|<\delta$. Restricting to those for which either $P\subseteq P'$ or $P\subseteq P'$ might miss some partitions... $\endgroup$ freishahiri – freishahiri 2014-06-09 21:12:58 +00:00 Commented Jun 9, 2014 at 21:12 Add a comment \| 6 $\begingroup$ Appendix to Chapter 13 in: CALCULUS, by M. Spivak. Yes, uniform continuity is the key. Share answered Aug 9, 2011 at 2:43 GEdgarGEdgar 118k99 gold badges128128 silver badges274274 bronze badges $\endgroup$ 16 1 $\begingroup$ In the same book, M. Spivak a very amazing and marvelous proof without the use of uniform continuity. In case you don't have access to this book, you can read the same proof online at paramanands.blogspot.com/2012/07/… $\endgroup$ Paramanand Singh – Paramanand Singh ♦ 2013-07-15 03:29:12 +00:00 Commented Jul 15, 2013 at 3:29 $\begingroup$ @ParamanandSingh In your notes, on the Second proof taken from Spivak's Calculus, can you explain what the notations $h {f(c)-\epsilon} , h {f(c)+\epsilon}, {J(c+h)-J(c)}/h$ mean? $\endgroup$ Zero Pancakes – Zero Pancakes 2017-07-19 07:02:33 +00:00 Commented Jul 19, 2017 at 7:02 $\begingroup$ @ZeroPancakes: $J(x), j(x) $ denote upper and lower Darboux integrals of $f$ on $[a, x] $ respectively. These functions are defined for $x\in[a, b] $. And the difference $J(c+h) - J(c) $ represents the upper Darboux integral of $f $ in interval $[c, c+h] $. This difference is needed to find the derivative of $J$ at $c$. $\endgroup$ Paramanand Singh – Paramanand Singh ♦ 2017-07-19 07:07:18 +00:00 Commented Jul 19, 2017 at 7:07 $\begingroup$ @ParamanandSingh Yes, but I haven't seen before multiplication and division with sets. What does this mean? $\endgroup$ Zero Pancakes – Zero Pancakes 2017-07-19 07:09:18 +00:00 Commented Jul 19, 2017 at 7:09 $\begingroup$ @ZeroPancakes: All the notation details are available in the blog post and if you read it carefully you will not have problem. $\endgroup$ Paramanand Singh – Paramanand Singh ♦ 2017-07-19 07:10:10 +00:00 Commented Jul 19, 2017 at 7:10 \| Show 11 more comments 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 calculus See similar questions with these tags. 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10742	https://drops.dagstuhl.de/storage/00lipics/lipics-vol224-socg2022/LIPIcs.SoCG.2022.50/LIPIcs.SoCG.2022.50.pdf	An (ℵ0, k + 2)-Theorem for k-Transversals Chaya Keller # Ariel University, Israel Micha A. Perles # Einstein Institute of Mathematics, Hebrew University, Jerusalem, Israel Abstract A family F of sets satisfies the (p, q)-property if among every p members of F, some q can be pierced by a single point. The celebrated (p, q)-theorem of Alon and Kleitman asserts that for any p ⩾q ⩾d + 1, any family F of compact convex sets in Rd that satisfies the (p, q)-property can be pierced by a finite number c(p, q, d) of points. A similar theorem with respect to piercing by (d −1)-dimensional flats, called (d −1)-transversals, was obtained by Alon and Kalai. In this paper we prove the following result, which can be viewed as an (ℵ0, k + 2)-theorem with respect to k-transversals: Let F be an infinite family of sets in Rd such that each A ∈F contains a ball of radius r and is contained in a ball of radius R, and let 0 ⩽k < d. If among every ℵ0 elements of F, some k + 2 can be pierced by a k-dimensional flat, then F can be pierced by a finite number of k-dimensional flats. This is the first (p, q)-theorem in which the assumption is weakened to an (∞, ·) assumption. Our proofs combine geometric and topological tools. 2012 ACM Subject Classification Theory of computation →Computational geometry Keywords and phrases convexity, (p, q)-theorem, k-transversal, infinite (p, q)-theorem Digital Object Identifier 10.4230/LIPIcs.SoCG.2022.50 Funding Chaya Keller: Research partially supported by the Israel Science Foundation (grant no. 1065/20). Acknowledgements The authors are grateful to Andreas Holmsen for valuable suggestions and information. 1 Introduction 1.1 Background Helly’s theorem and the (p, q)-theorem. The classical Helly’s theorem asserts that if F is a family of compact convex sets in Rd and every d + 1 (or fewer) members of F have a non-empty intersection, then the whole family F has a non-empty intersection. For a pair of positive integers p ⩾q, a family F of sets in Rd is said to satisfy the (p, q)-property if \|F\| ≥p, none of the sets in F is empty, and among every p sets of F, some q have a non-empty intersection, or equivalently, can be pierced by a single point. A set P ⊂Rd is called a transversal for F if it has a non-empty intersection with every member of F, or equivalently, if every member of F is pierced by an element of P. In this language, Helly’s theorem states that any family of compact convex sets in Rd that satisfies the (d + 1, d + 1)-property, has a singleton transversal. One of the best-known generalizations of Helly’s theorem is the (p, q)-theorem of Alon and Kleitman (1992), which resolved a 35-year old conjecture of Hadwiger and Debrunner . ▶Theorem 1 (the (p, q)-theorem ). For any triple of positive integers p ⩾q ⩾d + 1 there exists c = c(p, q, d) such that if F is a family of compact convex sets in Rd that satisfies the (p, q)-property, then there exists a transversal for F of size at most c. © Chaya Keller and Micha A. Perles; licensed under Creative Commons License CC-BY 4.0 38th International Symposium on Computational Geometry (SoCG 2022). Editors: Xavier Goaoc and Michael Kerber; Article No. 50; pp. 50:1–50:14 Leibniz International Proceedings in Informatics Schloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany 50:2 An (ℵ0, k + 2)-Theorem for k-Transversals In the 30 years since the publication of the (p, q)-theorem, numerous variations, generalizations and applications of it were obtained (see, e.g., the surveys [13, 20]). We outline below three variations to which our results are closely related. (p, q)-theorems for k-transversals. The question whether Helly’s theorem can be general-ized to k-transversals – namely, to piercing by k-dimensional flats (i.e., k-dimensional affine subspaces of Rd) – goes back to Vincensini , and was studied extensively. Santaló observed that there is no Helly-type theorem for general families of convex sets, even with respect to 1-transversals in the plane. Subsequently, numerous works showed that Helly-type theorems for 1-transversals and for (d −1)-transversals in Rd can be obtained under addi-tional assumptions on the sets of the family (see and the references therein). A few of these results were generalized to k-transversals for all 1 ⩽k ⩽d −1 (see [5, 6]). Concerning (p, q)-theorems, the situation is cardinally different. In , Alon and Kalai obtained a (p, q)-theorem for hyperplane transversals (that is, for (d −1)-transversals in Rd). The formulation of the theorem involves a natural generalization of the (p, q)-property: For a family G of objects (e.g., the family of all hyperplanes in Rd), a family F is said to satisfy the (p, q)-property with respect to G if among every p members of F, some q can be pierced by an element of G. A set P ⊂G is called a transversal for F with respect to G if every member of F is pierced by an element of P. ▶Theorem 2 (). For any triple of positive integers p ⩾q ⩾d + 1 there exists c = c(p, q, d) such that if F is a family of compact convex sets in Rd that satisfies the (p, q)-property with respect to piercing by hyperplanes, then there exists a hyperplane transversal for F of size at most c. As an open problem at the end of their paper, Alon and Kalai asked whether a similar result can be obtained for k-transversals, for 1 ⩽k ⩽d −2. The question was answered on the negative by Alon, Kalai, Matoušek and Meshulam , who showed by an explicit example that no such (p, q)-theorem exists for line transversals in R3. (p, q)-theorems without convexity. Numerous works obtained variants of the (p, q)-theorem in which the convexity assumption on the sets is replaced by a different (usually, topological) assumption. Most of the results in this direction base upon a result of Alon et al. , who showed that a (p, q)-theorem can be obtained whenever a fractional Helly theorem can be obtained, even without a convexity assumption on the elements of F. In particular, the authors of obtained a (p, q)-theorem for finite families of sets which are a good cover, meaning that the intersection of every sub-family is either empty or contractible. Matoušek showed that bounded VC-dimension implies a (p, q)-theorem, and Pinchasi proved a (p, q)-theorem for geometric hypergraphs whose ground set has a small union complexity. Recently, several more general (p, q)-theorems were obtained for families with a bounded Radon number, by Moran and Yehudayoff , Holmsen and Lee , and Patáková . (p, q)-theorems for infinite set families. While most of the works on (p, q)-theorems concentrated on finite families of sets, several papers studied (p, q)-theorems for infinite set families. It is well-known that Helly’s theorem for infinite families holds under the weaker assump-tion that all sets are convex and closed, and at least one of them is bounded. In 1990, Erdős asked whether a (p, q)-theorem can be obtained in this weaker setting as well. Specifically, his conjecture – which was first published in – was that a (4, 3)-theorem holds for infinite families of convex closed sets in the plane in which at least one of the sets is bounded. C. Keller and M. A. Perles 50:3 Following Erdős and Grünbaum, who refuted Erdős’ conjecture and replaced it by a weaker conjecture of his own, several papers studied versions of the (p, q)-theorem for infinite families (see [25, 27]). These papers aimed at replacing the compactness assumption (which can be removed completely for finite families) by a weaker assumption. 1.2 Our contributions In this paper we study variants of the (p, q)-theorem for infinite families F of sets in Rd. Our basic question is whether the assumption of the theorem can be replaced by the following weaker infinitary assumption, which we naturally call an (ℵ0, q)-property: Among every ℵ0 elements of F, there exist some q that can be pierced by a single point (or more generally, by an element of G). We show that despite the apparently weaker condition, (p, q)-theorems can be obtained in several settings of interest. An (ℵ0, 2)-theorem for closed balls in Rd. Our first result concerns the classical setting of point transversals and considers families of closed balls in Rd. For such families, Danzer obtained in 1956 a (2, 2)-theorem in the plane, answering a question of Gallai. Grünbaum obtained a (2, 2)-theorem in Rd, Kim et al. obtained a (p, 2) theorem in the plane for all p ⩾2, and finally, Dumitrescu and Jiang obtained a (p, 2)-theorem in Rd for all p ⩾2. We show that an (ℵ0, 2)-theorem holds as well. ▶Theorem 3. Let F be an infinite family of closed balls in Rd. If among every ℵ0 elements of F, some two intersect, then F can be pierced by a finite number of points. We note that unlike the standard (p, q)-theorems, there does not exist a universal constant c = c(d) such that every family of closed balls in Rd can be pierced by at most c points. Indeed, for any m ∈N, if the family consists of ℵ0 copies of m pairwise disjoint balls then it satisfies the (ℵ0, 2)-property (and actually, even the much stronger (ℵ0, ℵ0)-property), yet it clearly cannot be pierced by less than m points. An (ℵ0, k + 2)-theorem for “fat” sets in Rd, with respect to k-transversals. Our main result concerns (p, q)-theorems with respect to k-transversals. In this setting, the construction presented in [2, Sec. 9] suggests that no (ℵ0, k + 2)-theorem with respect to k-transversals can be obtained for general families of convex sets in Rd where k < d −1, since even the stronger (d + 1, d + 1)-property does not imply a bounded-sized k-transversal. However, we show that if the convexity assumption is replaced by an assumption that the elements of the family are “fat”,1 then an (ℵ0, k + 2)-theorem can be obtained. ▶Definition 4. Let 0 < r ⩽R. A family F of sets in Rd is called (r, R)-fat if any A ∈F contains a ball of radius r and is contained in a ball of radius R. ▶Theorem 5. Let 0 < r ⩽R, 0 ⩽k ⩽d −1, and let F be an infinite (r, R)-fat family of sets in Rd. If among every ℵ0 elements of F, some k + 2 can be pierced by a k-flat, then F can be pierced by a finite number of k-flats. 1 We note that a “fatness” assumption was considered in the context of (p, q)-theorems for families of convex sets in the plane, by Gao and Zerbib . SoCG 2022 50:4 An (ℵ0, k + 2)-Theorem for k-Transversals Theorem 5 allows significantly weakening the (p, q)-property assumption of “classical” (p, q)-theorems into an (∞, q)-property assumption, it applies to k-transversals for all 0 ⩽k ⩽d−1 (while the (p, q)-theorem for k-transversals holds only for k = 0, d−1), and it does not require the sets in the family to be convex. On the other hand, it requires a significant additional assumption – namely, that the elements of the family are “fat”. We show by an explicit construction that this assumption is essential. ▶Proposition 6. There exists an infinite family F of open discs in the plane that satisfies the (3, 3)-property (and so, also the (ℵ0, 3)-property) with respect to 1-transversals (i.e., piercing by lines), but cannot be pierced by a finite number of lines. Note that such a strong example could not be obtained for families of closed discs in the plane, since by Theorem 2, a family of compact convex sets in the plane that satisfies the (3, 3)-property with respect to piercing by lines, admits a bounded-sized line transversal. An infinite Ramsey-type theorem. In , Larman et al. observed that every (p, 2)-theorem can be used to obtain a Ramsey-type theorem. Using a similar argument (presented in Sec. 7), Theorem 5 can be used to obtain the following Ramsey-type result. ▶Corollary 7. Let 0 < r ⩽R, 0 ⩽k ⩽d −1, and let F be an infinite (r, R)-fat family of sets in Rd. Denote α = \|F\|. Then one of the following holds: There exists S ⊂F with \|S\| = ℵ0 such that no k+2 elements of S can be pierced by a k-flat. There exists S′ ⊂F with \|S′\| = α, such that every k + 2 elements of S′ can be pierced by a k-flat. For α > ℵ0 and k ⩾1, the assertion of Corollary 7 is significantly stronger than the best possible “generic” Ramsey theorem that can be obtained in the same setting. Indeed, the corresponding Ramsey-type theorem concerns (blue, red)-colorings of all r-element subsets of a set with cardinality α, for r ⩾3. In this setting, Erdős and Rado [14, Thm. 28] showed that in general, one cannot guarantee even the existence of either a set of r + 1 elements all of whose r-tuples are blue or a set of cardinality α all of whose r-tuples are red. Corollary 7 provides either an “all-blue” set with cardinality ℵ0 or an “all-red” set with cardinality α (of course, for the specific coloring in which a (k + 2)-tuple is colored blue if it can be pierced by a k-flat). This provides yet another example of the phenomenon that graphs and hypergraphs arising in geometry satisfy much stronger forms of Ramsey’s theorem than arbitrary graphs and hypergraphs. This phenomenon was demonstrated in several works in the finite setting (see [4, 7, 15, 23]), and our result provides an infinitary example. Organization of the paper. In Section 2 we present some definitions, notations, and basic observations. In Section 3 we prove a lemma which shall be used in the proof of Theorem 5. Then, in Section 4 we prove Theorem 5. The construction of Proposition 6 is presented in Section 5, and the proof of Theorem 3 is given in Section 6. A more detailed comparison of Corollary 7 with generic Ramsey results is presented in Section 7. We conclude the paper with an open problem in Section 8. C. Keller and M. A. Perles 50:5 2 Definitions, Notations, and Basic Observations 2.1 Definitions and notations We use the following classical definitions. For 0 ⩽k ⩽d −1, a k-flat in Rd is a k-dimensional affine subspace of Rd (namely, a translation of a k-dimensional linear subspace of Rd). In particular, a 0-flat is a point, an 1-flat is a line, and a (d −1)-flat is a hyperplane. The direction of a k-flat (k > 0) in Rd is defined as follows. First, the k-flat is translated such that it will pass through the origin. Then, its direction is defined as the great (k −1)-sphere in which the k-flat intersects the sphere Sd−1. (This definition follows .) A k1-flat and a k2-flat are called parallel if the direction of one of them is contained in the direction of the other. (Equivalently, this means that if both are translated so that they will pass through the origin, then one translation will be included in the other. Note that this relation is not transitive, and that two flats of the same dimension are parallel, if and only if one of them is a translation of the other.) For ϵ > 0, an (open) ϵ-neighborhood of a point x ∈Sd−1 on the sphere is B◦(x, ϵ) ∩Sd−1, where B◦(x, ϵ) is the open ball with radius ϵ centered at x. A family F = {Bα}α of sets in Rd, is independent w.r.t. k-flats if no k-flat π ⊂Rd intersects k + 2 Bα’s or more. In the proofs of the theorems in the sequel, we mostly consider families F of closed unit balls in Rd, d ⩾1, no two of them are equal. We always assume w.l.o.g. that F does not contain a ball centered at the origin, since all such balls are pierced by a single point, and hence by a single k-flat. We use the following definitions and notations: For B = B(x, 1) ∈F, the direction of B is the point ˆ x = x/\|\|x\|\|2. Of course, ˆ x ∈Sd−1. For any ˆ x ∈Sd−1 (which is not necessarily a direction of a ball in F) and any ϵ > 0, the (open) ϵ-neighborhood of ˆ x in F is Fˆ x,ϵ = {B(y, 1) ∈F : ˆ y ∈B◦(ˆ x, ϵ) ∩Sd−1}, that is, the set of all elements of F whose directions are in an ϵ-neighborhood of ˆ x. For convenience, we often focus on the point ˆ x = (0, 0, . . . , 0, 1) ∈Sd−1, on the line ℓ= {tˆ x : t ∈R}, and on projections onto the hyperplane orthogonal to ℓ(i.e., projections onto the first d −1 coordinates.) For each B = B(x, 1) ∈F, we denote by B′ ⊂Rd−1 and x′ ∈Rd−1 the projections of B and x, repsectively. The d’th coordinate of x, omitted in the projection, is denoted by x(d). 2.2 Basic claims and observations We use the two following simple claims. ▷Claim 8. Let ˆ x = (0, 0, . . . , 0, 1), and let {B(xn, 1)}n=1,2,... be a sequence of pairwise disjoint unit balls in Rd such that limn→∞ˆ xn = ˆ x. Then limn→∞xn(d) = ∞. Proof. Let 0 < M ∈R, ϵ > 0. There exists n1 ∈N such that for any n > n1, ˆ xn is in the ϵ-neighborhood of ˆ x. The set Fˆ x,ϵ ∩{B(xn, 1) : n > n1, xn(d) < M} is contained in a finite area (which is a function of ϵ, d and M). By the disjointness of the balls in F, there exists n2 > n1 such that for any n > n2, xn(d) ⩾M. ◁ SoCG 2022 50:6 An (ℵ0, k + 2)-Theorem for k-Transversals ▷Claim 9. Let F ⊂Rd be a family of balls of radius 1, and let G be a family of balls of radius r > 0, with the same centers. Then for any 0 ⩽k ⩽d −1, F can be pierced by a finite set of k-flats if and only if G can be pierced by a finite set of k-flats. Proof. Assume w.l.o.g. that r > 1. If F can be pierced by finitely many k-flats, then the same clearly holds for G as well, as the elements of F are contained in corresponding elements of G. Assume that G can be pierced by finitely many k-flats, and take a finite family H of k-flats that pierces it. Replace each k-flat π in H by a sufficiently dense net of k-flats parallel to it, whose distance from π is at most 2r. It is clear that the resulting finite family of k-flats pierces F. ◁ 3 A Technical Lemma In the proof of Theorem 5, we shall need the following lemma. ▶Lemma 10. Let F be a family of closed unit balls in Rd, let 0 ⩽k ⩽d −1, and let ˆ x = (0, 0, . . . , 0, 1). Assume that for any ϵ > 0, the set Fˆ x,ϵ cannot be pierced by a finite collection of k-flats. Then there exists a sequence of balls, {B(xn, 1)}n=1,2,3,... ⊂F such that limn→∞ˆ xn = ˆ x and the sequence cannot be pierced by a finite family of k-flats. We derive Lemma 10 from the following proposition. ▶Proposition 11. Let F be a family of closed unit balls in Rd, let 0 ⩽k ⩽d −1 and m ∈N. If any finite subfamily of F can be pierced by at most m k-flats, then F can be pierced by at most m k-flats. We first derive Lemma 10 from Proposition 11, and then present the proof of the proposition. Proof of Lemma 10, assuming Proposition 11. Let F, ˆ x be as in the statement of the lemma, and assume that for any ϵ > 0, the set Fˆ x,ϵ cannot be pierced by a finite collection of k-flats. We construct the sequence of balls {B(xn, 1)}n=1,2,3,... ⊂F as follows. We take a sequence {ϵm}m=1,2,3,..., where ϵm = 1/m. For each m ∈N, we find in Fˆ x,ϵm a finite family Gm of balls that cannot be pierced by m k-flats (this is possible by Proposition 11). We define the sequence {B(xn, 1)}n=1,2,3,... as S m∈N Gm. Namely, we arbitrarily order the balls in each Gm and add them to the sequence, allowing repetitions, starting with m = 1, proceeding to m = 2, etc.. We have limn→∞ˆ xn = ˆ x, since for any ϵ > 0, only a finite number of B(xn, 1)’s do not belong to Fˆ x,ϵ. Furthermore, {B(xn, 1)}n=1,2,3,..., cannot be pierced by m k-flats (for any m ∈N) since it contains the family Gm that cannot be pierced by m k-flats by its construction. Hence, {B(xn, 1)}n=1,2,3,... cannot be pierced by a finite number of k-flats, as asserted. ◀ Proof of Proposition 11. Any k-flat π ⊂Rd can be represented as π = {c + λ1v1 + . . . + λkvk : λi ∈R}, where c is the point on π closest to the origin, and {v1, . . . , vk} is an orthonormal basis of the vector subspace π −π = {x −y : x, y ∈π} which is parallel to π. (This actually means that the vector ⃗ 0c is orthogonal to each vi.) C. Keller and M. A. Perles 50:7 Assign to each k-flat π all the corresponding (k + 1)-tuples of the type {c, v1, . . . , vk}. Note that while c is uniquely determined by π, the orthogonal basis is not. We obtain a representation of all k-flats in Rd as (k + 1)-tuples of d-vectors A = {(c, v1, . . . , vk) : c, vi ∈Rd ∧∀i ̸= j, vi ⊥vj ∧vi ⊥c ∧\|\|vi\|\| = 1} ⊂Rd(k+1). By the conditions of the proposition, we can assume w.l.o.g. that there exists a finite sub-family F0 ⊂F that cannot be pierced by m −1 k-flats. Let Am = {(c1, v1 1, . . . , v1 k, . . . , cm, vm 1 , . . . , vm k ) : ∀1 ⩽j ⩽m, (cj, vj 1, . . . , vj k) ∈A} ⊂Rd(k+1)m represent m-tuples of k-flats in Rd. Note that Am is not compact (as a subset of Rd(k+1)m), since \|\|cj\|\| may be arbitrarily large. However, for any fixed closed unit ball B = B(x0, 1) ⊂Rd, the subset ΠB ⊂Am that represents all m-tuples of k-flats intersecting B ∪F0, is a compact subset of Am. Indeed, all the m coordinates cj, satisfy \|\|cj\|\| ⩽max{\|\|x0\|\| + 1, maxB′∈F0 dist(B′, 0) + 2}. Consider the family {ΠB}B∈F where ΠB represents all m-tuples of k-flats that pierce B ∪F0. (For each such m-tuple of k-flats, we take all possible (d(k + 1)m)-tuples that represent it.) Each ΠB is compact, and by the assumption, any finite sub-family {ΠBi}n i=1 has non-empty intersection (that contains the representation of some m-tuple of k-flats that together intersect F0, B1, . . . , Bn). Therefore, by the finite intersection property of compact sets, the whole family {ΠB}B∈F has non-empty intersection. Any element in this non-empty intersection represents an m-tuple of k-flats that pierce together all the balls in F. ◀ ▶Remark 12. Proposition 11 holds not only when F is a family of unit balls, but actually for any family F of non-empty compact sets in Rd. 4 Proof of the Main Theorem We restate Theorem 5, in a formulation that will be more convenient for the proof: ▶Theorem 5 (restated). Let R, r > 0 and let F be a family of sets in Rd such that each S ∈F contains a ball of radius r and is contained in a ball of radius R. Let 0 ⩽k ⩽d −1. Then one of the two following conditions must hold: F can be pierced by a finite number of k-flats. F contains an infinite sequence of sets that are independent w.r.t. k-flats (i.e., no k-flat pierces k + 2 of them). First, we observe that it is sufficient to prove Theorem 5 for families of closed unit balls in Rd. ▷Claim 13 (Reduction to closed unit balls). If the assertion of Theorem 5 holds for all families of closed unit balls in Rd, then it holds in the full generality stated in the theorem. Proof. Let F be a family as in the assumption. Construct a family F1 by taking, for each S ∈F, a closed ball of radius r contained in S. (Note that we can make sure that the centers of these balls are distinct, possibly at the price of reducing their radii to r/2.) Then, construct another family F2 by taking, for each S ∈F, a closed ball of radius 2R that contains S, with the same center as the corresponding ball in S1. Apply Theorem 5 to F2. If it contains an infinite sequence of balls that are independent w.r.t. k-flats, then so does F, since for each element of the sequence, we can take the element of F that corresponds to it, and the resulting sequence of elements of F will clearly be independent as well. SoCG 2022 50:8 An (ℵ0, k + 2)-Theorem for k-Transversals Otherwise, F2 can be pierced by a finite number of k-flats. Hence, by Claim 9, F1 can be pierced by a finite number of k-flats as well. This implies that F can be pierced by a finite number of k-flats, since any element of F contains an element of F1. Therefore, the assertion of the theorem holds for F. ◁ By Claim 13, it is sufficient to prove Theorem 5 for families of closed unit balls in Rd. A second reduction, before procceeding to the proof, is passing to pairwise disjoint unit balls. ▷Claim 14 (Reduction to pairwise disjoint balls). If the assertion of Theorem 5 holds for all families of pairwise disjoint closed unit balls in Rd, then it holds in the full generality stated in the theorem. Proof. By Claim 13, it is sufficient to prove that if Theorem 5 holds for all families of pairwise disjoint closed unit balls in Rd, then it holds for any family of closed unit balls. Indeed, assume correctness for all families of pairwise disjoint closed unit balls in Rd, and let F be a family of arbitrary closed unit balls in Rd. First, we pass to a subfamily ¯ F ⊂F of pairwise disjoint balls, which is maximal under inclusion: Consider the family G of all subsets of F in which all balls are pairwise disjoint. View G as a poset with respect to inclusion. As each chain in G has a maximal element (which is the union of its elements), by Zorn’s lemma G has a maximal element. This maximal element ¯ F ⊂F is a set of pairwise disjoint balls, which is maximal under inclusion, among all the pairwise disjoint subfamilies. By assuming correctness of Theorem 5 for families of pairwise disjoint closed unit balls, either ¯ F contains an infinite sequence { ¯ Fn}n∈N of balls that are independent w.r.t. k-flats, or ¯ F can be pierced by a finite number of k-flats. In the first case, { ¯ Fn}n∈N ⊂¯ F ⊂F satisfies the second assertion of Theorem 5. In the second case, by the maximality of ¯ F, any ball in F intersects some ball in ¯ F. Therefore, by replacing each k-flat in the finite piercing set of ¯ F, by a sufficiently dense net of k-flats surrounding it and parallel to it, we obtain a finite piercing set of k-flats for F, that satisfies the first assertion of Theorem 5. ◁ The proof of Theorem 5 is by induction, passing from (k−1, d−1) to (k, d). The induction basis is the case k = 0 of Theorem 5, reduced to a family of closed unit balls, by Claim 13. (The reduction to disjoint balls is not needed here.) We observe: ▶Observation 15. Let F be a family of (not neccessarily disjoint) closed unit balls in Rd. Then one of the two following conditions must hold: F can be pierced by a finite number of points. F contains an infinite sequence of pairwise disjoint balls. Proof. Consider the set A = {x ∈Rd\|B(x, 1) ∈F} of all centers of balls in F. If A is bounded in some B(0, R) ⊂Rd, then clearly a finite set of points pierces all elements of F. Otherwise, A is unbounded, hence F contains an infinite sequence of pairwise disjoint balls, that can be obtained inductively. ◀ For d = 1, the assertion of Theorem 5, after applying the reduction of Claim 13, is exactly Observation 15. For d ⩾2, we shall prove the following version, which is sufficient due to the reductions of Claims 13 and 14: C. Keller and M. A. Perles 50:9 ▶Theorem 16. Let d ⩾2 and 0 ⩽k ⩽d−1. Let F be a family of pairwise disjoint closed unit balls in Rd, and assume w.l.o.g. that F does not contain a ball centered at the origin. Then: 1. If for any ˆ x ∈Sd−1, there exists ϵ(ˆ x) = ϵ > 0 such that Fˆ x,ϵ can be pierced by finitely many of k-flats, then F can be pierced by finitely many of k-flats. 2. If the condition of (1) does not hold, then F contains an infinite sequence of balls that are independent w.r.t. k-flats (i.e., no k-flat pierces k + 2 of them). Proof. First, we give the proof of the first assertion. Assume that for any ˆ x ∈Sd−1, there exists ϵ = ϵ(ˆ x) > 0 such that Fˆ x,ϵ can be pierced by a finite number of k-flats. Pick such an ϵ(ˆ x) for each ˆ x ∈Sd−1, and obtain an open covering of Sd−1 by open balls B(ˆ x, ϵ(ˆ x)), for all ˆ x ∈Sd−1. By the compactness of the sphere, we can find a finite sub-cover, generated by balls around ˆ x1, . . . , ˆ xn. As each Fˆ xi,ϵ(ˆ xi) can be pierced by a finite number of k-flats, we can pierce all elements of F by a finite collection of k-flats (which is the union of the k-flats that pierce Fˆ xi,ϵ(ˆ xi), for i = 1, . . . , n). Now we move to the second assertion. Assume that for some ˆ x ∈Sd−1 and for any ϵ > 0, the family Fˆ x,ϵ cannot be pierced by a finite number of k-flats. We assume w.l.o.g. that ˆ x = (0, 0, . . . , 1). We shall construct a sequence of elements of F that is independent w.r.t. k-flats. The construction goes by induction, which reduces from k-flats in Rd to (k −1)-flats in Rd−1. Induction basis: k = 0. This case, which concerns piercing by points, follows by the argument of Observation 15. Induction step: From (k −1, d −1) to (k, d). Assume that we proved the assertion for families in Rd−1, with respect to piercing by (k −1)-flats, and consider a family F ⊂Rd of pairwise disjoint closed unit balls. First, we use Lemma 10 to find a sequence G = {B(xn, 1)}n=1,2,... of elements of F such that limn→∞ˆ xn = ˆ x and the sequence cannot be pierced by a finite number of k-flats. From now on, we restrict ourselves to this sequence. We project each B(xn, 1) ∈G onto its first d −1 coordinates. Let the resulting set be G′, and similarly to the proof of Claim 14, let G′′ ⊂G′ be a subset of pairwise disjoint balls, maximal under inclusion in G′. By the induction hypothesis, either G′′ (and therefore G′) can be pierced by a finite number of (k −1)-flats in Rd−1, or else it contains a sequence of (d −1)-dimensional balls that are independent w.r.t. (k −1)-flats. The first option cannot happen, as otherwise, one could pierce G with a finite number of k-flats (which are the pre-images of the (k−1)-flats-transversal in Rd−1 under the projection), contrary to the choice of G. Hence, there exists a sub-sequence ¯ G = {B(xnl, 1)}l=1,2,... ⊂ G of balls whose projections are independent w.r.t. (k −1)-flats in Rd−1. Note that as limn→∞ˆ xn = ˆ x, we have liml→∞ˆ xnl = ˆ x. From now on, we restrict ourselves to this sequence and construct inductively a subsequence of it that will be independent w.r.t. k-flats in Rd. We construct the subsequence {B(xn, 1)}∞ n=1 inductively. ({xn}∞ n=1 is a subsequence of {xnl}∞ l=1.) The first k + 1 elements can be chosen arbitrarily. Assume that we already chose the balls B(x1, 1), . . . , B(xm, 1), for m ⩾k + 1. To choose B(xm+1, 1), we first look at each (k+1)-tuple of balls (B(xi1, 1), . . . , B(xik+1, 1)) separately. By assumption, the corresponding projections on the first d −1 coordinates cannot be pierced by a (k −1)-flat in Rd−1. This implies that no k-flat that is parallel to the line ℓ= {tˆ x : t ∈R} can pierce all the k + 1 balls B(xi1, 1), . . . , B(xik+1, 1). SoCG 2022 50:10 An (ℵ0, k + 2)-Theorem for k-Transversals t0 Cϵ0 x y B(x1, 1) B(x3, 1) B(x2, 1) Figure 1 An illustration for the proof of Theorem 16 for d = 2,k = 1. Consider the family U of all k-flats that pierce (B(xi1, 1), . . . , B(xik+1, 1)). As none of them is parallel to ℓ, neither of their directions2 contains the point (0, 0, . . . , 0, 1) = ˆ x ∈Sd−1. By compactness of the elements of F, this implies that there exists ϵ0 > 0, such that all these directions are disjoint with the ϵ0-neighborhood of ˆ x on Sd−1. Now, let Cϵ0 be the unbounded cone whose vertex is the origin and whose intersection with Sd−1 is the boundary of ϵ0/2-neighborhood of ˆ x on Sd−1. (Informally, this is a cone of small aperture around the positive direction of the d’th axis.) We claim that there exists t0 ∈R such that for any t > t0, the translation (0, 0, . . . , 0, t) + Cϵ0 is disjoint from all k-flats in U (see Figure 1). To see this, for each k-flat L ∈U we define a function fL : R →R⩾0 by fL(t) = min{\|\|x′\|\| : x ∈L ∧x(d) = t} (for all the relevant notations, see the end of Section 2.1). It is clear that fL attains a minimum, and that since L is not parallel to ℓ, this minimum is attained in a single point, t = argmin(fL) ∈R. Now, we define a function g : U →R by g(L) = argmin(fL). By compactness of the elements of F, this function attains a maximum, t0. As the direction of any L ∈U is disjoint with the ϵ0-neighborhood of ˆ x on Sd−1, it follows that L ∩((0, 0, . . . , 0, t) + Cϵ0) = ∅, for all t > t0. We are now ready to choose the ball B(xm+1, 1). We go over all (k + 1)-tuples of balls (B(xi1, 1), . . . , B(xik+1, 1)) with 1 ⩽i1 < i2 < . . . < ik+1 ⩽m. For each of them, we find (ϵ0, t0) such that for any t > t0, any k-flat that pierces (B(xi1, 1), . . . , B(xik+1, 1)) is disjoint with the cone (0, 0, . . . , 0, t) + Cϵ0, where Cϵ0 is as defined above. Let ϵ1 be the minimum of the ϵ0 values, and let t1 be the maximum of the t0 values. It is clear that if we make sure that B(xm+1, 1) is entirely included in the cone (0, 0, . . . , 0, t1 +1)+ Cϵ1, then no k-flat will pierce both B(xm+1, 1) and a (k +1)-tuple (B(xi1, 1), . . . , B(xik+1, 1)). We can indeed choose B(xm+1, 1) in this way, by Claim 8. This completes the proof. ◀ 2 See Section 2.1 for the needed definitions. C. Keller and M. A. Perles 50:11 Figure 2 An illustration for Section 5. 5 Proof of Proposition 6 In this section we prove Proposition 6. Namely, we construct an infinite family of open discs in the plane that satisfies the (3, 3)-property with respect to line transversals, but cannot be pierced by a finite number of lines. Proof of Proposition 6. Let F = {Fn}∞ n=1 ⊂R2, where Fn = B(n, 1/n) is an open disc centered at (n, 1 n) with radius 1 n. The family F does not admit a finite line transversal, since the x-axis meets no element of F, any line that is parallel to the x-axis meets finitely many elements of F, and any line that forms a positive angle with the x-axis, intersects a finite subfamily of F. On the other hand, any F′ ⊂F which is independent w.r.t. lines, satisfies \|F′\| ⩽2. Indeed, consider the two leftmost discs B1, B2 ∈F′. The right wedge that the two common inner tangents of B1 and B2 form, contains all elements of F that are to the right of B1 and B2 (see Figure 2). Therefore, any element of F that lies to the right of B1 and B2 is pierced by a line that passes through B1 and B2, and hence cannot be contained in F′. ◀ We note that no similar example could be constructed with closed balls, since by the Danzer-Grünbaum-Klee theorem , such a family would be pierced by a single line. 6 An 0-flat Transversal With no Restriction on the Radii In this section we prove Theorem 3, which is a much stronger version of Observation 15. This stronger version holds with no restriction on the radii. Let us restate the theorem in a formulation which is more convenient for us: ▶Theorem 3 (restated). Let F be a family of closed balls in Rd (with no restriction on the radii). Then one of the two following conditions must hold: F can be pierced by a finitely many points. F contains an infinite sequence of pairwise disjoint balls. Before proceeding into the proof, we prove a reduction to the case where all elements of F are contained in a closed bounded ball B(0, R) ⊂Rd. ▷Claim 17. Let R > 0 and assume we proved Theorem 3 where any ball in F is contained in B(0, R). Then Theorem 3 holds. Proof. Define the distance of a closed ball B ⊂Rd from the origin, dist(B, 0), as the Euclidian distance between the origin and the point x ∈B which is closest to the origin. If the set {dist(B, 0) : B ∈F} is unbounded in Rd, then one can inductively construct an infinite sequence of pairwise disjoint balls in F, whose distance from the origin tends to infinity. SoCG 2022 50:12 An (ℵ0, k + 2)-Theorem for k-Transversals From now on we assume that there exists some 0 < R ∈R such that for any B ∈F, dist(B, 0) ⩽R −2. Replace each B ∈F whose radius r(B) > 1, by some closed smaller ball B′ ⊂B with r(B′) = 1, such that dist(0, B) = dist(0, B′). Let F′ be the obtained family. Any ball in F′ is contained in B(0, R). By the assumption of our claim, either F′ can be pierced by finitely many points, or F′ contains an infinite sequence F′′ ⊂F′ of pairwise disjoint balls. In the first case, the finite piercing set of F′ pierces F as well. In the second case, remove from F′′ all balls with radius 1. There are only finitely many such balls, since F′′ ⊂F′ ⊂B(0, R), and the elements of F′′ are pairwise disjoint. After removing from F′′ all balls with radius 1, we are left with an infinite subfamily of balls each of which belongs to F (since the transition from F to F′ involved only the radius-1 balls of F′), which are pairwise disjoint. ◁ Proof of Theorem 3. By Claim 17 we can assume that there exists R > 0 such that each ball in F is contained in B(0, R). We can assume w.l.o.g. that F contains no ball of radius 0. Indeed, if F contains finitely many such balls, we can remove them without changing the assertion. Otherwise, F contains an infinite sequence of radius-0 balls, and then we are done again. Each x ∈B(0, R) is of exactly one of the two following types: Type (a): For each δ > 0, there exists some B ∈F, B ∩B◦(x, δ) ̸= ∅, with r(B) < δ and B ∩{x} = ∅. Type (b): There exists 0 < δ = δ(x) such that for any B ∈F with B ∩B◦(x, δ) ̸= ∅, the following holds: Either r(B) ⩾δ or B ∩{x} ̸= ∅. If B(0, R) contains some point x of type (a), then there exists an infinite sequence of pairwise disjoint balls in F (that tends to {x}). Indeed, start with δ0 = 1 and pick some B0 ∈F, b0 ∩B◦(x, δ0) ̸= ∅, with r(B0) < δ0 and B0 ∩{x} = ∅. Since B0 is closed, it has a positive distance ϵ from x. Let δ1 = ϵ 10 and pick some B1 ∈F, B1 ∩B◦(x, δ1) ̸= ∅, r(B1) < δ1 and B1 ∩{x} = ∅. Continue in the same manner to construct an infinite sequence of pairwise disjoint balls in F. The remainig case is where each x ∈B(0, R) is of type (b). Then for each x ∈B(0, R) there exists 0 < δ = δ(x) such that any B ∈F that that intersects B◦(x, δ) can be pierced by finitely many points, say, by f(x) points. (Note that the exact value of f(x) depends on the choice of δ = δ(x).) By the finite intersection property of compact sets in Rd, the open cover S x∈B(0,R) B◦(x, δ(x)) of B(0, R) has a finite sub-cover B(0, R) ⊂Sk i=1 B◦(xi, δ(xi)). Since all the balls in F that intersect B◦(xi, δ(xi)) can be pierced by f(xi) points, it follows that all the elements of F can be pierced by at most Σk i=1f(xi) points. ◀ 7 Comparison of Corollary 7 with Generic Infinite Ramsey-type Theorems We begin with a restatement of Corollary 7. ▶Corollary 7 (restated). Let 0 < r ⩽R, 0 ⩽k ⩽d −1, and let F be an infinite (r, R)-fat family of sets in Rd. Denote α = \|F\|. Then one of the following holds: There exists S ⊂F with \|S\| = ℵ0 s.t. no k + 2 elements of S can be pierced by a k-flat. There exists S′ ⊂F with \|S′\| = α, s.t. every k+2 elements of S′ can be pierced by a k-flat. Proof. If the first condition does not hold, then F satisfies the (ℵ0, k + 2) property, and hence by Theorem 5, F can be pierced by a finite number of k-flats L1, L2, . . . , Ln. Denote Fi = {A ∈F : A ∩Li ̸= ∅}. At least one of the families Fi is of cardinality α, and every k + 2 elements of it can be pierced by a k-flat. ◀ C. Keller and M. A. Perles 50:13 For α = ℵ0, Corollary 7 is not interesting, as it follows directly from the infinite Ramsey theorem . For α > ℵ0 and k = 0 (i.e., piercing by points), Corollary 7 is already significantly stronger than the conclusion of the “diagonal” Ramsey’s theorem, which guarantees only a countable monochromatic subset. However, it is still uninteresting since it follows from the Erdős-Dushnik-Miller theorem , which asserts that for any infinite α, any (blue, red)-coloring of a graph on α vertices contains either a monochromatic blue set of cardinality ℵ0 or a monochromatic red set of cardinality α. The interesting case is k ⩾1 – i.e., piercing by k-flats with k ⩾1, which is the hard case in Theorem 5. Here, the corresponding Ramsey-type theorem concerns (blue, red)-colorings of all r-element subsets of a set with cardinality α, for r ⩾3. In this setting, Erdős and Rado [14, Thm. 28] showed that in general, one cannot guarantee even the existence of either a set of r + 1 elements all of whose r-tuples are blue or a set of cardinality α all of whose r-tuples are red. Corollary 7 provides either an “all-blue” set with cardinality ℵ0 or an “all-red” set with cardinality α (of course, for the specific coloring in which a (k + 2)-tuple is colored blue if it can be pierced by a k-flat). Therefore, in its “main” setting of k ⩾1, Theorem 5 provides an infinite Ramsey theorem which is significantly stronger than the best possible “generic” Ramsey theorems. Moreover, the assertion of Corollary 7 cannot be strengthened to obtain a first possibility with \|S\| > ℵ0, since once no k + 2 elements of S can be pierced by a k-flat, all elements of S must be pairwise disjoints; hence \|S\| ⩽ℵ0. 8 Open Problem A natural open problem which arises in light of Theorem 3 and Proposition 6 is, whether an (ℵ0, k + 2)-theorem (like Theorem 5) can be obtained for families of closed balls, without the “fatness” assumption. For 1 ⩽k < d −1, such a theorem cannot be obtained for general families of compact convex sets, as shown by the construction of Alon et al. . However, it still might hold for families of balls. References 1 N. Alon and G. Kalai. Bounding the piercing number. Discrete Comput. Geom., 13:245–256, 1995. 2 N. Alon, G. Kalai, J. Matoušek, and R. Meshulam. Transversal numbers for hypergraphs arising in geometry. Adv. Appl. Math., 29:79–101, 2002. 3 N. Alon and D. J. Kleitman. Piercing convex sets and the Hadwiger-Debrunner (p,q)-problem. 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Klee, editor, Convexity, Proceedings of Symposium in Pure Mathematics, volume 7, pages 100–181. American Mathematical Society, Providence, RI, 1963. SoCG 2022 50:14 An (ℵ0, k + 2)-Theorem for k-Transversals 11 A. Dumitrescu and M. Jiang. Piercing translates and homothets of a convex body. Algorithmica, 61(1):94–115, 2011. 12 B. Dushnik and E. W. Miller. Partially ordered sets. American Journal of Mathematics, 63(3):600–610, 1941. 13 J. Eckhoff. A survey of the Hadwiger-Debrunner (p, q)-problem. In B. Aronov, S. Basu, J. Pach, and M. Sharir, editors, Discrete and Computational Geometry, volume 25 of Algorithms and Combinatorics, pages 347–377. Springer Berlin Heidelberg, 2003. 14 P. Erdős and R. Rado. A partition calculus in set theory. Bulletin of the American Mathematical Society, 62(5):427–489, 1956. 15 J. Fox, J. Pach, and C. D. Tóth. Intersection patterns of curves. J. London Math. Society, 83:389–406, 2011. 16 S. Gao and S. Zerbib. The (2,2) and (4,3) properties in families of fat sets in the plane. SIAM Journal of Discrete Math., 33(3):1326–1337, 2019. 17 B. Grünbaum. On intersections of similar sets. Portugaliae Mathematica, 18:155–164, 1959. 18 H. Hadwiger and H. Debrunner. Über eine variante zum Hellyschen satz. Archiv der Mathematik, 8(4):309–313, 1957. 19 E. Helly. Uber mengen konvexer körper mit gemeinschaftlichen punkte. Jahresbericht der Deutschen Mathematiker-Vereinigung, 32:175–176, 1923. 20 A. Holmsen and R. Wenger. Helly-type theorems and geometric transversals. In J. O’Rourke J. E. Goodman and C. D. Tóth, editors, Handbook of Discrete and Computational Geometry, 3rd Edition, pages 91–123. CRC Press LLC, Boca Raton, FL, 2017. 21 A. F. Holmsen and D. Lee. Radon numbers and the fractional Helly theorem. Isr. J. Math., 241:433–447, 2021. 22 S. J. Kim, K. Nakprasit, M.J. Pelsmajer, and J. Skokan. Transversal numbers of translates of a convex body. Discrete Math., 306:2166–2173, 2006. 23 D. Larman, J. Matou˘ sek, J. Pach, and J. Töröcsik. A ramsey-type result for planar convex sets. Bulletin of London Math. Soc., 26:132–136, 1994. 24 J. Matoušek. Bounded VC-dimension implies a fractional Helly theorem. Discrete Comput. Geom., 31(2):251–255, 2004. 25 A. Montejano, L. Montejano, E. Roldán-Pensado, and P. Soberón. About an Erdős–Grünbaum conjecture concerning piercing of non-bounded convex sets. Discrete Comput. Geom., 53(4):941– 950, 2015. 26 S. Moran and A. Yehudayoff. On weak ϵ-nets and the Radon number. Discret. Comput. Geom., 64(4):1125–1140, 2020. 27 T. Müller. A counterexample to a conjecture of Grünbaum on piercing convex sets in the plane. Discrete Math., 313(24):2868–2871, 2013. 28 Z. Patáková. Bounding Radon number via Betti numbers. In 36th International Symposium on Computational Geometry, SoCG 2020, volume 164 of LIPIcs, pages 61:1–61:13. Schloss Dagstuhl - Leibniz-Zentrum für Informatik, 2020. 29 R. Pinchasi. A note on smaller fractional Helly numbers. Discrete Comput. Geom., 54:663–668, 2015. 30 F. P. Ramsey. On a problem of formal logic. Proc. London Math. Soc. (Ser. 2), 30(1):264–286, 1930. 31 L. Santaló. Un teorema sobre conjuntos de paralelepipedos de aristas paralelas. Publ. Inst. Mat. Univ. Nac. Litoral, 2:49–60, 1940. 32 P. Vincensini. Figures convexes et variétés linéaires de l’espace euclidien à n dimensions. Bull Sci. Math., 59:163–174, 1935.
10743	https://www.aan.com/Guidelines/home/GetGuidelineContent/689	©2015 American Academy of Neurology AAN.com Summary of Evidence-based Guideline for PATIENTS and their FAMILIES MANAGING AN UNPROVOKED FIRST SEIZURE IN ADULTS This information sheet is provided to help you understand the treatment options for adults who have had a first unprovoked seizure. It also looks at risk factors for a second seizure. This information is a service of the American Academy of Neurology (AAN) and the American Epilepsy Society (AES). The AAN is the world’s largest association of neurologists and neuroscience professionals. Neurologists are doctors who identify and treat diseases of the brain and nervous system. The AAN is dedicated to promoting the highest quality patient-centered neurologic care. The AES is one of the oldest professional organizations for neurologists in the United States. The AES promotes multispecialty communications, scientific research, and exchange of clinical information about epilepsy and seizure disorders. Experts from the AAN and AES carefully reviewed the available scientific studies on managing adults who have had a first unprovoked seizure. The following information is based on evidence from those studies. The information summarizes the main findings of the 2015 AAN and AES guideline on managing a first seizure in adults. There are separate guidelines on managing a first seizure in children. There also are separate guidelines on evaluating a first seizure in adults. To read these earlier guidelines, or to read the full 2015 guideline, visit AAN.com/guidelines .After a first seizure, it is important to understand the risk of another seizure. This risk is greatest within the first two years. Immediate treatment with an epilepsy drug can lower this risk. Patients should work with their doctors to decide if immediate treatment is right for them. What is a seizure? Seizures are events caused by abnormal firing of neurons (brain cells). Brain cells use electrical energy to “fire” messages to one another. When the cells fire abnormally, a seizure may happen. This causes a surge of electrical activity in the brain. For some people, a seizure is a one-time event. Other people may have recurrent seizures. These may occur the same day or over longer periods of time. Having a seizure can be frightening. The person may become confused and lose the ability to communicate. The person’s muscles may contract repeatedly. In some cases, the person may lose consciousness (black out). Injuries may result from muscle spasms or from falls. In very rare cases, death may occur. But usually seizures are brief events, and the person recovers completely. This guideline looks specifically at first seizures that are “unprovoked.” These happen for no immediately known reason. I recently had a seizure for the first time, and I don’t know why. How do I know if I am at risk of another seizure? An adult who has had a first seizure may be at risk for another seizure. The amount of risk depends on certain factors. These include a person’s background and health history. The risk also differs from person to person. Strong evidence shows that the risk is greatest within the first two years. This risk ranges from 21 percent to 45 percent. In other words, depending on background and health history, the risk: • May be as low as about a one-in-five chance (or about one in every five people) • May be as high as almost a one-in-two chance (or almost one in every two people) There are several risk factors that affect this risk. Strong evidence shows that the risk is greater for people: • Who have had a previous (pre-existing) brain problem, such as a head or brain injury, a stroke, or a brain tumor • Who have EEG test results that show signs of epilepsy There is moderate evidence that the risk is greater in people: • Whose results from brain imaging tests show a significant abnormality • Whose first seizure happened during sleep (a nocturnal seizure) ©2015 American Academy of Neurology AAN.com What can I do to prevent another seizure? Epilepsy drugs are drugs used to manage seizures in epilepsy. These drugs also are known as antiepileptic drugs, or AEDs. In some cases, doctors prescribe these drugs immediately after a first seizure. This is done in order to prevent more seizures. Studies have examined whether immediate treatment is helpful. The studies compared immediate treatment with waiting to treat until after another seizure happened. The results show moderate evidence that immediate treatment can lower the risk of another seizure within the first two years. However, the evidence is different for treatment over the longer term. According to the studies, beyond three years, getting immediate treatment after a first seizure: • Is unlikely to increase or decrease the likelihood of long-term improvement or seizure freedom (moderate evidence) • May not improve quality of life (weak evidence) How can I know if treatment is right for me? What are the risks? The decision whether to treat immediately after a first seizure is complex and controversial. Epilepsy drugs can cause side effects. There is moderate evidence that a person’s risk of having side effects ranges from seven percent to 31 percent. In other words, depending on background and health history, the risk: • May be as low as about a one-in-fifteen chance (or about one in every fifteen people) • May be as high as almost a one-in-three chance (or almost one in every three people) However, these side effects usually are mild and are not permanent. At the same time, risking another seizure can be dangerous. Seizures can lead to physical injury and can harm the brain. The use of epilepsy drugs also can affect lifestyle. For example, a person taking an epilepsy drug may be more likely to be given approval for driving. For people who have had a first seizure, it is very important to: • Understand their individual risk levels • Discuss care options with their doctors or care teams What other information is important to know? What should I ask my doctor or care team? More research is needed to better understand: • Health outcomes for people after a first seizure over the long term • How lifestyle choices are affected by seizures • Clearer information on risks of side effects from epilepsy drugs • If and when a person who starts an epilepsy drug and has no more seizures can safely stop taking the drug It is important to get answers to these questions from a doctor or care team: • What is my risk of another seizure? • If I take an epilepsy drug, what side effects might I have? • What is my risk of having side effects from an epilepsy drug? • What are my health outcomes long-term if I take an epilepsy drug? This guideline was endorsed by the American Neurological Association and the World Federation of Neurology. This statement is provided as an educational service of the American Academy of Neurology and the American Epilepsy Society. It is based on an assessment of current scientific and clinical information. It is not intended to include all possible proper methods of care for a particular neurologic problem or all legitimate criteria for choosing to use a specific procedure. Neither is it intended to exclude any reasonable alternative methodologies. The AAN and the AES recognize that specific patient care decisions are the prerogative of the patient and the physician caring for the patient, based on all of the circumstances involved. After the experts review all of the published research studies, they describe the strength of the evidence supporting each recommendation: Strong evidence = more than one high-quality scientific study Moderate evidence = at least one high-quality scientific study or two or more studies of a lesser quality Weak evidence = the studies, while supportive, are weak in design or strength of the findings Not enough evidence = either different studies have come to conflicting results or there are no studies of reasonable quality The AAN develops these summaries as educational tools for neurologists, patients, family members, caregivers, and the public. You may download and retain a single copy for your personal use. Please contact guidelines@aan.com to learn about options for sharing this content beyond your personal use. American Academy of Neurology, 201 Chicago Avenue, Minneapolis, MN 55415 Copies of this summary and additional companion tools are available at AAN.com or through AAN Member Services at (800) 879-1960. American Epilepsy Society, 135 South LaSalle Street, Suite 2850, Chicago, IL 60603 • (312) 883-3800 • AESnet.org
10744	https://www.apple.com/tv-pr/originals/shantaram/news/	Shantaram - News - Apple TV+ Press Apple Store Shop Shop the Latest Mac iPad iPhone Apple Watch Apple Vision Pro AirPods Accessories Quick Links Find a Store Order Status Apple Trade In Financing Personal Setup College Student Offer Shop Special Stores Certified Refurbished Education Business Veterans and Military Government Mac Explore Mac Explore All Mac MacBook Air MacBook Pro iMac Mac mini Mac Studio Mac Pro Displays Compare Mac Switch from PC to Mac Shop Mac Shop Mac Help Me Choose Mac Accessories Apple Trade In Financing College Student Offer More from Mac Mac Support AppleCare macOS Tahoe Apple Intelligence Apps by Apple Continuity iCloud+ Mac for Business Education Accessibility iPad Explore iPad Explore All iPad iPad Pro iPad Air iPad iPad mini Apple Pencil Keyboards Compare iPad Why iPad Shop iPad Shop iPad iPad Accessories Apple Trade In Financing College Student Offer More from iPad iPad Support AppleCare iPadOS 26 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10745	https://msgrantsphysics.weebly.com/uploads/2/1/3/3/21339258/polarization_ppt.pdf	Explain what causes the phenomena in the picture. Polarization What kind of waves can be polarized? • All transverse waves such as – Wave in string – Light • Plane polarized: wave is always in a single fixed plane • This string wave is vertically polarized; it can pass through a vertical “filter” but not a horizontal one Light waves are transverse waves • Light consists of perpendicular oscillating electric and magnetic fields • We can concentrate on the electric field in this discussion since it is the one our eyes perceive Polarizing Light by Absorption • Unpolarized light has electric fields in all directions • Passing unpolarized light through a filter reduces its intensity by 50% Polarization by Transmission • What happens when polarized light is passed through additional filters? • According to Malus’s Law: I = I0cos2 θ where I = transmitted intensity I0 = incident intensity θ = angle between e-field and filter’s transmission axis Malus’ Law Applet Polarizers and Analyzers • Polarizer: a polarizing filter used to produce polarized light • Analyzer: a polarizing filter used to determine if light is polarized • Unpolarized light’s intensity is reduced by 50% when passing through a polarizer/analyzer regardless of the orientation of the polarizer/analyzer • Already polarized light’s intensity is reduced depending on orientation of polarizer/analyzer according to Malus’s Law Polarization by Reflection When unpolarized light reflects off a non-metallic surface, the reflected ray is partially polarized Plane of polarization is parallel to the reflecting surface Brewster’s Angle • At a particular incident angle the reflected ray is 100% polarized along plane parallel to surface • This occurs when angle between reflected and refracted rays is 90° • Applying Snell’s Law: n1 sin θp = n2 sin (90°- θp) = n2 cos (θp) tan θp = n2/n1 tan ϕ = n Brewster’s Angle Video • zEKg • Note: This same David Brewster also invented the kaleidoscope! Optical Activity • Optical activity is the rotation of the plane of polarization • Materials showing this phenomenon are called optically active –Sugar solutions – Tartaric acid –Turpentine – Quartz crystals –Liquid crystals • Used in –Measuring solution concentrations –Liquid crystal displays A Liquid Crystal Display 1. Polarizing filter 2. Glass with shaped electrodes 3. Twisted liquid crystal 4. Glass with common electrode film 5. Polarizing filter 6. Reflective surface Polarimeter: used to determine conc. of optically active substances How Polarimeter Works Amount of rotation depends on (1) length of sample tube and (2) solution concentration Some compounds rotate the polarized light to the right (dextrorotary), others to the left (levorotary) Homemade Polarizers • 9BY • eQU&NR=1 Stress analysis using polarizers
10746	https://www.studypug.com/calculus-help/polar-coordinates	Converting Polar to Cartesian Equations: A Comprehensive Guide Unlock the power of coordinate conversion! Learn to effortlessly transform polar equations into Cartesian form. Master this crucial math skill and enhance your problem-solving abilities. Get the most by viewing this topic in your current grade. Pick your course now. Free to Join! Easily See Your Progress Make Use of Our Learning Aids Last Viewed Practice Accuracy Suggested Tasks Get quick access to the topic you're currently learning. See how well your practice sessions are going over time. Stay on track with our daily recommendations. Earn Achievements as You Learn Create and Customize Your Avatar Introduction to Polar Coordinates Welcome to the fascinating world of polar coordinates! Unlike the familiar Cartesian coordinates you've been using, polar coordinates offer a unique way to describe points in a plane. Instead of using x and y values, we use distance from a central point (called the pole) and an angle. This system is incredibly useful for describing circular motion or spiral patterns in mathematics and real-world applications. Our introduction video is designed to give you a clear, visual understanding of how polar coordinates work. It's a great starting point for grasping this concept, which might seem a bit tricky at first. As we explore polar coordinates together, you'll see how they can simplify certain problems that might be complex in Cartesian coordinates. From describing planetary orbits to analyzing circular motion, polar coordinates have a wide range of applications. So, let's dive in and discover how this powerful mathematical tool can expand your problem-solving toolkit! Here are some frequently asked questions about polar coordinates: 1. How do you convert polar coordinates to Cartesian coordinates? To convert polar coordinates (r, θ) to Cartesian coordinates (x, y), use these formulas: x = r cos(θ) y = r sin(θ) Where r is the radius and θ is the angle in radians. 2. What are the rules for converting polar equations to Cartesian equations? The general steps are: 1. Replace r with (x² + y²) 2. Replace cos(θ) with x/r, which becomes x/(x² + y²) 3. Replace sin(θ) with y/r, which becomes y/(x² + y²) 4. Simplify and rearrange the equation 3. How do you graph polar equations? To graph polar equations: 1. Create a table of r and θ values 2. Plot points on a polar grid using (r, θ) pairs 3. Connect the points smoothly 4. Pay attention to symmetry and periodicity 4. What are some real-world applications of polar coordinates? Polar coordinates are used in: 1. Navigation and GPS systems 2. Radar and sonar technology 3. Planetary motion and astronomy 4. Engineering, especially for circular or rotating systems 5. Weather mapping, particularly for cyclones and hurricanes 5. How do you convert Cartesian equations to polar form? To convert Cartesian equations to polar form: 1. Replace x with r cos(θ) 2. Replace y with r sin(θ) 3. Use the identity r² = x² + y² if needed 4. Simplify and solve for r or θ as required Mastering polar coordinates requires a solid foundation in several key mathematical concepts. One of the most crucial prerequisites is integration using trigonometric identities. This skill is essential because polar coordinates heavily rely on trigonometric functions to describe points and curves in a circular system. Understanding how to manipulate and integrate these identities will greatly enhance your ability to work with polar equations and solve complex problems in this coordinate system. Another fundamental skill is converting between degrees and radians. Polar coordinates often express angles in radians, so being comfortable with both units and the conversion process is crucial. This knowledge allows you to seamlessly switch between different representations of angles, which is frequently necessary when working with polar equations or graphing in the polar plane. The concept of horizontal circular motion is also closely related to polar coordinates. Understanding how objects move in circular paths provides a practical context for polar coordinates and helps visualize the relationship between angular displacement and radial distance. This physical interpretation can make abstract polar concepts more tangible and easier to grasp. Proficiency in derivatives of inverse trigonometric functions is another crucial skill. In polar coordinates, you'll often encounter these functions when dealing with complex curves or when converting between polar and Cartesian forms. Being able to differentiate and manipulate these functions is essential for analyzing the behavior of polar curves and solving related calculus problems. When it comes to graphing polar equations, understanding how to determine the number of solutions to equations becomes particularly important. This skill helps you predict the shape and behavior of polar curves, including whether they form closed loops, spirals, or other intricate patterns. It's a critical step in visualizing and interpreting polar graphs accurately. Lastly, proficiency in solving polar equations is essential. This skill draws on your ability to handle distance and time-related questions in linear equations, but applies them to the circular nature of polar coordinates. Being able to solve these equations efficiently is crucial for finding intersection points, determining curve characteristics, and solving real-world problems expressed in polar form. By mastering these prerequisite topics, you'll build a strong foundation for understanding and working with polar coordinates. Each concept contributes uniquely to your ability to navigate this circular coordinate system, from the fundamental trigonometric relationships to the more advanced calculus applications. As you progress in your study of polar coordinates, you'll find that these prerequisite skills continually resurface, reinforcing their importance and demonstrating the interconnected nature of mathematics. Become a member to get more! Try Free © 2015 - 2025 StudyPug No Javascript It looks like you have javascript disabled. You can still navigate around the site and check out our free content, but some functionality, such as sign up, will not work. If you do have javascript enabled there may have been a loading error; try refreshing your browser.
10747	https://www.cuemath.com/numbers/3-digit-subtraction/	LearnPracticeDownload 3-Digit Subtraction 3-digit subtraction is the subtraction of numbers in which the minuend consists of 3 digits and the subtrahend can be of either 1, 2, or 3 digits. We know that subtraction is an operation in Math, that gives the difference between two numbers. The number from which we subtract is called the minuend and the number which is subtracted from the minuend is called the subtrahend. While addition of numbers can be done with any set of numbers, in subtraction, we need to remember that the minuend as a whole should be larger than the subtrahend. 3-digit subtraction can be done with regrouping or without regrouping depending upon the value of the digits. Let us learn more about 3-digit subtraction in this article. \| \| \| --- \| \| 1. \| What is 3-Digit Subtraction? \| \| 2. \| 3-Digit Subtraction Without Regrouping \| \| 3. \| 3-Digit Subtraction With Regrouping \| \| 4. \| 3-Digit Addition and Subtraction \| \| 5. \| FAQs on 3-Digit Subtraction \| What is 3-Digit Subtraction? In 3-digit subtraction, we need to subtract the given numbers after placing them correctly according to their place values and we need to ensure that the bigger number is placed in the upper row, while the smaller number is placed below it. After aligning them into columns of ones, tens and hundreds, we start the process of subtraction. We know that the number from which the other number is subtracted is called the minuend, and the number which is subtracted from the minuend is called the subtrahend. The result of the subtraction of the two given numbers is called their difference. Mathematically, this can be expressed as: Minuend - Subtrahend = Difference Let us see the basic steps that explain how to do 3-digit subtraction. Step 1: Write the given numbers one below the other in such a way that the bigger number is placed up and the smaller number is placed below it. They should be correctly placed under the columns of ones, tens, and hundreds. Step 2: Start subtracting the numbers from the ones column, followed by the tens column and then the hundreds column. Step 3: After all the columns are subtracted, we get the difference of the given numbers. 3-Digit Subtraction Without Regrouping 3-digit subtraction without regrouping means subtracting a set of numbers without borrowing any number from the preceding digit. While subtracting 3-digit numbers, if all the digits in the minuend are bigger than the digits in the subtrahend, the subtraction can be easily done by subtracting each column one by one. Once all the columns are subtracted, the final answer is obtained. This is called subtraction without regrouping. In such a case, there is no regrouping or borrowing of numbers because all the digits of the minuend are bigger in value than the subtrahend. For example, let us subtract two numbers to understand this better. Let us subtract 342 from 754. Let us understand 3-digit subtraction without regrouping using the following steps: Step 1: Subtract the numbers under the ones column. 4 - 2 = 2. Write the difference (2) in ones column. Step 2: Subtract the numbers under the tens column. 5 - 4 = 1. Write the difference (1) in the tens column. Step 3: Subtract the numbers under the hundreds column. 7 - 3 = 4. Write the difference (4) in the hundreds column. Hence, the difference between the given numbers is 412. 3-Digit Subtraction With Regrouping 3-digit subtraction with regrouping means we need to borrow a number from the preceding digit. In subtraction, regrouping is also known as borrowing. When we subtract 3-digit numbers, sometimes, a digit in the upper row is smaller than the digit in the lower row. In this case, we borrow a number from the preceding column so that the smaller minuend becomes bigger than the subtrahend. This is known as regrouping or borrowing, For example, let us subtract 167 from 283. Let us understand this three-digit subtraction with regrouping using the following steps: Step 1: Write the given numbers according to their place values, one below the other in such a way that 283 is placed up and 167 is placed below it. They should be correctly placed under the columns of ones, tens, and hundreds. Step 2: Start subtracting the numbers from the ones column. It can be seen that 3 is smaller than 7. So, let us borrow 1 from the tens column which will make it 13. This is known as borrowing or regrouping in subtraction. So, 13 - 7 = 6. Now, we will write the difference (6) under the ones column. Step 3: After giving 1 to the ones column in the previous step, the '8' in the tens column changes to 7. Now, let us subtract the digits at the tens place and write the difference under the tens column (7 - 6 = 1). Step 4: In the hundreds column, we will subtract 1 from 2 and write the difference (1) in the column. (2 - 1 = 1). Thus, after subtracting all the 3 digits, we get the difference as 116. 3-Digit Addition and Subtraction 3-digit addition and subtraction is similar to 2-digit addition and subtraction. When we add 3-digit numbers, we write the given numbers (addends) one below the other such that they are placed correctly under the columns of ones, tens, and hundreds. Then, we add each column starting from the right-hand side. After all the columns are added, we get the sum of the given numbers. In a similar way, we subtract 3-digit numbers by writing the bigger number on top and the smaller number below it correctly. Then, we subtract each column one by one starting from the ones column and moving on to the tens column, and then the hundreds column. Once all the columns are subtracted, we get the difference of the numbers. ☛ Related Articles 2-Digit Addition 2-Digit Subtraction 2-Digit Multiplication 3-Digit Addition 3-Digit Multiplication 4-Digit Addition 4-Digit Subtraction Addition and Subtraction of Decimals Addition and Subtraction of Integers Subtraction With Regrouping Subtraction on Number Line Examples on 3-Digit Subtraction Example 1: Subtract the 3-digit numbers: 769 - 245. Solution: Let us place the numbers one below the other and subtract them. Step 1: Subtract the numbers under the ones column. 9 - 5 = 4. Write the difference (4) in ones column. Step 2: Subtract the numbers under the tens column. 6 - 4 = 2. Write the difference (2) in the tens column. Step 3: Subtract the numbers under the hundreds column. 7 - 2 = 5 Write the difference (5) in the hundreds column. This was three-digit subtraction without regrouping and the difference between the given numbers is 524. 2. Example 2: Find the difference between the 3-digit numbers: 463 - 174 Solution: Let us subtract 174 from 463. Step 1: Start subtracting the numbers under the ones column. We can see that 3 is less than 4. So, let us borrow 1 from the tens column which will make it 13. Now, 13 - 4 = 9. Write the difference (9) in ones column. Step 2: Moving on to the tens column, we know that after giving 1 to the ones column in the previous step, the '6' in the tens column changes to 5. But 5 is again smaller than 7. So, let us borrow 1 from the hundreds column, which will make it 15. Now, 15 - 7 = 8. We will write the difference (8) in the tens column. Step 3: Now, let us subtract the numbers under the hundreds column. Since 1 was given to the tens column, the '4' in the hundreds column changes to 3. Now, 3 - 1 = 2, so we will write the difference (2) in the hundreds column. Thus, after subtracting all the 3 digits, we get the difference as 289. This was 3-digit subtraction with regrouping and the difference between the given numbers is 289. 3. Example 3: Write true or false for the following statements. a.) 3-digit subtraction with regrouping means we can borrow a number from the preceding column. b.) Three-digit subtraction means both the minuend and the subtrahend should be of 3 digits. Solution: a.) True, 3 digit subtraction with regrouping means we can borrow a number from the preceding column. b.) False, in three-digit subtraction, the minuend should be of 3 digits but the subtrahend can be of 1, 2 or 3 digits. View Answer > go to slidego to slidego to slide Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Practice Questions on 3-Digit Subtraction Check Answer > go to slidego to slide FAQs on 3-Digit Subtraction What is 3-Digit Subtraction in Math? Three-digit subtraction is the subtraction of numbers in which the minuend consists of 3 digits and the subtrahend can be of either 1, 2, or 3 digits. In subtraction, we need to remember that the minuend as a whole should be larger than the subtrahend. We subtract 3-digit numbers column-wise, starting from the ones column, moving on to the tens column and then to the hundreds column. 3 digit subtraction can be done with regrouping or without regrouping depending upon the value of the digits. What is 3-Digit Subtraction Without Regrouping? 3-digit subtraction without regrouping means we need not borrow any number when we subtract the given numbers. While subtracting 3-digit numbers, if all the digits in the minuend are bigger than the digits in the subtrahend, the subtraction can be easily done. We start subtracting the numbers in the ones column, followed by the tens column, and then the hundreds column. Once all the columns are subtracted, the final answer is obtained. This is called subtraction without regrouping. In such a case, there is no regrouping or borrowing of numbers because all the digits of the minuend are bigger in value than the subtrahend. How to do 3-Digit Subtraction? In 3-digit subtraction, we need to subtract the given numbers after placing them correctly according to their place values. We also need to ensure that the bigger number is placed in the upper row, while the smaller number is placed below it. After aligning them into columns of ones, tens and hundreds, we start the process of subtraction. We start subtracting the numbers in the ones column, then we move on to the tens column and then to the hundreds column. After the digits in all the columns are subtracted, we get the difference of the given numbers. What is 3-Digit Subtraction With Borrowing? 3-digit subtraction with borrowing means we need to regroup the numbers in the minuend because they are smaller than the respective subtrahend. In subtraction, regrouping is also known as borrowing. When we subtract 3-digit numbers, sometimes, a digit in the upper row is smaller than the digit in the lower row. In this case, we borrow a number from the preceding column so that the smaller minuend becomes bigger than the subtrahend. This is known as regrouping or borrowing. For example, let us subtract 167 from 485. Step 1: After arranging the numbers according to their place value, we start subtracting the digits from ones place. We can see that 5 is smaller than 7. So, we will borrow 1 from the tens column which will make it 15. So, 15 - 7 = 8. This is known as regrouping or borrowing in subtraction. Now, we will write 8 under the ones column. Step 2: After giving 1 to the ones column in the previous step, the '8' in the tens column changes to 7. Now, let us subtract the digits at the tens place and write 1 under the tens column (7 - 6 = 1). Step 3: In the hundreds column, we will subtract 1 from 4 and write 3 in this column. (4 - 1 = 3). Thus, after subtracting all the columns, we get the difference as 318. Is 3-Digit Subtraction with Regrouping the same as 3 Digit Subtraction With Borrowing? Yes, 3-digit subtraction with regrouping means 3-digit subtraction with borrowing. Whenever a number in the minuend is smaller than the respective number in the subtrahend, this number is regrouped by borrowing 1 from the preceding column, so that it becomes larger than the number in the subtrahend and then it can be subtracted easily. How to do 3-Digit Subtraction with Zeros? In 3-digit subtraction, when there are zeros in the subtrahend, it can be easily subtracted. For example, 543 - 100 = 443. However, when there are zeros in the minuend, we need to regroup and borrow 1 from the preceding column and then subtract the given numbers. For example, let us subtract 500 - 237. After placing the numbers in order, we start from the ones column and we can see that 0 is smaller than 7. So, we will borrow 1 from the tens column which will convert the 0 to 10. So, 10 - 7 = 3. It should be noted here that when we had borrowed 1 from the tens column, the digit in the tens column was also 0, so this 0 had to borrow 1 from the hundreds column and then give 1 to the ones column. Now, getting back to the subtraction in the tens column, we know that this 0 had changed to 10 when it had borrowed 1, and after giving 1 to the ones column, it became 9. So, we will subtract 9 - 3 = 6. Now, let us move to the hundreds column. Since 5 had given 1 to the tens column, it becomes 4, and 4 - 2 = 2. After all the columns have been subtracted one by one, we get the difference as, 500 - 237 = 263. 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10748	https://arxiv.org/pdf/1706.07380	Published Time: Sun, 22 Jan 2023 18:38:45 GMT Alexander Kalmynin Intervals between numbers that are sums of two squares Abstract. In this paper, we improve the moment estimates for the gaps between numbers that can be represented as a sum of two squares of integers. We consider certain sum of Bessel functions and prove the upper bound for its weighted mean value. This bound provides estimates for the γ-th moments of gaps for all γ ⩽ 2. § 1. Introduction Let S = {s1 < s 2 < . . . < s n < . . . } = {1, 2, 4, 5, 8, . . . } ⊂ N be the set of all natural numbers that are expressible as the sum of two squares of integers. The classical area in the research of the properties of this set is the study of gaps between consecutive elements, i.e. the quantity sn+1 − sn or, equivalently, the value distribution of the distance function of S R(x) = min n∈S \|x − n\| for x → +∞.The following fact is well known (see ,): Theorem 1. The inequality R(x) ≪ x1/4 holds. Proof. Let us note that if f (x) = x − [√x]2, where [y] is an integer part of the number y, then 0 ⩽ f (x) ≪ √x for x ≫ 1 and x − f (x) = [ √x]2 is a square of integer. Therefore, x1/4 ≫ f (f (x)) = x − (x − f (x) + f (x) − f (f (x))) ⩾ 0. But x − f (x) and f (x) − f (f (x)) are squares of integers, so for some integers a and b we have \|x − a2 − b2\| ≪ x1/4, which was to be proved. This estimate was probably known to L. Euler and, unfortunately, was not improved after that (it is still unknown if the identity R(x) = o(x1/4) is true). Conjecturally, the correct order of growth of R(x) is much smaller: c⃝ Alexander Kalmynin, 2018 arXiv:1706.07380v2 [math.NT] 13 Jul 2018 2ALEXANDER KALMYNIN Conjecture 1. For any ε > 0 the inequality R(x) ≪ε xε is true. As for the large values of R(x), by the work of I. Richards, we have the following: Theorem 2. For any ε > 0 there exist infinitely many positive integers x such that R(x) > 14 − ε ln x. In addition to the the upper and lower bounds, one can also consider the mean values of our function. The best result of this type was proved by C. Hooley : Theorem 3. For any 0 < γ < 53 we have X sn+1 ⩽x (sn+1 − sn)γ ≪ x(ln x)(γ−1) /2. Therefore, for almost all n the inequality sn+1 − sn ≪ √ln sn holds. More precisely: Corollary 1. Let g(x) be any function that tends to infinity. Then the number of sn ⩽ x with sn+1 − sn ≫ g(x)√ln x is O( xg(x)γ √ln x ) for any γ < 5/3. The main goal of the present work is to improve Hooley’s theorem, i.e. to prove analogous estimate for the wider range of values of γ. Theorem 4. For any 1 < γ ⩽ 2 the inequality X sn+1 ⩽x (sn+1 − sn)γ ≪ x(ln x) 32 (γ−1) δ(x, γ ) is true, where δ(x, γ ) = ( 1, if γ < 2; ln x, if γ = 2 . Remark 1. In 1986 V. A. Plaksin has published an article in which he pre-sented the proof of the result similar to the Theorem 3, but for larger range of γ,namely 0 < γ < 2. However, later his paper was shown to contain some mistakes (see ) and attempts to fix the arguments were not successful (see ). § 2. Proof of the main theorem In this section, we will prove Theorem 4 by reducing the original problem to the question about the distribution of values of certain sum of Bessel functions. To do this, we need some lemmas. Let us start with the transformation formula for the theta-function. INTERVALS BETWEEN SUMS OF TWO SQUARES 3 Lemma 1. Let M be a positive real number. For any real x the equality ϑM (x) := X n∈Z e−πM (x+n)2 = 1 √M X m∈Z e2πimx e−πm 2/M is true. Proof. Consider the function g(x) = e−πM x 2 on the real line. It is easy to show that g is a Schwartz function and ϑM (x) = X n∈Z g(x + n). By the Poisson summation formula, for any x ∈ R we get X n∈Z g(x + n) = X m∈Z e2πimx bg(m), where bg(ξ) = R R g(x)e−2πiξx dx is a Fourier transform of our function. On the other hand, it is well known that Z R e−πM x 2−2πiξx dx = 1 √M e−πξ 2/M . Using this relation, we obtain the required result. With the help of Lemma 1 we will prove the following identity, which will be crucial for the subsequent considerations: Lemma 2. Let M and N be some positive real numbers. Then I(N, M ) := 12π Z π −π ϑM (√N sin φ)ϑM (√N cos φ)dφ == 1 M X n⩾0 r2(n)J0(2 π√N n )e−πn/M =: 1 M S(N, M ), where J0(x) = 12ππR −π eix cos φdφ is the Bessel function of the first kind of order zero and r2(n) is the number of pairs (a, b ) of integers such that a2 + b2 = n. Proof. Using lemma 1, we find I(N, M ) = 12πM Z π −π X a∈Z e2πia √N sin φe−πa 2/M ! X b∈Z e2πib √N cos φe−πb 2/M ! dφ. Both series are absolutely and uniformly convergent, so we can replace the prod-uct of their sums by the double sum and interchange summation and integration. Consequently, 4 ALEXANDER KALMYNIN I(N, M ) = 12πM X (a,b )∈Z e−(a2+b2)π/M Z π −π e2πi √N (a sin φ+b cos φ)dφ. Let us now compute the inner integral for all integers a and b. If (a, b ) = (0 , 0) , then the integrand is equal to 1 and so the integral equals 2π. If, in contrast, a2 + b2̸ = 0 ,then there exists some θ ∈ [−π, π ] such that sin θ = a √a2 + b2 , cos θ = b √a2 + b2 . Therefore, a sin φ + b cos φ = √a2 + b2 cos( φ − θ). Applying the change of variables φ = φ1 + θ, we obtain Z π −π e2πi √N (a sin φ+b cos φ)dφ = Z π−θ −π−θ e2πi √N (a2+b2) cos φ1 dφ 1. Since the integrand is periodic with the period 2π, the last integral is equal to the integral of the same function over the interval [−π, π ]. Thus, we finally find Z π −π e2πi √N (a sin φ+b cos φ)dφ = Z π −π e2πi √N (a2+b2) cos φ1 dφ 1 = 2 πJ 0(2 πpN (a2 + b2)) . Substituting the obtained result into the formula for I(N, M ), we deduce the iden-tity I(N, M ) = 1 M X (a,b )∈Z J0(2 πpN (a2 + b2)) e−π(a2+b2)/M = 1 M S(N, M ), which was to be proved. Remark 2. Computing the integral I(N, M ) without using Lemma 1 yields I(N, M ) = e−πN M X n⩾0 r2(n)I0(2 πM √N n )e−πnM . Therefore, for all N, M > 0 we have M e −πN M X n⩾0 r2(n)I0(2 πM √N n )e−πnM = X n⩾0 r2(n)J0(2 π√N n )e−πn/M . Now, if for all z, τ ∈ C, Im τ > 0 we have f (τ, z ) = X n⩾0 r2(n)J0(4 π√nz )eπinτ then we get S(N, M ) = f iM , √N 2 and by identity principle f − 1 τ , zτ = −iτ exp 4πiz 2 τ f (τ, z ). So, the function f (τ, z ) behaves like a Jacobi form. This phenomenon can be gen-eralized to the Cohen-Kuznetsov series and provide some results on Rankin-Cohen brackets of modular forms (see and ). INTERVALS BETWEEN SUMS OF TWO SQUARES 5 Now we want to show that if R(N ) is large, then the quantity S(N, M ) is rather small. To prove this, we will need some more additional lemmas. Lemma 3. Let N > 0, M > 1 and denote by C(N ) ⊂ R2 the circle of radius √N centered in the point (0 , 0) . Let d(N ) be the distance between the sets C(N ) and Z2, that is d(N ) = inf x∈C(N) y∈Z2 ρ(x, y ), where ρ is a standard euclidean distance on the plane. Then the inequality S(N, M ) ⩽ M e −πM d (N )2 O(M e −πM/ 4) holds. Proof. Let us notice first that for any φ ∈ [−π, π ] the inequality \|\| √N cos φ\|\| 2 + \|\| √N sin φ\|\| 2 ⩾ d(N )2 holds. Indeed, for some integers a and b we have \|\| √N cos φ\|\| 2 + \|\| √N sin φ\|\| 2 = ( √N cos φ − a)2 + ( √N sin φ − b)2 == ρ2(( √N cos φ, √N sin φ), (a, b )) ⩾ d(N )2, because (√N cos φ, √N sin φ) ∈ C(N ) and (a, b ) ∈ Z2. Then from the Lemma 2 we obtain S(N, M ) = M 2π Z π −π ϑM (√N cos φ)ϑM (√N sin φ)dφ == M 2π Z π −π e−πM \|\| √N cos φ\|\| 2 O(e−πM/ 4) e−πM \|\| √N sin φ\|\| 2 O(e−πM/ 4) dφ. Therefore, from the previous observation we get the desired inequality. It turns out that the distance between C(N ) and Z2 is closely related to the distance from N to the nearest sum of two squares. More precisely, the following proposition holds: Lemma 4. For any N > 0 we have d(N ) ⩾ 2R(N )5√N Proof. Let O = (0 , 0) and A ∈ C(N ), B ∈ Z2 be the points such that d(N ) = ρ(A, B ) (points with this condition certainly exist because C(N ) is compact) Then the points A, B and O are collinear, because if this is not the case then the triangle ABO is nondegenerate and thus 6 ALEXANDER KALMYNIN ρ(A, B ) > \|ρ(A, O ) − ρ(B, O )\|. But then if we take D to be the intersection point of the line BO and the circle C(N ), which lies closer to B than the second point, we get ρ(D, B ) = \|ρ(D, O ) − ρ(B, O )\| = \|ρ(A, O ) − ρ(B, O )\| < ρ (A, B ), which is a contradiction. Therefore, d(N ) = ρ(A, B ) = \|ρ(A, O ) − ρ(B, O )\| = \|√N − pa2 + b2\| for some integers a and b. So we get d(N ) = \|N − a2 − b2\|√N + √a2 + b2 and we clearly can assume that a2 + b2 ⩽ 2N , as otherwise we have d(N ) ⩾ (√2 − 1) √N > 25 √N ⩾ 2R(N )5√N , because R(N ) ⩽ \|N − 0\| = N . Now, if a2 + b2 ⩽ 2N then √N + √a2 + b2 ⩽ (√2 + 1) √N < 5√N / 2. By the definition of R(N ) we also have \|N − a2 − b2\| ⩾ R(N ) and so d(N ) ⩾ R(N )5√N / 2 = 2R(N )5√N . Combination of Lemma 3 and Lemma 4 gives us Lemma 5. Let H > 1, N > 3 and R(N ) ⩾ H. If M ⩾ 2N ln NH2 then S(N, M ) ≪ N −1/200 Proof. From the Lemma 3 we get S(N, M ) ≪ M e −πM d (N )2 M e −πM/ 4. As R(N ) ≪ N 1/4, the inequality M ≫ N ln NR(N )2 ≫ √N ln N holds and so M e −πM/ 4 ≪ N −1/200 . Now, Lemma 4 gives us d(N ) ⩾ 2R(N )5√N ⩾ 2H 5√N .INTERVALS BETWEEN SUMS OF TWO SQUARES 7 Consider the function t(M ) = M e −πd (N )2M . We have t′(M ) = (1 − πd (N )2M )t(M )/M, so the quantity t(M ) is decreasing when M > 1 πd (N )2 . Furthermore, M ⩾ 2N ln NH2 ⩾ 25 N 4πH 2 ⩾ 1 πd (N )2 . Therefore, t(M ) ⩽ t 2N ln NH2 ≪ N e −2πN d (N )2 ln N/H 2 . And now we have by the previous estimates 2πN d (N )2 ln NH2 ⩾ 8π 25 ln N > 1.005 ln N, so t(M ) ≪ N −0.005 = N −1/200 . Consequently, S(N, M ) ≪ t(M ) + M e −πM/ 4 ≪ N −1/200 , which was to be proved. So, if R(N ) is sufficiently large, then the quantity S(N, M ) is close to 0. But the function J0(x) is oscillating for x → ∞ and so S(x, M )−1 is an infinite sum of certain oscillating functions. Therefore, because of possible cancellation it is reasonable to expect that S(x, M ) is close to 1. More precisely, the following L2-estimate holds: Lemma 6. Let N ⩾ 2, H ⩾ 40(ln N )3/2 and M = 2N ln NH2 ⩾ 1. Then the inequality J(N, M ) = Z N 0 (S(x, M ) − 1) 2dx ≪ √N M ln N holds. Remark 3. The constant 40 here is not optimal, but this has no effect on sub-sequent considerations. To prove this lemma, we need three more propositions. First of them is very well known: 8 ALEXANDER KALMYNIN Lemma 7. For any x ⩾ 1 we have X 0<n<x r22 (n) ≪ x ln x We also need Weber’s second exponential integral: Lemma 8. For arbitrary α, β, γ > 0 the formula Z +∞ 0 e−αx J0(2 β√x)J0(2 γ√x)dx = 1 α I0 2βγ α exp − β2 + γ2 α is true, where I0(x) = J0(ix ) is the modified Bessel function. Proof. See , p. 395, section 13.31. Furthermore, we will use the asymptotic formula for the modified Bessel function. Lemma 9. For any positive real x we have the following asymptotic formula: I0(x) ∼ ex √2πx Proof. See , p. 203, section 7.23. Proof of the Lemma 6. Let us define J∗(N, M ) = Z +∞ 0 (S(x, M ) − 1) 2e−πx/N dx. For any x ⩽ N we have e−πx/N ⩾ e−π , so for all N and M we get J(N, M ) ⩽ eπ J∗(N, M ) and so it is enough to prove an analogous estimate for J∗(N, M ).Now, by the definition of S(x, M ), we have S(x, M ) − 1 = X n⩾1 r2(n)J0(2 π√nx )e−πn/M and the series converge absolutely and uniformly for x ⩾ 0. Therefore we obtain J∗(N, M ) = Z +∞ 0 (S(x, M ) − 1) 2e−πx/N dx == X n,m ⩾1 r2(n)r2(m)e−π(n+m)/N Z +∞ 0 J0(2 π√mx )J0(2 π√nx )e−πx/N dx. The integrals can be computed using the change of variables x = Nπ y and Lemma 8 as follows: Z +∞ 0 J0(2 π√mx )J0(2 π√nx )e−πx/N dx = Nπ Z +∞ 0 J0(2 pπnN y )J0(2 pπmN y )e−y dy =INTERVALS BETWEEN SUMS OF TWO SQUARES 9 = Nπ e−πN (n+m)I0(2 πN √nm ). Therefore, J∗(N, M ) = Nπ X n,m ⩾1 r2(n)r2(m)e−πN (n+m)−π(n+m)/m I0(2 πN √nm ). We can split this sum into «diagonal» and «non-diagonal» summands as follows: J∗(N, M ) = Nπ (S0 + S1), where S0 = X n⩾1 r2(n)2e−2πn/M I0(2 πnN )e−2πnN and S1 = X n,m ⩾1 n̸=m r2(n)r2(m)e−πN (n+m)−π(n+m)/M I0(2 πN √nm ). Let us estimate S1 first. Note that all the summands are positive and for for any pair (n, m ) at least one of three pairs of inequalities is true: 1 ⩽ n, m ⩽ 100 M ln N, n ⩾ 1, m > 100 M ln N or m ⩾ 1, n > 100 M ln N. Thus we deduce S1 ⩽ X 1⩽n,m ⩽100 Mln Nn̸=m r2(n)r2(m)e−πN (n+m)−π(n+m)/M I0(2 πN √nm )+ +2 X n⩾1,m> 100 Mln N r2(n)r2(m)e−πN (n+m)−π(n+m)/M I0(2 πN √nm ) = S2 + 2 S3. To estimate the sum S3, let us notice that Lemma 9 gives us e−πN (n+m)I0(2 πN √nm ) ≪ e−πN (n+m)+2 πN √nm = e−πN (√n−√m)2 ⩽ 1. And so we obtain the following inequality: S3 ≪ X n⩾1,m> 100 Mln N r2(n)r2(m)e−π(n+m)/M =10 ALEXANDER KALMYNIN = X n⩾1 r2(n)e−πn/M  X n> 100 Mln N r2(n)e−πn/M ! . Now, for any x > 0 we have X n⩽x r2(n) ≪ x, so X n⩾1 r2(n)e−πn/M ⩽ +∞ X m=0 e−πm X n⩽(m+1) M r2(n) ≪ +∞ X m=0 (m + 1) e−mM ≪ M and X n> 100 Mln N r2(n)e−πn/M ⩽ +∞ X m=100 ln N e−πm X n⩽(m+1) M r2(n) ≪ X m=100 ln N e−πm (m+1) M ≪≪ M ln N e −100 π ln N ≪ N −313 so we finally get S3 ≪ M N −313 ≪ N −312 . Now we turn to S2. For any integer pair n, m with conditions 1 ⩽ n, m ⩽ 100 M ln N we get by the Lemma 9 e−πN (n+m)I0(2 πN √nm ) ≪ e−πN (n+m−2√nm ). Let us notice that as n̸ = m we have n + m − 2√nm = ( √n − √m)2 = (n − m)2 (√n + √m)2 ⩾ 1400 M ln N . Consequently, −πN (n + m) + 2 πN √nm ⩽ − πN 400 M ln N . From this inequality and conditions M = 2N ln NH2 and H ⩾ 40(ln N )3/2 we obtain −πN (n + m) + 2 πN √nm ⩽ − πH 2 800 ln 2 N ⩽ − 1600 π ln 3 N 800 ln 2 N ⩽ −2π ln N. Therefore, S2 ≪ X n,m ⩽100 Mln N r2(n)r2(m)N −2π =  X n⩽100 Mln N r2(n)  2 N −2π ≪ N 2−2π ≪ N −4. And we deduce for the «non-diagonal» summand the estimate INTERVALS BETWEEN SUMS OF TWO SQUARES 11 S1 ≪ N −4 + N −312 ≪ N −4. It remains to prove that N S 0 ≪ √N M ln N. From the Lemma 9 we deduce e−2πN n I0(2 πN n ) ≪ 1 √N n . Consequently, S0 ≪ X n⩽100 Mln N r2(n)2 √N n e−πn/M . Let K be the smallest integer with 2K+1 > 100 M ln N . Then we obviously have S0 ≪ K X k=0 S0(k), where S0(k) = X 2k⩽n< 2k+1 r2(n)2 √N n e−πn/M . From the Lemma 7 we get S0(k) ⩽ X 2k⩽n< 2k+1 r2(n)2 √N 2k/ 2 e−2k π/M ⩽ e−2k π/M √N 2k/ 2 X n< 2k+1 r2(n)2 ≪ (k + 1)2 k/ 2e−2k π/M √N . Summing this over all k ⩽ K, we obtain S0 ≪ r MN ln N. Consequently, J∗(N, M ) = Nπ (S0 + S1) ≪ √N M ln N + N −3 ≪ √N M ln N and J(N, M ) ⩽ eπ J∗(N, M ) ≪ √N M ln N, which concludes the proof. Lemma 10. Let x, H > 3 and denote by M (H, x ) the set of all real y ⩽ x with R(y) ⩾ H. Then the inequality μ(M (H, x )) ≪ x(ln x)3/2 H holds, where μ is the Lebesgue measure. 12 ALEXANDER KALMYNIN Proof. As this bound is trivial for H ⩽ 40(ln x)3/2, we can assume that H ⩾ 40(ln x)3/2.Then from the Lemma 3 we get \|S(y, M )\| ≪ y−1/200 for any y ∈ M (H, x ). Therefore for large enough y ∈ M (H, x ) we have \|S(y, M ) − 1\| ≫ 1. Thus, J(x, M ) ⩾ Z M(x,H ) (S(y, M ) − 1) 2e−πy/N dy ≫ μ(M (x, H )) − O(1) . Consequently, by the Lemma 5, μ(M (x, H )) ≪ √xM ln x + O(1) = √2x(ln x)3/2 H + O(1) ≪ x(ln x)3/2 H , which was to be proved. From this last lemma we deduce the Theorem 4. Proof of Theorem 4. Observe that P sn+1 ⩽x (sn+1 − sn)γ ≪ x R 0 R(t)γ−1dt .Indeed, for any positive integer n we have Z sn+1 sn R(t)γ−1dt = Z sn+1 sn min( t − sn, s n+1 − t)γ−1dt = (sn+1 − sn)γ 2γ−1γ . Summation over all n with condition sn+1 ⩽ x gives the desired result. Let k be some positive integer. Consider the set Bk ⊂ [0 , x ] of all real y with 2k ⩽ R(y) ⩽ 2k+1 .Due to the Lemma 9, μ(Bk) ≪ x(ln x)3/22−k. Therefore, Z Bk R(t)γ−1dt ⩽ 2(k+1)( γ−1) μ(Bk) ≪ x(ln x)3/22(γ−2) k. Let B be the set of y ∈ [0 , x ] such that R(y) ⩽ c(ln x)3/2 with some sufficiently large constant c. Trivially, we have B ⊂ [0 , x ] and so μ(B) ⩽ x. Consequently, Z B R(t)γ−1dt ≪ μ(B)(ln x) 32 (γ−1) ⩽ x(ln x) 32 (γ−1) . Due to the fact that on the interval [0 , x ] the inequality R(x) ⩽ x holds, we have [0 , x ] = B ∪ [ x⩾2k⩾c(ln x)3/2 Bk. Therefore, INTERVALS BETWEEN SUMS OF TWO SQUARES 13 Z x 0 R(t)γ−1dt = Z B R(t)γ−1dt + X x⩾2k⩾c(ln x)3/2 Z Bk R(t)γ−1dt. Furthermore, we have Z x 0 R(t)γ−1dt ≪ x(ln x) 32 (γ−1) + X x⩾2k⩾c(ln x)3/2 x(ln x)3/22(γ−2) k ≪ x(ln x)3/2( γ−1) δ(x, γ ). This concludes the proof of Theorem 4. § 3. Conclusion In this work, we constructed certain sum of Bessel functions that is unusually small in the points that are far from numbers that are sums of two squares. The estimate for some mean value of this sum allowed us to prove the upper bound for the measure of the set of points with this property. However, we were not able to prove sharp enough bound for the sum S(N, M )−1. Nontrivial estimates for this quantity would allow us to improve the exponent in the inequality R(x) ≪ x1/4. One can show that our construction works not only for the sums of two squares, but also for the set of values of arbitrary positive definite quadratic form with integer coefficients. It would be also interesting to generalize this construction to the case of indefinite forms. § 4. Acknowledgments The author thanks Maxim Aleksandrovich Korolev for the useful comments and discussion. The author is partially supported by Laboratory of Mirror Symmetry NRU HSE, RF Government grant, ag. №14.641.31.0001, the Simons Foundation, the Moebius Contest Foundation for Young Scientists, the «Young Russian Mathematics» con-test and the Program of the Presidium of the Russian Academy of Sciences №01 «Fundamental Mathematics and its Applications» under grant PRAS-18-01. References R. P. Bambah, Chowla, S., «On numbers which can be expressed as a sum of two squares», Proc. Nat. Acad. Sci. India, 13, 1947, 101-103. H. Cohen, «Sums Involving the Values at Negative Integers of L-Functions of Quadratic Characters», Mathematische Annalen 217 (1975): 271-285 G. Harman, «Sums of two squares in short intervals», Proceedings of the London Mathematical Society, 62, 1991, 225-241 C. Hooley, «On the intervals between numbers that are sums of two squares I», Acta Math., 127, 1971, 279-297 A. A. Karatsuba, «Euler and number theory», Proc. Steklov Inst. Math., 274, suppl. 1 (2011), 169-179 N. V. Kuznetsov, «A new class of identities for the Fourier coefficients of modular forms» (in Russian), Acta. Arith. 27 (1975) 505-519 14 ALEXANDER KALMYNIN V. A Plaksin, «The distribution of numbers representable as a sum of two squares», Mathematics of the USSR-Izvestiya(1988),31(1):171 V. A Plaksin, «Letter to the editor: correction to the paper ”The distribution of num-bers representable as a sum of two squares”», Russian Academy of Sciences. Izvestiya Mathematics(1993),41(1):187 I. Richards, «On the gaps between numbers which are sums of two squares», Advances in Math, 46, 1982, 1-2 G. N. Watson, «A Treatise on the Theory of Bessel Functions» (2nd.ed.), Cambridge University Press, 1966 Alexander Kalmynin National Research University Higher School of Economics, Russian Federation, Math Department, International Laboratory of Mirror Symmetry and Automorphic Forms E-mail : alkalb1995cd@mail.ru
10749	https://byjus.com/ncert-solutions-class-11-maths-chapter-1-sets-ex-1-5/	The NCERT Solutions for Class 11 Maths Chapter 1 Exercise 1.5 are created by subject experts according to the latest CBSE Syllabus 2023-24. Here, we have provided the solutions for the questions in the fifth exercise of the chapter. Exercise 1.5 of NCERT Solutions for Class 11 Maths Chapter 1 – Sets are based on the following topics: Complement of a Set: If U is the universal set and A a subset of U. Then the complement of A is the set of all elements of U which are not the elements of A. De Morgan’s laws: The complement of the union of two sets is the intersection of their complements, and the complement of the intersection of two sets is the union of their complements. Some Properties of Complement Sets: Complement laws De Morgan’s law Law of double complementation Laws of empty set and universal set Students can access these NCERT Solutions for Class 11 Maths Chapter 1 and kickstart their exam preparations. NCERT Solutions for Class 11 Maths Chapter 1 – Sets Exercise 1.5 Download PDF Previous Next Class 11 Maths Chapter 1 – Sets Exercise 1.5 Solutions 1. Let U = {1, 2, 3; 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, and C = {3, 4, 5, 6}. Find (i) A’ (ii) B’ (iii) (A U C)’ (iv) (A U B)’ (v) (A’)’ (vi) (B – C)’ It is given that U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 2, 3, 4} B = {2, 4, 6, 8} C = {3, 4, 5, 6} (i) A’ = {5, 6, 7, 8, 9} (ii) B’ = {1, 3, 5, 7, 9} (iii) A U C = {1, 2, 3, 4, 5, 6} So, we get (A U C)’ = {7, 8, 9} (iv) A U B = {1, 2, 3, 4, 6, 8} So, we get (A U B)’ = {5, 7, 9} (v) (A’)’ = A = {1, 2, 3, 4} (vi) B – C = {2, 8} So, we get (B – C)’ = {1, 3, 4, 5, 6, 7, 9} 2. If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets: (i) A = {a, b, c} (ii) B = {d, e, f, g} (iii) C = {a, c, e, g} (iv) D = {f, g, h, a} Solution: (i) A = {a, b, c} So, we get A’ = {d, e, f, g, h} (ii) B = {d, e, f, g} So, we get B’ = {a, b, c, h} (iii) C = {a, c, e, g} So, we get C’ = {b, d, f, h} (iv) D = {f, g, h, a} So, we get D’ = {b, c, d, e} 3. Taking the set of natural numbers as the universal set, write down the complements of the following sets: (i) {x: x is an even natural number} (ii) {x: x is an odd natural number} (iii) {x: x is a positive multiple of 3} (iv) {x: x is a prime number} (v) {x: x is a natural number divisible by 3 and 5} (vi) {x: x is a perfect square} (vii) {x: x is perfect cube} (viii) {x: x + 5 = 8} (ix) {x: 2x + 5 = 9} (x) {x: x ≥ 7} (xi) {x: x ∈ N and 2x + 1 > 10} Solution: We know that U = N: Set of natural numbers (i) {x: x is an even natural number}´ = {x: x is an odd natural number} (ii) {x: x is an odd natural number}´ = {x: x is an even natural number} (iii) {x: x is a positive multiple of 3}´ = {x: x ∈ N andx is not a multiple of 3} (iv) {x: x is a prime number}´ ={x: x is a positive composite number andx = 1} (v) {x: x is a natural number divisible by 3 and 5}´ = {x: x is a natural number that is not divisible by 3 or 5} (vi) {x: x is a perfect square}´ = {x: x ∈ N and xis not a perfect square} (vii) {x: x is a perfect cube}´ = {x: x ∈ N and xis not a perfect cube} (viii) {x: x + 5 = 8}´ = {x: x ∈ N and x ≠ 3} (ix) {x: 2x + 5 = 9}´ = {x: x ∈ N and x ≠ 2} (x) {x: x ≥ 7}´ = {x: x ∈ N and x < 7} (xi) {x: x ∈ N and 2x + 1 > 10}´ = {x: x ∈ N and x≤ 9/2} 4. If U = {1, 2, 3, 4, 5,6,7,8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that (i) (A U B)’ = A’ ∩ B’ (ii) (A ∩ B)’ = A’ U B’ Solution: It is given that U = {1, 2, 3, 4, 5,6,7,8, 9} A = {2, 4, 6, 8} B = {2, 3, 5, 7} (i) (A U B)’ = {2, 3, 4, 5, 6, 7, 8}’ = {1, 9} A’ ∩ B’ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9} Therefore, (A U B)’ = A’ ∩ B’. (ii) (A ∩ B)’ = {2}’ = {1, 3, 4, 5, 6, 7, 8, 9} A’ U B’ = {1, 3, 5, 7, 9} U {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9} Therefore, (A ∩ B)’ = A’ U B’. 5. Draw an appropriate Venn diagram for each of the following: (i) (A U B)’ (ii) A’ ∩ B’ (iii) (A ∩ B)’ (iv) A’ U B’ Solution: (i) (A U B)’ (ii) A’ ∩ B’ (iii) (A ∩ B)’ (iv) A’ U B’ 6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’? Solution: A’ is the set of all equilateral triangles. 7. Fill in the blanks to make each of the following a true statement. (i) A U A’ = …….. (ii) Φ′ ∩ A = ……. (iii) A ∩ A’ = ……. (iv) U’ ∩ A = ……. Solution: (i) A U A’ = U (ii) Φ′ ∩ A = U ∩ A = A So, we get Φ′ ∩ A = A (iii) A ∩ A’ = Φ (iv) U’ ∩ A = Φ ∩ A = Φ So, we get U’ ∩ A = Φ Access other exercise solutions of Class 11 Maths Chapter 1 – Sets The NCERT Class 11 Solutions of Chapter 1 can be accessed using the links below. Exercise 1.1 Solutions 6 Questions Exercise 1.2 Solutions 6 Questions Exercise 1.3 Solutions 9 Questions Exercise 1.4 Solutions 12 Questions Exercise 1.6 Solutions 8 Questions Miscellaneous Exercise on Chapter 1 Solutions 16 Questions Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
10750	http://eceweb1.rutgers.edu/~gajic/solmanual/slides/chapter7_CON.pdf	7.1 Solving Linear Differential Equations Dynamics of time invariant, linear, continuous-time systems is described by th order linear differential equations with constant coefﬁcients where and represent, respectively, the system input and output signals, and are constant coefﬁcients. This differential equation corresponds to a real physical system for . The initial conditions that carry information about and its derivatives at the initial time , that is, are assumed to be known. The reason for specifying the system initial conditions at instead of at is twofold. Firstly, we have to be able to ﬁnd the The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–1 system impulse response due to the impulse delta signal located at the time origin, , which requires integration from in order to be able to completely include the impulse delta signal within the integration limits. Secondly, the system zero-input response is contributed by the system initial conditions that are present in the system before the input is applied to the system. Assuming that the initial time at which the system input is applied is , the system initial conditions must be deﬁned before , that is at . It is known from elementary theory of differential equations that the solution of a differential equation has two components: –particular solution and –homogeneous solution, that is The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–2 The homogeneous solution satisﬁes the corresponding homogeneous differential equation, obtained by setting , that is The particular solution is any solution that satisﬁes the given differential equation regardless of the system initial conditions. Homogeneous Solution The characteristic equation of the given differential equation is deﬁned by The solutions of the characteristic equation (roots of the characteristic polynomial ) are called the system characteristic values. The characteristic values are The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–3 also known under the name system eigenvalues. Depending on the nature of the solutions of the characteristic equation, we can get in a systematic way the expression for the homogeneous solution. Let us ﬁrst assume that the solutions of the characteristic equation are distinct, that is ! " with ! " , then the homogeneous differential equation has a solution given by # $&%(' ! $)+' " $-,.' The constant coefﬁcients / have to be determined from the system initial conditions in the manner explained later on in this section. It can be The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–4 shown that the form of the given homogeneous solution guarantees the existence of the unique constants 0 . Example 7.1: Consider a continuous-time linear system whose mathematical model is described by the following homogeneous linear differential equation 1 2 1 2 2 The characteristic equation of this differential equation is 1 which implies the following characteristic values 3 1 . The corresponding homogeneous solution is given by 2 3 465 1 475 The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–5 Note that a pair of complex characteristic values, say 89;: , represents two distinct characteristic values. Hence, we can use the previously given formula to ﬁnd the corresponding homogeneous solution, that is < 8 =?>@ACBEDGF : =?>IHJA.BEDGF >EF 8 ACBKF : HJA.BKF However, this solution can be further simpliﬁed by using Euler’s formula and determining the constants 8 and : from the initial conditions. Since < is a real function of time, its ﬁnal expression can not contain complex numbers. In general, a pair of complex conjugate characteristic values implies oscillatory system response so that we must obtain sine and cosine functions in the ﬁnal expression for < . That is why, a more convenient form for the homogeneous solution in the case of complex conjugate characteristic values is < 8 >EF : >EF The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–6 Example 7.2: The homogeneous linear differential equation given below has the complex conjugate characteristic values L M L M M that is L which implies N L . The corresponding homogeneous solution is given by M N O6P L O6P The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–7 In the case when we have a multiple root, say the characteristic value Q has multiplicity , and the remaining characteristic values are distinct, the homogeneous solution has the form R Q S T S U UWV Q XZY[ U^] Q X_(a Y [ b XcC[ In general, a characteristic value d of multiplicity d generates a term of the form d d ] Q d ]UfegV S U e V S d ]UhegV Q U e V Q X e [ in the homogeneous solution of the corresponding differential equation. It can be shown that the chosen form for R facilitates the existence of unique coefﬁcients i . The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–8 In the case of multiple complex characteristic values, for example the charac-teristic value j j j has multiplicity , the corresponding part of the homogeneous solution has the form j k lnm\o j p j j q j lrmso j t j j u j q l m o j v j j w j xWy p lrmzo j q xWy p j j q x lrmzo j The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–9 Example 7.3: The homogeneous linear differential equation { \| { } \| } \| \| has the characteristic equation { } } and the characteristic values ~ } { . The corresponding homogeneous solution is given by \| ~ 6 } } { } The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–10 Example 7.4: Consider the homogeneous linear differential equation with a triple characteristic value The characteristic equation is given by The triple characteristic values are so that the corresponding homogeneous solution is 6 6 6 The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–11 Particular Solutions In Chapters 4 and 6, we have already dealt with a special form of a particular solution given by the convolution integral ^ where , representing the system impulse response, is the solution of the considered differential equation for . The lower integration limit in the convolution integral is taken at in order to accommodate the impulse delta function at . Note that when , the system impulse response contains the impulse delta function at the origin, as demonstrated in the next example. The delta impulses can also be generated due to the system differentiation of input signals as demonstrated in Example 4.21. It should be emphasized that the particular solution deﬁned by the convolution integral represents the system The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–12 zero-state response, . In general, the particular solution, , is any solution that satisﬁes the given differential equation regardless of the system initial conditions. For any at the initial time the following relations must be satisﬁed G ? This formula requires a bit more explanation, since the system initial conditions are speciﬁed at . Let us solve a simple ﬁrst-order example in order to get a better understanding of a potential problem caused by the system differention of the input signal. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–13 Example 7.5: Consider the following linear differential equation We ﬁrst solve this differential equation by using the Laplace transform method from Chapter 4, which leads to ¡ ¡ ¡ Note that at ¢ we have ¢ . Hence, an instant jump occurs at the initial time from to ¢ . This jump in the initial condition is contributed by the system differentiation of the input signal, which in general generates the impulse delta function as an additional input to The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–14 the system. The impulse delta function is able to change system initial conditions instantaneously. Note that £ ¤¥ £ ¤¥ £ ¤¥ . Let us now ﬁnd the homogeneous and particular solutions of the given differential equation. The homogeneous differential equation is given by ¦ ¦ Its solution is easily obtained as ¦ § £ ¨¥ . In order to ﬁnd a particular solution, we ﬁrst guess its form as © £ ¤¥ . Substituting this particular solution in the original differential equation, we ﬁnd that , hence © £ ¤¥ . It follows that ¦ © § £ ¨¥ £ ¤¥ The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–15 If we use information about ª we will get either « (if we assume that the particular solution is not present at ª ) or « , which implies « . Obviously, both values for « are different from and hence they produce wrong expressions for . The way to get the correct answer for « is to use information about ¬ , which is (in this particular example) equal to ¬ (see the solution obtained using the Laplace transform). That information will produce « « , leading to the correct answer for . However, information about the value of ¬ is not given, in general. It seems that we are in a trouble while using the classical method for solving a linear differential equation that corresponds to a real physical linear system since we need to know the system initial condition at ¬ . In general, it is difﬁcult to determine the system initial conditions at ¬ based on information about The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–16 the system differential equation and the system initial conditions at . However, that problem can be completely avoided if we use the convolution formula to obtain the particular solution. In such a case, the particular solution represents the zero-state solution (it does not depend on the system initial conditions) and the homogeneous solution represents the zero-input solution (it depends only on the system initial conditions). In this example, the system impulse response is easily found as ® ® ¯° so that the particular solution obtained from the convolution integral is given by ± ° ²^³ ° ²´³ ¯¶µ ·^¸°fµ¹ ¯° ·° º» The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–17 We can ﬁnd the constant ¼ in the expression for the homogeneous solution as ½ ¾ ¼ ¾ ¼ ½ ¾ ¿À Á´Â which leads to the result obtained using the Laplace transform ½ Ã Á´Â ÁÄ ¾ ¿À ¾ ¿À ¾ÅÀ ¾ ¿À ¾ ÅÀ We can conclude from this example that in the case when the system differentiates the input signal, the classical method for solving linear differential equations with constant coefﬁcients can be used for ﬁnding the response of time invariant linear systems only if either the particular solution is obtained by using the convolution formula or if the system initial conditions are known at Æ (which is hardly known and difﬁcult to determine from information about system’s initial conditions at ¾ ). The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–18 Hence, in the process of ﬁnding the system response, we have to use the particular solution obtained through the convolution formula. In such a case, Ç È´É so that all constants in Ç are obtained from Ê ÈË Ç Ì Ì ÍÎÏ Ç Ì ÍÎÏ Ì Í?Ð Ì ÎÏ Ç Ì ÍÐ Ì ÎÏ Ì In this course, we ﬁnd the particular solution of linear differential equations representing continuous-time linear systems through the convolution procedure. In that direction, we will present in Section 7.4 a time domain method for ﬁnding the continuous-time system impulse response. Note that the frequency domain technique for ﬁnding the continuous-time system impulse response by using the inverse Laplace transform has been already presented in Chapter 4. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–19 Note that a procedure for obtaining a particular solution of a linear differential equation with constant coefﬁcients known as the method of undetermined coefﬁ-cients is studied in detail in elementary differential equations courses. That method is based on guessing a particular solution, usually as a linear combination of poly-nomial, exponential, and sinusoidal functions (a linear combination of the forcing function and its derivatives). However, that procedure is not systematic and some-times has difﬁculties ﬁnding a particular solution. Another systematic method for ﬁnding particular solutions of linear differential equations with constant coefﬁcients is the method of variation of parameters. That method is computationally much more involved than the method of undetermined coefﬁcients. In the next two examples we indicate difﬁculties that one is faced with while ﬁnding a particular solution using the method of undetermined coefﬁcients. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–20 Example 7.6: In this example, we demonstrate difﬁculties one is faced with in the process of guessing a particular solution of a second-order differential equation. Consider ﬁrst a differential equation given by Ñ Ñ Ñ A natural choice will be to take for a candidate for a particular solution the following expression Ò Ñ (a linear combination of the forcing function and its derivatives) and to try to determine unknown constant coefﬁcients . By taking the ﬁrst and second derivatives of Ò and plugging them into the differential equation, it can be checked easily that no exist such that the guessed solution Ñ is a particular solution of this differential equation. We have to try a more complex form that also includes an integral of the forcing function, that is Ò Ñ Ó . The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–21 This will produce the required solution as Ô Õ Ö Õ Ö × Õ Ø Ù , with being an arbitrary constant. For simplicity, we may set so that Ô × Ù The particular solution obtained through the convolution formula is given by Ú Ô × Ù Û ×Ü In the last formula we have used the upper script to indicate that the particular solution is obtained through the convolution method. Note that the system impulse response is Û ×Ü . It should be emphasized that the particular solution obtained through the convolution formula by integrating a product of the system input and the system impulse response carries information about both the actual system input and the internal system dynamics represented in the system The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–22 impulse response by a weighted sum of ÝÞàß terms (note the term á â ãä ß in the particular solution obtained using the convolution formula, and its absence from the particular solution obtained using the method of undetermined coefﬁcients). The previous example was not so difﬁcult. However, if we change the original differential equation slightly into ä ä ä ã ß then the problem of obtaining a particular solution by guessing its form becomes much more difﬁcult. The students may try several choices. However, we can obtain the particular solution of this differential equation using the convolution formula in a pretty straight forward manner (see Problem 7.2). The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–23 Example 7.7: Consider a continuous-time linear system represented by the differential equation deﬁned in Example 7.6, that is å å å and the initial conditions given by æ çèé æ . The characteristic values of this differential equation are è å so that its homogeneous solution is ê è å æ åë . Using the particular solution obtained in Example 7.6 via the method of undetermined coefﬁcients, the solution of this differential equation has the form ê ì è å æ åë å í The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–24 From the given initial conditions, we obtain a set of linear algebraic equations for determining the constants î and ï ð î ï ñ îò ð ï which leads to ï and î ï . The required solution, obtained via the method of undetermined coefﬁcients, is given by ð ïó ï ô If we use the particular solution obtained through the convolution, then the constants î and ï have to be determined from the following set of linear algebraic equations The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–25 õ ö õ ÷ ø ù ÷ú õ ù ÷ú ö õ ø which produces ø and ÷ . The solution obtained this way is identical to the solution obtained using the method of undetermined coefﬁcients since û ö û ü õ øfý ø þ õ øý ø þ õ øý In conclusion, in the case when the linear time invariant system does not differentiate the input signal, we can ﬁnd the system response using any method for determining a particular solution of the corresponding linear constant coefﬁcient differential equation. However, in the case when the system differentiates the input signal the particular solution need to be found through the convolution procedure. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–26 7.4 Continuous-Time System Impulse Response The continuous-time system impulse response is obtained using the impulse delta function as the system input signal and assuming that the system initial conditions are zero. It is important to observe that the impulse delta function is present only at the initial time and that during that time it sets the system zero initial conditions to nonzero values. The linear continuous-time system impulse response is deﬁned by ÿ ÿ ÿ ÿ ÿ with ÿ The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–27 For , we have the homogeneous differential equation (an input free system) whose initial conditions have to be determined. Finding the initial conditions can be done similarly like in the discrete-time domain by using either the direct method or the method based on system linearity and time invariance. In the continuous-time domain, in general, the direct method is pretty cumbersome so that we will limit our presentation to the method based on the linearity and time invariance principles. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–28 We deﬁne ﬁrst the elementary continuous-time system impulse response by Since the delta impulse function sets at some of the system initial con-ditions to non zero values and then it disappears, the above system becomes the homogeneous one for , that is The initial conditions ! " for this system can be obtained by using the following reasoning. The right-hand side of the equation for at is equal to the delta impulse function so that the left-hand side must be also equal to the impulse delta function. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–29 Recalling the notion of generalized derivatives from Chapter 2, the above observation means that none of the terms # $% & # $')(+& # can have a jump discontinuity at since the derivative of a function with a jump discontinuity produces the impulse delta function. Hence, we must have for all of them $,& # ( $,& # -. However, in order to get the delta impulse function on the left-hand side, the term $'(% & # must contain a jump discontinuity at . In addition that jump must be equal to one since after being differentiated in the $'.& # term, it produces the impulse delta function multiplied by one on the left-hand side so that the equality between the left- and right-hand sides holds. We conclude that the initial conditions for the elementary system impulse response at -are given by # -$% & # -$'( & # -$'(% & # -The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–30 Using linearity and time invariance, the original impulse response becomes / 0132 0 0 2 0 Note that we have assumed that . The case will be discussed soon. The obtained result is graphically presented in Figure 7.3. ( t) b0δ h0 ( t) bm (m) ( t) 1h b 0 (1) ( t) b0h0 i=0 Σ m ( t) h i b 0 (i) =h t ( ) b bm ( t) bi i=0 Σ m . . . . Linear System Linear System Linear System Linear System 1 t) δ ( δ δ ( t) (1) (m) (i) Figure 7.3: Determination of the continuous-time linear system impulse response The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–31 We need to justify a general claim that the linear time invariant system at rest under the input signal equal to the th derivative of the system input produces on its output the th derivative of the original system output. This can be justiﬁed as follows. Let 4 be the solution of the elementary impulse response differential equation. If we take the th derivative of the corresponding differential equation we will obtain 5 5 6 6 4 5)78 578 578 6 6 4 8 6 6 4 4 6 6 4 6 6 Assuming that all initial conditions are zero, which is the case for the impulse response, it follows that 6 4 6 is the solution (the system output) of the above differential equation due to the forcing function given by 6 6 . The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–32 The above derivations are also valid when is any differentiable signal (in the generalized sense) so that we have the following general result valid for continuous time invariant linear system 9;: 9;: < < < 9;: < = = = 9;: = Note that in the case when , the obtained formula for the impulse response has to be slightly modiﬁed. Namely, integrating -times, we conclude that in such a case at we have > . The modiﬁed formula for becomes The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–33 ? @AB @)CED CEF D B G ? @AB @)CED CEF D B Example 7.19: Consider the problem of ﬁnding the system impulse response solved in Example 4.20 by using the Laplace transform. The system is deﬁned by H H so that the system impulse response satisﬁes H H I I The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–34 The impulse response is given by J J where the elementary system impulse response J is obtained from K J K J J J L J L The solution for J is given by J M NPO K N K O Using the initial conditions at L set by the impulse delta function, we obtain M K . Hence, the elementary system impulse response is given by The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–35 Q RPS RT S so that the system impulse response sought is Q Q RUS RTVS RUS RTVS RUS RTVS which is identical to the result obtained in Example 4.20. Example 7.20: Consider now the case when . The impulse response of the following system T T T T The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–36 satisﬁes the differential equation W W W W X X It follows that W Y W Y Y where the elementary system impulse response Y is obtained from W Y W Y Y Y Z Y Z The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–37 The solution for [ is given by [ \ ]^ ] ^ Using the initial conditions at a , we obtain \ . Hence, the elementary system impulse response is given by [ ]U^ so that the system impulse response is [ [ ` [ ]U^ ]U^ ]U^ ]U^ ]U^ ]U^ ]U^ ]U^ The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–38 7.5 Complete Continuous-Time System Response In Chapter 1, we have initiated discussion about the relationship between the zero-state and zero-input components of the system response and the particular and homogeneous solutions of the corresponding differential equation. Now, we know that the particular solution, obtained through the convolution procedure, is equal to zero-state system response. In such a case, the homogeneous solution that satisﬁes the system initial conditions is the zero-input system response. Based on the discussion presented at the end of Section 7.1, we conclude that the classical method for solving linear differential equations with constant coefﬁcients can be used for ﬁnding the complete response of linear time invariant systems even in the case when the system differentiate the input signal. Even more, the same method can be used for ﬁnding the zero-state and zero-input components of the complete system response. To do so, we must perform the following steps. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–39 Step 1: Find the impulse response in terms of the elementary system impulse response. Step 2: Find the particular solution using the convolution formula, that is b . That particular solution is the system zero-state response, b c;d . Step 3: Find the general form of a homogeneous solution and determine its constants using the following initial conditions b c;d e f f gh i e f gjhki f gl f hVi e f gml f h i f The homogeneous solution obtained represents the system zero-input response, that is, e con . The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–40 Example 7.21: The above steps are demonstrated on the second-order system considered in Example 7.20, that is p p p p qUr s q tu v q The impulse response of this system (Step 1) has been already found in Example 7.20 as qUs qUs , so that the particular solution (Step 2) obtained using the convolution formula is given by w s xzy s x y qU{ q\|{ q\|r}t~sq\|{v The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–41 \| z + \| V V V \| U U In Step 3, we ﬁnd the homogeneous solution that satisﬁes the system initial conditions. The homogeneous solution is given by U U Using the system initial conditions, we have The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–42 and jk Hence o U The complete system response is now given by ; o U U \| Very often in linear system theory and practice, we use the notions of the system transient response and the system steady state response. Sometimes, they are identiﬁed, respectively, as the system zero-input and zero-state responses, which is only partially true as explained below. The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–43 The system transient response is determined by –the complete system response during the short time interval immediately following the initial time, that is for some small , see Figure 1.4. How small should be depends on the system. The transient response is usually characterized with fast signal changes mostly contributed by o . The system steady state response is deﬁned theoretically as follows ¡ £¢ Practically, this means that the system response after some large period of time is considered as the steady state system response, V . Usually, the zero-input response decays to zero rapidly, hence the steady state response is predominantly determined by the system zero-state response (the forced response). The slides contain the copyrighted material from Linear Dynamic Systems and Signals, Prentice Hall 2003. Prepared by Professor Zoran Gajic 7–44
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AI generated content may present inaccurate or offensive content that does not represent Symbolab's view. Verify your Answer Subscribe to verify your answer Subscribe Save to Notebook! Sign in to save notes Sign in Verify Save Show Steps Hide Steps Number Line Related Functions Inverse Examples inverse\:y=\frac{x^2+x+1}{x} inverse\:f(x)=x^3 inverse\:f(x)=\ln (x-5) inverse\:f(x)=\frac{1}{x^2} inverse\:y=\frac{x}{x^2-6x+8} inverse\:f(x)=\sqrt{x+3} inverse\:f(x)=\cos(2x+5) inverse\:f(x)=\sin(3x) Function Inverse Calculator: A Comprehensive Guide In mathematics, the functions are the system like machines that takes an input, process it according to the rule, and give an output as per the system design (output). But sometimes, we need to work backward: as given the output, we need to find the input that is produced. This is where the inverse functions find application. They really "undo" the original purpose. Particularly in disciplines like engineering, physics, computer science, and economics, inverse functions are very valuable. Actually, they enable us to simulate real-world events, solve equations, and grasp system behavior. Finding the inverse of a function, meanwhile, may be challenging—especially for complex functions. That’s where an Inverse Function Calculator becomes a lifesaver. In this article, we’ll discuss on what inverse functions are, how an inverse function calculator works, and why this is a very useful tool for us. We’ll also look at the examples, practical applications, and even some advanced topics. 1. What is an Inverse Function? An inverse function $( f^{-1} )$ accepts ( y ) and returns ( x ), if a function ( f ) takes an input ( x ) and creates an output ( y ). Like the reverse of another function. Simply put: $ \text{If } f(x) = y, \text{ then } f^{-1}(y) = x. $ The function should be one-to- one only if it has an inverse, so every output will perfectly match one input. When a function is not one-to- one, we may usually limit its domain—the set of potential inputs—to enable invertibility. Example 1: Linear Function Let’s take the function f(x) = 2x + 7. To find its inverse: Replace f(x) with y: y = 2x + 7. Swap (x) and (y): x = 2y + 7. Solve for y: y = $(\frac{x - 7}{2} )$. So, the inverse function is $( f^{-1}(x) = \frac{x - 3}{2} )$. Example 2: Quadratic Function Now, consider $( f(x) = x^2 )$ this function is not one-to-one over all real numbers because both ( x = 2 ) and ( x = -2 ) give the same output (( y = 4 )). To make it invertible, we can restrict the domain to $( x \geq 0 )$. The inverse function then becomes $( f^{-1}(x) = \sqrt{x} )$. 3. What is an Inverse Function Calculator? Performing calculations manually in order to get the inverse of a function may be a time-consuming process that is also prone to errors, particularly when dealing with complicated functions. An instrument that is capable of automating this procedure is known as a "Inverse Function Calculator." By taking a function as input and providing you with its inverse in a short amount of time, it helps you save both time and effort. 4. Why Use an Inverse Function Calculator? Speed: It gives results in seconds. Accuracy: It eliminates human errors. Ease of Use: You don’t need to be a math expert to use it. Versatility: It can handle a wide range of functions, from simple linear ones to more complex trigonometric or exponential functions. 5. Features of an Inverse Function Calculator The following are some of the most important features of inverse function calculators that help them to be useful: Simple instructions and input areas abound on most math calculators, which facilitates their usage. Some calculators will lead you through the process of finding the inverse, therefore increasing your familiarity with the method it uses. Furthermore, sophisticated calculators can graph not just the original function but also its inverse, which simplifies the link between the two greatly. These calculators can manage many kinds of numerical expressions, including polyn equations, logarithms, and trigonometric functions. Should you engage in a function without an inverse, the calculator will notify you of a mistake you have committed. Configurable parameters: To satisfy your particular need, you may change settings like domain restrictions. 6. How to Use an Inverse Function Calculator Simple use an inverse function calculator. Here is a methodical guide: Enter the function you want to invert here. ( f(x) = 3x + 5 ), for instance, • Indicate the domain (should the function call for a limited scope). The calculator will analyze the data and show the inverse function when you click "Calculate". • View the outcomes: Usually accompanied with other information like graphs or stages, the inverse function will show. • For $( f(x) = e^x )$ input $( f(x) = e^x )$. The calculator comes back with $( f^{-1}(x) = \ln(x) )$ • For $( f(x) = \sin(x) )$ use a calculator. With a limited domain of $( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} )$ input $( f(x) = \sin(x) )$. • The calculator finds $( f^{-1}(x) = \arcsin(x) )$. 7. Types of Inverse Functions Inverse functions can be categorized based on the type of the original function: Linear Inverse Functions: These are inverses of linear functions, like $( f(x) = mx + b )$. Quadratic Inverse Functions: These require domain restrictions to ensure they’re invertible. Exponential and Logarithmic Inverse Functions: Exponential functions like $( f(x) = a^x )$ have logarithmic inverses. Trigonometric Inverse Functions: Functions like $( \sin(x) ), ( \cos(x) )$, and $( \tan(x) )$ have inverses like $( \arcsin(x) ), ( \arccos(x) )$, and $( \arctan(x) )$. Rational Inverse Functions: These are inverses of rational functions, like $( f(x) = \frac{1}{x} )$. 8. Advanced Topics in Inverse Functions Composition of Functions and Inverses When you compose a function with its inverse, you get the identity function: $[ f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x ]$ Derivatives of Inverse Functions The derivative of an inverse function can be found using this formula: $[ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} ]$ Example: Derivative of $( f(x) = \ln(x) )$ The inverse of $( f(x) = \ln(x) )$ is $( f^{-1}(x) = e^x )$. The derivative of $( f^{-1}(x) )$ is: $[ (f^{-1})'(x) = e^x. ]$ Example: Derivative of $( f(x) = \sqrt{x} )$ The inverse of $( f(x) = \sqrt{x} )$ is $( f^{-1}(x) = x^2 )$. The derivative of $( f^{-1}(x) )$ is: $[ (f^{-1})'(x) = 2x. ]$ 9. Practical Applications of Inverse Functions Numerous situations that occur in the real world make use of inverse functions: For the purpose of securing data, cryptography makes use of them in the form of encryption and decryption techniques. In the field of physics, they contribute to the modeling of processes like as wave propagation and heat transport. Through the use of inverse demand and supply functions, economics may examine the behavior of the market. Control systems and signal processing are two areas of engineering that include their use. In the field of medicine, inverse functions are used to simulate medication dose and response. Example: Cryptography In RSA encryption, the inverse of a modular exponentiation function is used to decrypt messages. Example: Physics In kinematics, the inverse of the position-time function gives the velocity-time function. Detailed Examples and Case Studies Example: Inverse Function in Economics Consider the demand function ( Q = 100 - 2P ), where ( Q ) is the quantity demanded and ( P ) is the price. To find the inverse demand function: Replace ( Q ) with ( y ) and ( P ) with ( x ): ( y = 100 - 2x ). Swap x and y: x = 100 - 2y. Solve for y: y = $(\frac{100 - x}{2} )$. So, the inverse demand function is $( P = \frac{100 - Q}{2} )$. Example: Inverse Function in Medicine In pharmacokinetics, the concentration ( C ) of a drug in the bloodstream over time ( t ) can be modeled by $( C(t) = C_0 e^{-kt} )$, where ( C0 ) is the initial concentration and ( k ) is the elimination rate constant. The inverse function $( t(C) = -\frac{1}{k} \ln\left(\frac{C}{C_0}\right) )$ helps determine the time at which a specific concentration is reached. 10. Limitations and Challenges Although inverse functions and calculators are very useful, they do have several drawbacks, including the following: There are some functions that do not have inverses, such as the function $( f(x) = x^2 )$ over all real numbers. These functions do not have inverses unless their domains are constrained. Finding an inverse analytically may be difficult for really complicated functions, so numerical techniques are necessary. Users should provide suitable domains to guarantee invertibility. 11. Future Directions With mathematical and technological developments, the topic of inverse functions keeps changing. Future events may consist: Calculators Driven by Artificial Intelligence improved calculators solving increasingly difficult tasks by use of machine learning. Integration with Different Instruments Combining educational platforms, graphing tools, symbolic computing software, and inverse function calculators. Real-Time Programs In real-time systems like robots and autonomous cars, using inverse functions. 12. Conclusion Inverse functions are a powerful tool in mathematics, with applications in many fields. An Inverse Function Calculator makes finding inverses quick and easy, whether you’re a student, researcher, or professional. By understanding inverse functions and how to use these calculators, you can save time, avoid errors, and gain deeper insights into mathematical problems. 13. Additional Examples and Exercises Here are more examples to help you practice: Example: Inverse of a Cubic Function Find the inverse of $( f(x) = x^3 + 2 )$. Replace ( f(x) ) with $( y ): ( y = x^3 + 2 )$. Swap ( x ) and $( y ): ( x = y^3 + 2 )$. Solve for $( y ): ( y = \sqrt{x - 2} )$. So, the inverse function is $( f^{-1}(x) = \sqrt{x - 2} )$. Example: Inverse of a Rational Function Find the inverse of $( f(x) = \frac{2x + 1}{x - 3} )$. Replace ( f(x) ) with $( y ): ( y = \frac{2x + 1}{x - 3} )$. Swap ( x ) and $( y ): ( x = \frac{2y + 1}{y - 3} )$. Solve for $( y ): ( y = \frac{3x + 1}{x - 2} )$. So, the inverse function is $( f^{-1}(x) = \frac{3x + 1}{x - 2} )$. 15. Glossary of Terms Function: A rule that maps each input to exactly one output. Inverse Function: A function that reverses the effect of another function. One-to-One Function: A function where each output corresponds to exactly one input. Domain: The set of all possible input values for a function. Range: The set of all possible output values for a function. Frequently Asked Questions (FAQ) How do you calculate the inverse of a function? To calculate the inverse of a function, swap the x and y variables then solve for y in terms of x. What are the 3 methods for finding the inverse of a function? There are 3 methods for finding the inverse of a function: algebraic method, graphical method, and numerical method. What is the inverse of a function? The inverse of a function f is a function f^(-1) such that, for all x in the domain of f, f^(-1)(f(x)) = x. Similarly, for all y in the domain of f^(-1), f(f^(-1)(y)) = y Can you always find the inverse of a function? Not every function has an inverse. A function can only have an inverse if it is one-to-one so that no two elements in the domain are matched to the same element in the range. A non-one-to-one function is not invertible. Why users love our Functions Inverse Calculator \| \| \| --- \| \| 🌐 Languages \| EN, ES, PT & more \| \| 🏆 Practice \| Improve your math skills \| \| 😍 Step by step \| In depth solution steps \| \| ⭐️ Rating \| 4.6 based on 20924 reviews \| en Related Symbolab blog posts A function basically relates an input to an output, there’s an input, a relationship and an output. For every input... 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10752	https://medium.com/@dorjderemdorj202/quantguide-square-shade-solution-6f70b2331d6e	Quantguide — Square shade solution \| by Dorjderem Namsraijav \| Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in Quantguide — Square shade solution Dorjderem Namsraijav Follow 2 min read · Jun 30, 2024 2 Listen Share This is the start of a series where I solve every single hard brainteaser problems on QuantGuide. Here is the link to the original problem: But in short, in a 2023x2023 grid, the squares in the middle row are shaded. The task is to determine the probability of selecting a rectangle that contains at least one shaded square. Solution: Get Dorjderem Namsraijav’s stories in your inbox Join Medium for free to get updates from this writer. Subscribe Subscribe Let’s first visualise the problem so that it’s easy to follow. Press enter or click to view image in full size Now we need to choose a random rectangle inside this big grid. Let’s assume we have a random rectangle with position (x, y) and size (w, h). Then let’s create another rectangle with position (p, y) and size (q, h). Think about how this changes the probability of that square having a shaded square. The answer is that it does not, as you can see in the picture. Press enter or click to view image in full size So we can simplify the problem into 1 dimensional. Press enter or click to view image in full size Now we just count the number of rectangles including the middle cell. First, we can count the number of rectangles we can form. If we set the starting position of the rectangle to 1, then we can have 2023 different rectangles. If we set the starting position of the rectangle to 2, then we can have 2022 different rectangles, and so on. Thus, n = 2023 + 2022 + 2021 + … + 1 = 2024 2023 / 2. To find the number of rectangles containing a shaded square, we can use the same method as before. If we set the starting position of the rectangle to 1, then we can have 1012 different rectangles. If we set the starting position of the rectangle to 2, then we can have 1012 different rectangles, and so on. So the final answer becomes: (1012 1012) / (2024 2023 / 2) = 1012 / 2023. Quant Quant Finance Brain Teaser Brain Teasers Puzzles Puzzle 2 2 Follow Written by Dorjderem Namsraijav ------------------------------- 2 followers ·1 following Follow No responses yet Write a response What are your thoughts? Cancel Respond More from Dorjderem Namsraijav Dorjderem Namsraijav Flutter vs React Native Expo. (Android Development) --------------------------------------------------- ### Performance Test: Jan 12, 2023 Dorjderem Namsraijav Quantguide — Cats and Mice -------------------------- Jun 30, 2024 See all from Dorjderem Namsraijav Recommended from Medium Abhinav Docker Is Dead — And It’s About Time ------------------------------------ ### Docker changed the game when it launched in 2013, making containers accessible and turning “Dockerize it” into a developer catchphrase. Jun 9 182 In Long. Sweet. Valuable. by Ossai Chinedum I’ll Instantly Know You Used Chat Gpt If I See This --------------------------------------------------- ### Trust me you’re not as slick as you think May 16 1441 Devlink Tips Apple is quietly rewriting iOS and it’s not in Swift or Objective-C ------------------------------------------------------------------- ### The hidden language shift happening inside Cupertino, why it matters, and what it means for your future apps. Sep 15 42 The Latency Gambler I Interviewed 20+ Engineers. Here’s Why Most Can’t Code ------------------------------------------------------- ### Over the past year as a Senior Software Engineer at a B2B SaaS company, I’ve conducted 20+ technical interviews for roles ranging from… Sep 9 66 Will Lockett The AI Bubble Is About To Burst, But The Next Bubble Is Already Growing ----------------------------------------------------------------------- ### Techbros are preparing their latest bandwagon. Sep 14 342 In Coding Nexus by Civil Learning I Handed ChatGPT $100 to Trade Stocks — Here’s What Happened in 2 Months. ------------------------------------------------------------------------- ### What happens when you let a chatbot play Wall Street? It’s up 29% while the S&P 500 lags at 4%. Sep 3 103 See more recommendations Help Status About Careers Press Blog Privacy Rules Terms Text to speech
10753	http://dslavsk.sites.luc.edu/courses/phys301/classnotes/symmetry2.pdf	FUNCTIONAL SYMMETRY AND INTEGRALS Overview We have already seen that understanding the basics of symmetry of functions can help us in computing and understanding various coefficients we obtain as part of Fourier analysis.In this note, we will show how symmetry arguments can be used to simplify our work. Odd and Even Functions We are familiar with fundamental definitions of odd and even functions. A function is odd if f (x) = -f (-x), and a function is even if f (x) = f (-x). Not surprisingly, polynomials consisting only of odd powers of x are odd, and polynomials consisting of only even powers of x are even. Odd and even functions also have characteristic graphs. Fig. 1 shows a plot of the odd function f (x) = x3 - x from -3 to 3: In:= Plot@x^3 −x, 8x, −3, 3<, Filling →AxisD Out= Fig. 1 Graph of x3 - x on [-3,3] Notice also the shading between the curve and the graph. This indicates nicely that the area above the x axis is matched by the area below the x axis, so that the total integral of this function is zero between the limits [-3, 3]. This illustrates one of the key concepts of odd functions : the integral of an odd function is zero if it is evaluated by limits that are symmetric across the origin. We can verify this statement by explicitly calculating the integral: In:= IntegrateAx3 −x, 8x, −3, 3<E Out= 0 We can even show this symbolically; if we integrate this function between any limits that are ±a, we find: In:= Integrate@x^3 −x, 8x, −a, a<D Out= 0 Let' s consider now the properties of even functions. An even function has the basic property that f (x) = f (-x). Fig. 2 shows a plot of the function f HxL = x4 - x2+3which is an even function since each power of x is even. In:= Plot@x^4 −x^2 + 3, 8x, −3, 3<, Filling →AxisD Out= Fig. 2 Graph of x4 - x2 + 3 on [-3,3] We can see that that the graph is symmetric with respect to the y axis, and that the area of the curve in the left half plane is equal to the area of the curve in the right half plane. In fact, even functions have the property that : (1) ‡ -a a feven HxL dx = 2‡ 0 a feven HxL dx We can verify this property for this function by calculating these two integrals and taking their ratio : In:= HIntegrate@x^4 −x^2 + 3, 8x, −a, a<DL ê H2 Integrate@x^4 −x^2 + 3, 8x, 0, a<DL Out= 6 a −2 a3 3 + 2 a5 5 2 J3 a −a3 3 + a5 5 N In:= Simplify@%D Out= 1 The final step showing that the ratio of the two integrals is one, proving their equality. Odd and Even Trig functions We can extend our definitions of odd and even functions to the familiar trig functions, sine and cosine. We can consider simple plots of each function to determine their behavior : 2 symmetry.nb In:= Plot@Sin@xD, 8x, −π, π<, Filling →AxisD Out= Fig. 3 Graph of Sin (x) And similarly : In:= Plot@Cos@xD, 8x, −π, π<, Filling →AxisD Out= Fig. 4 Graph of Cos (x) Not only do the graphs of sin and cos reveal their symmetry, but we can also rely on the Taylor series expansions of these functions to determine their symmetry properties : (2) Sin x = x - x3 3! + x5 5! - x7 7! + ... Cos x = 1 - x2 2! + x4 4! - x6 6! + ... Clearly, sin is an odd function since its series expansion involves only odd powers of x, and cos is an even function since its series expansion involves only even powers of x. Products of even and odd functions We can draw some helpful conclusions by considering products of even and odd functions. Consider the following possibilities : symmetry.nb 3 • even function x even function -> even function egs : x2 ÿ x4 = x6; x2 ÿ cos HxL • even function x odd function -> odd function egs: x2 ÿ x = x3; cos x ÿ sinx = 1 2 sinH2 xL(and remember sin is an odd function) • odd function x odd function -> even function egs: xÿ x3 = x4; sin x ÿ sin x = sin2 x = 1 - cos2 x (1 and cos2 are even) Thus, we can use combine this information with our understanding of symmetry properties of integrals to show : • The integral of an odd function times an even function over the interval [-a, a] is zero : In:= IntegrateA9x x2, Cos@xD x3, Exp@−x^2D x=, 8x, −a, a<E Out= 80, 0, 0< The input line above consists of a set of three products of an even function times an odd function; the output line shows each integral has a value of zero when evaluated on the interval [-a, a]. • The integral of an even function times an even function, or the integral of the product of odd functions, satisfies the relationship : (3) ‡ -a a f1 HxL f2 HxL dx = 2‡ 0 a f1 HxL f2 HxL dx Examples : In:= Integrate@x^2 Cos@xD, 8x, −π, π<D Out= −4 π In:= Integrate@x^2 Cos@xD, 8x, 0, π<D Out= −2 π In:= Integrate@x Sin@xD, 8x, −π, π<D Out= 2 π In:= Integrate@x Sin@xD, 8x, 0, π<D Out= π 4 symmetry.nb
10754	https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/Discrete_Structures/05%3A_Set_Theory/5.04%3A_Cartesian_Products	5.4.1 5.4.1 A B C A×(B∪C) (A×B)∪(A×C) A×(B∪C)⊆(A×B)∪(A×C) u∈A×(B∪C) x∈A y∈B∪C u=(x,y) y∈B∪C y∈B y∈C y∈B u=(x,y) x∈A y∈B u∈A×B u∈(A×B)∪(A×C) y∈C u=(x,y) x∈A y∈C u∈A×C u∈(A×B)∪(A×C) u∈(A×B)∪(A×C) u A×(B∪C) u∈(A×B)∪(A×C) A×(B∪C)⊆(A×B)∪(A×C).(5.4.5) (A×B)∪(A×C)⊆A×(B∪C) v∈(A×B)∪(A×C) v∈(A×B) v∈(A×C) v∈(A×B) s∈A t∈B v=(s,t) t∈C t∈B∪C Skip to main content 5.4: Cartesian Products Last updated : Sep 29, 2021 Save as PDF 5.3: Properties of Set Operations 5.5: Indexed Families of Sets Page ID : 86125 Ted Sundstrom Grand Valley State University via ScholarWorks @Grand Valley State University ( \newcommand{\kernel}{\mathrm{null}\,}) PREVIEW ACTIVITY 5.4.15.4.1: An Equation with Two Variables In Section 2.3, we introduced the concept of the truth set of an open sentence with one variable. This was defined to be the set of all elements in the universal set that can be substituted for the variable to make the open sentence a true statement . In previous mathematics courses, we have also had experience with open sentences with two variables. For example, if we assume that x and y represent real numbers, then the equation 2x+3y=122x+3y=12 is an open sentence with two variables. An element of the truth set of this open sentence (also called a solution of the equation) is an ordered pair (aa, bb) of real numbers so that when aa is substituted for xx and bb is substituted for yy, the open sentence becomes a true statement (a true equation in this case). For example, we see that the ordered pair (6, 0) is in the truth set for this open sentence since 2⋅6+3=122⋅6+3=12 is a true statement . On the other hand, the ordered pair (4, 1) is not in the truth set for this open sentence since 2⋅4+3⋅1=122⋅4+3⋅1=12 is a false statement . Important Note: The order of the of the two numbers in the ordered pair is very important. We are using the convention that the first number is to be substituted for xx and the second number is to be substituted for yy. With this convention, (3, 2) is a solution of the equation 2x+3y=122x+3y=12, but (2, 3) is not a solution of this equation. List six different elements of the truth set (often called the solution set ) of the open sentence with two variables 2x+3y=122x+3y=12. 2. From previous mathematics courses, we know that the graph of the equation 2x+3y=122x+3y=12 is a straight line. Sketch the graph of the equation 2x+3y=122x+3y=12 in the xyxy-coordinate plane. What does the graph of the equation 2x+3y=122x+3y=12 show? 3. Write a description of the solution set SS of the equation 2x+3y=122x+3y=12 using set builder notation. PREVIEW ACTIVITY 5.4.15.4.1: The Cartesian Product of Two Sets In Preview Activity 5.4.15.4.1, we worked with ordered pairs without providing a formal definition of an ordered pair. We instead relied on your previous work with ordered pairs, primarily from graphing equations with two variables. Following is a formal definition of an ordered pair. Definition: ordered pair Let AA and BB be sets. An ordered pair (with first element from AA and second element from BB) is a single pair of objects, denoted by (aa, bb), with a∈Aa∈A and b∈Bb∈B and an implied order. This means that for two ordered pairs to be equal, they must contain exactly the same objects in the same order. That is, if a,c∈Aa,c∈A and b,d∈Bb,d∈B, then (aa, bb) = (cc, dd) if and only if a=ca=c and b=db=d. The objects in the ordered pair are called the coordinates of the ordered pair. In the ordered pair (aa, bb), aa is the first coordinate and bb is the second coordinate. We will now introduce a new set operation that gives a way of combining elements from two given sets to form ordered pairs. The basic idea is that we will create a set of ordered pairs. Definition: Cartesian product If AA and BB are sets, then the Cartesian product, A×BA×B, of AA and BB is the set of all ordered pairs (xx, yy) where x∈Ax∈A and y∈By∈B. We use the notation A×BA×B for the Cartesian product of AA and BB, and using set builder notation, we can write A×B={(x,y) \| x∈A and y∈B}A×B={(x,y) \| x∈A and y∈B}. We frequently read A×BA×B as "AA cross BB." In the case where the two sets are the same, we will write A2A2 for A×AA×A. That is, A2=A×A={(a,b) \| a∈A and b∈A}A2=A×A={(a,b) \| a∈A and b∈A}. Let A =A = {1, 2, 3} and B =B = {aa, bb}. Is the ordered pair (3, aa) in the Cartesian product A×BA×B? Explain. Is the ordered pair (3, aa) in the Cartesian product A×AA×A? Explain. Is the ordered pair (3, 1) in the Cartesian product A×AA×A? Explain. Using the roster method to specify all the elements of A×BA×B. (Remember that the elements of A×BA×B will be ordered pairs. Use the roster method to specify all of the elements of the set A×A=A2A×A=A2. 6. For any sets CC and DD, explain carefully what it means to say that the ordered pair (xx, yy) is not in the Cartesian product C×DC×D. Cartesian Products When working with Cartesian products, it is important to remember that the Cartesian product of two sets is itself a set . As a set , it consists of a collection of elements. In this case, the elements of a Cartesian product are ordered pairs. We should think of an ordered pair as a single object that consists of two other objects in a specified order. For example, If a≠1a≠1, then the ordered pair (1, aa) is not equal to the ordered pair (aa, 1). That is, (1, aa) ≠≠ (aa, 1). If A =A = {1, 2, 3} and B =B = {aa, bb}, then the ordered pair (3, aa) is an element of the set A×BA×B. That is, (3, aa) ∈A×B∈A×B. If A =A = {1, 2, 3} and B =B = {aa, bb}, then the ordered pair (5, aa) is not an element of the set A×BA×B since 5∉A5∉A. That is, (5,a)∉A×B(5,a)∉A×B. In Section 5.3, we studied certain properties of set union, set intersection, and set complements, which we called the algebra of sets. We will now begin something similar for Cartesian products. We begin by examining some specific examples in Progress Check 5.23 and a little later in Progress Check 5.24. progress check 5.23 (relationships between Cartesian products) Let A=A= {1, 2, 3}, T=T= {a, b}, and C=C= {a, c}. We can then form new sets from all of the set operations we have studied. For example, B∩C=B∩C= {aa}, and so A×(B∩C)={(1,a),(2,a),(3,a)}.A×(B∩C)={(1,a),(2,a),(3,a)}. Use the roster method to list all of the elements (ordered pairs) in each of the following sets: (a) A×BA×B (b) T×BT×B (c) A×CA×C (d) A×(B∩C)A×(B∩C) (e) (A×B)∩(A×C)(A×B)∩(A×C) (f) A×(B∪C)A×(B∪C) (g) (A×B)∪(A×C)(A×B)∪(A×C) (h) A×(B−C)A×(B−C) (i) (A×B)−(A×C)(A×B)−(A×C) (j) B×AB×A 2. List all the relationships between the sets in Part (1) that you observe. Answer : Add texts here. Do not delete this text first. The Cartesian Plane In Preview Activity 5.4.15.4.1, we sketched the graph of the equation 2x+3y=122x+3y=12 in the xyxy-plane. This xyxy-plane, with which you are familiar, is a representation of the set R×RR×R or R2R2. This plane is called the Cartesian plane. The basic idea is that each ordered pair of real numbers corresponds to a point in the plane, and each point in the plane corresponds to an ordered pair of real numbers. This geometric representation of R2R2 is an extension of the geometric representation of RR as a straight line whose points correspond to real numbers. Since the Cartesian product R2R2 corresponds to the Cartesian plane, the Cartesian product of two subsets of RR corresponds to a subset of the Cartesian plane. For example, if AA is the interval [1, 3], and BB is the interval [2, 5], then A×B={(x,y)∈R2 \| 1≤x≤3 and 2≤y≤5}.A×B={(x,y)∈R2 \| 1≤x≤3 and 2≤y≤5}. A graph of the set A×BA×B can then be drawn in the Cartesian plane as shown in Figure 5.6. This illustrates that the graph of a Cartesian product of two intervals of finite length in RR corresponds to the interior of a rectangle and possibly some or all of its boundary. The solid line for the boundary in Figure 5.6 indicates that the boundary is included. In this case, the Cartesian product contained all of the boundary of the rectangle. When the graph does not contain a portion of the boundary, we usually draw that portion of the boundary with a dotted line. Note: A Caution about Notation. The standard notation for an open interval in RR is the same as the notation for an ordered pair, which is an element of R×RR×R. We need to use the context in which the notation is used to determine which interpretation is intended. For example, If we write (sqrt2sqrt2, 7) ∈R×R∈R×R, then we are using (sqrt2sqrt2, 7) to represent an ordered pair of real numbers. If we write (1, 2) ×× {4}, then we are interpreting (1, 2) as an open interval. We could write (1, 2) ×× {4} = {(xx, 4) \| 1 < xx < 2}. The following progress check explores some of the same ideas explored in Progress Check 5.23 except that intervals of real numbers are used for the sets. Progress Check 5.24: Cartesian Products of Intervals We will use the following intervals that are subsets of RR. A=A= [0, 2] T=T= (1, 2) B=B= [2, 4) C=C= (3, 5] Draw a graph of each of the following subsets of the Cartesian plane and write each subset using set builder notation. (a) A×BA×B (b) T×BT×B (c) A×CA×C (d) A×(B∩C)A×(B∩C) (e) (A×B)∩(A×C)(A×B)∩(A×C) (f) A×(B∪C)A×(B∪C) (g) (A×B)∪(A×C)(A×B)∪(A×C) (h) A×(B−C)A×(B−C) (i) (A×B)−(A×C)(A×B)−(A×C) (j) B×AB×A 2. List all the relationships between the sets in Part (1) that you observe. Answer : Add texts here. Do not delete this text first. One purpose of the work in Progress Checks 5.23 and 5.24 was to indicate the plausibility of many of the results contained in the next theorem. Theorem 5.25 Let AA, BB. and CC be sets. Then A×(B∩C)=(A×B)∩(A×C)A×(B∩C)=(A×B)∩(A×C) A×(B∪C)=(A×B)∪(A×C)A×(B∪C)=(A×B)∪(A×C) (A∩B)×C=(A×C)∩(B×C)(A∩B)×C=(A×C)∩(B×C) (A∪B)×C=(A×C)∪(B×C)(A∪B)×C=(A×C)∪(B×C) A×(B−C)=(A×B)−(A×C)A×(B−C)=(A×B)−(A×C) (A−B)×C=(A×C)−(B×C)(A−B)×C=(A×C)−(B×C) If T⊆AT⊆A, then T×B⊆A×BT×B⊆A×B. If T⊆BT⊆B, then A×Y⊆A×BA×Y⊆A×B. We will not prove all these results; rather, we will prove Part (2) of Theorem 5.25 and leave some of the rest to the exercises. In constructing these proofs, we need to keep in mind that Cartesian products are sets, and so we follow many of the same principles to prove set relationships that were introduced in Sections 5.2and 5.3. The other thing to remember is that the elements of a Cartesian product are ordered pairs. So when we start a proof of a result such as Part (2) of Theorem 5.25, the primary goal is to prove that the two sets are equal. We will do this by proving that each one is a subset of the other one. So if we want to prove that A×(B∪C)⊆(A×B)∪(A×C)A×(B∪C)⊆(A×B)∪(A×C), we can start by choosing an arbitrary element of A×(B∪C)A×(B∪C). The goal is then to show that this element must be in (A×B)∪(A×C)(A×B)∪(A×C). When we start by choosing an arbitrary element of A×(B∪C)A×(B∪C), we could give that element a name. For example, we could start by letting u be an element of A×(B∪C). u be an element of A×(B∪C).(5.4.1) We can then use the definition of "ordered pair" to conclude that there exists x∈A and there exits y∈B∪C such that u=(x,y). there exists x∈A and there exits y∈B∪C such that u=(x,y).(5.4.2) In order to prove that A×(B∪C)⊆(A×B)∪(A×C)A×(B∪C)⊆(A×B)∪(A×C), we must now show that the ordered pair uu from (5.4.1) is in A×(B∪C)⊆(A×B)∪(A×C)A×(B∪C)⊆(A×B)∪(A×C). In order to do this, we can use the definition of set union and prove that u∈(A×B) or u∈(A×C). u∈(A×B) or u∈(A×C).(5.4.3) Since u=(x,y)u=(x,y), we can prove (5.4.3) by proving that (x∈A and y∈B) or (x∈A and y∈C). (x∈A and y∈B) or (x∈A and y∈C).(5.4.4) If we look at the sentences in (5.4.2) and (5.4.4), it would seem that we are very close to proving that A×(B∪C)⊆(A×B)∪(A×C)A×(B∪C)⊆(A×B)∪(A×C). Following is a proof of Part (2) of Theorem 5.25. Theorem 5.25 (Part (2)). Let AA, BB. and CC be sets. Then A×(B∪C)=(A×B)∪(A×C)A×(B∪C)=(A×B)∪(A×C) Proof : Let A, B. and C be sets. We will prove that A×(B∪C) is equal to (A×B)∪(A×C) by proving that each set is a subset of the other set . To prove that A×(B∪C)⊆(A×B)∪(A×C), we let u∈A×(B∪C). Then there exists x∈A and there exists y∈B∪C such that u=(x,y). Since y∈B∪C, we know that y∈B or y∈C. In the case where y∈B, we have u=(x,y), where x∈A and y∈B. So in this case, u∈A×B, and hence u∈(A×B)∪(A×C). Similarly, in the case where y∈C, we have u=(x,y), where x∈A and y∈C. So in this case, u∈A×C and, hence, u∈(A×B)∪(A×C). In both cases, u∈(A×B)∪(A×C). Hence, we may conclude that if u is an element of A×(B∪C), then u∈(A×B)∪(A×C), and this proves that A×(B∪C)⊆(A×B)∪(A×C). We must now prove that (A×B)∪(A×C)⊆A×(B∪C). So we let v∈(A×B)∪(A×C). Then v∈(A×B) or v∈(A×C). In the case where v∈(A×B), we know that there exists s∈A and there exists t∈B such that v=(s,t). But because t∈C, we can conclude that t∈B∪C and, hence, v∈A×(B∪C). In both cases, v∈A×(B∪C). Hence, we may conclude that if v∈(A×B)∪(A×C), then v∈A×(B∪C), and this proves that (A×B)∪(A×C)⊆A×(B∪C). The relationships in (5.4.5) and (5.4.6) prove that A×(B∪C)=(A×B)∪(A×C). Final Note. The definition of an ordered pair in Preview Activity 5.4.2 may have seemed like a lengthy definition, but in some areas of mathematics, an even more formal and precise definition of “ordered pair” is needed. This definition is explored in Exercise (10). Exercises for Section 5.4 Let A= {1, 2}, B= {a, b, c, d}, and C= {1, a, b}. Use the roster method to list all of the elements of each of the following sets: (a) A×B (b) B×A (c) A×C (d) A2 (e) A×(B∩C) (f) (A×B)∩(A×C) (g) A×∅ (h) B×{2} 2. Sketch a graph of each of the following Cartesian products in the Cartesian plane. (a) [0, 2] × [1, 3] (b) (0, 2) × (1, 3] (c) [2, 3] × {1} (d) {1} × [2, 3] (e) R × (2, 4) (f) (2, 4) × R (g) R × {-1} (h) {-1} × [1, +∞) 3. Prove Theorem 5.25, Part (1): A×(B∩C)=(A×B)∩(A×C). 4. Prove Theorem 5.25, Part (4): (A∪B)×C=(A×C)∪(B×C). 5. Prove Theorem 5.25, Part (5): A×(B−C)=(A×B)−(A×C). 6. Prove Theorem 5.25, Part (7): If T⊆A, then T×B⊆A×B. 7. Let A= {1}, B= {2}, and C= {3}. (a) Explain why A×B≠B×A. (b) Explain why A×B)×C≠A×(B×C). 8. Let A and B be nonempty sets. Prove that A×B=B×A if and only if A=B. 9. Is the following proposition true or false? Justify your conclusion. Let A, B and C be sets with A≠∅. If A×B=A×C, then B=C. Explain where the assumption that A≠∅ is needed. Explorations and Activities 10. (A Theoretic Definition of an Ordered Pair) In elementary mathematics, the notion of an ordered pair introduced at the beginning of this section will suffice. However, if we are interested in a formal development of the Cartesian product of two sets, we need a more precise definition of ordered pair. Following is one way to do this in terms of sets. This definition is credited to Kazimierz Kuratowski (1896 – 1980). Kuratowski was a famous Polish mathematician whose main work was in the areas of topology and set theory. He was appointed the Director of the Polish Academy of Sciences and served in that position for 19 years. Let x be an element of the set A, and let y be an element of the set B. The ordered pair (x, y) is defined to be the set {{x},{x,y}}. That is, (x,y)={{x},{x,y}}. (a) Explain how this definition allows us to distinguish between the ordered pairs (3, 5) and (5, 3). (b) Let A and B be sets and let a,c∈A and b,d∈B. Use this definition of an ordered pair and the concept of set equality to prove that (a,b)=(c,d) if and only if a=c and b=d. An ordered triple can be thought of as a single triple of objects, denoted by (a, b, c), with an implied order. This means that in order for two ordered triples to be equal, they must contain exactly the same objects in the same order. That is (a,b,c)=(p,q,r) if and only if a=p, b=q and c=r. (c) Let A, B and C be sets, and let x∈A, y∈B, and z∈C. Write a set theoretic definition of the ordered triple (x,y,z) similar to the set theoretic definition of “ordered pair.” Answer : Add texts here. Do not delete this text first. 5.3: Properties of Set Operations 5.5: Indexed Families of Sets
10755	https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem?srsltid=AfmBOoqFJAQURvSK4Mj0nDHLLz_OwfvGX28GFFG6dGh_OU8T-lPOqe4u	Art of Problem Solving Angle Bisector Theorem - 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 Bisector Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Angle Bisector Theorem This is an AoPSWiki Word of the Week for June 6-12 Contents [hide] 1 Introduction & Formulas 2 Proof 3 Examples & Problems 4 See also Introduction & Formulas The Angle bisector theorem states that given triangle and angle bisector AD, where D is on side BC, then . It follows that . Likewise, the converse of this theorem holds as well. Further by combining with Stewart's theorem it can be shown that Proof By the Law of Sines on and , First, because is an angle bisector, we know that and thus , so the denominators are equal. Second, we observe that and . Therefore, , so the numerators are equal. It then follows that Examples & Problems Let ABC be a triangle with angle bisector AD with D on line segment BC. If and , find AB and AC. Solution: By the angle bisector theorem, or . Plugging this into and solving for AC gives . We can plug this back in to find . 2. In triangle ABC, let P be a point on BC and let . Find the value of . Solution: First, we notice that . Thus, AP is the angle bisector of angle A, making our answer 0. 3. Part (b), 1959 IMO Problems/Problem 5 See also Angle bisector Geometry Stewart's theorem Retrieved from " Categories: Geometry 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.
10756	https://resources.pcb.cadence.com/blog/2021-what-is-nodal-analysis-in-circuit-design	Skip to main content What is Nodal Analysis in Circuit Design? Published Date Author Cadence PCB Solutions Key Takeaways Nodal analysis is a mathematical method that is used to analyze circuits in terms of voltage drops between nodes in a circuit diagram. Nodal analysis can be performed in the time domain or frequency domain, but it is limited to linear time-invariant (LTI) systems. Although the solution algorithm used in nodal analysis is only defined for linear systems, nonlinear systems can be approximated around an operating point and used in nodal analysis. Circuit simulation results need to be compared with in-situ measurements during prototyping and production. Analyzing electrical circuits is vital to ensuring modern technology works as expected. Most modern electronics go through some sort of simulation and evaluation process to ensure designs work as intended, and to provide a set of reference calculations for comparison with in-circuit tests. SPICE simulations are the workhorses for circuit design and analysis, and today’s commercial solutions provide many built-in simulations. Among the simulations found in SPICE simulators, nodal analysis is a fundamental technique used to examine voltage and current distribution in a circuit. This technique effectively combines both of Kirchhoff’s laws and Ohm’s law into a single matrix equation. Keep reading to learn more about what nodal analysis is and how it’s used in circuit design and analysis. What is Nodal Analysis? Nodal analysis is a mathematical method for calculating the distribution of voltages between nodes in a circuit. Also referred to as the node-voltage method, this strategy uses Ohm’s law, Kirchhoff’s voltage law, and Kirchhoff’s current law to define an equation relating the voltage measured between each circuit node and some reference (usually ground). The voltage drops measured between neighboring nodes are taken as the variables in a set of linear equations and the system can be solved using a standard algorithm (e.g., Gauss-Jordan elimination). Constructing a Matrix Equation in Nodal Analysis The general process for constructing a matrix equation for nodal analysis is to use Kirchoff’s laws between each node in a circuit diagram to write out a set of equations relating the voltage drop across different components to the currents flowing into each node. Use the following process: Draw out the circuit diagram with currents defined at each node. Pick a reference node (usually ground) and write a variable for the voltage at each node with respect to the reference node. Write out Kirchoff’s current law for each node in terms of the circuit impedances and the voltages at neighboring nodes. Rewrite the system of equations for the node voltages in matrix form. Solve the matrix equation using an inverse matrix. To see how this works, it helps to look at a simple example. An Example of Nodal Analysis The example circuit diagram below shows 4 impedances with 1 voltage source and 4 distinct nodes. Node A is just the input voltage, leaving the voltages at the 3 other nodes to be determined. By applying steps 1-3, we arrive at the following set of equations for the voltages at each node: The circuit diagram and the system of equations used to determine the voltages at each node. We can then rewrite the above 3 equations as a matrix equation: Rewriting the circuit diagram’s system of equations as a matrix equation. This matrix equation is then easily solved by multiplying by the inverse of the coefficient matrix. In the above circuit diagram, we’ve defined a “reference node” at the common ground connection. The voltage variables are all measured with respect to this point. This is important for more than just creating a closed circuit; it sets the measurement reference for all voltage variables that appear in the above matrix equation. Reference Nodes One key to doing nodal analysis successfully is to properly define reference nodes, also called datum nodes. In nodal analysis, a reference node is the point that is used for comparison when measuring voltages, because a voltage is only defined between two points in space. Imagine you were using a voltmeter on your circuit diagram; the reference node would be the point where the negative lead was connected, and the positive lead would connect to any other node. This is why voltage measurement probes in SPICE simulations are normally shown as a single probe; the measured voltage is taken with respect to a reference node in the circuit. In this PSpice simulation result, the voltage measured at this probe is just the voltage measured between R4 and GND; this is determined using nodal analysis. In SPICE simulations, the reference node is normally taken as ground. If there are two grounds, the reference node is taken as the nearest connected ground potential. In nodal analysis, any node can be taken as a reference point. This gives you an advantage in that you can define a potential difference between two different ground points, such as you might find in a galvanically isolated PCB with two different ground planes. How Does Nodal Analysis Work on Nonlinear Circuits? Because the current and voltage in a nonlinear component is a nonlinear function (meaning the relationship does not obey Ohm’s law), the matrix equation developed in nodal analysis will have terms that are functions of the voltages at different nodes. The solved voltage column vector in this equation is defined in terms of a nonlinear impedance matrix and the current in each portion of the circuit. This transcendental matrix equation may include coupled terms, which require simultaneous iterative numerical techniques to determine solutions. In this general formulation, applying nodal analysis to nonlinear circuits requires solving a transcendental matrix equation for multiple unknowns. This can be a very difficult numerical problem to solve, depending on the form of the derived matrix equation. Therefore, in order to examine nonlinear circuits using nodal analysis, the current vs. voltage relationship measured between adjacent nodes in the circuit needs to be approximated around some operating point. The typical method to do this is to use one of the following general approximation techniques for a nonlinear time-invariant system: Taylor series: For a system of N components, define 2N new variables that are shifted away from the intended operating point with a linear transformation. Now, expand the current-voltage function for each nonlinear component as a Taylor series in terms of these new variables. Maclaurin series: Expand the current-voltage function for each nonlinear component directly around the operating point without defining new variables as a Maclaurin series. By keeping only the first-order terms (i.e., linear terms), you now have a linearized circuit that includes nonlinear components. The results from this procedure are only valid within a small deviation from the operating point, depending on the impedance function for each nonlinear component in the system. Circuit simulators that use iterative solution techniques will apply these approximations for you and can solve operating point matrix equations in nodal analysis for nonlinear circuits. Vital node voltage and nodal voltage analysis requires a strong understanding of current source in your electronic circuit but when tasked with unknown node voltages or independent voltage source, your nodal equation can get difficult. Looking at principal nodes for your current source and branch current references can encourage a better understanding of nodal voltage. But generally for electric circuit performance, circuit analysis through SPICE simulation will unlock all types of circuit variables from transient analysis to parallel circuits to various matrices and beyond. Use your node equations wisely in your next electrical circuit design. If you’re still unclear as to what nodal analysis is, you can use an industry-standard SPICE simulator to calculate voltage and current throughout your circuits. Once you’ve got your results and you’re ready to create your physical design, use the best set of PCB design and analysis software to capture your schematics as a blank circuit board. Allegro PCB Editor from Cadence gives you the layout and routing features you need to create your circuit boards and prepare them for production. You can then use nodal analysis results as a comparison for electrical tests. If you’re looking to learn more about how Cadence has the solution for you, talk to us and our team of experts.
10757	https://www.rcboe.org/cms/lib/GA01903614/Centricity/Domain/8243/12.1.pdf	© Houghton Mifflin Harcourt Publishing Company Name Class Date Explore Constructing Similar Triangles In the following activity you will see one way to construct a triangle similar to a given triangle. A Do your work for Steps A–C in the space provided. Draw a triangle. Label it ABC as shown. B Select a point on _ AB . Label it E. C Construct an angle with vertex E that is congruent to ∠B. Label the point where the side of the angle you constructed intersects _ AC as F. D Why are ‹ − › EF and _ BC parallel? E Use a ruler to measure _ AE , _ EB , _ AF , and _ FC . Then compare the ratios AE ___ EB  and AF ___ FC  . Resource Locker B A C B E A C \192.168.9.251\07Macdata\07Vol1Data\From Graphics\02192014\From Ramesh\Econ 2016\W B E A C F Module 12 631 Lesson 1 12.1 Triangle Proportionality Theorem Essential Question: When a line parallel to one side of a triangle intersects the other two sides, how does it divide those sides? © Houghton Mifflin Harcourt Publishing Company Reflect 1. Discussion How can you show that △AEF ∼ △ABC? Explain. 2. What do you know about the ratios AE ___ AB and AF ___ AC ? Explain. 3. Make a Conjecture Use your answer to Step E to make a conjecture about the line segments produced when a line parallel to one side of a triangle intersects the other two sides. Explain 1 Proving the Triangle Proportionality Theorem As you saw in the Explore, when a line parallel to one side of a triangle intersects the other two sides of the triangle, the lengths of the segments are proportional. Triangle Proportionality Theorem Theorem Hypothesis Conclusion If a line parallel to a side of a triangle intersects the other two sides, then it divides those sides proportionally. AE _ EB = AF _ FC Example 1 Prove the Triangle Proportionality Theorem A Given: ‹ − › EF ∥ _ BC Prove: AE ___ EB = AF ___ FC Step 1 Show that △AEF ∼ △ABC. Because ‹ − › EF ∥ _ BC , you can conclude that ∠1 ≅ ∠2 and ∠3 ≅ ∠4 by the Theorem. So, △AEF ∼ △ABC by the . B E F A C EF ∥ BC B E 1 2 3 4 F A C Module 12 632 Lesson 1 © Houghton Mifflin Harcourt Publishing Company Step 2 Use the fact that corresponding sides of similar triangles are proportional to prove that AE ___ EB = AF ___ FC . AB _ AE = Corresponding sides are proportional. AE + EB _ AE = Segment Addition Postulate 1 + EB _ AB = Use the property that a + b ____ c = a __ c + b __ c . EB _ AE = Subtract 1 from both sides. AE _ EB = Take the reciprocal of both sides. Reflect 4. Explain how you conclude that ▵AEF ∼▵ABC without using ∠3 and ∠4. Explain 2 Applying the Triangle Proportionality Theorem Example 2 Find the length of each segment. A _ CY It is given that _ XY ∥ _ BC so AX ___ XB = AY ___ YC by the Triangle Proportionality Theorem. Substitute 9 for AX, 4 for XB, and 10 for AY. Then solve for CY. 9 _ 4 = 10 _ CY Take the reciprocal of both sides. 4 _ 9 = CY _ 10 Next, multiply both sides by 10. 10 ( 4 _ 9 ) = ( CY _ 10 ) 10 → 40 _ 9 = CY, or 4 4 _ 9 = CY B Find PN. It is given that _ PQ ∥ _ LM , so NQ ___ QM = by the Triangle Proportionality Theorem. Substitute for NQ, for QM, and 3 for . 5 _ 2 = NP _ 3 Multiply both sides by : ( 5 _ 2 ) = ( NP _ 3 ) → = NP B X Y C 10 4 9 A L P Q N M 3 5 2 Module 12 633 Lesson 1 © Houghton Mifflin Harcourt Publishing Company Your Turn Find the length of each segment. 5. _ DG 6. _ RN Explain 3 Proving the Converse of the Triangle Proportionality Theorem The converse of the Triangle Proportionality Theorem is also true. Converse of the Triangle Proportionality Theorem Theorem Hypothesis Conclusion If a line divides two sides of a triangle proportionally, then it is parallel to the third side. ‹ − › EF \|\| _ BC Example 3 Prove the Converse of the Triangle Proportionality Theorem A Given: AE _ EB = AF _ FC Prove: ‹ − › EF ∥ _ BC Step 1 Show that △AEF ∼ △ABC. It is given that AE _ EB = AF _ FC , and taking the reciprocal of both sides shows that . Now add 1 to both sides by adding AE _ AE to the left side and AF _ AF to the right side. This gives . Adding and using the Segment Addition Postulate gives . Since ∠A ≅ ∠A, △AEF ∼ △ABC by the Theorem. Step 2 Use corresponding angles of similar triangles to show that ‹ − › EF ∥ _ BC . ∠AEF ≅ ∠ and are corresponding angles. So, ‹ − › EF ∥ _ BC by the Theorem. E F C D G 24 32 40 P M N Q R 10 8 5 A B E C F AE EB AF FC = A B E C F Module 12 634 Lesson 1 © Houghton Mifflin Harcourt Publishing Company Reflect 7. Critique Reasoning A student states that _ UV must be parallel to _ ST . Do you agree? Why or why not? Explain 4 Applying the Converse of the Triangle Proportionality Theorem You can use the Converse of the Triangle Proportionality Theorem to verify that a line is parallel to a side of a triangle. Example 4 Verify that the line segments are parallel. A _ MN and _ KL JM _ MK = 42 _ 21 = 2 JN _ N L = 30 _ 15 = 2 Since JM _ MK = JN _ N L , _ MN \|\| _ KL by the Converse of the Triangle Proportionality Theorem. B _ DE and _ AB (Given that AC = 36 cm, and BC = 27 cm) AD = AC - DC = 36 - 20 = 16 BE = BC - = - = CD _ DA = _ = _ CE _ EB = _ = Since CD _ DA = _ , ― DE \|\| ― AB by the Theorem. Reflect 8. Communicate Mathematical Ideas In △ABC, in the example, what is the value of AB ___ DE ? Explain how you know. Your Turn 9. Verify that _ TU and _ RS are parallel. R S T U V 90 67.5 54 72 K L M 42 30 15 21 N J A B D E C 15 cm 20 cm R S T U V Module 12 635 Lesson 1 © Houghton Mifflin Harcourt Publishing Company Elaborate 10. In △ABC, _ XY \|\| _ BC . Use what you know about similarity and proportionality to identify as many different proportions as possible. 11. Discussion What theorems, properties, or strategies are common to the proof of the Triangle Proportionality Theorem and the proof of Converse of the Triangle Proportionality Theorem? 12. Essential Question Check-In Suppose a line parallel to side ― BC of ▵ABC intersects sides ― AB and ― AC at points X and Y, respectively, and AX ___ XB = 1. What do you know about X and Y? Explain. 1. Copy the triangle ABC that you drew for the Explore activity. Construct a line ‹ − › FG parallel to _ AB using the same method you used in the Explore activity. 2. _ ZY \|\| ‹ − › MN . Write a paragraph proof to show that XM ___ MZ = XN ___ NY . • Online Homework • Hints and Help • Extra Practice Evaluate: Homework and Practice M N X Y Z A B C X Y Module 12 636 Lesson 1 © Houghton Mifflin Harcourt Publishing Company Find the length of each segment. 3. _ KL 4. _ XZ 5. _ VM Verify that the given segments are parallel. 6. _ AB and _ CD 7. _ MN and _ QR 8. _ WX and _ DE 9. Use the Converse of the Triangle Proportionality Theorem to identify parallel lines in the figure. 10. On the map, 1st Street and 2nd Street are parallel. What is the distance from City Hall to 2nd Street along Cedar Road? G 6 4 8 H m n J K L U V T M N 14 8 49 X Y U V 30 18 30 Z A B C D E 12 14 4 4 2 3 P Q R N M 3 2.7 9 10 D W X E F 1.5 2.5 3.5 2.1 A B C L M N 20 16 12 15 8 16 2.4 mi 2.1 mi 2.8 mi Cedar Rd. Aspen Rd. 1st St. 2nd St. City Hall Library Module 12 637 Lesson 1 © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Stuart Walker/Alamy 11. On the map, 5th Avenue, 6th Avenue, and 7th Avenue are parallel. What is the length of Main Street between 5th Avenue and 6th Avenue? 12. Multi-Step The storage unit has horizontal siding that is parallel to the base. a. Find LM. b. Find GM. c. Find MN to the nearest tenth of a foot. d. Make a Conjecture Write the ratios LM ___ MN and HJ __ JK as decimals to the nearest hundredth and compare them. Make a conjecture about the relationship between parallel lines ‹ − › LD , ‹ − › ME , and ‹ − › NF and transversals ‹ − › GN and ‹ − › GK. 13. A corollary to the Converse of the Triangle Proportionality Theorem states that if three or more parallel lines intersect two transversals, then they divide the transversals proportionally. Complete the proof of the corollary. Given: Parallel lines ‹ − › AB ∥ ‹ − › CD, ‹ − › CD ∥ ‹ − › EF Prove: AC ___ CE = BX ___ XE , BX ___ XE = BD ___ DF , AC ___ CE = BD ___ DF Statements Reasons 1. ‹ − › AB ∥ ‹ − › CD , ‹ − › CD ∥ ‹ − › AF 1. Given 2. Draw ‹ − › EB intersecting ‹ − › CD at X. 2. Two points 3. AC ___ CE = BX __ XE 3. 4. BX __ XE = BD ___ DF 4. 5. AC ___ CE = BD ___ DF 5. Property of Equality 0.3 km 0.5 km Spring St. 0.4 km Main St. 7th Ave. 6th Ave. 5th Ave. The storage unit has horizontal siding that is parallel to the base. G L K D H 11.3 ft 10.4 ft 2.6 ft 2.2 ft J M E N F A B C D E F X Module 12 638 Lesson 1 © Houghton Mifflin Harcourt Publishing Company 14. Suppose that LM = 24. Use the Triangle Proportionality Theorem to find PM. 15. Which of the given measures allow you to conclude that ― UV ∥ ― ST ? Select all that apply. A. SR = 12, TR = 9 B. SR = 16, TR = 20 C. SR = 35, TR = 28 D. SR = 50, TR = 48 E. SR = 25, TR = 20 H.O.T. Focus on Higher Order Thinking 16. Algebra For what value of x is _ GF ∥ ― HJ ? 17. Communicate Mathematical Ideas John used △ABC to write a proof of the Centroid Theorem. He began by drawing medians ― AK and ― CL , intersecting at Z. Next he drew midsegments ― LM and ― NP , both parallel to median ― AK . Given: △ABC with medians ― AK and ― CL , and midsegments ― LM and ― NP Prove: Z is located 2 _ 3 of the distance from each vertex of △ABC to the midpoint of the opposite side. a. Complete each statement to justify the first part of John’s proof. By the definition of , MK = 1 _ 2 BK. By the definition of , BK = KC. So, by , MK = 1 _ 2 KC, or KC ___ MK = 2. Consider △LMC. ― LM ∥ ― AK (and therefore ― LM ∥ ― ZK ), so ZC ___ LZ = KC ___ MK by the Theorem, and ZC = 2LZ. Because LC = 3LZ, ZC ___ LC = 2LZ ___ 3LZ = 2 _ 3 , and Z is located 2 _ 3 of the distance from vertex C of △ABC to the midpoint of the opposite side. b. Explain how John can complete his proof. K N 10 15 M L P U S T V R 20 16 A B C M L N Z K P 45 40 5x + 1 4x + 4 G F J E H L M K P C Z Module 12 639 Lesson 1 © Houghton Mifflin Harcourt Publishing Company 18. Persevere in Problem Solving Given △ABC with FC = 5, you want to find BF. First, find the value that y must have for the Triangle Proportionality Theorem to apply. Then describe more than one way to find BF, and find BF. Shown here is a triangular striped sail, together with some of its dimensions. In the diagram, segments BJ, CI, and DH are all parallel to segment EG. Find each of the following: 1. AJ 2. CD 3. HG 4. GF 5. the perimeter of △AEF 6. the area of △AEF 7. the number of sails you could make for $10,000 if the sail material costs $30 per square yard Lesson Performance Task 0 1 2 3 4 5 6 7 8 9 1 2 6 3 4 5 7 8 9 x y A (1, 2) E (4, 2) C (10, 2) F (7, 6) B (5.5, y) E A F G D C B J I H 3.5 ft 6.5 ft 6 ft 2.25 ft 2.5 ft 1.8 ft 1.2 ft Module 12 640 Lesson 1
10758	https://math.stackexchange.com/questions/2886818/number-of-way-5-people-can-be-divided-into-3-groups	Skip to main content Number of way 5 people can be divided into 3 groups Ask Question Asked Modified 7 years ago Viewed 8k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. The question asks the number of ways to divide 5 people into 3 groups with no conditions attached. (In case people think this is a duplicate). I have studied both grouping and distribution in combinatorics but I often get confused where to apply what. Anyway, I digress. My First Approach: The first thing I did was to assign variables. Let the first, second and third groups be xa,xb and xc. Now xa+xb+xc=5 I solved this by partitioning 5 in 3 ways, that is (n+r−1r−1) where n=5,r=3 giving us, (72)=21 The answer, however, is given as 25. Please tell me where I'm going wrong and also could someone advise me on when to use partitioning and when to make groups if the group size is not given. combinatorics permutations combinations Share CC BY-SA 4.0 Follow this question to receive notifications edited Aug 18, 2018 at 15:38 Especially Lime 47.1k99 gold badges6464 silver badges109109 bronze badges asked Aug 18, 2018 at 15:25 Prakhar NagpalPrakhar Nagpal 1,3601212 silver badges2525 bronze badges Add a comment \| 3 Answers 3 Reset to default This answer is useful 3 Save this answer. Show activity on this post. Assume that the groups are interchangeable, so that the division (A,B),(C,D),(E) is the same as the division (C,D),(A,B),(E) and assume that you don't allow empty groups. The maximal group has to have either 2 or 3 members. Case I: (maximal group has 3 members). Then the counts must be {3,1,1} and the division is determined by the members in the group of 3. Hence there are (53)=10 divisions of this type. Case II: (maximal group has 2 members). Then the counts must be {2,2,1}. Then there are (52)=10 ways to populate one group of two, and then (32)=3 to populate the other. As switching the two groups of two does not change the division we see that there are 10×32=15 divisions of this type. Combining we see that there are 10+15=25 possible divisions. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Aug 18, 2018 at 15:36 lulululu 77.4k66 gold badges8989 silver badges140140 bronze badges 1 Alright, I think I understand. Thank you Prakhar Nagpal – Prakhar Nagpal 2018-08-18 15:40:30 +00:00 Commented Aug 18, 2018 at 15:40 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. The formula you've applied is for labelled groups of unlabelled things (i.e. all that matters is the numbers of things in each group, but having groups of size 3,1,1 is different to 1,3,1. This is exactly the opposite of what you want, which is unlabelled groups of labelled things. (If you have groups of size 3,1,1 it matters which people go in the same group, but not what order you put the groups in.) I think for your problem there is also another restriction: no group is allowed to be empty. One way to do this is to first consider labelled people in labelled groups, then divide by the number of ways to label the groups. Since there are three groups, there are 3!=6 ways to label them. Now the number of ways to put 5 labelled people into 3 labelled groups is 35 - for every person you can choose group A, B or C, and these choices are independent. However, this counts some divisions that aren't allowed (where some group is empty), so you have to allow for this. The best way to do that is the inclusion-exclusion formula. In this case, inclusion-exclusion is relatively simple as there are only three teams. You start from all ways to assign A, B or C to each person (243 ways). Then for each team you subtract off the ways that don't use that team. So there are 25=32 ways that don't use A, and the same for B and C. So this gives 243−3×32=147. However, there is one way to assign the teams that doesn't use A or B (all C), and you subtracted it twice (once because it doesn't use A, and once because it doesn't use B), so you need to add it back on. The same applies to the other two possibilities with two empty teams, so in total you have 243−3×32+3=150 assignments. Now every way to split the people into three teams is counted 6 times (because there are 6 ways to label the teams), so the final answer is 150/6=25. If there were n people the same approach would give 3n−3×2n+36 possibilities. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Aug 18, 2018 at 16:19 answered Aug 18, 2018 at 15:37 Especially LimeEspecially Lime 47.1k99 gold badges6464 silver badges109109 bronze badges 3 I haven't done inclusion-exclusion in depth, is there any other way to solve this, or perhaps you could explain it to me yourself, or provide a link. Prakhar Nagpal – Prakhar Nagpal 2018-08-18 15:38:57 +00:00 Commented Aug 18, 2018 at 15:38 I've edited with more explanation. Especially Lime – Especially Lime 2018-08-18 16:19:45 +00:00 Commented Aug 18, 2018 at 16:19 I'm surprised about all these answers and nobody even mentions the keyword "Stirling"! Diger – Diger 2018-08-18 17:11:26 +00:00 Commented Aug 18, 2018 at 17:11 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. First Approach: Note that these people are distinguishable. You cannot treat them like stars and bars. Thinking about it, there must be one person in each group. Now, we have some casework. We have two cases. WLOG first number ≥ second number ≥ third number. Our cases are 3-1-1 and 2-2-1. 3-1-1: The first leads to 5!3!2!=10 solutions. 2-2-1: This leads to 5!2!2!2!=15 solutions. This is a total of 10+15=25 solutions. Partitions are all the addition of positive integers that are equal to a number. For example, partitions of 4 are 4,3+1,1+3,2+2,1+1+2,1+2+1,2+1+1,1+1+1+1. Correct me if I'm wrong, but I believe you mean distribution. Use them when there are items to be handed out and items that can receive them. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Aug 18, 2018 at 16:08 answered Aug 18, 2018 at 15:28 Jason KimJason Kim 94266 silver badges2020 bronze badges 5 I agree I figured that that might be what was wrong with partitioning. So can you tell me how to go about it then, since the group size is not given, should I take cases? Prakhar Nagpal – Prakhar Nagpal 2018-08-18 15:30:03 +00:00 Commented Aug 18, 2018 at 15:30 Are the groups distinguishible? Jason Kim – Jason Kim 2018-08-18 15:32:45 +00:00 Commented Aug 18, 2018 at 15:32 As in they cannot be of the same size? Prakhar Nagpal – Prakhar Nagpal 2018-08-18 15:36:26 +00:00 Commented Aug 18, 2018 at 15:36 "... and also could someone advise me on when to use partitioning and when to make groups if the group size is not given?" amWhy – amWhy 2018-08-18 15:48:50 +00:00 Commented Aug 18, 2018 at 15:48 Thinking about it, there must be one person in each group. Now, we have some casework. We have two cases: 3-1-1 2-2-1 3-1-1: This leads to 5!/3!2!=10 solutions. 2-2-1: This leads to 5!/2!2!2!=15 solutions. This is a total of 10+15=25 solutions. Now I guess I'll experiment for other cases. Jason Kim – Jason Kim 2018-08-18 15:51:42 +00:00 Commented Aug 18, 2018 at 15:51 Add a comment \| You must log in to answer this question. 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10759	https://www.youtube.com/watch?v=Nkh8eRSkM0M	(cos(pi + x) cos(-x))/(sin(pi - x) cos(pi/2 + x)) = cot^2 x Ravi Ranjan Kumar Singh 22000 subscribers 74 likes Description 4093 views Posted: 7 Nov 2020 In this video, we will learn to prove the equation (cos(pi + x) cos(-x))/(sin(pi - x) cos(pi/2 + x)) = cot^2 x. The link of the video explaining, the proof of the identity cos(pi + x) = - cos x has been given below The link of the video explaining, the proof of the identity cos(-x) = cos x has been given below The link of the video explaining the proof of the identity sin(pi-x) = sin x has been given below The link of the video explaining the proof of the identity cos(pi/2 + x) = -sin x has been given below Other topics of this video are: Class XI exercise 3.3 problem 8 NCERT Class XI exercise 3.3 problem 8 I, Ravi Ranjan Kumar Singh, have produced this video. All credits of this video belong to me. You can email me at raviranjans@gmail.com You can follow me on facebook. My facebook account is 13 comments Transcript: In this video, we will learn to prove the equation cosine of pi plus x times cosine of minus x upon sine of pi minus x times cosine of pi upon 2 plus x is equal to square of cotangent of x. Left hand side of the equation to be proved is cosine of pi plus x times cosine of minus x upon sine of pi minus x times cosine of pi upon 2 plus x. We know cosine of pi plus x is equal to minus cosine of x. The proof the trigonometry identity is given in the video appearing in the upper right corner of this video. The link of the video is also given in the description section. We also know cosine of minus x is equal to cosine of x. The proof the trigonometry identity is given in the video appearing in the upper right corner of this video. The link of the video is also given in the description section. We also know sine of pi minus x is equal to sine of x. The proof the trigonometry identity is given in the video appearing in the upper right corner of this video. The link of the video is also given in the description section. And we also know cosine of half of pi plus x is equal to minus sine of x. The proof the trigonometry identity is given in the video appearing in the upper right corner of this video. The link of the video is also given in the description section. Application of the above identities gives us left hand side of the equation is equal to minus cosine of x times cosine of x upon sine of x times minus sine of x. Cancelling out the common minus sign in the numerator and the denominator gives us left hand side of the equation is equal to square of cosine of x upon square of sine of x. As, cosine of x upon sine of x is cotangent of x. So, left hand side of the equation is equal to square of cotangent of x which is equal to the right hand side of the given equation. So, we have proved the given equation. Thank you for being with me. I hope you like this video. Please click the like button and subscribe button. Do not forget to press the notification bell so that, you get the notifications of my new videos. Share the video and do write in the comments. I am eager to listen to your thoughts. If you have any question related to video or if you think I should make a video on any other topics in which you are interested please write in the comments. Thank you.
10760	https://www.mathworksheets4kids.com/vertical-angles.php	Child Login Main Menu Math Language Arts Science Social Studies Interactive Worksheets Browse By Grade Become a Member Become a Member Math Interactive Worksheets Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Worksheets by Grade Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade Number Sense and Operations Number Recognition Counting Skip Counting Place Value Number Lines Addition Subtraction Multiplication Division Word Problems Comparing Numbers Ordering Numbers Odd and Even Prime and Composite Roman Numerals Ordinal Numbers Properties Patterns Rounding Estimation In and Out Boxes Number System Conversions More Number Sense Worksheets Measurement Size Comparison Time Calendar Money Measuring Length Weight Capacity Metric Unit Conversion Customary Unit Conversion Temperature More Measurement Worksheets Financial Literacy Writing Checks Profit and Loss Discount Sales Tax Simple Interest Compound Interest Statistics and Data Analysis Tally Marks Pictograph Line Plot Bar Graph Line Graph Pie Graph Scatter Plot Mean, Median, Mode, Range Mean Absolute Deviation Stem-and-leaf Plot Box-and-whisker Plot Factorial Permutation and Combination Probability Venn Diagram More Statistics Worksheets Geometry Positions Shapes - 2D Shapes - 3D Lines, Rays and Line Segments Points, Lines and Planes Angles Symmetry Transformation Area Perimeter Rectangle Triangle Circle Quadrilateral Polygon Ordered Pairs Midpoint Formula Distance Formula Slope Parallel, Perpendicular and Intersecting Lines Scale Factor Surface Area Volume Pythagorean Theorem More Geometry Worksheets Pre-Algebra Factoring GCF LCM Fractions Decimals Converting between Fractions and Decimals Integers Significant Figures Percents Convert between Fractions, Decimals, and Percents Ratio Proportions Direct and Inverse Variation Order of Operations Exponents Radicals Squaring Numbers Square Roots Scientific Notations Logarithms Speed, Distance, and Time Absolute Value More Pre-Algebra Worksheets Algebra Translating Algebraic Phrases Evaluating Algebraic Expressions Simplifying Algebraic Expressions Algebraic Identities Equations Quadratic Equations Systems of Equations Functions Polynomials Inequalities Sequence and Series Matrices Complex Numbers Vectors More Algebra Worksheets Trigonometry Calculus Math Workbooks English Language Arts Summer Review Packets Science Social Studies Holidays and Events Support Worksheets> Math> Geometry> Angles> Vertical Angles Vertical Angles Worksheets When two lines intersect, two pairs of congruent angles are formed. Our printable vertical angles worksheets for grade 6, grade 7, and grade 8 take a shot at simplifying the practice of these congruent angles called vertically opposite angles. Upon close observation, it's revealed that two intersecting lines give rise to four linear pairs too. Bring into play the appropriate properties of these angles formed by intersecting lines and crack the pdf exercises on finding the unknown angle measures and solving variable equations involving angle measures. Take a look at our free worksheets! Vertical Angles \| Vertically Opposite Angles Bank on success in identifying vertically opposite angles in multiple rays and finding the size of angles that share the same vertex, with our vertical angles worksheets. Remember vertical angles are congruent. Finding Unknown Angles Before you hand out our printable vertical angles worksheets to 6th grade and 7th grade students, drill them on the congruent and supplementary properties of the angles formed by intersecting lines. Algebra in Vertical Angles \| One-Step This set of angles formed by intersecting lines worksheets depicts vertical angles with one angle measure and the opposite angle as a variable expression. Simply equate both the angles and solve for x. Algebra in Vertical Angles \| Two-Step Hurray! The solution is just two steps away! Set up the equation using the expression on the left and measure on the right; rearrange it to make x the subject; perform the operations; solve for the variable. Algebra in Vertical Angles \| Multi-Step Let your 8th grade students test themselves with these pdfs. Equate the expressions and solve for x, plug in the value and find a pair of opposite angles, subtract the measure from 180°, and determine the other pair. Related Worksheets »Complementary and Supplementary Angles »Adjacent Angles »Angles Formed by a Transversal »Angles Around a Point Home Become a Member Membership Information Login Printing Help FAQ How to Use Interactive Worksheets How to Use Printable Worksheets About Us Privacy Policy Terms of Use Contact Us Follow us Copyright © 2025 - Math Worksheets 4 Kids ×
10761	https://www.quora.com/How-is-it-that-the-function-%F0%9D%91%93-%E2%84%9D-%E2%84%9D-given-by-%F0%9D%91%93-%F0%9D%91%A5-%F0%9D%91%A5-3-%F0%9D%91%A5-is-onto	How is it that the function 𝑓: ℝ → ℝ given by 𝑓(𝑥) = 𝑥^3 + 𝑥 is onto? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics On to Function The Real Numbers Types of Functions Proofs (mathematics) Onto Functions Operations on Functions Functions and Mappings Algebra 5 How is it that the function 𝑓: ℝ → ℝ given by 𝑓(𝑥) = 𝑥^3 + 𝑥 is onto? All related (41) Sort Recommended Paul Bankston PhD in Mathematical Logic and Foundations&General Topology, University of Wisconsin - Madison (Graduated 1974) · Upvoted by BowTangey , PhD Mathematics, Iowa State University (1988) · Author has 920 answers and 504.3K answer views ·2y In addition to Sohel Zibara’s argument—which uses the fact that R R is a real-closed field (all odd-degree polynomials have at least one root)—you could use facts from elementary Calculus: for x≥0 x≥0, we have f(x)≥x f(x)≥x, and for x≤0 x≤0 we have f(x)≤x f(x)≤x. Hence f f is unbounded both above and below. Given any real number a a, there is a c c such that f(c)≤a f(c)≤a, and there is a d d such that f(d)≥a f(d)≥a. Because f f is continuous, the Intermediate Value Theorem tells us that there is some b b (with b b between c c and d d) such that f(b)=a f(b)=a. Thus f f is onto. BTW, since f′(x)=3 x 2+1 f′(x)=3 x 2+1 is always positive, you k Continue Reading In addition to Sohel Zibara’s argument—which uses the fact that R R is a real-closed field (all odd-degree polynomials have at least one root)—you could use facts from elementary Calculus: for x≥0 x≥0, we have f(x)≥x f(x)≥x, and for x≤0 x≤0 we have f(x)≤x f(x)≤x. Hence f f is unbounded both above and below. Given any real number a a, there is a c c such that f(c)≤a f(c)≤a, and there is a d d such that f(d)≥a f(d)≥a. Because f f is continuous, the Intermediate Value Theorem tells us that there is some b b (with b b between c c and d d) such that f(b)=a f(b)=a. Thus f f is onto. BTW, since f′(x)=3 x 2+1 f′(x)=3 x 2+1 is always positive, you know that f f is strictly increasing. Hence f f is also one-to-one. Something to ponder: You know that if a a is rational, then so is a 3+a a 3+a. Hence f f may also be viewed as a function from Q Q to itself. Is this new f f onto? One-to-one? (Try the same question, but using the integers Z Z instead.) Upvote · 9 3 9 2 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. 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. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. 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. Upvote · 999 485 999 103 99 17 Related questions More answers below How do I determine 𝑓: ℝ → ℝ,𝑤𝑖𝑡ℎ 𝑦 = 𝑓(𝑥) = 𝑥 √𝑥+1 is bijective? Is the function ℎ: ℝ \ {2020} → ℝ, ℎ (x) = (2021 − x) / (x − 2020) bijections function? Give an explanation complete How can we find a real function 𝑓:ℝ→ℝ such that 𝑓(𝑓(𝑥)) =−𝑥? If 𝑓: ℝ → ℝ be a continuous function satisfying 𝑓(𝑥) + 𝑓 (3 − 𝑥) = 4, then? What is the value of--- 0 to 3 ∫ 𝑓(𝑥) 𝑑x Explain: If f:[-2,2] A is given by f(X) = 3^3 then f is onto, if A is what? Math Questions Answered by Enrico Gregorio · Author has 18.4K answers and 16M answer views ·Dec 4, 2022 Because the function is continuous and lim x→−∞f(x)=−∞lim x→−∞f(x)=−∞ lim x→∞f(x)=∞lim x→∞f(x)=∞ The intermediate value theorem finishes the argument. Don’t you see why? Take r∈R.r∈R. Then, by the limits above, you can find x 1 x 1 and x 2 x 2 such that f(x 1)<r<f(x 2)f(x 1)<r<f(x 2) so you can invoke the IVT on the interval [x 1,x 2].[x 1,x 2]. The function is also one-to-one. Indeed, both f 1(x)=x 3 f 1(x)=x 3 and f 2(x)=x f 2(x)=x are strictly increasing functions, so also their sum f f is. We can even write down the inverse function. Indeed, if y∈R y∈R, then the only solution to f(x)=y f(x)=y is given by Cardano’s formula x=\sqrt[x=\sqrt[ Continue Reading Because the function is continuous and lim x→−∞f(x)=−∞lim x→−∞f(x)=−∞ lim x→∞f(x)=∞lim x→∞f(x)=∞ The intermediate value theorem finishes the argument. Don’t you see why? Take r∈R.r∈R. Then, by the limits above, you can find x 1 x 1 and x 2 x 2 such that f(x 1)<r<f(x 2)f(x 1)<r<f(x 2) so you can invoke the IVT on the interval [x 1,x 2].[x 1,x 2]. The function is also one-to-one. Indeed, both f 1(x)=x 3 f 1(x)=x 3 and f 2(x)=x f 2(x)=x are strictly increasing functions, so also their sum f f is. We can even write down the inverse function. Indeed, if y∈R y∈R, then the only solution to f(x)=y f(x)=y is given by Cardano’s formula x=3√y 2+√1 27+y 2 4+3√y 2−√1 27+y 2 4 x=y 2+1 27+y 2 4 3+y 2−1 27+y 2 4 3 Upvote · Sohel Zibara Studied at Doctor of Philosophy Degrees (Graduated 2000) · Upvoted by BowTangey , PhD Mathematics, Iowa State University (1988) · Author has 5.1K answers and 2.6M answer views ·2y Let a be an arbitrary real number. Thus a = x³ + x implies that x³ + x – a = 0 where the left hand side is a real polynomial of degree 3 and has therefore at least one real root. In other words, for every real number a, there exists at least one real number b such that f(b) = a which means that f is surjective. Upvote · 9 8 Narinder Kumar Assistant Professor ·2y From graph of x^3+x Every line parallel to x axis passes through the graph which implies x^3+x is onto. Continue Reading From graph of x^3+x Every line parallel to x axis passes through the graph which implies x^3+x is onto. Upvote · 9 1 Sponsored by StealthGPT All your schoolwork, one app. Trusted by millions of students, writing countless essays and answering any question. Learn More 99 83 Related questions More answers below How can we find all functions 𝑓:ℝ→ℝ which are continuous and satisfy 𝑓(𝑥+𝑦) =𝑓(𝑥) +𝑓(𝑦)? Consider the following relation on R:x∼y ℝ:𝑥∼𝑦 if and only if x 3−y 3∈Z 𝑥 3−𝑦 3∈ℤ. How is this is an equivalence relation? How do I compute the dualizing sheaf of $X \times Y$? What is the value of (.1 𝑥 .01 𝑥 .001) / (.2 𝑥 .02 𝑥 .002)? How can you tell if a function 𝑓:ℝ → ℝ is a bijection without using tables/graphs (or even a formula)? Mitch Cohen Former Retired (1978–2010) · Author has 2.8K answers and 873.8K answer views ·Updated 2y Set g(x) =DEF x^3 for all real numbers, x. Set h(x) =DEF x for all real numbers, x. Then define f by f =DEF (g + h) =DEF (g(x) + h(x)) for all real x. Clearly, g is an increasing bijection from R to R. Also, h is clearly an increasing bijection from R to R. Claim: The set of increasing bijections from R to R is closed under addition. Proof: Take any point y in R. If g, h are increasing bijections, then f = g + h is also an increasing injection, since x < y ===> f(x) = (g(x) + h(x)) < (g(y) + h(y)) = f(y), as g(x) < g(y) and h(x) < h(y). To show f is onto, f, as the sum of two continuous function Continue Reading Set g(x) =DEF x^3 for all real numbers, x. Set h(x) =DEF x for all real numbers, x. Then define f by f =DEF (g + h) =DEF (g(x) + h(x)) for all real x. Clearly, g is an increasing bijection from R to R. Also, h is clearly an increasing bijection from R to R. Claim: The set of increasing bijections from R to R is closed under addition. Proof: Take any point y in R. If g, h are increasing bijections, then f = g + h is also an increasing injection, since x < y ===> f(x) = (g(x) + h(x)) < (g(y) + h(y)) = f(y), as g(x) < g(y) and h(x) < h(y). To show f is onto, f, as the sum of two continuous functions, g and h, is itself continuous so has the Intermediate Value Property. So f = (g + h) takes on every the value of every real number between - Infinity and + Infinity. So, f = g + h as defined above is itself an increasing bijection from R to R, hence is onto, since a bijection is always a surjection, which means it is onto. QED. Upvote · 9 1 Sam Hill Math was my favorite subject in school · Author has 452 answers and 80.2K answer views ·2y Look at the limit as x approaches +infinity and -infinity. Since f is continuous, this will be sufficient. Another way, notice that f(x) is the sum of two functions each of which map (-infinity, infinity) to itself, and both are increasing—done. Note: g(x) = x + (-x)=0 satisfies the first part, but -x is not increasing. A third way, notice that f(x) is a polynomial of “odd” degree. Actually, you can you the first technique to show that “Every polynomial having real coefficients of odd degree maps R onto R”. Hope that helps. Upvote · Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 35 Bernd Leps Former scientific official; retired · Upvoted by BowTangey , PhD Mathematics, Iowa State University (1988) · Author has 5.8K answers and 1.3M answer views ·2y … or another answer: For big and bigger x, x³+x tends towards infinity. For negative x, f(x) is negative too and for x tending to minus infinity f(x) tends to minus infinity. That is the result of odd powers only in the function: The function is continuos and the results span f(x) from minus infinity to plus infinity. So all possible y = f(x) are represented by (at least) a point of the curve: f(x) is surjective. Upvote · 9 1 9 2 Michael Lamar PhD in Applied Mathematics · Upvoted by Tim Farage , Former Professor of Mathematics and Justin Rising , PhD in statistics · Author has 3.7K answers and 17.5M answer views ·Updated 1y Related Why does my book says that y=x 3 y=x 3 is not an onto function? To me, it is 100% onto. I doubt your book is trying to make this point, but I will tell you why y=x 3 y=x 3 isn’t an onto function. It isn’t a function at all. It’s a rule specifying how one might map some number x x to some number y y, but without a domain and a codomain being specified, it isn’t a function. Why does this matter? Let the domain of function y y be {0,1}{0,1} and the codomain be the set of real numbers with the mapping given by y(x)=x 3 y(x)=x 3. That’s a perfectly well defined function. It isn’t onto because 0 0 maps to 0 0 and 1 1 maps to 1 1 and those are the only two points in the image of the domain while the codomain includes al Continue Reading I doubt your book is trying to make this point, but I will tell you why y=x 3 y=x 3 isn’t an onto function. It isn’t a function at all. It’s a rule specifying how one might map some number x x to some number y y, but without a domain and a codomain being specified, it isn’t a function. Why does this matter? Let the domain of function y y be {0,1}{0,1} and the codomain be the set of real numbers with the mapping given by y(x)=x 3 y(x)=x 3. That’s a perfectly well defined function. It isn’t onto because 0 0 maps to 0 0 and 1 1 maps to 1 1 and those are the only two points in the image of the domain while the codomain includes all real numbers. Let the domain of function y y be the set of all real numbers and the codomain be the set of real numbers with the mapping given by y(x)=x 3 y(x)=x 3. That’s a perfectly well defined function. It is onto because the image of the domain is the set of all real numbers which is the codomain. Without completely specifying your function by including the domain, codomain, and mapping rule, there’s now way to understand its properties. Upvote · 999 131 99 30 Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder ·Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix. Don’t waste your time Continue Reading Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix. Don’t waste your time browsing insurance sites for a better deal. 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It is a cubic equation in x & x is teal , therefore, it must have areal root that is there is a real no. r such that r³ + r - y = 0 or y = f(r) . Upvote · Silvio Capobianco Senior Researcher at Tallinn University of Technology (2009–present) · Author has 3.4K answers and 634.3K answer views ·2y For every real number k k, the polynomial p(x)=x 3+x−k p(x)=x 3+x−k satisfies lim x→±∞p(x)=±∞lim x→±∞p(x)=±∞, so there exist a,b∈R a,b∈R such that p(x)<−1 p(x)<−1 for every x1 p(x)>1 for every x>b x>b. Now apply the intermediate value theorem. Upvote · Subra Studied Mathematics at Indian Institute of Technology, Bombay (IITB) (Graduated 1979) · Author has 3.5K answers and 658.2K answer views ·2y let y=x^3+x , x^3+x ∈ ℝ===> ∃y ∈ ℝ st y=x^3+x ∀ x ∈ ℝ, hence f(x)=x^3+x is onto function Upvote · 9 1 Michael Ph.D. in Mathematics, University of Manchester (Graduated 1982) · Upvoted by Max Gretinski , PhD Mathematics and David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 79 answers and 64.8K answer views ·4y Related Is F(x) = x\|x\| onto? I'll assume you mean F:R→R,F(x)=x\|x\|F:R→R,F(x)=x\|x\|. Then the answer is yes. First note that since \|x\|={x for x≥0−x for x<0\|x\|={x for x≥0−x for x<0, F(x)={x 2 for x≥0−x 2 for x<0 F(x)={x 2 for x≥0−x 2 for x<0 Now suppose y∈R y∈R. To show F F is onto, we must find x∈R x∈R such that F(x)=y F(x)=y. If y≥0 y≥0, then let x=√y≥0 x=y≥0, so F(x)=x 2=(√y)2=y F(x)=x 2=(y)2=y. If y<0 y<0, then let x=−√−y<0 x=−−y<0, so F(x)=−x 2=−(√−y)2=−(−y)=y F(x)=−x 2=−(−y)2=−(−y)=y. So for all y∈R y∈R we have found x∈R x∈R such that F(x)=y F(x)=y, which proves tha Continue Reading I'll assume you mean F:R→R,F(x)=x\|x\|F:R→R,F(x)=x\|x\|. Then the answer is yes. First note that since \|x\|={x for x≥0−x for x<0\|x\|={x for x≥0−x for x<0, F(x)={x 2 for x≥0−x 2 for x<0 F(x)={x 2 for x≥0−x 2 for x<0 Now suppose y∈R y∈R. To show F F is onto, we must find x∈R x∈R such that F(x)=y F(x)=y. If y≥0 y≥0, then let x=√y≥0 x=y≥0, so F(x)=x 2=(√y)2=y F(x)=x 2=(y)2=y. If y<0 y<0, then let x=−√−y<0 x=−−y<0, so F(x)=−x 2=−(√−y)2=−(−y)=y F(x)=−x 2=−(−y)2=−(−y)=y. So for all y∈R y∈R we have found x∈R x∈R such that F(x)=y F(x)=y, which proves that F F is onto. Here is the graph of the function F(x)=x\|x\|F(x)=x\|x\|: It is an odd function, with a graph at first sight similar to that of y=x 3 y=x 3. The function is continuous (product of the continuous functions \|x\|\|x\| and x x), and once differentiable. Indeed, F′(x)=2\|x\|F′(x)=2\|x\| (a straight forward exercise: consider separately the cases x<0 x<0, x>0 x>0 and x=0 x=0; in this last case use the definition of the derivative as a limit), so the first derivative exists but is not itself differentiable (not differentiable at x=0 x=0, differentiable elsewhere). On the other hand, the indefinite integral of \|x\|\|x\| is 1 2 F(x)+C=1 2 x\|x\|+C,C∈R 1 2 F(x)+C=1 2 x\|x\|+C,C∈R. Upvote · 99 10 9 1 Gaurav Kumar Jha Gold Medalist in Civil Engineering I AIR 560 - GATE'25 · Author has 389 answers and 1.2M answer views ·5y Related How do l prove f(x)=x 3+1∀x∈R f(x)=x 3+1∀x∈R is one to one and onto? Upvote · 9 2 Related questions How do I determine 𝑓: ℝ → ℝ,𝑤𝑖𝑡ℎ 𝑦 = 𝑓(𝑥) = 𝑥 √𝑥+1 is bijective? Is the function ℎ: ℝ \ {2020} → ℝ, ℎ (x) = (2021 − x) / (x − 2020) bijections function? Give an explanation complete How can we find a real function 𝑓:ℝ→ℝ such that 𝑓(𝑓(𝑥)) =−𝑥? If 𝑓: ℝ → ℝ be a continuous function satisfying 𝑓(𝑥) + 𝑓 (3 − 𝑥) = 4, then? What is the value of--- 0 to 3 ∫ 𝑓(𝑥) 𝑑x Explain: If f:[-2,2] A is given by f(X) = 3^3 then f is onto, if A is what? How can we find all functions 𝑓:ℝ→ℝ which are continuous and satisfy 𝑓(𝑥+𝑦) =𝑓(𝑥) +𝑓(𝑦)? Consider the following relation on R:x∼y ℝ:𝑥∼𝑦 if and only if x 3−y 3∈Z 𝑥 3−𝑦 3∈ℤ. How is this is an equivalence relation? How do I compute the dualizing sheaf of $X \times Y$? What is the value of (.1 𝑥 .01 𝑥 .001) / (.2 𝑥 .02 𝑥 .002)? How can you tell if a function 𝑓:ℝ → ℝ is a bijection without using tables/graphs (or even a formula)? What's the difference between ℂ and ℝ²? How do we prove that arccos(𝑥) +arccos(−𝑥) =𝜋 when 𝑥∈ [−1,1]? What's the range of the function 𝑓(𝑥) = 5 sin [2 (𝑥 + pi/3)] − 4? What does the function f: x ↦ y mean? Let 𝐴 = {1, 2, 3, 4, 5, 6, 7} be a set. 𝑓(𝑥) is a function A function that maps a set 𝐴 to a set 𝐴. How many 𝑓(𝑥) functions satisfies the equality 𝑓(𝑓(𝑥)) = 𝑥 for all 𝑥 ∈ 𝐴? Related questions How do I determine 𝑓: ℝ → ℝ,𝑤𝑖𝑡ℎ 𝑦 = 𝑓(𝑥) = 𝑥 √𝑥+1 is bijective? Is the function ℎ: ℝ \ {2020} → ℝ, ℎ (x) = (2021 − x) / (x − 2020) bijections function? Give an explanation complete How can we find a real function 𝑓:ℝ→ℝ such that 𝑓(𝑓(𝑥)) =−𝑥? If 𝑓: ℝ → ℝ be a continuous function satisfying 𝑓(𝑥) + 𝑓 (3 − 𝑥) = 4, then? What is the value of--- 0 to 3 ∫ 𝑓(𝑥) 𝑑x Explain: If f:[-2,2] A is given by f(X) = 3^3 then f is onto, if A is what? How can we find all functions 𝑓:ℝ→ℝ which are continuous and satisfy 𝑓(𝑥+𝑦) =𝑓(𝑥) +𝑓(𝑦)? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10762	https://studyfinance.com/ti-84-plus-tvm-tutorial/	TI-84 Plus TVM Tutorial Updated: April 12, 2022 When compared to other financial calculators used on finance courses, the TI-84 is fairly easy. If you want to learn how to use its financial functions in order to handle the time value of money problems and want to easily calculate anything financial, then make sure you keep on reading. If you’re looking for a quick recommendation for a finance calculator that will be suitable for most finance and accounting students, the calculator we recommend is the Hewlett Packard 10bII+. (click here to find out why) This calculator is usually around $25 on Amazon and is a good choice for business and finance needs to calculate loan repayments, interest rates, TVM, NPV, IRR, cash flows and more. This calculator is suitable for use in the CFP exam but not the CFA exam. If you need a calculator for CFA, go for the HP 12C or the TI BAII Plus instead. How to set up the TI 84 Plus Before you start using TI 84 Plus, there is just one thing that you need to adjust. Since the displays of this calculator only have two decimal places by default, it’s highly recommended that you set it up into having five decimal places. You can change the display by pressing the MODE key and then the down arrow key just once. You can then use the right arrow key so you can highlight the 5 and then press Enter. Lastly, exit the menu by pressing 2nd MODE. If you’ve done all of that successfully, congratulations! You can now start using your calculator. In this tutorial, we will be using a lot of the TVM Solver. The TI 84 Plus, however, offers more financial functions in the Finance menu. Be sure to check the FAQ section as well. Sample Calculations Let’s start by working on a very simple problem that will teach you some skills and will help you solve financial math on the TI-84: Example 1: Mark has $100 to invest for a period of 5 years with a 10% interest rate per year. Once this time period ends, how much money would Mark have accumulated? In this problem, here are the things we know: The present value (PV) is $100 N is 5 i is 10% Make sure that you put the calculator into the TVM Solver mode before you enter the data. Then, press the Apps button, choose the Finance menu (or you may also press the 1 key), and then choose the TVM Solver (or you may also press the 1 key). You can then start entering the data as shown in the table below: \| Field \| Entry \| --- \| \| N \| 5 \| \| I% \| 10 \| \| PV \| -100 \| \| PMT \| 0 \| \| FV \| 0 \| \| P/Y \| 1 \| \| C/Y \| 1 \| You then just have to scroll to the FV line and press Alpha Enter in order to find the future value. You should get 161.05 as an answer. A couple of things to keep in mind: There are either 4 or 5 variables in every time value of money. They correspond to the 5 basic financial keys. These variables will always give you 3 or 4 and will be asked to find a solution for the other. You have a 4-variable problem and were given 3 of them (N, I%, and PV) and were required to find a solution for the 4th (FV), in this case. You can solve these problems by simply entering the variables that you know on the correct lines and then scroll to the line for the variable you wish to find a solution with. You just have to press Alpha Enter so you can get the answer. Don’t forget to set any variables not in the problem to 0 because if you don’t, they will be included in the calculation. It doesn’t matter whether you entered the numbers in order or not. Ensure that the payments per year (P/Y) and compounding periods per year (C/Y) is set to 1. We prefer it this way. It’s not really necessary because these can be seen on the screen at all times. It will be much easier for you if you can remember to change these (1 for annual compounding, 12 for monthly compounding, etc) to the appropriate values for each problem. We placed 10 rather than 0.10 when we entered the interest rate. The reason for this is that the calculator automatically divides any entered number on the I% line by 100. The future would have come out to 100.501 had you entered 0.10 which is obviously not correct. Did you notice that we entered the 100 in PV as a negative number? We actually did that on purpose. Financial calculators mostly follow the Cash Flow Sign Convention. The reason for this is to keep the direction of the cash flow straight. Positive numbers being entered are cash inflows while negative numbers being entered are cash outflows. In the problem above, the investment was $100 and in five years, the future value of $161.05 would be a cash inflow. No harm would have been done even if you entered the $100 as a positive number but a negative number would have been the returned answer. Had you borrowed $100 today (cash inflow) and said yes to repaying $161.05 (cash outflow) in five years, this would have been correct. Changing the sign of a number using – or the “minus” key is wrong. Use the (-) instead. You don’t have to re-enter all of the data in order to change any of the variables in this problem. For instance, you want to know what the future value is going to be if you left the money invested for 10 years instead of 5. You only need to enter 10 on the N line and find the solution for FV. The answer that you will find is 259.37. Solving for the Present Value of Lump Sums The calculations for the present value of a lump sum and the future value is very similar. Don’t forget that the present value will always be less than the future value unless the interest is negative. Keeping this in mind will help you identify wrong answers because of the wrong input. Here’s another problem we can try: Jim is planning to send his daughter to college in 18 years. He assumed that he will need $100,000 at that time so he can pay for her tuition, room and board, party supplies, and other expenses. How much money will Jim need to invest today as a lump sum to achieve his goal of earning an average 8% annual rate of return per year? In this problem, here are the facts we already know: The future value is $100,000 The number of periods is 18 years The per-period rate is 8% per year What you need to find now is the present value. To do this, go to the TVM Solver and enter the data as follows: 18 into N 8 into I% 100,000 into FV Notice that the $100,000 is entered as a positive number because in 18 years, Jim will be withdrawing that amount. The next step is to move to PV and press ALPHA ENTER and you will see the amount of $25,024.90 which Jim will need to invest today in order to meet his goal. You might think it’s a lot of money to invest all at once but later on, you’ll learn to lessen the pain by investing each year in small amounts. How to Solve for the Number of Periods There are times that you’re fully aware of how much money you have now and how much you need to have at an undetermined future time period. You can solve the amount of time that it will take for the present value to grow to the future value if you know the interest rate and by solving for N. For example, Steph has $1,250 today and she’s interested in finding out how long it will take for her to double her money to $2,500. Her assumption is that she can earn 9% per year on her investment. Using the Rule of 72, you can quickly approximate this type of problem but if you want to easily find the exact answer, you can use the TI 84 Plus calculator. All you have to do is enter 9 into I%, -1250 into PV, and 2500 into FV. The next step is to scroll up to N and press ALPHA ENTER and you will see that in order for your money to double, you have to wait for 8.04 years. Another important thing to note is that entering your number must be based on the cash flow sign convention. You will get ERR: DOMAIN on the screen if you don’t make either the PV or FV a negative number (and the other one positive). The reason why this happens is that positive numbers makes the calculator think that you are getting a benefit without making any investment. In case this error occurs to you, just press 2 (Goto) to return to the TVM Solver and then change the sign of either PV or FV to fix the problem. How to Solve for the Interest Rate Solving the interest rates is especially helpful for people who recently sold an investment and would like to know what their compound average annual rate of return is. Calculating the interest rate is also helpful for people who want to make an investment and would like to know what rate of return they need in order to reach a certain future value. Let’s look back at the college savings problem we solved above but we’ll tweak it a bit. Let’s say Jim plans to send her daughter to college in 18 years and assumed that he already knows that he will need $100,000 at that time in order to pay for tuition, room and board, party supplies, and other expenses. If Jim is investing $20,000 today, what compound average annual rate of return will Jim need to earn to reach his goal? Like we previously did, we need to be cautious when entering the PV and FV into the calculator. In this case, here are the things we know: Jim is invest $20,000 today which is a cash outflow He will receive $100, 000 in 18 years which is cash inflow This means that we have to enter -20,000 into PV and 100,000 into FV. We need to type 18 into N and then solve for I% in order to get 9.35% average earnings per year. Again, if you didn’t follow the cash flow sign convention, you’ll get ERR: NO SIGN CHNG instead of an answer. You can then press 2 to return to the TVM Solver and have the problem fixed. Notice that in our original problem, our assumption was Jim would earn 8% per year, and found that Jim needed to invest about $25,000 to reach his goal. In this case, though, we assumed that Jim started with only $20,000. You would, therefore, need to earn a higher interest rate to reach the same goal. Make sure you give the answer a second look after you solved the problem and check if it’s accurate. You must understand the calculations made by the calculator and the relationships between the variables. If you don’t, you’ll quickly learn that entering the wrong numbers will give you wrong answers. PV and FV of Annuities We previously learned how to look at the basic time value of money keys and how they are used to calculate the present and future value of lump sums. In this section, we will learn how to use the TI 84 Plus to calculate the present and future values of regular annuities and annuities due. A regular annuity is an equal cash flow series happening at equally spaced time periods. The first cash flow happens at the end of the first period when it comes to a regular annuity. An annuity due and a regular annuity is similar except that the first cash flow occurs immediately. How to Find Present Value of Annuities For example, Mark decided to make an investment that will pay him $1,000 per year for 10 years. If Mark earns a rate of 9% per year on similar investments, how much would he be willing to pay for this annuity? What we need to do in this problem is to solve for the present value of this annuity since that is the amount Mark is willing to pay right now. You have to enter the following numbers to their appropriate lines: 10 into N 9 into I% 1000 (cash inflow) into PMT 0 for FV To solve the problem, move to the PV line and press Alpha Enter. You’ll find that the answer is -6,417.6577. As we previously mentioned, the result becomes negative because it represents the amount Jim would have to pay (cash outflow) today so he can buy this annuity. How to Solve for Future Value of Annuities For example, Jane borrows $1000 each year for 10 years on a 9% rate and pays back the loan immediately after she receives the last payment. How much will Jane have to repay? The solution is simple. We just have to put a 0 into PV to clear it out and then solve for FV to get -15,192.92972 (a cash outflow). How to Solve for the Payment Amount Solving for annuity payment is oftentimes necessary. If you want to know how much a mortgage or auto loan payment will be or you if you want to know how much money you need to save yearly to reach a particular goal, solving for annuity payment is the only way. We previously used saving for college as an example so let’s use that again, only this time, we’ll look at it as an annuity problem instead of a lump sum. In 18 years, Jim is planning to send his daughter to college. He already assumed that he needed $100,000 at that time in order to pay for all the college expenses. If Jim thinks he can earn an average annual rate of return of 8% yearly, how much money would Jim need to invest at the end of each year to achieve his goal? As you may recall, Jim would have to invest $25, 024.90 if he were to make a lump sum investment today. That change is quite huge. Saving for college, in this case, will be fairly easy because the investment is spread over 18 years rather than all at once. The data that we will enter is the following: 18 into N 8 into I% 100,000 into FV Next, we will solve for PMT which will then lead us to a result of $2,670.21 investment per year for the next 18 years in order for Jim to meet his goal of having $100,000. How to Solve for the Number of Periods When you solve for N, you will know the answer to the question, “How long will it take..” Here’s an example: John has just retired. He has a nest egg of $1,000,000. For the rest of John’s life, he will be drawing down money from that nest egg. If John expects he will earn 6% per year on average and withdraw $70,000 per year, how long will John be able to afford to live? The assumption here is that John’s first withdrawal will occur one year from today. The data that you have to enter are as follows: 6 into I% -1,000,000 into PV (you are investing this amount so it’s negative) 70,000 into PMT The next step is to solve for N which will then result in 33.40 withdrawals. John can, therefore, afford to live for about another 34.40 years assuming that he can live for about a year on the last withdrawal. How to Solve for the Interest Rate Finding the I% is similar to solving any other variables. This tutorial mentioned several times that paying attention to the signs of the numbers that you enter into the TVM keys is important. You will get a potentially wrong answer or an error message if you don’t get the signs right when solving for N, I%, or PMT. For solving the interest rate, here’s an example we can take a look at: Suppose that Helen is offered an investment that will cost $925 and for the next 20 years, will pay her an interest of $80 per year. Additionally, the investment will pay $1000 at the end of the 20 years. What will be Helen’s compound average annual rate of return if she purchased this investment? In this problem, here are the things we already know: Present value is $925 Future value is $1,000 Annuity payment is $80 per year You need to be especially careful to get the signs right as mentioned above. In this case, your cash inflows are both the annuity payment and the future value which means they should be entered as positive numbers. The cost of the investment and the cash outflow is the present value which means it should be entered as a negative number. Note that you will get the wrong answer if you made a mistake by entering the payment as a negative number. Now if you were to enter all three with the same sign, an error message will appear. Here are the numbers we need to enter: 20 into N -925 into PV 80 into PMT 1000 into FV If you solve for I%, it should give you 8.81% per year as the average return of investment. How to Solve for Annuities Due In the previous examples, our assumption was that the first payment would be made at the end of the year which is usually expected. But what if you plan to make or receive the first payment today? The cash will, therefore, change from a regular annuity into an annuity due. The calculator is normally working in End Mode. The cash flows are assumed to occur at the end of the period. However, in this case, the payments occur at the beginning of the period. So the calculator must be put into Begin Mode. You can scroll down to the bottom of the TVM Solver if you want to change to Begin Mode. You will notice that END is currently highlighted. Highlight BEGIN by pressing the right arrow key and then press ENTER. You’ll see that nothing will change about how you enter the numbers. The cash flows will simply be shifted for you by the calculator. It’s obvious that you’ll get a different answer. Let’s have a look at the college savings problem but this time, let’s assume that Jim start’s investing immediately: Jim plans to send his daughter to college in 18 years and he assumes that he will need $100,00 in order to pay for the tuition, room and board, party supplies, and other college expenses. How much would Jim need to invest at the beginning of the year starting today if his goal is to earn an average annual rate of return of 8% per year? Just like what we previously did, let’s enter the following data: 18 into N 8 into I% 100,000 into FV Since we are now treating this as an annuity due, just solve for PMT once you changed to Begin Mode. The result should give you $2,472.42 as the amount you need to invest if you start investing today. Because of the extra time for your investments to compound, you get about $200 per year less than if you make the first payment a year from now. After you solve the problem, make sure you switch back to End Mode. Always practice switching back since you may always find yourself needing to be in End Mode. Next, you have to scroll down to the bottom of the TVM Solver, highlight END and press Enter. How to Solve for Perpetuities We occasionally have to deal with annuities that pay forever (theoretically) instead of for a finite period of time. This type of cash flow is called perpetuity, sometimes perpetual annuity or infinite annuity. TI 84 Plus, however, has no way to specify an infinite number of periods for N, and that could become a problem. It’s easy to calculate the present value of perpetuity using a formula. We do it by dividing the payment per period by the interest rate per period. The payment is $1,000 per year in our example while the interest rate is 9% annually. If that was perpetuity, the present value would therefore be: $11,111.11 = 1,000 ÷ 0.09 Did you know that you can “trick” the calculator into getting the correct answer if you happen to forget the formula? Cool, right? You can do this because the present value of a cash flow far enough into the future is going to be approximately $0. This means that the cash flows no longer add anything to the present value beyond some future point in time. We can, therefore, get a very close approximation if we have a specific suitably large number of payments to perpetuity. Let’s apply this with our perpetuity. Just enter the following data: 500 into N 9 into I% 1000 into PMT If you scroll to PV and press Alpha Enter, the answer you will get is $11,111.11. Since the cash flows never end (period infinity never arrives), there will be no future value of a perpetuity. The third part of this tutorial will involve uneven cash flow streams, net present value, internal rate of return, and modified internal rate of return. Calculating NPV, IRR and MIRR So now that we’re done with the basic time value of money keys and have learned how to use them to calculate present and future value of lump sums and regular annuities, it’s time to learn how to calculate the present and future values of uneven cash flow streams using the TI 84 Plus. In this page, we will also learn how to calculate the following: Net present value (NPV) Internal rate of return (IRR) Modified internal rate of return (MIRR) How to Solve Present Value of Uneven Cash Flows You might find it more difficult to use the TI-84 Plus here than most other financial calculators. However, once you get used to it, it will no longer be as hard. Let’s get started by exiting from the TVM Solver menu and pressing 2nd MODE and then pressing APPS, then return to the finance menu. We need to use the NPV function if we want to find the present value of an uneven stream of cash flows. The definition of this function is: NPV( Rate, Initial Outlay, {Cash Flows}, {Cash Flow Counts}) Remember that the {Cash Flow Counts} part is optional and we won’t deal with it here. If for example, Malcolm is offered investment and at the end of the next five years, it will give him the following cash flows: \| Period \| Cash Flow \| --- \| \| 0 \| 0 \| \| 1 \| 100 \| \| 2 \| 200 \| \| 3 \| 300 \| \| 4 \| 400 \| \| 5 \| 500 \| If Malcolm’s assumed rate of return is 12% per year, how much would he be willing to pay for this investment? To solve this problem, we have to find the present value of each of these cash flows. We will do it individually and then we will sum up the results. But this route can be hard. What we can do instead is to use the NPV function. Let’s start by scrolling down in the finance menu until we get the line that reads NPV(. To select that function, just press Enter and the beginning of the NPV function will appear on the screen. We will then complete the function like this: NPV(12,0,{100,200,300,400,500}) As soon as we press ENTER to solve the function, the present value should be $1,000.17922. Remember that if you press the 2nd Enter, we can easily change the interest rate because we retrieved the function, and then we can edit it by using the arrow keys. Here’s an example: Change the rate to 10% by pressing 2nd Enter and then use the arrow keys to transfer to the interest rate and press DEL to delete the 12. You will then press 2nd DEL which is the INS function and then enter 10. You will find that the answer is now $1,065.25883 once you press Enter. Before you continue, make sure you reset the interest rate to 12. How to Solve Future Value of Uneven Cash Flows In order to find the future value of these cash flows instead of the present value, we will need to use a bit of genius approach since there is a function to do this. Note that one way to find the future value of any set of cash flows is to first solve for the present value. The next step is to find the future value of that present value and that’s how you can get your solution. In this case, we already know that the present value is $1,000.17922. We will then recall the NPV function by pressing 2nd Enter (pressing this twice might be necessary to get back to the original 12% interest rate). Next, let’s add 1.12 ^ 5 to the end of the function. The end result will be: NPV(12,0,{100,200,300,400,500})1.12^5 Once you press Enter, $1,762.65754 will appear as the future value of these cash flows. Well, that was easy! At least easier than adding up the future values of each of the individual cash flows. It will, however, require you to know the equation for the future value of a lump sum but that’s something you already know anyway. Solving for Net Present Value (NPV) When you calculate the net present value (NPV) and/or internal rate of return (IRR), you will find that it’s kind of similar to solving for the present value of an uneven cash flow stream as we have done in Example 3. If you were offered the investment in Example 3 at a cost of $800, what would your NPV and IRR be? To solve this problem, here are some things to keep in mind: The calculator must not only have the annual cash flows but the cost as well. Remember that we set the cost to 0 since we only needed the present value of the cash flows. In general, an investment must be paid first before you can receive its benefits so the cost or the initial outlay is said to occur at a time period (today for example). We will find the NPV by recalling the NPV function and editing it so that the initial outlay becomes -800. This is how the function is going to look like on the screen: NPV(12,-800,{100,200,300,400,500}) You will get the solution by pressing Enter. The answer will then be $200.17922 as the NPV. How to Solve Internal Rate of Return Finding the IRR is done the same way as above except that we’ll use the IRR function. The definition of this function is: IRR(Initial Outlay, {Cash Flows}, {Cash Flow Counts}) For this problem, the function is: IRR(-800, {100,200,300,400,500}) Again, we will ignore the {Cash Flow Counts} because it’s optional. We have to press APPS, return to the finance menu, and scroll down until we see the IRR(. We will enter the function as shown above and then press Enter to get (19.5382%) as the answer. How to Solve the Modified Internal Rate of Return Since it’s simple and can be interpreted, the IRR has been a popular metric for evaluating investments for many years. However, there are some serious problems with the IRR. One of its problems is that it completely assumes that the cash flows will be invested again for the life of the project at a rate that equals the IRR. A project that is considered good may have an IRR that is significantly greater compared to any reasonable reinvestment assumption. This is why at times, the IRR can be misleading. In order to solve this problem, the modified internal rate of return (MIRR) may use a straightforward reinvestment rate. The issue here is that financial calculators don’t have a MIRR key as they have in the IRR key. So we will have to use a smart approach to calculate the MIRR. Good thing it’s not difficult. We will use the following steps in the algorithm: We have to calculate each of the cash flows’ total present value starting from period 1. Don’t forget to set the initial outlay to 0. Just like we did in Example above, we have to use the calculator’s NPV function. We will find the present value by using the reinvestment rate as your discount rate. Next, we have to calculate the future value as of the end of the project life of the present value from step 1. We will use the reinvestment rate as our interest rate to find the future value. Lastly, we will look for the discount rate equal to the initial cost of the investment with the future value of the cash flows. MIRR will be the discount rate which can be interpreted as the compound average annual rate of return earned on investment if you invest the cash flows again at the reinvestment rate. Now, let’s say you were offered the investment in Example 3 at a cost of $800 and in the reinvestment rate is 10% per year, what will your MIRR be? Here’s a step-by-step procedure when going through our algorithm: Example 3 shows the present value of the cash flows. NPV(10,0,{100,200,300,400,500}) We will discover that the present value is $1,065.26. In order to find the future value of the cash flows, here’s what you need to do: Go to the TVM Solver and enter 5 into N 10 into I% -1065.26 into PV When you solve for the FV, you will get $1,715.61 as the answer. The problem has now been transformed into an $800 investment with a lump sum cash flow of $1,715.61 in period 5. The discount rate (I%) that equates these two numbers is the MIRR. If you enter -800 into PV and then solve for I%, you’ll get 16.48% per year as the MIRR. Remember that we can combine steps 1 and 2. We can calculate the future value using our 10% reinvestment rate as we did in Example 3 like this: NPV(10,0,{100,200,300,400,500})1.10^5 Similar to what we found above, $1,715.61 is our future value. Let’s go to the TVM Solver this time and enter 5 into N, -800 into PV, and into FV enter 1715.61. When you solve the I%, you will get 16.48% as the MIRR just like before. So now, we found out that at a cost of $800, our project is acceptable. We found that the NPV is positive, the required return is 12% which is less than the IRR, and the required return is 12% which is lesser than the MIRR. Solve Non-Annual Periods on TI-84 Plus Annual time periods is probably one of the most used calculations in the real world. For instance, most consumer loans need monthly payments such as mortgages, car loans, and credit cards. In fact, we have used annual time periods in all the previous examples for simplicity. Now, it’s time to dive through the non-annual problems. Considerations You must understand all the annual problem principles you have learned still apply for non-annual problems. Nothing has changed at all, actually. You’ll understand it better if you think in terms of “periods” rather than years. Any amount of time can be a period. Daily, monthly, quarterly, semiannually, or annually are the most common. A time period, however, can be any imaginable amount of time. You must first consider the number of periods in a year when dealing with non-annual periods. This is because consistency in entering data into the 84 Plus is important. Whatever numbers you enter into the N, I% and PMT keys must agree as to the length of the time periods you used. Therefore, if you are dealing with a monthly problem, then the total number of months should be your N, the monthly interest rate should be you I%, and the monthly annuity payment should be your PMT. Example It’s very common for you to face a problem that involves annual numbers but then told that “payments are made on a monthly basis,” or that “interest is compounded daily.” You must, therefore, adjust the numbers given if you ever face this kind of problem. Let’s have a look at this example: Leo is thinking about buying a new home for $250,000. Leo was told by his banker that they are willing to offer him a 30-year fixed-rate loan at 7% with monthly payments. What will Leo’s required monthly payment be if he borrows the entire $250,000? From that problem, here are the things we know: The loan term is 30 years Interest rate is 7% per year (merely implied, not stated explicitly) In this case, you might be forgiven for expecting that a period is one year but you will, later on, learn that the payments must be made every month. This means the length of a period is one, and in order to get the correct answer, you must convert the variables to a monthly basis. We calculate the total number of periods by multiplying 30 years by 12 months per year (since there are 12 months in a year). This means that N is 360 months and not 30 years. Similarly, when you divide the 7% annual rate by 12, you get 0.5833% per month as the interest rate. Since it occurs at a single point in time, not repeatedly, we will not make any adjustments to the PV ($250,000). If there was an FV in this problem the same logic would apply. The calculator will give you the monthly payment amount automatically (per period to be exact) when you solve for the payment. In this problem, we have to enter 360 in N, 0.5833 into I%, and 250,000 into PV to solve for the payment amount. The monthly payment will be $1,663.26 when you solve for PMT. Rounding is another thing you need to be cautious of. For instance, you should do the calculation in the I% line in the TVM Solver when calculating the monthly interest rate. It’s wrong to do the calculation and then write down the answer for a later entry. Doing this will shorten the interest rate to the number of decimal places that appear on the screen and your answer will be impacted by the rounding. It doesn’t matter whether the difference only involves a few pennies, every penny matters. If you really want to see what happens, try sending your lender a payment that is consistently three cents less than required. We bet you’ll get a nasty letter the next day. Things to Remember If you think it’s okay to treat the problem as an annual one and then, later on, adjust your answer to be monthly, then you’re wrong. Mathematics doesn’t work that way. If you want proof, try to input annual numbers, and then convert the annual payment to monthly by dividing by 12. Next, enter 30 into N, 7 into I%, and 250,000 into PV. You will find that when you solve for PMT, the answer will be $20,146.60. However, monthly payments must be made so if we divide that by 12, a monthly payment of $1,678.88 appears. As you can see, if the problem is solved this way, you’ll get a $15.63 answer every month which is too high. Therefore, you must always adjust your variables before solving the problem. The compounding of interest is the reason why there’s a difference. The more frequently interest is compounded, the smaller the payment needs to be for it to grow to a specific future value. TI-84 Plus Payments per Year Setting You probably noticed that by using the P/Y setting at the bottom of the TVM Solver, the TI 84 Plus can semi-automatically adjust for payment frequency. For situations where the compounding frequency is different from the payment frequency, it can also be adjusted by using C/Y. This feature, however, causes more problems than it solves. It’s because when you do the next problem, you will more likely forget to change it and will probably get the wrong answer. This setting can give you problems that cannot be easily identified. Let’s discuss how to use this setting anyway. If for example, you set the P/Y value or “payments per year” to 12, the calculator will assume monthly compounding and appropriately adjust the interest rate. However, the number of periods or the payment amount will not be adjusted by it and take note, this is very important. This is what makes the feature worthless. C/Y, on the other hand, means “compounding periods per year” and is normally similar to P/Y. If you change P/Y you will find that C/Y will also change to the same value. If the compounding frequency differs from the payment frequency, C/Y can be changed. For instance, the interest rate of your quarterly payments is compounded monthly, you would then set P/Y to 4 and C/Y to 12. Now, let’s do the problem again. This time by using this feature: First, set P/Y to 12 (monthly) Enter the data: 360 into N (again, you still have to enter the total number of periods), 7 into I%, and 250,000 into PV. Next, let’s solve for the payment by pressing Alpha ENTER. The answer will be $1,663.26 as the monthly payment. As you can see, we got the correct answer but we didn’t save anything at all by using that “shortcut.” What actually happened was that it took an extra keystroke or two to use this feature. Additionally, you will get the wrong answer if you forget to change the setting when you do the next problem. Unless of course the problem also happens to use monthly compounding. We highly recommend that you follow the simple steps outlined above. Just set P/Y to 1 and you don’t have to think about it anymore. Always make the total number of periods as the N, the interest rate per period as the I%, and the payment per period as the PMT. Frequently Asked Questions Can you explain the optional [cash flow counts] part when using the NPV and IRR functions? The purpose of that feature is to save some data entry in specific situations. It’s also an identical purpose to the frequency prompt in several other calculators. You can include the “cash flow counts” array if you find that some of your cash flows repeat. For instance: If the discount rate is 8% per year, what will the present value of the following series of cash flows be? 100, 200, 200, 200, 300. This problem is usually solved through the NPV function as follows: NPV(8,0,{100,200,200,200,300}) However, as you can see, you will be required to type “200” three times in a row. We can do it like this when you use the “cash flow counts” array: NPV(8,0,{100,200,300},{1,3,1}) Either way, you’ll get $774.01 as the present value. But you won’t have to put in too much effort typing and the chance of getting data entry errors is lesser. In this example, it doesn’t really save much time but in terms of other problems, it can. Why does my TI-84 Plus manual say we have to use lists in the NPV and IRR functions, yet you never use them? It’s because we want to create on the fly arrays the quicker and easier way instead of using lists. Also, we don’t recommend you to reuse a list more than once or twice. If you do end up reusing it, just use the editing functions to get it done quickly. I’m getting an error when I use the NPV and/or IRR function. What’s happening? There are different reasons as to why this is happening. Typically, it’s because someone is not being careful in doing variable separation in the function by commas. It is absolutely necessary to use every comma you see. Another reason why this happens is that you neglected to make the initial outlay a negative number. When using the TVM Solver or the functions, you must obey the cash flow sign convention. Is it possible to calculate the number of days between two dates using the TI 84 Plus? Yes. There are two date formats supported by TI-84 Plus. They are MM.DDYY or DDMM.YY. Use the dbd (days between dates) function from the Finance menu when calculating the difference between two days. For instance, get the number of days between 25 July 2007 and 31 December 2007 by typing: dbd(08.2507,12.3107) if you are using MM.DDYY or dbd(2508.07,3112.07) if you are using DDMM.YY Whichever way you do it, the answer you will get is 128 days between those two dates. Remember that you cannot specify a day count basis when using the TI 84 Plus. The basis will always be actual/actual. Is there any other way to find roots other than using square roots of the TI-84 Plus? It may seem like calculating various roots is all we do in finance. The good news is that it’s actually very simple to do. You only need to remember simple mathematical rules. In order to calculate the 5th root of 100, just raise 100 to the 1/5th power. You can do this in the TI-84 Plus by typing: 100^(1/5) ENTER. The 5th root of 100 is equal to 2.51189 in this example. Keep in mind that the parentheses are necessary otherwise you would raise 100 to the 1st power, divide by 5, and the answer you will get is 20. You can calculate any root using this strategy. I only have a key to calculate natural logarithms in my calculator. How can I calculate logarithms to other bases? We would usually use natural logarithms (base e) in finance which is typically abbreviated as Ln(x). But there are times we need to use other bases. You can covert base e to any other base by doing the following formula (say we are converting to base 10): Thus, you can compute the natural log of your number and then divide it by the new base’s natural log. For instance, Log10(3) = Ln(3)/Ln(10) = 0.478. If you want to calculate base 10 logarithms, you can do so by using the LOG key of TI-84 Plus. What should I do to clear the memory of all variables and lists? It’s actually easier and safer to do the following compared to a complete reset: Press 2nd MEM (that is the second function of the + key) Select 2 (Mem Mgmt/Delete) Select 1 (All) Scroll through the list and delete anything that isn’t important using the DEL button. To exit, press 2nd QUIT. Just remember to be careful not to delete any lists or variables that you need to save. I want to completely reset my TI 84 Plus back to the factory default settings. How can I do that? Be careful when resetting because it will remove any programs that you may store in the calculator. This is how you can reset it: Press 2nd MEM (that is the second function of the + key) Choose 7 (Reset) Scroll right so that ALL is selected Press 1 Press 2 (Reset, and read the warnings) Keep it in mind that resetting may impact the contrast of your screen. If you really have to reset, then you can change the contrast of the TI 84 Plus screen by pressing the 2nd and then the up arrow or down arrow button. There are times you will be required to do this repeatedly. Link To or Reference This Page If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support! TI-84 Plus TVM Tutorial "TI-84 Plus TVM Tutorial". StudyFinance.com. Accessed on September 28, 2025. "TI-84 Plus TVM Tutorial". StudyFinance.com, Accessed 28 September, 2025 TI-84 Plus TVM Tutorial. 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10763	https://www2.math.upenn.edu/~kazdan/508F14/Notes/compactness.pdf	Math 508 Fall 2014 Jerry Kazdan Compactness In these notes we will assume all sets are in a metric space X. These proofs are merely a rephrasing of this in Rudin – but perhaps the diﬀerences in wording will help. Intuitive remark: a set is compact if it can be guarded by a ﬁnite number of arbitrarily nearsighted policemen. Theorem A compact set K is bounded. Proof Pick any point p ∈K and let Bn(p) = {x ∈K : d(x, p) < n}, n = 1, 2, . . .. These open balls cover K. By compactness, a ﬁnite number also cover K. The largest of these is a ball that contains K. Theorem 2.34 A compact set K is closed. Proof We show that the complement Kc = X −K is open. Pick a point p ̸∈K. If q ∈K, let Vq and Wq be open balls around p and q of radius 1 2d(p, q). Observe that if x ∈Wq then d(q, p) ≤d(q, x) + d(x, p) < 1 2d(p, q) + d(x, p) so d(x, p) > 1 2d(p, q), that is, all the points in this ball are at least 1 2d(p, q) from p . By compactness, a ﬁnite number of these balls, Wq1, . . . WqN cover K. Look at the corre-sponding balls Vq1, . . . VqN . They are all centered at p. The smallest (their intersection) is a neighborhood of p that contains no points of K. Theorem 2.35 Closed subsets of compact sets are compact. Proof Say F ⊂K ⊂X where F is closed and K is compact. Let {Vα} be an open cover of F. Then F c is a trivial open cover of F c. Consequently {F c} ∪{Vα} is an open cover of K. By compactness of K it has a ﬁnite sub-cover – which gives us a ﬁnite sub-cover of F. Theorem 2.38 Let In be a sequence of nested closed intervals in R, so In ⊇In+1, n = 1, 2, . . .. Then ∩∞ n=1In is not empty. Proof Say In = {x ∈R : an ≤x ≤bn}. The nested property means a1 ≤an ≤an+1 ≤bn+1 ≤bn ≤b1. Let a = sup an and b = inf bn. It is clear that ∩∞ n=1In = {a ≤x ≤b}. It is clear that this immediately extends to closed cells (“rectangles”) in R2 and Rk. We use it to show Theorem 2.40 Closed and bounded intervals x ∈R : {a ≤x ≤b} are compact. Proof Idea: keep on dividing a ≤x ≤b in half and use a microscope. Say there is an open cover {Gα} that has no ﬁnite sub-cover. Divide the interval in half. Then one (or both) halves are closed sets with an open cover that has no ﬁnite cover. Keep on repeating this. At the nth step we have a closed interval In of length (b −a)/2n where there is no ﬁnite sub-cover of our {Gα}. 1 By the previous theorem, the intersection of these (nested) intervals ∩∞ n=1In has at point p. Since p is contained in at least one of the {Gα} so there is some interval around p. This shows that for n large In is covered by one of the sets Gα. Contradiction. Theorem 2.37 In any metric space, an inﬁnite subset E of a compact set K has a limit point in K. [Bolzano-Weierstrass] Proof Say no point of K is a limit point of E. Then each point of K would have a neighborhood containing at most one point q of E. A ﬁnite number of these neighborhoods cover K – so the set E must be ﬁnite. Theorem 2.41 Let {E ∈Rk}. The following properties are equivalent: (a) E is closed and bounded. (b) E is compact. (c)Every inﬁnite subset of E has a limit point in E. [Bolzano-Weierstrass Property] Proof We do this for sets E ∈R1. The ore general case is then straightforward. (a) implies (b): Since E is bounded it is contained in some closed interval I. This interval is compact (Theorem 2.40). But then E is a closed subset of a compact set so it is compact (Theorem 2.35). (b) implies (c): Theorem 2.37. (c) implies (a). If E is not bounded, then for each n = 1, 2, . . . there is a point xn ∈E with \|xn\| > n. This inﬁnite set has no limit point, a contradiction. If E ⊂R is not closed then there is a point p ∈R which is a limit point of E but not in E. Thus, for each n = 1, 2, 3, . . . there is a point xn ∈E with \|xn −p\| < 1/n. This set S = {x1, x2, . . .} has p ̸∈E as its only limit point. Contradiction. Example Let K be a compact set in a metric space X and let p ∈X but p ̸∈K. Then there is a point x0 in K that is closest to p. In other words, let α = infx∈K d(x, p). then there is at least one point x0 ∈K with d(x0, p) = α, Remark: There may be many such points, for example if K is the unit circle x2 + y2 = 1 in the plane and p = (0, 0), then every point on the circle minimizes the distance to the origin. Solution: For any n = 1, 2, . . . there is at least one point xn ∈K with d(xn, p) ≤α + 1 n. If this set {x1, x2, . . .} is ﬁnite (for instance if K only has a ﬁnite number of points), pick the point closest to p. If the set has inﬁnite many points, by the Bolzano-Weierstrass property it has a limit point q in K. This is the desired point in K that is closest to p. Example In ℓ2 the set of unit vectors e1 = (1, 0, 0, . . .), e2 = (0, 1, 0, 0 . . .), . . . is closed and bounded but not compact. 2
10764	https://physics.umd.edu/courses/Phys375/GammonSpring10/lab3.pdf	Experiment 3 – The Polarization of Light 1 Experiment 3 The Polarization of Light 1 Introduction In this experiment, we will study various polarizations of light. These will include linearly polarized, circularly polarized, elliptically polar-ized and unpolarized light. The polarization of light will be described quantitatively by the degree of polarization. We will also study the re-ﬂection of linearly polarized light from a plane surface. In these studies we will use a photodetector and a computer to acquire the data. 2 Background - see Hecht, Chap. 8 2.1 Types of Polarization The electric ﬁeld of a propagating transverse electromagnetic wave lies in a plane normal to the direction of propagation. Consider a wave propagating along ˆ z. For most of this discussion, we shall consider a wave at a particular instant in time and examine the orientation of the electric ﬁeld vector in our x −y coordinate system. 2.1.1 Polarized Light A general expression for the E-ﬁeld is: E = ˆ xE0x cos(kz −ωt) + ˆ yE0y cos(kz −ωt + φ). (1) where φ is a phase shift between the ˆ x and ˆ y directions. If φ = 0, the electric ﬁeld oscillates along a line. This is linearly polarized light. The angle of the polarization is determined by the relative magnitudes of E0x and E0y. If φ = ±π/2, and E0x = E0y the electric ﬁeld vector traces out a circle in the x −y plane. This describes circularly polarized light. Experiment 3 – The Polarization of Light 2 If φ ̸= 0, ±π/2 and/or E0x ̸= E0y, the light is elliptically polarized. The electric ﬁeld traces out an ellipse, that in general can have any major to minor axis ratio and an arbitrary angle. 2.1.2 ”Unpolarized” Light If light is monochromatic, it is polarized. It must be some form of elliptically polarized light (where we consider linear and circular to be limiting cases.) For light to be ”unpolarized” requires a random phase variation between the x and y components of the ﬁeld, and some time averaging. All light is always fully polarized at any instant in time. Unpolarized light is only possible if there are multiple frequencies present. Our HeNe lasers may or may not be fully polarized. They tend to lase in a few ”modes” with orthogonal polarizations, producing light separated in frequency by ∼1 GHz (note this is only a part in 105 of the optical frequency, so it still looks monochromatic). We will learn more about lasers later in the course, but at the moment, what you need to know is that over time (minutes to hours) the amount of light in the two modes can drift. If it is all in one mode, the laser will be linearly polarized. If it lases in both modes with equal intensity, it will appear unpolarized (because any measurement we make will average over many cycles of the diﬀerence frequency). This drift will be an impediment to your measurements, and some of you will be lucky and will have much more stable lasers than others. To do experiments we will put one polarizer in front of the laser to make it linearly polarized, but how much light is transmitted through this polarizer will depend on the state of the laser (and may change in time!). 2.2 Malus’s Law Consider polarized light of intensity I(0) incident on an ideal linear polarizer. Let the transmission axis of this polarizer be at an angle of θ with respect to the initial polarization. Only light with it’s electric ﬁeld vector parallel to the axis of the polarizer will be transmitted through the polarizer. The intensity of the light emerging from the second polarizer is given by Malus’ Law: I(θ) = I(0) cos2(θ). (2) To go one step further, consider the situation where light from the ﬁrst polarizer is incident on a second polarizer whose transmission axis is at an angle of ψ with respect to the initial polarization. The intensity of light emerging from the second polarizer is given by I(ψ) = I(0) cos2(θ) cos2(θ −ψ). (3) Experiment 3 – The Polarization of Light 3 2.3 Degree of Polarization The degree of polarization is basically a measure of the extent to which the light we are studying is polarized and is given by: V = Imax −Imin Imax + Imin , (4) where Imin and Imax are the minimum and maximum intensities of the light, respectively. For partially polarized light, this can also be written as: V = Ip Ip + Iup , (5) where Ip is the intensity of the polarized portion and Iup is the intensity of the unpolarized portion. Sometimes V is called the fringe visibility. 2.4 Reﬂection of Linearly Polarized Light When linearly polarized light is incident on a plane surface, the amount of light reﬂected, the reﬂectance (R), depends on the angle of incidence and the orientation of the polarization vector relative to the scattering plane. This is described by the Fresnel equations: R∥= tan (θi −θt) tan (θi + θt) 2 (6) R⊥= sin (θi −θt) sin (θi + θt) 2 . (7) One consequence of Eq. 6 is that when θi +θt = π/2, tan (θi + θt) → ∞and R∥→0. This occurs when θi = θB where θB is called Brewster’s angle and θt = β. Substituting θB and β into Eq. 6, subject to θB+β = π 2 leads to tan θB = nt n0 . (8) At θB only light with its E-ﬁeld perpendicular to the scattering plane will be reﬂected. Consequently, it is possible to produce polarized light from unpolarized light by allowing it strike an interface at Brewster’s angle. 3 Experiment You are supplied with 3 polarizers, one on a motorized rotation stage. You will have to deal with the laser drifts, so you will need to make stability measurements to disentangle eﬀects due to the laser from your Experiment 3 – The Polarization of Light 4 measurements. Using the data acquisition system, measure the inten-sity stability of your laser, and then the time-dependent polarization state of your laser. Measure Brewster’s angle for the block of glass supplied. The polar-izers’ axes are not marked, so you will need to ﬁgure out how to assure that the polarization of the beam striking the block is parallel to the plane of incidence. Verify Malus’ Law for one and two polarizers. A demonstration that mystiﬁes many observers is that placing a polarizer in between two crossed polarizers allows light to be transmitted that was previously blocked. Set up this system and ﬁgure out why this is so. Go in the hallway and look at the light reﬂected oﬀthe ﬂoor. Is it polarized, and if so, along what direction? If you have or can borrow polarizing sunglasses, measure their po-larization properties. Multiple scattering of light can aﬀect the polarization state. Waxed paper makes a good multiple scatterer. Measure the polarization of the light after passing through a piece of waxed paper. An optically active medium has a diﬀerent index of refraction for right- and left-circularly polarized light, which has the eﬀect of rotating linearly polarized light. Sugar molecules are chiral - they have a left-handed twist, so a sugar solution is optically active. Measure the eﬀect of transmitting polarized light through a cell containing corn syrup. We only have two cells, so they need to be shared.
10765	https://helpingwithcalculus.com/Derivatives/X-Axis-Intersect.htm	X-axis intersect of a Tangent to a line OverviewDifferentiation - Slope of a TangentIntegration - Area Under a Line Pre-reqsDistance Between PointsSlope Between PointsEquation for LinesAnglesFunctions and Graphs of Functions IntroTangentsDerivativesFormulas & NotationRulesMore on DerivativesSymbols & Notation DerivativesChain Rule Part 1Chain Rule Part 2Parametric EquationsX-axis intersect of a TangentLinear ApproximationsApproximation AccuracyImplicit DifferentiationLogarithmic Differentiation Graphs & CurvesMaximums and MinimumsRolle's TheoremMean Value TheoremShaping of a FunctionShaping a Curve ExamplesSketch a TangentPower RuleMore Rules for DerivativesPower-Rule-FunctionsChain RuleHigher DerivativesLinear ApproximationsTrigonometric DerivativesSketching a Curve X-axis intersect of a Tangent to a line Brief Explanation In this lesson, we will introduce Newton's method for finding roots. First let's do a couple practice questions which may or may not have new concepts in them to you. X-axis intersect of a Tangent to a line For the line f(x) = 3x 3 + 2x - 1, find where the tangent line at point (3,4) intersects the x-axis. So, to be clear, we want to know where the tangent line at the point (1,4) intersects the x-axis. We do not want to know where the curve f(x) = 3x 3 + 2x - 1 intersects the x-axis. So, to find the x-axis intercept for a line, we must find the point where x=0. To do this we must get an equation for that line. In this case we need the equation for the tangent line at point (1 ,4). So first calculate the equation for the tangent line to f(x) = 3x 3 + 2x - 1 by differentiation - this gives us the equation for the slope of the tangent at any given point. f ' (x) = (3)(3x 3 - 1) + 2x 1 - 1) - 0 = 9x 2 + 2 We now have the equation f'(x) so that gives us the slope at a point. Input our point and we will find the slope at that point! For (1, 4) , f'(x) = 9x 2 + 2 f'(1) = 9(1)2 + 2 = 11 Now, to find the equation of the tangent line at point (1, 4), we have a point on that line and we know the slope, so we calculate the equation of the tangent using y = mx + b. sub in: x = 1, y = 4, &m = 11 into y = mx + b 4 = (11)(1) + b 4 = 11 + b b = -7 Use m = 11&b = -7 to get the equation of our tangent (at point (1, 4)) line. y = 11x - 7 Lastly, to find the x-axis intercept, plug in y = 0 as the is the y-value for any point on the x-axis. 0 = 11x - 7 11x = 7 x = 7 11 So the x-axis intercept of the tangent line at point (1, 4) of f(x) = 3x 3 + 2x - 1 is (7⁄11, 0) Prev: Parametric Equations Next: Linear Approximations Calculus Help - Homepage \| © HelpingWithCalculus.com 2025 All rights reserved \| Contact\| Privacy\| About Us\| Site Map
10766	https://www.quora.com/How-would-one-draw-a-rhombus-of-sides-6-cm-with-one-angle-of-60-then-draw-a-right-triangle-of-the-same-area	How would one draw a rhombus of sides 6 cm with one angle of 60° then draw a right triangle of the same area? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Drawing Sketch Geometric Properties of P... Area of Figures Rhombus (geometry) Right Triangles Calculation of Shapes Euclidean Geometry Area of Triangles 5 How would one draw a rhombus of sides 6 cm with one angle of 60° then draw a right triangle of the same area? All related (35) Sort Recommended Assistant Bot · 1y To draw a rhombus with sides of 6 cm and one angle of 60°, and then to draw a right triangle with the same area, follow these steps: Step 1: Draw the Rhombus Draw the First Side: Start by drawing a horizontal line segment of 6 cm. Label the endpoints as A A and B B. Construct the 60° Angle: At point A A, use a protractor to measure a 60° angle and draw a line segment of 6 cm from A A. Label the endpoint as C C. Complete the Rhombus: Now, from point B B, draw a line segment of 6 cm at a 120° angle (which is 180° - 60°) to line A B A B. Label the endpoint as D D. Connect points C C and D D to complete Continue Reading To draw a rhombus with sides of 6 cm and one angle of 60°, and then to draw a right triangle with the same area, follow these steps: Step 1: Draw the Rhombus Draw the First Side: Start by drawing a horizontal line segment of 6 cm. Label the endpoints as A A and B B. Construct the 60° Angle: At point A A, use a protractor to measure a 60° angle and draw a line segment of 6 cm from A A. Label the endpoint as C C. Complete the Rhombus: Now, from point B B, draw a line segment of 6 cm at a 120° angle (which is 180° - 60°) to line A B A B. Label the endpoint as D D. Connect points C C and D D to complete the rhombus A B C D A B C D. Step 2: Calculate the Area of the Rhombus The area A A of a rhombus can be calculated using the formula: A=a 2 sin(θ)A=a 2 sin⁡(θ) where a a is the length of a side and θ θ is any angle. For our rhombus: a=6 a=6 cm θ=60°θ=60° Calculating the area: A=6 2⋅sin(60°)=36⋅√3 2=18√3 cm 2≈31.18 cm 2 A=6 2⋅sin⁡(60°)=36⋅3 2=18 3 cm 2≈31.18 cm 2 Step 3: Draw a Right Triangle with the Same Area To find a right triangle with the same area, use the area formula for a triangle: A=1 2×base×height A=1 2×base×height Set the Area Equal: Set the area of the right triangle equal to the area of the rhombus: 1 2×b×h=18√3 1 2×b×h=18 3 where b b is the base and h h is the height. 2. Choose Dimensions: You can choose a base b b and then solve for the height h h. For example, if you choose b=6 b=6 cm: 1 2×6×h=18√3 1 2×6×h=18 3 3 h=18√3⟹h=6√3≈10.39 cm 3 h=18 3⟹h=6 3≈10.39 cm 3. Draw the Triangle: Draw a horizontal line segment of 6 cm (the base). At one endpoint, draw a vertical line segment of 6√3 6 3 cm (the height). Connect the endpoints of these two segments to form the right triangle. Summary You have now drawn a rhombus with sides of 6 cm and one angle of 60°, and a right triangle with the same area of approximately 31.18 cm 2 31.18 cm 2 using a base of 6 cm and a height of 6√3 6 3 cm. Upvote · Jayanta Mukherjee B Tech IEE in Instrumentation Engineering, Jadavpur University (Graduated 1990) · Author has 43.6K answers and 11.1M answer views ·5y Originally Answered: How do you draw a rhombus of side 5 cm and one angle 50° then draw a right triangle of the same area? · One side of 5 cm length and one angle of 50° measure are given. Step -1: Draw a line segment AB of length 5 cm. Step-2: Draw another line AP such that angle PAB = 50°. Cut off AD = 5 cm from AP. Step-3: Taking centre at B and D, draw two arcs each of radius 5 cm - that cut one another at point C. Thus, ABCD is the rhombus having each side of length 5 cm and angle 50°. Step-4: Connect diagonals AC and BD of the rhombus. Let, AC = m and BD = n. We know, diagonals of a rhombus are perpendicular bisectors. Area of the rhombus = {(1 / 2) m n}. Thus, if we draw a right angled triangle with perpendicul Continue Reading One side of 5 cm length and one angle of 50° measure are given. Step -1: Draw a line segment AB of length 5 cm. Step-2: Draw another line AP such that angle PAB = 50°. Cut off AD = 5 cm from AP. Step-3: Taking centre at B and D, draw two arcs each of radius 5 cm - that cut one another at point C. Thus, ABCD is the rhombus having each side of length 5 cm and angle 50°. Step-4: Connect diagonals AC and BD of the rhombus. Let, AC = m and BD = n. We know, diagonals of a rhombus are perpendicular bisectors. Area of the rhombus = {(1 / 2) m n}. Thus, if we draw a right angled triangle with perpendicular sides of m and n long; area of that right angled triangle equals the rhombus. Step-5: Draw a line segment PQ = m. Now, draw another straight line at P, PL such that PL is perpendicular to PQ. Cut off PR = n from PL. Join QR to get a right angled triangle PQR, whose perpendicular sides are PQ (= m) and PR (= n). Area of the right angled triangle PQR = area of the rhombus ABCD. Upvote · 9 2 9 1 Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 201 99 34 9 3 Anca Horsfall likes Maths, books, languages. · Author has 988 answers and 1.8M answer views ·7y Like this but hopefully without accidentally slicing off the top of your drawing (like I somehow managed to do). Start with segment AB of length 6. With a protractor get a 60 deg. angle at A, then draw segment AD of length 6. Draw segments parallel to AB and AD to finish the rhombus (shown in orangle, below); you can use your protractor to put 120 degree angles at B and D, for example (as the opposite sides of a rhombus are parallel, the angles made by these parallel sides with the adjacent side between them add up to 180 degrees). The idea is to notice that your rhombus ABCD is actually made o Continue Reading Like this but hopefully without accidentally slicing off the top of your drawing (like I somehow managed to do). Start with segment AB of length 6. With a protractor get a 60 deg. angle at A, then draw segment AD of length 6. Draw segments parallel to AB and AD to finish the rhombus (shown in orangle, below); you can use your protractor to put 120 degree angles at B and D, for example (as the opposite sides of a rhombus are parallel, the angles made by these parallel sides with the adjacent side between them add up to 180 degrees). The idea is to notice that your rhombus ABCD is actually made of 2 equilateral triangles (look at triangle ABD: if it’s isosceles, with 2 sides = 6, and has an angle at the tip of 60 degrees, it follows that the other 2 angles are equal and of measure 60 degrees as well). And if you build a perpendicular to AB in point B, which intersects AD at E, you create a right-angled triangle ABE that can fit inside it all of the orange area. Upvote · 9 3 Related questions More answers below Draw a rhombus of sides 6 cm and one angle is 60 degreeThen draw a right triangle of the same area? How do you draw a rhombus of side 5 cm and one angle 50° then draw a right triangle of the same area? One angle of rhombus is given to be 60 degrees and its side is given 10 cm. Then, what will be the length of longer diagonal? How do I construct a rhombus with one side that is 6cm and an angle of 60°? How would you draw a rhombus with one side of 4 cm and an acute angle of 60°? Phlyn Heubach BS in Mechanical Engineering, California Polytechnic State University, San Luis Obispo (Graduated 1997) · Author has 479 answers and 886.2K answer views ·7y Originally Answered: Draw a rhombus of sides 6 cm and one angle is 60 degreeThen draw a right triangle of the same area? · Question: Draw a rhombus of sides 6 cm and one angle is 60 degree Then draw a right triangle of the same area? There are many right triangles that have an area of 18√2 c m 2 18 2 c m 2. I just drew 1. A rhombus is a special parallelogram, so I used the formula for the area of a parallelogram (A=b h A=b h). When the height of this rhombus is drawn a 30–60–90 right triangle is formed therefore the short leg and long leg (height) can be found. The area of a triangle is A=1 2 b h A=1 2 b h since this is a right triangle the legs can be used as the base (b) and the height (h). Continue Reading Question: Draw a rhombus of sides 6 cm and one angle is 60 degree Then draw a right triangle of the same area? There are many right triangles that have an area of 18√2 c m 2 18 2 c m 2. I just drew 1. A rhombus is a special parallelogram, so I used the formula for the area of a parallelogram (A=b h A=b h). When the height of this rhombus is drawn a 30–60–90 right triangle is formed therefore the short leg and long leg (height) can be found. The area of a triangle is A=1 2 b h A=1 2 b h since this is a right triangle the legs can be used as the base (b) and the height (h). Upvote · 9 8 9 3 Ansh Keer Former Owner · Author has 6K answers and 1.7M answer views ·4y Originally Answered: Draw a rhombus of sides 6 cm and one angle is 60 degreeThen draw a right triangle of the same area? · Angles are 60 and 120 Each of four triangles formed by 1/2 D1 and 1/2 D2 have angles 30–60–90 1/2 D1 = 6 cos 60 = 3 1/2 D2 = 6 sin 60 = 3√3 So diagonals are 6, 6√3 and hence area = 18√3 Right triangle = 1/2 bh = 18√3 bh = 36√3 Sides of right triangle are 6, 6√3 and 12 Upvote · 9 1 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 616 Ravi Sharma Former Group A Officer From Indian Railways (1973–2009) · Author has 15.1K answers and 3.8M answer views ·3y Related How do you draw an equilateral triangle with sides 6cm and draw a right angled triangle of the same area? DRAW A LINE SEGMENT IF LENGTH 6 CMS. WITH CENTERS B AND C DRAW ARCS WHOSE RADIUS IS 6 CMS TO INTERSECT AT A. ABC IS EQUILATERAL TRIANGLE. ABC IS THE EQUILATERAL TRIANGLE. DRAW AP PERPANDICULAR TO BC. THROUGH B AND C DRAW LINES PARALLEL TO AP. THROUGH A DRAW LINE PARALLEL TO BC. THESE LINES WILL INTERSECT AT D AND E. JOIN DC. BCD IS REQUIRED RIGHT ANGLED TRIANGLE. YOU CAN ALSO JOIN BE. IN THAT CASE BCE IS THE REQUIRED RIGHT ANGLED TRIANGLE. Continue Reading DRAW A LINE SEGMENT IF LENGTH 6 CMS. WITH CENTERS B AND C DRAW ARCS WHOSE RADIUS IS 6 CMS TO INTERSECT AT A. ABC IS EQUILATERAL TRIANGLE. ABC IS THE EQUILATERAL TRIANGLE. DRAW AP PERPANDICULAR TO BC. THROUGH B AND C DRAW LINES PARALLEL TO AP. THROUGH A DRAW LINE PARALLEL TO BC. THESE LINES WILL INTERSECT AT D AND E. JOIN DC. BCD IS REQUIRED RIGHT ANGLED TRIANGLE. YOU CAN ALSO JOIN BE. IN THAT CASE BCE IS THE REQUIRED RIGHT ANGLED TRIANGLE. Upvote · Related questions More answers below What is the area of a rhombus with an angle of 90 degrees between its sides? Sides 5cm 6cm 4.5cm 8.5cm inside line 7.5cm draw a rhombus of sides 6cm and 1 angle 60 digree then draw a right triangle of same side? How do you draw a rhombus with side lengths of 6 units and angle measures of 60 degrees, 120 degrees and 120 degrees? If a side of a rhombus is 30, what would the other sides be? How can I draw a rhombus with a diagonal length of 3cm and one angle 40? Satyanarayana Mudunuri Worked at ITC Limited (1979–2002) · Author has 3.4K answers and 1.8M answer views ·4y Related How do you draw an equilateral triangle with sides 6cm and draw a right angled triangle of the same area? Draw a horizontal line BC of length 6 cm. With B as centre with radius 6 cm draw an arc above BC. With C as centre and with radius 6 cm draw aw another arc likewise. Let these two arcs intersect at A. Then ABC is required equilateral triangle Draw a line parallel to BC through A, call it XAY. At B (with set squares) draw a line perpendicular to BC. Let it intersect XAY at P. Then PBC is a required rt. angled triangle with the same area as the equilateral triangle. Upvote · Sponsored by LPU Online Career Ka Turning Point with LPU Online. 100% Online UGC-Entitled programs with LIVE classes, recorded content & placement support. Apply Now 999 255 Robert Nichols Upvoted by Richard Enison , M.A. Mathematics & Mathematical Logic and Foundations, University of California, Berkeley (1974) · Author has 5K answers and 15.6M answer views ·7y Related How do I draw a rhombus with one angle being 60 degrees and the sides are 6 cm? Join together two equilateral triangle each having sides of 6 cm. Continue Reading Join together two equilateral triangle each having sides of 6 cm. Upvote · 99 11 9 1 Minerva Hilsinger Author has 158 answers and 233.2K answer views ·5y Related How do you construct a rhombus with a side of 6 cm and an angle equal to 60 degrees with the figure and figure figure? A rhombus with a 60 degree angle is simply 2 equilateral triangles attached at 1 side. Start with a 6 cm line AB Use a compass to make an arc centered at A to B close to 90 degrees With the compass unchanged in size, center at B from A and cross the 1st arc and continue about double in arc angle where the 2 arcs cross is point C. This is the 1st equilateral triangle Again, with the compass unchanged, center at C, make an arc to cross the over extended 2nd arc. This is point D at the opposite side of the rhombus of point A connect the out lines AB, AC, DB, DC. BC not needed but shows the 2 triangles. Continue Reading A rhombus with a 60 degree angle is simply 2 equilateral triangles attached at 1 side. Start with a 6 cm line AB Use a compass to make an arc centered at A to B close to 90 degrees With the compass unchanged in size, center at B from A and cross the 1st arc and continue about double in arc angle where the 2 arcs cross is point C. This is the 1st equilateral triangle Again, with the compass unchanged, center at C, make an arc to cross the over extended 2nd arc. This is point D at the opposite side of the rhombus of point A connect the out lines AB, AC, DB, DC. BC not needed but shows the 2 triangles. Upvote · 9 2 Promoted by HP HP Tech Takes Tech Enthusiast \| Insights, Tips & Guidance ·Updated Sep 18 Which is a good affordable wireless laser printer that prints both sides of paper automatically? Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal f Continue Reading Finding an affordable wireless laser printer that supports automatic duplex printing can be a smart investment if you want to save paper and streamline your workflow. Duplex printing allows the printer to print on both sides of the page without manual intervention, which is especially useful for producing professional documents, reports, or booklets. Wireless connectivity adds convenience by enabling printing from multiple devices, including smartphones, tablets, and laptops. HP offers several models that combine these features with reliable performance and cost-efficiency, making them ideal for home offices or small businesses. One strong recommendation is the HP LaserJet Pro MFP 3102fdw. This model is priced around £229 to £329 and includes automatic duplex printing, wireless connectivity, and multifunction capabilities such as scanning and copying. It’s designed for users who need consistent, high-speed printing with minimal maintenance. The compact design makes it suitable for smaller workspaces, while the efficient toner system helps keep running costs low. Its compatibility with the HP Smart app allows for easy mobile printing, adding flexibility to your workflow. For a more budget-conscious alternative, the HP LaserJet M234sdw is another excellent option. Priced from £136 to £210, it offers automatic duplex printing and wireless functionality, along with fast print speeds of up to 29 pages per minute. This model is ideal for users who primarily need high-volume monochrome printing without the added features of scanning or copying. It supports mobile printing through platforms like Apple AirPrint and the HP Smart app, making it easy to print from various devices. It uses toner really efficiently, so you won’t be spending loads on refills, thus great for everyday printing. LaserJet Printers - Black & White or Color Document Printers In conclusion, if you need a multifunction printer with duplex printing and wireless capabilities, the HP LaserJet Pro MFP 3102fdw offers excellent value and performance. If your focus is on fast, reliable black and white printing with duplex support at a lower price point, the HP LaserJet M234sdw is a dependable and cost-effective solution. Both models can give you strong results without exceeding your budget, and the choice depends on whether you require additional features beyond printing. For more duplex printer model recommendations, check out this blog post: Top Multifunction Printers for Small Businesses Features And Recommendations \| HP® Tech Takes By Lizzie - HP Tech Expert Upvote · 99 15 9 5 Ernest Leung B.Sc. (Hons.) in Chemistry Honors&Mathematics, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views ·Jul 11 Related If triangle XYZ,/XY/=6cm, YXZ=60° and XYZ is a right angle, how do you calculate /XZ/ in cm? If triangle XYZ, \|XY\| = 6cm, YXZ = 60° and XYZ is a right angle, how do you calculate \|XZ\| in cm? Refer to the figure of ΔXYZ below. In ΔXYZ: cos∠YXZ = XY / XZ cos60ᵒ = (6 cm) / XZ XZ = 6 / cos60ᵒ cm XZ = 6 / (1/2) cm XZ = 12 cm Continue Reading If triangle XYZ, \|XY\| = 6cm, YXZ = 60° and XYZ is a right angle, how do you calculate \|XZ\| in cm? Refer to the figure of ΔXYZ below. In ΔXYZ: cos∠YXZ = XY / XZ cos60ᵒ = (6 cm) / XZ XZ = 6 / cos60ᵒ cm XZ = 6 / (1/2) cm XZ = 12 cm Upvote · 9 7 Richard Spence B.S. in Mathematics, Massachusetts Institute of Technology · Author has 190 answers and 38.7K answer views ·Jan 28 Related How do you draw and label a right-angled triangle where the length of two sides are 5cm and 12cm? Here is a right triangle with sides 5 and 12, and hypotenuse 13 (by the Pythagorean theorem), generated using LaTeX/TikZ. LaTeX/TikZ code: \documentclass{standalone} \usepackage{tikz, tkz-euclide} \usepackage{mathpazo} \begin{document} \begin{tikzpicture} \coordinate (A) at (12,0); \node[below right] at (A) {\large $A$}; \coordinate (B) at (0,5); \node[above left] at (B) {\large $B$}; \coordinate (C) at (0,0); \node[below left] at (C) {\large $C$}; \draw (A)--(B)--(C)--cycle; \path (A)--(C) node[midway,below] {\large 12}; \path (B)--(C) node[midway,left] {\large 5}; Continue Reading Here is a right triangle with sides 5 and 12, and hypotenuse 13 (by the Pythagorean theorem), generated using LaTeX/TikZ. LaTeX/TikZ code: \documentclass{standalone} \usepackage{tikz, tkz-euclide} \usepackage{mathpazo} \begin{document} \begin{tikzpicture} \coordinate (A) at (12,0); \node[below right] at (A) {\large $A$}; \coordinate (B) at (0,5); \node[above left] at (B) {\large $B$}; \coordinate (C) at (0,0); \node[below left] at (C) {\large $C$}; \draw (A)--(B)--(C)--cycle; \path (A)--(C) node[midway,below] {\large 12}; \path (B)--(C) node[midway,left] {\large 5}; \path (A)--(B) node[midway,above] {\large 13}; \tkzMarkRightAngle(B,C,A) \end{tikzpicture} \end{document} Upvote · Avirat Dhodare Jr.Executive - Exam Services at Coempt Eduteck (2022–present) · Author has 3.4K answers and 7.2M answer views ·2y Related How do we prove that the area of a rhombus on the hypotenuse of a right triangle with one of the angles of 60 degrees is equal to the sum of the areas of the rhombus with one of the angles as 60 degrees drawn on the other two sides? Solution: ΔABC is a right angle with ∠B as right ∠. let AB = a and BC = b By Pythagoras theorem AC = √ (a2 + b2 ) ACDE is rhombus with 60° inclination Draw a perpendicular EF from E on base AC. EF = AE sin 60° AE = AC = √ (a^2 + b^2 ) EF = √ (a^2 + b^2 ) sin 60° = √ (a^2 + b^2 ) √3 /2 Area of rhombus ( ACDE ) = base height ACDE = AC EF = √ (a^2 + b^2 ) √ (a^2 + b^2 ) √3/2 = √3/2 (a^2 + b^2 ) ............ Continue Reading Solution: ΔABC is a right angle with ∠B as right ∠. let AB = a and BC = b By Pythagoras theorem AC = √ (a2 + b2 ) ACDE is rhombus with 60° inclination Draw a perpendicular EF from E on base AC. EF = AE sin 60° AE = AC = √ (a^2 + b^2 ) EF = √ (a^2 + b^2 ) sin 60° = √ (a^2 + b^2 ) √3 /2 Area of rhombus ( ACDE ) = base height ACDE = AC EF = √ (a^2 + b^2 ) √ (a^2 + b^2 ) √3/2 = √3/2 (a^2 + b^2 ) ....................( Equation 1) similarly in rhombus BCJG Side BC = a angle BCJ = 60° (GIVEN) height = a sin 60° = a √3/2 area of rhombus BCGJ = a^2 √3/2 .................( equation 2) similarly area o... Upvote · Dave Benson trying to make maths easy. · Author has 6.1K answers and 2.1M answer views ·Jul 11 Related If triangle XYZ,/XY/=6cm, YXZ=60° and XYZ is a right angle, how do you calculate /XZ/ in cm? Recognise 1:√3:2 triangle and xz is hypotenuse = 26 = 12 cm answer Continue Reading Recognise 1:√3:2 triangle and xz is hypotenuse = 26 = 12 cm answer Upvote · 9 6 Related questions Draw a rhombus of sides 6 cm and one angle is 60 degreeThen draw a right triangle of the same area? How do you draw a rhombus of side 5 cm and one angle 50° then draw a right triangle of the same area? One angle of rhombus is given to be 60 degrees and its side is given 10 cm. Then, what will be the length of longer diagonal? How do I construct a rhombus with one side that is 6cm and an angle of 60°? How would you draw a rhombus with one side of 4 cm and an acute angle of 60°? What is the area of a rhombus with an angle of 90 degrees between its sides? Sides 5cm 6cm 4.5cm 8.5cm inside line 7.5cm draw a rhombus of sides 6cm and 1 angle 60 digree then draw a right triangle of same side? How do you draw a rhombus with side lengths of 6 units and angle measures of 60 degrees, 120 degrees and 120 degrees? If a side of a rhombus is 30, what would the other sides be? How can I draw a rhombus with a diagonal length of 3cm and one angle 40? What is the area of a rhombus with side of 8 cm at an acute angle between one pair of sides 30°? What is the area of a rhombus with one side of 4 cm and one angle of 60 degrees? Can the sides of a rhombus be equal to its sides? One side of rhombus ABCD is 8 cm in length. If angle A=64, what are BD and AC? What are the other angles of a rhombus if the interior angle is 60 degrees and the sides of length is 4cm? Related questions Draw a rhombus of sides 6 cm and one angle is 60 degreeThen draw a right triangle of the same area? How do you draw a rhombus of side 5 cm and one angle 50° then draw a right triangle of the same area? One angle of rhombus is given to be 60 degrees and its side is given 10 cm. Then, what will be the length of longer diagonal? How do I construct a rhombus with one side that is 6cm and an angle of 60°? How would you draw a rhombus with one side of 4 cm and an acute angle of 60°? What is the area of a rhombus with an angle of 90 degrees between its sides? 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10767	https://softmath.com/algebra-word-problems/show.php?id=20719	The total mass of a basket, 4 large... Algebrator Algebrator Home Desktop Math Solver Why Algebrator? Guarantee Testimonials Press Comparison Online Demo Order Who We Are FAQs Resources Privacy All Solvers Math Tutorials Word Problem Solver Get it on Google PlayGet it on Apple Store Problema Solution The total mass of a basket, 4 large pebbles, and 2 small pebbles is 7/12 kilogram. The total mass of the basket, 2 large pebbles and 1 small pebble is 5/12 kilogram. Find the mass of the empty basket Answer provided by our tutors let l = the mass of one large pebble s = the mass of one small pebble b = the mass of the basket The total mass of a basket, 4 large pebbles, and 2 small pebbles is 7/12 kilogram b + 4l + 2s = 7/12 The total mass of the basket, 2 large pebbles and 1 small pebble is 5/12 kilogram b + 2l + s = 5/12 Multiply both sides of the equation b + 2l + s = 5/12 by 2 and subtract the equation b + 4l + 2s = 7/12 2(b + 2l + s) - (b + 4l + 2s) = 2(5/12) - 7/12 2b + 4l + 2s - b - 4l - 2s = 5/6 - 7/12 b = 5/6 - 7/12 by solving we find b = 0.25 kg click here to see the step by step solution Click to see all the steps the mass of the empty basket is 0.25 kg. ← Previous Page Next Page →
10768	https://zhuanlan.zhihu.com/p/349330406	S1-2 逻辑量词 - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册 S1-2 逻辑量词首发于从零开始的数学教育切换模式 S1-2 逻辑量词咸鱼晓孔调参机器，一条曾经的数院咸鱼，ACM退役选手收录于 · 从零开始的数学教育 5 人赞同了该文章前言在进行证明时，非常重要的是命题的转换，以及搞清楚命题的范围，是描述普遍性质还是特殊性质，即任意和存在。逆否命题对于「若p，则q」的命题，即 p⇒q p\Rightarrow q ，我们有三种衍生的命题。否命题，把条件和结论同时否定，得到 (¬p)⇒(¬q)(\neg p)\Rightarrow (\neg q) 。逆命题，把条件和结论进行交换，得到 q⇒p q\Rightarrow p 。逆否命题，即逆命题的否命题，得到 (¬q)⇒(¬p)(\neg q)\Rightarrow (\neg p) ，也等于否命题的逆命题。要注意，否命题和否定不同，复合命题的否定为 ¬(p⇒q)=¬((¬p)∨q)=p∧(¬q)\neg (p\Rightarrow q)=\neg( (\neg p)\vee q)= p\wedge (\neg q) 。在保证条件p为真的情况下，可以转化为 p⇒(¬q)p\Rightarrow (\neg q) ，即只否定结论。可以证明，原命题和逆否命题等价： p⇒q=(¬q)⇒(¬p)p\Rightarrow q = (\neg q)\Rightarrow (\neg p) 练习：证明这个结论。所以有时候我们可以证明逆否命题来证明原命题。探索：有哪些利用逆否命题完成的证明？通常来说，命题逻辑是不考虑可能性的，例如三角形内角和一定是180度，否定是三角形内角和一定不是180度，而非三角形内角和不一定是180度。探索：考虑可能性的逻辑是什么样的？（模态逻辑）任意与存在考虑以下命题，没有最大的自然数。所谓的没有最大的自然数，就是任意的意思，是所有自然数的普遍性质，需要所有自然数都满足才行。我们用F(x)判断x是最大的自然数，可以写作： \forall x(F(x)=0) 而命题的否定，有最大的自然数，则是存在的意思，是某个自然数的特殊性质，只需要有一个满足即可，可以写作： \exists x(F(x)=1 ) 存在和任意被称为逻辑量词，相应命题叫做存在命题和全称命题。我们可以选择证明任意自然数都不是最大的，即对于任意自然数x，有更大的自然数x+1。也可以使用反证法，设存在最大的自然数是x，则这与x小于自然数x+1矛盾。有时候用反证法会是更好的选择，反证需要能正确写出命题的否定。可以证明，在对命题否定时，存在和任意会相互转换： \neg(\exists x(p))=\forall x(\neg p),\neg(\forall x(p))=\exists x(\neg p) 探索：如何证明这个结论？探索：还有哪些其他词语表示存在或任意？三段论接下来介绍一种经典的推理方法，三段论。任意整数都是有理数，1是整数，所以1是有理数。用逻辑符号来表示如下： A=\forall x(P(x)\Rightarrow Q(x))),B=P(y),C=Q(y)\ (A\wedge B)\Rightarrow C A是大前提，B是小前提，C是结论。大前提，任意P(x)为真可以得到Q(x)为真，例如A是整数可以得到A是有理数。小前提，P(y)为真，例如1是整数。那么可以得到结论，Q(y)为真，即1是有理数。探索：如何证明三段论？探索：还有那些推理方法？后记关于任意与存在，表面看起来简单，但其实这是个很复杂的问题。尤其是复合命题拥有多层的任意与存在之后，在数学分析的学习中，就会遇到多层的任意与存在，熟练的命题转换是基本功。编辑于 2021-07-24 18:13 逻辑学逆否命题三段论赞同 5添加评论分享喜欢收藏申请转载写下你的评论... 还没有评论，发表第一个评论吧关于作者咸鱼晓孔调参机器，一条曾经的数院咸鱼，ACM退役选手回答 554文章 138关注者 3,113 关注他发私信推荐阅读高中数学————逻辑用语 ============ 1.命题 1）定义在数学中，把用语言、符号或式子表述的，可以判断真假的陈述句叫做命题。 2）分类真命题假命题3）常见形式若 p，则 q 。4）四种命题及相互关系一旦一个命题被定义为“原… 瓜皮发表于高中数学知...集合论基础(2)---谓词逻辑极简概述 =================== 1. 前置知识命题逻辑的相关知识锤同学LikeMath：命题逻辑极简概述2. 为什么需要谓词逻辑？命题逻辑高度概括，不能表达具象一些的事物关系和特征。 a. 命题逻辑表达“所有”的概念存在困难… 锤同学LikeMath一阶谓词逻辑完全性的证明 ============ 一、引言：命题逻辑的完全性证明很早就被逻辑学家所证明，然而关于一阶谓词逻辑的完全性则直到1930年，哥德尔才给出了第一个证明。王浩在书中谈到了，哥德尔（ Kurt Gödel）认为完全性定理… 小晨小哥关于逆否命题的理解 ========= 逆否命题与原命题正确性相同，这是个很有用的结论。但是高中的我一直没有去想为什么。然后发动了一下独立思考，试着用可视化解决这一问题。如果A→B，那么用集合的思想，可以表示为其中U… 巨鶸RD 想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
10769	https://app.prepanywhere.com/prep/textbooks/stewart-calculus-7e-early-trans/chapters/chapter-4-applications-of-differentiation/materials/4-4-indeterminate-forms-and-l-hospital-s-rule	PrepAnywhere Log InGet Started Books>University>Unit 4.4 Stewart Calculus 7E Early Trans ISBN: / Video Solutions 025. Q25044. Q44077. Q77084. Q84085. Q85086. Q86088. Q88 Lectures 001. Introduction of L'Hospital's Rule002. Rough Proof of L'Hospital's Rule003. ex2 e^x over polynomial004. ex3 with log005. ex4 with Trig006. ex5 Make sure it is a L'Hospital's Rule!007. ex6 Indeterminate form look like a product008. ex7 Difference of function indeterminate form009. Indetermine Powers010. ex1 of exponential indeterminate form011. ex2 of exponential intederminate form Ready to get started? Sign up or contact us Get Started Free Privacy Policy\|Terms & Conditions
10770	https://en.wikipedia.org/wiki/Kinetic_isotope_effects_of_RuBisCO	Kinetic isotope effects of RuBisCO - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Reaction details and energetics 2 Ecological trade-offs influence isotope effectsToggle Ecological trade-offs influence isotope effects subsection 2.1 Specificity (S C/O) 2.2 Rate of carboxylation (V C) and Michaelis-Menten constant (K C) 2.3 Phylogenetic patterns 2.4 Isotope effects 3 See also 4 References Kinetic isotope effects of RuBisCO [x] Add languages Add links Article Talk [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 Add interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia The Calvin-Benson Cycle. The KIE of RuBisCO is associated with the step where RuBisCO catalyzes the fixation of carbon dioxide to Ribulose-1,5-bisphosphate. The kinetic isotope effect (KIE) of ribulose-1,5-bisphosphate carboxylase oxygenase (RuBisCO) is the isotopic fractionation associated solely with the step in the Calvin-Benson cycle where a molecule of carbon dioxide (CO 2) is attached to the 5-carbon sugar ribulose-1,5-bisphosphate (RuBP) to produce two 3-carbon sugars called 3-phosphoglycerate (3 PGA). This chemical reaction is catalyzed by the enzyme RuBisCO, and this enzyme-catalyzed reaction creates the primary kinetic isotope effect of photosynthesis. It is also largely responsible for the isotopic compositions of photosynthetic organisms and the heterotrophs that eat them. Understanding the intrinsic KIE of RuBisCO is of interest to earth scientists, botanists, and ecologists because this isotopic biosignature can be used to reconstruct the evolution of photosynthesis and the rise of oxygen in the geologic record, reconstruct past evolutionary relationships and environmental conditions, and infer plant relationships and productivity in modern environments. Reaction details and energetics [edit] Carboxylation of RuBP catalyzed by RuBisCO. Each step is shown in two panels: 1) The upper panel shows how each molecule is coordinated to the active site, while 2) The lower panel shows specifically how RuBP is being modified. Overall, the carboxylation of RuBP is a multi-step process. The fixation of CO 2 by RuBisCO is a multi-step process. First, a CO 2 molecule (that is not the CO 2 molecule that is eventually fixed) attaches to the uncharged ε-amino group of lysine 201 in the active site to form a carbamate. This carbamate then binds to the magnesium ion (Mg 2+) in RuBisCO's active site. A molecule of RuBP then binds to the Mg 2+ ion. The bound RuBP then loses a proton to form a reactive, enodiolate species. The rate-limiting step of the Calvin-Benson cycle is the addition of CO 2 to this 2,3-enediol form of RuBP. This is the stage where the intrinsic KIE of Rubisco occurs because a new C-C bond is formed. The newly formed 2-carboxy-3-keto-D-arabinitol 1,5-bisphosphate molecule is then hydrated and cleaved to form two molecules of 3-phosphoglycerate (3 PGA). 3 PGA is then converted into hexoses to be used in the photosynthetic organism's central metabolism. The difference in activation energy required for a heavy or light molecule of carbon dioxide. The isotopic substitutions that can occur in this reaction are for carbon, oxygen, and/or hydrogen, though currently only a significant isotope effect is seen for carbon isotope substitution.Isotopes are atoms that have the same number of protons but varying numbers of neutrons. "Lighter" isotopes (like the stable carbon-12 isotope) have a smaller overall mass, and "heavier" isotopes (like the stable carbon-13 isotope or radioactivecarbon-14 isotope) have a larger overall mass. Stable isotope geochemistry is concerned with how varying chemical and physical processes preferentially enrich or deplete stable isotopes. Enzymes like RuBisCO cause isotopic fractionation because molecules containing lighter isotopes have higher zero-point energies (ZPE), the lowest possible quantum energy state for a given molecular arrangement. For this reaction, 13 CO 2 has a lower ZPE than 12 CO 2 and sits lower in the potential energy well of the reactants. When enzymes catalyze chemical reactions, the lighter isotope is preferentially selected because it has a lower activation energy and is thus more energetically favorable to overcome the high potential-energy transition state and proceed through the reaction. Here, 12 CO 2 has a lower activation energy so more 12 CO 2 than 13 CO 2 goes through the reaction, resulting in the product (3 PGA) being lighter. Ecological trade-offs influence isotope effects [edit] The observed intrinsic KIEs of RuBisCO have been correlated with two aspects of its enzyme kinetics: 1) Its "specificity" for CO 2 over O 2, and 2) Its rate of carboxylation. Specificity (S C/O) [edit] The reactive enodiolate species is also sensitive to oxygen (O 2), which results in the dual carboxylase / oxygenase activity of RuBisCO. This reaction is considered wasteful as it produces products (3-phosphoglycerate and 2-phosphoglycolate) that must be catabolized through photorespiration. This process requires energy and is a missed-opportunity for CO 2 fixation, which results in the net loss of carbon fixation efficiency for the organism. The dual carboxylase / oxygenase activity of RuBisCO is exacerbated by the fact that O 2 and CO 2 are small, relatively indistinguishable molecules that can bind only weakly, if at all, in Michaelis-Menten complexes. There are four forms of RuBisCO (Form I, II, III, and IV), with Form I being the most abundantly used form. Form I is used extensively by higher plants, eukaryotic algae, cyanobacteria, and Pseudomonadota (formerly proteobacteria). Form II is also used but much less widespread, and can be found in some species of Pseudomonadota and in dinoflagellates. RuBisCOs from different photosynthetic organisms display varying abilities to distinguish between CO 2 and O 2. This property can be quantified and is termed "specificity" (S c/o). A higher value of S c/o means that a RuBisCO's carboxylase activity is greater than its oxygenase activity. Rate of carboxylation (V C) and Michaelis-Menten constant (K C) [edit] A generalized Michaelis-Menten curve. The rate of carboxylation (V C) is the rate that RuBisCO fixes CO 2 to RuBP under substrate saturated conditions. A higher value of V C corresponds to a higher rate of carboxylation. This rate of carboxylation can also be represented through its Michaelis-Menten constant K C, with a higher value of K C corresponding to a higher rate of carboxylation. V C is represented by V max, and K C is represented as K M in the generalized Michaelis-Menten curve. Although the rate of carboxylation varies among RuBisCO types, RuBisCO on average fixes only three molecules of CO 2 per second. This is remarkably slow compared to typical enzyme catalytic rates, which usually catalyze reactions at the rate of thousands of molecules per second. Phylogenetic patterns [edit] Relationship between specificity and carboxylation rate of varying photosynthetic organisms. It has been observed among natural RuBisCOs that an increased ability to distinguish between CO 2 and O 2 (larger values of S c/o) corresponds with a decreased rate of carboxylation (lower values of V C and K C). The variation and trade-off between S c/o and K C has been observed across all photosynthetic organisms, from photosynthetic bacteria and algae to higher plants. Organisms using RuBisCOs with high values of V C / K C, and low values of S c/o have localized RuBisCO to areas within the cell with artificially high local CO 2 concentrations. In cyanobacteria, concentrations of CO 2 are increased using a carboxysome, an icosahedral protein compartment about 100 nm in diameter that selectively uptakes bicarbonate and converts it to CO 2 in the presence of RuBisCO. Organisms without a CCM, like certain plants, instead utilize RuBisCOs with high values of S c/o and low values of V C and K C. It has been theorized that groups with a CCM have been able to maximize K C at the expense of decreasing S c/o, because artificially enhancing the concentration of CO 2 would decrease the concentration of O 2 and remove the need for high CO 2 specificity. However, the opposite is true for organisms without a CCM, who must optimize S c/o at the expense of K C because O 2 is readily present in the atmosphere. This trade-off between S c/o and V C or K C observed in extant organisms suggest that RuBisCO has evolved through geologic time to be maximally optimized in its current, modern environment. RuBisCO evolved over 2.5 billion years ago when atmospheric CO 2 concentrations were 300 to 600 times higher than present day concentrations, and oxygen concentrations were only 5-18% of present-day levels. Therefore, because CO 2 was abundant and O 2 rare, there was no need for the ancestral RuBisCO enzyme to have high specificity. This is supported by the biochemical characterization of an ancestral RuBisCO enzyme, which has intermediate values of V C and S C/O between the extreme end-members. It has been theorized that this ecological trade-off is due to the form that 2-carboxy-3-keto-D-arabinitol 1,5-bisphophate in its transient transition state before cleaving into two 3PGA molecules. The more closely the Mg 2+-bound CO 2 moiety resembles the carboxylate group in 2-carboxy-3-keto-D-arabinitol 1,5-bisphophate, the greater the structural difference between the transition states of carboxylation and oxygenation. The larger structural difference allows RuBisCO to better distinguish between CO 2 and O 2, resulting in larger values of S c/o. However, this increasing structural similarity between the transition state and the product state requires strong binding at the carboxyketone group, and this binding is so strong that the rate of cleavage into two product 3PGA molecules is slowed. Therefore, an increased specificity for CO 2 over O 2 necessitates a lower overall rate of carboxylation. This theory implies that there is a physical chemistry limitation at the heart of Rubisco's active site, and may preclude any efforts to engineer a simultaneously more selective and faster Rubisco. Isotope effects [edit] Specificity of RuBisCO for CO 2 vs O 2 determines the extent of Carbon isotope fractionation. S c/o has been positively correlated with the magnitude of carbon isotope fractionation (represented by Δ 13 C), with larger values of S c/o corresponding with a larger values of Δ 13 C. It has been theorized that because increasing S c/o means the transition state is more like the product, the O 2 C---C-2 bond will be shorter, resulting in a higher overall potential energy & vibrational energy. This creates a higher energy transition state, which makes it even harder for 13 CO 2 (lower in the potential energy well than 12 CO 2) to overcome the required activation energy. The RuBisCOs used by varying photosynthetic organisms vary slightly in their enzyme structure, and this enzyme structure results in varying transition states. This diversity in enzyme structure is reflected in the resulting Δ 13 C values measured from different photosynthetic organisms. However, overlap exists between the Δ 13 C values of different groups because the carbon isotope values measured are generally of the entire organism, and not just its RuBisCO enzyme. Many other factors, including growth rate and the isotopic composition of the starting substrate, can affect the carbon isotope values of whole organism and cause the spread seen in C isotope measurements. See also [edit] Isotope geochemistry Fractionation of carbon isotopes in oxygenic photosynthesis Isotopes of carbon Isotopic signature References [edit] ^Farquhar, Graham D.; O'Leary, Marion H.; Berry, Joe A. (1982). "On the relationship between carbon isotope discrimination and the intercellular carbon dioxide concentration in leaves". Functional Plant Biology. 9 (2): 121–137. Bibcode:1982FunPB...9..121F. doi:10.1071/PP9820121. ^Zanden, M. Jake Vander; Rasmussen, Joseph B. (November 2001). "Variation in δ15N and δ13C trophic fractionation: Implications for aquatic food web studies". Limnology and Oceanography. 46 (8): 2061–2066. Bibcode:2001LimOc..46.2061Z. doi:10.4319/lo.2001.46.8.2061. ISSN0024-3590. ^McCutchan, James H.; Lewis, William M.; Kendall, Carol; McGrath, Claire C. (August 2003). "Variation in trophic shift for stable isotope ratios of carbon, nitrogen, and sulfur". Oikos. 102 (2): 378–390. Bibcode:2003Oikos.102..378M. doi:10.1034/j.1600-0706.2003.12098.x. ISSN0030-1299. ^Farquhar, G. D.; Hubick, K. T.; Condon, A. G.; Richards, R. A. (1989), "Carbon Isotope Fractionation and Plant Water-Use Efficiency", Stable Isotopes in Ecological Research, Springer New York, pp.21–40, doi:10.1007/978-1-4612-3498-2_2, ISBN9781461281276 ^Körner, Ch.; Farquhar, G. D.; Wong, S. C. (September 1991). "Carbon isotope discrimination by plants follows latitudinal and altitudinal trends". Oecologia. 88 (1): 30–40. Bibcode:1991Oecol..88...30K. doi:10.1007/bf00328400. ISSN0029-8549. PMID28312728. S2CID36219207. ^Hayes, John M.; Strauss, Harald; Kaufman, Alan J. (September 1999). "The abundance of 13C in marine organic matter and isotopic fractionation in the global biogeochemical cycle of carbon during the past 800 Ma". Chemical Geology. 161 (1–3): 103–125. 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Series B, Biological Sciences. 363 (1504): 2629–40. doi:10.1098/rstb.2008.0023. PMC2606765. PMID18487131. ^ Jump up to: abcdShih PM, Occhialini A, Cameron JC, Andralojc PJ, Parry MA, Kerfeld CA (January 2016). "Biochemical characterization of predicted Precambrian RuBisCO". Nature Communications. 7 10382. Bibcode:2016NatCo...710382S. doi:10.1038/ncomms10382. PMC4735906. PMID26790750. ^Halliwell B, Stumpf PK, Conn EE, Hatch MD, Boardman NK (1982-02-08). "The Biochemistry of Plants: A Comprehensive Treatise". FEBS Letters. 138 (1): 153. doi:10.1016/0014-5793(82)80429-8. ^Pierce J, Lorimer GH, Reddy GS (1986). "Kinetic mechanism of ribulosebisphosphate carboxylase: Evidence for an ordered, sequential reaction". Biochemistry. 25 (7): 1636–1644. doi:10.1021/bi00355a029. ^ Jump up to: ab"PDB101: Molecule of the Month: Rubisco". RCSB: PDB-101. Retrieved 2018-05-25. ^ Jump up to: abcdeSavir Y, Noor E, Milo R, Tlusty T (February 2010). "Cross-species analysis traces adaptation of Rubisco toward optimality in a low-dimensional landscape". Proceedings of the National Academy of Sciences of the United States of America. 107 (8): 3475–80. arXiv:1007.4461. Bibcode:2010PNAS..107.3475S. doi:10.1073/pnas.0911663107. PMC2840432. PMID20142476. ^Yeates TO, Kerfeld CA, Heinhorst S, Cannon GC, Shively JM (September 2008). "Protein-based organelles in bacteria: carboxysomes and related microcompartments". Nature Reviews. Microbiology. 6 (9): 681–91. doi:10.1038/nrmicro1913. PMID18679172. S2CID22666203. ^Laws, Edward A.; Popp, Brian N.; Bidigare, Robert R.; Kennicutt, Mahlon C.; Macko, Stephen A. (March 1995). "Dependence of phytoplankton carbon isotopic composition on growth rate and [CO2)aq: Theoretical considerations and experimental results". Geochimica et Cosmochimica Acta. 59 (6): 1131–1138. Bibcode:1995GeCoA..59.1131L. doi:10.1016/0016-7037(95)00030-4. ISSN0016-7037. ^Popp, Brian N.; Laws, Edward A.; Bidigare, Robert R.; Dore, John E.; Hanson, Kristi L.; Wakeham, Stuart G. (January 1998). "Effect of Phytoplankton Cell Geometry on Carbon Isotopic Fractionation". Geochimica et Cosmochimica Acta. 62 (1): 69–77. Bibcode:1998GeCoA..62...69P. doi:10.1016/s0016-7037(97)00333-5. ISSN0016-7037. Retrieved from " Categories: Chemical kinetics Photosynthesis Isotope separation This page was last edited on 23 August 2025, at 06:03(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Kinetic isotope effects of RuBisCO Add languagesAdd topic
10771	https://artofproblemsolving.com/wiki/index.php/2010_USAMO_Problems/Problem_2?srsltid=AfmBOoo5Ek4kKVHD4Ug4IqACmFoS_cV-O33rQ8C4rNeQQQHr9NljbwAJ	Art of Problem Solving 2010 USAMO Problems/Problem 2 - 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 2010 USAMO Problems/Problem 2 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2010 USAMO Problems/Problem 2 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 4 Comment 5 See also Problem There are students standing in a circle, one behind the other. The students have heights . If a student with height is standing directly behind a student with height or less, the two students are permitted to switch places. Prove that it is not possible to make more than such switches before reaching a position in which no further switches are possible. Solution 1 We adopt the usual convention that unless . With this, the binomial coefficients are defined for all integers via the recursion: It is clear that the circle is oriented and all the students are facing in same direction (clockwise or counterclockwise). We'll call this direction forward. In any switch consider the taller student to have moved forward and the shorter student to have remained stationary. No backward motion is allowed. With this definition of forward motion, the first two students with heights and are always stationary, while other students potentially move past them. For , the student with height can never switch places with the student with height , and the former can make at most more forward moves than the latter (when all the students of heights are between and in the forward direction). Therefore, if the student can make forward steps, the student can make at most steps. With and , and a constant second difference of , we quickly see that . With students in all, the total number of steps is therefore at most , by the Hockey Stick Theorem Solution 2 WLOG, let . Now, we find the end arrangement with the most switches possible. We claim that the arrangement will be , where the left to right direction is the "backwards" direction. To prove this makes the most switches, we show that there is always at least one more switch that can be done for any other arrangement. This is elementary to show. There will always be one height such that the number to its right is less than , unless every number has to the right of (other than and ). The exception occurs at our claim, so our claim is proven. Now we want to find the maximum ways we can "undo" our arrangement. But undoing a switch is just doing a switch from our arrangement in the opposite direction. So, the start arrangement with the most possible switches is the reverse of the end arrangement or . We want to find how many switches must be done to get from the start arrangement to the end arrangement. We start by switching around until it cannot be switched anymore. We find that we can switch times. When we are switching (or, in other words, moving to the right) the number , we can switch it times before it is to the left of . But then we can also switch each of (in that order) times before they get stopped again. Let . Then, the sum of the switches is . Rearranging and summing, we get Comment This process of switching is very similar to Bubble Sort, except that the terms must be in consecutive decreasing order, and wraps around . This answer seems to disagree, though, because the worst case efficiency is , not . See also 2010 USAMO (Problems • Resources) Preceded by Problem 1Followed by Problem 3 1•2•3•4•5•6 All USAMO Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " 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.
10772	https://concord.org/ngss/	Follow Us Concord Consortium on Twitter Find Your Path through the NGSS Want to print the NGSS Pathfinder as a poster or handout? Download a printer-friendly PDF. The graphics are licensed under CC BY 4.0, so you’re welcome to use them. · NGSS Pathfinder without paths · NGSS Pathfinder with paths Find your path through the Next Generation Science Standards with help from the Concord Consortium. Start in the center with a core idea. What do you teach? Add a science and engineering practice—or two! Add a crosscutting concept. You’ve created one path through the NGSS. Use the Concord Consortium resources available for that path below. Practices Asking Questions Defining Problems Using Models ConductingInvestigations AnalyzingData Using Mathematics & Computational Thinking Constructing Explanations Designing Solutions Arguing fromEvidence CommunicatingInformation Core IdeasWhat do you teach? Life Sciences Engineering & Technology Earth & Space Sciences Physical Sciences Crosscutting Concepts Patterns Causation Scale Systems Energy Structure & Function Stability & Change Start by selecting a Core Idea. 3D Exploration of Bound Antibody and Antigen Explore the role of shape in how antibodies and antigens interact. Air Pollution Model (aerial) Explore the connections between point-source pollution, geography, and wind on regional air quality. Air Pollution Model (cross-section) Explore the connections between pollution sources, weather, geography, and air quality. Atom and Ion Builder Create multiple versions of helium atoms and make observations of how changing protons, electrons, and neutrons affect atoms. Atom and Ion Builder (with table) Create multiple versions of various atoms and record the number of protons, electrons, and neutrons in a table. Atomic Structure Explore ion formation, isotopes, and electron orbital placement using interactive models of atomic structure. Atoms and Conservation of Energy How does the Law of Conservation of Energy (the First Law of Thermodynamics) apply to atoms? Baggie Chemistry Household chemicals mixed in a baggie produce dramatic results. Boiling Point This model allows you to explore why polar and non-polar substances have very different boiling points. Boiling Point of Polar & Nonpolar Substances Explore the relationships between properties of molecules, temperature, and movement of particles. Breaking a Molecular Bond (conceptual version) Adjust the initial velocity of a third atom as it hits two bonded atoms and track the changes in energy during this interaction. Chain Reaction Between Hydrogen and Oxygen Observe how a chemical reaction evolves over time and affects the balance of potential and kinetic energy in the system. Bridges Design, build, and test a bridge made with file folders to see which design is the strongest. Build and Test a Model Solar House Construct and test the energy efficiency and solar heat gain of a cardboard model house. Building a Bungee Jump Build and study a bungee jump that is safe and fun to use. Building a Zip Line Design, build, and study a zip line that is safe and fun to use. Buildings and Earthquakes Build model buildings and test them for earthquake resistance using a shake table. Can We Feed the Growing Population? Explore the interconnected resources that make up our agricultural system as you consider food production. Catalysts Explore the effects of homogeneous catalysts. Cellular Respiration Explore how your body converts the chemical energy of glucose into the chemical energy of ATP. Ceramic Forces Explore what happens when a force is exerted on a ceramic material. Changes in the Environment Students learn that different species can arise from a common ancestor if different groups have different selection pressures. Charge Intensity and Electric Force Explore the relationship between charge, electric fields, and forces on objects by manipulating charge. Charge Intensity and Electric Force 2 Explore the relationship between charge, electric fields, and forces on objects by manipulating charge. Charged and Neutral Atoms Explore the role of charge in interatomic interactions. Chemical Bonds Explore the different kinds of chemical bonds that can form, ranging from non-polar covalent to ionic. Chemical Reactions and Stoichiometry Control the concentrations of molecules and temperatures of reactions to explore reaction rate, and learn to balance chemical equations. Clouds Why are there clouds? Explore the water cycle. Comparing Attractive Forces Between Molecules Explore the difference in attractive forces between polar and nonpolar molecules. Comparing Dipole-Dipole to London Dispersion Investigate the difference in the attractive force between polar and non-polar molecules. Comparing Potential Energy of a Bond Compare the change in potential energy when you separate molecules from each versus when you break molecules apart. Competition Use a model to study the effect of a consumer (rabbits) on two species of producers (grass and weeds). Concentrating Charge and Electric Fields Simulate an analogy to Rutherford’s gold foil experiment to distinguish between the plum pudding and hard nucleus model of the atom. Conduction: Combine Heat Capacity and Conductivity This model shows the combined effects of heat capacity and conductivity. Conduction: Heat Conduction Through Materials Observe the rate of heat flow through materials with different conductivities. Conduction: The Effect of Temperature Difference Observe the effect of temperature difference on the flow of heat. Conduction: The Effect of Wall Thickness on Heat Conduction Compare the flow of heat through different thicknesses of material. Convection: Blowing Wind Observe the motion of air around and through a building when the wind is blowing. Convection: Forced Convection Observe how forced convection can change an object’s surface temperature and cool it down rapidly. Convection: Natural Convection This model illustrates the difference between conduction and convection in the transfer of heat. Convection: Natural Convection Inverted This model shows how convection works when the heat source is at the top of a space rather than at the bottom. Convection: Slow Down Convection This model explores how natural convection can be reduced with barriers. Convection: The Stack Effect Observe the effect of leaks in the walls and roof on the air circulation in a heated house. Conflicting Selection Pressures Observe how heredity and natural selection allow a population to adapt to a changing environment. Conversion of Electric Potential Energy Explore how potential energy created by particles of varying charge is converted to thermal energy. Describing Velocity Learn about velocity-time graphs and their connection to corresponding position-time graphs, and determine velocity during different intervals. Deformed Electron Cloud Explore the effect of an electric field on the electron distribution of an atom. Dew Point Figure out the dew point temperature for your classroom. Diffusion Investigate how the random motion and collisions of particles results in diffusion. Diffusion Across a Semipermeable Membrane Explore the role of pore size in the diffusion of a substance across a membrane. Diffusion and Molecular Mass Explore the role of a molecule’s mass with respect to its diffusion rate. Diffusion and Temperature Explore the role of temperature in the rate of diffusion of a substance. Diffusion of a Drop Explore the random molecular motion of a dye in water. Diffusion, Osmosis and Active Transport Explore how water and ions can diffuse both passively and actively through cell membranes. Direction and Strength of Force in Electric Fields Explore the strength and direction of forces between two charged objects by observing the color and direction of the force pointers. Direction of Force Around a VDG (Negatively Charged) Observe the direction of forces between a negatively charged Van de Graaff generator and a positively charged object. Direction of Force on Charged Objects Drag around a stationary charged object and observe the force on the stationary object when it is positive and negative. Dissolving Observe the impact on potential energy when a substance dissolves in water. Dissolving Experimental Explore how the polarity of molecules affects how they mix (or don’t) when combined. DNA to Protein Explore what DNA is and how proteins are synthesized from the genetic information stored in it. Earthquakes Around the World In this activity you will use a computer model to investigate earthquakes and volcanic eruptions. Electric Current Explore the relationships between voltage, current, and resistance that make up Ohm’s Law using molecular models of circuits. Electric Potential Energy and Charge Intensity Set the amount and type of charge on particles and compare the potential energy of the electric field that is generated. Electric Potential Energy and Type of Charge Set the charge of two particles and compare the potential energy of the electric field they generate as the particles are moved around. Electrons in Atoms and Molecules The interactions of electrons with matter are central to many technologies from transistors to sophisticated quantum computing. Electrostatics Discover how atoms can be charged, and manipulate charge and distance to examine Coulomb’s Law. Electrostatics: Maze Game Apply knowledge of the interactions between charged particles to guide an object through a maze. Elements and Polarity Compare the surface charges on various molecules and explore which atom types tend to cause uneven sharing of electrons. Energy of a Pendulum Set the initial height of a pendulum and observe how potential, kinetic, and thermal energy change during pendulum swings. Energy of a Spring Set the initial position of a mass on a spring and observe how potential, kinetic, and thermal energy change when the spring is released. Energy of Bond Formation (conceptual version) Explore how different elements come together to form bonds and compare changes in potential energy and forces. Evaporative Cooler How can the cooling effect of evaporating water be used for air conditioning? Excited States and Photons Investigate how atoms can be excited to give off radiation. Experiment with Ecosystems Experiment with virtual ecosystems and test your hypotheses about producer/consumer and predator/prey relationships. Exploring Electron Properties Compare the behavior of charged atoms and cathode ray particles (electrons). Exploring Hydrophobic Core Explore the structure of various proteins and see how the nonpolar amino acids form the core of many protein structures. Exploring Protein 3D Structure Explore a protein and its components using a simplified representation to see structure; or a view of all atoms to see full details. Factors Affecting London Dispersion Attractions Explore the role of size and shape in the strength of London dispersion attractions. Forming a Molecular Bond (conceptual version) Explore the potential and kinetic energy as two atoms form a covalent bond. Forming a Molecule Explore the balance of forces and electron distribution as two atoms are moved closer and further apart. Gas Laws Explore the interrelationships of pressure, temperature, and volume with atomic models. Gas Pressure in a Syringe Explore how a particle model of gases works to predict the behavior of a syringe under various conditions. Global Climate Change Model: Making Predictions About Future Climate Explore how changing human emissions may affect Earth’s temperature in the future. Greenhouse Effect in a Greenhouse Relate changes in sunlight to the temperature of air trapped in a container. Greenhouse Gases Use a computer model to explore how the earth’s atmosphere affects the energy balance between incoming and outgoing radiation. Greenhouse Light and Temperature By using a temperature sensor, we can relate changes in sunlight to the temperature of the air being trapped in a container. Hands-On Experiment with Sensor Data Collection Design and run your own hands-on experiment using a sensor. Analyze results in CODAP. Hands-On Experiment with Dual Sensor Data Collection Use two sensors to design and run your own hands-on experiment. Analyze results in CODAP. Heat and Light from Electricity How does energy change from one form to another? Heat Storage Depends on Size Measure how size affects the flow of heat energy from a warm to a cold object. Heat Storage Depends on Specific Heat Measure how specific heat affects the flow of heat energy from a warm to a cold object. How Can a Small Spark Start a Huge Explosion? Explore connections between electric forces and molecules using simulations, and explain energy transfers using the conservation of energy. How Does an Object Become Charged? Develop atomic-level causal models to explain observations of electrostatic interactions. How Electrons Move Discover the forces affecting the movement of electrons, including electric and magnetic fields. How Loud, How High? Explore the frequency, wavelength, amplitude and velocity of sound waves. Hydraulic Fracturing Model Explore how hydraulic fracturing is used to extract oil and natural gas and how the process may affect local aquifers. Hydrogen Bonds: A Special Type of Attraction Explore the polar molecule interactions known as hydrogen bonds. Intermolecular Attractions Explore how London dispersion attraction and dipole-dipole interactions explain the different boiling points of materials. Intermolecular Attractions and States of Matter Explore how states of matter are related to the strength of intermolecular attractions. Introduction to Quantum Mechanics Discover the quantum nature of electrons including their wave nature, tunneling abilities, and their bound and excited states. Is There Life in Space? Explore the question: Can there be life outside of Earth? Land Management Model Explore the effects of different land management strategies, terrain, and climate on erosion rate and soil quality. Launching a Satellite Study how to fire something into space. Leaf Photosynthesis This NetLogo model of leaf photosynthesis shows the macroscopic outcome of the reaction. Maintaining Steady Temperature Explore how temperature swings in a house can be moderated. Making and Breaking Bonds Explore the association and dissociation of diatomic molecules. Making and Breaking Bonds: The Effect of Temperature Observe the effects of temperature on chemical reactions. Making Heat Does change in concentration change the amount of heat released in a chemical reaction? Making Molecules Modify an existing molecule and observe how different atoms affect the electron distribution within the model. Measuring Heat Transfer Observe and measure the flow of heat energy from a warm to a cold object. Meiosis Learn how meiosis and fertilization shuffle the alleles that offspring inherit. Melting Ice What is the temperature of ice as it melts? Metal Forces Explore what happens when a force is exerted on a metallic material. Micelles Observe changes in potential energy as molecules self-assemble into micelles. Mixing Liquids Explore how mixing two different liquids together can result in less total volume by investigating at the molecular level. Modeling Transcription Explore how an mRNA copy is made of DNA. Modeling Translation Explore how a protein is made from an mRNA sequence. Molecular Geometry Use models of electron arrangement around atoms to discover how molecules form linear, trigonal planar, and trigonal pyramidal shapes. Molecular Self-Assembly Explore how molecules assemble themselves into defined patterns, a process called molecular self-assembly. Molecular Sorting Add various unknown molecules to oil and water, and observe how the molecules sort themselves in response to interactions with the surrounding environment. Molecular View of a Gas Explore the structure of a gas at the molecular level. Molecular View of a Liquid xplore the structure of a liquid at the molecular level. Molecular View of a Solid Explore the structure of a solid at the molecular level. Motion on a Ramp Study the motion of an object as it moves up and down a ramp. Mutations Explore how changing the DNA sequence can change the amino acid sequence of a protein. Mystery Plant Adaptation Learn that species are adapted to their environments. Variation in a species can help the species adapt to changes in the environment. Mystery Plants Mystery Learn about intraspecific differences and how variation in a population can help a species adapt to living in different environments. Natural Selection Students explore how changes in the environment affect both plants and animals in a simple ecosystem. Non-Bonding (conceptual version) Compare the electron distribution, potential energy, and forces of two interacting hydrogen atoms (which can bond) with two helium atoms (which do not). Oil and Water Explore the interactions that cause water and oil to separate from a mixture. Opposites Attract Change the charge on spheres to positive or negative and observe how charges affect the interaction between them. Parachute and Terminal Velocity Explore the factors that affect an object’s speed as it falls through the atmosphere. Parachute Experiment Measure the terminal velocity of a falling parachute using coffee filters and a motion sensor. Parachute Model Measure the terminal velocity of a virtual falling parachute and analyze data in CODAP. Pendulum Explore the factors that affect a pendulum’s motion. Pendulum and Spring Explore the motion of a pendulum suspended by a spring. Phase Change Explore what happens at a molecular level as substances change phase. Planet Hunting Model Explore how a star’s movement and light intensity are affected by an orbiting planet, and the characteristics of stars and planets important to habitability potential. Plants What do plants eat? This unit explores plants and how they make food. Plastic Forces Explore what happens when a force is exerted on a polymeric plastic material. Polar and Nonpolar Interface Observe how molecules with hydrophilic and hydrophobic regions move in a mixture of oil and water, and the effects on potential energy. Polarity and Attractive Strength Explore the role of polarity in the strength of intermolecular attractions. Polarization Explore how the types of atoms forming a bond affect the distribution of electrons and overall shape of the molecule. Population Explosion Study how populations stay in balance with their environment and respond to various factors such as food supply and predators Predators and Prey Students learn that selection pressure can lead to a change in the characteristics of a population. Probability Clouds Investigate the probability map of electron orbitals. Protein Folding Explore how hydrophobic and hydrophilic interactions cause proteins to fold into specific shapes. Protein Folding Exploring Generate proteins with different molecular properties and observe how their folding changes the potential energy of the system. Protein Partnering and Function Build “partnerships” between a protein and small molecules and explore the effects of surface charge, polarity, and shape on partnering. Quantum Tunneling Explore the unique concept of quantum tunneling and its importance to modern technology. Radiant Energy Flow Explore the energy balance between incoming and outgoing radiation on the earth. Ramp Game Race a virtual car down a ramp, create graphs and analyze data in CODAP (Common Online Data Analysis Platform). Reaction Between Hydrogen and Oxygen Atoms Observe a reaction between hydrogen and oxygen atoms, and watch how potential and kinetic energy change. Reaction Between Hydrogen and Oxygen Molecules Carefully observe changes in kinetic and potential energy as hydrogen and oxygen molecules react. Relative Humidity Measurement Measure relative humidity in the air by comparing the wet bulb and dry bulb temperatures Scanning Tunneling Microscopy Use a virtual scanning tunneling microscope to explore the quantum tunneling effect. Seeing Intermolecular Attractions Explore different types of attractions between molecules. Seeing Motion Investigate simple, straight-line motion. Seismic Explorer This model allows you to explore the pattern of earthquakes on Earth, including magnitude, depth, location, and frequency. Semiconductors Explore the structure and behavior of natural and doped semiconductors. Solar Oven Design, build, and test a solar oven. Solubility Explore molecular views of solvents and solutes to explain how substances dissolve. Spectroscopy Explore why excited atoms emit different wavelengths of radiation and learn how to identify atoms based on their unique atomic spectra. Spring Model Explore the factors that affect a spring’s motion. Spring and Mass Experiment Explore mass as an independent variable in a hands-on experiment with a spring-mass system. Spring and Mass Model Explore the motion of a mass hanging on a spring and how the amount of mass affects the period. States of Matter How do the forces and attractions differ between the states of matter? Sticking a Balloon to a Wall Explore the interactions between a charged balloon and a neutral wall at the atomic-level. Structure of an Atom Map the probable locations of electrons around a nucleus to understand probability distributions and the electron cloud model. Sunlight, Infrared, CO2 and the Ground Explore how solar radiation interacts with Earth’s surface and atmosphere. Target Game (Charge Magnitude / Force Relationship) Manipulate the magnitude of charges on two objects to get a third positively charged particle to hit a target. Target Game (Distance/Force Relationship) Drag the location of charges to get a positively charged particle to the target while observing forces and fields. Target Game (Free Play) Manipulate the location and magnitude of charges to get a positively charged particle to hit a target. The Temperature-Pressure Relationship Explore the relationship between the temperature of a gas and the pressure it exerts on its container. The Temperature-Volume Relationship Explore the relationship between the temperature of a gas and its volume. Tire Forces Explore what happens when a force is exerted on a rubber tire. Transistors: The Field Effect The field effect transistor is the most common type of transistor. Tree of Life Zoom down from what we can see with our own eyes to the macromolecules from which they are made Understanding Probability Maps Use a dart board simulation to understand probability distributions. Van de Graaff (VDG) Discharge Explore how the charging and discharging of a Van de Graaff generator occurs and changes in potential energy. Variations and Adaptations Students determine how climate can affect ecosystems. Vertical Temperature Gradients Explore the vertical temperature variations in a house due to natural convection. The Virtual Ecosystem Students learn that organisms with similar needs compete with one another for resources. The Virtual Field Learn the life cycle of organisms, variation within a species, and heritability of traits. The Virtual Greenhouse Students learn about basic needs of organisms. An organism thrives in specific environments that match its specific needs. Visualizing Electric Fields and Forces Model how electric fields change due to the number and placement of charged objects. The Volume-Pressure Relationship Investigate the relationship between the volume of a gas and the pressure it exerts on its container. Was Galileo Right? Explore the effect of gravity on objects of various mass during free fall. Water Model Explore how water moves through Earth’s layers and determine if wells can produce sustainable amounts of water while maintaining the underlying aquifer’s health. Well and Poorly Insulated Houses Compare the heating of two houses with different insulation values. What Are All Materials Made of? Evaluate particle nature of matter and use particle nature of matter to explain observations of liquids and gasses. What Are Factors that Affect the Interactions Between Objects? Explore and use models of electric fields to explain how charged objects interact with other objects. What Are Nature’s Building Blocks? Develop, revise, and interpret representations of atomic structure and interactions of charges within atoms. What Are Our Energy Choices? Explore the advantages and disadvantages of using renewable and non-renewable sources to generate electricity. What Happens to the Energy of Water Molecules During Hurricanes? Add energy to a model of how molecules interact by exploring phase changes of water. What Is Happening When a Spark Occurs? Students define potential and kinetic energy, explore energy transfer and energy conservation, and connect energy to atomic structure. What Is Pressure? Explore pressure at the atomic level. What Is the Future of Earth’s Climate? Examine climate data and models to predict Earth’s future climate. What Makes Water Special? Develop explanations connecting observations of water with properties that water exhibits. Where Does All the Energy in an Explosion Come From? Construct a model of chemical reactions involving energy and electrostatic interactions and compare reactions and changes in energy. Where Does the Energy of a Spark Come From? Define potential energy and explore the relationship between potential energy and fields. Where Is the Most Heat Lost? Use movable thermometers to discover where a house has poor insulation. Will the Air Be Clean Enough to Breathe? With more of the world becoming industrialized, will the air be clean enough to breathe? Will There Be Enough Fresh Water? As the human population has grown, water use has increased. Explore water movement and predict water availability. Wind Generator Build a wind turbine and test different blade designs to generate electricity. About the Next Generation Science Standards You can also download ourNGSS “fortune tellers” (PDF) for another fun way to find resources available for different paths. The Next Generation Science Standards provide a framework and examples for STEM learning. Grounded in the National Academy of Science’s thoughtful Framework for K-12 Science Education, these new K-12 science standards have been developed to provide students an internationally benchmarked science education, and signify a new direction for STEM education. They elevate the importance of Earth science, present engineering education as coequal with science education for the first time and emphasize a key set of Scientific and Engineering Practices and Crosscutting Concepts that should buttress all learning in these disciplines. The NGSS hold the potential for helping focus the current national concern for improving STEM education. They will undoubtedly help bring clarity and unity to the patchwork of state standards developed throughout the standards movement in the past decades. As this occurs, innovative educational technology will be a critical component in this STEM education revolution. The Concord Consortium and the NGSS Our NGSS Pathfinder provides numerous examples of how the NGSS—and especially its Practices and Crosscutting Concepts—are central to our work in STEM education. Over nearly two decades, we’ve been demonstrating how technology can make complex concepts more approachable, underscore important crosscutting ideas and engage students in the practices of science and engineering. Using computational models and probe-based activities, elementary students can watch biological evolution, middle school students can analyze and interpret data to understand the genetic basis of inheritance and high school and college students can argue from evidence in discussing interactions between molecules. With these technology-supported activities, students can engage in doing real science as they plan and carry out investigations, use models, analyze data and design solutions. Students also gain wide experience with crosscutting concepts—from scales in space and time to energy and systems—across domains in science, math and engineering. More about the NGSS Next Generation Science Standards Website Improving STEM Education with Next Generation Science Standards Meet NGSS with Concord Consortium Activities MW and the New Science Standards Next Generation Science Assessment project Using Technology to Enhance NGSS-Aligned Assessment Tasks for Classroom Formative Use
10773	https://www.sciencedirect.com/science/article/abs/pii/S0038092X19302956	Analytical formulation of effective heat transfer coefficient and extension of lumped capacitance method to simplify the analysis of packed bed storage systems - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (23) Cited by (17) Solar Energy Volume 183, 1 May 2019, Pages 606-618 Analytical formulation of effective heat transfer coefficient and extension of lumped capacitance method to simplify the analysis of packed bed storage systems Author links open overlay panel Vaibhav Ranmode, Manmeet Singh, Jishnu Bhattacharya Show more Add to Mendeley Share Cite rights and content Highlights •Design of thermal energy storage is important for CSP plants. •Simplification of the analysis of the packed bed storage system is attempted. •We extend the validity range of lumped capacitance method to larger Biot numbers. •An effective heat transfer coefficient is derived from the exact analytical solution. •The developed method is generalizable to three dimensions. Abstract Analysis of the transient temperature evolution during charging or discharging of the packed bed thermal storage systems is immensely simplified with the formulation of an effective heat transfer coefficient between the solid storage materials and the heat transfer fluid. It can cut significant computational cost which is otherwise required for a complete numerical simulation. The lumped capacitance method, the simplest of the available options, is restricted only to the low Biot number scenarios and hence is seldom applicable to any real system. The formulation of the effective heat transfer coefficient allows the extension of the lumped capacitance method for moderate Biot numbers. The present work develops such a formulation to simplify the analysis of packed bed storage systems through analytical route. Earlier attempts in this direction were made through weighted average time method which is inherently restricted to the simple one-dimensional heat conduction problems. On the other hand, we find out the effective heat transfer coefficient starting from a general three dimensional analytical solution of the transient heat conduction and proceed with the well-known one term approximation which acts as the basis of the Heisler charts. The method is not dimensionally restricted and hence we can include realistic three dimensional shapes such as cuboid, short cylinder etc. in the formulation. We validate our method by comparing the resulting temperature profiles for the one-dimensional geometries which have been attempted earlier by the weighted average time method. We also provide the accuracy estimation (as a function of increasing Biot number) against the full numerical simulation for the geometries where the weighted average time is not applicable. Therefore, the current study provides the tool for an inexpensive theoretical estimation for the transient heat transfer behaviour in the thermal storage tanks which has long term design implications particularly for the large scale concentrated solar thermal power plants. Introduction Solar energy harvesting is an attractive option for a greener future; nevertheless, the inevitable diurnal and seasonal variability associated with it makes the storage an indispensable part for large scale stand-alone units (Vijayakumar et al., 2006). In case of the solar photovoltaic panels, which convert sunrays into electricity, rechargeable battery packs are used for the storage. On the other hand, the thermal route of solar energy conversion requires thermal storage system to be utilized in off-sun hours (Köberle et al., 2015). Thermodynamically, storage of heat is more prudent compared to the storage of electricity as the self-discharge loss costs less as heat is a lower grade of energy. Moreover, other than conversion of heat into electricity through some heat-engine, thermal energy can be used for space or water heating, drying or as the process heat in industrial applications (Köberle et al., 2015, Pelay et al., 2017). The most wide-scale method of storing heat is through the sensible storage in packed bed storage tanks (Alva et al., 2017, John et al., 2013, Tiskatine, 2017). Therefore, several studies regarding thermal properties of the heat transfer fluid (Grosu et al., 2018, Chang, 2018), the effects of conduction, convection and radiation in packed bed storage tanks have been reported in recent years (Qian et al., 2018, Chang et al., 2017, Wang et al., 2018,, Geissbühler et al., 2019).Fig. 1 shows a schematic arrangement for such a tank during the charging and discharging operations (Reddy et al., 2017). The storage bed is usually packed with naturally occurring, low-cost, high specific heat solid materials such as rocks, pebbles, sand, wood etc. (Alva et al., 2017, John et al., 2013, Tiskatine, 2017). The heat transfer fluid (HTF) passes through the bed of such solid materials to dispose of the extra heat energy (charging) or to regain the heat (discharging) as shown in Fig. 1. In order to quantitatively evaluate the storage capacity or the thermal inertia during charge or discharge process, understanding the heat transfer behaviour between the packing materials and the heat transfer fluid is imperative. The variation of the temperature of the packing materials and the heat transfer fluid along the flow direction (vertically up or down) is schematically represented in Fig. 2 and the heat balance in an infinitesimally thin control volume is represented by the following equations (Bradshaw et al., 1970, Xu et al., 2012).ρ f C p,f V f∂T f∂t+U∂T f∂z=h S(T LM-T f)f o r h e a t t r a n s f e r f l u i d.ρ s C s V s∂T LM∂t=-h S(T LM-T f)f o r s t o r a g e b e d p a c k i n g s o l i d. Above equations were originally derived by Schumann (Flowing). Van Lew et. al (Van Lew et al., 2011, Li et al., 2011) has developed very efficient techniques to solve these equations (Eqs. (1), (2)) using the method of characteristics, which significantly reduces computation time without significant variation on accuracy. However, the heat balance in equations, i.e. Eqs. (1), (2) assume uniform spatial temperature distribution for the packing solids throughout the transient phase. The assumption is formally called a lumped capacitance assumption and it is valid only when the value of the Biot number is less than 0.1 (Ghajar, 1986). For larger Biot number, the spatial temperature variation in the packing material cannot be assumed to be uniform. It holds true only when the internal thermal resistance of solid filler material is large. Therefore, for the realistic scenarios where the Bi > 0.1, the above model can only be applied after crucial modification. In the current study, we attempt such a modification through analytical route so that Eqs. (1), (2) can still be used for the analysis of the packed bed storage tanks. Fully transient numerical simulation of the packed bed storages is one option which can provide accurate temperature evolution for both the heat transfer fluid and the packing material. However, the method is computationally expensive and hence not applicable at the design stage when the choice of the materials and dimensions go through iterations each of which requires quick estimates rather than accurate temperature history. Moreover, the complete transient simulation is required to be performed for each alteration of the operating parameter and geometry. Therefore, it is clear that a complete numerical simulation resulting in the accurate temperature evolution history is no substitute in the design stage of the packed bed storage systems. One effective bypass route for the numerical simulation is to estimate some modified value of the heat transfer coefficient from either the experimental correlations or through some analytical methods. This modified heat transfer coefficient can be used instead of surface heat transfer coefficient in Eqs. (1), (2) to extend its validity for larger Biot number scenarios. The concept of modifying the lumped capacitance method by finding an effective heat transfer coefficient was first introduced by Bradshaw et al. (1970). They carried out the analysis for Biot numbers higher than the lumped capacitance limit, through the method of weighted average time. The analysis was done for a solid sphere in fluid bath. The concept of modified heat transfer coefficient which takes care of the internal thermal resistance of the solid material was further extended by Xu et al. (2012) for other one dimensional geometries such as an infinite plate and an infinite cylinder. The weighted average time method is inherently restricted to the one dimensional geometries only. However, in case of a packed bed storage systems, the storage materials come in geometries which can hardly be considered one dimensional in terms of the heat transfer characteristics. The naturally occurring solids which are used as the packing materials can be of wide range of shape and size which can be approximated as spheres, cuboids, short cylinders etc. Singh et al., 2013, Singh et al., 2008) carried out experimental study on packed bed storage systems having large sized elements of different geometric shapes such as spheres, cubes and rectangular blocks of concrete. Packed beds of short cylindrical particles are important in chemical engineering applications (Zhang et al., 2006). Dorai et al. (2015) carried out fully resolved simulations of the flow through packed beds of short cylinders to study the energy and pressure loss. Therefore, to study the thermal performance of such beds through Eqs. (1), (2), a method of finding the effective heat transfer coefficient is needed which can be extended to the three dimensions. Through this work, we show that the route through the exact solution provides us with the flexibility of applying it for such three dimensional geometries. In the current study, we apply the one-term approximate solution of the transient heat conduction to obtain the expression for the effective heat transfer coefficient for more realistic three dimensional geometries. We compare the accuracy of the present method for the one-dimensional geometries with that of the weighted average time method (Xu et al., 2012). For the three-dimensional geometries, where no analytical alternative is available till date, we benchmark the current method against full numerical simulation of the heat transfer and fluid flow and highlight the current method’s effectiveness as an extremely inexpensive way of estimating the effective heat transfer coefficient in packed bed storage tanks. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Basic lumped capacitance model Consider a solid object initially at uniform temperature T i which is suddenly immersed in a fluid bath of constant temperature T∞. Let T be the temperature of solid at time t. Biot number (Bi) of the object is defined as,Bi=h L c k where h is the heat transfer coefficient, k is the thermal conductivity of material and L c is the characteristic dimension defined as the ratio of volume of the solid to its surface area. When the Bi is less than 0.1, the temperature gradient in the solid is neglected and Calculation of the effective heat transfer coefficient for a cuboid: Now we consider, a cuboid (2 L×2 H×2 W) shaped material for the analysis. Heat is transferred by convection from cuboid surface to the fluid which is maintained at a uniform temperature T∞. Therefore, governing equation for three dimensional transient heat conduction in cuboid is given as,ρ C∂T∂t=k∂2 T∂x 2+∂2 T∂y 2+∂2 T∂z 2 At time t=0, the whole object is at a uniform temperature T i.H e n c e,i n i t i a l c o n d i t i o n i s T x,y,z,0=T i Due to the symmetry and convection from surfaces of cuboid, the boundary conditions can Results and discussions The effective heat transfer coefficients as tabulated in Table 1, provides an inexpensive and easy method to estimate the heat interactions between the packing material and the heat transfer fluid in a packed bed sensible heat storage system. For very low Biot number, the approximation of lumped capacitance method provides a good estimate but fails for moderately high Biot numbers. We test the validity of the method formulated in the current study against the lumped capacitance model at the low Conclusions From the above discussions, we can conclude the following. •The current study focuses on extending the validity of the lumped capacitance method by finding an effective heat transfer coefficient for solid energy storage materials. The analysis is based on one term approximation of exact series solutions for the transient heat conduction in solids. •The work is of great significance to the simplification and analysis of heat transfer and energy storage in packed bed storage systems. •Unlike previous Acknowledgement The authors gratefully acknowledge the financial support from IMPRINT project number 7911 (Internal allocation: MHRD/ME/16408AB). Recommended articles References (23) C. Chang Enhanced heat transfer in a parabolic trough solar receiver by inserting rods and using molten salt as heat transfer fl uid ☆ Appl. Energy (2018) C. Chang et al. Comparative study of the transient natural convection in an underground water pit thermal storage Appl. Energy (2017) L. Geissbühler et al. An assessment of thermocline-control methods for packed-bed thermal- energy storage in CSP plants, Part 2: Assessment strategy and results Sol. Energy (2019,) Y. Grosu et al. Solar energy materials and solar cells unexpected effect of nanoparticles doping on the corrosivity of molten nitrate salt for thermal energy storage Sol. Energy Mater. Sol. Cells (2018) P. Li et al. Generalized charts of energy storage effectiveness for thermocline heat storage tank design and calibration Sol. Energy (2011) Y. Qian et al. International journal of heat and mass transfer comparative evaluation of heat conduction and radiation models for CFD simulation of heat transfer in packed beds Int. J. Heat Mass Transf. (2018) H. Singh et al. Performance of a packed bed solar energy storage system having large sized elements with low void fraction Sol. Energy (2013) R. Tiskatine Suitability and characteristics of rocks for sensible heat storage in CSP plants Sol. Energy Mater. Sol. Cells (2017) C. Wang et al. Numerical investigation of radiation transfer in packed bed of semitransparent coated spherical particles Int. J. Therm. Sci. (2018) B. Xu et al. Extending the validity of lumped capacitance method for large Biot number in thermal storage application Sol. Energy (2012) G. Alva et al. Thermal energy storage materials and systems for solar energy applications Renew. Sustain. Energy Rev. (2017) View more references Cited by (17) A performance analysis of the spray-type packed bed thermal energy storage for concentrating solar power generation 2022, Journal of Energy Storage Show abstract The concept of spray-type packed bed thermal energy storage (medium and high temperature thermal energy storage, from 200°C to 350°C, Alumina pellets+T66 fluid) has been proposed in our previous investigation. Its low-cost and high-efficiency shows its promising application prospect. Further investigation of the new concept is carried out both numerically and experimentally in this paper. First, the “charging-standby-discharging” cycle of a pilot spray-type packed bed is investigated experimentally. Second, the dynamic liquid holdup of the spray-type packed bed is correlated with the experiment data and is applied to build the mass balance equations. Based on the energy balance equations and mass balance equations, a three-phase one-dimensional transient model for the spray-type packed bed is established. Results from the experiment confirms the correctness of the numerical model. Based on the model, the liquid holdup of the packed bed during the “charging-standby-discharging” cycle is studied. At last, sensitivity analysis was carried out to study the influence of the parameters of the “charging-standby-discharging” process on the thermal performance of the spray-type packed bed. Values of these parameters can be found to achieve the best thermal performance of a spray-type packed bed. ### Wall impact on efficiency of packed-bed thermocline thermal energy storage system 2022, Energy Show abstract Packed-bed single-tank thermocline system with reduced cost is an alternative to the conventional two-tank system for thermal energy storage. This work systematically explores the wall impact on thermocline behavior of packed-bed tanks. For this purpose, adapted transient models were developed and fully exploited for the first time. Two tank configurations were investigated and compared: a high-temperature pilot-scale tank with a steel wall and a low-temperature lab-scale tank with a polycarbonate wall, both tanks being insulated by mineral wool. Results showed that the maximum energy stored in the wall at fully charged state can be up to 10% of the total stored energy. This part of stored energy has a negative impact on the discharging, causing up to 15% increase of the thermocline thickness. The energy stored in the insulation is very small so that this phase can be simplified as a thermal resistance in the modeling. The optimal wall parameters for packed-bed TES tanks were obtained that a thinner wall has a smaller impact on the energy and exergy efficiencies at the discharging cutoff time. The findings of study could provide useful design guideline for pack-bed thermocline TES tanks for different industrial applications. ### Application of numerical methods for the design of thermocline thermal energy storage: Literature review and critical analysis 2022, Journal of Energy Storage Citation Excerpt : Moreover, the effect of surface roughness was found to be non-negligible at high particle-to-fluid conductivity ratios . The analytical solution for different cases relevant to packed bed sensible energy storage (sphere, short and long cylinders, cuboids, plates) is given in . However, in most cases, the use of empirical equations is preferred . Show abstract Sensible or hybrid sensible-latent thermal energy storage systems are becoming mature systems and are the most common choice in high temperature applications. Among the different configurations for these systems, thermocline storages are a well-established one, presenting some advantages such as low cost and simple layout. The design of such systems is often achieved through numerical modeling and only a few experimentally validated systems are available. Considering the wide recognition of this technology also in other new fields of applications, such as new generations district heating and cooling or advanced power cycles, a deep knowledge of the numerical methods available and their main features is needed. In the present paper, an overview of numerical methods and codes applied for the design of thermocline storages is presented, discussing the constituent equations, additional correlations needed and the software and solvers employed. The effect of operating parameters according to the parametric analyses available is reported and a critical discussion is made, regarding the selection of the most suitable model for the various applications, as well as the proper range of design parameters. ### Thermocline packed bed thermal energy storage system: a review 2022, Renewable Energy Production and Distribution Recent Developments Show abstract Solar energy as a promising and renewable energy source has been applied in diverse filed for human activities. Thermal energy storage (TES) is applied to overcome the intrinsic deficiency of solar energy by migrating the dispatching between the energy supply and demand. The thermocline packed-bed TES system acted as dual-media is alternative to conventional two-tank system, exhibiting excellent cost and heat capacity advantages. It has been widely used, not only for high-temperature applications such as for the concentrated solar power, but also for industrial waste heat recovery at medium-temperature and for low temperature applications in built-environment. This book chapter provides an overview of thermocline packed-bed TES systems, covering the storage mechanisms, packing configurations, and the application temperature ranges. The influencing factors, that is, operational, geometrical, and thermophysical parameters, on the thermal performance of storage system are also surveyed. Some numerical simulation models are presented based on dimensional numbers, phase numbers, ignoring/nonignoring temperature gradient inside particles, etc. In all, the thermocline packed-bed tank with latent-sensible fillers is prospective to be promoted in future industries due to the advantages of low cost, stable performance, and high energy density. The existing challenges are also identified, including the thermocline stability, the packing topology, and the control strategy under variable operational parameters. ### A review on sensible heat based packed bed solar thermal energy storage system for low temperature applications 2020, Solar Energy Citation Excerpt : Using both approaches, it was obtained that the grille sphere has high overall heat transfer efficiency as random packing having a drawback of high pressure drop and simple cubic packing provides low value of Nusselt number. Ranmode et al. (2019) modified the lumped capacitance method through investigating the effective HTC using one term approximation of exact series solution for unsteady heat conduction under solid. The proposed method is suitable to assess the effective HTC of plate, cube, cuboid, spheres, cylinders and even of complicated 3-D solid geometries. Show abstract Solar thermal energy is one of the categories of renewable energy source and it is quantitative abundant and qualitative superior. It is capable to fulfil the global thermal energy demand and it emerges as a competitive option with the conventional equipment’s if these systems are incorporated with storage units. The solar thermal storage unit can also improve the equipment performance in terms of a smooth supply of energy with fluctuated solar energy collection as solar radiation varies throughout a day. Packed bed storage system is one of the feasible techniques to store the solar thermal energy which can be assembled with various solar thermal applications of low temperature as well as high temperature. The present review covers the sensible heat based packed bed solar thermal energy storage systems for low temperature applications. It includes a brief discussion about packed bed, its thermodynamic background, temperature distribution within packed bed, various design parameters affecting its performance and its analysis based on the energy as well as exergy efficiency. Various experimental and numerical investigations for performance analysis of PBSS have also been reported. The economic feasibility of the PBSS and comparison of sensible heat based PBSS with latent heat based PBSS has also been discussed in detail. ### Numerical modeling of transient heat transfer in a phase change composite thermal energy storage (PCC-TES) system for air conditioning applications 2020, Applied Thermal Engineering Citation Excerpt : 2 Table 1 summarizes the major parameters of the mathematical numerical model. For rough, order-of-magnitude estimates of the thermal transient in the PCC-TES, one might consider a lumped-capacitance model (i.e. [28–33]) in which the entire stack of PCC slabs is assumed to be reside at a spatially uniform temperature. Here the number of transfer units accounts for cooling of the EG from inlet to outlet. Show abstract This paper presents a detailed numerical analysis to describe the transient heat transfer in a phase change composite-thermal energy storage (PCC-TES) system exchanging heat with a heat transfer fluid. The “PCC-TES” is precisely composed of 78% low temperature paraffin, namely n-Tetradecane (C14H30) and 22% expanded graphite) and a heat transfer fluid (namely, ethylene glycol). Such detailed numerical approach to precisely describe the heat exchange between this specifically proposed phase change composite thermal energy storage (PCC-TES) and the heat transfer fluid has not been yet addressed in literature. The mathematical modeling of the PCC-TES involves formulating two PDEs that represent conduction within the PCC (phase 1) and advection within the heat transfer fluid (phase 2). The two coupled PDEs were solved with a finite difference method. Numerics were implemented using Fortran. Results were validated using experimental data and demonstrated acceptable agreement and an accurate representation of this specific transient heat transfer problem. The variations between experimental time traces of temperature and numerically calculated data are estimated at approximately (4–9%) and attributed to the heat loss in the actual experiment. Accordingly, the slabs in the actual experiment needed a longer time to reach the temperature of the Ethylene Glycol entering the PCC-TES structure from the inlet at the top. The paper also demonstrated that a lumped-parameter ODE model can perform an incredibly good job of predicting the enthalpy uptake of the PCC precisely matching the PDE results for the enthalpy uptake of the PCC due to the high thermal conductivity of the graphite. The proposed PCC-TES system will be integrated with air conditioning systems to efficiently meet cooling demand and reduce emissions. Previous work by our research group has demonstrated that the [AC + PCC-TES] hybrid design can reduce compressor size by 50%, double the compressor efficiency (COP) during mid/off-peak hours, reduce electricity consumption by 30%, and cut CO2 emissions by 30% – all without compromising overall system performance or building comfort level. View all citing articles on Scopus View full text © 2019 International Solar Energy Society. Published by Elsevier Ltd. All rights reserved. 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10775	https://www.youtube.com/watch?v=ApNbPjOAQ_g	How to divide a circle into six equal parts Arthur Geometry 92400 subscribers 648 likes Description 63877 views Posted: 1 Mar 2020 Learn how to split a circle into 6 equal sectors and how to construct a regular hexagon inscribed in. This YouTube channel is dedicated to teaching people how to improve their technical drawing skills. It focusses on drawing figures from the geometric plane to descriptive geometry and also different systems of technical drawing representation. If you subscribe, click, like or leave a comment you will be helping us to grow our channel and help more people with their technical drawing skills. Thank you in advance. Dubbed by Frank Shaw. Music by Antonio Fernández Ruiz. Geometry #HowtoDraw Make a Donation: . 14 comments Transcript: [Music] hello today we are going to learn how to divide any circle into six equal parts the first step we are going to draw a vertical diameter which costs the circumference of the circle at two points a and D as can be seen here the next step with our compass set to the length of the radius of the circle and centered on the point a we're going to scribe an arc until it cuts the circumference at two points as can be seen here and now we are going to repeat this process this time setting our compass on point D and with the same radius we scribe another arc until it cuts our circle at two more points as can be seen in the video are two arcs pass through our center point O the next step is the name the final four divisions B C F and E and if you join these points a b c d e and f as i have done with a red line you will get a rectangular hexagon inscribed in a circle and joining these six points to the center point 0 with the blue line we will get six equal divisions or circular sectors which can be seen here highlighted in the video so I hope this video has helped you please press like or subscribe to our YouTube channel thank you until the next time [Music] you
10776	https://hexacalculator.com/calculators/mathematics/geometry/rhombus-area-calculator	Rhombus Area Calculator \| hexacalculator hexacalculator.com Home Finance Mathematics Physics Health Daily Help Home Mathematics Geometry Rhombus Area Calculator Rhombus Area Calculator Diagonal 1 (D1) Units Diagonal 2 (D2) Units Area Units Share this calculator! Get widget to embed on your site! On this page Introduction How to use the Rhombus Area Calculator? What is a Rhombus? How is the area of the Rhombus Calculated? Examples Introduction Welcome to our Rhombus Area Calculator! which will help you calculate the Rhombus’ area quickly and accurately. A Rhombus is a two-dimensional closed figure with four sides, where opposite sides are parallel, and all sides are also of equal length. The rhombus is considered a special case of a parallelogram, i.e. an equilateral parallelogram. The area of the rhombus is the space that the shape occupies, and it is dependent on the length of the diagonals of the rhombus. We measure The rhombus’ area in square units, and generally we use units including sq cm, sq m, sq feet, sq inch,etc. How to use the Rhombus Area Calculator? Using the Rhombus Area calculator, you can calculate the area of the rhombus by inputting the values of the length of both diagonals of the rhombus. The variables in the Rhombus Area Calculator include: Diagonal (D1) The length of the first diagonal of the Rhombus. Diagonal (D2) The length of the second diagonal of the Rhombus Area of the Rhombus (A) We can calculate the area of the rhombus using the formula A=1 2×d 1×d 2\text{A} = \normalsize \dfrac{1}{2} \times d_1 \times d_2 Where, d1 = Length of Diagonal 1 of Rhombus d2 = Length of Diagonal 2 of Rhombus What is a Rhombus? A rhombus is a closed polygon with four sides, and these four sides are of equal length. So, the opposite sides of the Rhombus are parallel to each other. So, you might ask what is the difference between a square and a Rhombus? The answer is that the angles of a rhombus need not necessarily have to be right angles, but for the square, the angles have to be right angles. You could also say that the square is a case of Rhombus where all the angles of the Rhombus happen to be equal to 90 degrees. In summary, let’s look at the properties of the Rhombus: All sides are equal Opposite sides are parallel Opposite angles of a rhombus are equal The sum of all interior angles in 360 degrees Diagonals bisect each other The sum of two adjacent angles is equal to 180 degrees. In addition, there are different names for the rhombus, like diamond or lozenge. How is the area of the Rhombus Calculated? The total space occupied by the Rhombus in a two-dimensional plane is the area of the rhombus. It could also be thought of as the number of square units required to fill the region inside the rhombus. Hence the measurement will be in square units. We can calculate the area of the Rhombus using the following formula. A=1 2×d 1×d 2\text{A} = \normalsize \dfrac{1}{2} \times d_1 \times d_2 Where, d1 = Length of Diagonal 1 of Rhombus d2 = Length of Diagonal 2 of Rhombus Examples Let’s say there is a Rhombus, that has diagonals with lengths d1 = 6cm and d2 = 10cm. Find the area of the rhombus. We can calculate the area of the rhombus using the following formula A=1 2×6 c m×10 c m=60 c m 2 2=30 c m 2\begin{aligned} \text{A} &= \normalsize \dfrac{1}{2} \times 6cm \times 10cm \[10pt] &= \normalsize \dfrac{60\;cm^2}{2} \[10pt] &= 30\; cm^2 \end{aligned} Therefore, the area of the Rhombus is 30 square cm. Share this calculator! Get widget to embed on your site! Author hexacalculator design team Our team blends expertise in mathematics, finance, engineering, physics, and statistics to create advanced, user-friendly calculators. We ensure accuracy, robustness, and simplicity, catering to professionals, students, and enthusiasts. Our diverse skills make complex calculations accessible and reliable for all users. You might also like Mathematics Geometry ### Rhombus Perimeter Calculator Quickly and easily calculate the perimeter of the Rhombus using our Rhombus perimeter calculator! Try it out now! Mathematics Geometry ### Square Perimeter Calculator Calculate the square's perimeter quickly and easily using our nifty and user-friendly Square Perimeter Calculator! Try it out now! Mathematics Geometry ### Pentagon Area Calculator Calculate the pentagon's area quickly and accurately using our user-friendly Pentagon Area Calculator! Try it out now! 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10777	https://www.khanacademy.org/math/geometry/hs-geo-circles/xff63fac4:hs-geo-proofs-with-inscribed-shapes/v/right-triangles-inscribed-in-circles-proof	Proof: Right triangles inscribed in circles (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content High school geometry Course: High school geometry>Unit 8 Lesson 9: Proofs with inscribed shapes Proof: Right triangles inscribed in circles Inscribed quadrilaterals proof Proof: radius is perpendicular to a chord it bisects Proof: perpendicular radius bisects chord Math> High school geometry> Circles> Proofs with inscribed shapes © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Proof: Right triangles inscribed in circles SCCCR.Math: 8GCM.GS.MGSR.7.2, GS.MGSR.7.1, GS.MGSR.7.2 Google Classroom Microsoft Teams About About this video Transcript Proof showing that a triangle inscribed in a circle having a diameter as one side is a right triangle.Created by Sal Khan. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Celine 13 years ago Posted 13 years ago. Direct link to Celine's post “At 1:20, why does sal use...” more At 1:20 , why does sal use theta instead of a, b, c, etc? Answer Button navigates to signup page •1 comment Comment on Celine's post “At 1:20, why does sal use...” (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Paul Mullan 13 years ago Posted 13 years ago. Direct link to Paul Mullan's post “We use theta for angles i...” more We use theta for angles in math. It is not so important now, but when you take trigonometry, you will use it all the time. Comment Button navigates to signup page (16 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Levi Jesus 5 years ago Posted 5 years ago. Direct link to Levi Jesus's post “Couldn't we demonstrate i...” more Couldn't we demonstrate in a more simplified way? Given that central angle / 2 = inscribed angle The angle of the diameter (180 °) is the central angle that subtends the arc represented by half the circumference. Tracing a triangle with the diameter being one of the sides, we would automatically form an inscribed angle that also subtends the same arc as the angle of the diameter. Thus, that inscribed angle would be half of 180 ° (90 °), that is, a right angle Answer Button navigates to signup page •Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer HZWang 5 years ago Posted 5 years ago. Direct link to HZWang's post “Hi Levi, the answer is, Y...” more Hi Levi, the answer is, YES! That is a great proof. There doesn't always have to be only one proof for a theorem, so knowing and understanding different ideas and concepts that build different proofs is always good. Sorry for the late reply, but I hope that helped! ~Hannah Comment Button navigates to signup page (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more gameboyjustin 8 years ago Posted 8 years ago. Direct link to gameboyjustin's post “I don't think this video ...” more I don't think this video described what an Inscribed Angle is at all. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer A Dumb Nerd 3 years ago Posted 3 years ago. Direct link to A Dumb Nerd's post “Wouldn't that make all tr...” more Wouldn't that make all triangles right triangles? Because you could draw a circle around the triangle with the longest side being the diameter, and according to this proof, the opposite angle would be right? Or am I going crazy? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer David Severin 3 years ago Posted 3 years ago. Direct link to David Severin's post “No, this only proves that...” more No, this only proves that the triangle is a right triangle if and only if one of the sides is a diameter. There are other triangles that can be drawn that are inscribed in circles that are not right triangles. If you draw a circle and put any three points such that any two points do not form a diameter, then the triangle will not be a right triangle. 1 comment Comment on David Severin's post “No, this only proves that...” (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Jus do 7 years ago Posted 7 years ago. Direct link to Jus do's post “I have a question. A regu...” more I have a question. A regular hexagon with area 54√3 (inch square) is inscribed in a circle. What is the circumference of the circle? Express answer in terms of π. The answer is 25π. Can you explain the answer. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer kubleeka 7 years ago Posted 7 years ago. Direct link to kubleeka's post “A regular hexagon can be ...” more A regular hexagon can be be cut into 6 equilateral triangles. In your case, each triangle will have area 54√3/6=9√3. The area of an equilateral triangle with side length s is s²√3/4. Since we know the areas of these triangles, we can solve for their side lengths: s²√3/4=9√3 s²/4=9 s²=36 s=6 So the triangles have sides of length 6. And when follow a diameter of the circumcircle, we trace two sides of equilateral triangles. So the circle has diameter 6·2=12, and radius 6. So its circumference is 2·π·6=12π. If your circle had circumference 25π, it would have area of roughly 490. 54√3 is roughly 93, so the inscribed hexagon would have to occupy just under 1/5th of the circles area, which doesn't sound reasonable. Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more esadaltindis123 a year ago Posted a year ago. Direct link to esadaltindis123's post “couldn't you just say tha...” more couldn't you just say that the 90 degree angle is the inscribed angle that subtends the same arc as the 180 degree diameter and prove it like that? Answer Button navigates to signup page •2 comments Comment on esadaltindis123's post “couldn't you just say tha...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Timothy 13 years ago Posted 13 years ago. Direct link to Timothy's post “He mentions that in anoth...” more He mentions that in another video he proved something about the angles subtending another angle or something like that... I'm new to inscribed triangles, can anyone help? What video was this introduced in? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Paolo Miguel Bartolo 13 years ago Posted 13 years ago. Direct link to Paolo Miguel Bartolo's post “Try "Inscribed and Centra...” more Try "Inscribed and Central Angles", also in the topic "Circles" Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Megha 7 years ago Posted 7 years ago. Direct link to Megha's post “how do we know that the 2...” more how do we know that the 2(theta) lets call it "b" is twice theta? And don't we already know that theta is 45 degrees because he bisected the diameter which was 180 degrees? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer David Severin 7 years ago Posted 7 years ago. Direct link to David Severin's post “I do not know if you have...” more I do not know if you have gone through this video yet, but that is the inscribed angle theorem, "The measure of the central angle is double the measure of the inscribed angle" which Sal covers in It would only be 45 if Sal perpendicularly bisected the diameter to form a right angle, not just any bisector - Sal does not want to limit it to this even though it "appears" close, but the purple arc does look slightly smaller than the other part of the arc. The point of the video is that it does not matter what the angle theta is, the angle opposite the diameter will always be 90 degrees. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more George dionne 12 years ago Posted 12 years ago. Direct link to George dionne's post “So any triangle within a ...” more So any triangle within a circle where one side of the triangle is the diameter length of a circle and the other sides form a vertex on the circumference of the circle will always be a right triangle? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer benbsellslv 3 years ago Posted 3 years ago. Direct link to benbsellslv's post “why does the vertex have ...” more why does the vertex have to be in the center. cant the vertex be to one side so that the height of the triangle is not the radius? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer David Severin 3 years ago Posted 3 years ago. Direct link to David Severin's post “While it is drawn in the ...” more While it is drawn in the middle, Sal's proof does not rely on the particular placement of the point on the semicircle, In fact, your assumption is that the height of the triangle is the radius, but this is neither assumed nor part of the proof itself. That would only be true if the two formed triangles were both right triangles (the point is 90 degrees from each of the two ends of the diameter). He shows the middle angle as 2 theta, not 90 degrees as you appear to be assuming. Thus, the radius that divides it into two triangles does not have to be the height of the triangle, it just may be easier to draw when it is approximately 90 degrees. If it were too close to one of the ends of the diameter, it would be harder to write all the info into the triangle, but it would in fact still work the same no matter where the point is. Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Video transcript Let's say we have a circle, and then we have a diameter of the circle. Let me draw my best diameter. That's pretty good. This right here is the diameter of the circle or it's a diameter of the circle. That's a diameter. Let's say I have a triangle where the diameter is one side of the triangle, and the angle opposite that side, it's vertex, sits some place on the circumference. So, let's say, the angle or the angle opposite of this diameter sits on that circumference. So the triangle looks like this. The triangle looks like that. What I'm going to show you in this video is that this triangle is going to be a right triangle. The 90 degree side is going to be the side that is opposite this diameter. I don't want to label it just yet because that would ruin the fun of the proof. Now let's see what we can do to show this. Well, we have in our tool kit the notion of an inscribed angle, it's relation to a central angle that subtends the same arc. So let's look at that. So let's say that this is an inscribed angle right here. Let's call this theta. Now let's say that that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here. This is a radius. This is the same radius -- actually this distance is the same. But we've learned several videos ago that look, this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago. So this is going to be 2theta. It's the central angle subtending the same arc. Now, this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I flipped it over it would look like that, that, and then the green side would be down like that. And both of these sides are of length r. This top angle is 2theta. So all I did is I took it and I rotated it around to draw it for you this way. This side is that side right there. Since its two sides are equal, this is isosceles, so these to base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle. Now let me see, I already used theta, maybe I'll use x for these angles. So this has to be x, and that has to be x. So what is x going to be equal to? Well, x plus x plus 2theta have to equal 180 degrees. They're all in the same triangle. So let me write that down. We get x plus x plus 2theta, all have to be equal to 180 degrees, or we get 2x plus 2theta is equal to 180 degrees, or we get 2x is equal to 180 minus 2theta. Divide both sides by 2, you get x is equal to 90 minus theta. So x is equal to 90 minus theta. Now let's see what else we could do with this. Well we could look at this triangle right here. This triangle, this side over here also has this distance right here is also a radius of the circle. This distance over here we've already labeled it, is a radius of a circle. So once again, this is also an isosceles triangle. These two sides are equal, so these two base angles have to be equal. So if this is theta, this is also going to be equal to theta. And actually, we use that information, we use to actually show that first result about inscribed angles and the relation between them and central angles subtending the same arc. So if this is theta, that's theta because this is an isosceles triangle. So what is this whole angle over here? Well it's going to be theta plus 90 minus theta. That angle right there's going to be theta plus 90 minus theta. Well, the thetas cancel out. So no matter what, as long as one side of my triangle is the diameter, and then the angle or the vertex of the angle opposite sits opposite of that side, sits on the circumference, then this angle right here is going to be a right angle, and this is going to be a right triangle. So if I just were to draw something random like this -- if I were to just take a point right there, like that, and draw it just like that, this is a right angle. If I were to draw something like that and go out like that, this is a right angle. For any of these I could do this exact same proof. And in fact, the way I drew it right here, I kept it very general so it would apply to any of these triangles. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. 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10778	https://www.scribd.com/document/526346596/5-continuous-discrete-compounding	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 Upload 0 ratings0% found this document useful (0 votes) 1K views3 pages Continuous & Discrete Compounding The document discusses continuous and discrete compounding. Discrete compounding calculates interest at the end of fixed periods like months or years. 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10779	https://www.quora.com/What-is-the-difference-between-proportional-limit-and-yield-point	Something went wrong. Wait a moment and try again. Yield Strength Proportional Limit Stress Strain Curve Material Science Research Mechanics of Materials Mechanical Engineering Stress and Strain Strength of Materials 5 What is the difference between proportional limit and yield point? · The terms "proportional limit" and "yield point" are both related to the mechanical behavior of materials, particularly in the context of stress-strain curves. Here are the key differences: Proportional Limit Definition: The proportional limit is the maximum stress at which a material behaves elastically and follows Hooke's Law. Up to this point, the relationship between stress and strain is linear. Behavior: Beyond the proportional limit, the material may still deform elastically, but the stress-strain relationship becomes nonlinear. Measurement: It is determined from the initial linear portion o The terms "proportional limit" and "yield point" are both related to the mechanical behavior of materials, particularly in the context of stress-strain curves. Here are the key differences: Proportional Limit Definition: The proportional limit is the maximum stress at which a material behaves elastically and follows Hooke's Law. Up to this point, the relationship between stress and strain is linear. Behavior: Beyond the proportional limit, the material may still deform elastically, but the stress-strain relationship becomes nonlinear. Measurement: It is determined from the initial linear portion of the stress-strain curve. Yield Point Definition: The yield point is the stress level at which a material begins to deform plastically. Beyond this point, the material will not return to its original shape when the load is removed. Behavior: The yield point indicates the transition from elastic deformation (where the material can return to its original shape) to plastic deformation (where permanent deformation occurs). Measurement: It can be defined in two ways: Upper Yield Point: The maximum stress reached during the yield phase. Lower Yield Point: The stress at which the material continues to deform plastically without an increase in stress. The proportional limit is the end of the linear elastic region, while the yield point marks the beginning of plastic deformation. In many materials, especially ductile ones, the yield point occurs after the proportional limit. Ala'a Al-Afghani PhD student in Mechanical Engineering, University of California, Merced · Author has 144 answers and 378.8K answer views · 7y Proportional limit is the point at which the linear relationship: stress = modulus strain stops being true. The yield point is the point after permanent deformation will occur and the part if unloaded will not return to its original shape. Usually the proportional limit occurs on the stress strain diagram slightly before the yield point. Sometimes they are so close that people use them interchangeably. Piyush Patil Studied Materials Science and Engineering at University of Stuttgart · Author has 62 answers and 315.3K answer views · Updated 8y Related What is the difference between yield point and elastic limit? Elastic Limit and Yield Point may convey the same state of material that is but the latter is used for an engineering convenience. Elastic limit is a value of stress upto which material can be deformed elastically under load, after unloading it will return to it's original dimension. Beyond elastic limit, material will start deform plastically which is characterized with permanent deformation. It is marked by the breakage of bonds. Technically, it is a stress at which first of such bond breaks. Limitations of measuring such delicate phenomena has made engineers to define another point at whic Elastic Limit and Yield Point may convey the same state of material that is but the latter is used for an engineering convenience. Elastic limit is a value of stress upto which material can be deformed elastically under load, after unloading it will return to it's original dimension. Beyond elastic limit, material will start deform plastically which is characterized with permanent deformation. It is marked by the breakage of bonds. Technically, it is a stress at which first of such bond breaks. Limitations of measuring such delicate phenomena has made engineers to define another point at which permanent deformation can be measured which is known as Yield Point (Offset). The yield strength is defined as the stress which will produce a small amount of permanent deformation, generally equal to a strain of 0.002 or 0.2%. This is an Engineering stress-strain curve for a ductile material. Source : Mechanical metallurgy by G.E. Dieter. So, for your question, in the region between the elastic limit and the yield point, material just deforms plastically, although on an extremely small scale. P.S. Source that I have mentioned also contains explanation about stress-strain curve from the perspective of molecular level. It is quite an interesting read. Related questions What is the difference between yield point and elastic limit? What's the difference between elastic limit, tensile strength and proportional limit? What is the relationship between proportional limit and yield strength? Are they different concepts altogether? If not, why aren't they related to each other? For the usual steel, when the yield stress is reached but not yet entirely throughout the cross section, at what point in the stress strain curve is this located? Does this occur between the proportional limit and the yield point? How would you define the terms proportional limit, elastic limit, yield strength, proof stress and fracture limit of the stress- strain graph of ductile materials? Deva Uppara struggling · Upvoted by Vishnu Karath , B.Tech Mechanical Engineering, Government Engineering College, Thrissur (2016) · 9y Related What is the difference between upper yield point and lower yield point of a stress-strain curve for mild steel? Upper yield point is the max load at which deformation starts, starting of deformation means dislocations are started moving in the material. So this type of phenomenon is called permanent deformation by slip ( slip mechanism). As the slip is taking place in the material , it offers less resistance to the material and hence curve falls slightly ( stress is the measurement of resistance offered by the material during the application of load). And it reaches to some stress ( lower yiled point stress) which is the minimum stress required to maintain the deformation in the mateial.. And at the lower Upper yield point is the max load at which deformation starts, starting of deformation means dislocations are started moving in the material. So this type of phenomenon is called permanent deformation by slip ( slip mechanism). As the slip is taking place in the material , it offers less resistance to the material and hence curve falls slightly ( stress is the measurement of resistance offered by the material during the application of load). And it reaches to some stress ( lower yiled point stress) which is the minimum stress required to maintain the deformation in the mateial.. And at the lower yield point for the low carbon steels ( mild steels) the stress strain cure is in some wave nature , this is because to break bonds with impurites while dislocations are moving out of the material , hence resistance increases and decreases periodically after that strain hardening takes place which increases resistance slowly by increasing of dislocations in the material... Hope it will help.. Nikhil Nayak Student of life · Author has 139 answers and 999.9K answer views · 11y Related What is the difference between yield point and elastic limit? They are pretty much the same but also not the same.And encyclopedia britannica agrees with me,if I say they are the same : "ELASTIC LIMIT is the maximum stress within a solid material that can arise before the onset of permanent deformation. When stresses up to the elastic limit are removed, the material resumes its original size and shape. Stresses beyond the elastic limit cause a material to yield or flow. For such materials the elastic limit marks the end of elastic behaviour and the beginning of plastic behaviour. YIELD POINT is the amount of stress in a solid at the onset of permanent defo They are pretty much the same but also not the same.And encyclopedia britannica agrees with me,if I say they are the same : "ELASTIC LIMIT is the maximum stress within a solid material that can arise before the onset of permanent deformation. When stresses up to the elastic limit are removed, the material resumes its original size and shape. Stresses beyond the elastic limit cause a material to yield or flow. For such materials the elastic limit marks the end of elastic behaviour and the beginning of plastic behaviour. YIELD POINT is the amount of stress in a solid at the onset of permanent deformation. The yield point, alternatively called the elastic limit, marks the end of elastic behaviour and the beginning of plastic behaviour. When stresses less than the yield point are removed, the material returns to its original shape." Yet a difference arises because of the definition...The elastic point marks the END OF ELASTICITY, while the yield point marks the BEGINNING OF PLASTICITY..So obviously the yield point is outside the elastic region.But since the jump between the 2 is so small,they're almost the same thing.Almost.When a material is at the EL,and you apply a force (F→0),it jumps to the YP. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. 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When mild steel material is subjected to tension test, then it passes through various stages before fracture. Discussion is limited to Plastic region only.. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus permanent deformation or permanent set when load is removed. These two points are termed as upper and lower yield points respectively. From figure you can easily understand that, When ductile material is stretched beyond elastic point, material starts to show plastic behavior. A point at which Maximum load or stress required to init When mild steel material is subjected to tension test, then it passes through various stages before fracture. Discussion is limited to Plastic region only.. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus permanent deformation or permanent set when load is removed. These two points are termed as upper and lower yield points respectively. From figure you can easily understand that, When ductile material is stretched beyond elastic point, material starts to show plastic behavior. A point at which Maximum load or stress required to initiate the plastic deformation of material such point is called as Upper yield point. And a point at which minimum load or stress required to maintain the plastic behavior of material such a point is called as Lower yield point. Upper yield point is unstable in nature but lower yield point is stable in nature. so while designing the components we consider the lower yield point of material. Related questions What happens in between elastic limit point and yield point? Are yield point and elastic limit the same point in a stress-strain graph? What is the difference between flash point and ignition tempertaure? What is a major difference among science, engineering and technology? Why does the proportional limit increase for 2nd loading after prior yield? Vishvendra Singh Tomar Stress \| IC Engine \| CFD \| R&D \| Engineer \| DTU (DCE) · Upvoted by Akshith Gowda , Master’s Mechanical Engineering & Automobile Engineering, Technische Hochschule Deggendorf (2023) · Author has 113 answers and 1.5M answer views · 7y Related What is the difference between ultimate stress and yield stress? Suppose, I’ve a mild steel specimen (Long Cylindrical Piece), when I apply tensile force to it using UTM (Universal Testing (Tensile) Machine), it will start deforming (elongating) in direct proportion to the force’s magnitude. Fig 1: UTM with a mild steel specimen. Now, if I plot stress-strain curve for the above specimen, as per the physical changes observed in the specimen. I’ll get the following curve Fig. 2. If in the above curve, you observe, there are two points P1 & P2. These are very crucial points. From here, we obtain the definition of ‘Yield’ & ‘Ultimate’ Stress. For a moment, forget a Suppose, I’ve a mild steel specimen (Long Cylindrical Piece), when I apply tensile force to it using UTM (Universal Testing (Tensile) Machine), it will start deforming (elongating) in direct proportion to the force’s magnitude. Fig 1: UTM with a mild steel specimen. Now, if I plot stress-strain curve for the above specimen, as per the physical changes observed in the specimen. I’ll get the following curve Fig. 2. If in the above curve, you observe, there are two points P1 & P2. These are very crucial points. From here, we obtain the definition of ‘Yield’ & ‘Ultimate’ Stress. For a moment, forget about any concept. Just think about, what these words mean in an English dictionary. Yield means: Surrender! & Ultimate means: Final! Yield Stress or Surrender Stress is the point where material surrenders! When we apply force initially, metal remains in proportional range, it means it can regain its size. But as soon as it crosses P1, it can not regain its size back. It surrenders! This is called Yield Stress! Ultimate or Final Stress is, when metal crosses P2, the metal size was already elongated after P1, now the shape of cross-section also changes. Neck-Formation will start from P2 itself. Metal is saying, this is final load Man! I can not take it any more. This is Ultimate Stress! For example, the values of Yield and Ultimate Stress for AISI 1020 (Mild) Steel are: 350 MPa and 420 MPa. ~ Concepts should be called concepts only when you get them easily. ~ VST :) Sponsored by CDW Corporation How do updated videoconference tools support business goals? Upgrades with CDW ensure compatibility with platforms, unlock AI features, and enhance collaboration. Vishakh Rajendran M.S. in Aerospace and Aeronautical Engineering, Nanyang Technological University · Author has 682 answers and 3M answer views · 8y Related How would you define the terms proportional limit, elastic limit, yield strength, proof stress and fracture limit of the stress- strain graph of ductile materials? Proportional limit - Limit till which stress is Proportional to strain . Linear relationship. Slope of graph in this region is a constant and is the young’s modulus. Hookes law is obeyed here. Not the same as elastic limit. This occurs first in the stress strain curve. Elastic limit - Limit within which body reaches original configuration on load removal. Slope of graph is not a constant. This is not the Proportional limit. This occurs after the Proportional limit. Thus, there is a small region between elastic limit and Proportional limit. Yield limit - Point which signifies the start of plastic Proportional limit - Limit till which stress is Proportional to strain . Linear relationship. Slope of graph in this region is a constant and is the young’s modulus. Hookes law is obeyed here. Not the same as elastic limit. This occurs first in the stress strain curve. Elastic limit - Limit within which body reaches original configuration on load removal. Slope of graph is not a constant. This is not the Proportional limit. This occurs after the Proportional limit. Thus, there is a small region between elastic limit and Proportional limit. Yield limit - Point which signifies the start of plastic deformation. But this does not signify the end of limit of elastic deformation. End of elastic limit is signified by elastic limit. Thus, there is a small elasto plastic region between the elastic and yield limit where deformation is neither true elastic nor true plastic. Resilience - Area under the stress strain curve within the proportionality limit (energy absorbed during the linear limit) Proof resilience - Maximum energy that can be absorbed during the elastic limit , that is till the Proportional limit. Proof stress - Stress corresponding to the Proportional limit. Vivek Dhakal Studied at IOE, Pulchowk Campus · Updated 5y Related What is the difference between the elastic limit and proportional limit? Proportionality limit : It is the limit beyond which linear variation of steel ceases. Elastic limit: It is the limit up to which a specimen regains its original shape on removal of applied load. Upto this point, Hook’s law is applicable. Actually, elastic limit comes just after the proportionality limit in stress - strain curve of but it is difficult to determine the two limits separately in elastic range near the yield point of steel. So, elastic limit is assumed to coincide with proportionality limit. 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How I got started At the start of the year, I set myself a simple challenge: make $930 in one month without picking up another job. I was tired of side hustle lists that required selling stuff, investing money I didn’t have, or waiting months for results. That’s when I stumbled on a site called Freecash. Signing up took less than a minute, and I even got a $5 bonus just for creating my account. Honestly, I was skeptical - but since it didn’t cost anything, I gave it a try. If you want to test something similar, creating an account is simple and low-risk - you could see if it fits your routine too. What Freecash really is In plain words, Freecash is a platform that rewards you for completing online offers. Some are quick (like surveys), while others are longer but pay more (like reaching levels in mobile games). The big difference I noticed compared to other apps: everything is upfront. 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Final thoughts For me, Freecash wasn’t about becoming rich - it was about hitting a very specific goal: $930 in a month. I wasn’t expecting it to work, but it turned out to be exactly what I needed at the right time. If you ever want to test something like this for yourself, you can explore Freecash casually and see if it works with your daily routine. Syed Hasan Shirazi Civil Engineer · Author has 1.6K answers and 8M answer views · 7y Related What is the difference between proportional point and the elastic point in elasticity? For steel, stress-strain curve has various stages as follows: Elastic Stage Plastic Stage Strain Hardening Stage Necking & Rupture In the elastic stage there are four distinct points of interest, as follows: Proportional Limit: This is the point up to which the stress-strain curve is linear. The slope of this linear line is known as Young’s Modulus . Elastic Limit: After Proportional Limit comes the Elastic Limit. This is the point on the stress-strain curve up to which the deformation is elastic in nature i.e. within this stress range, steel will not undergo any permanent deformation due applied load For steel, stress-strain curve has various stages as follows: Elastic Stage Plastic Stage Strain Hardening Stage Necking & Rupture In the elastic stage there are four distinct points of interest, as follows: Proportional Limit: This is the point up to which the stress-strain curve is linear. The slope of this linear line is known as Young’s Modulus. Elastic Limit: After Proportional Limit comes the Elastic Limit. This is the point on the stress-strain curve up to which the deformation is elastic in nature i.e. within this stress range, steel will not undergo any permanent deformation due applied load i.e. when the load is removed, strain will become zero. This point is beyond the point of proportionality which means that the stress-strain curve is no longer linear. Upper Yield Point: This point comes after Elastic Point. At this point steel begins to yield i.e. flow plastically without any increase in load. Upper Yield Point can be considered as the stress required to initiate yielding. This is a momentary point in the stress-strain curve hence it does not have much significance. Lower Yield Point: After Upper Yield Point is reached, the stress rapidly drops to a lower value i.e. Lower Yield Point. It can be considered as the stress required to sustain yielding. Plastic deformation continues beyond this point at this constant stress. The stress corresponding to Lower Yield Point is known as Yield Stress. The above points are visually depicted in the figure below: A = Proportionality Point B = Elastic Limit C = Upper Yield Point D = Lower Yield Point (Yield Stress) E = Start of Strain Hardening Stage F = Ultimate Stress G = Fracture Point Admin at · Author has 310 answers and 847.3K answer views · 5y Related What is 'yield point phenomenon'? Hello everyone. Thank you everyone. Hello everyone. Thank you everyone. Brian Chattillon Author has 152 answers and 30K answer views · 7y Proportional limit= set vale that can NOT be over breeched. Yield Point= the event horizon of ones goals and aspirations to grandeur. Feby Philip Abraham a technology enthusiast fascinated by how materials behave · Upvoted by Hitul Ramani , M.tech. Mechanical Engineering, Gujarat Technological University · 9y Related What is the difference between upper yield point and lower yield point of a stress-strain curve for mild steel? Upper yield point corresponds to the load that is required to initiate yielding. Lower yield point corresponds to the min load that is required to maintain yield. Normally we use the lower yield point to determine the yield strength of the material being tested, cause the upper yield is momentary. In the above fig it can be seen that when the stress-strain curve for a low-carbon steel is plotted the upper yield value is almost limited to a point. But the value of lower yield is more defined. Here is a stress strain curve which has a better defined yield value (lower). I would like to humbly Upper yield point corresponds to the load that is required to initiate yielding. Lower yield point corresponds to the min load that is required to maintain yield. Normally we use the lower yield point to determine the yield strength of the material being tested, cause the upper yield is momentary. In the above fig it can be seen that when the stress-strain curve for a low-carbon steel is plotted the upper yield value is almost limited to a point. But the value of lower yield is more defined. Here is a stress strain curve which has a better defined yield value (lower). I would like to humbly disagree with Kaustav Sen. Yielding starts at the upper yield point. PS: If you are a student I would recommend two books for reference: Mechanics of materials - Beer, Johnston, Dewolf and Mazurek Mechanics of materials - Gere and Goodno if available in your college library. Related questions What is the difference between yield point and elastic limit? What's the difference between elastic limit, tensile strength and proportional limit? What is the relationship between proportional limit and yield strength? Are they different concepts altogether? If not, why aren't they related to each other? For the usual steel, when the yield stress is reached but not yet entirely throughout the cross section, at what point in the stress strain curve is this located? Does this occur between the proportional limit and the yield point? How would you define the terms proportional limit, elastic limit, yield strength, proof stress and fracture limit of the stress- strain graph of ductile materials? What happens in between elastic limit point and yield point? Are yield point and elastic limit the same point in a stress-strain graph? What is the difference between flash point and ignition tempertaure? What is a major difference among science, engineering and technology? Why does the proportional limit increase for 2nd loading after prior yield? What is the difference between elastic limit and yield point and what happens when the stress is between these two values? What is elastic limit for S235 steel and S355? Why does yield point elongation occur (elongation at constant stress)? What is the difference between the elastic limit and proportional limit? What is the difference between yield stress and yield strength? What is the difference between limit order and stoploss limit order? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10780	https://www.freemathhelp.com/forum/threads/the-binomial-theorem-example-problem-proof-explanation-needed.134042/	New posts Search forums Menu Log in Register Install the app Forums Free Math Help Intermediate/Advanced Algebra You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. The Binomial Theorem example problem/proof explanation needed Thread starter alexapprend Start date A alexapprend New member Joined : Apr 7, 2022 Messages : 5 #1 Hello, I am struggling through a HL IB textbook independently and don't understand the parts circled- I feel like I have pretty good non-calculator maths so I can manipulate algebra but I'm very lost. From the textbook:My notes (mostly just ended up copying from the textbook, the bits I circled show me understanding some of the steps they skipped, but whilst I can see for example 2(2-1) does in fact equal 1!(3-1)! I don't understand it as a general rule, will it work for all numbers? What maths theory is it using? The asterisk is where I stopped understanding. You can also see this circled above. Thanks in advance A AvgStudent Full Member Joined : Jan 1, 2022 Messages : 256 #2 alexapprend said: Hello, I am struggling through a HL IB textbook independently and don't understand the parts circled- I feel like I have pretty good non-calculator maths so I can manipulate algebra but I'm very lost.From the textbook:View attachment 32111 My notes (mostly just ended up copying from the textbook, the bits I circled show me understanding some of the steps they skipped, but whilst I can see for example 2(2-1) does in fact equal 1!(3-1)! I don't understand it as a general rule, will it work for all numbers? What maths theory is it using? The asterisk is where I stopped understanding. You can also see this circled above. View attachment 32112 Thanks in advance Click to expand... Hi alexapprend, They're using this fact: [imath]n!=n(n-1)![/imath] So [imath]2!=2\times 1 = 2\times 1![/imath] Going backwards [imath] 2\times 1!=2\times 1 = 2![/imath] You can try for higher n to see it more clearly. Hope this helps Last edited: pka Elite Member Joined : Jan 29, 2005 Messages : 11,990 #3 alexapprend said: Hello, I am struggling through a HL IB textbook independently and don't understand the parts circled- I feel like I have pretty good non-calculator maths so I can manipulate algebra but I'm very lost.From the textbook:View attachment 32111 My notes (mostly just ended up copying from the textbook, the bits I circled show me understanding some of the steps they skipped, but whilst I can see for example 2(2-1) does in fact equal 1!(3-1)! I don't understand it as a general rule, will it work for all numbers? What maths theory is it using? The asterisk is where I stopped understanding. You can also see this circled above. Click to expand... First you need to practice evaluating binomial coefficients. If [imath]N~\&~k[/imath] are non-negative integers such that [imath]0\le k\le N[/imath] then [imath]\dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}[/imath] That symbol is read [imath]N\text{ choose }k[/imath]. Here are three to practice on: 1)[imath]\dbinom{10}{3}[/imath]; 2)[imath]\dbinom{5}{4}[/imath; & 3)[imath]\dbinom{7}{1}[/imath] EXAMPLE to follow: [imath]\dbinom{15}{9}=\dfrac{15!}{9!(15-9)!}[/imath] To see if you did it correctly SEE HERE A alexapprend New member Joined : Apr 7, 2022 Messages : 5 #4 AvgStudent said: Hi alexapprend,They're using this fact: [imath]n!=n(n-1)![/imath]So [imath]2!=2\times 1 = 2\times 1![/imath]Going backwards [imath] 2\times 1!=2\times 1 = 2![/imath]You can try for higher n to see it more clearly.Hope this helps Click to expand... Thank you for responding I do know and understand this rule, you can see I used it in the example above in the bit I did understand (red box). I've shown what this would still mean for me even if I were to accept that 10!=1 and ignore it (obviously this is true but I just worry it's meant to be used to group or simplify things): A alexapprend New member Joined : Apr 7, 2022 Messages : 5 #5 pka said: First you need to practice evaluating binomial coefficients.If [imath]N~\&~k[/imath] are non-negative integers such that [imath]0\le k\le N[/imath] then [imath]\dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}[/imath]That symbol is read [imath]N\text{ choose }k[/imath].Here are three to practice on: 1)[imath]\dbinom{10}{3}[/imath]; 2)[imath]\dbinom{5}{4}[/imath; & 3)[imath]\dbinom{7}{1}[/imath]EXAMPLE to follow: [imath]\dbinom{15}{9}=\dfrac{15!}{9!(15-9)!}[/imath]To see if you did it correctly SEE HERE Click to expand... Thank you for the examples although a few didn't come out I do know how to do this (and I double checked on your examples) it was in the previous exercise. I've posted pictures below: Although I didn't understand 4b as the shows, maybe you can explain that too ? Some similar sort of factorial simplification questions from the exercise two before this one so you can see what I know: Thank you for taking the time to respond. I'm guessing I must be missing something obvious because both you and the other person have replied with a concept I already understand but for some reason I don't understand the working in the original question ? BigBeachBanana Senior Member Joined : Nov 19, 2021 Messages : 2,277 #6 alexapprend said: Thank you for the examples although a few didn't come out I do know how to do this (and I double checked on your examples) it was in the previous exercise. I've posted pictures below:View attachment 32117 Although I didn't understand 4b as the shows, maybe you can explain that too ? View attachment 32118 View attachment 32119 Some similar sort of factorial simplification questions from the exercise two before this one so you can see what I know: View attachment 32120 Thank you for taking the time to respond. I'm guessing I must be missing something obvious because both you and the other person have replied with a concept I already understand but for some reason I don't understand the working in the original question ? Click to expand... For 4b) Consider the different scenarios when the team must have at least 1 boy and 1 girl. You will have either: 1) three boys and one girl 2) two boys and two girls, or 3) one boy and three girls Count the number of ways for each scenario then sum them up. You must log in or register to reply here. 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10781	https://www.mathsisfun.com/algebra/quadratic-equation-graph.html	Explore the Quadratic Equation A Quadratic Equation a, b, and c can have any value, except that a can't be 0. Try changing a, b and c below to see how they change the curve. Notice the "roots" (the solutions to the equation, shown as dots on the graph). ../geometry/images/parabola-ball.js?mode=pts Read more about the Quadratic Equation. Explore Move the a, b and c slider bars to explore the properties of the quadratic graph. Look at The effect of changes in a The effect of changes in b The effect of changes in c The effect of negative values of a The effect of positive values of a What happens when a=0 ? See if you can get the curve to just touch the x-axis (y=0) Can you get the "roots" −1.0 and 1.0 ? Roots The "roots" are the solutions to the equation. When the curve crosses the x-axis (y=0) you will have: two solutions or ONE solution (if it just touches) When the curve does not cross the line there are still solutions, but: the two solutions include Imaginary Numbers Quadratic Equation Explore the Properties of a Straight Line Graph Algebra Index Copyright © 2024 Rod Pierce
10782	https://articles.outlier.org/what-are-critical-points	Skip to content If you could change one thing about college, what would it be? Save money Graduate faster Better quality online classes Flexible schedule Access to top-rated instructors Outlier Articles Home Calculus Critical Points - Your Calculus 101 Guide 08.14.2023 • 6 min read Rachel McLean Subject Matter Expert In this article, learn what are critical points, the different types, and follow a step-by-step explanation of how to find them. Share In This Article What Is a Critical Point? 3 Types of Critical Points How To Find Critical Points Examples of Critical Points Practice With the Best What Is a Critical Point? What are critical points? In calculus, a critical point is a point on a function where the derivative of the function is either zero or undefined. We say that x=c is a critical number of the function f if either f′(c)=0, or f′(c) is undefined. We say that (c,f(c)) are the critical points of the function. The critical point definition is sometimes confused with the definition of critical numbers. The critical numbers of a function f are the x-values c in the domain of the function for which f’(c)=0 or f’(c) is undefined. The critical points of a function are the points on a graph whose coordinates are (c,f(c)). Critical points are a big deal because they can help us ‌identify relative extrema, like relative minima and maxima. These are the peaks and valleys on a graph. All local extrema occur at critical points, but not all critical points are local extrema. To determine which critical points are local minima or maxima, we use the First Derivative Test or Second Derivative Test. When thinking about critical points note that functions can have both increasing and decreasing intervals. Suppose a function f is differentiable on the interval I. Then: If f’(x)>0 for each x on I, then f is increasing on I. If f’(x) < 0 for each x on I, then f is decreasing on I. If f’(x)=0 for each x on I, then f is constant on I. Critical points that are relative extrema occur at the points where the curve changes direction from increasing to decreasing, or decreasing to increasing. In other words, the first derivative f’(x) can only change sign from positive to negative, or negative to positive, by crossing points where f’(x)=0 or points where f’(x) does not exist. 3 Types of Critical Points Three types of critical points exist in single-variable calculus: Relative Maxima For a function f, relative maxima occur at the critical points (c,f(c)) where both of the following criteria are satisfied: f’(c)=0 or f’(c) does not exist f’(x) > 0 on the left of x = c, and f’(x) < 0 on the right of x = c Relative Minima For a function f, relative minima occur at the critical points (c,f(c)) where both of the following criteria are satisfied: f’(c)=0 or f’(c) does not exist f’(x) < 0 on the left of x = c, and f’(x) > 0 on the right of x = c Vertical Tangents Some critical points are neither relative maxima nor relative minima. At vertical tangents, the derivative is undefined. These are ‌inflection points, which are the points where the curve changes concavity from concave up to concave down, or concave down to concave up. Take the second derivative on either side of the critical point to determine if the sign of the second derivative changes from positive to negative or negative to positive at this point. If it does, then the critical point is an inflection point. Note: not all inflection points are critical points! How To Find Critical Points In this section, we will learn how to find critical numbers, how to find critical points, and how to classify them. Take the first derivative of the function. Set the first derivative function equal to zero and solve for x. Also solve for any points where the first derivative function is undefined. These x-values c are the critical numbers of the function. Plug c into f(x) to obtain the y-values. The coordinates (c,f(c)) are the critical points of the function. To determine if the critical point is a relative maxima, relative minima, or neither, you can use the First Derivative Test. This test involves analyzing the sign of the first derivative of the function on either side of the critical point. If the sign of the first derivative changes from positive to negative at the critical point, it's a relative maximum. The function changes from increasing to decreasing at this point. If the sign of the first derivative changes from negative to positive at the critical point, it's a relative minimum. The function changes from decreasing to increasing at this point. You can also use the Second Derivative Test to determine if the critical point is a local maximum or local minimum. Find each stationary point of the function, which are the critical points c where f’(c)=0. For each stationary point c: If f"(c) is positive, then the function has a relative minimum at c. If f"(c) is negative, then the function has a relative maximum at c. If f"(c) is zero or non-existent, then the test is inconclusive. To determine if a critical point is an inflection point, use test points to establish if the second derivative f’’(x) changes sign from positive to negative, or negative to positive, on either side of x=c. If f”(x) > 0 on the left of x = c, and f”(x) < 0 on the right of x = c, then the function is changing from concave up to concave down. This is an inflection point. If f”(x) < 0 on the left of x = c, and f”(x) > 0 on the right of x = c, then the function is changing from concave down to concave up. This is an inflection point. If f”(x) is the same sign on either side of x=c, then the function does not change concavity at c, and c is not an inflection point. Examples of Critical Points Let’s walk through some critical point examples. Critical Points of a Function Identify all critical points of the polynomial f(x)=x3+2x2. Step 1 The first step is to take the first derivative of the given function. Using the power rule for derivatives, f’(x)=3x2+4x. Step 2 The second step is to set the first derivative equal to zero and solve. By factoring, this gives us our critical numbers. Note there are no points where the first derivative is undefined in this case. 3 x 2 + 4 x = 0 x ( 3 x + 4 ) = 0 x 0 , x = − 3 4 Step 3 The third step is to plug our critical numbers into our original function. This gives us the y-coordinates of our critical points. So, our critical points are (0,0) and (−34,2732). Critical Points on a Graph We can also identify critical points on the graph of a function. Consider the below graph of the function y=2x4+2x3+4. Can you identify its critical points? A local minimum is at x=−3, and so we have a critical point at (−3,−9.5). While there are no further relative extrema, we can make another horizontal tangent line at x=0 where the function changes concavity. So, we have another critical point at (0,4). This is an inflection point. Note that there is another inflection point at x=−2. However, this is not a critical number since there is no horizontal or vertical tangent here. Critical Points of Multivariable Functions In multivariable calculus, functions have more than one variable. The critical points of the function z=f(x,y) are the points (a,b) such that the partial derivatives fx(a,b)=0 and fy(a,b)=0. Just like one-variable functions, finding a critical point of a multivariable function is not a guarantee relative extrema is at that point. As with single-variable calculus, there are 3 kinds of critical values: Relative Maxima: If f(a,b)>=f(x,y) for all points (x,y) in the region, then f(a,b) is a relative maximum. Relative Minima: If f(a, b) <= f(x, y) for all points (x, y) in the region, then f(a, b) is a relative minimum. Saddle Point: Saddle points are not local extrema. They appear as a relative maximum when approached from one direction, and a relative minimum when approached from the other direction. Below is the saddle point of z=x2−y2. Practice With the Best Experience game-changing results in your critical point calculus with Outlier’s Calculus I course. Instruction is 100% online and Zoom-free! Choose from three world-class instructors, including Tim Chartier of Davidson College, Hannah Fry of University College London, and John Urschel of MIT. Earn 3 college credits towards your degree for every course you complete. Credits are received from the University of Pittsburgh, a top 50 school. And, transferring credits to your school of choice is easy! Exam windows are flexible, and lectures are viewable on-demand, anywhere. If you do the work and don’t pass, you’ll receive a full refund. Explore Outlier's Award-Winning For-Credit Courses Outlier (from the co-founder of MasterClass) has brought together some of the world's best instructors, game designers, and filmmakers to create the future of online college. Check out these related courses: Calculus I The mathematics of change. Explore course Calculus I The mathematics of change. Explore course Intro to Statistics How data describes our world. Explore course Intro to Statistics How data describes our world. Explore course Intro to Microeconomics Why small choices have big impact. Explore course Intro to Microeconomics Why small choices have big impact. Explore course Share Related Articles Calculus Rolle's Theorem: Definition, Examples & Conditions Learn about Rolle's Theorem conditions, Lagrange’s Mean Value Theorem, and differentiable and continuous functions. We’ll also practice with solved examples. Rachel McLean Subject Matter Expert Calculus Understanding the Squeeze Theorem Learn what the squeeze theorem is, how to prove it, and practice with some examples and tips. Rachel McLean Subject Matter Expert Calculus Fraction Integration Techniques Learn how to integrate a fraction using partial fractions, trigonometric identities, and substitution. There’s also solved examples and FAQs. Rachel McLean Subject Matter Expert Further Reading ### What Is the Difference Between Calculus and Algebra? ### Best Integral Shortcuts: Solve & Master Integrals ### Methods of Integration: Techniques & Examples ### Integration by Parts Explained ### Disk Method: Definition, Formula & Examples ### The Arc-Length Formula: A Step-By-Step Guide
10783	https://limnu.com/sketch-easy-90-degree-rotate-vectors/	Sketch: Easy 90-degree Rotation of Vectors – Limnu Home Features Teams Tutors - Teachers - Schools Whiteboard API Pricing FAQ & Support Educator Plan About Customer Stories Professional Services Blog Contact Us Press Room Login SIGN UP FREE FROM THE BLOG Sketch: Easy 90 degree vector rotation June 12, 2015 Tags: Math, Using LimnuBy David Hart I was putting clickable name tags into Limnu the other day – they show up when multiple people are drawing, and someone draws somewhere you can’t see. A name tag appears on the border of your screen, and you can click it to zoom over to their view. As part of figuring out where the name tag goes, I needed to rotate a vector by 90°. Normally rotating vectors involves matrix math, but there’s a really simple trick for rotating a 2D vector by 90° clockwise: just multiply the X part of the vector by -1, and then swap X and Y values. TRY LIMNU You can try Limnu with no sign up or credit card required. Even better, get a free account and start brainstorming with your team. Try Limnu Recent Posts Get everyone on the same page with Limnu's Leading Feature Limnu enterprise plan enables a test prep company to tutor online Video conferencing now available on Limnu whiteboards Introducing Presentations on Limnu Whiteboards Customer Spotlight: Team uses Limnu to collaborate on image processing algorithms Limnu is being used by thousands of people around the world. Check out our customer stories for some unique ways that people are using Limnu. Customer Stories Categories Communication (4) Customer Stories (6) data visualization (2) Features (11) In the News (22) Limnu for Teams (1) Math (3) Pins (1) Remote Teams (8) Sketch (4) Slack Whiteboard (2) Stylus (2) Using Limnu (16) Archives November 2022(1) September 2022(1) May 2019(1) April 2019(1) October 2018(2) August 2018(4) January 2017(1) December 2016(1) November 2016(2) October 2016(1) September 2016(2) July 2016(4) June 2016(9) May 2016(2) April 2016(3) March 2016(8) February 2016(8) January 2016(2) December 2015(6) November 2015(3) October 2015(1) September 2015(2) August 2015(1) July 2015(3) June 2015(7) Limnu: The online whiteboard you've been looking for. © 2018, Limnu. \| Privacy Policy \| Terms of Service \| Press Room ACCEPT Limnu uses cookies to give you the best online experience. By using our website you agree to our use of cookies in accordance with our cookie policy. Privacy Policy
10784	https://www2.gov.bc.ca/gov/content/health/practitioner-professional-resources/bc-guidelines/sleep-apnea	Skip to main navigation Accessibility Statement Home/Health/Practitioner & Professional Resources/BC Guidelines/Obstructive Sleep Apnea: Assessment and Management in Adults More topics BC Guidelines About the Guidelines External Review of Guidelines Continuing Professional Development (CPD) Credits Guidelines by Alphabetical Listing Partner Guidelines Guidelines Eligible for Incentive Payments Addictions and Substance Use Guidelines by Topic Cardiovascular Diagnostic Imaging High Ferritin and Iron Overload Emergency Endocrine System Gastrointestinal System Geriatric Medicine Head and Neck Laboratory Mental Health Oncology Pediatric Palliative Care Preventative Health Respiratory System Rheumatological and Musculoskeletal Systems Urological System Chronic Pain Trauma Informed Practice Tobacco Use Disorder Chronic Kidney Disease – Diagnosis and Management Obstructive Sleep Apnea: Assessment and Management in Adults Last updated on September 19, 2023 Effective Date: October 21, 2021 Recommendations and Topics Scope Key Recommendations Types of Sleep Apnea Epidemiology Risk Factors Risks of Untreated or Undertreated OSA History and Physical Examination Testing and Referral Interpretation of Results OSA Management Follow-Up Methodology Resources Scope This guideline applies to adult patients 19 and older with suspected obstructive sleep apnea (OSA). While sleep apnea may occur in 1-4% of children1, pediatric diagnosis and management is beyond the scope of this guideline. For a brief pediatric overview, see Appendix A: Sleep Apnea in Children. Other conditions that may contribute to daytime sleepiness are also out of the scope of this guideline. Please see Appendix B: Other Conditions Associated with Daytime Sleepiness. TOP Key Recommendations History If OSA is suspected, conduct a detailed history and a physical examination, focused on the upper airway.2 The STOP-Bang questionnaire can be used to help determine if a patient is at increased risk of moderate to severe OSA.2 While the most common symptom of OSA is excessive daytime sleepiness, the clinical presentation can vary. Completion of the Epworth Sleepiness Scale is recommended.2 Patients with untreated OSA may have increased perioperative morbidity.3 Consider appropriate screening when referring patients for potential surgery (e.g. STOP-Bang Questionnaire). See also the BC Surgical Rehabilitation Toolkit for further information. [Expert opinion] Testing and Referral Home Sleep Apnea Test (HSAT) should not be used to screen asymptomatic patients. HSAT is only recommended for the diagnosis of OSA in symptomatic patients who are determined to be at an increased risk of moderate-to-severe OSA, and who have no exclusion criteria (see Requisition).2 A negative or equivocal HSAT does not exclude OSA. If an HSAT is negative, inconclusive or technically inadequate, and OSA is suspected, polysomnography is recommended.2 BC now has a Standard Requisition for HSAT that referring practitioners are required to use. Symptomatic patients for which the HSAT is not the appropriate diagnostic test should be referred for a sleep disorder consultation for polysmnography.2 [Expert Opinion] Management and Follow-up When assessing whether a patient should be treated, it is important to consider the severity of the symptoms, presence and severity of any comorbid disease, presence of any safety-critical occupation and the results of all sleep studies. OSA is a serious chronic disease that warrants regular follow-up, short-term to ensure initial compliance and response to treatment and long-term to confirm continued effectiveness.3 Patients with OSA may be prone to drowsiness while driving. Physicians caring for these patients should be familiar with BC’s Driver Medical Fitness Information for Medical Professionals. [Expert Opinion] Surgery for OSA, including minimally invasive techniques to reduce tissue volume, incurs typical surgical risks. Since there may be effective medical alternatives to surgery, prior referral to a sleep disorder physician is recommended. TOP Types of Sleep Apnea There are four types of sleep apnea: Obstructive sleep apnea (OSA) is the most common type of sleep apnea encountered in primary care. It is characterized by complete cessation or transient reduction in breathing with maintained or increasing respiratory effort. Home Sleep Apnea Testing (HSAT) should only be performed on patients who present with an increased risk of moderate-to-severe OSA. Other less common types of sleep apnea include: Central sleep apnea is characterized by complete cessation or transient reduction in breathing with absent respiratory effort. This occurs in patients with neurological disease or in association with drug/substance abuse or high altitude. Cheyne-Stokes breathing is characterized by a crescendo-decrescendo pattern of respiration between central apnea. This frequently occurs in patients with heart failure or cerebrovascular disease. Complex sleep apnea is characterized by the persistence or emergence of central apneas or hypopneas when treated with Continuous Positive Airway Pressure (CPAP or BiPAP) and obstructive apnea has been resolved. This may occur in up to 10% of patients with OSA treated with CPAP. Sleep hypoventilation is a sleep disorder characterized by significant hypercapnia and hypoxemia during sleep. This usually occurs in patients with OSA and morbid obesity or obstructive or restrictive lung disease. These patients usually require urgent evaluation.4 TOP Epidemiology OSA is associated with poor quality of life 6,8,9 and has been linked to severe chronic health conditions such as obesity, diabetes, metabolic syndrome 3,10 and neuro-psychiatric problems. 10 Moderate–severe OSA is associated with an increased risk of cardiovascular disease, resulting in hypertension, coronary disease, stroke, heart failure, and atrial fibrillation.5 OSA is widely underdiagnosed; 86% to 95% of individuals found in population surveys with clinically significant OSA report no prior OSA diagnosis.6 The Canadian Health Measures Survey results from 2016 and 2017 showed:7 6.4% of Canadian adults had been diagnosed with OSA by a health care professional Those at particular risk include: Older adults (60 to 79 years, 3 times more likely) Males (2 times more likely to be diagnosed compared to females) Males reporting snoring, trouble breathing or high neck circumference (≥17 inches) Females reporting fatigue, insomnia, or high body mass index (>35kg/m2) At the clinically relevant Apnea Hypopnea Index (AHI) level of 15 or greater per hour, the prevalence of OSA in the elderly population can be very high - up to 49% in some studies.8,9 TOP Risk Factors10 Risk factors for OSA include, but are not limited to: Down syndrome11 family history of obstructive sleep apnea Mandibular hypoplasia (retrognathia, micrognathia) low-lying soft palate (i.e., high Mallampati Score) large tongue tonsillar hypertrophy upper body obesity with a large neck size East Asian origin12 Parkinson’s disease13 traumatic brain injury14 nasal and nasopharyngeal obstruction15 neuromuscular disease16 Marfan syndrome17 polycystic ovarian disease18 TOP Risks of Untreated or Undertreated OSA 2,19-21 Untreated or undertreated OSA pose serious risks and have been associated with: daytime sleepiness, impaired quality of life, motor vehicle crashes, occupational injury, systemic hypertension, type 2 diabetes, cardiac arrhythmia, aortic dilatation/dissection, coronary artery disease, heart failure, stroke, depression, cognitive impairment, cancer, ocular disease, pneumonia, renal dysfunction, dementia, seizures, hypogonadism, maternal/fetal health, post-operative complications, and premature death. Patients with untreated OSA may have increased perioperative morbidity (postoperative cardiac events, acute respiratory failure, desaturation and reintubation);21 consider appropriate screening when referring patients for potential surgery (e.g. using the STOP-Bang Questionnaire, Appendix C: STOP-Bang Questionnaire). See also the BC Surgical Rehabilitation Toolkit for further information. TOP History and Physical Examination History and physical examination are crucial first steps towards the recognition and diagnosis of OSA.2 Patient history should focus on nocturnal breathing abnormalities, daytime sleepiness, and family and personal medical history. Consider using the STOP-Bang questionnaire to help determine if a patient is at increased risk of moderate to severe OSA (Appendix C: STOP-Bang Questionnaire).2 Determine the patient’s risk factors for OSA (see list above). While the clinical presentation of OSA can vary, the most common symptom is excessive daytime sleepiness. Inquire about the impact of daytime sleepiness (e.g., “Have you had any accidents or near misses related to sleepiness while driving?” or “Have you ever had to stop an activity due to sleepiness?”). Request completion of the Epworth Sleepiness Scale. A score of greater than 10 suggests significant daytime sleepiness, although a score of 10 or less does not exclude daytime sleepiness or OSA. Other OSA symptoms include: fatigue habitual loud snoring choking, gasping, or pauses in breathing during sleep morning headache recurrent night-time awakenings unrefreshing or restless sleep impaired concentration nocturia All patients should be questioned about driving or safety critical occupation (e.g. truck, taxi, bus drivers, railway engineers, commercial pilots22) where sleepiness could be a hazard, whether they operate heavy equipment, the class of their driver’s license and whether they have fallen asleep at the wheel or have come close to doing so in the past 5 years. Physical examination: Head and neck examination are important. Measure the neck circumference if feasible Examine the upper airway and it may be useful to evaluate the Mallampati score - Mallampati Classification (Appendix D: Mallampati Classification). A higher Mallampati classification has been associated with OSA diagnosis and increased AHI. For every 1- point increase in the Mallampati score, the odds of having OSA increased by more than 2-fold.23 TOP Testing and Referral To determine if a patient requires a diagnostic test for OSA, it is important to first identify if they are at increased risk of moderate-to-severe OSA. This is indicated by the presence of excessive daytime sleepiness or fatigue and at least two of the following three criteria: witnessed apneas or gasping or choking. habitual loud snoring. diagnosed hypertension. If the patient is at an increased risk of moderate-to-severe OSA, then the Home Sleep Apnea Test (HSAT) is an appropriate diagnostic test, rather than a polysomnography. However, HSAT is not the appropriate test if one or more of the following exclusion criteria apply (any one item precludes HSAT): concern for non-respiratory sleep disorder (e.g., chronic insomnia, sleep walking/talking). risk of hypoventilation (e.g., neuromuscular disease, BMI≥ 40 kg/m2). chronic/regular opiate medication use. significant cardiopulmonary disease (e.g., history of stroke, heart failure, moderate-to-severe lung disease). previous negative or equivocal HSAT. age < 16 years. inability to complete necessary steps for self-administered HSAT (e.g., cognitive, physical, or other barriers). Recognizing that some patients may not have a bed partner, the referring physician may, while conducting a physical examination, assess whether there is a reasonable suspicion of nocturnal breathing events such as apneas, gasping, choking or habitual loud snoring. HSAT is not recommended for pediatric patients, however, physician evaluation, rather than age, may be the best way to determine whether an adolescent presents as an adult. Patients at increased risk of moderate-to-severe OSA but with one of more of the exclusion criteria above should be referred to a sleep disorder physician for potential polysomnography (PSG). Referral for HSAT is limited to: Registrants of the College of Physicians and Surgeons of British Columbia Nurse practitioners or other designated health professional as authorized by the College of Physicians and Surgeons of British Columbia. If an HSAT is negative, equivocal, or the patient has been mis-referred: The diagnostic facility is to notify the referring physician that the patient did not/does not meet the requirements for an HSAT and recommend a sleep disorder consultation and/or Level 1 polysomnogram study (preferred), or The diagnostic facility’s Medical Director or interpreting physician may refer the patient for a sleep disorder consultation and/or Level 1 study and notify the originating/referring physician, ensuring the originating/referring physician receives a copy of the HSAT result. In this case the Medical Director or interpreting physician is responsible for patient follow up as the ordering physician and cannot assume the original referring physician will provide patient follow up on the test. Facilities must have a policy and procedure for notifying the referring physician of critical results immediately. To order HSAT or to request a referral for a sleep disorder consultation, see Associated Documents. For patients with suspected OSA, Figure 1. outlines the patient pathway. Figure 1. Patient Pathway for Patients with Suspected OSA Interpretation of Results The diagnostic sleep report should provide an assessment as to whether the patient has OSA and whether it is mild, moderate, or severe. Apnea Hypopnea Index (AHI) : Is defined by the number of apnea and hypopnea per hour of measured sleep. In addition to symptoms and presence of comorbid disease, AHI is one of the criteria used to determine the severity of sleep apnea. AHI values should be considered in conjunction with clinical presentation, including a patient’s symptoms, co-morbidities and occupation, and all sleep monitoring abnormalities such as degree of hypoxemia (see Table 1: AHI Interpretation below). Table 1: AHI Interpretation \| Result \| Interpretation \| --- \| \| AHI Values \| Severity of OSA:2 in conjunction with symptoms and comorbid disease Normal (< 5 events per hour) Mild (5-14 events per hour) Moderate (15-29 events per hour) Severe (≥ 30 events per hour) \| TOP OSA Management When assessing whether a patient should be treated it is important to consider the severity of the symptoms, presence and severity of any comorbid disease, presence of any safety-critical occupation and all the sleep monitoring abnormalities. Patients with OSA and daytime sleepiness and/or sleep related arterial oxygen desaturation appear more likely to develop vascular complication if left untreated. It is important to treat most patients with OSA if they have comorbid disease or work in a safety critical occupation such as a commercial driver. Treatment should be considered in patients with fatigue and/or sleepiness even if the AHI is in the milder range. Patients with OSA may be prone to drowsiness while driving. Physicians caring for these patients should be familiar with BC’s Driver Medical Fitness Information for Medical Professionals.[Expert opinion] Surgical options are primarily for patients with tonsillar hypertrophy or craniofacial abnormalities and require referral to a sleep disorder physician and to the appropriate surgeon for more comprehensive evaluations. Healthy Behaviours and Diet: The importance of healthy behaviours and diet, including exercise and weight loss, should be recommended for patients with elevated BMI, although it is not adequate as a stand-alone therapy if the patient is sleepy or has moderate to severe OSA. Weight loss is more successful if the patient uses definitive therapy to control their OSA and then retest after their weight goal is achieved to see if definitive therapy is still required. Continuous Positive Airway Pressure (CPAP): CPAP is generally the most effective treatment for OSA. If a patient is diagnosed with OSA and is unwilling or intolerant of CPAP, consider alternate therapies or refer to sleep disorder physician. Positional Therapy: Avoiding sleeping in the supine position is an option if testing shows that OSA is primarily in the supine position. There are electronic devices that can monitor and record sleeping positions. Other devices (oral, etc.) Oral appliances may be useful for mild to moderate OSA or those intolerant/unwilling to use CPAP and can be effective for some patients. For additional OSA management information, see Appendix F: OSA Management Options. For additional information on oral and other devices, seeAppendix G: Oral and Other Devices. TOP Follow-Up OSA is a serious chronic disease that warrants regular follow-up; short-term to ensure initial compliance and response to treatment and long-term to confirm continued effectiveness.3 There is no need to re-test if the patient is doing well on long-term CPAP therapy and there is no change in clinical status. Annual follow-up is recommended with CPAP download Indications for retesting: Persistent daytime symptoms, abnormal overnight oximetry or residual sleep disordered breathing based on CPAP download Whatever treatment is used, the patient should be followed until the AHI is normal (less than 5 events per hour), the Epworth Sleepiness Scale score is 10 or less, the patient feels rested, and a bed partner reports no residual snoring. Ideally, the patient is using the treatment every night, all night. CPAP machines also provide a patient accessible read out of ‘events per hour’. This is an index of residual sleep apnea and not a true AHI as it is not derived from sleep monitoring but is a useful daily indication of therapy effectiveness. Problem Solving CPAP therapy After treatment, non-adherence to CPAP therapy, CPAP intolerance and persistent sleepiness after therapy may occur. Considerations include: CPAP non-adherence: Lack of interest/understanding of OSA – importance of treatment can be reinforced; depends on indications for treatment and severity of disease.24 CPAP intolerance – is common and should be addressed by the CPAP provider.24 See table below for common causes of intolerance. See Table 2: CPAP Intolerance and Treatment Recommendations below. Table 2: CPAP Intolerance and Treatment Recommendations \| CPAP Intolerance/Problem \| Recommended Treatment \| --- \| \| Nasal congestion \| Saline nasal rinses +/- intranasal steroid, increase humidity, consider full face mask \| \| Difficulty exhaling against pressure \| Use Expiratory Pressure Release (EPR) option \| \| Mouth/nasal dryness \| Adequate hydration and heated humidity on CPAP machine; chin strap for mouth leak \| \| Claustrophobia \| Trial of CPAP while sitting or awake and supine before using during sleep; consider nasal mask or pillows (vs full face mask) \| \| Persistently elevated AHI \| Make sure any leak is well controlled, obstructive events generally respond to pressure increase \| \| Persistent fatigue \| Ensure CPAP is used for full night (7 hours or more recommended); possible other sleep disorder, refer for consultation \| TOP Methodology These guideline recommendations are tailored to support practice in British Columbia and are based on the ADAPTE Collaboration guideline adaption methodology.25 Clinical recommendations were developed based on the sourced guidelines,2,3,19,26–29 a Primary Care Pathway: Uncomplicated Obstructive Sleep Apnea,24 as well as expert clinical consensus where evidence was insufficient or unavailable. The source guidelines were chosen following an environmental scan of internationally available guidelines. Guidelines were chosen for adaptation following an evaluation using the Appraisal of Guidelines for Research and Evaluation II (AGREE II) tool.30 In situations where there is a lack of rigorous evidence, best clinical opinion has been provided to support decision making and high-quality patient care. The guideline development process includes significant engagement and consultation with primary care providers, sleep disorder physicians, and key stakeholders. For more information about Guidelines and Protocols Advisory Committee (GPAC) guideline development processes, refer to the GPAC handbook available at BCGuidelines.ca. The level and quality of evidence for the key recommendations are based on those used by the AASM: Clinical Practice Guideline for Diagnostic Testing for Adult Obstructive Sleep Apnea (2017).2 Please refer to those guidelines for a summary of the levels of evidence (table 5). Where evidence was insufficient or unavailable expert clinical consensus was used and is indicated in the key recommendations as “expert opinion”. TOP Resources References Kirk V, Baughn J, D’Andrea L, Friedman N, Galion A, Garetz S, et al. American Academy of Sleep Medicine Position Paper for the Use of a Home Sleep Apnea Test for the Diagnosis of OSA in Children. J Clin Sleep Med. 2017 Oct 15;13(10):1199–203. Kapur Vishesh K., Auckley Dennis H., Chowdhuri Susmita, Kuhlmann David C., Mehra Reena, Ramar Kannan, et al. Clinical Practice Guideline for Diagnostic Testing for Adult Obstructive Sleep Apnea: An American Academy of Sleep Medicine Clinical Practice Guideline. Journal of Clinical Sleep Medicine. 2017;13(03):479–504. Collop NA, Anderson WM, Boehlecke B, Claman D, Goldberg R, Gottlieb DJ, et al. Clinical guidelines for the use of unattended portable monitors in the diagnosis of obstructive sleep apnea in adult patients. Portable Monitoring Task Force of the American Academy of Sleep Medicine. Journal of clinical sleep medicine: JCSM: official publication of the American Academy of Sleep Medicine. 2007 Dec 15;3(7):737–47. Hypoventilation syndromes - Symptoms, diagnosis and treatment \| BMJ Best Practice US [Internet]. [cited 2020 Oct 26]. Available from: Tietjens JR, Claman D, Kezirian EJ, De Marco T, Mirzayan A, Sadroonri B, et al. Obstructive Sleep Apnea in Cardiovascular Disease: A Review of the Literature and Proposed Multidisciplinary Clinical Management Strategy. JAHA [Internet]. 2019 Jan 8 [cited 2021 Jul 23];8(1). Available from: Yeghiazarians Y, Jneid H, Tietjens JR, Redline S, Brown DL, El-Sherif N, et al. Obstructive Sleep Apnea and Cardiovascular Disease: A Scientific Statement From the American Heart Association. Circulation [Internet]. 2021 Jul 20 [cited 2021 Jul 23];144(3). Available from: Government of Canada SC. Sleep Apnea in Canada, 2016 and 2017 [Internet]. 2018 [cited 2021 Feb 10]. Available from: Heinzer R, Vat S, Marques-Vidal P, Marti-Soler H, Andries D, Tobback N, et al. Prevalence of sleep-disordered breathing in the general population: the HypnoLaus study. Lancet Respir Med. 2015 Apr;3(4):310–8. Benjafield AV, Ayas NT, Eastwood PR, Heinzer R, Ip MSM, Morrell MJ, et al. Estimation of the global prevalence and burden of obstructive sleep apnoea: a literature-based analysis. Lancet Respir Med. 2019 Aug;7(8):687–98. Young T. Risk Factors for Obstructive Sleep Apnea in Adults. JAMA. 2004 Apr 28;291(16):2013. Simpson R, Oyekan AA, Ehsan Z, Ingram DG. Obstructive sleep apnea in patients with Down syndrome: current perspectives. Nat Sci Sleep. 2018;10:287–93. Mirrakhimov AE, Sooronbaev T, Mirrakhimov EM. Prevalence of obstructive sleep apnea in Asian adults: a systematic review of the literature. BMC pulmonary medicine. 2013 Feb 23;13:10. Elfil M, Bahbah EI, Attia MM, Eldokmak M, Koo BB. Impact of Obstructive Sleep Apnea on Cognitive and Motor Functions in Parkinson’s Disease. Mov Disord. 2021 Mar;36(3):570–80. Zuzuárregui JRP, Bickart K, Kutscher SJ. A review of sleep disturbances following traumatic brain injury. Sleep Science Practice. 2018 Dec;2(1):2. Michels D de S, Rodrigues A da MS, Nakanishi M, Sampaio ALL, Venosa AR. Nasal involvement in obstructive sleep apnea syndrome. Int J Otolaryngol. 2014;2014:717419. Albdewi MA, Liistro G, El Tahry R. Sleep-disordered breathing in patients with neuromuscular disease. Sleep Breath. 2018 May;22(2):277–86. Sedky K, Gaisl T, Bennett DS. Prevalence of Obstructive Sleep Apnea in Joint Hypermobility Syndrome: A Systematic Review and Meta-Analysis. J Clin Sleep Med. 2019 Feb 15;15(2):293–9. Kahal H, Kyrou I, Tahrani AA, Randeva HS. Obstructive sleep apnoea and polycystic ovary syndrome: A comprehensive review of clinical interactions and underlying pathophysiology. Clin Endocrinol (Oxf). 2017 Oct;87(4):313–9. Douglas JA, Chai-Coetzer CL, McEvoy D, Naughton MT, Neill AM, Rochford P, et al. Guidelines for sleep studies in adults – a position statement of the Australasian Sleep Association. Sleep Medicine. 2017 Aug 1;36:S2–22.. Somboon T, Grigg-Damberger MM, Foldvary-Schaefer N. Epilepsy and Sleep-Related Breathing Disturbances. CHEST. 2019 Jul 1;156(1):172–81. Kaw R, Chung F, Pasupuleti V, Mehta J, Gay PC, Hernandez AV. Meta-analysis of the association between obstructive sleep apnoea and postoperative outcome. British Journal of Anaesthesia. 2012 Dec 1;109(6):897–906. British Columbia Medical Services Commission. Policies and Guidelines of the Medical Services Commission’s Advisory Committee on Diagnostic Facilities [Internet]. British Columbia Ministry of Health; 2020. Available from: Nuckton TJ, Glidden DV, Browner WS, Claman DM. Physical examination: Mallampati score as an independent predictor of obstructive sleep apnea. Sleep. 2006 Jul;29(7):903–8. Clinical Pathways + Specialty Access \| Specialist Link [Internet]. [cited 2021 Feb 22]. Available from: ADAPTE Resource Toolkit for Guideline Adaptation Version 2.0 — Guidelines International Network [Internet]. [cited 2020 Mar 13]. Available from: Blackman A, McGregor C, Dales R, Driver HS, Dumov I, Fleming J, et al. Canadian Sleep Society/Canadian Thoracic Society position paper on the use of portable monitoring for the diagnosis of obstructive sleep apnea/hypopnea in adults. Can Respir J. 2010;17(5):229–32. Fleetham J, Ayas N, Bradley D, Fitzpatrick M, Oliver TK, Morrison D, et al. Canadian Thoracic Society 2011 guideline update: Diagnosis and treatment of sleep disordered breathing. Can Respir J. 2011;18(1):25–47. Rosen IM, Kirsch DB, Carden KA, Malhotra RK, Ramar K, Aurora RN, et al. Clinical Use of a Home Sleep Apnea Test: An Updated American Academy of Sleep Medicine Position Statement. Journal of clinical sleep medicine: JCSM: official publication of the American Academy of Sleep Medicine. 2018 15;14(12):2075–7. Mysliwiec V, Martin JL, Ulmer CS, Chowdhuri S, Brock MS, Spevak C, et al. The Management of Chronic Insomnia Disorder and Obstructive Sleep Apnea: Synopsis of the 2019 U.S. Department of Veterans Affairs and U.S. Department of Defense Clinical Practice Guidelines. Annals of Internal Medicine. 2020 03;172(5):325–36. AGREE Next Steps Consortium. The AGREE II instrument [Electronic version]. [Internet]. [cited 2020 Mar 13]. Available from: Marcus CL, Brooks LJ, Draper KA, Gozal D, Halbower AC, Jones J, et al. Diagnosis and management of childhood obstructive sleep apnea syndrome. Pediatrics. 2012 Sep;130(3):576–84. Leung TN, Cheng JW, Chan AK. Paediatrics: how to manage obstructive sleep apnoea syndrome. Drugs Context. 2021;10:2020-12–5. Chai-Coetzer CL, Antic NA, Hamilton GS, McArdle N, Wong K, Yee BJ, et al. Physician Decision Making and Clinical Outcomes With Laboratory Polysomnography or Limited-Channel Sleep Studies for Obstructive Sleep Apnea: A Randomized Trial. Ann Intern Med. 2017 Mar 7;166(5):332–40. Patil SP, Ayappa IA, Caples SM, Kimoff RJ, Patel SR, Harrod CG. Treatment of Adult Obstructive Sleep Apnea With Positive Airway Pressure: An American Academy of Sleep Medicine Systematic Review, Meta-Analysis, and GRADE Assessment. J Clin Sleep Med. 2019 Feb 15;15(2):301–34. Abbreviations AHI BMI CPAP EPR HSAT OSA PSG Apnea Hypopnea Index Body Mass Index Continuous Positive Airways Pressure Expiratory Pressure Release Home Sleep Apnea Testing Obstructive Sleep Apnea Polysomnography Appendices Appendix A: Sleep Apnea in Children Appendix B: Other Conditions Associated with Daytime Sleepiness Appendix C: STOP-Bang Questionnaire Appendix D: Mallampati Classification Appendix E: Sleep Study Levels Appendix F: OSA Management Options Appendix G: Oral and Other Devices Associated Documents Standard Requisition for Home Sleep Apnea Test (without Sleep Disorder Physician consultation) – HLTH 1944 Referral Request – Sleep Disorder Consultation – HLTH 1945 TOP This guideline is based on scientific evidence current as of the effective date. The guideline was developed by the Guidelines and Protocols Advisory Committee and adopted by the Medical Services Commission. For more information about how BC Guidelines are developed, refer to the GPAC Handbook available at BCGuidelines.ca: GPAC Handbook. THE GUIDELINES AND PROTOCOLS ADVISORY COMMITTEE \| \| \| The principles of the Guidelines and Protocols Advisory Committee are to: encourage appropriate responses to common medical situations recommend actions that are sufficient and efficient, neither excessive nor deficient permit exceptions when justified by clinical circumstances Contact Information: Guidelines and Protocols Advisory Committee PO Box 9642 STN PROV GOVT Victoria, BC V8W 9P1 Email: hlth.guidelines@gov.bc.ca Website: www.BCGuidelines.ca Disclaimer The Clinical Practice Guidelines (the “Guidelines”) have been developed by the Guidelines and Protocols Advisory Committee on behalf of the Medical Services Commission. The Guidelines are intended to give an understanding of a clinical problem, and outline one or more preferred approaches to the investigation and management of the problem. The Guidelines are not intended as a substitute for the advice or professional judgment of a health care professional, nor are they intended to be the only approach to the management of clinical problem. We cannot respond to patients or patient advocates requesting advice on issues related to medical conditions. If you need medical advice, please contact a health care professional. \| Guideline Related Resources Download the following: Full Guideline (PDF, 906KB) Appendix A: Sleep Apnea in Children (PDF, 91KB) Appendix B: Other conditions Associated with Daytime Sleepiness (PDF, 91KB) Appendix C: STOP-Bang Questionnaire (PDF, 92KB) Appendix D: Mallampati Classification of Airways (PDF, 331KB) Appendix E: Sleep Study Level (PDF, 91KB) Appendix F: OSA Management Options (PDF, 95KB) Appendix G: Oral and Other Devices (PDF, 96KB) Stay Informed Keep current on BC Guidelines by signing up for our email notification service. Notification Service Get Involved! Interested in contributing to BC Guidelines? BC Guidelines is always looking for knowledgeable practitioners to chair and serve on our working groups. Email BC Guidelines at hlth.guidelines@gov.bc.ca and ask for an application package today. Contact Us Questions, comments or suggestions? We would love to hear from you. Note: We cannot respond to patients or patient advocates requesting advice on issues related to medical conditions. If you need medical advice, please contact a health care professional. Did you find what you were looking for? The B.C. Public Service acknowledges the territories of First Nations around B.C. and is grateful to carry out our work on these lands. We acknowledge the rights, interests, priorities, and concerns of all Indigenous Peoples - First Nations, Métis, and Inuit - respecting and acknowledging their distinct cultures, histories, rights, laws, and governments.
10785	https://www.sciencedirect.com/science/article/abs/pii/S1389945799000064	A five week, polysomnographic assessment of zaleplon 10 mg for the treatment of primary insomnia - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract Introduction Section snippets References (19) Cited by (126) Sleep Medicine Volume 1, Issue 1, 1 February 2000, Pages 41-49 Original article A five week, polysomnographic assessment of zaleplon 10 mg for the treatment of primary insomnia Author links open overlay panel James K.Walsh a b, Gerald W.Vogel c, Martin Scharf d, Milton Erman e, C.William Erwin f, Paula K.Schweitzer a, Richard M.Mangano g, Thomas Roth h Show more Add to Mendeley Share Cite rights and content Abstract Objective: To examine the hypnotic efficacy of zaleplon 10 mg, a selective benzodiazepine receptor agonist, over a period of 35 nights in primary insomniacs. Methods: A double-blind, parallel-group, placebo-controlled design was employed. Subjects were 113 men and women, ages 18 to 65 years. Polysomnographic and subjective sleep data were collected during baseline, on two nights during each of five treatment weeks, and on the first two nights after discontinuation of active medication. Results: Sleep latency was significantly shortened with zaleplon 10 mg for all 5 weeks of treatment as assessed by polysomnography and by subjective sleep measures. Total sleep time, whether evaluated with polysomnography or with subjective estimates, was inconsistently affected. Sleep architecture was similar with zaleplon and placebo. There was no evidence of tolerance to the sleep promoting effects of zaleplon during the five weeks of administration, and there was no rebound insomnia upon discontinuation. Adverse events occurred with equal frequency in the zaleplon and placebo groups. Conclusions: Zaleplon 10 mg is effective in the treatment of sleep onset insomnia over a period of 35 nights, with minimal evidence of undesired effects. Introduction Zaleplon, a pyrazolopyrimidine hypnotic, binds selectively to the benzodiazepine type I site on the type A, gamma-aminobutyric acid (GABA) receptor complex. Pharmacokinetic studies indicate a plasma t max and an elimination t 1/2, both, of about 1 h . This pharmacokinetic profile predicts a rapid onset and a relatively short duration of sedative activity. Clinical studies of zaleplon, administered for 2–14 days to either adult or elderly , , insomnia patients, demonstrate sleep-promoting effects which are quite consistent with this prediction. That is, both objective and subjective measures of sleep latency are reduced with zaleplon doses ranging from 2 to 10 mg. Total sleep time is less consistently increased, showing positive results in studies involving older patients , . The current investigation assessed the efficacy of zaleplon during 5 weeks of nightly administration for primary insomnia. We hypothesized that both polysomnographic (PSG) and subjective sleep latency would be significantly shorter with zaleplon 10 mg (Z10) than with placebo (PBO) over the entire study duration. Additionally, PSG and subjective total sleep time were hypothesized to be increased with Z10 as compared to PBO at each time point. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Experimental design A double-blind, parallel group design was used to compare Z10 with PBO at eleven study sites. Early in the study, zaleplon 20 mg was a third treatment condition; however, the sponsor discontinued randomization to that condition because subject enrollment was slower than expected and because the 20 mg dose has not been shown to be substantially superior to 10 mg (data on file, Wyeth-Ayerst Research). Thus, the present paper includes data only from those subjects randomly assigned to PBO or Z10. Polysomnographic efficacy measures Fig. 1 displays median LPS for both treatment groups during baseline, each week of double-blind treatment, and on the first night following discontinuation of active drug. LPS for Z10 was significantly shorter than for PBO during all five treatment weeks. Median LPS with Z10 was 22–25 min shorter than baseline values during each of the treatment weeks. In comparison, median LPS for the PBO group was 9.5 to 16.0 min shorter than baseline values. Medians, means, and results of ANCOVAs are Discussion Z10 significantly decreased both PSG and subjective measures of sleep latency during all five weeks of treatment in this investigation. PSG and subjective variables that relate more to sleep maintenance were inconsistently improved or unaffected by Z10 relative to PBO. These findings are consistent with the shorter term studies of zaleplon , , , . This study also extends to five weeks the duration of nightly use for which sleep latency is significantly reduced by Z10 in adult Acknowledgements The authors thank Gail Koshorek of Henry Ford Hospital, Detroit, and Mary Kay McPherson, Noreen Scherer, Timothy Whitaker, MD, and Carmen Wickland, Ph.D. of Wyeth-Ayerst Research, Radnor, PA, for their contributions to the study. This research was supported by Wyeth-Ayerst Research. Recommended articles References (19) B. Beer et al. A review of the preclinical development of zaleplon, a novel non-benzodiazepine hypnotic for the treatment of insomnia CNS Drug Rev (1997) J.K. Walsh et al. Efficacy and safety of 14-day administration of zaleplon 5 mg and 10 mg for the treatment of primary insomnia Clin Drug Invest (1998) S. Ancoli-Israel et al. Zaleplon, a novel non-benzodiazepine hypnotic, effectively treats insomnia in elderly patients without causing rebound effects Prim Care Companion J Clin Psychiat (1999) J. Hedner et al. Improvement in sleep latency and sleep quality with zaleplon in elderly patients with primary insomnia J Sleep Res (1998) C.W. Erwin Zaleplon effectively and safely treats chronic insomnia in elderly patients J Am Geriatrics Soc (1999) American Psychiatric Association Diagnostic and statisical manual of mental disorders (1994) A. Rechtschaffen et al. A manual of standardized terminology, techniques and scoring systems for sleep stages of human subjects (1968) W.B. Mendelson Medications in the treatment of sleep disorders T. Roth et al. Issues in the use of benzodiazepine therapy J Clin Psychiat (1992) There are more references available in the full text version of this article. Cited by (126) An evaluation of the efficacy and safety of eszopiclone over 12 months in patients with chronic primary insomnia 2005, Sleep Medicine Show abstract A double-blind placebo-controlled study of eszopiclone found significant, sustained improvement in sleep and daytime function. The 6-month open-label extension phase is described herein. Adults (21–64) with primary insomnia who reported sleep duration <6.5 h/night or sleep latency >30 min/night were included. Patient-reported endpoints included sleep and daytime function. Safety and compliance were assessed at monthly clinic visits. The final double-blind month was used as the baseline for efficacy analyses of the open-label period. Patients who were initially randomized to double-blind placebo and then switched to open-label eszopiclone (n=111) significantly reported the following: (1) decreased sleep latency, wake time after sleep onset, and number of awakenings; (2) increased total sleep time and sleep quality; and (3) improved ratings of daytime ability to function, alertness and sense of physical well-being compared to baseline (P≤0.0001 all monthly endpoints). There was no evidence of tolerance on any measure in either group. These subjects (n=360) sustained the double-blind treatment gains for all sleep and daytime parameters, with further significant improvement in a number of measures. Eszopiclone was well tolerated in both groups; unpleasant taste was the only undesirable effect reported by >5% of patients. The significant improvements in sleep and daytime function were evident in those switched from double-blind placebo to 6 months of open-label eszopiclone therapy and were sustained during the 6 months of open-label treatment for those receiving prior double-blind eszopiclone. During 12 months of nightly treatment, eszopiclone 3 mg was well tolerated; tolerance was not observed. ### The assessment and management of insomnia: an update 2019, World Psychiatry ### The impact of stress on sleep: Pathogenic sleep reactivity as a vulnerability to insomnia and circadian disorders 2018, Journal of Sleep Research ### Clinical Practice Guideline for the Pharmacologic Treatment of Chronic Insomnia in Adults: An American academy of sleep medicine clinical practice guideline 2017, Journal of Clinical Sleep Medicine ### Efficacy and safety of doxepin 1 mg and 3 mg in a 12-week sleep laboratory and outpatient trial of elderly subjects with chronic primary insomnia 2010, Sleep ### New drugs for insomnia: Comparative tolerability of zopiclone, zolpidem and zaleplon 2003, Drug Safety View all citing articles on Scopus View full text Copyright © 2000 Elsevier Science B.V. All rights reserved. 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10786	https://www.quora.com/What-is-the-relationship-between-period-frequency-and-wavelength-for-waves	Something went wrong. Wait a moment and try again. Periodic Wave Properties of Wave Frequency (physics) Wave Sciences Wave Frequencies 5 What is the relationship between period, frequency, and wavelength for waves? Mark Dixon Citizen of the world; Student of life. · Author has 128 answers and 35K answer views · 3y Frequency and wavelength are inversely proportional to each other, so as one increases the other decreases. The general formula is Frequency = Wave-speed / Wave-length. Wave-speed is important because it is vastly different when considering something like light traveling through space vs waves in a body of water. The wave-speed for light or any other electromagnetic wave in a vacuum is generally the “speed of light (c)” which is about 300 million meters per second. Period is the amount of time for a complete wave cycle, i.e. how long in seconds from one wave peak till the next one. This is clos Frequency and wavelength are inversely proportional to each other, so as one increases the other decreases. The general formula is Frequency = Wave-speed / Wave-length. Wave-speed is important because it is vastly different when considering something like light traveling through space vs waves in a body of water. The wave-speed for light or any other electromagnetic wave in a vacuum is generally the “speed of light (c)” which is about 300 million meters per second. Period is the amount of time for a complete wave cycle, i.e. how long in seconds from one wave peak till the next one. This is closely related to wavelength in that period is the time for one complete wave cycle while wavelength is the distance a wave travels in one complete wave cycle. So the velocity of the wave is also important for comparing those two. Period and wavelength are directly proportional, so if the period is longer, the wavelength is also longer. Period is usually stated in seconds. Wavelength is usually stated in meters. Frequency is usually stated in cycles per second, also called Hertz. Related questions What is the relationship between frequency, period, and wavelength of light waves? Can you explain the difference between wavelength, frequency, and period of a wave? What is the relationship between amplitude and wavelength, period, and frequency in a wave? What is the relationship between frequency and wavelength? What is the relationship between frequency, wavelength and speed (of a travelling wave)? Kip Ingram PhD in Electrical Engineering, The University of Texas at Austin Cockrell School of Engineering (Graduated 1992) · Author has 20.1K answers and 21.6M answer views · 3y Related What is the relationship between frequency and wavelength? Does it depend on the medium through which the waves are passing? Generally speaking, the product of frequency and wavelength gives you the speed at which the wave propagates. And yes, that does depend on the medium; for example, sound travels through water faster than it does through air, and faster still through steel (railroad tracks, for example). The atomic structure of the medium governs how quickly molecular motion moves along from molecule to molecule. Y Generally speaking, the product of frequency and wavelength gives you the speed at which the wave propagates. And yes, that does depend on the medium; for example, sound travels through water faster than it does through air, and faster still through steel (railroad tracks, for example). The atomic structure of the medium governs how quickly molecular motion moves along from molecule to molecule. You can see the same effect by creating an electrical transmission line comprised of two parallel wires. If you apply a voltage step to one end, it will propagate down the line with some sp... Assistant Bot · 1y The relationship between period, frequency, and wavelength for waves can be described by the following concepts: Frequency (f) : This is the number of cycles (or oscillations) of a wave that pass a given point in one second. It is measured in hertz (Hz), where 1 Hz = 1 cycle per second. Period (T) : This is the time it takes for one complete cycle of the wave to pass a given point. It is the reciprocal of frequency and is measured in seconds. The relationship is given by: T = 1 f 3. Wavelength (λ) : This is the distance between successive crests (or troughs) of a wave. It is typically measur The relationship between period, frequency, and wavelength for waves can be described by the following concepts: Frequency (f): This is the number of cycles (or oscillations) of a wave that pass a given point in one second. It is measured in hertz (Hz), where 1 Hz = 1 cycle per second. Period (T): This is the time it takes for one complete cycle of the wave to pass a given point. It is the reciprocal of frequency and is measured in seconds. The relationship is given by: T=1f Wavelength (λ): This is the distance between successive crests (or troughs) of a wave. It is typically measured in meters. Wave Speed (v): This is the speed at which the wave travels through a medium. The relationship between wave speed, frequency, and wavelength is given by the equation: v=f⋅λ Summary of Relationships: Period and Frequency: T=1f and f=1T Wave Speed: v=f⋅λ Example: If a wave has a frequency of 2 Hz, its period would be: T=12Hz=0.5seconds If the wave travels at a speed of 340 m/s, the wavelength can be calculated as: λ=vf=340m/s2Hz=170meters These relationships are fundamental in understanding wave behavior in various contexts, including sound, light, and water waves. Richard Herman Alsenz inventing and physics are my passion. · Author has 2K answers and 794.9K answer views · 2y Related What is the relationship between frequency, period, and wavelength of light waves? A human can not measure space. Humans see light that propagates from an object observed and later enters one's eye. Then, it is processed through the human nervous system and brain to make a conclusion. The transmission takes place in the Electric Field. Light does not enter the observation of a human through space: (space) is missing that complete conviction of necessity (thus of absolute truth) Gauss to Bessel Goettingen 9 April 1830 … The ease with which you delved into my views on geometry gives me real joy, given that so few have an open mind for such. My innermost conviction is that the study A human can not measure space. Humans see light that propagates from an object observed and later enters one's eye. Then, it is processed through the human nervous system and brain to make a conclusion. The transmission takes place in the Electric Field. Light does not enter the observation of a human through space: (space) is missing that complete conviction of necessity (thus of absolute truth) Gauss to Bessel Goettingen 9 April 1830 … The ease with which you delved into my views on geometry gives me real joy, given that so few have an open mind for such. My innermost conviction is that the study of space is a priori completely different than the study of magnitudes; our knowledge of the former (space) is missing that complete conviction of necessity (thus of absolute truth) that is characteristic of the latter; we must, in humility, admit that if number is merely a product of our mind. Related questions What is the relationship between the frequency and the wavelength of a wave? What is the relationship between frequency and wavelength? Does it depend on the medium through which the waves are passing? What is the relationship between wavelength, wave number, and frequency? What is the relationship between frequency and wavelength of radio waves? What is the relationship between a wave’s frequency and its wavelength? Mourits de Beer Studied Electrical and Electronics Engineering & Telecommunications (Graduated 2017) · Author has 60 answers and 86K answer views · Updated 9mo Related What is the relationship between amplitude and wavelength, period, and frequency in a wave? In general, the amplitude of a wave is not typically related with those other characteristics: wavelength, period, and frequency. Amplitude is the “height” of the wave — graphically denoted as the distance between the zero-line and the crest or trough of the wave: In electromagnetic waves, the amplitude is proportional to the power of the wave. In sound, the amplitude is the loudness of the wave. You can see that the amplitude is generally independent of the wavelength or frequency. “Generally” because in practice there are limits: you can't have a very high amplitude with a very high frequency — In general, the amplitude of a wave is not typically related with those other characteristics: wavelength, period, and frequency. Amplitude is the “height” of the wave — graphically denoted as the distance between the zero-line and the crest or trough of the wave: In electromagnetic waves, the amplitude is proportional to the power of the wave. In sound, the amplitude is the loudness of the wave. You can see that the amplitude is generally independent of the wavelength or frequency. “Generally” because in practice there are limits: you can't have a very high amplitude with a very high frequency — there will eventually be a limit to how fast the wave can rise. But, generally speaking, they're independent. Period, frequency and wavelength are indeed related: The period of a wave is how long it takes, in seconds, before it repeats — or, how many seconds per wave. The frequency of a wave is how many times it repeats in a second — or, how many waves per second. Therefore, the period is 1 / frequency: P=1/f. And frequency is 1 / period: f=1/P. The wavelength, λ, is the length of a full wave, let's say in meters. It is determined by the speed, v, say meters per second, and the frequency: λ=v/f. Other relations can be derived from the above. Promoted by TruthFinder Carl Madden Finance Security Expert · May 20 Have you ever googled yourself? I got curious one night, and Googled myself. Most of the websites out there gave me only a scrap of information or were wrong entirely, but I did find 1 site that ended up being the HOLY GRAIL of information on anyone (including myself!). When I typed my name into TruthFinder, I was simply CREEPED out by how much they knew! 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Bradley Bobbs Ph.D. in Physics, University of California, Los Angeles (Graduated 1988) · Author has 7.9K answers and 1M answer views · 2y Originally Answered: Can you explain the relationship between frequency, period and wavelength for waves like sound or light? · Yes, I can. Question answered. As a bonus, I’ll go ahead and DO it! The period is one divided by the frequency, and the wavelength is the wave’s speed divided by the frequency. Raunak Nayak Digital Marketing Consultant at Freelancing (2018–present) · Author has 374 answers and 1.5M answer views · 7y Related What is the relationship between wavelength, wave number, and frequency? What is the relationship between frequency and wavelength? The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency for electromagnetic waves is: λν=c where λ is the wavelength, ν is the frequency and c is the speed of l What is the relationship between frequency and wavelength? The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency for electromagnetic waves is: λν=c where λ is the wavelength, ν is the frequency and c is the speed of light. What is the relationship between wavelength, frequency and energy? The greater the energy, the larger the frequency and the shorter (smaller) the wavelength. Given the relationship between wavelength and frequency — the higher the frequency, the shorter the wavelength — it follows that short wavelengths are more energetic than long wavelengths. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. 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Wave Number - It is the repetency of a wave i.e. how many times it repeats Given as - K (Wave Number) = 2pi/lambda Frequency is just reciprocal of wavelength - f = 1/lambda So , K = 2pif= 2pi/lambda Wavelength - It is the length of wave till it starts t repeat again. OR It is the difference between two crests and troughs. Wave Number - It is the repetency of a wave i.e. how many times it repeats Given as - K (Wave Number) = 2pi/lambda Frequency is just reciprocal of wavelength - f = 1/lambda So , K = 2pif= 2pi/lambda Renzo Carlo Avanzo Former Director of Environmental Physics · Author has 448 answers and 107.7K answer views · 2y Originally Answered: What is the relationship between the period and wavelength/frequency of electromagnetic radiation (or other waves)? · Wavelength=vperiod=v/frequency Being period=1/frequency and v the speed of radiation Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Eric Bittner Professor of Chemical Physics · Author has 2.5K answers and 9M answer views · 3y Related What is the relationship between the wavelength and the frequency of a wave? Depends on the wave. Generally one needs to define the “dispersion” relation which relates frequency, ω(k), to wavenumber, k, which in turn is related to the wavelength by k=2π/λ Wavenumber is simply the number of periods in a given unit of length…for example 3000cm−1 For E&M waves and other waves arising from the wave-equation: ω(k)=c\|k\| Which gives the usual relation between frequency, wavelength, and speed of propagation, c. νλ=c However, you can have more complex relations well. Tirtha Chakrabarti Studying Physics · Upvoted by Ahmed Khalil , Master's degree Physics, University of Cologne (2024) · Author has 690 answers and 3.5M answer views · 8y Related What is the relationship between wavelength, wave number, and frequency? Wavelength is the distance over which the wave's shape (a cycle) repeats . Wavenumber is the number of full cycles in a unit distance. wavenumber= 1/wavelength Frequency is the number of full cycles per unit time. Frequency=1/time period= wave velocity/wavelength MUKESH SAINI M.A. from University of Rajasthan (Graduated 2008) · 2y Related What is the relationship between frequency, period, wavelength & speed for electromagnetic waves (EMW)? Note: speed of any periodic wave is the product of its wavelength and frequency. v = λf. The speed of any electromagnetic waves in free space is the speed of light c = 3108 m/s. Electromagnetic waves can have any wavelength λ or frequency f as long as λf = c.Spectroscopy is the precise study of colour, broadened from visible light to all bands of the electromagnetic spectrum; indeed, spectroscopy began as a study of the wavelength dependence of visible light scattered by a prism absorption by gas phase materials. For spectroscopy, nearly all frequencies and wavelengths of electromagnetic rad Note: speed of any periodic wave is the product of its wavelength and frequency. v = λf. The speed of any electromagnetic waves in free space is the speed of light c = 3108 m/s. Electromagnetic waves can have any wavelength λ or frequency f as long as λf = c.Spectroscopy is the precise study of colour, broadened from visible light to all bands of the electromagnetic spectrum; indeed, spectroscopy began as a study of the wavelength dependence of visible light scattered by a prism absorption by gas phase materials. For spectroscopy, nearly all frequencies and wavelengths of electromagnetic radiation can be employed Hint: Energy, wavelength, and frequency can all be used to describe the electromagnetic spectrum. Each approach of thinking about the electromagnetic spectrum is mathematically related to the others. The relationships between energy, wavelength, and frequency can be stated as wavelength equals the speed of light divided by the frequency. Complete answer: Light's wavelength and frequency are inextricably linked. The shorter the wavelength, the higher the frequency. Hence, frequency and wavelength are inversely proportional to each other. Because all light waves move at the same speed in a vacuum, the number of wave crests passing at a given spot in one second is determined by the wavelength. For a short-wavelength wave, this number, also known as the frequency, will be higher than for a long-wavelength wave. The wavelength and frequency of electromagnetic waves are related by the equation λ ν = c , where λ λ is the wavelength, ν ν is the frequency, and c c is the speed of light. The wavelength and frequency are inversely proportional. c = ν λ c = ν ⋅ λ (The speed of light is directly proportional to ν ν λ λ c ν = λ c ν = λ (The speed of light is directly proportional to ν ν , and ν ν is inversely proportional to λ λ c λ = ν c λ = ν (The speed of light is directly proportional to λ λ and λ λ is inversely proportional to ν ν Nikhil Panikkar Communications & Signal Processing Engineer · Author has 1.1K answers and 2.8M answer views · Updated 7y Related What is the relationship between wavelength, wave frequency, and wave velocity? When analysing a wave(longitudnal or transverse), there are two different kinds of properties to keep track of: Properties of the oscillation(the disturbance that generates the energy propagation). The frequency of oscillation is one such property. Properties of the energy propagation itself. The wavelength and velocity of the wave are two such properties. Let’s first try to understand what these three terms mean. Wave velocity Wave velocity refers to the phase velocity i.e the velocity at which a particular phase of oscillation is replicated in space. For example, in the figure below, you see two When analysing a wave(longitudnal or transverse), there are two different kinds of properties to keep track of: Properties of the oscillation(the disturbance that generates the energy propagation). The frequency of oscillation is one such property. Properties of the energy propagation itself. The wavelength and velocity of the wave are two such properties. Let’s first try to understand what these three terms mean. Wave velocity Wave velocity refers to the phase velocity i.e the velocity at which a particular phase of oscillation is replicated in space. For example, in the figure below, you see two identical phases of oscillation at x0 and x. The phase velocity is the velocity with which the phase at x0 ‘travels’ to x. υ=dxdt Wavelength Next lets look at wavelength.Wavelength is the spatial frequency of phase. For example, in the figure below you can see two peaks of energy propagation. The wavelength is the frequency, with which these peaks are distributed in space. Mathematically, λ=dxdθ The inverse of wavelength(i.e the spatial variation of phase k=dθdx)is called the wavenumber. Next lets look at frequency.Frequency is the time rate of change of the phase of the oscillation. For example, in the figure below, you can see a spring propagating a transverse wave and a longitudnal wave. For the transverse wave, the frequency is the rate at which the hand(disturbance) goes up and down, while for the longitudnal wave, it is the rate at which the hand moves back and forth. Mathematically, f=dθdt Multiplying the spatial frequency of phase with the time rate of change of the phase of oscillation, gives us the phase velocity of the wave: f.λ=dθdt.dxdθ=dxdt=υ υ=f.λ which is the equation relating the phase velocity, frequency and wavelength of a wave. Related questions What is the relationship between frequency, period, and wavelength of light waves? Can you explain the difference between wavelength, frequency, and period of a wave? What is the relationship between amplitude and wavelength, period, and frequency in a wave? What is the relationship between frequency and wavelength? What is the relationship between frequency, wavelength and speed (of a travelling wave)? What is the relationship between the frequency and the wavelength of a wave? What is the relationship between frequency and wavelength? Does it depend on the medium through which the waves are passing? What is the relationship between wavelength, wave number, and frequency? What is the relationship between frequency and wavelength of radio waves? What is the relationship between a wave’s frequency and its wavelength? Can you give some examples of how the relationship between wavelength, period, and frequency can be used to identify a type of wave? What is the relationship between wavelength, wave frequency, and wave velocity? What is the relationship between frequency, period and wavelength of a transverse EM wave? What is the relationship between wavelength frequency and the speed of a wave? Can you explain the relationship between frequency, period and wavelength for waves like sound or light? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10787	https://www.khanacademy.org/a/ee-loop-current-method	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
10788	https://www.pw.live/school-prep/exams/angle-formulas	Angle Formulas: Definition, Different Types, Solved Examples All Courses Competitive Exams IIT JEE, NEET, ESE, GATE, AE/JE, Olympiad Only IAS UPSC, State PSC School Preparation Foundation (Class 6-12), Commerce, Arts, CuriousJr (1st - 8th), Science, International Boards Govt Exam Judiciary, SSC, Defence, Teaching, JAIIB & CAIIB, BIHAR EXAMS WALLAH, UP Exams, Railway, Nursing Exams, Banking, WB Exams UG & PG Entrance Exams MBA, IPMAT, IIT JAM, LAW, CUET UG, UGC NET, GMAT, Design & Architecture, Pharma, CSIR NET, CUET PG, NEET PG FINANCE CA, CS, Finance Courses, ACCA Others Online Degrees English Proficiency Test IELTS, TOEFL Agriculture Agriculture IIT JEE NEET ESE GATE AE/JE Olympiad VidyapeethPW SkillsPW StoreReal TestClass 1st - 8thPower Batch Login/Register SCHOOL EXAMS Angle Formulas Angle Formulas: Definition, Different Types, Solved Examples Angle formulas are fundamental mathematical expressions used to calculate and describe various aspects of angles in geometry and trigonometry. These formulas play a crucial role in solving problems related to angles, both in theoretical mathematics and practical applications. Anchal Singh 8 Sept, 2023 Share Angle formulas are employed to determine the measurements of angles. An angle materializes when two intersecting rays, referred to as the arms of the angle, meet at a common endpoint. The point where these arms intersect is recognized as the vertex of the angle. An angle quantifies the degree of rotation between these two lines. Radians and degrees are both units of measurement for angles. In this discussion, we will explore the formulas for central angles in a circle, as well as formulas for multiple angles and double angles, accompanied by a selection of solved examples. What Are Angle Formulas? In this context, we've covered angle formulas related to the central angle formed by two radii and the corresponding arc in a circle. Now, let's delve into trigonometric formulas for multiple angles and double angles. Also Check - Rational Formula Multiple Angle Formulas Multiple angles typically arise within trigonometric functions. Calculating the precise values of these multiple angles directly can be challenging, but it's achievable by expanding each trigonometric function into its extended form. Expressions for multiple angles like sin(nx), cos(nx), and tan(nx) are established exclusively in terms of sin(x) and cos(x), thanks to Euler's formula and the Binomial Theorem. Mathematics employs a set of multiple angle formula identities for these purposes. Formula 1: The trigonometric expression for multiple angles in terms of sine is as follows: where n=1,2,3,…… General formulas are, Sin2θ =2 × Cosθ.Sinθ Sin3θ =3Sinθ - 4Sin 3 θ Formula 2: The formula for expressing multiple angles in terms of cosine is presented below: where n = 1,2,3 The general formula goes as: Cos2θ = Cos 2 θ – Sin 2 θ Cos3θ = 4Cos 3 θ – 3Cosθ Formula 3: The formula for expressing multiple angles in terms of tangent is as follows: Tan nθ= Sin nθ/ Cos nθ Where n = 1,2,3.... Double Angle Formulas Double angle formulas serve to express trigonometric ratios of double angles (2θ) in terms of those for single angles (θ). These formulas are derived from the sum formulas of trigonometry and can also be derived using Pythagorean identities. To obtain the double angle formulas for sine, cosine, and tangent, we substitute A = B into the sum formulas. Additionally, we derive alternative formulas using Pythagorean identities. Also Check - Algebra Formulas The double angle formulas for sine, cosine, and tangent are as follows: sin 2A = 2 sin A cos A (or) (2 tan A) / (1 + tan 2 A) cos 2A = cos 2 A - sin 2 A (or) 2cos 2 A - 1 (or) 1 - 2sin 2 A (or) (1 - tan 2 A) / (1 + tan 2 A) tan 2A = (2 tan A) / (1 - tan 2 A) What is Central Angle of Circle Formula? The central angle formula for a circle allows us to determine the angle formed between two radii or the angle subtended by the circle's arc and the two radii emanating from the circle's center. The central angle is defined by the radius vectors serving as its arms. To compute the central angle, you need two pieces of information: the measurement of the arc length that the central angle subtends at the circle's center and the radius of the circle. The formula for the central angle of a circle is expressed as follows: Central angle, θ = (Arc length × 360º) / (2πr) degrees or Central angle, θ = Arc length / r radians, where r represents the radius of the circle. Also Check - Absolute Value Formula Angle Formulas Examples Example 1: Jill possesses a circular segment with an arc length of 7π units and a radius of 9 units. Let's determine the segment's angle using the angle formula. Solution: Arc length = 7π (given) Radius = 9 units (given) Utilizing the angle formula: Angle = (Arc Length × 360°) / (2πr) Angle = (7π × 360°) / (2π × 9) = 140° Hence, the angle of the segment is 140 degrees. Example 2: Given that tan A = 3/4, let's determine the values of sin 2A, cos 2A, and tan 2A. Solution: We can use the double angle formulas to find sin 2A, cos 2A, and tan 2A. First, we need to find sin A and cos A using the given information: tan A = 3/4 We can find sin A and cos A using the Pythagorean identity: sin^2 A + cos^2 A = 1 (3/4)^2 + cos^2 A = 1 9/16 + cos^2 A = 1 cos^2 A = 1 - 9/16 cos^2 A = 7/16 cos A = ±√(7/16) Since A is an acute angle, cos A is positive: cos A = √(7/16) cos A = √7/4 cos A = √7/4 2/2 cos A = √7/8 Now that we have sin A and cos A, we can calculate sin 2A, cos 2A, and tan 2A using the double angle formulas: sin 2A = 2 sin A cos A sin 2A = 2 (3/5) (√7/8) sin 2A = (3√7) / 20 cos 2A = cos^2 A - sin^2 A cos 2A = (√7/8)^2 - (3/5)^2 cos 2A = (7/8) - (9/25) cos 2A = (35/40) - (32/40) cos 2A = (3/40) tan 2A = (2 tan A) / (1 - tan^2 A) tan 2A = (2 (3/4)) / (1 - (3/4)^2) tan 2A = (3/2) / (1 - 9/16) tan 2A = (3/2) / (7/16) tan 2A = (3/2) (16/7) tan 2A = 24/7 So, the values are: sin 2A = (3√7) / 20 cos 2A = 3/40 tan 2A = 24/7 Example 3: We aim to prove the identity: 3sinθ - 4sin^3θ cos^3θ - 3cosθ = tan^3θ Solution: We'll use multiple angle formulas to prove this identity. First, we'll express sin^3θ and cos^3θ in terms of sin3θ and cos3θ, and then we'll apply the formulas. Multiple angle formulas state: sin3θ = 3sinθ - 4sin^3θ cos3θ = 4cos^3θ - 3cosθ Now, we can rewrite the identity using these expressions: 3sinθ - 4(sin3θ)cosθ - 3cosθ = tan^3θ Substituting the multiple angle formulas: 3sinθ - 4(3sinθ - 4sin^3θ)cosθ - 3cosθ = tan^3θ Now, simplify the expression: 3sinθ - 12sinθ cosθ + 16sin^3θ cosθ - 3cosθ = tan^3θ Factor out sinθ cosθ: 3sinθ(1 - 4cos^2θ + 16sin^2θ) - 3cosθ = tan^3θ Now, use the Pythagorean identity sin^2θ + cos^2θ = 1: 3sinθ(1 - 4(1 - sin^2θ) + 16sin^2θ) - 3cosθ = tan^3θ Simplify the expression inside the parentheses: 3sinθ(1 - 4 + 4sin^2θ + 16sin^2θ) - 3cosθ = tan^3θ Combine like terms: 3sinθ(20sin^2θ - 3) - 3cosθ = tan^3θ Now, factor out 3sinθ: 3sinθ(20sin^2θ - 3) - 3cosθ = 3sinθtan^3θ Divide both sides by 3sinθ: 20sin^2θ - 3 - cosθ = tan^3θ Now, use the Pythagorean identity sin^2θ = 1 - cos^2θ: 20(1 - cos^2θ) - 3 - cosθ = tan^3θ Distribute the 20: 20 - 20cos^2θ - 3 - cosθ = tan^3θ Combine like terms: -20cos^2θ - cosθ + 17 = tan^3θ Now, we have the identity: -20cos^2θ - cosθ + 17 = tan^3θ So, we've successfully proven the given identity using multiple-angle formulas. Angle Formulas FAQs Define the term Angle Formula. Angle formulas are employed to determine the measurements of angles. An angle materializes when two intersecting rays, referred to as the arms of the angle, meet at a common endpoint. Define the term Angle. An angle materializes when two intersecting rays, referred to as the arms of the angle, meet at a common endpoint. Define the Angle for double angle. cos(2a) = cos2(a)–sin2(a) = 2cos2(a) −1 = 1 − 2sin2(a) sin(2a) = 2sin(a) cos(a) tan(2a) = 2tan(a) ÷ 1−tan2(a) Using the formula for arc length, calculate the length of the arc when the radius (r) is 8 units and the angle (θ) measures π/2 radians. To find the length of the arc with a radius (r) of 8 units and an angle (θ) of π/2 radians, you can use the arc length formula: Arc Length (s) = r θ In this case: r = 8 units θ = π/2 radians Now, plug these values into the formula: Arc Length (s) = 8 (π/2) Simplify the expression: Arc Length (s) = 4π units So, the length of the arc with a radius of 8 units and an angle of π/2 radians is 4π units. 🔥 Trending Blogs NCERT Solutions for Class 6 English A Pact With the Sun NCERT Solutions for Class 6 English A Pact With the Sun NCERT Solutions for Class 9 English Chapter 5 The Snake and the Mirror NCERT Solutions for Class 9 English Chapter 5 The Snake and the Mirror NCERT Solutions Class 9 English Chapter 4 A Truly Beautiful Mind NCERT Solutions Class 9 English Chapter 4 A Truly Beautiful Mind NCERT Solutions Class 9 English Poem Chapter 4 The Lake Isle of Innisfree NCERT Solutions Class 9 English Poem Chapter 4 The Lake Isle of Innisfree NCERT Solutions Class 9 English Poem Chapter 2 Wind NCERT Solutions Class 9 English Poem Chapter 2 Wind Talk to a counsellor Have doubts? 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10789	https://www.metric-conversions.org/length/millimeters-to-meters.htm	Millimeters to Meters (mm to m) Millimeters to Meters (mm to m) conversion calculator 1mm = 0.001m Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select "decimal" from the options above the result. Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result. Note: For a pure decimal result please select "decimal" from the options above the result. Meters to Millimeters (Swap units) Millimeters to Meters formula Meters = Millimeters / 1000 Millimeters to Meters calculation Meters = Millimeters / 1000 Meters = 1 / 1000 Meters = 0.001 What is a millimeter? A millimeter is a thousandth of a meter (1/1000) which is the SI (International System of Units) unit of length. It is normally used to measure small lengths like the thickness of a sheet of paper or the dimensions of a small object. One millimeter is approximately equal to 0.03937 inches (about 1/25th). Presicely there are 25.4 millimeters in an inch and it is often used in science and engineering. It is used in countries that have adopted the metric system. You may come across millimeters when measuring the size of electronic components, jewelry or even the thickness of a fingernail. What is a meter? A meter is a unit of length in the metric system, and it is equivalent to 100 centimeters or 1,000 millimeters. It is the base unit of length in the International System of Units (SI) and is widely used around the world for measuring distances. The meter was originally defined as one ten-millionth of the distance from the North Pole to the equator along a meridian passing through Paris, France. However, in 1983, the meter was redefined as the distance traveled by light in a vacuum during a specific time interval. How do you convert millimeters to meters? To convert millimeters to meters, you need to divide the number of millimeters by 1000. This is because there are 1000 millimeters in one meter. The millimeter is a unit of length in the metric system, commonly used for measuring small distances or dimensions. On the other hand, the meter is the fundamental unit of length in the metric system, used for larger measurements. To perform the conversion, simply take the number of millimeters and divide it by 1000. For example, if you have 5000 millimeters, you would divide 5000 by 1000 to get 5 meters. Similarly, if you have 250 millimeters, dividing it by 1000 would give you 0.25 meters. This conversion is straightforward and can be done using a calculator or mentally by moving the decimal point three places to the left. Millimeters to Meters table Starting value Increment Accuracy 0.00100m 2mm 0.00200m 3mm 0.00300m 4mm 0.00400m 5mm 0.00500m 6mm 0.00600m 7mm 0.00700m 8mm 0.00800m 9mm 0.00900m 10mm 0.01000m 11mm 0.01100m 12mm 0.01200m 13mm 0.01300m 14mm 0.01400m 15mm 0.01500m 16mm 0.01600m 17mm 0.01700m 18mm 0.01800m 19mm 0.01900m Millimeters Meters 20mm 0.02000m 21mm 0.02100m 22mm 0.02200m 23mm 0.02300m 24mm 0.02400m 25mm 0.02500m 26mm 0.02600m 27mm 0.02700m 28mm 0.02800m 29mm 0.02900m 30mm 0.03000m 31mm 0.03100m 32mm 0.03200m 33mm 0.03300m 34mm 0.03400m 35mm 0.03500m 36mm 0.03600m 37mm 0.03700m 38mm 0.03800m 39mm 0.03900m Millimeters Meters 40mm 0.04000m 41mm 0.04100m 42mm 0.04200m 43mm 0.04300m 44mm 0.04400m 45mm 0.04500m 46mm 0.04600m 47mm 0.04700m 48mm 0.04800m 49mm 0.04900m 50mm 0.05000m 51mm 0.05100m 52mm 0.05200m 53mm 0.05300m 54mm 0.05400m 55mm 0.05500m 56mm 0.05600m 57mm 0.05700m 58mm 0.05800m 59mm 0.05900m Millimeters Meters 60mm 0.06000m 61mm 0.06100m 62mm 0.06200m 63mm 0.06300m 64mm 0.06400m 65mm 0.06500m 66mm 0.06600m
10790	https://mathmonks.com/inequalities	Shapes Rectangle Square Circle Triangle Rhombus Squircle Oval Hexagon Pentagon Trapezoid Kite Parallelogram Quadrilateral Polygon Nonagon Heptagon Decagon Octagon Ellipse Parallelepiped Tetrahedron Cylinder Prism Sphere Pyramid Frustum Polyhedron Dodecagon Dodecahedron Octahedron Torus Cube Cone Hyperbola Rectangular Prism Fibonacci Sequence Golden Ratio Parabola Worksheets Calculators Fraction Calculator Mixed Fraction Calculator Greatest Common Factor Calulator Decimal to Fraction Calculator Angle Arithmetic Whole Numbers Rational Numbers Place Value Irrational Numbers Natural Numbers Binary Operation Numerator and Denominator Decimal Order of Operations (PEMDAS) Scientific Notation Symmetry Fractions Triangular Number Complex Number Binary Number System Logarithm Binomial Theorem Quartic Function Mathematical Induction Group Theory Modular Arithmetic Eulers Number Inequalities Sets De Morgans Laws Transcendental Numbers About Us Table of Contents Last modified on June 8th, 2024 chapter outline Inequality is the mathematical symbol used to compare two values or expressions that are not equal. Symbols Here are the four inequality notations or symbols used to write mathematical statements: \| Symbols \| Words \| Examples \| --- \| < \| Less than \| 5x < 2 \| \| > \| Greater than \| 7x > 16 \| \| ¤ \| Less than or equal to \| x – 7 ¤ 24 \| \| ¥ \| Greater than or equal to \| x + 7 ¥ 12 \| Strict Inequality The symbols < and > are known as strict inequalities since the expression on the left of the symbol must be less/greater than the expression on the right. From the above table, 5x < 2 and 7x > 16 are the strict inequalities. Slack Inequality The symbols ¤ and ¥ are slack (weak) inequalities since the expression on the left of the symbols may be equal or less/greater than the expression on the right. From the above table, the slack inequalities are x – 7 ¤ 24 and x + 7 ¥ 12. Moreover, if two or more symbols are present in an expression, they are compound inequalities (sometimes, double inequalities). Properties Transitive It states that if one quantity is less than another, and the second quantity is less than a third quantity, then the first quantity is also less than the third quantity. Mathematically, if a < b and b < c, then a < c Also, we conclude that if a > b and b > c, then a > c For example, if Marco is younger than Jacob, and Jacob is younger than Lucas, then Marco must be younger than Lucas. Addition It states that adding the same quantity on both sides does not change the direction of the inequality. Mathematically, If a > b, then a + c > b + c If a < b, then a + c < b + c For example, if Marco is younger than Jacob, then after 5 years, Marco will still be younger than Jacob. Subtraction It states that subtracting the same quantity from both sides does not change the direction of the inequality. Mathematically, If a > b, then a – c > b – c If a < b, then a – c < b – c For example, if Jacob is younger than Lucas, then 5 years ago, Jacob was still younger than Lucas. Multiplication It states that multiplying both sides by the same positive quantity does not change the direction of the inequality. Mathematically, If a > b and c > 0, then ac > bc If a < b and c > 0, then ac < bc For example, Marco is younger than Lucas. If they double their ages, Marco will still be younger than Lucas. However, multiplying both sides by the same negative quantity flips the direction of the inequalities, which means that: If a > b and c < 0, then ac < bc Also, if a < b and c < 0, then ac > bc Division It states that dividing both sides by the same positive quantity does not change the direction of the inequality. Mathematically, If a > b and c > 0, then ${\dfrac{a}{c} >\dfrac{b}{c}}$ If a < b and c > 0, then ${\dfrac{a}{c} <\dfrac{b}{c}}$ However, dividing both sides by the same negative quantity flips the direction of the inequalities, which means: If a > b and c < 0, then ${\dfrac{a}{c} <\dfrac{b}{c}}$ If a < b and c < 0, then ${\dfrac{a}{c} >\dfrac{b}{c}}$ For example, if x + 5 > -2, then on dividing both sides by (-1), we get ${\dfrac{x+5}{-1} <\dfrac{-2}{-1}}$ ${-\left( x+5\right) <2}$ Inversion It states that taking the inverse of both sides changes the direction of the inequality. Mathematically, If a > b, then ${\dfrac{1}{a} <\dfrac{1}{b}}$ If a < b, then ${\dfrac{1}{a} >\dfrac{1}{b}}$ For example, if 5 > 2, then ${\dfrac{1}{5} <\dfrac{1}{2}}$ However, if a > b, then -a < -b, and if a < b, then -a > -b For example, since the numbers on the left are the smaller numbers on the number line. Thus, from 2 to 9 is an increase, but from -2 to -9 is a decrease, as shown. Solving We solve inequalities to find the value of an unknown variable in an expression. If the variable is already independent, solving those basic inequalities is unnecessary. For example, x ¤ -11, y < 2, and z > 21 are the solved inequalities. Now, let us solve the inequality 3x – 7 < 2 + 8x On subtracting 3x from both sides (subtraction property), we get 3x – 7 – 3x < 2 + 8x – 3x -7 < 2 + 5x On subtracting 2 from both sides (subtraction property), we get -7 – 2 < 2 + 5x – 2 -9 < 5x On dividing both sides by 5 (division property), we get ${\dfrac{-9}{5} <\dfrac{5x}{5}}$ ${\dfrac{-9}{5} ${x >\dfrac{-9}{5}}$ : Simplify ${\dfrac{a+5}{-3}\leq a+2}$ Solution: Here, ${\dfrac{a+5}{-3}\leq a+2}$Using multiplication property, we get${\left( -3\right) \left( \dfrac{a+5}{-3}\right) \geq \left( -3\right) \left( a+2\right)}$ ${\left( a+5\right) \geq \left( -3a-6\right)}$Using addition property, we get${a+5+3a\geq -3a-6+3a}$ ${4a+5\geq -6}$Using the subtraction property, we get${4a+5-5\geq -6-5}$ ${4a\geq -11}$Using the division property, we get${\dfrac{4a}{4}\geq \dfrac{-11}{4}}$ ${a\geq \dfrac{-11}{4}}$Thus, ${a\geq \dfrac{-11}{4}}$ Graphing For One Variable To plot the inequalities with one variable, we use a number line. Now, graphing the inequalities ¤ -8 and b > -5, we get: We observe that the solutions of inequalities are the rays graphed on the number line. Since the variables are already independent, these inequalities can be considered solved. Here, The closed circle denotes that a can equal -8. Thus, for slack inequalities (¤ and ¥), a closed dot indicates the endpoint to be a part of the solution. The open circle denotes that b is not equal to -5. Thus, for the strict inequalities (< or >), an open dot indicates the endpoint is not part of the solution. For Two Variables Let us plot the inequality 3x + 1 > y on the graph. First, we plot y = 3x + 1, as shown. Now, shading the area, we get Here, the dashed line is formed since 3x + 1 > y (less than symbol). However, we make it a solid line for y ¤ or ¥ y. Now, let us solve the inequality 5x – 4 < 0 very easily using graphs. First, we sketch a graph of y = 5x – 4 as shown. We observe that the line has a slope of 5 and an intercept on the y-axis of 4. If y = 0, x = ${\dfrac{4}{5}}$ By finding the values of x for which 2x + 3 is less than zero, we get those points on the graph where y is less than zero, which corresponds to the values of x less than ${\dfrac{4}{5}}$. This is the solution of the inequality. We mark this range on the graph, using the x-axis as the number line. : Plot the graph of the inequality ${2y-\dfrac{x}{2}\leq 2}$ Solution: Here, ${2y-\dfrac{x}{2}\leq 2}$ ${4y-x\leq 4}$ (by multiplication property) ${4y\leq x+4}$ (by addition property) ${y\leq \dfrac{x}{4}+1}$ (by division property)Since the given one is a weak inequality, a solid line is formed. Now, shading the area of the graph ${y=\dfrac{x}{4}+1}$, we get the solution as shown. More Resources Graphing Inequalities Compound Inequality Solving Inequalities Absolute Value Inequalities Linear Inequalities System of Inequalities Graphing Inequalities on a Number Line Inequality Word Problems Quadratic Inequalities Cauchy Schwarz Inequality Markov and Chebyshevs Inequality Jensens Inequality HÃ¶lders Inequality Reverse Triangle Inequality AM-GM Inequality Hoeffding Inequality Chernoff Inequality Rational Inequalities Taylors Inequality Minkowski Inequality Youngs Inequality Bernoulli Inequality Bernstein Inequality Bonferroni Inequality Triangle Inequality for Vectors Last modified on June 8th, 2024 Related Materials Absolute Value Inequalities Worksheets Compound Inequalities Worksheets Graphing Inequalities Worksheets Inequalities Worksheets Inequality Word Problems Worksheets Linear Inequalities Worksheets Multi-Step Inequalities Worksheets One Step Inequalities Worksheets Quadratic Inequalities Worksheets Rational Inequalities Worksheets Solving Equations and Inequalities Worksheets Systems of Inequalities Worksheets Two-Step Inequalities Worksheets Writing Inequalities Worksheets About us Contact us Privacy Policy Categories Algebra Arithmetic Geometry Statistics Trigonometry Grades 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade 7th Grade 8th Grade 9th Grade 10th Grade 11th Grade 12th Grade Join Our Newsletter © 2025 Mathmonks.com. 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Sep 22 Special Offer: Get 25% Off Target Test Prep GMAT Plans 12:00 PM EDT 11:59 PM EDT The Target Test Prep GMAT Flash Sale is LIVE! Get 25% off our game-changing course and save up to $450 today! Use code FLASH25 at checkout. This special offer expires on September 30, so grab your discount now! Back to Forum Create Topic Reply Mixture Alligation Sahil15 Sahil15 Joined: 18 Jul 2017 Last visit: 05 May 2018 Posts: 5 GMAT 1:660 Q49 V31 GMAT 1:660 Q49 V31 Posts: 5 Post URL18 Jul 2017, 11:46 Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C? Please explain how to solve it by alligation method 4 Kudos Add Kudos 6 Bookmarks Bookmark this Post Most Helpful Reply DeeptiManyaExpert DeeptiManyaExpert Joined: 13 Jul 2019 Last visit: 12 Jun 2022 Posts: 49 Posts: 49 Post URL16 Jul 2019, 05:34 This is a simple mixture and alligation question, which is made complicated by providing the different quantities of the Alloys. Keep few basic rules in your mind and you will become pro in solving similar questions. Rule 1: Remember that the quantity does not matter to get the mixture ratio. So, forget about the quantities at first place. Rule 2: If the material to be mixed is also made up of different materials, convert the ratio into fraction by picking one of the components. If you have understood this well, you are ready to go. Step 1: Let's start by converting ratios into fractions. You can pick either copper or tin (It's all your wish). I am taking Tin in order to show that you will still get the same answer. Tin in First Alloy: 7/8 Tin in Second Alloy: 4/7 Step 2: We do not know anything about the weightage of tin in the mixture. Let it be X. Step 3: Final ratio is 1:2. Set an alligation. 7/8 4/7 \ / \ / \ / X / \ / \ / \ 1 2 Use formula for alligation: 7/8 - X = 2 X - 4/7 1 Solve to get, X = 113/168 => Quantity of Tin in the mixture = 113 Therefore, Copper = 168 - 113 = 55 Thus, ratio of Cu to Tin in the mixture will be 55:113. Give kudos if you liked the solution. Deepti Singh Manya - The Princeton Review 8 Kudos Add Kudos 3 Bookmarks Bookmark this Post General Discussion mikemcgarry mikemcgarry Magoosh GMAT Instructor Joined: 28 Dec 2011 Last visit: 06 Aug 2018 Posts: 4,480 Expert Expert reply Posts: 4,480 Post URL18 Jul 2017, 16:41 Sahil15 Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C? Please explain how to solve it by alligation method Show more Dear Sahil15, I'm happy to respond. My friend, you are relatively new to GMAT Club. Welcome. A few comments. This subforum is not really the appropriate forum for this particular forum. The "Ask GMAT Experts" subforum is for general questions--strategies, retake advice, rule changes on the GMAT itself, etc. What you have asked is an individual math question. An individual math question should be posted in the Quant subforum. I will also say, it's a particularly bad approach to post a problem and say that you want to see it solved in a way that's familiar to you. Instead of clinging to the familiar, you should welcome as many different ways to think about a problem as possible. The whole point of the GMAT is to be intellectually flexible. Here's a blog you may find germane. GMAT Solution and Mixing Problems Here's how I would think about the problem. In what follows, I will use the elemental abbreviations (Copper = Cu, Tin = Sn) Alloy A of 24 kilograms contains Copper and Tin in 1:7. Alloy B of 28 kilograms contains Copper and Tin in 3:4. Hold on to those two facts for a moment. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. Well, that's 6 kg of A and 12 kg of B. Let's look at those individual. First, look at 6k of A. In Alloy A, Cu:Sn = 1:7, so copper is 1/8 of the sample. We know 1/8 of 6 kg would be 6/8 = 3/4 kg. That's the amount of copper, and tin would be the rest, 5.25 kg = 21/4 kg. Now, look at 12 kg of B. In Alloy B, Cu:Sn = 3:4, so copper is 3/7 and tin is 4/7. These are the really ugly ratios in this problem. amount of copper = 12(3/7) = 36/7 kg amount of tin = 12(4/7) = 48/7 kg I will leave those as improper fractions for the moment. In Alloy C, the total amounts we have, in kg, are amount of copper = 3/4 + 36/7 = 21/28 + 144/28 = 165/28 amount of tin = 21/4 + 48/7 = 147/28 + 192/28 = 339/28 Thus, in Alloy C, Cu:Sn = 165/339 = 55/113 That's a particularly ugly problem. Does all this make sense? Mike Mike McGarry Magoosh Test Prep Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939) Signature Read More 2 Kudos Add Kudos 2 Bookmarks Bookmark this Post Bunuel Bunuel Math Expert Joined: 02 Sep 2009 Last visit: 29 Sep 2025 Posts: 104,360 Products: Expert Expert reply Active GMAT Club Expert! Tag them with @ followed by their username for a faster response. Posts: 104,360 Post URL19 Jul 2017, 03:30 Sahil15 Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C? Please explain how to solve it by alligation method Show more Moved to Quantitative forum. P.S. Please read carefully and follow our RULES OF POSTING. Thank you. New to the GMAT Club? • Questions: Question Bank \| Bunuel's Signature Collection • Tests: GMAT Club Tests \| Forum Quiz • Guides: Quantitative \| Verbal \| Ultimate Quantitative Collection \| All You Need for Quant • Rules: Quantitative \| Verbal Signature Read More Kudos Add Kudos Bookmarks Bookmark this Post IanStewart IanStewart GMAT Tutor Joined: 24 Jun 2008 Last visit: 28 Sep 2025 Posts: 4,150 Expert Expert reply Posts: 4,150 Post URL19 Jul 2017, 06:42 Sahil15 Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C? Please explain how to solve it by alligation method Show more I can show you how to solve this using alligation (and the below might not make much sense to anyone reading this who doesn't know that method), but this isn't like a real GMAT weighted average problem. Alligation is normally really easy on a real GMAT problem, but here it's really messy because the fractions become a bit horrible to work with. The actual amounts (24 kg, 28 kg, 18 kg) are completely irrelevant, so we can ignore those numbers. The first mixture is 1/8 copper, the second is 3/7 copper. When we mix them in a 1:2 ratio, we'll be finding the fraction that is 2/3 of the way along between 1/8 and 3/7, so we want the value of X on the number line below: ---1/8--------------X-------3/7--- Here the distance between 1/8 and X is twice as big as the distance between X and 3/7. In other words, this equation is true: X - 1/8 = 2 (3/7 - X) and if you solve that, you'll find X. Or you can get the same denominator for both fractions, and then multiply by 3 on the top and bottom (so that the distance between them is easier to divide in a 2 to 1 ratio) to get this number line, which is identical to the one above but easier to work with: ---21/168-------------X-----72/168---- The total distance between 21 and 72 is 51, so 2/3 of that distance is 34, and X = 21/168 + 34/168 = 55/168. X is the proportion of the final mixture that is copper, but we want the ratio of copper to tin, so the answer is 55 to 168-55, or 55 to 113. now teaching gmat focus ianstewartgmat.com Signature Read More 1 Kudos Add Kudos 1 Bookmarks Bookmark this Post Sahil15 Sahil15 Joined: 18 Jul 2017 Last visit: 05 May 2018 Posts: 5 GMAT 1:660 Q49 V31 GMAT 1:660 Q49 V31 Posts: 5 Post URL20 Jul 2017, 08:15 IanStewart Sahil15 Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C? Please explain how to solve it by alligation method I can show you how to solve this using alligation (and the below might not make much sense to anyone reading this who doesn't know that method), but this isn't like a real GMAT weighted average problem. Alligation is normally really easy on a real GMAT problem, but here it's really messy because the fractions become a bit horrible to work with. The actual amounts (24 kg, 28 kg, 18 kg) are completely irrelevant, so we can ignore those numbers. The first mixture is 1/8 copper, the second is 3/7 copper. When we mix them in a 1:2 ratio, we'll be finding the fraction that is 2/3 of the way along between 1/8 and 3/7, so we want the value of X on the number line below: ---1/8--------------X-------3/7--- Here the distance between 1/8 and X is twice as big as the distance between X and 3/7. In other words, this equation is true: X - 1/8 = 2 (3/7 - X) and if you solve that, you'll find X. Or you can get the same denominator for both fractions, and then multiply by 3 on the top and bottom (so that the distance between them is easier to divide in a 2 to 1 ratio) to get this number line, which is identical to the one above but easier to work with: ---21/168-------------X-----72/168---- The total distance between 21 and 72 is 51, so 2/3 of that distance is 34, and X = 21/168 + 34/168 = 55/168. X is the proportion of the final mixture that is copper, but we want the ratio of copper to tin, so the answer is 55 to 168-55, or 55 to 113. Show more Thank you so much for the answer Kudos Add Kudos Bookmarks Bookmark this Post Sahil15 Sahil15 Joined: 18 Jul 2017 Last visit: 05 May 2018 Posts: 5 GMAT 1:660 Q49 V31 GMAT 1:660 Q49 V31 Posts: 5 Post URL20 Jul 2017, 08:19 mikemcgarry Sahil15 Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C? Please explain how to solve it by alligation method Dear I'm happy to respond. My friend, you are relatively new to GMAT Club. Welcome. A few comments. This subforum is not really the appropriate forum for this particular forum. The "Ask GMAT Experts" subforum is for general questions--strategies, retake advice, rule changes on the GMAT itself, etc. What you have asked is an individual math question. An individual math question should be posted in the Quant subforum. I will also say, it's a particularly bad approach to post a problem and say that you want to see it solved in a way that's familiar to you. Instead of clinging to the familiar, you should welcome as many different ways to think about a problem as possible. The whole point of the GMAT is to be intellectually flexible. Here's a blog you may find germane. Here's how I would think about the problem. In what follows, I will use the elemental abbreviations (Copper = Cu, Tin = Sn) Alloy A of 24 kilograms contains Copper and Tin in 1:7. Alloy B of 28 kilograms contains Copper and Tin in 3:4. Hold on to those two facts for a moment. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. Well, that's 6 kg of A and 12 kg of B. Let's look at those individual. First, look at 6k of A. In Alloy A, Cu:Sn = 1:7, so copper is 1/8 of the sample. We know 1/8 of 6 kg would be 6/8 = 3/4 kg. That's the amount of copper, and tin would be the rest, 5.25 kg = 21/4 kg. Now, look at 12 kg of B. In Alloy B, Cu:Sn = 3:4, so copper is 3/7 and tin is 4/7. These are the really ugly ratios in this problem. amount of copper = 12(3/7) = 36/7 kg amount of tin = 12(4/7) = 48/7 kg I will leave those as improper fractions for the moment. In Alloy C, the total amounts we have, in kg, are amount of copper = 3/4 + 36/7 = 21/28 + 144/28 = 165/28 amount of tin = 21/4 + 48/7 = 147/28 + 192/28 = 339/28 Thus, in Alloy C, Cu:Sn = 165/339 = 55/113 That's a particularly ugly problem. Does all this make sense? Mike Show more Dear Mike, Thank you for the reply. Apologies for posting the question incorrectly. I was able to get the answer in this way, but I was having problem with the alligation method hence asked for that. I know both ways now. Thanks again Kudos Add Kudos Bookmarks Bookmark this Post Bismarck Bismarck Joined: 18 Jun 2018 Last visit: 15 Mar 2023 Posts: 217 Posts: 217 Post URL07 Oct 2018, 05:04 Sahil15 Alloy A of 24 kilograms contains Copper and Tin in 1:7 and Alloy B of 28 kilograms contains Copper and Tin in 3:4. Both alloys are mixed in 1:2 and Alloy C of 18 kilograms is created. What is the ratio of Copper and Tin in Alloy C? Please explain how to solve it by alligation method Show more Alloy C : 18 Kg contains 1 3∗18=6 1 3∗18=6 Kg Alloy A and 2 3∗18=12 2 3∗18=12 Kg Alloy B Alloy A : 24 Kg contains 1 8∗24=3 1 8∗24=3 Kg Copper and 7 8∗24=21 7 8∗24=21 Kg Tin 6 Kg contains 1 8∗6=3 4 1 8∗6=3 4 Kg Copper and 7 8∗6=21 4 7 8∗6=21 4 Kg Tin Alloy B : 28 Kg contains 3 7∗28=12 3 7∗28=12 Kg Copper and 4 7∗28=16 4 7∗28=16 Kg Tin 12 Kg contains 3 7∗12=36 7 3 7∗12=36 7 Kg Copper and 4 7∗12=48 7 4 7∗12=48 7 Kg Tin Alloy C contains 3 4+36 7 3 4+36 7 Kg Copper and 21 4+48 7 21 4+48 7 Kg Tin. Alloy C contains 165 28 165 28 Kg Copper and 339 28 339 28 Kg Tin. Ratio of Copper and Tin in Alloy C =165 28 339 28=165 339=55 113=165 28 339 28=165 339=55 113 1 Kudos Add Kudos Bookmarks Bookmark this Post ilepton ilepton Joined: 26 Nov 2018 Last visit: 09 Aug 2022 Posts: 6 Posts: 6 Post URL12 Jul 2019, 05:42 This Question can be done much easily , just ignore the kgs of a particular mixtures and imagine if both of them are mixed with these proportions then whatever i will take from resultant mixture either 18kg or x kg it will always be in that same proportion. Now A is given in 1:7 therefore let’s assume it is 1+7 = 8 units (you can assume it as 8x as well ) B is 3:4 therefore (3+4 = 7 units ) now let’s make them equal solutions and then add in ratios of 1:2 Therefor let’s multiply the 1st ratio by 7 and 2nd one by 8 making them both equal to 56 units and then again multiplying 2nd by 2 as the need to be mixed in 1:2 This will give me resultant soln as (1x7 + 3x4x2): (7x7 + 4x8x2) => 55:113 Now if the question asked what is 1st quantity in the solution then it will be (55x18)/113. Posted from my mobile device Kudos Add Kudos Bookmarks Bookmark this Post bumpbot bumpbot Non-Human User Joined: 09 Sep 2013 Last visit: 04 Jan 2021 Posts: 38,123 Posts: 38,123 Post URL24 Jan 2025, 21:16 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). 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10792	https://www.purplemath.com/modules/workprob.htm	Select a Course Below "Work" Word Problems Painting & PipesTubs & Man-HoursUnequal TimesEtc. Purplemath "Work" problems usually involve situations such as two people working together to paint a house. You are usually told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together. Many of these problems are not terribly realistic — since when can two laser printers work together on printing one report? — but it's the technique that they want you to learn, not the applicability to "real life". The method of solution for "work" problems is not obvious, so don't feel bad if you're totally lost at the moment. There is a "trick" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time. For instance: Content Continues Below MathHelp.com Suppose one painter can paint the entire house in twelve hours, and the second painter takes eight hours to paint a similarly-sized house. How long would it take the two painters together to paint the house? If the first painter can do the entire job in twelve hours and the second painter can do it in eight hours, then (this here is the trick!) the first guy can do 1/12 of the job per hour, and the second guy can do 1/8 per hour. The question then becomes, how much then can they do per hour if they work together? To find out how much they can do together per hour, I make the necessary assumption that their labors are additive (in other words, that they never get in each other's way in any manner), and I add together what they can do individually per hour. So, per hour, their labors are: 1/12 + 1/8 = 5/24 In other words, they can do 5/24 of the job per hour. But the exercise didn't ask me how much they can do per hour; it asked me how long they'll take to finish one whole job, working togets. So now I'll pick the variable "t" to stand for how long they take (that is, the time they take) to do the job together. Then they can do: 1/t per hour This gives me an expression for their combined hourly rate. I already had a numerical expression for their combined hourly rate. So, setting these two expressions equal, I get: 5/24 = 1/t I can solve by flipping the equation; I get: t = 24/5 = 4.8 hours An hour has sixty minutes, so 0.8 of an hour has forty-eight minutes. Then: They can complete the job together in 4 hours and 48 minutes. Content Continues Below The important thing to understand about the above example is that the key was in converting how long each person took to complete the task into a rate. Each person took a certain number of hours to complete the task: hours to complete job: first painter: 12 second painter: 8 together: t Since the unit for completion was "hours", I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour. To do this, I simply inverted each value for "hours to complete job": completed per hour: first painter: 1/12 second painter: 1/8 together: 1/t Then, assuming that their per-hour rates were additive, I added the portion that each could do per hour, summed them, and set this equal to the "together" rate: adding their labor: 1/12 + 1/8 = 1/t 5/24 = 1/t 24/5 = t As you can see in the above example, "work" problems commonly create rational equations. But the equations themselves are usually pretty simple to solve. One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used? My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together. In this case, I know the "together" time, but not the individual times. One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times. Affiliate Advertisement You might wonder how to work with that "1.25 times as fast" part. If you're not sure, try doing similar computations with simpler numbers. For instance, if the faster pipe filled the pool 2 times as fast as the second pipe, then it would take one-half as long to fill the pool; ½ is the reciprocal of 2. If the faster pipe filled the pool 3 times as fast as the second pipe, then it would take one-third as long to fill the pool; 1/3 is the reciprocal of 3. Following this same reasoning of using the reciprocal, since the faster pipe fills the pool 1.25 = 5/4 as fast as does the second pipe, then it takes 4/5 as long as does the second pipe to fill the pool. Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time: hours to complete job: slow pipe: s fast pipe: 4/5s together: 5 Next, I'll convert all of the completion times to per-hour rates: completed per hour: slow pipe: 1/s fast pipe: 5/(4s) together: 1/5 Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate: adding their labor: 1/s + 5/(4s) = 1/5 multiplying through by 20s (being the lowest common denominator of all the fractional terms): 20 + 25 = 4s 45 = 4s 45/4 = 11.25 = s They asked me for the time of the slower pipe, so I don't need to find the time for the faster pipe. My answer is: The slower pipe takes 11.25 hours. Affiliate Affiliate Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable. If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him. In this case, if he goes 1.25 times as fast, then you take 1.25 times as long. So the variables could have been "f" for the number of hours the faster pipe takes, and then the number of hours for the slower pipe would have been "1.25f". URL: Page 1Page 2Page 3Page 4 hours to complete job: first painter: 12 second painter: 8 together: t Since the unit for completion was "hours", I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour. To do this, I simply inverted each value for "hours to complete job": completed per hour: first painter: 1/12 second painter: 1/8 together: 1/t Then, assuming that their per-hour rates were additive, I added the portion that each could do per hour, summed them, and set this equal to the "together" rate: adding their labor: 1/12 + 1/8 = 1/t 5/24 = 1/t 24/5 = t As you can see in the above example, "work" problems commonly create rational equations. But the equations themselves are usually pretty simple to solve. One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used? My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together. In this case, I know the "together" time, but not the individual times. One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times. Affiliate Advertisement You might wonder how to work with that "1.25 times as fast" part. If you're not sure, try doing similar computations with simpler numbers. For instance, if the faster pipe filled the pool 2 times as fast as the second pipe, then it would take one-half as long to fill the pool; ½ is the reciprocal of 2. If the faster pipe filled the pool 3 times as fast as the second pipe, then it would take one-third as long to fill the pool; 1/3 is the reciprocal of 3. Following this same reasoning of using the reciprocal, since the faster pipe fills the pool 1.25 = 5/4 as fast as does the second pipe, then it takes 4/5 as long as does the second pipe to fill the pool. Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time: hours to complete job: slow pipe: s fast pipe: 4/5s together: 5 Next, I'll convert all of the completion times to per-hour rates: completed per hour: slow pipe: 1/s fast pipe: 5/(4s) together: 1/5 Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate: adding their labor: 1/s + 5/(4s) = 1/5 multiplying through by 20s (being the lowest common denominator of all the fractional terms): 20 + 25 = 4s 45 = 4s 45/4 = 11.25 = s They asked me for the time of the slower pipe, so I don't need to find the time for the faster pipe. My answer is: The slower pipe takes 11.25 hours. Affiliate Affiliate Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable. If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him. In this case, if he goes 1.25 times as fast, then you take 1.25 times as long. So the variables could have been "f" for the number of hours the faster pipe takes, and then the number of hours for the slower pipe would have been "1.25f". URL: Page 1Page 2Page 3Page 4 Select a Course Below Standardized Test Prep K12 Math College Math Homeschool Math Share This Page Visit Our Profiles © 2024 Purplemath, Inc. All right reserved. Web Design by
10793	https://math.stackexchange.com/questions/4547120/how-to-find-initial-guess-for-newton-raphson-method	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 How to find initial guess for Newton-Raphson method Ask Question Asked Modified 2 years, 11 months ago Viewed 3k times 0 $\begingroup$ I have a polynomial equation $x^3-2x^2-5=0$ for which I have to find solutions accurate to $10^{-4}$ in the interval $[1,4]$. How can I find initial guess $P_0$? newton-raphson Share edited Oct 7, 2022 at 8:12 Rócherz 4,27144 gold badges1515 silver badges3030 bronze badges asked Oct 7, 2022 at 5:40 the19thcenturydeveloperthe19thcenturydeveloper 2311 silver badge55 bronze badges $\endgroup$ Add a comment \| 2 Answers 2 Reset to default 1 $\begingroup$ As far as I know, there is no general criterion to decide the best initial point. The algorithm will (eventually) converge to one of the possible roots. As a rule of thumb, you should start from a point inside an interval $I=[a,b]$, such that the function is continuous in $I$ and $f(a)\cdot f(b)<0$. This ensures that at least a solution there exists in $I$. If you guarantee that in $I$ the function is smooth, namely with continuous high-order derivatives, the algorithm will run smoother. In your case, I suggest you choose the middle point $x_0=2.5$. FYI, cubic polynomial admits closed-form solution, have a look on the Cardano-Tartaglias Formula, e.g. Click here Remark: The algorithm just gives you one root, say $x_1$. If $p(x)=x^3-2x^2-5$ is your polynomial function, then you can find $p_1(x)=p(x)/(x-x_1)$ by using the polynomial division algorithm. Applying the Newton-method to $p_1(x)$ will give you another root, say $x_2$. Then, define $p_2(x)=p_1(x)/(x-x_2)$ gives another root $x_3$, and so on... In your case, upon finding the first root, $p_1(x)$ is just a parabola whose roots are easy to be found. Share edited Oct 7, 2022 at 8:10 Rócherz 4,27144 gold badges1515 silver badges3030 bronze badges answered Oct 7, 2022 at 6:05 GinoGino 37211 silver badge1010 bronze badges $\endgroup$ 1 $\begingroup$ I think that the question is $x_0=1$ or $x_0=4$ $\endgroup$ Claude Leibovici – Claude Leibovici 2022-10-07 06:31:26 +00:00 Commented Oct 7, 2022 at 6:31 Add a comment \| 1 $\begingroup$ Provided the continuity of the second derivative, if you want to avoid overshoot of the solution, you must select $x_0$ such that $$f(x_0)\times f''(x_0) >0$$ This is Darboux theorem. So, in your case, $x_0=???$ For illustration, run the two possible cases Share answered Oct 7, 2022 at 6:05 Claude LeiboviciClaude Leibovici 294k5555 gold badges130130 silver badges316316 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 newton-raphson 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 Finding reciprocal via Newton Raphson: How to determine initial guess? 1 Apply Newton-Raphson method to find the solutions to 0 Is there general formula for initial guess of Newton-Raphson method for finding roots 1 Initial guess in Newton-Raphson method. 0 Best initial guess for Newton-Raphson method to find a real root of the cubic polynomial 1 How to determine the almost accurate initial guess to calculate the $\ln(x)$ via Newton-Raphson method? 1 Choosing the initial seed point in Newton Raphson method 2 Convergence Interval for the Newton-Raphson Method Hot Network Questions Are credit card statements required for Greece Schengen visa application? 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10794	https://www.wolframcloud.com/obj/fb14b5dd-45e1-4848-b4f2-0a15879a6f83?src=CloudBasicCopiedContent	Binomial Theorem (Step-by-Step) WOLFRAM NOTEBOOK ---------------- Make Your Own Copy Download Share Info Sign In Binomial Theorem (Step-by-Step) n 2 3 4 5 6 4 (x+y)=4 ∑ k=0  4 k  k x 4-k y = 4 0  0 x 4-0 y + 4 1  1 x 4-1 y + 4 2  2 x 4-2 y + 4 3  3 x 4-3 y + 4 4  4 x 4-4 y =4 y +4 x 3 y +6 2 x 2 y +4 3 x y+ 4 x The binomial theorem says that for positive integer n, n (x+y) = n ∑ k=0  n k  k x n-k y ,where  n k = n! k!(n-k)! .This widely useful result is illustrated here through termwise expansion. External Links Binomial Theorem (Wolfram MathWorld) Permanent Citation Bruce Colletti,Jeff Bryant "Binomial Theorem (Step-by-Step)" Wolfram Demonstrations Project Published: December 12, 2007 AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA AAAAAAAAAA
10795	https://stemtc.scimathmn.org/frameworks/732b-translations-reflections-coordinate-grid	7.3.2B Translations & Reflections on a Coordinate Grid \| Minnesota STEM Teacher Center Skip to main navigationSkip to main content Minnesota STEM Teacher Center Contact Glossary Keyword(s) Operator Contains all of these words Content type [x] Download [x] Framework [x] Basic page Home Frameworks Standards Resources About Us 7.3.2B Translations & Reflections on a Coordinate Grid SharePrint Standard 7.3.2 Analyze the effect of change of scale, translations and reflections on the attributes of two-dimensional figures. Grade: 7 Subject: Math Strand: Geometry & Measurement Benchmark: 7.3.2.4 Translations & Reflections on a Coordinate Grid Graph and describe translations and reflections of figures on a coordinate grid and determine the coordinates of the vertices of the figure after the transformation. For example: The point (1, 2) moves to (-1, 2) after reflection about the y-axis. Overview Big Ideas and Essential Understandings Standard 7.3.2 Essential Understandings Students connect their work on proportionality with their work on measurement of two - and three-dimensional shapes by investigating similar objects. They understand that if a scale factor describes how corresponding lengths in two similar objects are related, then the square of the scale factor describes how corresponding areas are related, and the cube of the scale factor describes how corresponding volumes are related. Students apply their work on proportionality to measurement in different contexts, including converting among different units of measurement to solve problems involving rates, such as motion at a constant speed. The big idea of this standard is 'scale factor.' Can students identify similar objects and the scale factor that defines this relationship? Can students use this identified scale factor to calculate measurements? In addition to similarity, students will be expected to recognize and perform transformations on a coordinate grid. In 4th grade, students are introduced to transformation vocabulary and asked to identify types of transformations on objects. In 7th grade, students extend this work with reflections and translations to the coordinate plane and are introduced to and expected to use appropriate notation. In high school, students will extend the work further to rotations and dilations. All Standard Benchmarks 7.3.2.1 Describe the properties of similarity, compare geometric figures for similarity, and determine scale factors For example: Corresponding angles in similar geometric figures have the same measure. 7.3.2.2 Apply scale factors, length ratios and area ratios to determine side lengths and areas of similar geometric figures. For example: If two similar rectangles have heights of 3 and 5, and the first rectangle has a base of length 7, the base of the second rectangle has length 35 3 35 3. 7.3.2.3 Use proportions and ratios to solve problems involving scale drawings and conversions of measurement units. For example: 1 square foot equals 144 square inches. For example: In a map where 1 inch represents 50 miles, .5 inch represents 25 miles. 7.3.2.4 Graph and describe translations and reflections of figures on a coordinate grid and determine the coordinates of the vertices of the figure after the transformation. For example: The point (1, 2) moves to (-1, 2) after reflection about the y-axis. Benchmark Cluster 7.3.2 Benchmark Group B - Translations & Reflections on a Coordinate Grid 7.3.2.4 Graph and describe translations and reflections of figures on a coordinate grid and determine the coordinates of the vertices of the figure after the transformation. For example: The point (1, 2) moves to (-1, 2) after reflection about the y-axis. What students should know and be able to do [at a mastery level] related to these benchmarks: Be able to, given a rule or word description, reflect a point or translate a point to its new location; Given a point and its new image (transformation), be able to identify the rule, either in rule format (x- 6, y - 2) or in description format (6 left and 2 down); Be able to reflect over y-axis, x-axis, or the lines y =x or y = -x; Be able to identify the new coordinates of the transformed image. Work from previous grades that supports this new learning includes: Graph in the first quadrant; Identify coordinates; Draw simple geometric figures; Understand vertex/vertices. Correlations NCTM Standards Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships: Understand relationships among the angles, side lengths, perimeters, areas and volumes of similar objects; Specify locations and describe spatial relationships using coordinate geometry and other representational systems: Use coordinate geometry to represent and examine and properties of geometric shapes. Apply transformations and use symmetry to analyze mathematical situations: Describe sizes, positions and orientations of shapes under informal transformations such as flips, turns, slides and scaling; Apply appropriate techniques, tools and formulas to determine measurements: Solve problems involving scale factors, using ratio and proportion. Common Core State Standards (CCSS) 7.G (Geometry) Draw, construct and describe geometrical figures and describe the relationships between them. 7.G.1. Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale. 8.G (Geometry) Understand congruence and similarity using physical models, transparencies, or geometry software. 8.G.3. Describe the effect of dilations, translations, rotations and reflections on two-dimensional figures using coordinates. 8.G.4. Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them. 9.GSRT (Similarity, Right Triangles, and Trigonometry)Understand similarity in terms of similarity transformations. 9.GSRT.2.Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides. Misconceptions Student Misconceptions Student Misconceptions and Common Errors Students sometimes believe that a reflection over y =x or y = -x is like a rotation of it. The correct reflection will actually look wrong compared to the original drawing. One way to help with this is to have the students physically fold their paper to make sure it does reflect over the correct line Students may confuse specific math vocabulary (reflection, rotation, translation) with the more casual math vocabulary for these terms (flip, turn, slide) that they learned in elementary grades. Students may identify the incorrect line of symmetry (x-axis vs. y-axis) when reflecting. When labeling transformed images, students sometimes mistakenly apply the wrong label. Either they use the wrong letter, forget the prime notation (e.g., A→A A→A instead of A→A′A→A′) , or do the wrong number of primes if there is more than one transformation on the graph (ex. A→A′A→A′ instead of A→A′′A→A″) . In translations, students can sometimes count to their points incorrectly or switch the order of the coordinates when writing the point down. Vignette In the Classroom This activity explores transformations - specifically, reflections. Teacher: Today we are going to work on transformations. Can someone tell me what a transformation is? Student: A change. Teacher: That's a very basic definition of it, but yes, it is a change. Transformations are geometric figures that have been changed by reflection, rotation and translation. We won't be looking at rotations in 7th grade at all, and we are just going to be looking at reflections today. We'll save translations for another day. Does anyone know what a reflection is? Student: It's what you see when you look in the mirror. Teacher: That's right, but what about math? How can we use reflections in math? I didn't bring my mirror, so how are we going to do reflections in math class? Student: It's the same principle as with a mirror. It's taking an image and kind of like seeing the reverse of it, like a reflection. Teacher: That's right. It is like a reverse of the figure. Teacher: A reflection of a geometric figure is a mirror image of the object. If we were to place a mirror on a line of reflection, it will give the position of the reflected image. Teacher: So how could we do that with a piece of paper if we don't have a mirror? Student: We could just fold the paper and line up the figure and draw it that way? Teacher: You're right. That is a great way to get started. Everyone take your piece of paper with the image on it and fold it on the x-axis. This is our line of reflection. Which axis is the x-axis? Student: The one that goes across or horizontally because the y goes high, or up and down. Teacher: Right. So everyone fold your paper on the x-axis. Do you see where your new figure would be if we reflected it over the x-axis? Go ahead and draw in the corresponding vertices of the new figure. Student: So we just put a dot where the other dots line up? Teacher: Yes. That is right. This isn't perhaps the fastest way, but it works, and it is accurate. Anyone have any ideas how we could do it another way? Student: Well, point H is 4 units above the x-axis, so its reflection would be 4 units below the x-axis. G is also 4 units above the x-axis, so its reflection would also be 4 units below the x-axis. Point F is 7 units above the x-axis, so F' would be 7 units below the x-axis. Then once you have them all plotted, just connect the dots. Teacher: Wow. Nice job! That is a great explanation. What happens if one of the points is on the x-axis, where is its reflection? Student: It is itself. Since it is 0 units from the x-axis, then its reflection is also. Teacher: So when we are done making our new image, F'G'H', how could we double check that it is correct, that we reflected correctly? Student: Just fold the paper and it should line up with the FGH. Teacher: Nice job. That's all there is to reflecting across the x-axis. You would do the same across the y-axis, but instead of going above or below the line of reflection, you would go left or right of the y-axis. Resources Instructional Notes Teacher Notes Rotation is not a part of the transformations that students will be studying at this point. For now, students will make reflections over vertical lines, horizontal lines, and the lines y = x and y = -x. Sometimes students struggle with this topic because they need review on the coordinate system, plotting points and naming points, rather than the struggle with the concepts of transformations. A quick review of those topics might be beneficial. Teachers need to stress the point that students should physically fold their papers to ensure their new reflection is correct. Students may find that reflecting over the line y =x or the line y = -x is not as difficult as first thought. A simple way to do this is to count the number of units each point is from the line to be reflected over. If it is 3 units to the left of the line, then they count down 3 units from the line to get the reflection of point. By making a 90-degree angle from the original point over to the line and down (or up), they have reflected it. For example, reflect A over the line y = x. A is 4 units down from the reflection line so A' is 4 units to the left of the line. Alternatively, students might count 4 left and then 4 up; the order doesn't matter. Plot the following points on grid paper and translate (slide) the point and name the new coordinates. Use colored pencils to mark new points. For example, point A: (-2, 4). Slide 4 units right to point A1: (2, 4). point B:(4, 0). Slide 3 units down to point B1:(4, -3). Source: 2007 Mississippi Mathematics Framework Revised Strategies, p. 41. Make sure students know how to graph the equations y = x and y = -x. This standard does not include dilation, which is addressed in 7.3.1 with scaling and scalar drawings (scale factor, similar figures, etc.). Make sure students are aware that the original object (preimage) and its image are congruent - identical in every respect except for their position. To translate a figure is to simply slide it somewhere else. But in the move, you may not change the figure in any other way. You cannot rotate it, resize it, or flip it over. You may only slide it side to side, up and down. Instructional Resources Schielack, J. (2010).Focus in Grade 8, Teaching with Curriculum Focal Points. (p. 80-83). Reston, VA: National Council of Teachers of Mathematics. Transformation applets This site leads to applets related to visualizing transformations. Additional Instructional Resources GeogebraReflections GeogebraTranslations Interactive geometric transformations lessons Web-based student activities Translating and reflecting shapes This applet allows the user to translate triangles, squares and parallelograms on both the x and y-axes. The user can also reflect the figure around x values, y values and the line x = y. New Vocabulary Note from writer:The terms below are first introduced in the Minnesota State Standards in 4th grade. Students will need to be retaught these terms in the context of doing these transformations on a coordinate grid. reflection: a transformation that "flips" a figure over a mirror or reflection line. Example: Reflection of the object ABDC on the left over the line on the right. This line is called the reflection line. translation:moving without resizing; a 'slide';every point of the shape must move the same distance and in the same direction. Example : If the shape gets moved 30 units in the "x" direction, and 40 units in the "y" direction, we can write: ●(x,y) → (x + 30, y + 40) This expression says "all the x and y coordinates will become x + 30 and y + 40" Source: transformation:the movement of a figure in a plane from its original position, the preimage, to a new position, the image. Also called a map. Transformations can occur when there is a reflection, a rotation, a translation, or a glide reflection of the original image. notation:figures or symbols used to represent mathematical functions, objects, or ideas. Examples: J and J' (labels for points before and after transformation) translation notation (x,y) → (x + 3, y - 2) ABCD and A'B'C'D' (labels for a quadrilateral before and after transformation) Professional Learning Communities Reflection - Critical Questions regarding the teaching and learning of these benchmarks Can students distinguish between the original figure and the one after the transformation? Do students understand the difference between a reflection and a translation? If students are given the transformed image's coordinates and the rule used to transform it, can they come up with the original figure's coordinates? References Mississippi Department of Education. (2007). 2007 Mississippi Mathematics Framework Revised Strategies, (p. 39). Jackson, MS: Mississippi Department of Education. Schielack, J. (2010).Focus in Grade 8, Teaching with Curriculum Focal Points. Reston, VA: National Council of Teachers of Mathematics. Schifter, D. (February, 1999). Learning Geometry: Some Insights Drawn from Teacher Writing. In Teaching Children Mathematics, 5 (5), 360-366. Pintozzi, C. (2001). Chapter 20: Transformations and Symmetry. In Mathematics Review. (p. 272-80). Woodstock, GA: American Book. Translation of a polygon Assessment 1. Answer:a Source: MinnesotaGrade7MathematicsMCA-IIIItemSampler Item, 2011, Benchmark 7.3.2.4 2. The point (3, - 4) is translated 2 units to the right and 3 units up. What are the coordinates of the image of (3, - 4)? Answer:(5, -1) Source: New England Common Assessment Program, Released items, Grade 8, 3. Answer:d Source: California Geometry released exam questions 4. Answer:b Source: Texas 2008-09 Released TAKS exam 8th grade math Differentiation Emergent Learners Review the coordinate grid system and plotting points with students before starting to perform transformations. Advise students to fold the paper to get the correct reflection; use different colors for each new figure. Use fairly large grid paper (1/2 inch, or not less than 1 cm grid paper) to plot and to be able to see the new coordinates. Give students multiple choice options rather than open-ended problems to help them see choices, which they can eliminate as they work. Print pages from this website out and have the students compile them in a booklet. Have students trace the original figure to be reflected or translated. Have them cut it out. They can then take the cut-out figure and reflect it or translate it. By using a physical figure, the students will be less likely to make errors and be able to get the right shape in their image because they can reflect it or translate it and then trace the figure. (Make sure they have labeled the cut out shape with the correct letters, which they can then transfer to the new image.) Review reflection in the line y = x. A reflection in the line y = x can be seen in the picture below in which A is reflected to its image A'. The general rule for a reflection in the y-axis: (A, B) → (B, A) A way to remember what translation means is "tranSLate means SLide" Gliding through the Galaxy using Transformations This website offers an elementary geared lesson on transformations in the Solar System, along with activities and ideas appropriate for struggling learners. Have students highlight the line that the figure is to be reflected over. English Language Learners Review which axis is the x axis and which is the y-axis ('y' goes 'high). Print these pages out and have the students compile them in a booklet: A way to remember what translation means is "tranSLate means SLide" Extending the Learning Have students do multi-step transformations for one figure. For example, translate the image using the rule (x+2, y-4), and then reflect that new image about the line, y = x. Have students reflect over lines other than y = x or y = -x Extend the lesson for students to learn about rotation. Parents/Admin Classroom Observation Administrative/Peer Classroom Observation Students are:(descriptive list)Students are:(descriptive list) Teachers are:(descriptive list)Teachers are:(descriptive list) Students are:(descriptive list)identifying the line that is being reflected over; it is not necessarily to be able to name the equation for it. Teachers are:(descriptive list)reminding students to fold their papers to double check that the reflection was done correctly. Students are:(descriptive list)using a coordinate grid to reflect and translate figures. Teachers are:(descriptive list)requiring students to write the translations using proper notation (A'B'C'). Students are:(descriptive list)identifying the new points of reflection or translation. Teachers are:(descriptive list)having students perform two translations on the same figure. Students are:(descriptive list)folding their papers to double check that their reflections are done correctly. Teachers are:(descriptive list) Students are:(descriptive list)Teachers are:(descriptive list) identifying the line that is being reflected over; it is not necessarily to be able to name the equation for it.reminding students to fold their papers to double check that the reflection was done correctly. using a coordinate grid to reflect and translate figures.requiring students to write the translations using proper notation (A'B'C'). identifying the new points of reflection or translation.having students perform two translations on the same figure. folding their papers to double check that their reflections are done correctly. Parents Parent Resources Similarity videos Relations and sizes Reflections practice Framework Feedback X Related Frameworks 7.3.2.A Similarity & Scaling in 2 Dimensions Benchmarks: 7.3.2.1 Similarity & Scale Factors 7.3.2.2 Apply Scale Factors 7.3.2.3 Proportions & Ratios Your Support is Critical The STEM Teachers Center is a unique online resource for Minnesota teachers. Your support is critical to making sure it stays available and current. Thanks to the 3M foundation for their generous support in funding the recent technology upgrade, mobile enhancements, and accessibility improvements to the STEM Teacher Center. Annual support for this website provided by: Credits Terms and Conditions © 2023 SciMathMN and the Minnesota Department of Education Framework Feedback X We want your feedback! Tell us what's working, what's not, and how you are using these Frameworks. 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10796	https://stats.libretexts.org/Courses/Las_Positas_College/Math_40%3A_Statistics_and_Probability/04%3A_Probability_and_Counting/4.03%3A_The_Addition_and_Multiplication_Rules_of_Probability	4.3: The Addition and Multiplication Rules of Probability - Statistics 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 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"licenseversion:40", "source@ ] [ "article:topic", "independent events", "mutually exclusive", "Multiplying probabilities", "Adding probabilities", "authorname:openstax", "transcluded:yes", "showtoc:no", "license:ccby", "source-stats-733", "program:openstax", "licenseversion:40", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. Las Positas College 4. Math 40: Statistics and Probability 5. 4: Probability and Counting 6. 4.3: The Addition and Multiplication Rules of Probability Expand/collapse global location 4.3: The Addition and Multiplication Rules of Probability Last updated Jul 1, 2020 Save as PDF 4.2: Independent and Mutually Exclusive Events 4.3.1: Contingency Tables Page ID 10893 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. The Multiplication Rule 2. The Addition Rule 1. Example 4.3.1 2. Exercise 4.3.1 3. Example 4.3.2 4. Exercise 4.3.2 5. Example 4.3.3 6. Exercise 4.3.3 7. Example 4.3.4 8. Exercise 4.3.4 9. Example 4.3.5 10. Exercise 4.3.5 References Review Formula Review Exercise 4.3.5 Exercise 4.3.6 Exercise 4.3.7 Exercise 4.3.8 Exercise 4.3.9 Exercise 4.3.10 Exercise 4.3.11 Exercise 4.3.12 Exercise 4.3.13 Exercise 4.3.14 Glossary When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. The Multiplication Rule If A and B are two events defined on a sample space, then: (4.3.1)P⁡(A AND B)=P⁡(B)⁢P⁡(A\|B) This rule may also be written as: P⁡(A\|B)=P⁡(A AND B)P⁡(B) (The probability of A given B equals the probability of A and B divided by the probability of B.) If A and B are independent, then P⁡(A\|B)=P⁡(A). and Equation 4.3.1 becomes P⁡(A AND B)=P⁡(A)⁢P⁡(B). The Addition Rule If A and B are defined on a sample space, then: (4.3.2)P⁡(A OR B)=P⁡(A)+P⁡(B)−P⁡(A AND B) If A and B are mutually exclusive, then P⁡(A AND B)=0. and Equation 4.3.2 becomes P⁡(A OR B)=P⁡(A)+P⁡(B). Example 4.3.1 Klaus is trying to choose where to go on vacation. His two choices are: A=New Zealand and B=Alaska. Klaus can only afford one vacation. The probability that he chooses A is P⁡(A)=0.6 and the probability that he chooses B is P⁡(B)=0.35. P⁡(A AND B)=0 because Klaus can only afford to take one vacation Therefore, the probability that he chooses either New Zealand or Alaska is P⁡(A OR B)=P⁡(A)+P⁡(B)=0.6+0.35=0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05. Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A= the event Carlos is successful on his first attempt. P⁡(A)=0.65. B= the event Carlos is successful on his second attempt. P⁡(B)=0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90. What is the probability that he makes both goals? What is the probability that Carlos makes either the first goal or the second goal? Are A and B independent? Are A and B mutually exclusive? Solutions a. The problem is asking you to find P⁡(A AND B)=P⁡(B AND A). Since P⁡(B\|A)=0.90:P⁡(B AND A)=P⁡(B\|A)⁢P⁡(A)=(0.90)⁢(0.65)=0.585 Carlos makes the first and second goals with probability 0.585. b. The problem is asking you to find P⁡(A OR B). (4.3.3)P⁡(A OR B)=P⁡(A)+P⁡(B)−P⁡(A AND B)=0.65+0.65−0.585=0.715 Carlos makes either the first goal or the second goal with probability 0.715. c. No, they are not, because P⁡(B AND A)=0.585. (4.3.4)P⁡(B)⁢P⁡(A)=(0.65)⁢(0.65)=0.423 (4.3.5)0.423≠0.585=P⁡(B AND A) So, P⁡(B AND A) is not equal to P⁡(B)⁢P⁡(A). d. No, they are not because P⁡(A and B)=0.585. To be mutually exclusive, P⁡(A AND B) must equal zero. Exercise 4.3.1 Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C= the event that Helen makes the first shot. P⁡(C)=0.75. D= the event Helen makes the second shot. P⁡(D)=0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws? Answer (4.3.6)P⁡(D\|C)=0.85 (4.3.7)P⁡(C AND D)=P⁡(D AND C) (4.3.8)P⁡(D AND C)=P⁡(D\|C)⁢P⁡(C)=(0.85)⁢(0.75)=0.6375 Helen makes the first and second free throws with probability 0.6375. Example 4.3.2 A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly. What is the probability that the member is a novice swimmer? What is the probability that the member practices four times a week? What is the probability that the member is an advanced swimmer and practices four times a week? What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? Are being a novice swimmer and practicing four times a week independent events? Why or why not? Answer 28 150 80 150 40 150 P⁡(advanced AND intermediate)=0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time. No, these are not independent events. (4.3.9)P⁡(novice AND practices four times per week)=0.0667(4.3.10)P⁡(novice)⁢P⁡(practices four times per week)=0.0996(4.3.11)0.0667≠0.0996 Exercise 4.3.2 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? Answer (4.3.12)P=200−140−40 200=20 200=0.1 Example 4.3.3 Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25. Let: M= math class, S= speech class, M\|S= math given speech What is the probability that Felicity enrolls in math and speech? Find P⁡(M AND S)=P⁡(M\|S)⁢P⁡(S). 2. What is the probability that Felicity enrolls in math or speech classes? Find P⁡(M OR S)=P⁡(M)+P⁡(S)−P⁡(M AND S). 3. Are M and S independent? Is P⁡(M\|S)=P⁡(M)? 4. Are M and S mutually exclusive? Is P⁡(M AND S)=0? Answer a. 0.1625, b. 0.6875, c. No, d. No Exercise 4.3.3 A student goes to the library. Let events B= the student checks out a book and D= the student check out a DVD. Suppose that P⁡(B)=0.40,P⁡(D)=0.30 and P⁡(D\|B)=0.5. Find P⁡(B AND D). Find P⁡(B OR D). Answer P⁡(B AND D)=P⁡(D\|B)⁢P⁡(B)=(0.5)⁢(0.4)=0.20. P⁡(B OR D)=P⁡(B)+P⁡(D)−P⁡(B AND D)=0.40+0.30−0.20=0.50 Example 4.3.4 Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B= woman develops breast cancer and let N= tests negative. Suppose one woman is selected at random. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative? Given that the woman has breast cancer, what is the probability that she tests negative? What is the probability that the woman has breast cancer AND tests negative? What is the probability that the woman has breast cancer or tests negative? Are having breast cancer and testing negative independent events? Are having breast cancer and testing negative mutually exclusive? Answers P⁡(B)=0.143;P⁡(N)=0.85 P⁡(N\|B)=0.02 P⁡(B AND N)=P⁡(B)⁢P⁡(N\|B)=(0.143)⁢(0.02)=0.0029 P⁡(B OR N)=P⁡(B)+P⁡(N)−P⁡(B AND N)=0.143+0.85−0.0029=0.9901 No. P⁡(N)=0.85;P⁡(N\|B)=0.02. So, P⁡(N\|B) does not equal P⁡(N). No. P⁡(B AND N)=0.0029. For B and N to be mutually exclusive, P⁡(B AND N) must be zero Exercise 4.3.4 A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports? Answer Let A= student is a senior going to college. Let B= student plays sports. P⁡(B)=140 200 P⁡(B\|A)=50 140 P⁡(A AND B)=P⁡(B\|A)⁢P⁡(A) P⁡(A AND B)=(140 200)(50 140)=1 4 Example 4.3.5 Refer to the information in Example 4.3.4. P= tests positive. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P⁡(P\|B)=1−P⁡(N\|B). What is the probability that a woman develops breast cancer and tests positive. Find P⁡(B AND P)=P⁡(P\|B)⁢P⁡(B). What is the probability that a woman does not develop breast cancer. Find P⁡(B′)=1−P⁡(B). What is the probability that a woman tests positive for breast cancer. Find P⁡(P)=1−P⁡(N). Answer a. 0.98; b. 0.1401; c. 0.857; d. 0.15 Exercise 4.3.5 A student goes to the library. Let events B= the student checks out a book and D= the student checks out a DVD. Suppose that P⁡(B)=0.40,P⁡(D)=0.30 and P⁡(D\|B)=0.5. Find P⁡(B′). Find P⁡(D AND B). Find P⁡(B\|D). Find P⁡(D AND B′). Find P⁡(D\|B′). Answer P⁡(B′)=0.60 P⁡(D AND B)=P⁡(D\|B)⁢P⁡(B)=0.20 P⁡(B\|D)=P⁡(B AND D)P⁡(D)=(0.20)(0.30)=0.66 P⁡(D AND B′)=P⁡(D)−P⁡(D AND B)=0.30−0.20=0.10 P⁡(D\|B′)=P⁡(D AND B′)⁢P⁡(B′)=(P⁡(D)−P⁡(D AND B))⁢(0.60)=(0.10)⁢(0.60)=0.06 References DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at www.field.com/fieldpollonline...rs/Rls2443.pdf (accessed May 2, 2013). Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at www.thestar.com/news/gta/2011..._suggests.html (accessed May 2, 2013). “Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-Mayoral_Approval%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013). “Roulette.” Wikipedia. Available online at (accessed May 2, 2013). Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at www.census.gov/hhes/socdemo/l...acs/ACS-12.pdf (accessed May 2, 2013). Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013). Data from U.S. Census Bureau. Data from the Wall Street Journal. Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at www.ropercenter.uconn.edu/ (accessed May 2, 2013). Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013). Review The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common. Formula Review The multiplication rule:P⁡(A AND B)=P⁡(A\|B)⁢P⁡(B) The addition rule:P⁡(A OR B)=P⁡(A)+P⁡(B)−P⁡(A AND B) Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino. In this problem, let: C= Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder. L= Latino Californians Suppose that one Californian is randomly selected. Exercise 4.3.5 Find P⁡(C). Exercise 4.3.6 Find P⁡(L). Answer 0.376 Exercise 4.3.7 Find P⁡(C\|L). Exercise 4.3.8 In words, what is C\|L? Answer C\|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder. Exercise 4.3.9 Find P⁡(L AND C) Exercise 4.3.10 In words, what is L AND C? Answer L AND C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder. Exercise 4.3.11 Are L and C independent events? Show why or why not. Exercise 4.3.12 Find P⁡(L OR C). Answer 0.6492 Exercise 4.3.13 In words, what is L OR C? Exercise 4.3.14 Are L and C mutually exclusive events? Show why or why not. Answer No, because P⁡(L AND C) does not equal 0. Glossary Independent Events The occurrence of one event has no effect on the probability of the occurrence of another event. Events A and B are independent if one of the following is true: 1. P⁡(A\|B)=P⁡(A) 2. P⁡(B\|A)=P⁡(B) 3. P⁡(A AND B)=P⁡(A)⁢P⁡(B) Mutually Exclusive Two events are mutually exclusive if the probability that they both happen at the same time is zero. If events A and B are mutually exclusive, then P⁡(A AND B)=0. This page titled 4.3: The Addition and Multiplication Rules of Probability is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 4.2: Independent and Mutually Exclusive Events 4.3.1: Contingency Tables Was this article helpful? Yes No Recommended articles 4.5: Two Basic Rules of ProbabilityThe multiplication rule and the addition rule are used for computing the probability of A and B, and the probability of A or B for two given events A,... 3.4: Two Basic Rules of ProbabilityThe multiplication rule and the addition rule are used for computing the probability of A and B, and the probability of A or B for two given events A,... 3.3: Two Basic Rules of ProbabilityThe multiplication rule and the addition rule are used for computing the probability of A and B, and the probability of A or B for two given events A,... 10.4: Two Basic Rules of ProbabilityThe multiplication rule and the addition rule are used for computing the probability of A and B, and the probability of A or B for two given events A,... 2.4: Two Basic Rules of ProbabilityThe multiplication rule and the addition rule are used for computing the probability of A and B, and the probability of A or B for two given events A,... 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10797	https://www.quora.com/Why-are-logarithms-not-defined-for-0-and-negatives	Something went wrong. Wait a moment and try again. Logarithmic Function Functions (mathematics) Basic Algebra Mathematical Sciences Mathematical Functions 5 Why are logarithms not defined for 0 and negatives? Narayan Dalai Studied Mathematics & Science &Electrical Engg at Diploma in Electrical Engineering (IAF) · Author has 2.5K answers and 2.6M answer views · 6y Logarithm of a Number to a particular base by which the base is raised in order to get the very Number. Example :— Number 10 raised to Power 2 to get 100. Expressed in another way, logarithm to base 10 of NUMBER 100=2. Expressed logarithmically, log (base 10) 100= 2. Expressed more generally, log ( base a) N= x; Here, as per logarithm defined, N=Real positive Number; a>0 but not = 1. Let us now analyse as to why the Logarithm is not defined for 0 (ZERO ) LOGARITHM of 0 (ZERO ) does not exist. What is log ( base)10 0=?. log 10 0=X (say ) Then 10^X=0 . HERE X has no value such that 10^X=0 ; Perhaps X= Logarithm of a Number to a particular base by which the base is raised in order to get the very Number. Example :— 10^2=100 Number 10 raised to Power 2 to get 100. Expressed in another way, logarithm to base 10 of NUMBER 100=2. Expressed logarithmically, log (base 10) 100= 2. Expressed more generally, log ( base a) N= x; Here, as per logarithm defined, N=Real positive Number; a>0 but not = 1. Let us now analyse as to why the Logarithm is not defined for 0 (ZERO ) LOGARITHM of 0 (ZERO ) does not exist. What is log ( base)10 0=?. log 10 0=X (say ) Then 10^X=0 . HERE X has no value such that 10^X=0 ; Perhaps X=-Infiinity which may satisfy the above equation; Again - Infiinity is interminate. THEREFORE LOGARITHM FOR 0 IS NOT DEFINED. But the value of Logarithm not restricted to 0 ; it may be zero, even -ve ( log 0.3=-1+0.4771, log 2=0.3010) B) Why logarithm not defined for negative numbers? log (base10) -100= ? log (base10) -100=(+-)y (say ) Then 10^y=-100; y = not determinate. 10^0=1 (a positive Number; 10^-1=1/10 again it is positive. So,perhaps, for this reason, logarithm is not defined for NEGATIVE NUMBERS . C) Why a (the base of the Logarithm ) not =1? log (base1) 10= ?; log (base1) 10= y ; then, 1^ y= Always 1 irrespective of the (+ve )value of y. In other words, it will not yield an unique result. So base=1 is notionally avoided. Hope , this will suffice for the question asked for. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. 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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 are negative logarithms? What are some examples? Can logarithms have a negative base? Why? Is a logarithm with base-0 possible or not? And why isn’t log0 with base-0 1, unlike the fact that log10 with base-10 is 1? Why can't x be negative in logarithm? Why can't Mantissa be negative in logarithms? Jack Huizenga Math Professor, Penn State · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Sam Sinai , Ph.D student in Mathematical Biology · Author has 613 answers and 4.4M answer views · 11y Originally Answered: Why is log not defined for negative values? · In high school math, the natural logarithm logx of a positive number x>0 is defined to be the unique real number y such that x=ey. The logarithm is the "inverse function" of the exponential function. For any real number y, the exponential ey is always a positive real number. So, if x is negative, there is no real number that we could call the logarithm of x and have it satisfy the defining equation that works for positive reals. You may have seen Euler's famous identity eiπ=−1. This "should" tell us that log(−1)=iπ. There's a catch though: this identity is a special case In high school math, the natural logarithm logx of a positive number x>0 is defined to be the unique real number y such that x=ey. The logarithm is the "inverse function" of the exponential function. For any real number y, the exponential ey is always a positive real number. So, if x is negative, there is no real number that we could call the logarithm of x and have it satisfy the defining equation that works for positive reals. You may have seen Euler's famous identity eiπ=−1. This "should" tell us that log(−1)=iπ. There's a catch though: this identity is a special case of the more general formula eix=cosx+isinx. So we also have ei(2k+1)π=−1 for any integer k. That is, all of the numbers i(2k+1)π have an equal right to be called "the" logarithm of −1. In order to properly define the logarithm on negative numbers, it is necessary to consider multi-valued functions, and the values are complex numbers. Once you do this, it is equally easy to define not only the logarithm of −1, but the logarithm of any nonzero complex number! Assistant Bot · 1y Logarithms are not defined for 0 and negative numbers due to the properties of exponential functions, which serve as the foundation for logarithms. Here’s a detailed explanation: Definition of Logarithms The logarithm of a number b with base a (written as loga(b)) answers the question: "To what exponent must the base a be raised to produce b?" Mathematically, this is expressed as: loga(b)=x⟺ax=b Why Logarithms are Undefined for 0 and Negative Numbers Logarithm of Zero: If we set b=0, we want to find x such that ax=0. However, for any positive base a (where a>0), ax Logarithms are not defined for 0 and negative numbers due to the properties of exponential functions, which serve as the foundation for logarithms. Here’s a detailed explanation: Definition of Logarithms The logarithm of a number b with base a (written as loga(b)) answers the question: "To what exponent must the base a be raised to produce b?" Mathematically, this is expressed as: loga(b)=x⟺ax=b Why Logarithms are Undefined for 0 and Negative Numbers Logarithm of Zero: If we set b=0, we want to find x such that ax=0. However, for any positive base a (where a>0), ax is never equal to zero for any real number x. Therefore, there is no real exponent x that satisfies this equation, making loga(0) undefined. Logarithm of Negative Numbers: If we set b<0, we want to find x such that ax=b. For any positive base a, the expression ax is always positive, regardless of whether x is positive or negative. Thus, there is no real number x such that ax equals a negative number, resulting in loga(b) being undefined for b<0. Summary Logarithm of 0: Undefined because no exponent can yield zero. Logarithm of Negative Numbers: Undefined because no exponent can yield a negative number. Complex Numbers In the context of complex numbers, logarithms can be defined for negative numbers and zero, but this involves more advanced concepts like the complex logarithm, which includes a branch cut (usually along the negative real axis) and is not typically covered in basic logarithmic functions. Balazs Koren 11y The problem is with a global definition. 0 to the power of anything is 0. That works of course. But what about 0 based log of 5? Allowing negative number and 0 based logs for some numbers would make the definition of log an a4 page long with all the exceptions. :-) Related questions Why is ln not defined at 0? What is the same in log of numbers 0.002489, 0. 2489 and 0.02489? Why are the inputs to logarithms limited to a domain greater than 0? I understand not equaling zero, but I was pretty easily able to find a way to make negative numbers work using Euler's Identity. How do I check whether a logarithmic function is negative, positive or equal to 0? Can the base of a logarithm be zero? If we assume a logarithm with zero base exists, what problems would it existence imply? Aniruddh Ramanujam B.E. in Bachelor of Engineering in Production Engineering, Sri Sairam Engineering College (Graduated 2022) · Author has 132 answers and 194.1K answer views · 6y Logarithm of a value x , in general for a positive base a , is the power b to which a has to be raised to get the value of x. i.e. x=ab loga x=b So, as a is only raised to some power, it cannot give you a negative value. So, log of negative value is not defined. For x=ab to be equal to zero (0) it should have very high negative power, i.e. − ∞ This is why it is not defined 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. Yassine Alouini Studied Mathematics at École Centrale Paris (Graduated 2013) · Author has 1K answers and 5.1M answer views · 11y Originally Answered: Why is log not defined for negative values? · You can extend the log function to some negative values by using the appropriate extension of the Complex logarithm. An example would be: Let f be the principle value of the complex logarithm (the other values are obtained by adding a multiple of 2πi) then: f(−1)=f(i2)=f(exp(i∗π))=i∗π. I have used these fact: −1=i2 i=exp(i∗π/2) f(ex)=x for "appropriate" x Read the Wikipedia page for more information about the "appropriate" arguments you can input to the complex logarithm. I hope this helped! Danny Kodicek BA in Mathematics, University of Cambridge · Author has 6.5K answers and 6.2M answer views · 1y Originally Answered: Why is the natural logarithm not defined when x is less than 0? · In the world of real numbers, the logarithm of a number x is the value a such that ea=x. This doesn’t exist for a negative number, since a positive number raised to any power is always positive. It’s possible to extend the idea of logarithms into complex numbers (numbers of the form a+bi, where i is the square root of −1). If you do this, then you can take the logarithm of any number except 0, including negative real numbers. However, the logarithm isn’t unique: there are infinitely many complex numbers z that satisfy ez=x. This is because for any value z that satisfies this equation, you can In the world of real numbers, the logarithm of a number x is the value a such that ea=x. This doesn’t exist for a negative number, since a positive number raised to any power is always positive. It’s possible to extend the idea of logarithms into complex numbers (numbers of the form a+bi, where i is the square root of −1). If you do this, then you can take the logarithm of any number except 0, including negative real numbers. However, the logarithm isn’t unique: there are infinitely many complex numbers z that satisfy ez=x. This is because for any value z that satisfies this equation, you can add any multiple of 2πi to it and get another solution. To get around this, we can arbitrarily define the “principal logarithm”, which is the solution whose imaginary part is greater than −π and less than or equal to π. For any negative real number, its principal logarithm is of the form a+πi . For example, the principal logarithm of −e is 1+πi. This follows from Euler’s famous identity eiπ=−1. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? 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Michael Carroll Former Teacher (Mathematics - Secondary Level) · Author has 556 answers and 1.2M answer views · 6y For the same reason that they are not defined for frying pans and ostriches: Not all behaviors are possible by all things. Blue is a color, not a sound. Up is a direction, not a taste. Logarithm is a function that asks a question about its argument: It asks “what power of the base produces the argument?” Sometimes, the answer is “there is no answer to this question.” For example, what power of 10 is equal to zero? The answer is “there is no answer because it is impossible for any power of 10 to be equal to zero.” Things are what they are. Just because we are able to ask “why“, it does not mean the For the same reason that they are not defined for frying pans and ostriches: Not all behaviors are possible by all things. Blue is a color, not a sound. Up is a direction, not a taste. Logarithm is a function that asks a question about its argument: It asks “what power of the base produces the argument?” Sometimes, the answer is “there is no answer to this question.” For example, what power of 10 is equal to zero? The answer is “there is no answer because it is impossible for any power of 10 to be equal to zero.” Things are what they are. Just because we are able to ask “why“, it does not mean there is an answer. It is the job of the scientist to observe the universe and determine what is and what isn’t, what questions have answers and what questions do not. Sometimes, the answer is “just because”. Kartik Dolas Knows English · Author has 133 answers and 485.4K answer views · 7y Originally Answered: Why can't we have log of negative numbers? · Let us first understand meaning of log… If log of x to base y is k.. Then log(y,x) = k Where y is base and x is index then.. x=(y)^k So if we are taking consideration the log of negative number.. x<0 According to rules of log base should always be positive so to find value of k.. Where x is negative and y is positive.. There will be no solutions for k.. As any number raised to a positive number won't let you to the answer which is negative For ex. (2)^k = -1 Then k won't have any solution..!! Hence we can't calculate log for negative numbers. Happy learning..!! Promoted by Webflow Metis Chan Works at Webflow · Aug 11 What are the best AI website builders now? When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real When it comes to AI website builders, there are a growing number of options, but a few stand out for their power, flexibility, and ability to grow with your needs. Webflow’s AI Site Builder is a top choice for small businesses, in-house teams, and agencies who want the speed of AI and the freedom to fully customize every part of their site. With Webflow, you can: Describe your business or idea and instantly generate a unique, production-ready website—no coding required. Edit visually in a powerful no-code canvas, customize layouts, and add advanced interactions. Collaborate with your team in real time, streamline feedback, and manage all your content in one place. Publish instantly on enterprise-grade hosting with built-in SEO, security, and the flexibility to scale as you grow. Many other tools offer AI-powered templates or quick site launches, but Webflow stands out by letting you take control—so your site never feels generic, and you can easily update, expand, or redesign as your needs change. Want to see what AI-powered site building can really do? Try Webflow AI Site Builder for free today. Aadit Kamat B.Comp. in Computer Science (college major), National University of Singapore - School of Computing (Graduated 2021) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · 11y Originally Answered: Why is log not defined for negative values? · This is because a positive number raised to any exponent gives you a positive number. Even if we take an exponent such as 1/2,i.e square root of number, the positive root is considered. That is why the condition while taking log is that the base should be positive i.e, loga (b) where a>0. Vikas Kumar Senior System Engineer at IBM India (2016–present) · Author has 105 answers and 230.3K answer views · Updated 5y Can logarithms have a negative base? Why? Consider a hypothetical negative base of −4, so the undefined (non-existent) function y=log−4(x). This logarithm would be the inverse of the function y=(−4)x, which can only be evaluated for exponents that can be written as a fraction where the denominator is odd. Remember a rational exponent, such as (−4)a/b, represents a radical, namely b√(−4)a, and a negative number can only be evaluated for an odd root (using real numbers). For example, (−4)1/2 means √−4 which is a non-real answer. Thus, an exponential function with a negative base, such as y=(−4)x isn't much Consider a hypothetical negative base of −4, so the undefined (non-existent) function y=log−4(x). This logarithm would be the inverse of the function y=(−4)x, which can only be evaluated for exponents that can be written as a fraction where the denominator is odd. Remember a rational exponent, such as (−4)a/b, represents a radical, namely b√(−4)a, and a negative number can only be evaluated for an odd root (using real numbers). For example, (−4)1/2 means √−4 which is a non-real answer. Thus, an exponential function with a negative base, such as y=(−4)x isn't much of a function at all (it is not continuous), since it can only be evaluated at very specific x-values. So, a logarithm with a negative base, like y=log−4(x) would also only work for very specific arguments (due to its connection to the non-continuous y=(−4)x) and such a logarithmic function would also not be continuous. It is for such reasons that we only consider logarithms with positive bases, as negative bases are not continuous and generally not useful. Hope this insight makes sense and is somewhat helpful! Ashu Rawat Marketing Associate at Amazon (company) (2017–present) · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · 11y Originally Answered: Why is log not defined for negative values? · Log to the base something of a negative number would mean that the "something " raised to a power is negative. And an exponential function can never be negative. The least value it can attain approaches zero when exponent is -infinity. Was this worth your time? This helps us sort answers on the page. Absolutely not Ankeet Kumar Hanny Coder, Logical Thinker and Problem Solver from India. · Author has 212 answers and 1.1M answer views · 7y Originally Answered: Why can't we have log of negative numbers? · Let’s assume that this question is about log to the base of 10. Now, loga=b this means that 10b=a If we are finding log of negative numbers means there should be a number a whose logarithm value to base 10 is b and a is a negative number. According to the explanation given above about the log we can say that [math]10^b =... Related questions What are negative logarithms? What are some examples? Can logarithms have a negative base? Why? Is a logarithm with base-0 possible or not? And why isn’t log0 with base-0 1, unlike the fact that log10 with base-10 is 1? Why can't x be negative in logarithm? Why can't Mantissa be negative in logarithms? Why is ln not defined at 0? What is the same in log of numbers 0.002489, 0. 2489 and 0.02489? Why are the inputs to logarithms limited to a domain greater than 0? I understand not equaling zero, but I was pretty easily able to find a way to make negative numbers work using Euler's Identity. How do I check whether a logarithmic function is negative, positive or equal to 0? Can the base of a logarithm be zero? If we assume a logarithm with zero base exists, what problems would it existence imply? What is the logarithm of 0? Can we have negative logarithms? How? Is it possible for the base of a logarithm to be equal to zero? If so, what happens? How did Napier discover logarithms? What is the value of log(1/0)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
10798	https://www.vedantu.com/question-answer/calculate-the-molar-mass-of-ch4-class-11-chemistry-cbse-5f0b53237514cd7895253197	Talk to our experts 1800-120-456-456 Calculate the molar mass of $C{H_4}$. Repeaters Course for NEET 2022 - 23 © 2025.Vedantu.com. All rights reserved
10799	https://www.thesaurus.com/browse/intrigued	Daily Crossword Word Puzzle Word Finder All games Word of the Day Word of the Year New words Language stories All featured Slang Emoji Memes Acronyms Gender and sexuality All culture Writing tips Writing hub Grammar essentials Commonly confused All writing tips Games Featured Culture Writing tips Advertisement Advertisement View definitions for intrigued intrigued adjective as in attracted Strong matches absorbed captivated charmed entertained fascinated interested Discover More Example Sentences Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com. Glennie: The idea intrigued me, so I purchased a few of Raymond’s books, including his beautiful children’s book “Can Bears Ski?” just to get a feel for his voice and rhythm. FromLos Angeles Times But the Londoner was intrigued enough to take up a game that has now become the central part of his life as he pursues a professional career in America. FromBBC And the defense secretary’s behavior strongly suggests he’s at least intrigued by Wilson’s opposition to the 19th Amendment. FromSalon "No matter your politics, he would get so many young people involved and intrigued. He provided a path for people to get involved." FromBBC I was intrigued by her, but she lived in Florida. FromLos Angeles Times Advertisement Discover More Related Words Words related to intrigued are not direct synonyms, but are associated with the word intrigued. Browse related words to learn more about word associations. engrossed adjectiveas in preoccupied; attentive to absorbed all wound up assiduous bugged busy captivated caught up caught up in consumed deep diligent engaged enthralled fascinated fiend for gone gripped head over heels heavily into hooked hung up immersed industrious intent into intrigued lost monopolized occupied rapt really into riveted sedulous submerged taken up with tied-up turned on up to here in wrapped-up From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group. Advertisement Advertisement Advertisement Browse

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