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7200 | https://www.youtube.com/watch?v=5xy4GNZk3Wk | Mono Sebae Care Guide - Caring for Mono sebae. Tank mates, feeding, breeding, mono fish compare
Uri Shasha
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Mono Sebae Care Guide:
How to care, tank mates, gender, feeding and breeding Mono Sebae fish.
Mono Sebae fish are also known as Monodactylus sebae, African Moony, Guinean Fingerfish, Rambali Finger Fish and African Mono.
also the differences between mono sebae and mono argentus
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7201 | https://nsj.org.sa/content/nsj/11/4/322.full.pdf | Chronic suppurative otitis media with intracranial complication Abdulaziz A. Aldakhail, MD, KSF.
C hronic otitis media refers to an inflammatory process within the middle ear cleft associated with irreversible tissue pathology. It may be active with ongoing suppuration, or inactive demonstrating sequelae of a previous infection. Complications of chronic otitis media can be classified as intra temporal or intra cranial. In the pre antibiotic era the rate of intra cranial complications of otitis media was 2.3%.1,2 In 1935, Kafka3 studied 2100 patients with acute mastoiditis and 1125 patients with chronic mastoiditis. Of the 3225 patients, 209 (6.4%) developed an intra cranial complication, 48% developed meningitis, 22.5% developed lateral sinus thrombosis or subdural abscess, 16.5% developed brain abscess, and 13% developed other complications. The mortality was 76.4% in patients with intra cranial complications, or one in every 20 patients with mastoiditis. In the antibiotic era, intra cranial complications of otitis media still occur. In a more recent series of From the Department of ENT, Riyadh Medical Complex, Riyadh, Kingdom of Saudi Arabia.
Received 5th October 2005. Accepted for publication in final form 6th March 2006.
Address correspondence and reprint request to: Dr. Abdulaziz A. Aldakhail, Director General, Directorate of General Health Affairs, Riyadh, Kingdom of Saudi Arabia. Tel. +966 (1) 4504651. Fax. +966 (1) 4505066. E-mail: aldakhaila@yahoo.com 322 over 1400 patients with chronic ear disease and cholesteatoma followed over 15 years, nearly 7.5% developed intra cranial complications.4 Meningitis is still the most common complication followed by intra cranial abscess and lateral sinus thrombosis.2,5 Lateral sinus thrombosis was the result of chronic otitis media in 50% of cases. Early and aggressive surgical intervention of this otogenic complication can potentially minimize mortality, hospital stay, and length of medical treatment.6 In 1995, Kangsanarak et al2 conducted a review of 24,321 patients with otitis media. This review revealed an intracranial complication rate of 0.36% with a mortality rate of 18.9%. Any patient with an extra cranial complication from otitis media should have a neurological exam to rule out additional complications intra cranially. Lateral sinus thrombosis is now a rare complication of ear disease in the developed world, since the advent of the widespread use of antibiotics.6 The main objective Case Reports A complication of otitis media is defined as a spread of infection beyond the pneumatized area of the temporal bone and the associated mucosa. Complications can be classified as intra temporal or intra cranial, which includes extradural abscess, brain abscess, subdural abscess, sigmoid sinus thrombophlebitis, otic hydrocephalus, and meningitis. Since the introduction of antibiotics, intra cranial complications in otitis media are less common, however, the occurrence should not be underestimated due to their associated morbidity and mortality. Herein, we report a case of chronic suppurative otitis media (CSOM) with intra cranial complication (brain abscess) in a 25-year-old male patient with ear discharge, severe headache, and vomiting. Computerized tomography scan confirmed the diagnosis. The patient was managed by a combined neurosurgical and otologic approach. Surgery of the abscess includes aspiration through a bur hole or craniotomy and by radical mastoidectomy for CSOM. Neurosciences 2006; Vol. 11 (4): 322-325 ABSTRACT 323 Neurosciences 2006; Vol. 11 (4) Chronic suppurative otitis media ... Aldakhail of this case is to report on a patient with brain abscess discovered through a general hospital in the Riyadh area, and to highlight the diagnostic procedures, and therapy plan adopted.
Case Report. A 25-year-old male patient, resident of Yemen, presented with discharge from the left ear of 10 years duration, with severe headache, fever, vomiting, blurring of vision, and neck stiffness for the last 5 days. On examination, he was conscious, oriented with neck stiffness. A left ear examination revealed swelling of the external auditory canal, which was full of pus, but not blood stained, foul smelling in nature and thick in consistency. Aural pus was taken for culture and sensitivity. A suction clearance was carried out, which revealed a whitish mass in the attic region. The posterior meatal wall near the attic region was eroded; the right ear tympanic membrane and all cranial nerves were intact. Funduscopic examination revealed papilledema. Routine investigations carried out included complete blood count, urine analysis, blood glucose, bleeding profile, liver and renal function tests; all were within normal limits except for leucocytosis. The left aural pus culture revealed Citrobacters. An audiogram revealed an air bone gap on the left side. Lumber puncture was carried out, CSF was taken and sent for detailed report and culture sensitivity. According to D/R, the sugar level was raised, and according to culture, no growth was seen. An emergency CT scan of the temporal region and brain was carried out and showed that the left mastoid area was sclerotic, with soft tissue density in the attic area (Figure 1). The scan also revealed a single large well-defined low-density lesion seen with peripheral enhancement in the posterior fossa on the left side along with 2 small locules of abscess in the left cerebellopontine angle (Figure 2). The features were suggestive of left cerebellar abscess formation, secondary to left chronic suppurative otitis media. An emergency evacuation of the left cerebellar abscess was carried out by a neurosurgeon via craniotomy followed by left radical mastoidectomy by the ENT surgeon. During mastoid exploration, it was noted that the lateral wall of the left antrum was eroded, with cholesteatoma present, which was removed and sent for histopathology. Postoperatively no facial nerve weakness was seen. A repeat CT scan showed a successfully drained cerebellar abscess (Figures 3). On a regular monthly follow up visit, he was free from any ear discharge with a dry mastoid cavity.
Discussion. The incidence of intra cranial complication is higher in males, particularly in those younger than 20 years of age.4,5 Most complications Figure 1 - A CT scan showing sclerotic mastoid with soft tissue density in the attic area indicating granulation tissue.
Figure 2 - A CT scan showing a large cerebellar abscess with peripheral enhancement, and 2 small pus locules in the left cerebellopontine angle.
Figure 3 - Post-operative CT after drainage of the cerebellar abscess.
324 Chronic suppurative otitis media ... Aldakhail Neurosciences 2006; Vol. 11 (4) arise from chronic rather than acute otitis media. Factors causing complications include high virulence of organism, poor resistance of the patient, presence of chronic systemic diseases, and resistance of the organism to antibiotics. A report by Fisher et al,7 of their experiences during the 1970s described a mortality of 14%, and this is a typical figure in recently reported series. Taylor8 achieved a figure of 6% in 50 patients. It is still estimated to occur frequently, with the annual risk of an adult with active chronic otitis media developing an abscess approximately 1 in 10,000.9 Currently, brain abscess accounts for 6-66.4% of otogenic intra cranial complications; 0.5% of the patients with acute otitis, and 3% with chronic otitis media develop brain abscesses. In children, 25% of all brain abscesses are otogenic, while in the adults, with a greater predominance of chronic ear disease, it is greater than 50%. In one recent series,10 73% of intracranial abscess were caused by chronic middle ear disease. Temporal lobe abscesses usually occupy the middle third of the temporal lobe, cerebellar abscess occupy the anterior part of the lateral lobe of the cerebellum, and only 3% are multiple. Some authors have reported 33% of the intra cranial complications from their series to be brain abscess, 69% developed in the temporal lobe and 31% in the cerebellum. All abscesses developed in patients with chronic otitis media, and 90% of patients had cholesteatoma. This trend has been observed by others; Nunes and Browning9 reported that 95% of otogenic brain abscesses are from chronic otitis media. Of patients with otogenic brain abscesses, 41% were found to have cholesteatoma.9 Culture from brain abscesses yield no growth almost 50% of the time.11 Culture positive abscesses often show polymicrobial infection, with a high incidence of anaerobes,11 of which streptococci are the common organisms.12 Pyogenic staphylococci, streptococci, pneumococci and streptococci haemolytics are often found. Gram negative Bacilli, Proteus, Escherichia Coli and Pseudomonas are cultured with increasing frequency, reflecting the higher incidence of otogenic abscesses from chronic ear disease. There are reports of otogenic brain abscesses with intact tympanic membranes. where an acute infection has presumably been treated effectively with antibiotics.13 A brain abscess progresses through 3 clinical stages; initial encephalitis, which usually lasts a few days and may be attributed to a viral syndrome. A latent or quiescent stage, in which there are no localizing neurologic changes, which lasts several weeks or rarely several months. The third stage of manifestation, the expanding abscess, is caused by compression as a result of edema and encephalitis around the abscess rather than by the space occupying abscess itself. Symptoms of brain abscesses may be generalized or localized. Generalized symptoms include fever, irritability, drowsiness, headache, personality changes, and altered mental status. Continuous and severe headache is the most constant generalized symptom of brain abscess.14 Signs and symptoms associated with cerebellar abscess include suboccipital headache, vomiting, ataxia, gaze nystagmus, past pointing, intention tremor, dysdiadochokinesis, weakness, and incoordination of ipsilateral muscles. Ipsilateral ataxia is the most constant sign of a cerebellar abscess. The CT scan with and without intravenous contrast is no doubt the most important investigation in the diagnosis of brain abscess, and is also the most valuable method for observing the progress of an abscess during treatment. The MRI has further improved the diagnosis.15 The conditions to be differentiated from brain abscess include meningitis, subdural abscess, lateral sinus thrombophlebitis, and otic hydrocephalus.
Brain abscess is the ultimate otogenic complication, both in severity and difficulty of management.14Initial management consists of stabilization of the patient as continued in the management of the patient with meningitis. Management requires a combined neurosurgical and otologic approach, along with the use of large doses of systemic antibiotics. Surgery of the abscess includes aspiration through a burr hole or formal craniotomy, open drainage, or rarely total excision. This may occur simultaneously with the surgical approach to the ear, or it may precede management of the ear if the intracranial problem is of such severity that it should be managed first. Surgical management of associated chronic otitis media is based on the extent of the underlying disease. For patients with acute otitis media, wide myringotomy and drainage are performed, and mastoidectomy is performed in the presence of coalescent mastoiditis. Antibiotic therapy should be maintained for several weeks, and the response of the abscess should be followed by serial contrast enhanced CT.
Acknowledgment. We would like to thank Dr. Mohammad Wasi Ahmed, Specialist, ENT Department for his constructive contribution and critical and comprehensive revising. We would also like to thank Dr. Bander Alqhtani, Specialist, Department of ENT for his valuable help.
References 1. Bluestone CE, Klein JO, editors. Intracranial suppurative complications of otitis media and mastoiditis. In: Pediatric Otolaryngology. Philadelphia: WB Saunders; 1990. p. 537-546. 2. Kangsanarak J, Navacharoen N, Fooanant S, Ruckphaopunt K. Intracranial complications of suppurative otitis media: 13 years’ experience. Am J Otol 1995; 16: 104-109.
325 Neurosciences 2006; Vol. 11 (4) Chronic suppurative otitis media ... Aldakhail 3. Kafka MM. Mortality of mastoiditis and cerebral complications with review of 3225 cases with complications. Laryngoscope 1935; 45: 790-822. 4. Maksimovic Z, Rukovanjski M. Intracranial complications of cholesteatoma. Acta Otorhinolaryngol Belg 1993; 47: 33-36. 5. Gower D, McGuirt WF. Intracranial complications of acute and chronic infectious ear disease: a problem still with us. Laryngoscope 1983; 93: 1028-1033. 6. O’Connell JE. Lateral sinus thrombosis: a problem still with us. J Laryngol Otol 1991; 105: 398-399. 7. Fischer EG, McLennan JE, Suzuki Y. Cerebral abscess in children. Am J Dis Child 1981; 135: 746-749. 8. Taylor JC. The case for excision in the treatment of brain abscess. Br J Neurosurg 1987; 1: 173-178. Review. 9. Nunez DA, Browning GG. Risks of developing an otogenic intracranial abscess. J Laryngol Otol 1990; 104: 468-472.
10. Kulai A, Ozatik N, Topcu I. Otogenic intracranial abscesses. Acta Neurochir (Wien) 1990; 107: 140-146.
11. Bradley PJ, Manning KP, Shaw MD. Brain abscess secondary to otitis media. J Laryngol Otol 1984; 98: 1185-1191.
12. Maurice-Williams RS. Open evacuation of pus: a satisfactory surgical approach to the problem of brain abscess? J Neurol Neurosurg Psychiatry 1983; 46: 697-703.
13. Samuel J, Fernandes CM. Otogenic complications with an intact tympanic membrane. Laryngoscope 1985; 95: 1387-1390.
14. Miyamoto RT, Worth RM. Otogenic cerebellar abscess. Ann Otol Rhinol Laryngol 1986; 95: 647-648.
15. Haimes AB, Zimmerman RD, Morgello S, Weingarten K, Becker RD, Jennis R. MR imaging of brain abscesses. AJNR Am J Neuroradiol 1989; 10: 279-291. |
7202 | https://www.everlywell.com/blog/blood-test/tibc-blood-test-normal-range/?srsltid=AfmBOoqemCWso0lxSPqD1hPQf7TRhxcNPi_lk1Pyt_F_DmX88GqkUM3S | TIBC Blood Test Normal Range: What Your Results Mean
By Dr. Diana Rangaves, PharmD
When it comes to understanding your overall iron health, the total iron binding capacity (TIBC) blood test plays a key role . If you've recently had your iron levels tested, you might notice a TIBC result listed alongside other markers like serum iron and transferrin saturation.
But what exactly does TIBC measure, what is considered a normal range, and what might high or low levels indicate?
Let's break it all down so you can better understand your test results and what they might mean for your health.
What Is a TIBC Blood Test?
The TIBC blood test measures the blood's capacity to bind iron with transferrin, a protein that transports iron throughout your body . Essentially, it tells you how much transferrin is available to carry iron.
TIBC is often ordered alongside other iron tests, including serum iron and transferrin saturation, to provide a complete picture of iron metabolism. The total iron binding capacity (TIBC) test helps diagnose conditions like iron deficiency anemia, hemochromatosis (iron overload), and chronic illnesses that impact iron levels .
What Is the Normal Range for TIBC?
Typically, the normal TIBC range falls between 171 to 505 mcg/dL for males and 149 to 492 mcg/dL for females . However, this range may vary slightly depending on the laboratory and individual factors like age, sex, and overall health .
TIBC alone isn't enough to diagnose an issue. Healthcare providers interpret it alongside serum iron and transferrin saturation to understand how much iron is being transported.
Understanding Iron Saturation and Its Normal Range
Iron saturation, or transferrin saturation, measures the percentage of transferrin that is saturated with iron . It is calculated by dividing serum iron by TIBC and multiplying by 100 .
A normal transferrin saturation range is generally around 20% to 50% .
Together, TIBC and transferrin saturation help determine if there's too little or too much iron circulating in your blood.
Low TIBC: What It Means
A low TIBC result can suggest that your body's ability to transport iron is reduced. Common causes include :
Low TIBC often occurs alongside high iron saturation, indicating that although iron is present, the capacity to carry more is reduced.
High TIBC: What It Means
A high TIBC result suggests that your body has an increased capacity to transport iron, often because there's a lack of iron available.
Common causes of elevated TIBC include :
How to Interpret TIBC and Iron Panel Results
Healthcare providers rarely interpret TIBC results in isolation. Instead, they look at the whole iron panel :
| Test | Low | High |
---
| Serum Iron | Iron deficiency, a chronic disease | Hemochromatosis, iron poisoning |
| TIBC | Hemochromatosis, malnutrition | Iron deficiency anemia, pregnancy |
| Transferrin Saturation | Iron deficiency anemia | Iron overload disorders |
Understanding these patterns helps healthcare providers diagnose conditions more accurately and plan appropriate treatment.
Frequently Asked Questions
What Is a Normal TIBC Level for Women and Men?
The normal TIBC range falls between 171 to 505 mcg/dL for men and 149 to 492 mcg/dL for women .
What Happens If TIBC Levels Are High?
High TIBC typically points toward iron deficiency or increased iron demand, as in pregnancy . It signals the body is trying to maximize iron transport because iron stores are low.
How Do You Interpret TIBC and Iron Saturation Results Together?
If your TIBC is high but your transferrin saturation is low, it likely indicates iron deficiency. If both are low, it may point toward chronic disease or malnutrition .
What Iron Levels Are Concerning?
Very low serum iron (below 60 mcg/dL in adults) paired with high TIBC is concerning for iron deficiency anemia . Very high iron paired with low TIBC could suggest iron overload disorders.
How Is High TIBC Treated?
Treatment targets the underlying cause. For iron deficiency, healthcare providers typically recommend dietary changes and iron supplementation.
What Drinks Are High in Iron?
Iron-rich beverages include prune juice, fortified plant-based milks, and smoothies made with leafy greens. Vitamin C-rich drinks can also help enhance iron absorption .
Should You Worry About a High Ferritin Blood Test Result?
High ferritin levels could indicate inflammation, infection, or iron overload. It should be interpreted with other iron studies .
Can Medications Affect TIBC Levels?
Yes. Oral contraceptives, estrogen therapy, and certain anti-inflammatory medications can influence TIBC and transferrin levels .
Managing Abnormal TIBC Results
If your TIBC is abnormal, your healthcare provider may recommend:
Monitoring your iron levels over time is crucial. Untreated imbalances, whether deficiency or overload, can have serious health consequences.
Take Charge of Your Iron Health with Everlywell
Understanding your total iron-binding capacity (TIBC) levels is essential for assessing how well your body manages iron transport. When reviewed alongside other iron-related markers, TIBC testing can help identify conditions such as iron deficiency anemia or iron overload disorders.
If your TIBC levels fall outside the normal range, it's important to follow up with a healthcare provider. Early detection and intervention can make a significant difference in your overall health.
With Everlywell 360, you can take control of your health from the comfort of home.
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Everlywell offers health and wellness solutions including laboratory testing for wellness monitoring, informational and educational use. With the exception of certain diagnostic test panels, list available here, the tests we offer access to are not intended to diagnose or treat disease. None of our tests are intended to be a substitute for seeking professional medical advice, help, diagnosis, or treatment. At-home lab tests are not available in NY.
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7203 | https://math.stackexchange.com/questions/4592617/converting-between-arbitrary-real-valued-base-systems | Stack Exchange Network
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Converting between arbitrary real-valued base systems
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I watched a video from Combo Class on YouTube about non-integer base systems which is something I've expressed interest in before but this did get me thinking about them again. In the video we are given a look at bases Ï, e, 1/2, 3/2, $\sqrt2$, $\sqrt10$, $\sqrt{10}$, and ɸ.
In theory, any real number $n$ can be used as a radix for a positional number system, and we can easily represent such bases using the powers of $n$ multiplied by the place value. For example, $3Ï^2 + 7Ï^1 + 2Ï^0 + 6Ï^{-1} + 4Ï^{-2}$. Of course, we might not want this representation of a base-Ï number; we'd want to represent it the way we normally show numbers, which is with positional notation and a limited number of digits, with a radix point separating the integral and fractional parts. We'll almost certainly get an infinite amount of digits, which is acceptable, we just want to be able to calculate the digits up to an arbitrary point, and know exactly how many symbols we need to represent it.
Up until now, I had thought that for any given real base $b$, you only need a number of digits $d$ such that $ceil(b) = d$. For example, base $\sqrt2$ is less than 2, but it still uses two symbols - 0 and 1 - just like binary, because $ceil(\sqrt2)$ = 2. The same goes for base ɸ. Yet, once Domotro covers base 3/2, it turns out we actually need 3 symbols - 0, 1, and 2 - rather than just the expected 2, since 3/2 = 1.5 which is less than 2. And even weirder, 3/2 is between $\sqrt2$ and ɸ in value, so we go from needing 2 digits to 3, to going back to only needing 2. In other words, $d$ is not monotonically increasing as $b$ increases.
Additionally, we see in the video that we can convert to some of these real-valued bases, but it's done by exploiting properties of the number used for the base - for example, conversion from base 10 to ɸ is made possible by using algebraic properties of ɸ. However, is there a general formula that can convert between any two arbitrary real-numbered bases? The standard algorithm used for positive integer bases doesn't work since you get fractional remainders, so I'm looking for a generalized version of the base conversion algorithm that works with any real number $n$. And, if such an algorithm exists, then how do we determine how many digits $d$ we need, since the rule $ceil(b) = d$ does not hold. Or is it the case that there does not exist such an algorithm, and generalized representations of arbitrary real-numbered bases is an undecidable problem and thus the rules are something we have to make up ourselves, based on what we find most convenient?
For example, say I wanted to, for whatever strange reason, convert the base-ten number 12 to the equivalent number in base-$ln(10)$, without exploiting properties specific to $ln(10)$ and just using a generalized algorithm, such that I get the corresponding value in a typical positional notation, with only the digits needed to represent that base.
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7204 | https://www.youtube.com/watch?v=XMwnJUprid8 | Inverse property of multiplication | Arithmetic properties | Pre-Algebra | Khan Academy
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Transcript:
Let's say that I have five lemons so that's [counting to five] ... five lemons and I were to ask you: what do I have to multiply times five to get one? or in this case: what do I have to multiply times five lemons to get one lemon? and so, another question you might ask because really multiplication and division are two sides of the same coin is what do I have to divide five by to get to one lemon or yellow circle, or whatever I have drawn right over here Well, if you have five things and you divide by five, you're gonna have five groups of one so if you divide by five, you're gonna have [counting to five] ... five groups So you could say five divided by five is equal to one take five things and divide it into five groups then each group is going to have one in them or you could say five times one fifth is equal to one (and I use the dot for multiplication) I could also say five times one fifth is equal to one these are all really saying the same thing Maybe what's kind of interesting here (although it's not some huge learning) it's really just another way of writing what you already probably know, is this idea that if I have a number and I multiply times it's multiplicative inverse (and most of the time when people talk about inverses in mathematics they are talking about the multiplicative inverse) then I'm going to get one so five time one fifth is equal to one but that's just because five times one fifth is the same thing as five divided by five if you were to actually multiply this out you actually take five times one fifth this is equal to five-over-one times one-over-five you multiply the numerators: five times one is five multiply the denominators: one times five is five so you have five fifths, and five fifths is the exact same thing as one So if someone where to ask you a question, they say "Hey, I have the number 217 and I want to multiply it by something, and I want to get one after multiplying it by that something" Well then you say - Well look, If I took 217 and divided it by 217 that would get me to one and dividing by 217 is the exact same thing as multiplying by one-over-217 multiplying by its multiplicative inverse which is, once again, a word that is fancier than the actual concept you are just multiplying by the inverse of this number Another way to think about it is if I have five things and I take one fifth of those things, how many things do I have? Well, if I take one fifth of five things I have exactly one thing right over here But the general idea is super-duper-duper simple if I have some crazy number ... 8,345 that's actually not so crazy, let's turn it to something in the millions ... and 271 ... so 8,345,271 And I say, what do I have to multiply by (and now I use this multiplication symbol right now) what do I have to multiply that by in order to get one? I just have to multiply it by the inverse of this the multiplicative inverse of this so one-over-8,345,271 |
7205 | https://en.wikipedia.org/wiki/Yen%27s_algorithm | Jump to content
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1 Algorithm
1.1 Terminology and notation
1.2 Description
1.3 Pseudocode
1.4 Example
2 Features
2.1 Space complexity
2.2 Time complexity
3 Improvements
3.1 Lawler's modification
4 See also
5 References
6 External links
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From Wikipedia, the free encyclopedia
Concept in graph theory mathematics
In graph theory, Yen's algorithm computes single-source K-shortest loopless paths for a graph with non-negative edge cost. The algorithm was published by Jin Y. Yen in 1971 and employs any shortest path algorithm to find the best path, then proceeds to find K − 1 deviations of the best path.
Algorithm
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Terminology and notation
[edit]
| Notation | Description |
--- |
| | The size of the graph, i.e., the number of nodes in the network. |
| | The -th node of the graph, where ranges from to . This means that is the source node of the graph, and is the sink node of the graph. |
| | The cost of the edge between and , assuming that and . |
| | The -th shortest path from to , where ranges from to . Then , where is the 2nd node of the -th shortest path, is the 3rd node of the -th shortest path, and so on. |
| | A deviation path from at node , where ranges from to . Note that the maximum value of is , which is the node just before the sink in the shortest path. This means that the deviation path cannot deviate from the shortest path at the sink. The paths and follow the same path until the -th node, then edge is different from any path in , where ranges from to . |
| | The root path of that follows that until the -th node of . |
| | The spur path of that starts at the -th node of and ends at the sink. |
Description
[edit]
The algorithm can be broken down into two parts: determining the first k-shortest path, , and then determining all other k-shortest paths. It is assumed that the container will hold the k-shortest path, whereas the container will hold the potential k-shortest paths. To determine , the shortest path from the source to the sink, any efficient shortest path algorithm can be used.
To find the , where ranges from to , the algorithm assumes that all paths from to have previously been found. The iteration can be divided into two processes: finding all the deviations and choosing a minimum length path to become . Note that in this iteration, ranges from to .
The first process can be further subdivided into three operations: choosing the , finding , and then adding to the container . The root path, , is chosen by finding the subpath in that follows the first nodes of , where ranges from to . Then, if a path is found, the cost of edge of is set to infinity. Next, the spur path, , is found by computing the shortest path from the spur node, node , to the sink. The removal of previous used edges from to ensures that the spur path is different. , the addition of the root path and the spur path, is added to . Next, the edges that were removed, i.e. had their cost set to infinity, are restored to their initial values.
The second process determines a suitable path for by finding the path in container with the lowest cost. This path is removed from container and inserted into container , and the algorithm continues to the next iteration.
Pseudocode
[edit]
The algorithm assumes that the Dijkstra algorithm is used to find the shortest path between two nodes, but any shortest path algorithm can be used in its place.
function YenKSP(Graph, source, sink, K): // Determine the shortest path from the source to the sink. A = Dijkstra(Graph, source, sink); // Initialize the set to store the potential kth shortest path. B = []; for k from 1 to K: // The spur node ranges from the first node to the next to last node in the previous k-shortest path. for i from 0 to size(A[k − 1]) − 2: // Spur node is retrieved from the previous k-shortest path, k − 1. spurNode = A[k-1].node(i); // The sequence of nodes from the source to the spur node of the previous k-shortest path. rootPath = A[k-1].nodes(0, i); for each path p in A: if rootPath == p.nodes(0, i): // Remove the links that are part of the previous shortest paths which share the same root path. remove p.edge(i,i + 1) from Graph; for each node rootPathNode in rootPath except spurNode: remove rootPathNode from Graph; // Calculate the spur path from the spur node to the sink. // Consider also checking if any spurPath found spurPath = Dijkstra(Graph, spurNode, sink); // Entire path is made up of the root path and spur path. totalPath = rootPath + spurPath; // Add the potential k-shortest path to the heap. if (totalPath not in B): B.append(totalPath); // Add back the edges and nodes that were removed from the graph. restore edges to Graph; restore nodes in rootPath to Graph; if B is empty: // This handles the case of there being no spur paths, or no spur paths left. // This could happen if the spur paths have already been exhausted (added to A), // or there are no spur paths at all - such as when both the source and sink vertices // lie along a "dead end". break; // Sort the potential k-shortest paths by cost. B.sort(); // Add the lowest cost path becomes the k-shortest path. A[k] = B; // In fact we should rather use shift since we are removing the first element B.pop(); return A;
Example
[edit]
The example uses Yen's K-Shortest Path Algorithm to compute three paths from to . Dijkstra's algorithm is used to calculate the best path from to , which is with cost 5. This path is appended to container and becomes the first k-shortest path, .
Node of becomes the spur node with a root path of itself, . The edge, , is removed because it coincides with the root path and a path in container . Dijkstra's algorithm is used to compute the spur path , which is , with a cost of 8. is added to container as a potential k-shortest path.
Node of becomes the spur node with . The edge, , is removed because it coincides with the root path and a path in container . Dijkstra's algorithm is used to compute the spur path , which is , with a cost of 7. is added to container as a potential k-shortest path.
Node of becomes the spur node with a root path, . The edge, , is removed because it coincides with the root path and a path in container . Dijkstra's algorithm is used to compute the spur path , which is , with a cost of 8. is added to container as a potential k-shortest path.
Of the three paths in container , is chosen to become because it has the lowest cost of 7. This process is continued to the 3rd k-shortest path. However, within this 3rd iteration, note that some spur paths do not exist. And the path that is chosen to become is .
Features
[edit]
Space complexity
[edit]
To store the edges of the graph, the shortest path list , and the potential shortest path list , memory addresses are required. At worse case, the every node in the graph has an edge to every other node in the graph, thus addresses are needed. Only addresses are need for both list and because at most only paths will be stored, where it is possible for each path to have nodes.
Time complexity
[edit]
The time complexity of Yen's algorithm is dependent on the shortest path algorithm used in the computation of the spur paths, so the Dijkstra algorithm is assumed. Dijkstra's algorithm has a worse case time complexity of , but using a Fibonacci heap it becomes , where is the number of edges in the graph. Since Yen's algorithm makes calls to the Dijkstra in computing the spur paths, where is the length of spur paths. In a condensed graph, the expected value of is , while the worst case is . The time complexity becomes .
Improvements
[edit]
Yen's algorithm can be improved by using a heap to store , the set of potential k-shortest paths. Using a heap instead of a list will improve the performance of the algorithm, but not the complexity. One method to slightly decrease complexity is to skip the nodes where there are non-existent spur paths. This case is produced when all the spur paths from a spur node have been used in the previous . Also, if container has paths of minimum length, in reference to those in container , then they can be extract and inserted into container since no shorter paths will be found.
Lawler's modification
[edit]
Eugene Lawler proposed a modification to Yen's algorithm in which duplicates path are not calculated as opposed to the original algorithm where they are calculated and then discarded when they are found to be duplicates. These duplicates paths result from calculating spur paths of nodes in the root of . For instance, deviates from at some node . Any spur path, where , that is calculated will be a duplicate because they have already been calculated during the iteration. Therefore, only spur paths for nodes that were on the spur path of must be calculated, i.e. only where ranges from to . To perform this operation for , a record is needed to identify the node where branched from .
See also
[edit]
Yen's improvement to the Bellman–Ford algorithm
References
[edit]
^ Yen, Jin Y. (1970). "An algorithm for finding shortest routes from all source nodes to a given destination in general networks". Quarterly of Applied Mathematics. 27 (4): 526–530. doi:10.1090/qam/253822. MR 0253822.
^ a b c Yen, Jin Y. (Jul 1971). "Finding the k Shortest Loopless Paths in a Network". Management Science. 17 (11): 712–716. doi:10.1287/mnsc.17.11.712. JSTOR 2629312.
^ Fredman, Michael Lawrence; Tarjan, Robert E. (1984). Fibonacci heaps and their uses in improved network optimization algorithms. 25th Annual Symposium on Foundations of Computer Science. IEEE. pp. 338–346. doi:10.1109/SFCS.1984.715934.
^ Bouillet, Eric (2007). Path routing in mesh optical networks. Chichester, England: John Wiley & Sons. ISBN 9780470032985.
^ Brander, Andrew William; Sinclair, Mark C. A comparative study of k-shortest path algorithms. Department of Electronic Systems Engineering, University of Essex, 1995.
^ Lawler, EL (1972). "A procedure for computing the k best solutions to discrete optimization problems and its application to the shortest path problem". Management Science. 18 (7): 401–405. doi:10.1287/mnsc.18.7.401.
External links
[edit]
Open Source C++ Implementation
Open Source C++ Implementation using Boost Graph Library
| v t e Graph and tree traversal algorithms |
| Search | α–β pruning A + IDA + LPA + SMA Best-first search Beam search Bidirectional search Breadth-first search + Lexicographic + Parallel B Depth-first search + Iterative deepening D Fringe search Jump point search Monte Carlo tree search SSS |
| Shortest path | Bellman–Ford Dijkstra's Floyd–Warshall Johnson's Shortest path faster Yen's |
| Minimum spanning tree | Borůvka's Kruskal's Prim's Reverse-delete |
| List of graph search algorithms |
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Find the number of trailing zeros in 50! [duplicate]
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Asked 7 years, 1 month ago
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Number of zeroes at end of factorial (2 answers)
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My attempt:
50! = 50 49 48 ....
Even even = even number
Even odd = even number
odd odd = odd number
25 evens and 25 odds
Atleast 26 of the numbers will lead to an even multiple (24 evens + 1 even 1 odd) so at most 26 trailing zeros.
50 is divisible by 5: 10 times. Atleast 10 trailing zeros.
What is the answer?
algebra-precalculus
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edited Aug 10, 2018 at 15:17
Mike Pierce
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asked Aug 10, 2018 at 15:09
user9995331user9995331
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3 Answer would be 12 12. Hint: What about 25 25's ?Jakobian –Jakobian 2018-08-10 15:10:52 +00:00 Commented Aug 10, 2018 at 15:10
Isn't dividing by 5 taking into the 25's because it is a multiple?user9995331 –user9995331 2018-08-10 15:13:04 +00:00 Commented Aug 10, 2018 at 15:13
1 It is. But 25 25's give us two 5 5's. If the factorial was big enough, we would also check for 125 125's, or any powers of 5 5 Jakobian –Jakobian 2018-08-10 15:14:11 +00:00 Commented Aug 10, 2018 at 15:14
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Every factor of 5 will mate with a factor of 2 and produce a trailing 0. How do you count the factors of 5 in the product?
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answered Aug 10, 2018 at 15:14
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You need to find the highest power of 10 10 that divides 50!50!, which is same as the highest power of 5 5 that divides 50!50!, since 10=5×2 10=5×2, and there are fewer multiples of 5 5 in the first 50 50 numbers. It's easy to count the 10 multiples of 5 5, namely 5,10,15,⋯,50 5,10,15,⋯,50, but also add two more powers of 5 5 as 25 25 and 50 50 have 5 2 5 2 as the highest power of 5 5 as factor.
The answer is 12 12.
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answered Aug 10, 2018 at 15:16
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There are ⌊50 5⌋⌊50 5⌋ numbers between 1 1 and 50 50 that are divisible by 5 5. Similiarily ⌊50 5 2⌋⌊50 5 2⌋ and ⌊50 5 3⌋⌊50 5 3⌋ numbers divisible by 5 2 5 2 and 5 3 5 3 respectively.
Thus, the highest power of 5 5 dividing 50!50! is
⌊50 5⌋+⌊50 5 2⌋+⌊50 5 3⌋=10+2+0=12⌊50 5⌋+⌊50 5 2⌋+⌊50 5 3⌋=10+2+0=12
With similiar arguments, one can show the greatest power of 2 2 dividing 50!50! is greater then 12 12. Thus, the greatest power of 10=2⋅5 10=2⋅5 dividing 50!50! is 12 12.
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answered Aug 10, 2018 at 15:18
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5Number of zeroes at end of factorial
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4Is there a way to find the number of trailing zeroes in a factorial with a certain base?
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1Greatest n n is possible? in 10 n 10 n as a divisor of 1995!1995!
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1Why n!n!(when represented in base fourteen) can never have 488 488 trailing zeroes?
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1,
Tuanhui Ren
Tuanhui Ren
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1,
Ning Gao
Ning Gao
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Fengyun Wen
Fengyun Wen
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Xiaoxia Li
Xiaoxia Li
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1
College of Animal Science and Technology, Henan University of Science and Technology, Luoyang 471023, China
2
Henan Provincial Key Laboratory for Grass-Feeding Animals, Luoyang 471023, China
3
College of Animal Science and Technology, Hunan Agricultural University, Changsha 410125, China
Authors to whom correspondence should be addressed.
Animals 2025, 15(15), 2246;
Submission received: 10 June 2025 / Revised: 22 July 2025 / Accepted: 27 July 2025 / Published: 31 July 2025
(This article belongs to the Section Animal Reproduction)
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Simple Summary
Male fertility depends greatly on the quality and function of spermatozoa. Spermatozoa need a lot of energy to move and fertilize the egg, and this energy is mainly produced by mitochondria. However, mitochondria can also produce harmful substances called reactive oxygen species, especially when the balance in the cell is disturbed. While small amounts of ROS are necessary for normal sperm function, too much can damage sperm quality. In this review, we summarize how mitochondria produce energy in sperm, how excess ROS can lead to problems, and how damage to mitochondrial DNA can affect sperm health. We also discuss promising treatments by antioxidants that specifically target mitochondria to protect spermatozoa from damage. Understanding how to keep mitochondria healthy in spermatozoa could lead to better treatments for male infertility and help improve reproductive success in both humans and animals.
Abstract
Mitochondria are central to energy production and redox regulation in spermatozoa, supporting key functions such as progressive motility, capacitation, and the acrosome reaction. These processes are essential for successful fertilization and embryo development. However, species-specific differences exist in the reliance on oxidative phosphorylation versus glycolysis. Mitochondria also generate reactive oxygen species, which at physiological levels aid in sperm function but can cause oxidative stress and damage when overproduced. Mitochondrial dysfunction and excessive ROS can impair membrane potential, induce apoptosis, and damage nuclear and mitochondrial DNA, ultimately compromising sperm quality. Sperm mitochondrial DNA is highly susceptible to mutations and deletions, contributing to reduced motility and fertility. Targeted antioxidant strategies have emerged as promising therapeutic interventions to mitigate oxidative damage. This article provides a comprehensive overview of mitochondrial regulation in spermatozoa, the consequences of redox imbalance, and the potential of mitochondria-targeted antioxidants to improve sperm function and male fertility outcomes. The paper aims to deepen our understanding of mitochondrial roles in sperm physiology and contribute to the advancement of strategies for addressing male infertility.
Keywords:
mitochondria; spermatozoa; energy metabolism; reactive oxygen species; mtDNA; antioxidant
Graphical Abstract
1. Introduction
Infertility affects approximately 10% to 15% of couples worldwide, with male factors contributing to nearly 50% of cases. Male infertility is most commonly associated with poor semen quality, including reduced sperm count (e.g., oligozoospermia or azoospermia), decreased motility (asthenozoospermia), abnormal morphology (teratozoospermia), and increased DNA fragmentation . In livestock breeding, male fertility is also a key determinant of reproductive success, influencing artificial insemination efficiency and overall productivity. For example, the quality of boar spermatozoa significantly affects the conception rate of sows [2,3], and studies have revealed a considerable variation in acrosome integrity among bulls with differing fertility levels , indicating the crucial role of sperm quality in determining the reproductive success of male livestock.
Mitochondria play a central role in sperm function by generating ATP through oxidative phosphorylation (OXPHOS), regulating redox balance, and triggering apoptosis . OXPHOS can also generate reactive oxygen species (ROS) as natural byproducts, linking energy metabolism to oxidative stress (OS) regulation. However, excessive ROS production or insufficient antioxidant defenses can disrupt mitochondrial redox balance and lead to sperm damage [6,7]. Mitochondrial ROS are also involved in the fertilization processes such as sperm maturation, capacitation and acrosome reaction. Mitochondrial dysfunction can trigger intrinsic apoptotic pathways, leading to loss of mitochondrial membrane potential (MMP) and the activation of caspases, ultimately compromising sperm viability and fertilization potential. Additionally, mitochondrial DNA (mtDNA) lacks histone protection, making it more susceptible to oxidative damage from ROS, which can lead to mutations that impair OXPHOS and reduce sperm motility .
Although several studies have reviewed some aspects of mitochondria in sperm, a comprehensive synthesis that integrates species-specific energy metabolism strategies, mitochondrial redox imbalance, mtDNA vulnerabilities, and therapeutic interventions remains lacking. The article aims to give a comprehensive summary of sperm mitochondrial energy metabolism, outlining the key factors that affect it, discussing how mitochondrial ROS and mtDNA damage contribute to sperm dysfunction, and highlighting current advances in mitochondria-targeted antioxidants as a potential strategy to improve sperm quality and male fertility. This may help to enrich our understanding of mitochondria’s function in sperm, expanding our knowledge of male reproductive physiology, and inspiring new research ideas for addressing male infertility, as well as enhancing the reproductive performance of male animals.
2. Relevance of Mitochondrial Energy Metabolism in Spermatozoa
2.1. The Mitochondrial Energy Metabolism in Spermatozoa
Spermatozoa produce ATP mainly through glycolysis and OXPHOS . Glycolysis occurs in the fiber sheath of sperm flagella [9,10], converting glucose into pyruvic acid. This process mainly involves a series of enzyme-catalyzed reactions, including hexokinase, phosphofructokinase, and glyceraldehyde-3-phosphate dehydrogenase, etc. . Glycolysis is important in sperm ATP production; inhibition of glycolysis not only reduces ATP levels but also decreases protein tyrosine phosphorylation, a critical signaling event associated with sperm capacitation and hyperactivated motility . Studies have shown that mouse spermatozoa primarily produce ATP by glycolysis . In the medium mainly constituted of glucose, mouse spermatozoa are capable of producing a considerable amount of ATP, which is crucial for maintaining their motility. Notably, any disruption to glycolysis severely impairs mouse sperm motility, even when provided with alternative substrates necessary for OXPHOS . And also, Takei et al. analyzed the distribution of adenosine monophosphate (AMP) in mouse sperm flagella, they found that glycolysis may transport ATP from mitochondria to the distal end of the flagella, thereby maintaining ATP concentration at the distal end of the flagella, which means that glycolysis could also play a role in energy transfer in mouse spermatozoa . Glycolysis is also important in the energy metabolism of human spermatozoa [11,16]. Nascimento et al. performed inhibition treatments targeting both OXPHOS and glycolysis in human spermatozoa, which indicated that OXPHOS did not yield enough ATP to maintain sperm motility, and motility decreases progressively if glycolysis is inhibited . And also, the hyperactivation of human spermatozoa is marked by a requirement for glycolytic substrates such as glucose or fructose, highlighting that the glycolysis pathway is the predominant energy supplier for human sperm function . In the realm of livestock species, the energy metabolism of spermatozoa in ram is similar to that of mice and humans; the inhibition of the glycolysis pathway in ram spermatozoa would result in a decrease of sperm motility and compromised fertilization ability . These findings indicate the critical role of glycolysis across these species.
Another pathway for ATP production is OXPHOS, a mitochondria-dependent process in which ATP is generated via electron transfer through the respiratory chain complexes I–IV and ATP synthase. In this process, NADH+, H+ and FADH2 are oxidized through a series of enzyme-catalyzed reactions and a continuous electron transfer, resulting in the production of H2O and ATP . The glycolysis and OXPHOS pathways jointly contribute to ATP production in spermatozoa, but the primary means of ATP production differs among species. Davila et al. inhibited OXPHOS in stallion spermatozoa; the intervention led to a notable reduction in sperm viability, motility, and membrane integrity, illustrating the crucial role of OXPHOS in ATP synthesis for stallion spermatozoa . And also, the preference for OXPHOS over glycolysis in stallion sperm has been confirmed by the Agilent Seahorse XFp Technology . Intriguingly, there are studies that show that stallion spermatozoa also require glycolysis to maintain high sperm velocities, suggesting that stallion sperm not only rely on OXPHOS for the production of ATP but also maintain the capacity to generate ATP via glycolysis [23,24]. Bovine spermatozoa are capable of harnessing both OXPHOS and glycolysis for energy production . Bulkeley et al. have demonstrated that impeding the electron transport chain (ETC) resulted in a marked decline in both the motility and vitality of bovine spermatozoa , which highlights the significance of OXPHOS in ATP provision for these cells. Nevertheless, in circumstances where ample glycolytic substrates are available, bovine spermatozoa retain the ability to produce ATP via the glycolytic pathway, showing a flexible energy metabolism in these cells . In addition, pig spermatozoa were thought to predominantly rely on glycolysis for ATP production [28,29], while Prieto et al. found that mitochondrial OXPHOS is the primary source of ATP in fresh boar semen, however, with extended storage under in vitro conditions, the OXPHOS capacity of pig spermatozoa progressively deteriorates, which shifts their energy production towards an enhanced reliance on glycolysis to sustain ATP synthesis . This indicates a temporal adaptation of energy metabolism in pig spermatozoa in response to extrinsic storage conditions. Spermatozoa differ significantly from somatic cells in terms of energy metabolism. While most somatic cells rely predominantly on mitochondrial OXPHOS for steady ATP production , sperm cells demonstrate a compartmentalized and species-dependent balance between glycolysis and OXPHOS. Exploring the intricacies of sperm energy metabolism not only elucidates these disparities but also paves the way for refining semen preservation methodologies, thereby contributing to advancements in reproductive science and technology.
2.2. Proteins Affecting Mitochondrial Energy Metabolism in Spermatozoa
Many proteins are involved in the spermatozoa energy metabolism process . The oxidative respiratory chain, mainly composed of respiratory chain protein complexes I–IV, ATP synthase, cytochrome C, and coenzyme Q10, is located on the inner membrane of mitochondria and serves as the main site for ATP production and plays a key role in mitochondrial energy metabolism [33,34]. Complex I, which has been known as NADH dehydrogenase, primarily catalyzes the oxidation of NADH within mitochondria . It also serves as a proton pump that transports protons from the matrix into the intermembrane space as electrons pass through, thus forming a proton gradient, and then drives ATP synthase to produce ATP from ADP and inorganic phosphate . Complex II, also known as succinate dehydrogenase (SDH), consists of four subunits (SDHA, SDHB, SDHC, and SDHD) . This complex could transfer electrons from succinate to ubiquinone via its iron-sulfur (Fe–S) clusters, and oxidizes succinate to fumarate in the tricarboxylic acid (TCA) cycle [37,38,39], which is important in reprogramming of metabolic and respiratory adaptation. Complex III is the cytochrome reductase complex, which is responsible for transferring electrons to the cytochrome C receptor . Complex IV is a cytochrome oxidase complex that serves as the final component of the respiratory chain by converting oxygen molecules into water molecules . Complex III and complex IV also possess the function of proton transfer , which generates a transmembrane potential that promotes ATP synthase for ATP synthesis . Research has revealed that oligozoospermia patients exhibit significantly lower expression levels of mitochondrial cytochrome oxidase in their spermatozoa compared to normal individuals [43,44], suggesting a direct correlation between the complex III and IV function and sperm quality. Additionally, ATP synthase, a component of the mitochondrial respiratory chain, converts ADP and Pi into ATP by harnessing the transmembrane proton electrochemical gradient created by the electron transfer chain [45,46,47]. Coenzyme Q10, known as ubiquinone, can promote energy production and neutralize the ROS in sperm mitochondria. Deficiency of coenzyme Q10 can lead to decreased sperm motility [48,49,50]. These results demonstrated that the mitochondrial respiratory chain is indispensable in sperm function; impaired mitochondrial respiratory chain protein complexes would destroy sperm energy metabolism.
Besides the respiratory chain proteins, many other proteins have been confirmed to participate in the sperm mitochondria’s energy metabolism. We summarized these proteins and their function in energy metabolism in Table 1. Studying the effects of these molecules on sperm metabolism may help us to better understand the physiological processes of sperm energy metabolism.
3. Effects of Mitochondrial ROS on Spermatozoa
3.1. Sources of Mitochondrial ROS in Spermatozoa
Mitochondria are the primary source of ROS produced in spermatozoa . The main types of mitochondrial ROS are O2−, ·OH and H2O2; they are oxygen-containing substances with high activity . Mitochondrial ROS primarily occurs during OXPHOS in the inner membrane of mitochondria . Spermatozoa possess antioxidant mechanisms that help maintain ROS balance. Controlled levels of ROS regulate physiological functions and support normal sperm performance, while excessive ROS can cause OS, leading to sperm damage and reduced sperm quality. ROS production in sperm mitochondria is influenced by various factors [6,68]. Spermatogenesis, which includes mitosis, meiosis, and cell differentiation, significantly contributes to ROS generation . The energy metabolic activity of mitochondria also produces substantial ROS as a byproduct . Under stressful conditions, the electron transport during mitochondrial OXPHOS is imperfect, resulting in the formation of superoxide anion (O2−). These superoxide anions are then converted into hydrogen peroxide (H2O2) through the action of superoxide dismutase . Both superoxide anions and hydrogen peroxide have short half-lives and are harmless to sperm under normal circumstances. For example, leukocytes and immature or morphologically abnormal spermatozoa in semen contribute significantly to ROS generation, exacerbating OS and impairing mitochondrial function . Additionally, diseases, nutrient deficiencies, unhealthy lifestyle habits, and environmental pollution are notable contributors to ROS production .
3.2. Consequences of Mitochondrial ROS Overproduction and Redox Imbalance on Sperm Quality
ROS-mediated damage compromises the structural and functional integrity of spermatozoa . The mitochondrial ETC is involved in the phosphorylation of ADP into ATP . Notably, ROS are primarily generated in respiratory complexes I and III of the ETC , highlighting their direct impact on mitochondrial energy metabolism. Indeed, elevated ROS levels have been shown to negatively correlate with the activity of ETC complexes and sperm motility [75,76]. Excessive ROS can reduce cytochrome C oxidase activity in spermatozoa, causing impaired mitochondrial ATP synthesis and disrupted energy metabolism . Excessive ROS also causes lipid peroxide (LP), damaging the polyunsaturated fatty acids (PUFA) in spermatozoa , and resulting in mitochondrial electron transport dysfunction and disruption of MMP.
Apart from affecting ATP production, ROS affect the molecular components in spermatozoa. Spermatozoa carry a variety of macromolecules, among which proteins and lipids are especially susceptible to ROS (Chianese and Pierantoni, 2021). ROS regulates the sperm proteins, especially those within the nucleus. ROS can oxidize nuclear proteins in sperm, particularly protamines, which are essential for compacting DNA during spermatogenesis. Oxidative modifications, such as carbonylation or disulfide bond disruption, can impair chromatin packaging and compromise DNA integrity, potentially affecting fertilization and embryo development [79,80]. ROS also increases sperm histone methylation and impairs histone acetylation, resulting in double-stranded DNA breaks, reduced sperm quality, and epigenetic dysregulation . Additionally, some enzymes involved in ATP production and ion channel regulation, as well as sperm proteins modified by tyrosine nitration, exhibit physiological or pathological effects that are closely linked to the levels of ROS generated . In spermatozoa, disulfide bonds exist between cysteine residues of protamine, which contribute to the stability of chromatin. Moderate levels of mitochondrial ROS can facilitate the formation of disulfide bonds, which ensure chromatin stability, protecting DNA from damage, and safeguarding mitochondria against proteolytic hydrolysis . Subsequently, ROS can activate adenylate cyclase and induce intracellular cyclic adenosine monophosphate (cAMP) production, which in turn activates protein kinase A (PKA), extracellular signal-regulated kinase (MEK)-like proteins, threonine-glutamic acid-tyrosine, and fibronectin to enable spermatozoa to achieve final capacitation [84,85,86]. The capacitated spermatozoa undergo phosphorylation of tyrosine proteins, calcium influx, leading to an increase in intracellular cAMP and PKA levels. This triggers the release of proteolytic enzymes, which enable sperm to penetrate and fuse with the egg . Thus, it can be seen that ROS is involved in sperm maturation, capacitation and acrosome reaction, and plays an important role in these processes.
Excessive mitochondrial ROS can also cause epigenetic alterations, telomere shortening, Y chromosome microdeletions, and the activation of apoptotic pathways [87,88]. Due to the limited DNA repair capacity in sperm cells, DNA damage often remains unrepaired, leading to increased phosphorylation and activation of p53, which in turn triggers the mitochondrial-dependent apoptotic pathway . This process also induces the opening of a permeability transition pore, the extrusion of cytochrome c and the activation of a caspase cascade, ultimately resulting in apoptosis-like phenomena . The resulting apoptosis helps eliminate structurally or functionally compromised sperm, thereby preserving overall semen quality. While apoptosis is essential for maintaining sperm homeostasis, excessive activation can contribute to subfertility by reducing the viable sperm population . 8-hydroxy-2′-deoxyguanosine (8-OHdG) is a key indicator reflecting DNA OS damage . 8-OHdG induced incomplete base pairing, which disrupts the ribose-phosphate backbone of DNA and contributes to DNA fragmentation . Comparing the normal fertile men with those experiencing infertility revealed that higher levels of 8-OHdG were found in the infertile group. This elevation suggests that ROS induces oxidative damage to sperm DNA, potentially contributing to male infertility [93,94]. And also, telomeric DNA, composed of guanine-rich repeat sequences, is highly vulnerable to oxidative radical attacks, which can impede the repair and elongation processes of telomeres [95,96]. Additionally, the ROS-induced DNA mutations may transfer and exert effects on the offspring [97,98]. The effects of mitochondrial ROS on spermatozoa have been illustrated in Figure 1.
4. The Importance of mtDNA Stability in Mitochondrial Function Maintenance
4.1. The Characteristics of Sperm mtDNA
Sperm mtDNA possesses distinct genetic characteristics, notably its maternal inheritance pattern . Both humans and animals have evolved a series of mechanisms to eliminate sperm mitochondria and mtDNA. As a result, the number of mitochondria and mtDNA in spermatozoa is progressively reduced during spermatogenesis and fertilization . Consequently, mtDNA primarily functions in the metabolic processes of spermatozoa before fertilization. And also, the susceptibility to mutation of mtDNA is notable . Sperm mtDNA has a smaller molecular weight and lacks introns; they are in a constant state of synthesis throughout the entire cell cycle, making it less stable and more susceptible to interference from various factors . What is more, mtDNA replication enzymes have poor proofreading capabilities, leading to a higher probability of errors during the duplication process. And the absence of effective repair systems contributes significantly to an escalated risk of mtDNA mutations, which stands at multiple-fold higher frequencies compared to those observed in nuclear DNA . In addition, mtDNA lacks protection from histones and DNA-binding proteins, leaving it directly exposed to the mitochondrial matrix. This makes it susceptible to damage from ROS, leading to a decline of normal physiological functions such as sperm motility, as we described above .
4.2. Effects of mtDNA Mutation on Sperm Function
The mtDNA abnormalities associated with male infertility have gained much attention . MtDNA mutations in spermatozoa can be generally categorized into point mutations, insert/deletion mutations and mtDNA copy number variation, all of which are known to adversely affect sperm motility and morphology, impairing sperm function and leading to male infertility . There is increasing evidence suggesting that single-nucleotide polymorphisms (SNPs) in mtDNA may significantly influence male fertility . For instance, high levels of A3243G point mutation in mtDNA are strongly linked to decreased sperm motility . And also, some SNPs in mtDNA genes, such as MT-CYB, MT-CO3 and MT-ATP6, have been proven to be associated with male fertility [105,109].
In addition, there are large deletion mutations in mtDNA, which mostly occur in gene fragments with tandem repeat sequences on both sides . The presence of this specific mtDNA deletion in spermatozoa is linked with various male infertility conditions . A higher frequency of 4977, 7599 and 7491 bp mtDNA deletion mutations is associated with asthenozoospermia, asthenoteratospermia and oligasthenospermia in males [111,112,113]. The 4977 bp deletion in mtDNA is among the most extensively researched genetic mutations in spermatozoa. The spermatozoa harboring this mutation show significantly lower motility and may also display abnormalities in morphology, such as head deformities or tail twisting, which render them incapable of successfully fertilizing the eggs . The frequency of spermatozoa with 4977 bp deletion mutation in mtDNA was found to be negatively correlated with the fertilization rate during IVF , and the occurrence rate of the 4977bp deletion mutation in mtDNA was higher in males with asthenospermia, oligospermia, and primary infertility . For example, the rate of 4977 bp deletions mutation in patients with asthenospermia was 85.93%; whereas the deletion rate was 14% in normal controls, with a significant difference between the two groups . What is more, a comparative study of 60 infertile men and 60 healthy controls has shown significant associations between the 4977 and 7599 bp deletion mutations and male infertility . These findings indicated that the 4977 bp deletion mutation in sperm mtDNA is significantly higher in male infertility patients compared with the normal group. This prompts the consideration that the 4977 bp deletion mutation has the potential to become a biomarker of sperm fertility. What is more, the deletion mutation resulted in functional defects of respiratory chain proteins encoded by sperm mtDNA, and then impaired ATP generation. This, in turn, can increase the production of mitochondrial ROS or free radicals, ultimately damaging sperm mtDNA .
MtDNA copy number is recognized as a sensitive biomarker of mitochondrial integrity and function. In spermatozoa, abnormal mtDNA has been implicated in defective spermatogenesis, heightened mitophagy, and the accumulation of dysfunctional mitochondria. Elevated sperm mtDNA copy number has also been correlated with impaired semen quality, including reduced concentration, total count, viability, and abnormal morphology, and is significantly associated with an increased risk of clinical infertility, Nguyen et al. have summarized the quantitative variations in mitochondrial content and mtDNA copy number across different semen quality categories in various species [7,118]. Importantly, mtDNA copy number in sperm exhibits strong diagnostic potential for predicting male infertility across clinical evaluations . However, investigations into the specific types of sperm mtDNA mutations and their associations with sperm functionality remain limited. Further studies are needed to elucidate the mechanisms through which mtDNA mutations regulate sperm fertilization.
5. Mitochondria-Targeted Antioxidant Therapeutic Approaches That Improve Sperm Quality
Given that mitochondria are the origin of ROS, antioxidant-based strategies targeting mitochondria have emerged as promising interventions to restore redox balance and improve sperm function. The common mitochondria-target antioxidants include ubiquinone (MitoQ), melatonin, quercetin, MitoTEMPO, etc. Mitochondrial-target antioxidants block oxidative damage through covalent attachment of lipophilic cations such as lipophilic cation triphenylphosphonium (TPP+) in mitochondria . The TPP+ of mitochondrial-target antioxidants can easily penetrate the mitochondrial membrane’s lipid bilayer and deliver the antioxidant to the mitochondrial matrix . Once inside the mitochondria, these compounds can accumulate in large amounts, exhibiting significant mitochondrial antioxidant effects .
MitoQ is a widely studied mitochondrial-target antioxidant . It accepts electrons from mitochondrial respiratory chain protein complex I or II, and reduces to ubiquinol, which then transfers these electrons to mitochondrial complex III . Ubiquinol also acts as an antioxidant by providing hydrogen atoms to lipid peroxyl radicals, thereby preventing LP . Supplementation of MitoQ in human spermatozoa resulted in improved sperm function, attenuated DNA damage, and concurrently induced the upregulation of antioxidant gene expression, thereby conferring supplementary benefits for sperm cryopreservation . Studies have been conducted to assess the comparative effects of MitoQ and a cytoplasmic antioxidant RESV on buffalo semen subjected to cryopreservation. Results showed that MitoQ significantly improved sperm motility and viability prior to freezing compared to RESV (p ≤ 0.05), and MitoQ-treated sperm displayed heightened acrosome integrity and a greater proportion of viable spermatozoa post-freezing and thawing . Furthermore, MitoQ treatment was found to boost the motility of freshly collected fish sperm and decrease LP levels; it may be attributed to its effective reduction of ROS generation in fish sperm following the freeze-thaw process . These results indicate that MitoQ exerts a mitochondrial-targeted antioxidant effect and can mitigate ROS-induced sperm cryodamage in frozen-thawed sperm.
Melatonin is also an antioxidant that targets mitochondria . Melatonin is predominantly synthesized within mitochondria and exerts its principal effects . Melatonin maintains optimal MMP by scavenging ROS, activating uncoupling proteins, and inhibiting 1 methyl-4-phenyl-1,2,3,6- tetrahydropyridine formation . Melatonin can also optimize the distribution of enzymes required for OXPHOS in spermatozoa, enhancing the activity of respiratory chain protein complexes, ultimately boosting the oxygen utilization capacity of cryopreserved spermatozoa upon thawing . In human spermatozoa, melatonin can improve mitochondrial function by reducing mitochondrial OS . And melatonin demonstrates a protective effect against environmental toxicity and cold-induced damage . The addition of melatonin at varying concentrations has been shown to enhance sperm viability and motility, concomitant with a decrease in intracellular ROS levels. However, the optimal melatonin concentration for these beneficial effects varies across different studies [134,135,136]. Moreover, melatonin supplementation in cryopreservation medium for farm animals such as bovine , rams , buffalo , and pigs has produced similar positive outcomes. Thus, melatonin supplementation plays a protective role in safeguarding both human and livestock spermatozoa, preserving their viability, and diminishing abnormal morphology and DNA fragmentation during the cryopreservation process. As a result, it has gained widespread application as an antioxidant additive in sperm cryopreservation [124,141].
Quercetin is a mitochondria-targeted flavone; it can reduce lipid peroxide accumulation by neutralizing harmful free radicals and attaching to transition metal ions. This activity alters the fluidity of the mitochondrial membrane and affects the oxidative proteins within the mitochondrial matrix [142,143]. It also modulates mitochondrial biogenesis, MMP, OXPHOS and ATP production, mitochondrial redox states, and ultimately triggers mitochondria-mediated apoptosis . Quercetin could modulate sperm mitochondrial respiration efficiency in asthenozoospermic men , and decreases ram sperm motility within the first two hours by acidifying the incubation medium, while subsequently stimulates sperm motility during the following three to four hours by maintaining mitochondrial respiration . Quercetin is also beneficial for spermatozoa during cryopreservation due to its antioxidative properties, as the freezing and thawing process typically induces high levels of OS. The antioxidant capability of quercetin has been confirmed across various animal species, demonstrating its efficacy in safeguarding sperm from oxidative stress during the cryopreservation procedure . The addition of quercetin to the freezing solution was found to decrease oxidative damage and thereby enhance the sperm quality after thawing in human , pig , goat , rooster , dog and so on. Despite its benefits, the precise molecular mechanisms behind quercetin’s actions are yet to be fully understood, and there are also studies showing that quercetin has no significant effects on frozen sperm quality. Further research is necessary to elucidate its protective effects in semen cryopreservation.
MitoTEMPO is a ROS scavenger composed of piperidine nitroxide TEMPOL and lipophilic triphenylphosphine (TPP). As a cell-permeable novel antioxidant, its lipophilic properties enable it to rapidly permeate through the lipid bilayer membrane of mitochondria and accumulate in high concentrations within these organelles . Research has demonstrated that MitoTEMPO selectively targets mitochondrial superoxide, thereby enhancing mitochondrial function and the cell’s antioxidant capabilities . Currently, research on MitoTEMPO on sperm quality is predominantly centered on semen cryopreservation. This emphasis is placed because OS has been recognized as a key factor affecting sperm quality, which could lead to the increase of ROS and LP during the cryopreservation process . Research has shown that the addition of MitoTEMPO significantly enhanced the human post-thaw sperm motility, membrane integrity and MMP, and reduced the ROS level in sperm mitochondria, suggesting that MitoTEMPO could serve as an effective cryoprotectant for semen samples, and also, MitoTEMPO could alleviate cryodamage of asthenozoospermic spermatozoa after cryopreservation [155,156,157]. Additionally, research has extended to examine the impact of MitoTEMPO on post-thaw spermatozoa from various animal species, including bulls [158,159], rams [160,161,162,163], roosters [164,165], and tomcats . These studies revealed that supplementation of MitoTEMPO significantly improved sperm quality in comparison to the untreated control group. However, further studies are required regarding MitoTEMPO’s impact on non-cryopreserved semen, as well as its effects on sperm quality in species such as pigs. Moreover, in-depth exploration of the mechanisms by which MitoTEMPO influences sperm quality can enhance our comprehension of sperm antioxidant systems, thereby facilitating the discovery of more efficacious mitochondria-targeted antioxidants to boost sperm quality.
The mitochondrial-target antioxidants have gained considerable attention, highlighting the critical role of mitochondrial antioxidant activity in maintaining sperm function. However, their effects have yielded inconsistent results, which may be explained by a biphasic, concentration-dependent response of sperm cells to the antioxidants. In addition to the previously introduced mitochondria-targeted antioxidants, an increasing number of compounds with mitochondrial antioxidant properties have also been investigated for their roles in sperm function. It can be deduced that mitochondria-targeted antioxidants are potentially useful in ameliorating OS in spermatozoa, particularly demonstrating favorable outcomes in mitigating mitochondrial OS induced by cryopreservation. Future research and endeavors should be taken to examine the effectiveness of mitochondrial antioxidants, elucidating the underlying mechanisms of their action, and developing additional potent mitochondria-specific antioxidants. Table 2 provides a summary of studies that have investigated the use of mitochondria-targeted antioxidants in spermatozoa.
6. Conclusions
Mitochondria play a central role in regulating sperm energy metabolism, redox balance, and overall function. The role of mitochondria in sperm energy metabolism varies across species, and proteins within sperm are among the key molecules influencing mitochondrial function. While physiological levels of mitochondrial ROS are essential for normal sperm processes such as capacitation and acrosome reaction, excessive ROS production leads to oxidative stress, mitochondrial dysfunction, and damage to nuclear and mitochondrial DNA. Mitochondria-targeted antioxidants have shown promising results in preserving sperm motility, viability, and fertilization potential by restoring redox balance and protecting mitochondrial integrity. However, more research is needed to optimize antioxidant type, dosage, and delivery methods across species. Future studies should also focus on refining diagnostic tools for assessing mitochondrial health in sperm and evaluating antioxidant strategies in clinical and agricultural settings. Such approaches may provide effective interventions for improving male fertility and reproductive efficiency.
Author Contributions
Conceptualization, Z.X.; methodology, Z.X.; validation, Z.X., X.L. and F.W.; investigation, Z.X., K.Z., Q.Y., Y.L., X.L. and F.W.; writing—original draft preparation, Z.X., K.Z., Q.Y. and Y.L.; writing—review and editing, C.Z., T.R., F.W., N.G. and X.L.; supervision, Z.X., X.L. and F.W.; project administration, Z.X., X.L. and F.W.; funding acquisition, Z.X. All authors have read and agreed to the published version of the manuscript.
Funding
This work was supported by grants from the National Natural Science Foundation of China (32302742), the Doctoral Scientific Research Foundation of Henan University of Science and Technology (13480096), and the Science and Technology Key Project of Henan Provincial Education Department in China (Project No. 24A230002).
Data Availability Statement
There are no new data associated with this article.
Acknowledgments
Figures for this review were created with BioRender.com.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
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Figure 1. Effects of mitochondrial ROS on spermatozoa. Note: Mitochondria are the major source of ROS in sperm. Excessive ROS can damage sperm MMP (Δψ), as well as sperm nuclear DNA and mtDNA damage. At the same time, ROS also plays an important role in sperm maturation, capacitation and acrosome reaction.
Figure 1. Effects of mitochondrial ROS on spermatozoa. Note: Mitochondria are the major source of ROS in sperm. Excessive ROS can damage sperm MMP (Δψ), as well as sperm nuclear DNA and mtDNA damage. At the same time, ROS also plays an important role in sperm maturation, capacitation and acrosome reaction.
Table 1. Sperm proteins are involved in mitochondrial energy metabolism.
Table 1. Sperm proteins are involved in mitochondrial energy metabolism.
| Protein | Species | Function | Ref. |
--- --- |
| Ant2 | Drosophila | Disrupting mitochondrial morphogenesis and spermatid maturation | |
| MORN2 | Mouse | MORN2 functions in mitochondrial sheath formation and regulates mitochondrial respiratory activity | |
| ClpP/ClpX | Mouse | ClpP/ClpX deficiency disrupted energy supply during meiosis and impaired zygotene-pachytene transformation in male germ cells | |
| Rsrc1-161aa | Mouse | Regulating mitochondrial ribosome assembly and translation, and then affecting male fertility. | |
| SLC22A14 | Mouse | Shortage of SLC22A14 could decrease the energy metabolism of spermatozoa | |
| PRSS55 | Mouse | Deficiency of PRSS55 damages the mitochondrial structure, which leads to lower ETC molecules and ATP contents | |
| SLC7A11 antiporter | Stallion | Transporting cystine into the cell in exchange for glutamate | |
| DJ-1 | Rat | Interacting mitochondrial complex I subunits NDUFS3 and NDUFA4 affect mitochondrial energy metabolism | |
| GSK3 | Goat | Phosphorylation of GSK3 regulated sperm motility and acrosome reaction by glycolysis and OXPHOS | |
| TPPP2 | Mouse | Deficiency of TPPP2 altered the structure and function of sperm mitochondria and caused decreased sperm count, motility and ATP content | |
| AMPK | Mouse/Goat | Increasing AMPK activity maintains MMP and ATP levels in spermatozoa | [61,62] |
| LDHC | Mouse | Maintenance of energy metabolism in sperm | |
| GLUT8 | Mouse | Lack of GLUT8 leads to impaired mitochondrial function and sperm motility in spermatozoa | |
Note: Ant2 (Adenine nucleotide translocase 2), MORN2 (nexus-motif protein 2), ClpP (Caseinolytic protease proteolytic subunit), ClpX (caseinolytic protease X), PRSS55(serine protease 55), AMPK (AMP-activated protein kinase), MMP (mitochondrial membrane potential), GSK3 (Glycogen synthase kinase-3), TPPP2 (tubulin polymerization promoting protein family member 2), LDHC (lactate dehydrogenase C).
Table 2. Applications of mitochondria-targeted antioxidants on sperm quality.
Table 2. Applications of mitochondria-targeted antioxidants on sperm quality.
| Antioxidants | Species | Functions | Ref. |
--- --- |
| MitoQ | Rooster | Improve the sperm quality in cryopreserved and cooled rooster spermatozoa | [167,168] |
| | Yak | Increase the sperm quality and antioxidant capacity in low-temperature yak semen preservation | |
| | Human | Enhance functional sperm parameters, reduce DNA fragmentation, and upregulate the expression of antioxidant-related genes in human sperm cryopreservation medium | [125,170] |
| | Stallion | Improve the frozen-thawed stallion sperm motility and kinetics, while detrimental to the motility and viability at higher concentrations | |
| | Buck | Improve the motility, viability, membrane functionality, mitochondrial activity, acrosome integrity, and reduce OS during the cryopreservation process of buck sperm | |
| | Bull | Do not affect the Simmental bull sperm cryo-resistance | |
| | | Enhance post-thaw buffalo bull sperm function and cryo-tolerance | |
| | yellow catfish | Reduce ROS production and LP, and enhance post-thaw sperm viability of yellow catfish | |
| Melatonin | Mouse | Alleviate post-weaning NaAsO2 exposure-mediated decline of sperm counts in adult mice | |
| | Human | Improve human sperm quality and overall male fertility | |
| | | Counteract intracellular ROS and MDA, increase post-thaw viability, motility and membrane integrity of human spermatozoa | [134,135,136] |
| | Ram | Improve OXPHOS of frozen-thawed ram sperm | |
| | | Improve progressive motility and reduce DNA fragmentation of ram spermatozoa following cryopreservation | |
| | Bull | Improve sperm function parameters in the fresh and post-thaw mithun bull semen | |
| | | Improve sperm viability and motility of swamp buffalo during cryopreservation | |
| | Boar | Increase the number of live sperms with intact acrosome, but did not affect the other spermatic quality of boar semen | |
| Quercetin | Human | Improve the human sperm progressive motility, MMP, and decrease ROS levels, DNA fragmentation | |
| | | Stimulate the respiratory active state and enhance mitochondrial function in asthenozoospermic samples | |
| | Boar | Maintain sperm motility and decrease superoxide in the liquid preservation of boar semen, and improve the post-thaw sperm quality and reduce the OS of boar sperm | [149,174] |
| | Canine | Improve the post-thaw sperm quality and reduce the oxidative damage in the dog | |
| | Goat | Decrease the goat sperm MDA and ROS levels, enhance motility, viability, membrane integrity, and mitochondria activity | |
| | Rooster | Improve post-thawed sperm quality and fertility of roosters | |
| | Ram | Inhibit the ram spermatozoa motility during the first 2 h of incubation, but subsequently stimulate motility over the following 3–4 h by sustaining mitochondrial respiration | |
| MitoTEMPO | Human | Improve post-thaw human sperm motility, viability, membrane integrity, MMP and antioxidant enzymes activities decrease DNA fragmentation index, ROS and MDA level | [155,156,157] |
| | Bull | Improve post-thaw buffalo sperm progressive motility, viability, acrosomal and membrane integrity, cholesterol to phospholipids ratio, MMP, antioxidant enzymes activities and fertility | [158,159] |
| | Ram | Improve the post-thaw ram sperm motility, kinematics, viability, membrane functionality, MMP, antioxidant capacity, glucose transporter and fertility, decreasing MDA | [160,161,162,163] |
| | Rooster | Improve the post-thaw rooster sperm motility kinematics, membrane functionality, MMP, viability and fertility, decrease lipid peroxidation, late apoptotic-like changes, DNA fragmentation and hydrogen peroxide level | [164,165] |
| | Tomcat | Improve the acrosome integrity of frozen-thawed epididymal spermatozoa in tomcats | |
| MitoPBN | Ram | Mitigate OS and improve mitochondrial function in cryopreserved ram sperm | |
| Crocin | Human | Improve human sperm progressive motility, MMP, and decrease ROS levels, DNA fragmentation, enhance the asthenozoospermic sperm motility and reduce OS | [148,176] |
| | Bull | Improve the sperm post-thaw quality and fertility of buffalo bull and Yanbian yellow cattle sperm | [177,178,179] |
| | Rooster | Improve the post-thawing rooster sperm viability, membrane functionality, MMP and fertility, reduce sperm LP and apoptosis | |
| Naringenin | Bull | Improve buffalo bull post-thaw quality, fertility-associated gene expression and fertilization potential | |
| | Boar | Maintenance of sperm motility and decrease of superoxide in the liquid preservation of boar semen | |
| | Rooster | Improve the post-thawing rooster sperm viability, membrane functionality, MMP and fertility, reduce sperm LP and apoptosis | |
| Metformin | Equine | Improve total motility and reduce LP of thawed equine spermatozoa | |
| | Sheep | Improve the post-thawing sheep sperm quality and reduce sperm OS | |
| | S. prenanti | Improve the quality and fertility of S. prenanti sperm by increasing the ATP content | |
| | Human | Inhibit human sperm motility, PKA and protein tyrosine phosphorylation pathways, do not affect sperm viability, MMP, mitochondrial superoxide anion generation | |
| | Boar | The effects on boar sperm quality remain controversial | [186,187] |
| | Canine | Improve the frozen-thawed canine sperm motility, mitochondrial activity, NAD+ content, and reduce the OS level | |
| Rosiglitazone | Equine | Increase the ROS level of thawed equine spermatozoa. Increase motility, MMP, ATP content, glucose uptake capacity, and decrease ROS level of equine spermatozoa in vitro. | [182,189] |
| | Bull | Maintain MMP, total and progressive motility of bovine sperm | |
| | Ram | Enhance motility, membrane and acrosome integrity, MMP, and decrease ROS level in frozen-thawed ram spermatozoa | |
| | Boar | Enhance boar sperm motility, membrane and acrosome integrity, MMP, ATP production and reduce ROS during storage at 17 °C | |
| Pyrroloquinoline quinone | Boar | Enhance post-thaw boar sperm motility, viability, acrosome integrity, MMP, ATP levels, mtDNA stability, and decrease MDA, ROS levels and DNA damage | |
| | Ram | Improve sperm motility, MMP, membrane and acrosome integrity, ATP levels and fertility, decrease sperm MDA and ROS levels of ram sperm at 4 °C | |
| | Bull | Reduce mitochondrial ROS in fresh bull semen, improve sperm motility and reduce ROS in frozen-thawed bull sperm | |
| Astaxanthin | Rooster | Improve rooster freeze-thaw sperm viability, motility, plasma membrane and acrosome integrity, mitochondrial activities, ATP level and antioxidant ability | [196,197] |
| | Ram | Increase sperm motility, viability, plasma membrane and acrosome integrity, MMP and antioxidant ability of Hu ram spermatozoa at 4 °C | |
| | Canine | Increase sperm motility, viability, plasma membrane and acrosome integrity, and mitochondria activity of post-thaw dog spermatozoa | |
| Ergothioneine | Boar | Improve sperm motility parameters, MMP, ATP, antioxidant capacity, plasma membrane and acrosome integrity in in vitro liquid preservation of boar spermatozoa | |
| | Bull | Improve sperm motility and reduce ROS level in frozen-thawed bull sperm | |
| | Canine | Increase the total motility, acrosomal integrity and reduce the abnormal morphology and ROS production of cryopreserved canine spermatozoa | |
| | Rooster | Increase the total motility, membrane integrity and mitochondria activity, reduce the apoptotic and dead sperm of post-thawed rooster spermatozoa | |
| Losartan | Rooster | Improve the total and progressive motility, mitochondrial activity, membrane integrity, antioxidant levels, and reduce LP and apoptosis of post-thawed rooster spermatozoa | |
| Lagenaria siceraria seed oil | Rabbit | Improve sperm motility, viability, membrane integrity and antioxidant capacity during 72-h chilled storage of rabbit spermatozoa | |
| Urolithin A | Bull | Improve bovine sperm motility quality and reduce OS | |
| AntiOxBEN2 | Bull | Enhance the bovine sperm quality, fertility, and reduce ROS | |
| BGP-15 | Human | Improve sperm motility, mucous penetration and MMP, reduce sperm DNA oxidative damage of post-thawed human spermatozoa | |
| L-Carnitine | Human | Improve sperm motility, viability, and reduce sperm DNA oxidative damage of post-thawed human spermatozoa | |
| Cysteine | Ram | Improve the rates of ALH, membrane integrity, and mitochondrial activity of the post-thawed ram sperm | |
| Isoespintanol | Canine | Increase the acrosomal integrity and reduce the abnormal morphology and ROS production of cryopreserved canine spermatozoa | |
| Vitamin C | Bull | Reduce mitochondrial ROS in fresh bull semen, improve sperm motility and reduce ROS in frozen-thawed bull sperm | |
| EGCG@PDA NPs | Boar | Maintain sperm motility, acrosome integrity, MMP, extending semen storage time from 3 days to 10 days, reducing ROS and sperm apoptosis of boar sperm during storage at 4 °C | |
| |
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Xu Z, Yan Q, Zhang K, Lei Y, Zhou C, Ren T, Gao N, Wen F, Li X. Mitochondrial Regulation of Spermatozoa Function: Metabolism, Oxidative Stress and Therapeutic Insights. Animals. 2025; 15(15):2246.
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Helmholtz decomposition doubt
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On the Wiki page for Helmholtz decomposition, it is said that certain differentiable vector fields can be resolved into the sum of an irrotational (curl-free) vector field and a solenoidal (divergence-free) vector field.
And than it is stated as:
F=−∇Φ+∇×A F=−∇Φ+∇×A
I am just starting my fluid dynamics class, but there we defined that we can decompose a velocity vector field as:
u=∇φ+∇×ψ(2)(2)u=∇φ+∇×ψ
where φ φ is the velocity potential and ψ ψ is the stream function (a kind of vector potential).
Now I am confused on why is there a minus sign in the general definition of the Helmholtz decomposition.
Is it only a matter of convention and, since in fluid dynamics we define a velocity potential positive as u=∇φ u=∇φ, we end up with the positive sign and the velocity field defined as it it and also because the Helmholtz decomposition is not unique?
Second question
As I understand, the equation (2)(2) is the general vector field definition. So just to be sure, we can describe all kind of velocity fields possible with that kind of equation? Meaning even with rotation and nonpotential?
Addendum
On the Wiki page for the Stream function under "In terms of vector potential and stream surfaces"
they define a similar function:
u=∇ψ×∇ϕ(3)(3)u=∇ψ×∇ϕ
Is that a typo of the Helmholtz decomposition or are they defining something totally different? Are the two equations (2)(2) and (3)(3) connected somehow?
fluid-dynamics
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edited Oct 3, 2024 at 21:22
User198User198
asked Oct 3, 2024 at 20:49
User198User198
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Yes you do what you want with a sign or even a constant. Just call ϕ=−Φ ϕ=−Φ and you are good to go. 2. I believe you need a differentiable (you are taking derivative) field that vanishes sufficiently fast at infinity (the proof I think requires the existence of fourier transform). These 2 conditions are always verified for physical velocity fields. Except maybe for shock waves. That are in any case a little bit pathological Syrocco –Syrocco 2024-10-03 20:56:36 +00:00 Commented Oct 3, 2024 at 20:56
@Syrocco yes, Helmholtz decomposition as written on even the Wiki page requires that the field decay fast enough at infinity. There is a section on that in the Wiki page. In practice, it is easier to just extract the simple behaviour at infinity out for explicit handling.naturallyInconsistent –naturallyInconsistent 2024-10-03 23:37:13 +00:00 Commented Oct 3, 2024 at 23:37
Fluid dynamics's stream function is entirely different from Helmholtz decomposition. They are doing complex analysis, and it holds for a totally different and bigger domain, at the cost of giving up treating viscosity.naturallyInconsistent –naturallyInconsistent 2024-10-03 23:39:37 +00:00 Commented Oct 3, 2024 at 23:39
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Now I am confused on why is there a minus sign in the general definition of the Helmholtz decomposition. Is it only a matter of convention[?]
Yes.
On the Wiki page for the Stream function under "In terms of vector potential and stream surfaces"
they define a similar function:
𝐮=∇𝜓×∇𝜙
Is that a typo of the Helmholtz decomposition or are they defining something totally different?
Yes, that is something different.
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answered Oct 4, 2024 at 0:42
Math Keeps Me BusyMath Keeps Me Busy
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7210 | https://www.scribd.com/presentation/679753095/math-circular-mod-27-ppt | Math Circular Mod 27 | PDF
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Math Circular Mod 27
This document discusses permutation of identical objects and circular permutation. It provides definitions, examples, and step-by-step solutions to problems involving permutation of objects …
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Permutation of Identical Objects and Circular Permutation
Mathematics Quarter 3 – Module 27:
10 - Faraday (S.Y 2022 - 2023)
MEIKO B. DESANTORES REAN MAE C. MANUAL REA MAE VILLACENCIO SHAYNE F. MONTANTE DON ALE JANDRO BABARIN
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DEFINITION OF TERMS
•
Permutation:a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged
•
Distinguishable Permutations:are permutations that can be distinguished from one another.
•
Circular Permutation: The number of ways to arrange n distinct objects along a fixed (i.e., cannot be picked up out of the plane and turned over)
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Distinguishable Permutations
are permutations that can be distinguished from one another. In the case of a number of things where each is different from the other, such as the letter s in the word “BAGUIO”, there is no difference between the number of permutations and the number of distinguishable permutations. But if the original set of things has repetition, then the number of distinguishable permutations of
objects of which n1 are alike and one of a kind, n2 are alike and one of a kind, …, nk are alike and one of a kind, the number of distinguishable permutations is:
Permutation of Identical Objects:
n!
n
1,
n
2,
n
3
,,, n
k
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Let’s analyze this.
N
O N E There are four letters in the word “NONE” and there are two letter “Ns”.
N
O N
E If you interchange the two letter “Ns”, you form the same word. The seemingly two different arrangements N O
N
E a r e a c t u a l l y o n e a n d t h e s a m e a r r a n g e m e n t.Therefore, these are considered one arrangement only
.
Let’s solve the problem How many ways can the letters in the word
“NONE”
be arranged?S t e p 1
4
P
4
= 4!S i n c e t h e r e a r e f o u r l e t t e r s, g e t t h e permutation of the four letters.Note:When you get the permutation of the four letters, you will be counting some words twice.
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S t e p 2 2!4! T h e p e r m u t a t i o n o f t h e f o u r l e t t e r s w i l l b e divided by the number of ways to arrange the repeated letter.Note:The number of ways to arrange the two letter “Ns” is
2
P
2
= 2!S t e p 3 4 = 2 4 =
12
Simplify 2! 2 Therefore, there are
12 ways
to arrange the letters of the word “NONE”. T o verify the answer, let us list down all the possible arrangements of the letters of the word “NONE” and remove the repeated words
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7211 | http://www.bad.org.uk/pils/isotretinoin | Visit The BCD Education Hub
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Patient Information Leaflets
Isotretinoin
Patient Information Leaflets
Isotretinoin
Isotretinoin-patient-guide
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What are the aims of this patient guide?
This patient guide is to help you understand more about isotretinoin so you can consider whether it is the right treatment for you. It explains the benefits of treatment, possible side effects and why isotretinoin must not be taken in pregnancy.
It discusses possible mental health and sexual function effects which were raised in the April 2023 report by the Commission on Human Medicines’ Isotretinoin Expert Working Group.
There is a separate information leaflet about isotretinoin especially written for young people – Medicines for Children ‘Oral isotretinoin guide for young people. We recommend you look at this if you are under 18 and considering isotretinoin treatment.
What is isotretinoin and how does it work?
Isotretinoin is a retinoid drug which is part of the same family as Vitamin A. It may be known by brand names that include Roaccutane, Accutane (USA) and Reticutan. It reduces the production of sebum – an oily substance made by the skin. It also reduces the production of keratin – the outer layer of skin, which can block pores.
What is isotretinoin used for?
Isotretinoin is licensed (approved by drug regulators in the UK) to treat severe acne in people over 12 years when other treatments have not worked or if there is a risk of permanent scarring. Dermatology healthcare professionals may occasionally recommend isotretinoin outside of its licensed use. They should tell you it is being used in this way and discuss with you whether you still want to take it.
How can I decide whether I want to take isotretinoin?
You should read this patient guide carefully and ask any questions you have before deciding whether you want to take isotretinoin. You should give your dermatology healthcare professional all the information they need about treatments you have already tried, how the acne affects you, possible plans for pregnancy, other medical problems and any history of mental health problems. This will help your dermatology healthcare professional decide with you whether isotretinoin is the best treatment for you.
What should I expect from isotretinoin?
Most people find isotretinoin helps their acne and, for many, can clear their skin. There is a possibility that your acne might get worse for a few days or weeks before it starts to get better. Most people find their acne stays better for several years after they finish isotretinoin.
How is my dose and treatment length decided?
Your dermatology healthcare professional will work out your initial dose according to your weight and after discussion about possible side effects. Your dose will be reviewed at each follow up appointment. Treatment is usually continued until your acne has been clear for a few weeks. Some specialists may treat to a fixed total amount (generally 120-150 mg/kg). A course of treatment is generally around 24 weeks (6 months) but may be longer if low doses are used.
How should I take isotretinoin?
Isotretinoin is usually taken once a day. It works best if it is taken with a main meal or a drink which contains some fat (for example, milk, butter or oil) because this helps it to be taken into the body.
Swallow the whole capsules. Do not crush or split them.
Keep the capsules in a cool (below 25°C), dark place and away from children.
If you have swallowing difficulties, your specialist can advise on the best way to take it.
What important things do I need to know about isotretinoin?
Isotretinoin must not be taken in pregnancy or if you are trying to get pregnant as this could cause serious harm to the baby (see section below for people with child-bearing potential).
Isotretinoin capsules must never be shared with other people.
You must not donate blood whilst taking isotretinoin and for at least 1 month after stopping the medication. This is because the blood may be given to a pregnant person.
Isotretinoin and alcohol can affect the liver so you should not drink much alcohol while you are taking isotretinoin. Do not drink more than the recommended maximum 14 units per week of alcohol – see NHS website for more details.
Isotretinoin capsules contain highly processed soya. Please discuss with your dermatology specialist if you are allergic to peanut or soya and see the British Association of Dermatologists position statement on isotretinoin and soya allergy (www.bad.org.uk/pils/isotretinoin-and-soya-allergy).
It is best to avoid treatments such as waxing, epilation, or dermabrasion, as well as tattoos and piercings during treatment and for up to 6 months afterwards. They could cause scarring, skin irritation, or rarely, changes in the colour of your skin.
People whose jobs require them to see at night (such as drivers, airline pilots, people in the military and those who operate heavy machinery) are advised to have a discussion with their employer before starting isotretinoin. This is because it can reduce night vision.
Can I take other medications at the same time as isotretinoin?
Most medicines are safe to take during isotretinoin treatment. However, some medicines may not be. Medications to avoid while taking isotretinoin include:
Tetracycline antibiotics (includes lymecycline and doxycycline).
Vitamin A supplements (including multi-vitamins containing vitamin A).
Some acne creams, due to a higher risk of irritation.
This is not a complete list, and it is important that you always inform your doctor and pharmacist that you are taking isotretinoin before taking any new prescriptions or over the counter medications and read the information leaflet in the packet.
Side Effects
Oral isotretinoin may cause serious side effects. While these are rare, it is important that you understand the potential risks in order to make an informed decision about whether to take it.
What are the common side effects of isotretinoin?
Dry lips (sometimes chapped or split).
Dry skin, which may be more delicate than usual.
Dry eyes.
Increased skin sensitivity to the sun with more chance of getting sunburnt.
Back pain, muscles and joints aches, especially after exercise.
Changes in the blood tests which are monitored by your dermatology team. These include mildly raised liver enzymes and changes in blood fat levels, which in healthy people are not usually serious.
These side effects usually resolve a few weeks after stopping isotretinoin.
What can I do to help?
Use a non-greasy moisturiser designed for acne-prone skin, and a lip balm.
To prevent skin irritation, you should avoid using exfoliating or anti-acne products.
Use lubricating eye drops (non-prescription) for dry eyes, if needed. If contact lenses are uncomfortable you may need to wear glasses instead.
Avoid too much sun. If you go out in the sun, protect your skin with clothing and wear a hat and sunglasses. Use sunscreen with a sun protection factor (SPF) of 30 or above. You must not use a sun lamp or sun beds.
If you get muscle or joint pains, you are advised to cut down on intensive exercise and physical activity to avoid making it worse.
Paracetamol or non-steroidal anti-inflammatory drugs (for example, ibuprofen) may be taken for muscle/joint pains or headaches.
What other possible side effects can occur?
Serious immediate allergic reaction with sudden rash, swelling of the lips, mouth or throat, and difficulty breathing or swallowing. Call 999 and take an antihistamine straight away, if available. Do not take any more isotretinoin.
Sore or inflamed throat and nose and possible nose bleeds.
Hair loss. This is usually only mild and temporary. Your hair should return to normal after the treatment ends.
Vision problems, especially decreased night vision. If you develop difficulties seeing in dark conditions, you should avoid driving and/or operating heavy machinery. In rare instances, vision problems and dry eyes have persisted after the end of isotretinoin treatment.
Bowel (gut) problems. This may lead to feeling sick (nausea), being sick (vomiting) and/or diarrhoea (which might contain blood).
Feeling weak or dizzy; having more frequent infections or getting bruising/bleeding.
Inflammation of the liver. This can show as yellow skin or eyes and feeling tired.
Raised pressure in the brain. This causes a lasting headache, along with feeling sick (nausea), being sick (vomiting) and a change in your eyesight including blurred vision.
Please read the information leaflet in the packet for more details of possible side effects.
If you are concerned that you have developed any serious side effects, you should stop taking isotretinoin straight away and contact your GP and dermatology team.
Isotretinoin and potential mental health problems (psychiatric side effects)
Isotretinoin has been associated with mental health problems in some people. The current published research does not give a clear answer as to whether isotretinoin is the cause of these mental health problems in some people.
We know that people with acne are more likely to have anxiety and depression than people without acne. This is because acne may affect self-confidence and make people distressed about the way they look.
Several studies have shown that isotretinoin can improve negative mood changes caused by acne.
However, there are some reported experiences of worsening mood or changes in behaviour, including anxiety, low mood, depression, agitation, aggression towards others, or a loss of contact with reality (psychosis). We do not know how often these may occur.
Extremely rarely, isotretinoin may be associated with people hurting themselves (self-harm) or suicide.
If you have ever had low mood, suicidal thoughts or any other mental health issues, please talk with your dermatology team about this before starting treatment.
If you or your friends and family notice any changes in your feelings or behaviour whilst taking isotretinoin, stop taking it and contact your dermatology team to discuss whether isotretinoin is suitable for you.
If you have thoughts of harming yourself or if there are serious concerns about your mental health, you should stop taking isotretinoin and immediately seek medical help.
For further information, please also see the Royal College of Psychiatrists’ information leaflet on isotretinoin and mental health.
Isotretinoin and sexual function side effects
There are rare reports of people having sexual function problems with isotretinoin. We do not know how often these may occur.
We know that sexual function problems can be related to mental health issues or other medication (such as anti-depressants), which may be taken at the same time as isotretinoin.
There are some individual reports of patients experiencing a lack of interest in sex (low libido), vaginal dryness, difficulty getting/keeping an erection, reduced sensation in the genitals.
Rarely, men have reported breast tissue development (gynaecomastia).
Some people have reported that the problems have continued after stopping treatment.
If you have any sexual function problems, discuss this with your dermatology healthcare professional before taking isotretinoin.
Information only for people of child-bearing potential (people who may be able to get pregnant)
You are considered to have child-bearing potential if you have a uterus and at least one ovary, unless:
You have undergone surgical sterilisation (tubal ligation), confirmed by a healthcare professional.
You are post-menopausal, confirmed by a healthcare professional.
Why must I not get pregnant on isotretinoin?
Isotretinoin may cause serious harm to the development of a baby. It also increases the risk of miscarriage. You should avoid pregnancy during isotretinoin treatment and for 1 month after stopping.
Will isotretinoin affect future pregnancies?
There is no evidence that suggests isotretinoin affects fertility. One month after stopping isotretinoin you may get pregnant with no additional risk to the unborn baby.
What is the Pregnancy Prevention Programme?
The Pregnancy Prevention Programme is a set of rules which doctors, nurses and pharmacists must follow to protect unborn babies against the risks of exposure to isotretinoin:
If you are a person who may be able to get pregnant, you will need to be enrolled in the Pregnancy Prevention Programme (PPP).
Being on the PPP means you will be asked about contraception (birth control) and monitored for possibility of pregnancy. This will be before starting treatment, during treatment and 1 month after stopping.
You will need to discuss contraception before you start isotretinoin. This may be with the GP or nurse within your own surgery or with your local sexual health/contraception clinic before you are seen by the dermatology team.
Your dermatology healthcare professional will need to be certain you are not pregnant when you start treatment. You must have used suitable contraception or not had sex with someone who could make you pregnant for 4 weeks before starting isotretinoin. You will need to have a negative pregnancy test. If you are having periods, you should ideally start isotretinoin on day 2 or 3 of your period.
What is suitable contraception?
The Pregnancy Prevention Programme recommends EITHER:
A ‘highly effective’ form of contraception (failure rates less than 1%). This may be the coil (IUD), intra-uterine system (IUS), or contraceptive implant which has been in place for at least 4 weeks.
OR
Two forms of contraception together. These may be the hormonal contraceptive pill or contraceptive injection plus a barrier method (such as a condom, femidom, vaginal cap).
Please discuss with your healthcare professional if you are unsure about your contraception.
Further guidance on pregnancy testing and contraception for pregnancy prevention during treatment with medicines is available here.
What can interfere with contraception?
Hormonal pills and implants may not work in some situations such as:
If you are starting new medications, including antibiotics or herbal preparations, such as St John’s Wort.
If you have diarrhoea or vomiting.
If you do not take your contraceptive pill every day.
You should check with your GP, pharmacist or family planning clinic when taking any new medications to make sure they will not affect how well the contraception works.
Who does not need contraception?
You do not need contraception if there is expected to be no chance of you getting pregnant during isotretinoin treatment and for 1 month afterwards.
This must be for one of the following reasons:
Being medically unable to get pregnant (no child-bearing potential – see above).
Long-term sexual abstinence (not having sex and not planning to have sex) for the duration of isotretinoin treatment and for 1 month after stopping isotretinoin treatment.
Only having sex/sexual intercourse with a person who has no potential to make you pregnant. Examples include sex with a:
same-sex partner
person who has had a vasectomy with two confirmed tests of being sperm-free
transgender man.
Your need for contraception must be discussed with your dermatology healthcare professional before starting isotretinoin and reviewed at each clinic visit.
How often do I need to have pregnancy tests?
Pregnancy tests are usually done every month during treatment. Prescriptions are only for 30 days and must be collected and dispensed within 7 days.
If you are using a ‘highly effective’ method of contraception your pregnancy tests may be less frequent and your prescription may be for longer. You may still choose to do monthly pregnancy tests at home because no contraception is 100% effective.
If there is expected to be no chance of you getting pregnant (see ‘Who does not need contraception?’ section above), you do not need pregnancy tests unless your situation changes.
What if I have unprotected sex or think I am pregnant during treatment?
If you have unprotected sex whilst taking isotretinoin (or 1 month afterwards), you will need emergency contraception. Emergency contraception may be the ‘morning-after pill’ or having an emergency IUD fitted. Urgently discuss with your doctor, pharmacy or a sexual health clinic. Stop taking oral isotretinoin.
If you miss a period, think you may be pregnant or find you are pregnant you should:
Stop taking isotretinoin immediately.
Seek medical advice as soon as possible.
Inform your dermatology team.
As a female-to-male transgender person, do I need to follow the same instructions?
Female-to-male transgender people who have not had a hysterectomy can still get pregnant if they have sex with a person who can produce sperm. For this reason, female-to-male transgender individuals of child-bearing potential should be entered into the Pregnancy Prevention Programme.
Where can I find out more about isotretinoin?
This patient guide does not provide information on all the benefits and side effects which have been associated with isotretinoin. If you would like further information, or if you have any concerns about your treatment, you should discuss this with your GP, dermatology healthcare professional or pharmacist. You should also read the patient information leaflet that comes with your medicine package (please see below).
Web links to relevant information:
Medicines for Children ‘Oral isotretinoin guide for young people’
Royal College of Psychiatrists’ information leaflet on isotretinoin and mental health
British Association of Dermatologists position statement on isotretinoin and soya allergy www.bad.org.uk/pils/isotretinoin-and-soya-allergy
Acne information:
Further resources:
Guidance on pregnancy prevention and contraception for teratogenic drugs
Isotretinoin expert safety report – a plain-language summary of the recommendations
Roaccutane Soft Capsules – Patient Information Leaflets (PILs)
Jargon Buster:
Please note that the British Association of Dermatologists (BAD) provides web links to additional resources to help people access a range of information about their treatment or skin condition. The views expressed in these external resources may not be shared by the BAD or its members. The BAD has no control of and does not endorse the content of external links.
This patient guide aims to provide accurate information about the subject and is a consensus of the views held by representatives of the British Association of Dermatologists: individual patient circumstances may differ, which might alter both the advice and course of therapy given to you by your doctor.
This patient guide has been assessed for readability by patient representatives
BRITISH ASSOCIATION OF DERMATOLOGISTS PATIENT GUIDE
PRODUCED | OCTOBER 2023
NEXT REVIEW DATE | OCTOBER 2026 |
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Basic Algebra
5
How do I find the a,b, and c of a quadratic equation (y=ax^2+bx+c) given its max point is (-1,-1) and its y-intercept is -2?
Nancy Mitchell
used to be a teacher.
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Author has 3.4K answers and 8M answer views
·
7y
How do I find the a,b, and c of a quadratic equation (y=ax^2+bx+c) given its max point is (-1, -1) and its y-intercept is -2?
y=f(x)=ax2+bx+c⟹ the graph of y is a U-shaped parabola.
That the y-intercept is −2
⟹c=−2 (because the constant term c is the y-intercept of a U-shaped parabola)
⟹f(x)=ax2+bx−2.
That the maximum of y is at the point (−1, −1) means that this point is the vertex, that is, the turning point of the parabola.
The formula of the vertex
\text{(}h\text{,}\space k\text{)}=\left(-\dfrac{b}{2a}\text{,}\space f\left(-\dfrac{b}{2a}\r
How do I find the a,b, and c of a quadratic equation (y=ax^2+bx+c) given its max point is (-1, -1) and its y-intercept is -2?
y=f(x)=ax2+bx+c⟹ the graph of y is a U-shaped parabola.
That the y-intercept is −2
⟹c=−2 (because the constant term c is the y-intercept of a U-shaped parabola)
⟹f(x)=ax2+bx−2.
That the maximum of y is at the point (−1, −1) means that this point is the vertex, that is, the turning point of the parabola.
The formula of the vertex
(h, k)=(−b2a, f(−b2a))
⟹−b2a=−1
⟹b2a=1
⟹b=2a, and f(−1)=−1
⟹−1=a(−1)2+b(−1)−2
⟹1=a−b.
Then a−b=1 and b=2a
⟹a−2a=1
⟹−a=1⟹a=−1
⟹b=2(−1)=−2
⟹your equation is
y=−1x2−2x+−2
⟹y=−x2−2x−2.
That the coefficient of x2 is −1 means the graph of y opens downwards. This makes sense if the maximum of y is at (−1, −1).
Related questions
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In the equation of line y=mx+b, what is y and x as m=slope and b=y-intercept?
In a standard quadratic equation
a
x
2
+
b
x
+
c
=
0
, what is the meaning of
b
?
What’s the a, b, and c of the equation with a y intercept of (0,3) and a minimum of (2,-1)?
How do I find a quadratic equation given 2 points and a maximum (max y coordinate)?
Steven Smith
Former Algebra Teaching Assistant at University of Central Missouri (2017–2018)
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Author has 3.4K answers and 9.1M answer views
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7y
It is possible to set up 3 equations with 3 unknowns. One equation can be achieved by using the fact that (-1,-1) is a maximum, which means the x value of the maximum is -1. We also know that the x value where the maximum occurs for a quadratic is −b2a.
For the other 2 equations, plug in the x and y values that we know (-1,-1) and (0,-2) into the equation y=ax2+bx+c
That will leave 3 unknowns: a, b, and c. You will have 3 equations involving those unknowns, so you can use algebra to get the values of those unknowns.
Doug Dillon
Ph.D. Mathematics
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Author has 12.4K answers and 11.4M answer views
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7y
Immediately
c
=
−
2
. Taking the derivative and setting to
0
,
2
a
(
−
1
)
+
b
=
0
and so
b
=
2
a
. Since
(
−
1
,
−
1
)
is on the parabola,
a
−
2
a
−
2
=
−
1
and so
a
=
−
1
and
b
=
−
2
Assistant
Bot
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1y
To find the coefficients
a
,
b
, and
c
of the quadratic equation
y
=
a
x
2
+
b
x
+
c
given its maximum point and y-intercept, we can use the provided information:
Maximum Point
(
−
1
,
−
1
)
: This means that the vertex of the parabola is at
(
−
1
,
−
1
)
. For a quadratic in vertex form, the equation can be written as:
y
=
a
(
x
+
1
)
2
−
1
Here,
−
1
is the y-coordinate of the vertex, and
−
1
(the x-coordinate) indicates the horizontal shift.
2. Y-Intercept
(
0
,
−
2
)
: The y-intercept occurs when
x
=
0
. We can substitute
x
=
0
into the vertex form to find
a
:
−
2
=
a
(
0
+
1
)
2
−
1
Simplifying this gives:
−
2
=
To find the coefficients a, b, and c of the quadratic equation y=ax2+bx+c given its maximum point and y-intercept, we can use the provided information:
Maximum Point (−1,−1): This means that the vertex of the parabola is at (−1,−1). For a quadratic in vertex form, the equation can be written as:
y=a(x+1)2−1
Here, −1 is the y-coordinate of the vertex, and −1 (the x-coordinate) indicates the horizontal shift.
Y-Intercept (0,−2): The y-intercept occurs when x=0. We can substitute x=0 into the vertex form to find a:
−2=a(0+1)2−1
Simplifying this gives:
−2=a(1)−1
−2+1=a
a=−1
Now that we have a, we can rewrite the equation:
y=−1(x+1)2−1
Next, we can expand this to standard form:
y=−1(x2+2x+1)−1
y=−x2−2x−1−1
y=−x2−2x−2
Now, we can identify the coefficients:
- a=−1
- b=−2
- c=−2
Thus, the quadratic equation is:
y=−x2−2x−2
In summary:
- a=−1
- b=−2
- c=−2
Related questions
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If (1/4) ^(2-y) =1, what is y?
What is the solution of
d
2
y
1
d
x
2
=
√
2
y
1
2
+
y
2
2
−
1
4
y
1
2
,
d
2
y
2
d
x
2
=
√
2
y
1
2
+
y
2
2
−
1
4
y
2
2
?
Goh Kim Tee
Former Tutor at Private (non-agency) (2008–2017)
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Author has 4.2K answers and 2.4M answer views
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7y
c=y-intercept=-2
max.pt=(-1,-1)→
(0,-2) and (-2,-2) are symmetrical points about x=-1
(-1,-1)→-1=a-b+2→a-b=-3
(-2.–2)→+2=4a-b-2→4a-b=4→
a=7/3→b=a+3=16/3,c=-2
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Bhalchandra Khare
Former Self Employed Chemical Engineer (1975–2016)
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Author has 4.3K answers and 1.4M answer views
·
1y
dy/dx=0 at (-1,-1)
2ax+b=-2a-b=0
b=-2a
-2=c
y=ax^2–2ax-2.
-1=a+2a-2
a=1/3
b=-2/3
Raymond Beck
Former Infantry Sergeant
·
Author has 42.4K answers and 10.3M answer views
·
Jan 2
c=-2 = y intercept
a<0 if (-1,-1) is maximum = vertex (h,k), h=-1 =k
y =a(x-h)^2 +k = a(x+1)^2 -1
=a(x^2+2x+1) -1= ax^2 +2ax +a -1 where a-1 = -2, a= -2+1 = -1
y = -x^2 -2x -2, a=-1, b=-2, c= -2
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James Wright
BS in Physics & Mathematics, University of Illinois at Chicago (Graduated 1972)
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Author has 2.4K answers and 1.4M answer views
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5y
Related
The graph of y=ax^2+bx+c has a minimum at (5,-3) and passes through (4,0). How do I find the values of a, b, and c?
I think I can do this!
First, I will insert the (x,y) values for the 2 given points & if that isn’t enough, I will find dy/dx, set it equal to zero and insert (5,-3) for (x,y).
Inserting (5,-3) gives -3 = 25a + 5b + c (Eq 1)
Inserting (4,0) gives 0 = 16a + 4b + c (Eq 2)
We still have only 2 equations with 3 unknowns, but if we find a third independent equation, we can solve for a,b & c.
The slope (dy/dx) at a minimum point will = 0, so dy/dx = 2ax + b = 0 at (5,-3), which means that 10a + b = 0, or
b = -10a (Eq 3)
Now, if Eq 2 is subtracted from Eq 1, we get -3 = 9a + b And substituting for b using E
I think I can do this!
First, I will insert the (x,y) values for the 2 given points & if that isn’t enough, I will find dy/dx, set it equal to zero and insert (5,-3) for (x,y).
Inserting (5,-3) gives -3 = 25a + 5b + c (Eq 1)
Inserting (4,0) gives 0 = 16a + 4b + c (Eq 2)
We still have only 2 equations with 3 unknowns, but if we find a third independent equation, we can solve for a,b & c.
The slope (dy/dx) at a minimum point will = 0, so dy/dx = 2ax + b = 0 at (5,-3), which means that 10a + b = 0, or
b = -10a (Eq 3)
Now, if Eq 2 is subtracted from Eq 1, we get -3 = 9a + b And substituting for b using Eq 3 gives us: -3 = 9a - 10a or a = 3
This means that b (which equals -10a) is -30
Substituting these values into Eq 2 gives 0 = 48 - 120 + c, or c = 72
Answer: a = +3
…………….b = -30
…………….c = +72
Paul Grimshaw
IT Architect in the Computer Industry (1983–present)
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Author has 3.5K answers and 14.9M answer views
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6y
Related
If the curve y = ax^2 + bx + c touches the line y = 2x at the origin and has a maximum point when x = 1, what is a, b and c?
If the curve y = ax2 + bx + c touches the line y = 2x at the origin and has a maximum point when x = 1, what is a, b and c?
Touching y=2x at the origin tells us two things. Firstly the curve passes through the origin, which means that c=0. Secondly the gradient of the curve at the origin equals the gradient of y=2x, which is of course 2. Differentiating the equation for the curve with respect to x tells us its gradient:
dydx=2ax+b
As we know this is equal to 2 when x=0 this means that b=2.
Finally having a maximum at x=1 means that the gradient is 0 at x=1. Hence 2a+b=0, and as b=
If the curve y = ax2 + bx + c touches the line y = 2x at the origin and has a maximum point when x = 1, what is a, b and c?
Touching y=2x at the origin tells us two things. Firstly the curve passes through the origin, which means that c=0. Secondly the gradient of the curve at the origin equals the gradient of y=2x, which is of course 2. Differentiating the equation for the curve with respect to x tells us its gradient:
dydx=2ax+b
As we know this is equal to 2 when x=0 this means that b=2.
Finally having a maximum at x=1 means that the gradient is 0 at x=1. Hence 2a+b=0, and as b=2 then this means a=−1.
So the answer is y=−x2+2x:
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
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Author has 3.9K answers and 4M answer views
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4y
Related
What is a quadratic function f(x) = ax2 + bx + c that passes through the points (0, 6), (3, -6) and (-1, 14)?
From the point (0,6), we can see that c=6. Now, we have y=ax2+bx+6.
From the point (3,−6), we get:
−6=a(3)2+b(3)+6
or
b=−(3a+4)
From the point (−1,14) and the expression for b that we just found:
14=a(−1)2−(3a+4)(−1)+6
a=1
b=−(3(1)+4)=−7
Therefore, y=x2−7x+6
The plot looks like this:
From the point (0,6), we can see that c=6. Now, we have y=ax2+bx+6.
From the point (3,−6), we get:
−6=a(3)2+b(3)+6
or
b=−(3a+4)
From the point (−1,14) and the expression for b that we just found:
14=a(−1)2−(3a+4)(−1)+6
a=1
b=−(3(1)+4)=−7
Therefore, y=x2−7x+6
The plot looks like this:
Brady West
BS in Civil Engineering, Michigan State University (Graduated 2020)
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Author has 58 answers and 100.9K answer views
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4y
Related
How do I find the equation y = ax^2 + bx + c of a parabola given 2 points it passes through, and c?
A parabola needs a minimum of three unique points to define it. We’re given two outright, and the third point is at (0,c). So we can solve.
phew.
From the general equation y = ax^2 + bx + c, if we consider what we’re given (c, x, and y), we’re left with only two unknowns, a and b. We can create two independent equations to solve for these two unknowns by plugging in both of the given points (x1, y1) and (x2, y2). Then it becomes a 2x2 system of equations—completely solvable.
For instance:
A parabola passes through the points (-2,3) and (1,6). The value c is given to be -1. Find the equation of t
A parabola needs a minimum of three unique points to define it. We’re given two outright, and the third point is at (0,c). So we can solve.
phew.
From the general equation y = ax^2 + bx + c, if we consider what we’re given (c, x, and y), we’re left with only two unknowns, a and b. We can create two independent equations to solve for these two unknowns by plugging in both of the given points (x1, y1) and (x2, y2). Then it becomes a 2x2 system of equations—completely solvable.
For instance:
A parabola passes through the points (-2,3) and (1,6). The value c is given to be -1. Find the equation of this parabola.
Using the given information, construct the two equations:
3 = a(-2)^2 + b(-2) -1
6 = a(1)^2 + b(1) -1
Solving the second equation for b:
b = 7 - a
Plugging into the first equation:
3 = 4a -2(7 - a) -1
a = 3
b = 7 - 3
b = 4
Thus, the equation of the parabola is
y = 3x^2 + 4x - 1
If you’re given a third point that doesn’t lie on the y-axis (i.e. some (x3, y3) instead of c), the procedure would be the same, you would just build three equations to solve for the three unknowns (a, b, and c).
Graham Dolby
Author has 3.2K answers and 1.7M answer views
·
2y
Related
For the equation x^2+y^2 + ax + by + c=0, what are the conditions on a, b, and c so that the graph is a single point?
[math]\text{Condition: }a^2+b^2=4c[/math]
[math]\text{e.g. }x^2+y^2+2x-2y+2=0[/math]
[math]\text{Condition: }a^2+b^2=4c[/math]
[math]\text{e.g. }x^2+y^2+2x-2y+2=0[/math]
Subhamoy Chattaraj
99.9 percentile @JEE MAIN Mathematics (2020–present)
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Author has 99 answers and 384.7K answer views
·
4y
Related
What is the maximum value and minimum value of ax^2+bx+c?
The maximum and minimum value depends upon the sign of “a”,i.e. if the parabola opens upwards or downwards.
lets assume that a>0,then the graph of the quadratic,which is parabolic in nature , will open upwards.
here is an example,
x^2+ 2x +1 =0,
As you can see from this image above,the open side is upwards.
Thus what we can infer from this is that,the function has maximum value of infinity,as the greater value of “x” you will give as input,larger the output will be.
Another point of inference is the minima,that the minimum value is “0”,when x=-1.
The point (-1,0) is the vertex of the Parabola.
Thus th
The maximum and minimum value depends upon the sign of “a”,i.e. if the parabola opens upwards or downwards.
lets assume that a>0,then the graph of the quadratic,which is parabolic in nature , will open upwards.
here is an example,
x^2+ 2x +1 =0,
As you can see from this image above,the open side is upwards.
Thus what we can infer from this is that,the function has maximum value of infinity,as the greater value of “x” you will give as input,larger the output will be.
Another point of inference is the minima,that the minimum value is “0”,when x=-1.
The point (-1,0) is the vertex of the Parabola.
Thus the point which gives minima in ax^2+bx+c =0 is “-b/2a” and the minimum value is “-D/4a”.
Now comes the second case,if a<0 :-
For example let us consider “-x^2 -2x-1”,which is actually -(x+1)^2,
What you can see here is,the graph of the quadratic is inverted.
Thus the minimum value is negative infinity and the maximum value here is “0”.
The function gives its maxima at x=-1,where (-1,0) is the vertex of the Parabola.
Thus the maxima is obtained at x=(-b/2a) and the maximum value is “-D/4a”.
So,what you can observe from this answer is that the maxima and minima depends upon the sign of “a”.
Now there are various methods to find the maxima and minima,one such conventional method is using Calculus.
I’ll suggest you to go to any graphing software and draw and observe many quadratic graphs ,it will help you to obtain clarity.
I hope this answer clarified your doubt.
Thank You,
Subhamoy Chattaraj.
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7: Integrals
A Primer of Real Analysis (Sloughter)
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7.3: Integrability Conditions
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Contents
1. Home
2. Bookshelves
3. Analysis
4. A Primer of Real Analysis (Sloughter)
5. 7: Integrals
6. 7.3: Integrability Conditions
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A Primer of Real Analysis (Sloughter)
Front Matter
1: Fundamentals
2: Sequences and Series
3: Cardinality
4: Topology of the Real Line
5: Limits and Continuity
6: Derivatives
7: Integrals
8: More Functions
Back Matter
7.3: Integrability Conditions
Last updated Sep 5, 2021
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7.2: Integrals
7.4: Properties of Integrals
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Dan Sloughter
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( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Proposition 7.3.1
2. Example 7.3.1
3. Proposition 7.3.2
4. Exercise 7.3.1
5. Exercise 7.3.2
6. Exercise 7.3.3
Proposition 7.3.1
If a<b and f:[a,b]→R is monotonic, then f is integrable on [a,b].
Proof
Suppose f is nondecreasing. Given ϵ>0, let n∈Z+ be large enough that
(7.3.1)(f(b)−f(a))(b−a)n<ϵ.
For i=0,1,…,n, let
(7.3.2)x i=a+(b−a)i n.
Let P={x 0,x 1,…,x n}. Then
(7.3.3)U(f,P)−L(f,P)=∑i=1 n f(x i)(x i−x i−1)−∑i=1 n f(x i−1)(x i−x i−1)=∑i=1 n(f(x i)−f(x i−1))b−a n=b−a n((f(x 1)−f(x 0))+(f(x 2)−f(x 1))+⋯+(f(x n−1)−f(x n−2))+(f(x n)−f(x n−1)))=b−a n(f(b)−f(a))<ϵ.
Hence f is integrable on [a,b]. Q.E.D.
Example 7.3.1
Let φ:Q∩[0,1]→Z+ be a one-to-one correspondence. Define f:[0,1]→R by
(7.3.4)f(x)=∑q∈Q∩[0,1]q≤x 1 2 φ(q).
Then f is increasing on [0,1], and hence integrable on [0,1].
Proposition 7.3.2
If a<b and f:[a,b]→R is continuous, then f is integrable on [a,b].
Proof
Given ϵ>0, let
(7.3.5)γ=ϵ b−a.
Since f is uniformly continuous on [a,b], we may choose δ>0 such that
(7.3.6)|f(x)−f(y)|<γ
whenever |x−y|<δ. Let P={x 0,x 1,…,x n} be a partition with
(7.3.7)sup{|x i−x i−1|:i=1,2,…,n}<δ.
If, for i=1,2,…,n,
(7.3.8)m i=inf{f(x):x i−1≤x≤x i}
and
(7.3.9)M i=sup{f(x):x i−1≤x≤x i},
then M i−m i<γ. Hence
(7.3.10)U(f,P)−L(f,P)=∑i=1 n M i(x i−x i−1)−∑i=1 n m i(x i−x i−1)=∑i=1 n(M i−m i)(x i−x i−1)<γ∑i=1 n(x i−x i−1)=γ(b−a)=ϵ.
Thus f is integrable on [a,b]. Q.E.D.
Exercise 7.3.1
Suppose a<b,f:[a,b]→R is bounded, and c∈[a,b]. Show that if f is continuous on [a,b]{c}, then f is integrable on [a,b].
Exercise 7.3.2
Suppose a<b and f is continuous on [a,b] with f(x)≥0 for all x∈[a,b]. Show that if
(7.3.11)∫a b f=0,
then f(x)=0 for all x∈[a,b].
Exercise 7.3.3
Suppose a<b and f is continuous on [a,b]. For i=0,1,…,n, n∈Z+, let
(7.3.12)x i=a+(b−a)i n
and, for i=1,2,…,n, let c i∈[x i−1,x i]. Show that
(7.3.13)∫a b f=lim n→∞b−a n∑i=1 n f(c i).
In the notation of Exercise 7.3.3, we call the approximation
(7.3.14)∫a b f≈b−a n∑i=1 n f(c i)
a right-hand rule approximation if c i=x i, a left-hand rule approximation if c i=x i−1, and a midpoint rule approximation if
(7.3.15)c i=x i−1+x i 2.
These are basic ingredients in creating numerical approximations to integrals.
This page titled 7.3: Integrability Conditions is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform.
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Atomic Number Density
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Critical Energy
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Radioactivity
Radiation
Alpha Radiation
Alpha Particle
Beta Radiation
Beta Particle
Gamma Rays
X-Rays
Neutrons
Fission Fragments
Ionization
Interactions
Interaction of Alpha Particles
Interaction of Beta Radiation
Interaction of Gamma Rays
Interaction of X-rays
Interaction of Neutrons
Positron Interactions
Decay Mode
Stable – Unstable
Decay Constant
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Alpha Decay
Beta Decay
Electron Capture
Gamma Decay
Isomeric Transition
Internal Conversion
Neutron Decay
Proton Decay
Spontaneous Fission
Unit of Radioactivity
Curie
Becquerel
Rutherford
Decay Chain
Thorium Series
Uranium Series
Actinium Series
Neptunium Series
Radioactive Series
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Radiometric Dating
Carbon-14 Dating
Uranium-lead Dating
Source of Radiation
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Radon
Internal Sources
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Carbon-14
Radioactive Contamination
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Mass
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What is Macroscopic Cross-section – Definition
2020-11-23 2019-05-22 by Nick Connor
The macroscopic cross-section represents the effective target area of all of the nuclei contained in the volume of the material. The units are given in cm-1. Periodic Table
Macroscopic Cross-section
The difference between the microscopic cross-section and macroscopic cross-sectionis very important and is restated for clarity. The microscopic cross section represents the effective target area of a single target nucleus for an incident particle. The units are given in barns or cm 2.
While the macroscopic cross-section represents the effective target area of all of the nuclei contained in the volume of the material. The units are given in cm-1.
A macroscopic cross-section is derived from microscopic cross-section and the atomic number density:
Σ=σ.N
Here σ, which has units of m 2, is the microscopic cross-section. Since the units of N (nuclei density) are nuclei/m 3, the macroscopic cross-section Σ have units of m-1, thus in fact is an incorrect name, because it is not a correct unit of cross-sections. In terms of Σ t (the total cross-section), the equation for the intensity of a neutron beam can be written as
-dI = N.σ.Σ t.dx
Dividing this expression by I(x) gives
-dΙ(x)/I(x) = Σ t.dx
Since dI(x) is the number of neutrons that collide in dx, the quantity –dΙ(x)/I(x) represents the probability that a neutron that has survived without colliding until x, will collide in the next layer dx.It follows that the probability P(x) that a neutron will travel a distance x without any interaction in the material, which is characterized by Σ t, is:
P(x) = e-Σ t.x
From this equation, we can derive the probability that a neutron will make its first collision in dx. It will be the quantityP(x)dx. If the probability of the first collision in dx is independent of its past history, the required result will be equal to the probability that a neutron survives up to layer x without any interaction (~Σ t dx) times the probability that the neutron will interact in the additional layer dx (i.e. ~e-Σ t.x).
P(x)dx = Σ t dx . e-Σ t.x = Σ t e-Σ t.x dx
Mean Free Path
From the equation for the probability of thefirst collision in dx we can calculate the mean free path that is traveled by a neutron between two collisions. This quantity is usually designated by the symbol λ and it is equal to the average value of x, the distance traveled by a neutron without any interaction, over the interaction probability distribution.
whereby one can distinguishλ s, λ a, λ f, etc. This quantity is also known as the relaxation length, because it is the distance in which the intensity of the neutrons that have not caused a reaction has decreased with a factor e.
For materials with high absorption cross-section, the mean free path is very short and neutron absorption occurs mostlyon the surface of the material. This surface absorption is called self-shielding because the outer layers of atoms shield the inner layers.
Macroscopic Cross-section of Mixtures and Molecules
Most materials are composed of several chemical elements and compounds. Most of chemical elements contains severalisotopes of these elements (e.g. gadolinium with its six stable isotopes). For this reason most materials involve many cross-sections. Therefore, to include all the isotopes within a given material, it is necessary to determine the macroscopic cross section for each isotope and then sum all the individual macroscopic cross-sections.
In this section both factors (differentatomic densities and different cross-sections) will be considered in the calculation of the macroscopic cross-section of mixtures.
First, consider the Avogadro’s number N 0 = 6.022 x 10 23, is the number of particles (molecules, atoms)that is contained in the amount of substance given by one mole.Thus if M is the molecular weight, the ratio N 0/Mequals to the number of molecules in 1g of the mixture. The number of molecules per cm 3 in the material of density ρ and the macroscopic cross-section for mixtures are given by following equations:
N i = ρ i.N 0 / M i
Scattering of slow neutrons by molecules is greater than by free nuclei.
Note that, in some cases, the cross-section of the moleculeis not equal to the sum of cross-sections of itsindividual nuclei. For example the cross-section ofneutron elastic scattering of water exhibits anomalies for thermal neutrons. It occurs, because the kinetic energy of an incident neutron is of the order or less than the chemical binding energy and therefore the scattering of slow neutrons by water (H 2 O) is greater than by free nuclei (2H + O).
Example - Macroscopic cross-section for boron carbide in control rods
A control rod usually contains solid boron carbide with natural boron. Natural boron consists primarily of two stable isotopes,11 B (80.1%) and10 B(19.9%). Boron carbide has a density of 2.52 g/cm 3.
Determine the total macroscopic cross-section and the mean free path.
Density:
M B = 10.8
M C = 12
M Mixture = 4 x 10.8 + 1×12 g/mol
N B4C = ρ . N a / M Mixture
= (2.52 g/cm 3)x(6.02×10 23 nuclei/mol)/ (4 x 10.8 + 1×12 g/mol)
= 2.75×10 22 molecules of B4C/cm 3
N B = 4 x 2.75×10 22 atoms of boron/cm 3
N C = 1 x 2.75×10 22 atoms of carbon/cm 3
N B10 = 0.199 x 4 x 2.75×10 22 = 2.18×10 22 atoms of 10B/cm 3
N B11 = 0.801 x 4 x 2.75×10 22 = 8.80×10 22 atoms of 11B/cm 3
N C = 2.75×10 22 atoms of 12C/cm 3
the microscopic cross-sections
σ t 10B = 3843 b of which σ(n,alpha)10B = 3840 b
σ t 11B = 5.07 b
σ t 12C = 5.01 b
the macroscopic cross-section
Σ t B4C= 3843×10-24 x 2.18×10 22 + 5.07×10-24 x 8.80×10 22 + 5.01×10-24 x 2.75×10 22
= 83.7 + 0.45 + 0.14 = 84.3 cm-1
the mean free path
λ t= 1/Σ t B4C = 0.012 cm = 0.12 mm (compare with B4C pellets diameter in control rods which may be around 7mm)
λ a ≈ 0.12 mm
Example - Atomic number density of 235U in uranium fuel
It was written the macroscopic cross-section is derived from microscopic cross-section and the atomic number density (N):
Σ=σ.N
In this equation, the atomic number density plays the crucial role as the microscopic cross-section, because in the reactor core the atomic number density of certain materials (e.g. water as the moderator) can be simply changed leading into certain reactivity changes. In order to understand the nature of these reactivity changes, we must understand the term the atomic number density.
See theory: Atomic Number Density
Most of PWRs use the uranium fuel, which is in the form of uranium dioxide(UO 2). Typically, the fuel have enrichment of ω 235 = 4% [grams of 235 Uper gram of uranium] of isotope 235 U.
Calculate the atomic number density of 235 U(N235U), when:
the molecular weight of the enriched uranium M UO2 = 237.9 + 32 = 269.9 g/mol
the uranium density ⍴UO2 = 10.5 g/cm 3
N UO2 = ⍴UO2. N A/ M UO2
N UO2= (10.5 g/cm 3) x (6.02×10 23 nuclei/mol)/ 269.9
N UO2= 2.34 x 10 22 molecules of UO2/cm 3
N U = 1 x 2.34×10 22 atoms of uranium/cm 3
N O = 2 x 2.34×10 22 atoms of oxide/cm 3
N 235U = ω 235.N A.⍴UO2/M 235U x (M U/M UO2)
N 235U = 0.04 x 6.02×10 23 x 10.5 / 235 x 237.9 / 269.9 =9.48 x 10 20 atoms of 235U/cm 3
References:
Nuclear and Reactor Physics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading,MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering,Springer; 4th edition, 1994, ISBN:978-0412985317
5. W.S.C. Williams. Nuclear and Particle Physics.Clarendon Press; 1 edition, 1991, ISBN:978-0198520467
6. G.R.Keepin. Physics of Nuclear Kinetics.Addison-Wesley Pub. Co; 1st edition, 1965
7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
8. U.S. Department of Energy, Nuclear Physics and Reactor Theory.DOE Fundamentals Handbook,Volume 1 and 2.January 1993.
Advanced Reactor Physics:
K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.
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7215 | https://brainly.in/question/54064785 | Two sides of a regular polygon with sides n when extended they meet at an angle 28 degree what is the - Brainly.in
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Two sides of a regular polygon with sides n when extended they meet at an angle 28 degree what is the smallest value of n
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Given
Two sides of a regular polygon when extended, they meet at 28°
Now, as we know that the sum of exterior angles of polygon = 360°
Each exterior angle of polygon = 28°
Now,
the number of sides is given by,
n = 360°/28°
=12.85
Hence, the smallest value of n is 12.
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7216 | https://www.youtube.com/watch?v=zRop6A3DMdk | nets of cones (KristaKingMath)
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In this video we'll learn how to draw nets of right circular cones. A right circular cone is a cone whose base is a circle and that has a right angle (90-degree angle) between the base of the cone and it's altitude. A net is the two-dimensional shape that can be folded into the three-dimensional object.
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In this video, we're talking about how to draw nets of right circular cones. And remember that when we talk about a net, we're talking about starting with a three-dimensional object and transforming it into a two-dimensional shape. And you can think about the two-dimensional shape as the piece of paper that you could fold up into this three-dimensional object. So that two-dimensional piece of paper is going to be the net for this three-dimensional object. And we're going to be drawing nets for right circular cones. And so we're going to start with a cone. And the fact that we call it a right circular cone means two things. First of all, circular here or circle means that the base of the cone is going to be a perfect circle. And the fact that it is a right circular cone means that the angle between the vertex of the cone and the center of the base here. So the angle between this line here and the radius here, this angle is a 90° angle or a right angle. So this is going to be a right circular cone. And we want to draw the net of it. The easiest way to draw a net for a right circular cone is this way. If we want to try to make it somewhat precise, we're going to draw a circle. First of all, let's indicate here the center of our circle. And we're going to draw a circle around this center point. So maybe our circle looks something like that. And then what we want to do is go ahead and draw two lines roughly like this coming from the edge of the circle to the center of the circle. So, we have two radius lines like this. And what we can do is then go ahead and erase everything outside of this section like so. And what we're left with, this is going to be the lateral face or the edge of the cone. Then all we need to do is account for the base of the cone. And that again is going to be another perfect circle, which we can just go ahead and draw on the side over here. And just make sure that it touches this portion that we drew earlier. So what we end up with then is the base of the cone. So if we go ahead and draw in the radius here, if we say this is about the center and this is the radius, if we want to label its dimensions so that we indicate that this net represents this cone here, we can label the radius there with five. And then 13 was given to us as the slant height. So we'll label that also. Before we do, let's go ahead and copy this net so that we can use it for our second cone over here. But we're gonna have this first net and we want to go ahead and label the radius with five. Since we were told that the radius of the base was five, we were told that the slant height is 13. So we can say that this is 13 and this is also 13. So this then would be the net of this right circular cone. And if we did another example here, we're given the radius and the slant height. We would draw the same net and we would just call the radius 2 and the slant height here 10. And obviously to be as accurate as possible, what we want to do is just make sure that we draw that large circle such that the radius here can be 10. And then in this case, draw a smaller circle such that the radius is two and these dimensions match up. But in general, that's how you draw the net for a right circular cone. |
7217 | https://flexbooks.ck12.org/cbook/ck-12-cbse-math-class-10/section/9.2/related/lesson/angles-of-elevation-and-depression-trig/ | Angles of Elevation and Depression - Definition and Examples | CK-12 Foundation
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Back To Angle of Elevation and Angle of DepressionBack
9.2
Angles of Elevation and Depression
Written by:Bradley Hughes |Larry Ottman | +5 more
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 02, 2025
You have decided to go camping with some friends. While out on a hike, you reach the top of a ridge and look down at the trail behind you. In the distance, you can see your camp. You're thinking about how far you've traveled, and wonder if there is a way to determine it.
By using a small device called a clinometer, you're able to measure the angle between your horizontal line of sight and the camp as 37∘, and you know that the hill you just hiked up has a height of 300 m. Is it possible to find out how far away your camp is using this information? (Assume that the trail you hiked is slanted like the side of a triangle.)
Angles of Elevation and Depression
You can use right triangles to find distances, if you know an angle of elevation or an angle of depression.
The figure below shows each of these kinds of angles.
The angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object. For example, if you are standing on the ground looking up at the top of a mountain, you could measure the angle of elevation. The angle of depression is the angle between the horizontal line of sight and the line of sight down to an object. For example, if you were standing on top of a hill or a building, looking down at an object, you could measure the angle of depression. You can measure these angles using a clinometer or a theodolite. People tend to use clinometers or theodolites to measure the height of trees and other tall objects. Here we will solve several problems involving these angles and distances.
Finding the angle of elevation
You are standing 20 feet away from a tree, and you measure the angle of elevation to be 38∘. How tall is the tree?
The solution depends on your height, as you measure the angle of elevation from your line of sight. Assume that you are 5 feet tall.
The figure shows us that once we find the value of T, we have to add 5 feet to this value to find the total height of the triangle. To find T, we should use the tangent value:
tan38∘=o p p o s i t e a d j a c e n t=T 20 tan38∘=T 20 T=20 tan38∘≈15.63 Height of tree≈20.63 f t
You are standing on top of a building, looking at a park in the distance. The angle of depression is 53∘. If the building you are standing on is 100 feet tall, how far away is the park? Does your height matter?
Finding the angle of depression
If we ignore the height of the person, we solve the following triangle:
Given the angle of depression is 53∘, ∠A in the figure above is 37∘. We can use the tangent function to find the distance from the building to the park:
tan37∘=o p p o s i t e a d j a c e n t=d 100 tan37∘=d 100 d=100 tan37∘≈75.36 f t.
If we take into account the height if the person, this will change the value of the adjacent side. For example, if the person is 5 feet tall, we have a different triangle:
tan37∘=o p p o s i t e a d j a c e n t=d 105 tan37∘=d 105 d=105 tan37∘≈79.12 f t.
If you are only looking to estimate a distance, then you can ignore the height of the person taking the measurements. However, the height of the person will matter more in situations where the distances or lengths involved are smaller. For example, the height of the person will influence the result more in the tree height problem than in the building problem, as the tree is closer in height to the person than the building is.
Real-World Application: The Horizon
You are on a long trip through the desert. In the distance you can see mountains, and a quick measurement tells you that the angle between the mountaintop and the ground is 13.4∘. From your studies, you know that one way to define a mountain is as a pile of land having a height of at least 2,500 meters. If you assume the mountain is the minimum possible height, how far are you away from the center of the mountain?
tan13.4∘=o p p o s i t e a d j a c e n t=2500 d tan13.4∘=2500 d d=2500 tan13.4∘≈10,494 m e t e r s.
Examples
Example 1
Earlier, you were asked if it was possible to find out how far away your camp is using the information given.
Since you know the angle of depression is 37∘, you can use this information, along with the height of the hill, to create a trigonometric relationship:
[Figure 6]
Since the unknown side of the triangle is the hypotenuse, and you know the opposite side, you should use the sine relationship to solve the problem:
sin37∘=300 h y p o t e n u s e h y p o t e n u s e=300 sin37∘h y p o t e n u s e≈498.5
You have traveled approximately 498.5 meters up the hill.
Example 2
You are six feet tall and measure the angle between the horizontal and a bird in the sky to be 40∘. You can see that the shadow of the bird is directly beneath the bird, and 200 feet away from you on the ground. How high is the bird in the sky?
We can use the tangent function to find out how high the bird is in the sky:
tan40∘=h e i g h t 200 h e i g h t=200 tan40∘h e i g h t=(200)(.839)h e i g h t=167.8
We then need to add your height to the solution for the triangle. Since you are six feet tall, the total height of the bird in the sky is 173.8 feet.
Example 3
While out swimming one day you spot a coin at the bottom of the pool. The pool is ten feet deep, and the angle between the top of the water and the coin is 15∘. How far away is the coin from you along the bottom of the pool?
Since the distance along the bottom of the pool to the coin is the same as the distance along the top of the pool to the coin, we can use the tangent function to solve for the distance to the coin:
tan15∘=o p p o s i t e a d j a c e n t tan15∘=10 x x=10 t a n 15∘x≈37.32∘
Example 4
You are hiking and come to a cliff at the edge of a ravine. In the distance you can see your campsite at the base of the cliff, on the other side of the ravine. You know that the distance across the ravine is 500 meters, and the angle between your horizontal line of sight and your campsite is 25∘. How high is the cliff? (Assume you are five feet tall.)
Using the information given, we can construct a solution:
tan25∘=o p p o s i t e a d j a c e n t tan25∘=h e i g h t 500 h e i g h t=500 tan25∘h e i g h t=(500)(.466)h e i g h t=233 meters
This is the total height from the bottom of the ravine to your horizontal line of sight. Therefore, to get the height of the ravine, you should take away five feet for your height, which gives an answer of 228 meters.
Review
Round your answers to the nearest hundredths place for the following questions.
A 70 foot building casts an 50 foot shadow. What is the angle that the sun hits the building?
You are standing 10 feet away from a tree, and you measure the angle of elevation to be 65∘. How tall is the tree? Assume you are 5 feet tall up to your eyes.
Kaitlyn is swimming in the ocean and notices a coral reef below her. The angle of depression is 35∘ and the depth of the ocean, at that point is 350 feet. How far away is she from the reef?
The angle of depression from the top of a building to the base of a car is 60∘. If the building is 78 ft tall, how far away is the car?
The Leaning Tower of Pisa currently “leans” at a 4∘ angle and has a vertical height of 55.86 meters. How tall was the tower when it was originally built?
The angle of depression from the top of an apartment building to the base of a fountain in a nearby park is 72∘. If the building is 78 ft tall, how far away is the fountain?
You are standing 15 feet away from a tree, and you measure the angle of elevation to be 35∘. How tall is the tree? Assume you are 5 feet tall up to your eyes.
Bill spots a tree directly across the river from where he is standing. He then walks 18 ft upstream and determines that the angle between his previous position and the tree on the other side of the river is 55∘. How wide is the river?
A 50 foot building casts an 50 foot shadow. What is the angle that the sun hits the building?
Eric is flying his kite one afternoon and notices that he has let out the entire 100 ft of string. The angle his string makes with the ground is 60∘. How high is his kite at this time?
A tree struck by lightning in a storm breaks and falls over to form a triangle with the ground. The tip of the tree makes a 36∘ angle with the ground 25 ft from the base of the tree. What was the height of the tree to the nearest foot?
Upon descent an airplane is 15,000 ft above the ground. The air traffic control tower is 200 ft tall. It is determined that the angle of elevation from the top of the tower to the plane is 15∘. To the nearest mile, find the ground distance from the airplane to the tower. (Note that 1 mile = 5280 feet)
Tara is trying to determine the angle at which to aim her sprinkler nozzle to water the top of a 10 ft bush in her yard. Assuming the water takes a straight path and the sprinkler is on the ground 4 ft from the tree, at what angle of inclination should she set it?
Over 3 miles (horizontal), a road rises 1000 feet (vertical). What is the angle of elevation?
Over 4 miles (horizontal), a road rises 1000 feet (vertical). What is the angle of elevation?
Review (Answers)
Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
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7218 | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/pages/unit-3-the-definite-integral-and-its-applications/part-a-definition-of-the-definite-integral-and-first-fundamental-theorem/ | Browse Course Material
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Single Variable Calculus
The Definite Integral and its Applications
Part A: Definition of the Definite Integral and First Fundamental Theorem
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The definite integral of a function describes the area between the graph of that function and the horizontal axis. The First Fundamental Theorem of Calculus confirms that we can use what we learned about derivatives to quickly calculate this area.
» Session 43: Definite Integrals » Session 44: Adding Areas of Rectangles » Session 45: Some Easy Integrals » Session 46: Riemann Sums » Session 47: Introduction of the Fundamental Theorem of Calculus » Session 48: The Fundamental Theorem of Calculus » Session 49: Applications of the Fundamental Theorem of Calculus » Session 50: Combining the Fundamental Theorem and the Mean Value Theorem » Problem Set 6
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7219 | http://t8el.com/wp-content/uploads/2015/08/SimpleCascades.pdf | A Simple Model of Cascades in Networks∗ Yongwhan Lim† MIT Asuman Ozdaglar‡ MIT Alexander Teytelboym§ University of Oxford November 7, 2016 Abstract We consider a linear threshold model of cascades in networks. An agent switches (e.g. adopts an innovation) if the proportion of his neighbors who have already switched exceeds his threshold.
Agents’ thresholds are drawn randomly at the start of the cascade process. We present a result for the expected number of switches in arbitrary finite networks with any initial seeds. We define a new measure of an agent’s ability to influence a cascade in a given network, called cascade centrality, which is the expected size of the cascade when the agent is the only seed in the network. We then define contagion centrality, which is the probability that all agents switch when the node is a seed. For certain network topologies, we find analytic expressions for cascade centrality and contagion centrality and show that there may be tension between them. Yet nodes with high cascade and contagion centrality share an interesting property: they have many neighbors, but their neighbors have few neighbors. As an illustration of cascade centrality, we look at how the network structure affects optimal prices when a profit-maximizing firm tries to spread an innovation. Optimal price and seeding behavior can be counterintuitive when firms price irreversible cascades. Our tractable model can be extended in various ways.
∗First draft: 20 December 2014. We would like to thank Rediet Abebe, Daron Acemoglu, Elie Adam, Yann Bramoull´ e, Kimon Drakopoulos, Mikhail Drugov, Yuichiro Kamada, Michael Kearns, Jon Kleinberg, Daniel Margo, Stephen Morris, and Bassel Tarbush, as well as seminar and conference participants at Boston College, Harvard, Qu´ ebec, Queen Mary, LMU, Oxford, Cambridge, GREQAM, HSE, 20th CTN Workshop, Conference on Information Transmission in Networks, 2014 INFORMS, 2015 SAET Conference, Monash, University of Sydney, and the 11th World Congress of the Econometric Society for their comments.
†Operations Research Center and Laboratory for Information and Decision Systems, MIT. Email: yongwhan@mit.edu.
‡Department of Electrical Engineering and Computer Science and Laboratory for Information and Deci-sion Systems, MIT. Email: asuman@mit.edu.
§Institute for New Economic Thinking at the Oxford Martin School, University of Oxford.
Email: alexander.teytelboym@inet.ox.ac.uk.
1 1 Introduction Many phenomena that occur in social and economic networks are, at least temporarily, irreversible. Examples include the spread of incurable diseases (Klovdahl, 1985), infor-mational leaks and rumors (Moreno et al., 2004), systemic risk of bank failures (Elliott et al., 2014), platform adoption (David, 1985), drug addiction (Bauman and Ennett, 1996), patenting (Aghion et al., 2015), religious conversion (Stark and Bainbridge, 1980), and dropping out of high school (Staffand Kreager, 2008). These phenomena have two key features. First, they exhibit network externalities: agents are (heteroge-neously) affected by their neighbors in the network. The structure of the network is critical for the irreversible phenomena that exhibit network effects. Some processes re-main contained in isolated clusters and others spread to the whole network. Secondly, these phenomena are inherently path-dependent: their irreversibility means that early history matters for the overall outcome. Whether the goal is to reduce contagion risk or to maximize adoption of an innovation, understanding how early history and network structure affect cascades is important for good policies.
In order to analyze cascades in networks, we revisit the familiar linear threshold model introduced by Granovetter (1978). We call an irreversible transition to new state, such as adoption of a product, a switch (Jackson, 2008, p. 295). Initially, all agents in the network are switched off. Then some agents are randomly (or optimally) switched, i.e., seeded.
Every agent in the network is endowed with an individual threshold. In the following periods, once the proportion of neighbors that switches exceeds his threshold, the agent also switches (Granovetter, 1978; Schelling, 1978).1 This process propagates through the network, but once an agent switches, he remains switched forever.2 Our objective in this paper is to characterize how the structure of the network and the initial seed set affect the total number of switches.
In contrast to some of the previous work (Acemoglu et al., 2011; Yildiz et al., 2011), we do not look at a particular instance of a distribution of thresholds. Instead we as-sume that agents’ thresholds are randomly and independently drawn from the uniform distribution at the start of the cascade (Kempe et al., 2003). This is a reasonable as-sumption if the social planner has no reason to believe that some thresholds are more 1There is empirical evidence that adoption products, such as iPhone, is sensitive to the proportion of friends who have adopted it (Godinho de Matos et al., 2014). The linear threshold model may not be an ideal model for a variety of processes, such as learning about new products, where the number rather than the proportion of neighbors affects the probability of switching. For example, Banerjee et al. (2013) do not find an effect of the proportion of neighbor-adopters on the adoption of microfinance.
2Kleinberg (2007) and Adam et al. (2012) call these types of threshold models “progressive”.
2 likely than others (by the principle of insufficient reason). We show that under this assumption the probability that an agent switches equals to the average probability that his neighbors switch conditional on his not switching. This striking simplifica-tion allows us to derive a simple analytical expression for the probability of any agent switching, building on insights by Kempe et al. (2003). We introduce a new concept – cascade centrality of a node – which measures the expected size of a cascade when this node alone is a seed in a network. Our key theorem states that the cascade centrality of an agent is equal to his degree (plus one) less the contribution of loops (paths that bend back on themselves exactly once). Although calculating the number of paths or loops is computationally challenging, it offers us a lot of analytical tractability. For certain networks, such as trees, cycles, complete networks, and a large class of random graphs, we can give analytical expressions for cascade centrality. In a tree, the cascade centrality of any node is its degree (number of neighbors) plus one, whereas in a com-plete network the cascade centrality of any node is proportional to the square root of the number of nodes. We then introduce another concept called contagion centrality of a node which measures the probability that the cascade reaches every other node in network when this node is the only seed (Morris, 2000). We show that there is a tension between cascade centrality and contagion centrality. For example, while the cascade centrality of a node in a complete network increases at the rate of √n (where n is the number of nodes), the probability of contagion declines slowly at the rate 1 n.
The theoretical literature on cascades and diffusion in networks is vast. Irreversibil-ity of our cascade process sets out the present paper apart from the strand of the literar-ture which assumes that agents can switch multiple times (Blume, 1993; Ellison, 1993; Blume, 1995; Young, 2006; Montanari and Saberi, 2010; Adam et al., 2012). These papers usually assume that agents play a coordination game with their neighbors and analyze the dynamics using tools from evolutionary game theory.3 For certain prob-lems, such as the possibility of contagion, the models are essentially equivalent (Morris, 2000; Watts, 2002; Dodds and Watts, 2004; Lelarge, 2012; Adam et al., 2012). Our approach allows us to analyze precisely how the structure of the network affects the cascade process. Since the number of paths determines expected switches, we show that, at least theoretically, we cannot expect any straightforward comparative statics 3In evolutionary models, multiple switches can occur either because agents make an error or because their thresholds are redrawn during the cascade. The process is ergodic so the stationary distribution does not depend on the initial conditions. Our model can be seen as a coordination game with heterogeneous payoffs, but in which seeds have a dominant strategy to switch. Another interesting class of non-ergodic models is one in which the seeds can also switch back and forth, but where the thresholds are fixed and the agents always best-respond. Adam et al. (2012) show that in this class of models, the process either converges to an equilibrium or the system oscillates between exactly two states.
3 using most macroscopic network properties, such as clustering (Centola et al., 2007; Centola, 2010; Acemoglu et al., 2011).4 We illustrate cascade centrality with an economic application. We consider how a profit-maximizing firm tries to diffuse a product in a network. It needs to pick a seed and a price. Higher prices make agents less likely to adopt the product for any proportion of neighbors who have already adopted it. While this problem becomes analytically challenging, we are able to use our tools to derive some interesting propo-sitions. We show that the structure of the network dramatically affects the optimal price. For example, in a line network the optimal price is bounded away from zero as the line becomes long, whereas when a complete graph becomes large the optimal price goes to zero. While there have been several papers on optimal static pricing in social networks (Candogan et al., 2012; Campbell, 2013; Ajorlou et al., 2015), to the best of our knowledge, our work is the first one to analyze optimal pricing and seeding of cascades simultaneously in general finite networks. Among papers that focus on analytical results, Candogan et al. (2012) consider optimal pricing in a model with a divisible good and Campbell (2013) considers pricing in random networks. In Lim et al. (2014), we use cascade centrality to analyze competitive rumor spread.
Many strategic aspects of cascades in networks, such as attacks on networks (Ace-moglu et al., 2013) and the role of imperfect information in collective action (Bikhchan-dani et al., 1992; Banerjee, 1992), are beyond the scope of this paper.
We proceed as follows. Section 2 describes the “simple” linear threshold model and the dynamics of the cascade. Section 3 explains the role of the uniform threshold assumption, describes the analytical expression for the expected number of switches, introduces cascade centrality and states the main theorem linking cascade centrality to the number of loops. This section also introduces contagion centrality. Section 4 applies the main theorem and describes analytical results for cascade centrality and contagion centrality for certain topologies. In Section 5, we describe the application to optimal pricing. Section 6 briefly mentions a few extensions of the model, while Section 7 concludes and points to directions for further work. All the proofs and simulations are in the Appendix.
4In Appendix C, we show that higher clustering can both increase and decrease the number of expected switches. This is also true when we consider “average cascades” i.e. when the seed is random.
4 2 General model of cascades 2.1 Preliminaries Let G(V, E) be a simple (unweighted and undirected), connected graph with a set of n agents V := {1, . . . , n} and a set of m links E.5 We denote the neighbors of i ∈V as Ni(G) := {j|(j, i) ∈E} and the degree of i as di := |Ni(G)|. A threshold for agent i is a random variable Θi drawn independently from a probability distribution with support [0, 1]. The associated multivariate probability density function for all the nodes in the graph is f(θ). Each agent is i ∈V assigned a threshold θi. Let’s define the threshold profile of agents as θ := (θi)i∈V . A network Gθ is a graph endowed with a threshold profile.
2.2 Dynamics of a deterministic cascade First, let us consider dynamics of a deterministic cascade on a given network Gθ. We follow the exposition of Acemoglu et al. (2011).6 The binary state of agent i at time t is denoted xi(t) = {0, 1}, corresponding to “off” and “switched”. Denote by St(Gθ) the set of additional switches in network Gθ at time t. At time t = 0, a subset of agents S0 ⊆V is selected to be the seed set. We assume that at t = 0 agents switch if and only if they are in the seed set. Hence, at t = 1, any i ∈V \ S0(Gθ) will switch, i.e., i ∈S1(Gθ) if |S0(Gθ) ∩Ni(Gθ)| |Ni(Gθ)| ≥θi.
This means that at t = 1 agents switch only if the proportion of their neighbors who were seeds exceeds their threshold.
Then, for a given period t ≥0 and node i ∈V \ ∪t−1 τ=0Sτ will switch at t, i.e., i ∈St(Gθ) if |{∪t−1 τ=0Sτ(Gθ)} ∩Ni(Gθ)| |Ni(Gθ)| ≥θi.
This means that any agent who has not switched by some period t, switches in time period t + 1 if the proportion of his neighbors who switched is greater or equal to 5An extension of the model to a directed graph is fairly straightforward and does not substantially affect the analysis.
6The dynamics of our model can be described as modified local interaction game (Morris, 2000, Equation 2.1) in which critical probability q (payofffrom being switched off) is drawn from Θ for each agent, except the seeds for which q = θ = 0 i.e. being switched is a best response. In a Nash equilibrium, each agent best responds in his own-payofflocal interaction game.
5 his threshold θi.
For a given network Gθ, define the fixed point of the process as S0 = S(Gθ, S0) ⇒St(Gθ) = ∅for all t > 0. As Acemoglu et al. (2011) show, this fixed point always exists.
2.3 Expected size of a cascade Let us now consider the “average” cascade dynamics on a network when thresholds are drawn from f(θ). For a given graph G and S0, we can map each realization f(θ) to a set of switches S(Gθ, S0). Hence, we can treat S(Gθ, S0) as a random variable with a probability distribution f(θ). Let us compute the expected probability of any particular agent i switching in network G is given seed S0 by taking the expectation with respect to f(θ) Pi(G, S0) = ˆ Rn |S(Gθ, S0) ∩{i}|f(θ)dθ.
Hence, the expected number of switches in graph G is: E[S(G, S0)] := ˆ Rn |S(Gθ, S0)|f(θ)dθ = n X i=1 Pi(G, S0).
While our deterministic process mimics the one in Acemoglu et al. (2011), our focus on the expected size of a cascade by integrating over possible thresholds is similar to Kempe et al. (2003). In Section 6, we also show that our analysis can be extended to the case where agents do not switch immediately and have multiple switching attempts.
3 Analysis of a simple model 3.1 Uniform distribution We first consider how the probability of an agent switching depends on the probability of his neighbors switching. We show that if the threshold for this agent is drawn from uniform distribution then his probability of switching is simply equal to the average switching probability of his neighbors conditional on his not switching.
Proposition 1. Let {G(n)}n∈N+ be a set of networks in which di(G(n)) = n −1 and a non-empty seed set is S0 ⊆V \ {i} on each G(n), then Pi(G(n), S0) = X j∈Ni(G(n)) Pj(G(n), S0|i / ∈S) di(G(n)) 6 5/8 Figure 1: Uniform Threshold Rule for a Star for any seed set on every G(n) if and only if Θi is uniformly distributed on [0, 1] almost everywhere.
It is easy to illustrate the proposition with a example of star network, which also places a key role in the proof. Figure 1 shows that five of center’s neighbors have a probability of switching equals to 1 because they are seed. The other three have a probability of switching equal to zero, conditional on the center not switching. Hence, the probability of the center node switching is 5 8 whenever his threshold is drawn from the uniform distribution. In the rest of the paper we use the following assumption.
Assumption 1. For any Gθ and every i ∈V , Θi ∼U(0, 1) and independent.
We therefore drop θ subscript and henceforth G ≡Gθ.
From now on, we use graph/network, agent/node, and link/edge interchangeably.
It may be tempting to conclude that the uniform distribution of thresholds implies that any agent’s probability of switching is equal to the probability of switching of all agent’s neighbors. However, this is not the case precisely due to the path dependence of the switching process. Consider a line network of length three with one seed at a leaf node. It is easy to show that if thresholds are drawn from a uniform distribution, the probability of the center and other leaf node switching is 1 2. However, the switching probability of the middle node is not the average of the probabilities of his neighbors (which is 3 4).
7 i j Figure 2: A path between two nodes 3.2 Cascade centrality In order to capture path dependence of the cascade process, we first introduce paths.7 Definition 1. A sequence of nodes P = (i0, · · · , ik) on a graph G is a path if ij ∈ Nij−1(G) for all 1 ≤j ≤k and each ij ∈P is distinct.
Since every node in the path is distinct, so is every edge. In general, there may be multiple paths between any two nodes. Figure 2 shows one possible path between nodes i and j. There are three paths between i and j in total in this network.
Definition 2. For a path P in G, the degree sequence along P is denoted (di(G))i∈P .
The degree sequence product along P is χP := Y i∈P di(G) di0 .
The degree sequence product of a particular path multiplies the degree every node in the path except the first one. The degree sequence product of the path in Figure 2 is 2 × 3 × 3 × 1 = 12.
7Some graph theory textbooks refer to “paths” as “simple paths”. Since we only use “simple paths” in this paper, we refer to them as “paths” without any ambiguity.
8 For any G and S0, let Pji be the set of all paths beginning at j ∈S0 and ending at i ∈V \ S0 and P∗ ji ⊆Pji denote the subset of those paths that exclude any other node in S0. The following result follows from the equivalence of the linear threshold process and random selection of “live-edge paths” established by (Kempe et al., 2003, Claim 2.6).
Proposition 2. Given a graph G and seed S0, the probability that node i ∈V \ S0 switches is Pi(G, S0) = X j∈S0 X P∈P∗ ji 1 χP .
Proposition 2 provides a remarkable insight into the calculation of the size of the expected cascade in any network with any seed. It says that the expected probability of any node switching is equal to the sum of the degree sequence products along all the paths from each seed (avoiding any other seed) to the node. In Section 4, we show this result can be applied to analyze a variety of networks.
Figure 3 illustrates how to apply the proposition to a fairly general network with two seeds A and B. Each path is labelled in a different color (some paths may include paths in themselves). The numbers next to the nodes denote the probabilistic contributions – inverse of the degree sequence products – of various paths. Note that seed B does not affect the probability of switching of the nodes in the top triangle and of the leftmost node because any path from B to these nodes must pass through A.8 Using Proposition 2, we can define the expected size of the cascade when a particular node is the seed.
Definition 3. Cascade centrality of node i in graph G is the expected number of switches in that graph given i is the seed, namely Ci(G) := E[S(G, {i})] = 1 + X j∈V {i} Pj(G, {i}) = 1 + X j∈V {i} X P∈Pij 1 χP and the average cascade in a graph G is C(G) := P i∈V Ci(G) n .
Cascade centrality captures the importance of a node in the network by measuring how large a cascade it induces when it alone is the seed. In order to see how cascade centrality can be used to generate insights into cascade processes, we prove a useful 8In the language of Bayesian networks, we could say that A is a Markov blanket for these nodes.
9 1/12+1/6 1/12+1/6 1 1 1/4+1/2 1/2+1/4 1/3 1/2+1/2 1 A B Figure 3: Network with two seeds and all the relevant paths decomposition result. We first introduce a loop, which is a sequence of nodes (i.e. a walk) that bends back on itself exactly once.
Definition 4. A sequence of nodes L = (i0, . . . , ik) on a graph G is a loop if (i0, . . . , ik−1) is a path and ik ∈{i0, . . . , ik−2} for some k ≥2.
The degree sequence product χL for a loop L = (i0, . . . , ik) is equal to the degree sequence product of its corresponding path (i0, . . . , ik−1) (see Definition 2). Let Lij be the set of all loops beginning at node i and ending at node j. Figure 4 illustrates a loop beginning at node i.
Theorem 1. The cascade centrality of any node i in G is: Ci(G) = 1 + di − X j∈V X L∈Lij 1 χL Theorem 1 shows that cascade centrality of any node in any network is at most its degree plus one. Fixing a node’s degree, the presence of loops strictly reduces its cascade centrality. Long loops and loops that involve nodes with high degrees will reduce cascade centrality by less.
10 i Figure 4: Loop The theorem immediately suggests how to design a network that maximizes the maximum cascade centrality over all nodes in the graph. Let G be the set of all graphs with vertex set V (|V | = n) and Gm be the set of all graphs with vertex set V and exactly m edges (|E| = m). A star graph of order n is an acyclic graph (tree) on n nodes with one node having degree n −1 and the other n −1 nodes having degree 1.
Proposition 3. Let G∗∼ = arg max G∈G {max i∈V Ci(G)} Then, G∗is uniquely a star graph of order n.
Let G∗ m ∼ = arg max G∈Gm{max i∈V Ci(G)} Then max i∈V Ci(G∗ m) > max i∈V Ci(G∗ m−1) whenever 1 ≤m ≤n −1 and max i∈V Ci(G∗ m) > max i∈V Ci(G∗ m+1) whenever n ≤m ≤ n 2 11 Proposition 3 says that adding edges to and removing edges from a star graph can only reduce the cascade centrality of the node with the highest cascade centrality.
Moreover, it seems clear that the ideal node for seeding has high degree, but all of its neighbors (and the neighbors of their neighbors) have a low degree. In contrast to the literature on “key players”, nodes with high cascade centrality do not have influential friends - instead they have many unpopular friends.
Theorem 1 and our simulations (in the Appendix) also show that cascade centrality is closely related to degree centrality. However, the simulations also indicate cascade centrality can be very different from centrality notions based on counting the number of walks (i.e. sequences of connected nodes which are not necessarily distinct), such as eigenvector centrality and Katz-Bonacich centrality (Katz, 1953; Bonacich, 1987).
Indeed, our simulations make it clear that since the contribution of paths of greater length decreases geometrically in cascade centrality, even knowing the set of paths of lengths one, two, and three can provide an excellent approximation to the expected size of the cascade.
3.3 Contagion centrality In the previous section, we focused on the expected size of the cascade given a seed.
We now turn our attention to extreme outcomes: i.e. the probability that the cascade reaches every node in the network (or a given connected subnetwork). This would allow us to compare how susceptible to total contagion different networks can be.
Definition 5. Contagion centrality of a node i is Ki(G) = ˆ Rn 1n (|S(Gθ, {i})|) f(θ)dθ (where 1n is the indicator function taking value 1 when the argument is equal to n and zero otherwise) i.e. the probability that the entire network switches when i is the only seed.
Contagiousness of a graph G is K(G) = P i∈V Ki(G) n i.e. the probability that every agent in G switches when one agent is seeded uniformly at random.
12 Our key result in this section states that contagion centrality of a node depends on the product of the eigenvalues of the Laplacian matrix (the difference between the diagonal degree matrix and the adjacency matrix) and the degree of all other nodes rather than, like cascade centrality, directly on the degree of the node itself. The reason for the relationship between the Laplacian and contagion is that contagion (in any finite network) can be represented by a spanning tree and the product of the eigenvalues of the Laplacian counts the number of spanning trees in the network.
Theorem 2. Contagion centrality of a node i in G is Ki(G) = 1 nλ1 × . . . × λn−1 Q j∈V {i} dj and contagiousness of a network is K(G) = 1 nλ1 × . . . × λn−1 × hP j∈V dj i n × Q j∈V dj where λ1, . . . , λn−1 are non-zero eigenvalues of the Laplacian matrix of G.
Unlike cascade centrality, computing contagion centrality for any node in any net-work only involves finding the eigenvalues of the Laplacian (and is therefore computa-tionally tractable). Since contagion centrality only depends on the count of spanning trees and the degree sequence of the graph, comparative statics results can be derived straightforwardly from corresponding comparative statics on the number of spanning trees in the graph.9 4 Analytical results 4.1 Cascade centrality We now show how Proposition 2 and Theorem 1 can be used to derive analytical expressions for the expected number of switches in certain fixed and infinite networks.
Corollary 1.
1. Suppose that G is a tree. Then for any i ∈V , Ci(G) = di(G) + 1 9See, for example, Das et al. (2013) for an analytical upper bound on the number of spanning trees in the graph as a function of various network parameters, such as the number of vertices, the number of edges, maximum degree, second maximum degree, and so on.
13 C(G) = 1 + 2(n −1) n 2. Suppose that G is a cycle of order n. Then, for any i ∈V , Ci(G) = 3 − 1 2n−2 Consider {G(n)}n∈N+, a sequence of cycles of order n. Then lim n→∞Ci(G(n)) = 3 3. Suppose that G is a complete graph of order n. Then, for all i ∈V , Ci(G) = C(G) = 1 + (n −1) n−1 X i=1 P(n −2, i −1) 1 n −1 i!
where P(n, i) ≡ n!
(n−i)! is number of ways of obtaining an ordered subset of i elements from a set of n elements.
Consider {G(n)}n∈N+, a sequence of complete graphs of order n. Then lim n→∞ Ci(G(n)) √n = rπ 2 Our first result, an immediate corollary of Theorem 1, says that the expected num-ber of switches from a single seed node in any tree equals to the degree of the seed plus one. The tree is a particularly convenient topology to analyze because there is a unique path between any two nodes i.e. there are no loops. Hence, the probability of switching of any node simply equals to the degree sequence product along this path. From this we obtain that the size of cascade starting from a random node in a (connected) tree is independent of its structure. Figure 5 illustrates the probabilities of switching for all the nodes in a tree with one seed.
Due to their symmetry, the cycle and the complete graph are also easy to analyze.
In these networks, the identity of the seed node is irrelevant.
As Figure 6 shows, in a cycle with one seed, there are only two paths to each node around the cycle.
Alternatively, using Theorem 1, we can simply subtract the contribution of the unique loop from three since di + 1 = 3 for any node in a cycle. As the cycle becomes large, its behavior approaches that of a non-edge seed node (i.e. with degree 2) on a path.
In a complete graph, any node can be reached from any other node is one step. But, as Figure 7 shows, we must also keep track of the number of paths of every possible 14 1 1 1/2 1/8 1/8 1/8 1/16 1/16 Figure 5: Tree with one seed 1/2+1/8 1/4+1/4 1/8+1/2 Figure 6: Cycle with one seed 15 1/3+2/9+1/27 1/3+2/9+1/27 1/3+2/9+1/27 Figure 7: Complete graph with one seed length. Fortunately, the number of paths of a length k can the expressed as number of ways to select k out of n−1 objects without repetition (for example, lottery numbers).
The result also shows that the size of the expected cascade in a complete graph grows at the rate √n.
We now prove a general result for cascade centrality for large random graphs. In order to do this, we first introduce the configuration model, closely following Bollob´ as and Riordan (2015). The configuration model allows us to create random graphs with a given degree distribution. First, we generate a desired degree sequence. Then we pick randomly two elements of the degree sequence – nodes with “stubs” – and link them together. We then delete those two elements from the degree sequence. This may produce self-loops and multi-edges, but it can be shown that the degree distribution in the configuration model converges in probability to the desired degree distribution. To avoid excess notation, we delegate the technical description of what we call a “regular” configuration model G∗ dn to the Appendix and simply state the key result.
Proposition 4. Let G∗ dn be a regular configuration model and consider the cascade centrality of a node in G = G∗ dn chosen uniformly at random, then Pr(C(G) = 1 + E(D)) →1 as n →∞where E(D) is the expected degree of G∗ dn.
The configuration model is useful because it can be used to generate any random 16 graph model, including Erd˝ os-R´ enyi graphs and power-law graphs. Our result only relies on the fact that the expected degree of the random graph is finite as the network gets large. The startling result (using Corollary 5 in Bollob´ as and Riordan (2015)) is that any random graph generated from the configuration model behaves like a tree.
Finally, we turn to the analysis of infinite lattices, first considered in a cascade framework by Morris (2000). We now turn our attention to infinite network. Note that under the uniform random threshold assumption, cascade centrality is well-defined for any infinite lattice with a fixed degree. While we cannot obtain analytical expression for cascade centrality in these lattices, we can give reasonable bounds. We will consider hypercubic lattices of dimension r: in these networks, nodes are the integer points of Rr and edges are the unit line segments between the integer points. Hypercubic lattices include the square lattice (r = 2) and hexagonal lattice (r = 3) discussed in Morris (2000).
Proposition 5. Let G be an infinite hypercubic lattice of dimension r ≥2, then for a generic node i: 3 ≤Ci(G) ≤1 + 2r The upper bound in this proposition comes from Theorem 1. The classical square lattice gives us us 3≤C ≤5 (computationally, we get 3.5 up to paths of length 10), whereas for the hexagonal lattice we obtain tighter bounds of 3 ≤C ≤7 (computation-ally, we get 5.0 up to paths of length 10).10 4.2 Contagion centrality We now turn to the analysis of contagion centrality. As we pointed out, contagion centrality is a straightforward computational problem on finite networks. The following result illustrates the networks already covered by Corollary 1.
10A path on an infinite lattice is called a “self-avoiding walk” (Madras and Slade, 1993). “The enumeration of self-avoiding walks ... has for half a century, been among the most challenging problems in enumerative combinatorics.” (Guttmann and Conway, 2001). The complexity of the problem of calculating the number of self-avoiding walks on an infinite lattice is not known (but conjectured to be NP-hard). Very few analytical results are known, Duminil-Copin and Smirnov (2012) is a notable exception. Using Theorem 2 and the results on the number self-avoiding walks up to length 10 from Fisher and Sykes (1959) we calculate that the cascade centrality of a node in a: honeycomb lattice is 2.913; square lattice is 3.498; Kagome (trihexagonal) lattice is 3.242; triangular lattice is 4.039. Here lattices with higher degree have higher cascade centrality, but lattices with higher clustering (and the same degree i.e. square and Kagome) have lower centrality.
17 Corollary 2.
1. Let G be a tree. For any i ∈V Ki(G) = 1 Q j∈V \i dj and K(G) = 2(n −1) n × Q j∈V dj 2. Let G be a cycle. For any i ∈V Ki(G) = K(G) = n 2n−1 3. Let G be a complete graph. For any i ∈V lim n→∞ Ki(G) 1 n = lim n→∞ K(G) 1 n = e Nodes with high cascade and contagion centrality have a lot of neighbors, who have fewer neighbors. However, our results highlight the tradeoffbetween cascade centrality and contagion centrality in networks. This should not be surprising. Cascade centrality depends directly on the degree of the node (Theorem 1) whereas contagion centrality does not. Let’s compare trees and complete graphs. In trees, the maximum cascade centrality of a node is n. Cascade centrality of any node in a large complete graph, on the other hand, is roughly √n. However, contagion centrality declines exponentially fast in trees, it declines at rate 1 n is complete graphs. Therefore, while possible cas-cades are “smaller” in complete graphs, the probability of an extreme outcome i.e. a contagion is relatively greater. In this sense, our framework supports observations that complete graphs are, in some sense, “robust but fragile” (Acemoglu et al., 2015).
Finally, we can use contagion centrality to express the probability of switching for any subset of nodes that are reachable from the seed.
Proposition 6. Consider a cascade process on G(V, E). The probability that a con-nected set of nodes V ′ ⊂V switches when i ∈V ′ is the seed is Y j∈V \V ′ 1 −|Nj(G) ∩V ′| dj " 1 nλ1 × . . . × λn−1 Q j∈V ′{i} dj # where λ1, . . . , λn−1 are non-zero eigenvalues of the Laplacian matrix of G′, subgraph induced by V ′ and the degrees of the nodes are calculated from G.
The first term expresses the probability that “none of the nodes that border with 18 V ′ switch”. So, for example, no neighboring leaves can remain not switched. If none of the nodes that border with V ′ switch, this creates a “blanket” between V ′ and the V \ V ′ ∪j∈V ′ Nj(G) and ensures that none of these remaining nodes switch. The last term is simply the probability that all of V ′ switch that we derive using Theorem 2.
5 Application: Pricing cascades In this section, we apply our simple model of cascades to pricing. Suppose that the seeder is in fact a profit-maximizing firm that is trying to spread its product. The firm selects one agent i ∈V as the seed and picks a uniform price 0 ≤ρ ≤1 for its product.
The seed adopts the product immediately and the firm makes no profit offhim. We assume that subsequent agents adopt the product at period t + 1 if: [1 −ρ] × [fraction of neighbors who adopted by t] ≥θi.
When ρ = 0 we recover our simple linear threshold model and when ρ = 1 no agent, except the seed, adopts the product. More generally, each agent’s demand for the product is downward sloping: The higher the price ρ = 1, the less likely (compared to the simple linear threshold model) is any agent to adopt the product for a given threshold.
When the seed is i and the price is ρ, the firm’s profit is π(i, ρ) = ρ× X j∈V {i} X P∈Pij (1 −ρ)|P| χP where the term in brackets is the “stochastic” cascade centrality, which we will return to later, and |P| is the length of path P. The profit-maximization problem for the firm is now a lot harder: it needs to pick the optimal node i∗and the optimal price ρ∗ simultaneously i.e. it is no longer sufficient to simply pick the node with the highest cascade centrality in order to maximize the size of the cascade. In general, the node with the highest cascade centrality is not the node that will maximize profit. Moreover, analytically solving the pricing problem for a general finite network is not possible.
Nevertheless, our setting can provide a number of illuminating observations.
Proposition 7. In any network and for any seed, the optimal price satisfies ρ∗≤1 2.
First, we easily show that the highest possible profit for a network of size n can be obtained in a star where is optimal price and seed are always the same.
19 Proposition 8. Maximum profit in any network with n nodes is 1 4 (n −1) and it is obtained in a star of order n. The optimal price is 1 2 irrespective of the order of the star.
Next, we consider a line network. Recall that every node in a line, except the leaves has a cascade centrality of 3. Our next result shows that the strictly optimal seed is, in fact, a penultimate node. Moreover, we can rank all the nodes according to their profitability.
Proposition 9. Consider a line network G with ordered nodes {1, 2, 3, ..., n} 1. The optimal seeds on any line are nodes i∗∈{2, n −1}. As n →∞, ρ∗→≈ .465571 and π(i∗, ρ∗) →≈.418588.
2. π(i, ρ∗) > π(j, ρ∗) > π(1, ρ∗) = π(n, ρ∗) for any 2 ≤i < j ≤⌊n/2⌋ 3. As n →∞and i ∈{1, n}, ρ∗→ √ 2 −1 and π(1, ρ∗) →( √ 2 −1)2.
4. For any seed, maximum profit is increasing in n.
The reason for this result follows the same intuition as for nodes that have high cascade centrality. The firm profits the most from seeding the node that has a lot of neighbors, but also one whose neighbors have few neighbors. In the case of the line, this distinction is stark and analytically tractable. In cycles, the optimal seeding problem becomes symmetical and is similar to seeding a leaf on a line.
Proposition 10. Consider {G(n)}n∈N+, a sequence of complete cycles of order n.
Then for a generic seed i ρ∗→ √ 2 −1 and π(i, ρ∗) →1 as n →∞.
We can apply this to a cycle of 3 to get ρ∗= 4 3 − √ 7 3 ≈0.451 and π(i, ρ∗) = 1 27 7 √ 7 −10 ≈0.3156. While the optimal price in a cycle and in a complete graph of thee is the same, optimal pricing in complete graphs looks radically different as the grow.
Proposition 11. Consider {G(n)}n∈N+, a sequence of complete graphs of order n.
Then for a generic seed i ρ∗→0 20 and π(i, ρ∗) →1 as n →∞.
The complete graph result can, in fact, be generalized to an r-regular tree.
Proposition 12. Consider an r-regular infinite tree. Then for a generic seed i ρ∗= 1 1 + √r and π(i, ρ∗) = r (√r + 1)2 It should be obvious that the last two results are completely congruent. As the regularity of the infinite tree increases, its optimal price tends to zero and its profit tends to 1, just like in a complete graph. The general, testable predictions of this model is that optimal price should fall in denser and larger networks, but the profit should rise. However, our results also highlight the subtle effects of the network structure on pricing cascading innovations.
6 Extensions Although the model we presented in this paper is rather stylized, it provides a lot of insight into the effect of network structure on cascades. We now turn to several ways to enrich and parameterize both the simple and the competition model for empirical applications (see, for example, Hodas and Lerman (2014)).
6.1 Homophily In our simple model all agents were identical. However, in many social networks there is substantial agent heterogeneity beyond the network structure. For example, neighbors can have heterogeneous impact on the probability of switching. This can be captured easily by directed, weighted networks Kempe et al. (2003). On the other hand, agents with similar thresholds might be more likely to be neighbors – this phenomenon is known as homophily McPherson et al. (2001). Our model can be extended to capture homophily in the following way: assign a parameter ζi ∈(0, 1) to each agent in such a way that “similar” agents have similar ζ. The homophily parameter linearly scales 21 down the probability of switching given any number of neighbors who have switched.
In order to adjust the cascade centrality measure, it is simply required to replace di with ζidi in every degree sequence product that involves agent i. All the results can go through with that amendment.
6.2 Susceptibility and network formation Suppose that we are interested in the probabiliy that a node switches but we do not know where the cascade process begins. An application of this could be a which agent expect a cascade following a period of network formation (e.g. Blume et al. (2011); Erol and Vohra (2014); Farboodi (2014)). To keep things simple, suppose that one node is seeded randomly. Then we can define the susceptibility of i in G as φi(G) = P j∈V Pi(G, {j}) n remembering that Pi(G, {i}) = 1. It turns out that the susceptibility of any node is related to the cascade centrality of all other nodes. Let us define for convenience χPii = 1.
Proposition 13. Susceptibility of node i is φi(G) = 1 din X j∈N X Pij∈Pij dj χPij Moreover, in any regular graph G φi(G) = Ci(G) n While the relationship between susceptibility and centrality can be quite compli-cated (even in trees where susceptibility of a node depends on the structure of the whole tree), in regular graphs, such as cycles or complete graphs, they are perfectly related: graphs (of the same size) in which nodes have higher cascade centrality are more susceptible.11 Let us briefly consider how susceptibility changes the outcomes of simple network formation games. Consider an extension of the setting of Jackson and Wolinsky (1996) (similar to Blume et al. (2011)) in which agents form a pairwise stable network in the first period and in the second period a cascade starting from a random seed hits the 11In general regular graphs, nodes will have different cascade centrality.
22 network. The agents receive a benefit from the network only if they do not switch (“fail”) after the cascade, however, the network formation costs are sunk.
We can write the expected utility ui(G) of an agent in G ui(G) = (1 −φi(G)) X j̸=i δtij −cdi where tij is the geodesic distance between i and j so the cost and benefits of links are exactly as in Jackson and Wolinsky (1996). Recall that when φi(G) = 0 for all i there are conditions on c and δ under which star networks are stable (Jackson and Wolinsky, 1996). However, when agents anticipate cascades, we get a contrasting result: Proposition 14. A star is never stable in the network formation game with cascades.
The simple proof shows that any two leaves in a star can add a link between them without increasing their probability of failure. To prevent them from doing that we require that δ < c. However, the centre can also add and remove links without affecting his probability of failure, but in order to encourage him to keep his links we require that δ ≥c.
This model can be developed further to find conditions under which complete net-works are stable and under what conditions stable networks are inefficient (in general they will be). Since there are many options for specifying the utility function and the structure of the cascades, we leave this for future work.
7 Conclusions This paper provided a simple framework for analyzing cascades in networks. We showed that when agents’ thresholds are drawn from a uniform distribution the cascade process can be expressed as a count of paths attenuated by their degree sequence products.
We offered two new tools for the analysis of cascades in networks: cascade centrality (that measures the expected) and contagion centrality. We show that there is a tradeoff between the two measures. We then applied our models to a setting of cascades pricing.
Our simple model can also be used to explore other questions, such as network formation in the presence of cascade and optimal network design. The model can also be taken to data. We offered some guidance on possible parameterization in Section 6. Parameters δ, c and ζ can, in principle, be estimated. There are several obvious directions for future work. First, one could look at dynamic, rather than static, pricing (Ajorlou et al., 2015). Second, it would be informative to understand how network 23 formation would play out in rich economic settings under cascade and contagion threats described by our model (Babus, 2013; Erol and Vohra, 2014; Farboodi, 2014).
24 Appendix A: Proofs Proofs of Propositions in Section 3 Proof of Proposition 1. First, we prove the following lemma, illustrated in the main text.
Lemma 1. Let {G(n)}n∈N+ be a sequence of star networks of order n ∈N + in which i is the center and the seed set is S0 ⊆V \ {i}, then Pi(G(n), S0) = |S0| di(G(n)) for any seed 0 ≤|S0| ≤n −1 on every G(n) if and only if Θi is uniformly distributed on [0, 1] almost everywhere.
Proof of Lemma 1. The “if” direction is trivial since for any star G(n) and 0 ≤|S0| ≤ n −1, under the property of the uniform distribution FU |S0| n −1 = |S0| n −1 and the probability of the center switching is therefore Pi(G(n), S0) = FΘi |S0| di(G(n)) = FΘi |S0| n −1 = FU |S0| n −1 = |S0| n −1 For the “only if” part, fix some |S0| ∈N+ and G(n), then as before Pi(G(n), S0) = FΘi |S0| di(G(n)) = FΘi |S0| n −1 Indeed, for each 1 ≤|S0| ≤n −1 on every G(n), we require that FΘi |S0| n −1 = |S0| n −1 Hence, for each 1 ≤|S0| ≤n −1 on every G(n), we require that: FΘi |S0| n −1 −FΘi |S0| −1 n −1 = 1 n −1 Therefore we require that Pr(a ≤x ≤b) = b −a for any rational a, b ∈[0, 1] s.t.
a < b. Hence, we require that Θi is uniform over every open interval (a, b) for any rational a, b ∈[0, 1] s.t. a < b (there are countably infinitely many such intervals).
Therefore, Θi is not uniform over a set of measure 0 so the distribution is uniform almost everywhere on [0, 1].
25 It remains to show that in order to calculate the switching probability of i, we need to average the switching probability of Ni(G) conditional on i not switching. We build on Claim 2.6 in Kempe et al. (2003). Without loss of generality, we can restrict ourselves to cases in which i in node that switches in the last round of the cascade process (what happens after i switches is irrelevant to i’s probability of switching).
Any live-edge path from the seeds that activates i must pass through at least one of the neighbors without passing through i first. Therefore, the probability of i switching must only depend on the probability of his neighbors switching, conditional on i not switching before. The probability that the live-edge path takes a step from a neigbor j ∈Ni(G) to i is 1 di irrespective of the path by from Lemma 1. The proposition follows immediately.
Proof of Proposition 2. Again, we build on Kempe et al. (2003). By Claim 2.6, the distribution of switches in an LTM process is the same as the distribution of switches under the independent cascade process with live-edges.
In an independent cascade process, by Claim 2.3, a node switches if and only if there is a live-edge path from some node in S0. By construction of live-edge paths (i) this path avoids other nodes in S0 (because nodes in S0 are already activated) and (ii) the path is from exactly one node in S0 (otherwise two path will activate one node). A node is activated by its neighbor along the live-edge path P ∈P∗with probability 1 di . Therefore, the probability that the nodes switches along this live-edge path is 1 χP . Hence, the probability that the node is activated by a path from a particular j ∈S0 is P P∈P 1 χP and by Claim 2.3 the probability that the node is activated by any one j ∈S0 is P j∈S0 P P∈P 1 χP .
Proof of Theorem 1. We first show that Ci(G) = di(G) + 1 when G is a tree. To save on notation, we often refer to Ci(G) as Ci.
We can show this by an induction on the number of nodes in the graph. For n = 1, Ci = 1 since it is an isolated node and there is no path in the graph. Now, fix N ≥1.
For any n ≤N, we have Ci = 1 + di. Now, consider n = N + 1. Fix a node i ∈V .
Then, we can decompose Ci as: Ci = 1 + X j∈Ni 1 dj · (Cj(G) −1).
However, Cj(G) can be re-expressed as Cj(G(j)) where G(j) is a branch of G starting from j. Now, each G(j) is a tree that has strictly smaller than n nodes. Hence, the induction hypothesis applies to each G(j), resulting in Cj(G) = Cj(G(j)) = dj + 1.
Therefore, we conclude that: Ci = 1 + X j∈Ni 1 dj · (Cj(G) −1) = 1 + X j∈Ni 1 dj · ((dj + 1) −1) = 1 + X j∈Ni 1 = di + 1.
Now that we have shown that when G is a tree Ci(G) = di(G) + 1 holds, let us use it a base step for the next induction hypothesis. What we want to show is that the 26 contribution from loops and paths add up to 1 + di: X j∈V {i} X P∈Pij 1 χP + X j∈V X L∈Lij 1 χL = 1 + di Suppose that Theorem 1 holds for a graph with C > 0 or fewer cycles. We want to show that Theorem 1 holds for a graph with C + 1 cycles.
The first two cases will allow us to generalize to deal with the most general Case (iii): • Case (i): The seed is on a unique one cycle (i.e. a closed loop). Create a graph G′ by removing all edges between seed i and Ni(G) (as well as i) and attaching a leaf seed to every Ni(G). The probability of switching for every node in G′ that is also in G has not changed, but the size of the total contribution has increased by di −1 because of the extra seeds. Consider G′′ di are di copies of G′ in which a unique seed of G′ is a seed. The size of the contribution in G′ (and G) is equal to the sum of the cascades in G′′ di. By the induction hypothesis (since there are now fewer than C cycles), each G′′ di has a contribution of 2 (since each seed has degree 1). Since there are di copies of it, the contribution of G is 2di −(di −1) = di + 1 as required. Note this argument also works even if there are cycles elsewhere in the graph (i.e. they do not include i).
• Case (ii): The seed is not on a cycle (but there are cycles elsewhere i.e. loops from i). Hence, the seed since on the intersection of one or more branches, each of which may lead to a cycle (if they do not, the branch is a tree and it’s contribution is immediately 2). In this case note, that the contribution can be broken down into a the contribution of a unique path P (branch) to cycle and the contribution of a cycle. Subtracting the contribution of the path P, “move” the seed from i along the path to the cycle. The contribution of the cycle is 2 - the same as in case (i) - by the induction hypothesis (discounting by 1 χP ).
• Case (iii): The seed is on multiple cycles and multiple branches.
Apply case (i) to all cycles, creating di −1 extra leaf seeds and case (ii) to all branches simultaneously.
Using Case (i) or Case (ii), we have that the contribution of each di such G′′ di with a single leaf seed is 2 by the induction hypothesis (as we have broken at least one cycle) because none of the probabilities of switching of any node have changed using our construction. Hence, once again we have 2di −(di −1) = di + 1 as required.
Proof of Proposition 3. The first statement is completely obvious and can be argued by contradiction.
That the maximum cascade is increasing up to m = n −1 in m is also obvious (create the largest possible star at each stage).
To show the maximum cascade is decreasing from m = n −1 in m, first note that in order to maximize the cascade with m ≥n −1 edges it is necessary to have a node with n −1 edges. If there is no such node, then it must be possible to rewire some 27 edges to this node. At each such rewiring, we will be increasing the cascade centrality of this node by Theorem 1 because its degree will be increasing and the contribution of loops will be falling. Therefore, for any m ≥n −1, the node that has the highest cascade centrality has degree n −1. Hence, any extra edge will necessarily create more loops (even allowing for rewiring) and by Theorem 1 the maximum cascade centrality will fall in m.
Proof of Theorem 2. Any realization of the deterministic cascade process κ ∈K on Gθ is charactersized by a unique sequence of additional switches κ(Gθ, S0) = {S0, S1, S2, . . .}.
Any κ(Gθ, S0) can be represented by a directed tree τ(G, S0) ∈T (G, S0) rooted at s on a subgraph of Gθ, for which there is an edge from i to j if and only if i ∈St and j ∈St+1. There are many trees that may represent the same sequence, but a given tree can represent at most one sequence.12 V ′ switching can be represented by any such tree that includes V ′.
Now let’s consider the probability of this process when thresholds are uniform ran-dom.
Our aim is to show that the probability of precisely nodes V ′ switching can be decomposed into tree in which all incoming edges of i have weight 1 di for all i.
For a given κ, we say node is exposed in period t where t = mint∈0,1,2,...{t|St ∈ κ, St ∩Ni(G) ̸= ∅}. Given a node is exposed in t, there are two cases: 1. i ∈St+1: St ∩Ni(G) = l (l ≥1). Using Assumption 1, probability that i ∈St+1 is l di . τ(G, S0) representing κ can be extended with one of l possible edges. We attach a weight of 1 di any such extra edge.
2. i ∈St+r: St ∩Ni(G) = l (l ≥1) and St+r ∩Ni(G) = l + a. If a node switches when there are a additional neighbors who switch, then we know that most l di < θ ≤l+a di . Using Assumption 1, the probability of the threshold being in that region is a di . τ(G, j) representing κ in t + r −1 can be extended with one (otherwise there is a cycle) of l possible edges. We attach a weight of 1 di any such extra edge.
Therefore each incoming edge into i in the tree has weight 1 di and the root has no incoming edges. Therefore, the weight of the tree is 1 χτ Contagion is represented by a spanning tree rooted in i, the weight of which is 1 Q j∈V {i} dj .
(The corresponding probability of the cascade process represented by all these trees is this probability times the number of trees that represent this cascade process). By Kirchoff’s Theorem, the number of spanning trees of G (rooted in any node, since the 12Suppose not. A node switches if and only if its in the tree. A switched node can be member of only one tree τ by construction. Then suppose that a tree represents two cascade processes: in one agent i switches at time t and in the other i switches at time t′. But by construction the period in which i switches is equal to the length of the unique path in that tree from sτto i. Therefore, t = t′, a contradiction.
28 number of spanning trees is same) is equal to the product of non-zero eigenvalues of the Laplacian over n. The result follows immediately.
Proofs of Propositions in Section 4 Proof of Corollary 1. 1. See first part of proof of Proposition 2.
2. Fix a node i. Using Theorem 1, it suffices to subtract two loops from 1 + 2 = 3, each contributing 1/2n−1. Therefore, Ci(G) = 3 −2 · 1/2n−1 = 3 −1/2n−2. The limit result follows immediately.
3. Fix i and j ̸= i. There are P(n −2, k −1) distinct paths of length k from i to j, each of which has a degree sequence product of (n −1)k; the number of ways of permuting k −1 nodes upon selecting them out of n−2 possible candidates is precisely P(n −2, k −1).
We can therefore re-write the cascade centrality of any node in a complete graph G(n) as: Ci(G(n)) = 1 + n−1 X i=0 i Y j=0 (1 −j/n −1) One can recognize this as the Ramanujan Q(n)-function. Following previous results, Flajolet et al. (1995) show that Q(n) admits a full asymptotic expansion in descending powers of √n, so: Ci(G(n)) ∼1 + rπn 2 −1 3 + 1 12 r π 2n − 4 135n + . . .
This gives us the required result: lim n→∞ Ci(G(n)) √n = rπ 2 .
Note also that we actually obtain an even tighter results: as n →∞, we have that Ci(G(n)) →pπn 2 + 2 3.
Proof of Proposition 4. We define a “regular” configuration model as follows: Consider a degree sequence d = (d1, . . . , dn) of non-negative integers with an even sum and length n. Gd is a random (simple) graph with degree sequence d and its asso-ciated random configuration multigraph is G∗ d created by uniformly randomly match-ing on the degree half-edges (. It can be shown that the degree distribution in the configuration model converges in probability to the degree distribution of Gd.
Let D = (r0, r1, . . .) ∈D be a probability distribution on non-negative integers, such that 0 < E(D) < ∞. We say that the probability of realisation k denote random variable δ with probability distribution D is rk = Pr(δ = k). Finally, let nk(d) = |{j : dj = k}| denote the number of times a particular degree i occurs in d and m(d) = 1 2 Pn j=1 dj as number of edges.
29 We impose two regularity conditions on D. We say that dn →D if lim n→∞ nk(dn) n = rk for each k and m(dn) n →E(D) 2 = 1 2 ∞ X k=0 krk as n →∞.
We say that a configuration model is “regular” if we let D ∈D and assume that dn →D.
This is enough to give us a general result for cascade centrality on random graphs.
Given a graph G, for i ∈V (G) and λ ≥0, let Γ≤λ(i) = ΓG ≤λ(i) denote the subgraph of G induced by the vertices within (graph) distance λ of i, regarded as a rooted graph with root i.
Lemma 2 (Corollary 5 in Bollob´ as and Riordan (2015)). Suppose that dn →D and let λ ≥1 be constant. Let i be a vertex of G = G∗ dn chosen uniformly at random. Then with high probability13 the neighborhood Γ≤t(i) of i in G is a tree.
The results follows immediately from Corollary 1 and from Lemma 2 by observing that Lemma 2 holds for any constant λ ≥1.
Proof of Proposition 5. Loose bounds on the number of paths cn (self-avoiding walks) of length n in an infinite d −D hypercubic lattice, see equation 1.1.1 in Madras and Slade (1993): dn ≤cn ≤2d(2d −1)n−1 The degree of each node in this lattice is 2d giving us for a generic node i: 1 + r r −d ≤Ci ≤min 1 + 2d, 1 + 2d 2d −1 2d r −(2d −1) 1 + 2d 2d −d ≤Ci≤min 1 + 2d, 1 + 4d2 2d −1 3 ≤Ci ≤1 + 2d since d > 1.
Proof of Corollary 2. 1. Tree: Ci(G) follows immediately from Proposition 2.
K(G) = h 1 Πi∈V \s1di + . . . + 1 Πi∈V \sndi i n = d1 + . . . + dN n × Πi∈V di = P i∈V di n × Πi∈V di = 2(n −1) n × Πi∈V di 13Given a sequence En of events, En holds with high probability if Pr(En) →1 as n →∞.
30 2. For any cycle, there are n spanning trees rooted in any node. Since di = 2 for all i ∈V Ki(G) = K(G) = n × 2n n × 2n = n 2n−2 3. Using the Cayley formula: Ki(G) = K(G) = nn−2 (n −1)n−1 = = nn−1 n × (n −1)n−1 = 1 n n −1 + 1 n −1 n−1 = 1 n 1 + 1 n −1 n−1 and the limit as n →∞of the term in brackets is e.
Proofs of Propositions in Section 5 Proof of Proposition 7. Note that ρ(1−ρ) is strictly decreasing function for ρ ∈[1/2, 1].
Also, a1 +a2(1−ρ)+· · ·+an(1−ρ)n is strictly decreasing function for ρ ∈[1/2, 1].
Proof of Proposition 8. Fix a seed node i. Then, a profit can be written as π(i, ρ) = ρ(a1(1 −ρ) + · · · + aD(1 −ρ)D) ≤ρ(1 −ρ) PD i=1 ai = ρ(1 −ρ)(Ci −1).
Now, we already know that Ci(G) ≤n, with equality if and only if G is a star.
Hence, center i∗gives the highest profit. So, π(i∗, ρ, n) = (1 −ρ) (n −1) ρ ∂π ∂ρ = (n −1) (1 −ρ −ρ) = 0 ρ∗ = 1 2 Hence π(i∗, ρ∗, n) = 1 4(n −1) Proof of Proposition 9. We prove each item separately.
1. Profit ranking of nodes and optimal seed i ∈{1, 2, 3, . . .} Note that as n →∞ π(2, ρ) = ρ (1 −p) | {z } node 1 + (1 −ρ) 2 | {z } node 3 + 1 −ρ 2 2 | {z } node 4 + . . .
31 and π(3, ρ) = ρ (1 −ρ)2 2 | {z } node 1 + (1 −ρ) 2 | {z } node 2 + (1 −ρ) 2 | {z } node 4 + 1 −ρ 2 2 | {z } node 5 + . . .
Note that π(2, ρ) > π(3, ρ) for ρ ∈(0, 1). So, π(3, ρ∗ 3) < π(2, ρ∗ 3) ≤π(2, ρ∗ 2) where ρ∗ i is the optimal price when ith node is a seed. So, π(2, ρ∗ 2) > π(3, ρ∗ 3).
Similarly, π(4, ρ) = ρ (1 −ρ)3 4 | {z } node 1 + (1 −ρ)2 4 | {z } node 2 + (1 −ρ) 2 | {z } node 3 + (1 −ρ) 2 | {z } node 5 + 1 −ρ 2 2 | {z } node 6 + . . .
Hence, π(4, ρ) < π(3, ρ). So, similarly proceeding, we find that π(4, ρ∗ 4) < π(3, ρ∗ 3).
Therefore, π(i, ρ∗ i ) > π(i + 1, ρ∗ i+1) for each i < n/2.
Finally, π(1, ρ) = ρ (1 −ρ) 2 | {z } node 2 + 1 −ρ 2 2 + . . .
| {z } node 3 Hence, π(1, ρ) < π(i, ρ) for each i = 2, · · · , ⌊n/2⌋. Thus, π(1, ρ∗ 1) is the smallest possible profit.
2. Optimal price at the optimal seed Seeding i∗∈{2, n −1} and as n →∞ π(i∗, ρ) = ρ · ((1 −ρ) + (1 −ρ) 2 + 1 −ρ 2 2 + · · · ) = 2ρ 1 + ρ −ρ2.
Its derivative is: −2(ρ3 + 2ρ2 + ρ −1) (1 + ρ)2 .
Its real root is: ρ∗= 1 6(−4 + 3 √ 4 · 3 q 29 −3 √ 93 + 3 √ 4 · 3 q 29 + 3 √ 93 ≈.465571 Hence, the optimal profit is ≈.418588.
3. Optimal price at the leaf seed For a leaf seed i and a line of length n π(i, ρ, n) = ρ × 2 −2ρ 1−ρ 2 n+1 1 + ρ −1 32 As n →∞ π(i, ρ) = ρ × 2 1 + ρ −1 ∂π ∂ρ = −ρ2 + 2ρ −1 (ρ + 1)2 = 0 ρ∗ = (√ 2 −1 > 0 −1 − √ 2 < 0 SOC is < 0.
4. Immediate from the expression of π(i, ρ) that is increasing in n.
Proof of Proposition 10. For a generic seed i and a cycle of length n π(i, ρ, n) = ρ × 2 × 1 − 1−ρ 2 n 1 −1−ρ 2 −1 As n →∞ π(i, ρ) = ρ × " 2 1 −1−ρ 2 −1 # ∂π ∂ρ = −2(ρ2 + 2ρ −1) (ρ + 1)2 = 0 ρ∗ = (√ 2 −1 > 0 −1 − √ 2 < 0 SOC is < 0. Note that that asymtotically this it the same price at the optimal price for a leaf seed on an infinite line.
Proof of Proposition 11.
Lemma 3. Fix 0 < ρ < 1 and let Q(n, ρ) = ∞ X k=0 ak where a0 = 1, a1 = 1 −1 n (1 −ρ), a2 = 1 −1 n 1 −2 n (1 −ρ)2, a3 = 1 −1 n 1 −2 n 1 −3 n (1 −ρ etc. Then, Q(n, p) →1 ρ as n →∞.
Proof of Lemma 3. The kth term in the sum is: 33 ak = 1 −1 n · · · · · 1 −k n (1 −ρ)k.
First of all, ak ≤1 · · · · · 1 · (1 −ρ)k = (1 −ρ)k .
So, Q(n, ρ) ≤ ∞ X k=0 ak = ∞ X k=0 pk = 1 ρ .
Also, ak ≥ 1 −k n (1 −ρ)k Hence, Q(n, ρ) ≥ k X i=0 1 −k n i (1 −ρ)i.
Now, as n →∞, we get k X i=0 (1 −ρ)i for a fixed constant k. Now, Q(n, ρ) ≥ k X i=0 (1 −ρ)i for any k. Hence, because Q(n, p) ≤1 ρ, it follows that Q(n, ρ) →1 ρ by the sandwich theorem as n →∞.
By stochastic cascade centrality tends to set 1 ρ so there are 1 ρ −1 extra adopters i.e. the size of the cascade no longer depends on n, but only on the price. Hence, as n →∞the firm’s profit tends to ρ(1 ρ −1) = 1 −ρ and, since by Lemma 3 ρ is bounded away from 0, profit is maximized when ρ is arbitarily close to 0.
Proof of Proposition 12. For a generic seed i, π(i, ρ) = ρ · 1 −ρ r · r + (1 −ρ)2 r2 · r(r −1) + (1 −ρ)3 r3 · r(r −1)2 · · · = ρ · r r −1 · (1 −ρ) · r −1 r + (1 −ρ)2 · (r −1)2 r2 + · · · = ρ · r r −1 · 1 1 −(1 −ρ)(1 −1/r) −1 34 Hence, ∂π ∂p = r((r −1)ρ2 + 2ρ −1) ((r −1)ρ + 1)2 , which which means that ρ∗= 1 1 ± √r. Since ρ∗∈[0, 1], we have that ρ∗= 1 1 + √r.
And, π(i, ρ∗) = r (√r + 1)2 .
Proofs of Propositions in Section 6 Proof of Proposition 13 . Define 1 χPii = 1.
φi(G) = 1 n X j∈N X Pji∈Pji 1 χPji = 1 n X j∈N X Pji∈Pji 1 dk × . . . × di = 1 n X j∈N X Pji∈Pji dj dj × . . . × dk × . . . × di = 1 n X j∈N X Pji∈Pji 1 di × dj dk × . . . × dj = 1 din X j∈N X Pij∈Pij dj χPij If G is regular then di = dj = d hence φi(G) = 1 dn X j∈N X Pij∈Pij d χPij = d dn X j∈N X Pij∈Pij 1 χPij | {z } Ci(G) 35 Proof of Proposition 14 . Probability of failure for a center link where there are n nodes and m edges: φi(G) = 1 n + 1 n hm m i = 2 n which is independent of m. Hence, the center will want to keep all his edges if δ ≥c (otherwise none).
Probability of faiure of a leaf in a star with n −1 edges φi(G) = 1 n + 1 n + 1 n n −2 n −1 = 3n −4 n(n −1) Probability of failure of leaves in a star that have just added an edge between them: φi(G) = 1 n + 1 n 1 2 + 1 4 + n −3 n 1 2(n −1) + 1 4(n −1) + 1 n 1 2(n −1) + 1 2 = 3n −4 n(n −1) Hence, the leaves won’t form an edge only if c < δ. A contradiction.
36 (a) n = 10 (b) n = 20 (c) n = 50 Figure 8: Cascade centrality vs. degree centrality (a) n = 10 (b) n = 20 (c) n = 50 Figure 9: Cascade centrality vs. Katz centrality Appendix B: Simulations We considered connected networks of size n = 10, 20, 50. We uniformly randomly se-lected a number of edges n −1 ≤m ≤ n 2 for each graph size and then randomly selected a network with m edges with a nonsingular adjacency matrix. We then ran-domly selected one node from this network. We repeated this 1000 times. For Katz centrality, we fixed α to be half of the largest eigenvalue of the adjacency matrix. k-order approximation means we apply Theorem 1 only to paths of length k. In networks of size 10, cascade centrality is computed exactly; in networks of size 20, it is computed up to paths of length 5; in networks of size 50, it is computed up to paths of length 4.
37 (a) n = 10 (b) n = 20 (c) n = 50 Figure 10: Cascade centrality vs. eigenvector centrality (a) n = 10 (b) n = 20 (c) n = 50 Figure 11: Cascade centrality vs. 1st order approximation (a) n = 10 (b) n = 20 (c) n = 50 Figure 12: Cascade centrality vs. 2nd order approximation 38 (a) n = 10 (b) n = 20 (c) n = 50 Figure 13: Cascade centrality vs. 3rd order approximation 39 Appendix C: Macroscopic comparative statics Figure 14 illustrates that increased clustering that either increase or decrease average cascade centrality across all nodes (and therefore the cascade centrality of at least one node).
Higher clustering ⇒lower cascade centrality. Graphs A and B have 4 nodes and 4 edges. The global clustering coefficient in A is 0 and in B is 0.6. Average cascade centrality across all nodes is 2.75 in A and 22 3 in B.
Higher clustering ⇒higher cascade centrality. Graphs C and D have 5 nodes and 5 edges. The global clustering coefficient in C is 3 8 and in D is 3 7. Average cascade centrality across all nodes is 2.75 in C and 27 9 in D.
Higher clustering ⇒higher cascade centrality for a node of a fixed degree. Graphs E and F have 5 nodes and 5 edges. The global clustering coefficient in E is 3 7 and in F is 1 2. The cascade centrality of node i is 1 8 9 in E and 111 12 in F.
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7220 | https://fiveable.me/theoretical-statistics/unit-6/interval-estimation/study-guide/8rlSOuAt88CIpdQs | Interval estimation | Theoretical Statistics Class Notes | Fiveable | Fiveable
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📈Theoretical Statistics Unit 6 Review
6.4 Interval estimation
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Theoretical Statistics
Unit 6 – Statistical inference
Topic: 6.4
📈Theoretical Statistics Unit 6 Review
6.4 Interval estimation
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
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📈Theoretical Statistics
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Probability Theory Basics
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Multivariate distributions
Limit Theorems and Convergence in Statistics
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6.1 Population and sample
6.2 Sampling distributions
6.3 Point estimation
6.4 Interval estimation
6.5 Maximum likelihood estimation
Estimation theory
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Bayesian statistics
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Interval estimation is a crucial concept in Theoretical Statistics, providing a range of plausible values for population parameters based on sample data. It offers a more comprehensive approach than point estimates, accounting for uncertainty and sampling variability in statistical inference.
Confidence intervals, a key tool in interval estimation, quantify the precision of parameter estimates. By selecting appropriate confidence levels and considering factors like sample size and population variability, researchers can construct intervals that balance certainty with precision in their statistical analyses.
Definition of interval estimation
Interval estimation provides a range of plausible values for a population parameter based on sample data
Plays a crucial role in inferential statistics by quantifying uncertainty in parameter estimates
Bridges the gap between point estimates and the true population value in Theoretical Statistics
Point estimates vs interval estimates
Point estimates offer a single value to represent a population parameter
Interval estimates provide a range of values with a specified level of confidence
Interval estimates account for sampling variability and offer more informative conclusions in statistical inference
Interpretation of confidence intervals
Confidence intervals represent a range of values likely to contain the true population parameter
95% confidence interval interprets as 95% of similarly constructed intervals would contain the true parameter value
Provides a measure of precision and reliability for parameter estimates in statistical analyses
Confidence level selection
Determines the probability that the interval contains the true population parameter
Common confidence levels include 90%, 95%, and 99%
Higher confidence levels result in wider intervals, balancing certainty with precision
Confidence interval width
Reflects the precision of the estimate and inversely related to sample size
Narrower intervals indicate more precise estimates of the population parameter
Affected by factors such as sample variability and chosen confidence level
Construction of confidence intervals
Involves calculating a point estimate and adding/subtracting a margin of error
Utilizes the sampling distribution of the statistic to determine the interval bounds
Requires consideration of the underlying distribution and sample size
Normal distribution method
Applies when the sampling distribution is approximately normal or for large sample sizes
Uses the standard normal distribution (z-distribution) to calculate interval bounds
Formula: CI=θ^±z α/2⋅S E(θ^)\text{CI} = \hat{\theta} \pm z_{\alpha/2} \cdot SE(\hat{\theta})CI=θ^±z α/2⋅SE(θ^)
$\hat{\theta}$: point estimate
$z_{\alpha/2}$: critical value from standard normal distribution
$SE(\hat{\theta})$: standard error of the estimate
Student's t-distribution method
Used for small sample sizes or when population standard deviation is unknown
Accounts for additional uncertainty by using the t-distribution
Formula: CI=θ^±t α/2,d f⋅S E(θ^)\text{CI} = \hat{\theta} \pm t_{\alpha/2, df} \cdot SE(\hat{\theta})CI=θ^±t α/2,df⋅SE(θ^)
$t_{\alpha/2, df}$: critical value from t-distribution with degrees of freedom (df)
Bootstrap method
Non-parametric approach for constructing confidence intervals
Involves resampling with replacement from the original sample
Useful when distributional assumptions are not met or for complex statistics
Types of confidence intervals
Various types of confidence intervals exist for different statistical scenarios
Selection depends on the nature of the data and research question
Each type addresses specific inferential needs in statistical analysis
One-sample intervals
Used to estimate a single population parameter (mean, proportion, variance)
Compares sample statistics to hypothesized population values
Applies in scenarios such as estimating average height of a population
Two-sample intervals
Estimates the difference between two population parameters
Used for comparing means or proportions from two independent groups
Applications include comparing treatment effects in clinical trials
Paired sample intervals
Estimates the mean difference between paired observations
Useful for before-and-after studies or matched-pairs designs
Accounts for the correlation between paired measurements
Factors affecting interval width
Understanding these factors helps in interpreting and designing studies
Interval width directly impacts the precision of parameter estimates
Researchers can manipulate these factors to achieve desired levels of precision
Sample size
Larger sample sizes generally lead to narrower confidence intervals
Increases precision of estimates by reducing standard error
Relationship follows the square root of n (√n) in many cases
Population variability
Greater variability in the population leads to wider confidence intervals
Measured by standard deviation or variance of the population
Impacts the standard error of estimates and thus interval width
Confidence level
Higher confidence levels result in wider intervals
99% confidence interval is wider than a 95% interval for the same data
Reflects the trade-off between certainty and precision in estimation
Assumptions for interval estimation
Validity of interval estimates relies on certain statistical assumptions
Violation of assumptions can lead to inaccurate or misleading intervals
Important to verify assumptions before interpreting confidence intervals
Sampling distribution assumptions
Many interval estimation techniques assume a normal sampling distribution
Central Limit Theorem often invoked for large sample sizes
Normality assumption critical for z-intervals and t-intervals
Population distribution assumptions
Some methods assume the population follows a specific distribution (normal)
Robustness to violations varies depending on the estimation method
Non-parametric methods available when distributional assumptions are not met
Applications of interval estimation
Interval estimation finds wide application across various fields
Provides a measure of uncertainty crucial for informed decision-making
Enables researchers to make probabilistic statements about population parameters
Quality control
Used to establish tolerance limits for manufacturing processes
Helps in assessing whether products meet specified quality standards
Allows for estimation of defect rates within a specified confidence level
Medical research
Estimates treatment effects and drug efficacy in clinical trials
Provides ranges for important health indicators (blood pressure, cholesterol)
Aids in determining clinically significant differences between treatments
Political polling
Estimates voter preferences with a margin of error
Accounts for sampling variability in survey results
Helps predict election outcomes and gauge public opinion
Limitations of interval estimation
Understanding limitations crucial for proper interpretation and application
Awareness of potential pitfalls helps avoid misuse of confidence intervals
Researchers should consider these limitations when designing studies and analyzing data
Sensitivity to assumptions
Violation of underlying assumptions can lead to inaccurate intervals
Non-normality can affect the validity of traditional confidence intervals
Robust methods or transformations may be necessary when assumptions are violated
Misinterpretation risks
Common misconception that the interval contains the parameter with the stated probability
Incorrect interpretation of non-significant results as proof of no effect
Overemphasis on statistical significance rather than practical importance
Relationship to hypothesis testing
Confidence intervals and hypothesis tests are complementary inferential tools
CI can be used to conduct two-sided hypothesis tests at the corresponding significance level
Non-overlapping CIs for two groups indicate a significant difference at the chosen level
Bayesian vs frequentist intervals
Frequentist intervals (confidence intervals) based on long-run frequency interpretation
Bayesian intervals (credible intervals) incorporate prior information and posterior probabilities
Bayesian approach allows for direct probability statements about parameters
Advanced interval estimation techniques
Go beyond basic methods to address complex estimation problems
Provide more accurate or appropriate intervals in specific situations
Require deeper understanding of statistical theory and computational methods
Asymptotic methods
Based on large-sample approximations of sampling distributions
Utilize central limit theorem and other asymptotic results
Include techniques like the delta method for functions of parameters
Exact methods
Provide precise intervals without relying on asymptotic approximations
Useful for small sample sizes or discrete distributions
Include methods like exact binomial intervals for proportions
Nonparametric methods
Do not assume a specific underlying distribution for the population
Include techniques like bootstrap and jackknife for interval estimation
Robust to violations of distributional assumptions in parametric methods
6.3 BackNext 6.5
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7221 | https://www.geeksforgeeks.org/physics/calculating-stopping-distance-and-reaction-time/ | Calculating Stopping Distance and Reaction Time
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When an object gains kinetic energy and changes position, it is known to be in motion. There are three types of motion possible, one dimensional, two-dimensional, and three-dimensional motion. One dimensional motion is the object moving in a straight line, two-dimensional motion is when an object moves covering two axes (x-y, y-z, z-x axis), three-dimensional motion is when an object moves in all three x-y-z directions. The motion of a body is observed w.r.t a frame of reference, on a graph, the origin is denoted as the frame of reference. Let's learn about the Motion in a straight line in more detail,
Motion in a Straight Line
Motion in one direction or motion in 1-dimension is the motion in a straight line. This type of motion is also known as Rectilinear motion. When the object has no kinetic energy and no external force is applied on the object, it remains at rest. When some external force is applied to the object and the object gains motion, it starts to move, if the motion occurs in one direction, it is known as rectilinear motion.
Uniform Motion
When the object travels with the same velocity throughout its motion, the motion is uniform. The initial and final speeds are the same in this case and the speed is given as,
Speed = Distance/Time
Non-Uniform Motion
When the object does not have the same velocity throughout its motion, it may increase its velocity or decrease it, such type of motion is known as Non-uniform motion. If the velocity keeps on decreasing, the object is decelerating, and if the velocity keeps on increasing, the object is accelerating. The formulae for non-uniform motion are given by Newton's equation of motion.
First equation of motion ⇢ v = u + at
Second Equation of motion ⇢ S = ut + 1/2(at2)
Third Equation of motion ⇢ v2 = u2 + 2as
Some very interesting concepts to see in Non-uniform motion are Stopping distance and Reaction time. Let's learn about these two topics in more detail,
Reaction Time
Reaction time can be defined as the time taken by a human to react or respond to a sudden change or stimulus. When people drive/ride their vehicle on the road, and suddenly an object comes in front of the road, now in order to stop the vehicle, the brain first has to acknowledge the hurdle on the way and warn the body to respond to the danger, the time taken to do this process is nothing but Reaction time.
Not only on road, normally reaction time can be analyzed too. Imagine holding a pen in hand at a certain height and asking a friend to instantly catch the pen as soon as it is dropped, it is obvious that the pen will fall a little before the friend can catch it. It is due to the fact that the friend is not aware of the time of the drop of a pen, hence, when the pen is dropped, he may take some time to respond to that, this is nothing but reaction time taken by the friend.
Reaction time depends on,
The person taking the decision.
The situations/surroundings in which the reaction time is taking place.
Stopping Distance
When an object moves with a certain velocity and breaks are applied, the object takes some time to come to a stop, it is obvious that at this time, the object must have covered a certain distance, this distance is known as the Stopping distance. Imagine the same scenario as before, people driving/riding a vehicle and an object coming in front, after analyzing that brakes are needed to be applied, when the brakes are actually applied, it takes a certain time for the vehicle to come to rest, the distance covered in that time is Stopping Distance.
Stopping distance depends on:
The velocity of the vehicle.
The stopping capacity/braking capacity of the vehicle, that is, how fast the vehicle can decelerate.
Deriving Formula for Stopping distance
At the stopping point, the vehicle has applied the brakes, and now it has to come to a stop. The final velocity will become zero, hence, v = 0 is taken. Now, applying the third equation of motion in this case of non-uniform motion,
v2 = u2 + 2aS
0 = u2 + 2aS
S = -u2/2a
Now, it is important to notice that a is taken in negative since the vehicle is decelerating. Hence, the stopping distance will finally come out to be positive.
Deriving formula for Reaction time
Imagine the case of a free-falling object and the time to react and catch it in order to find the reaction time. Keeping in mind that the initial velocity will be zero since the object is freely falling. Applying the second equation of motion in order to find the Reaction time of the vehicle,
S = ut + 1/2 (gt2)
u = 0m/sec
S = 1/2 (9.8 × t2)
t2 = S × 0.204
t2 = 0.204S
Sample Problems
Question 1: A car has a velocity of 20m/sec and the braking capacity of the car is given as 10m/sec2. What will be the Stopping distance for the car?
Solution:
Since the acceleration occurs in negative direction, a = -10m/sec2
Applying formula for stopping distance,
S = -u2/2a
S = -(20)2/2 × (-10)
S = 400/20
S = 20 meters
Question 2: How will the stopping distance be affected if the velocity of the vehicle is doubled?
Solution:
Imagine in the first case, the velocity of the vehicle was v m/sec and the acceleration was a m/sec2.
Now, stopping distance in the first case is given as,
S1 = -v2/ 2a
In the second case, when the velocity is doubled, v'= 2v m/sec, acceleration = a m/sec2.
Stopping distance is given as,
S2 = - (2v)2/2a
S2 = -(4v)/ 2a
S2 = 4S1
Hence, if the velocity is doubled, the stopping distance increases 4 times.
Question 3: A vehicle has a velocity of 25.5m/sec and the braking capacity of the vehicle is given as 16.5m/sec2. What will be the Stopping distance for the car?
Solution:
Since the acceleration occurs in negative direction, a = -16.5m/sec2
Applying formula for stopping distance,
S = -u2/2a
S = -(25.5)2/2 × (-16.5)
S = 650.25/33
S = 19.704 meters
Question 4: A car has a velocity of 19m/sec and the braking capacity of the car is given as 11m/sec2. What will be the Stopping distance for the car?
Solution:
Since the acceleration occurs in negative direction, a = -11m/sec2
Applying formula for stopping distance,
S = -u2/2a
S = -(19)2/2 × (-11)
S = 361/22
S = 16.409 meters
Question 5: An object is dropped and is now free-falling under gravity. The Stopping distance is given as 0.5 meters for the object. Find out the reaction time neglecting other factors.
Solution:
The stopping distance, S = 0.5 m
The reaction time is given as,
t2 = 0.204S
t2 = 0.204 × 0.5
t2 = 0.102
t = 0.319 sec
A
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7222 | https://courses.lumenlearning.com/oldwestbury-wm-macroeconomics/chapter/defining-money-by-its-functions/ | 6 Money and Banking
Defining Money by Its Functions
Learning Objectives
Explain the functions of money
Contrast commodity money and fiat money
Barter and the Double Coincidence of Wants
Money for the sake of money is not an end in itself. You cannot eat dollar bills or wear your bank account. Ultimately, money is only useful because you can exchange it for goods and services. As the American writer and humorist Ambrose Bierce (1842–1914) wrote in 1911, money is a “blessing that is of no advantage to us excepting when we part with it.” Money is what people regularly use when purchasing or selling goods and services; thus for something to be considered money, it must be widely accepted by both buyers and sellers. This concept of money is intentionally flexible, because money has taken a wide variety of forms in different cultures.
To understand the usefulness of money, we must consider what the world would be like without money. How would people exchange goods and services? Economies without money typically use the barter system. Barter—literally trading one good or service for another—is highly inefficient for conducting transactions. In a barter economy, an exchange between two people requires a double coincidence of wants, which means that what one person wants to buy is exactly what the other person wants to sell. This is harder than it sounds.
Suppose an accountant wants a new pair of shoes. The accountant doesn’t just need to find someone who has a pair of shoes in the correct size to sell, but they have to find a person who will also be willing to exchange the shoes for what the accountant has to offer, namely accounting services. Trades like these are likely to be difficult to arrange. Now imagine how this would work in a complex, modern economy, with its extensive division of labor that involves thousands upon thousands of different jobs and different goods & services. The number of transactions that end up taking place is likely to be much smaller than in an economy with money.
Another problem with the barter system is that it does not allow us to easily enter into future contracts for the purchase of many goods and services. For example, if the goods are perishable it may be difficult to exchange them today for other goods in the future. Imagine a farmer wanting to buy a tractor in six months using a fresh crop of strawberries harvested today. Because the strawberries won’t last, such a transaction is unlikely to occur.
While a barter system might work adequately in small economies, it will limit these economies’ growth. The time that individuals would otherwise spend producing goods and services and enjoying leisure time would instead be spent bartering.
Functions of Money
Money solves the double coincidence of wants problem. First, since money is generally accepted as a means of payment (or medium of exchange), the accountant can pay for new shoes with money, which the shoe seller is willing to accept (even if they don’t need accounting services) since he or she can use the money to purchase something they do need.
Second, people are willing to sell something for money, even if they have no immediate need to purchase something else, because money serves as a store of value. A store of value is anything that holds value. Some things are better stores of value than other things. Real estate has traditionally been a good store of value since it tends to increase in value over time. Shoes are not a particularly good store of value, because they wear out as you wear them, and even if you don’t, styles change over time making what used to be a stylish pair of shoes, nothing special and thus worth less today. Money doesn’t have to be a perfect store of value to be acceptable. In an economy with inflation, money loses some buying power each year, but it remains money.
Third, money serves as a unit of account, which means that it is the ruler by which other economic values are measured. If there were no unit of account, the price of every good or service would have to be expressed in terms of the price of every other good and services. What you pay would depend on what you had to sell! Businesses would have to keep track of the value of everything someone might sell in order to be able to decide on a price for their products. Money solves the problem by acting as a common denominator, an accounting method that simplifies thinking about trade-offs.
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Learn more about the functions of money in this video clip.
One additional function of money is that it must serve as a standard of deferred payment. This means that if money is usable today to make purchases, it must also be acceptable for contracts signed today that will be paid in the future. Loans and future agreements are stated in monetary terms and the standard of deferred payment is what allows us to buy goods and services today and pay in the future.
What is Money?
Money is anything that can serve all of these functions— it is a medium of exchange, a store of value, a unit of account, and a standard of deferred payment.
Commodity versus Fiat Money
Money has taken a wide variety of forms in different cultures. Gold, silver, cowrie shells, cigarettes, and even cocoa beans have been used as money. These items are examples of commodity money, which means they also have a value from use as something other than money. Gold, for example, has been used throughout the ages as jewelry or art, as well as money. Gold is a good conductor of electricity and is used today in the electronics and aerospace industry. Gold is also used in the manufacturing of energy efficient reflective glass for skyscrapers and is used in the medical industry as well.
Figure 1. A Silver Certificate and a Modern U.S. Bill. Until 1958, silver certificates were commodity-backed money—backed by silver, as indicated by the words “Silver Certificate” printed at the top of the bill. Today, U.S. bills are backed by the Federal Reserve, but as fiat money. (Credit: “The.Comedian”/Flickr Creative Commons).
As commodity money, gold has historically served its purpose as a medium of exchange, a store of value, and as a unit of account. Commodity-backed currencies are dollar bills or other currencies with values backed up by gold or some other commodity held at a bank. During much of its history, the money supply in the United States was backed by gold and silver. Interestingly, antique dollars dated as late as 1957, have “Silver Certificate” printed over the portrait of George Washington, as shown in Figure 1. This meant that the holder could take the bill to the appropriate bank and exchange it for a dollar’s worth of silver. As economies grew and became more global in nature, the use of commodity monies became more cumbersome. Countries moved towards the use of fiat money. Fiat money has no intrinsic value, but is declared by a government to be the legal tender of a country. The United States’ paper money, for example, carries the statement: “THIS NOTE IS LEGAL TENDER FOR ALL DEBTS, PUBLIC AND PRIVATE.” In other words, by government decree, if you owe a debt, then legally speaking, you can pay that debt with the U.S. currency, even though it is not backed by a commodity. The only backing of our money is universal faith and trust that the currency has value, and nothing more.
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This video provides an overview of how money has evolved through the ages.
GLOSSARY
barter:
: literally, trading one good or service for another, without using money
commodity money:
: an item that is used as money, but which also has value from its use as something other than money
commodity-backed currencies:
: dollar bills or other currencies with values backed up by gold or another commodity
double coincidence of wants:
: a situation in which two people each want some good or service that the other person can provide
fiat money:
: something used as money, but which has no intrinsic value besides that
medium of exchange:
: whatever is widely accepted as a method of payment
money:
: whatever serves society in four functions: as a medium of exchange, a store of value, a unit of account, and a standard of deferred payment.
standard of deferred payment:
: money must also be acceptable to make purchases today that will be paid in the future
store of value:
: something that serves as a way of preserving economic value that one can spend or consume in the future
unit of account:
: the common way in which we measure market values in an economy
Candela Citations
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Defining Money by Its Functions. Authored by: OpenStax College. Provided by: Rice University. Located at: License: CC BY: Attribution. License Terms: Download for free at
(Macro) Episode 29: What is Money?. Authored by: Dr. Mary J. McGlasson. Located at: License: CC BY: Attribution
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Money through the Ages. Authored by: geobeats. Located at: License: Other. License Terms: Standard YouTube License
Licenses and Attributions
CC licensed content, Original
Modification, adaptation, and original content. Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Defining Money by Its Functions. Authored by: OpenStax College. Provided by: Rice University. Located at: License: CC BY: Attribution. License Terms: Download for free at
(Macro) Episode 29: What is Money?. Authored by: Dr. Mary J. McGlasson. Located at: License: CC BY: Attribution
All rights reserved content
Money through the Ages. Authored by: geobeats. Located at: License: Other. License Terms: Standard YouTube License |
7223 | https://math.stackexchange.com/questions/130663/estimating-values-of-x-from-the-functions-graph | calculus - Estimating values of X from the functions graph - Mathematics Stack Exchange
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Estimating values of X from the functions graph
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I have the graph below:
The question is asking to estimate values of x x between 0 0 and 6 6 where derivative of f f is 2 2. I also got the answer as well which says x x is 4.2 4.2 and 5.6 5.6:
My question is, what is the point of this question? What should I look for? How the answer is 4.2 4.2 and 5.6 5.6?
calculus
derivatives
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asked Apr 11, 2012 at 23:30
DumboDumbo
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The point of the question is to drive home the interpretation of derivative as slope, and to see if you understand what the slope of a graph is.
You should look for points on the curve where the slope (of the tangent line) seems to be about 2. Look at the higher red line on your diagram - it goes over 1 and up 2, which is slope 2, which is what you are looking for.
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answered Apr 12, 2012 at 0:48
Gerry MyersonGerry Myerson
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Thanks, so you mean this is a visual question? I have to see if a line with slope 2 can be drawn on the graph?! or is there a equation to use? maybe tangentline equation?!Dumbo –Dumbo 2012-04-12 00:53:15 +00:00 Commented Apr 12, 2012 at 0:53
If all you are given is a graph, then, yes, the natural assumption is that it is a visual problem and there are no equations available for you to use.Gerry Myerson –Gerry Myerson 2012-04-12 00:59:59 +00:00 Commented Apr 12, 2012 at 0:59
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7224 | https://www.youtube.com/watch?v=TJf62Omf7Cw | Steroidogenesis | The Biosynthesis of Steroids from Cholesterol
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Intro [Music] welcome back to catalyst University my name is Kevin toke off please make sure to like this video and subscribe to my channel for future videos and notifications in this video we're going DeroytoGenesis to discuss Deroy to genesis which is the processor pathway by which some cells can manufacture steroids such as aldosterone estradiol testosterone from the parent steroid which is known as cholesterol now cholesterol has a variety of other functions which we've mentioned periodically in other videos such as it's involved in stabilizing cell membranes over a wide range of temperatures it's also used to manufacture bile acids which are used in the emulsification of fats but arguably one of the most important functions of cholesterol is its conversion into all of these steroids that are shown in this diagram and that's what we're gonna discuss in detail in this video so there's a couple of nice things about this diagram first of all it's divided into the various types or classes of steroids for example in yellow right here we have the progestogen x' down here in blue we have the androgens these are the masculinizing hormones over here on the right in pink we have the estrogens or typically feminizing hormones and then in green the glucocorticoids or corticosteroids and then up here in purple the one general mineralocorticoid which is aldosterone the other thing is some of these enzymes are actually going to be in different cellular compartments the ones in red are going to be mitochondrial enzymes and then all the others for the most part in green these are going to be microsomal enzymes meaning they're embedded in the membrane of the smooth endoplasmic reticulum make sure you understand it's smooth not rough alright so let's go into the pathway so here's our parents darrid cholesterol and pay attention to the numbering system because some of these enzymes are going to be named based on the numbering system of cholesterol for example they're going to act on certain carbon atoms or hydroxyl groups so here's our parent steroid cholesterol notice its numbering system we have 1 2 3 4 and so on and so forth these numbers are going to be important because any of these enzymes out here that have a number of side them such as 21 hydroxylase this numbering system is based on the numbering system of cholesterol and so this number in front of this enzyme name 21 hydroxylase gives you an indication of which atom it's acting on okay based on the cholesterol numbering system so first the tail of cholesterol beginning at carbon number 22 and going through 27 this tail is going to be removed and this is done so by the mitochondrial enzyme cholesterol sidechain cleavage enzyme now some of these enzymes which are p450 enzymes are typically going to be referred to by their gene name so for example cholesterol sidechain cleavage enzyme is a p450 enzyme it's encoded by the gene site 11a 1 and so typically in most textbooks you'll either see this name or the gene name Sipho 11a 1 and so inside chain cleavage enzyme our site 11a 1 acts on cholesterol this 22 through 27 carbon tail is going to be removed as ISO cap or aldehyde not shown here and we get this ketone molecule which is going to be pregnenolone so in some sources you may see pregnenolone as the parent steroid but that's only because pregnenolone is the first steroid that we see where the tail has been for the most part removed okay now pregnenolone can go one of two directions we're gonna go to the right here first and see the development of the glucocorticoids and aldosterone okay so first of all pregnenolone can be hydroxylated into 17 alpha hydroxy pregnenolone notice at the 17 position right here which is on the five membered ring it's going to get hydroxylated by this enzyme and that gives us 17 alpha hydroxy pregnenolone now both pregnenolone and 17 alpha hydroxy pregnenolone can go to the right here and form corticosteroids so for example let's look at this first enzyme 3 beta hydroxy steroid dehydrogenase so this enzyme is going to oxidize this hydroxyl group at the 3 position thus the name 3 beta okay so this hydroxyl group is going to become a ketone as shown right here the other thing is we see a net isomerization of this double bond between positions 5 & 6 it's eventually going to be here between positions four and five and this is the case for both of these two reactions so pregnenolone will be oxidized into progesterone and progesterone is actually going to be a progestogen or progestin that we can recognize the functional mammalian hormone and then 17 alpha-hydroxy pregnenolone is going to be oxidized down here into 17 alpha hydroxy progesterone now in general this progesterone right here can act independently or the progesterone can be processed into aldosterone down here the 17 alpha hydroxy progesterone is going to be processed into cortisol now the next enzyme in this pathway regardless of which pathway you're taking whether it's progesterone or this hydroxylated version the enzyme is going to be 21 hydroxylase so this position right up here on the ketone right here this carbon which actually is going to be positioned number 21 as noted by this enzyme is going to be hydroxylated by 21 hydroxylase 21 hydroxylase is another p450 enzyme this is actually going to be encoded by site 21 a 2 typically oh there's another version of this enzyme as well but in any case progesterone will be hydroxylated into deoxy corticosteroid and then 17 alpha hydroxy progesterone will be hydroxylated into 11 deoxy cortisol okay now both of these enzymes a21 hydroxylase and the 3 beta hydroxy steroid dehydrogenase are mehar microsomal enzymes the next enzyme we're gonna see is actually going to be a mitochondrial enzyme so deoxy corticosteroid 11 deoxy cortisol must be transported into the mitochondria and they can just move mostly by a simple diffusion into the mitochondria because they're hydrophobic molecules now 11 beta hydroxylase what it's going to do is it's going to hydroxylate the 11 position right here okay so this is on the highest up six membered ring on the corner right here in my mouse so 11 beta hydroxylase is going to hydroxylate deoxy corticosteroids darrone as shown right here we'll come back to that in a minute and then 11 deoxy cortisol will be hydroxylated into cortisol now cortisol is hopefully what we know from an anatomy course and so forth the primary glucocorticoid or corticosteroid that's present in humans so this is gonna be the chronic stress hormone that we see in in many mammals such as humans and so cortisol can act independently however the corticosteroids formed up here is present in humans and does possess some of the glucocorticoid activity that cortisol does but corticosteroids going to be a minor corticosteroid in humans there are some organisms however such as mice and rats where corticosteroid becomes a primary glucocorticoid okay and cortisol more or less becomes a minor one but in general these are both glucocorticoids now corticosteroids four into the one general mineralocorticoid known as aldosterone which is going to be involved in salt and water retention at the kidneys so this is gonna be another Aldosterone synthase mitochondrial enzyme aldosterone synthase and what this enzyme is gonna do is going to do two things not shown here it's actually first gonna hydroxylate this position right here this is actually position number 18 and so sometimes this enzyme is actually called site 18 be to because hydroxyl aids initially at the 18 position but then there's another activity of this enzyme which will oxidize the hydroxyl group right here that's put here into an aldehyde that's shown right here in aldosterone and the reason aldosterone gets its name is because it's an aldehyde aldosterone okay and so aldosterone synthase generally isn't is going to have these two activities a hydroxylase activity which is the site 18 B 2 and then an 18 hydroxy steroid dehydrogenase to convert that hydroxyl into the aldehyde shown here now in general unless we're talking about some minor degradation pathways which will be involved in the liver aldosterone and cortisol are going to be sort of ends of the pathway they're dead ends because these two hormones are gonna have one of two things they're gonna do they're either gonna act independently at their corresponding receptors throughout the body or they're just gonna be metabolized as I said in the liver so we're done with this part of the pathway let's go back to the progestin for a moment because now we're gonna see how androgens and estrogens are made so actually in order to make androgens you have to first make two progestogen x' and these progestogen czar the hydroxylated versions of pregnenolone and progesterone respectively they are 17 alpha hydroxy pregnenolone and 17 alpha hydroxy progesterone now we have another enzyme right here this is called 1720 lyase now actually 1720 lyase is the same enzyme as 17 alpha hydroxylase okay so it actually does the first activity right here of the 17 alpha hydroxylation which we see right here okay but sometimes the steroid can be released from the enzyme prematurely and that's okay because that gives us the 17 alpha hydroxy pregnenolone and the 17 alpha hydroxy progesterone which we know now are going to be processed to aldosterone and cortisol respectively and so if 1720 liest performs its second activity on pregnenolone and progesterone it's actually going to use the hydroxyl group it just added to induce the removal of this ketone and so what you see here are the two first androgens dehydroepiandrosterone or DHEA and andros team diam these are two first recognizable androgens and notice the difference between the androgens and the progestogen the main difference is that the tail here which was really just a smaller version of the tail this ketone has been completely removed in all the androgens and notice in all the estrogens is completely removed as well that's a characteristic feature of both of these sex hormones now there's another enzyme at play here and it's one that we've seen before it's actually this 3 beta hydroxy steroid dehydrogenase what we can also do is oxidize this 3 beta hydroxy group into a ketone and notice again just as we saw up here in the progestogen x' we have a net isomerization of this double bond which is initially between positions 5 & 6 it gets I saw him rise between positions 4 & 5 and so what we can have is 3 beta hydroxy steroid dehydrogenase oxidizes DHEA into Andres teen dye own that can also occur and then if we've already reduced DHEA into Andres teen dial then 3 beta hydroxy stayer dehydrogenase can then we oxidize this hydroxyl group into testosterone and so what we see here - general pathways from DHA to get to testosterone we can either first react with the three beta hydroxy steroid dehydrogenase and then the 17 beta hydroxy steroid dehydrogenase or we can do it the other way around but in any case from these four androgens the goal really is to get to testosterone because of out of these four testosterone is the most potent of those four androgens now testosterone is out of these four the most potent androgen but it is not the most potent androgen the most potent androgen at least shown here is dihydrotestosterone and it's formed from testosterone by the action of this enzyme 5 alpha reductase which uses the reducing power of NADPH to reduce this 4 or 5 double bond right here notice in dihydrotestosterone that double bond is gone and a simple removal of that double bond actually increases the androgenic power of this hormone by about twenty to a hundred fold so this is a very very strong androgen okay now back to these two androgens androstenedione and testosterone both of these androgens can be processed directly into estrogens and so that brings up a really important point if you want to actually synthesize estrogens you have to first make progestogen x' then convert them into androgens and then finally into your estrogens the enzyme that's going to be involved in this conversion is going to be called aroma taste so notice what aroma tase's action on Andrews dean diode does this ketone has a net reduction into an alcohol and then the a ring becomes completely aromatic okay so one of the defining features of all the estrogens is the a ring is aromatic okay and that's actually what gives this enzyme encoded by site 1981 it's named aroma taste because it aromatize as the a ring okay so that's what this enzyme does to endorse deand I own it converts it into ester own which is actually going to be the less active estrogen but testosterone can also be aromatized in the same manner net reduction of this ketone into an alcohol and airman ization and that gives us this molecule which is going be generally the primary estrogen in pretty much every mammalian species and that is estradiol now both of these ester and estradiol can be metabolized in the liver and placenta to produce this molecule which is called s trial I won't go into its function here that'll be a separate video but in any case the conversion of estradiol to s trial is a one-step process it involves a hydroxylation at this position right here on the corner of the five membered ring from the D ring and then s drone it's gonna be a two-step process first this ketone is gonna have to be reduced that's gonna be ultimately by 17 hydroxy steroid dehydrogenase and then the same hydroxylation at this position which happens to be the sixteen position that's going to be catalyzed by another p450 hydroxylase okay it's a sixteen hydroxylase as it would be named all right so this hopefully gives you a really good idea of the steroid genic pathway we have really three of these enzymes are gonna be located in the mitochondria the vast majority of these are gonna be embedded in the membrane of the smooth endoplasmic reticulum therefore they're a microsomal and hopefully what you see is that really even though cholesterol is the parent steroid the progesterone estrogen is in general could also be considered a parent because to get mineralocorticoids you have to first synthesize glucocorticoids which come from progestogen x' on top of that in order to make estrogens you have to first make androgens which clearly come from progestogen x' okay and so that's going to become important when you're learning this pathway and making some sense of it i'm one thing we're going to talk about in the next video is that some of these steroids particularly ones that have a three beta hydroxy group such as cholesterol DHEA and ester own can actually be sulfated and it turns out that sulfation is a good way to first of all inactivate a steroid but also it prepares it for excretion in the urinary system which is actually how these steroids that steroids are going to be removed that's actually going to be part of Phase two metabolism in the liver and Outro so that's what we're actually going to look at in the next video so join us then but hopefully as for this video it made some sense to you and you learn something please make sure to like this video and subscribe to my channel for future videos and notifications thank you very much |
7225 | https://mathoverflow.net/questions/304300/a-generalization-of-the-sawayama-thebault-theorem | mg.metric geometry - A generalization of the Sawayama-Thebault theorem - MathOverflow
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A generalization of the Sawayama-Thebault theorem
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1. Introduction
The Sawayama-Thebault theorem is one of the best nice theorem in plane geometry. The theorem has a long history. It was published in AMM in 1938 the first solution appeared in 1973 with 24 pages), You can see detail in here and here.
Theorem 1(Sawayama-Thebault): Through the vertex A A of a triangle A B C A B C, a straight line A D A D is drawn, cutting the side B C B C at D D. I I is the center of the incircle of triangle A B C A B C. Let P P be the center of the circle which touches D C D C, D A D A at E E, F F, and the circumcircle of ABC, and let Q be the center of a further circle which touches D B D B, D A D A in G G, H H and the circumcircle of A B C A B C. Then P P, I I and Q Q are collinear
Source of the Figure 1 in here
I have worked for four years to find a real nice generalization of the Thebault theorem and publish in here since two years ago, but has no proof. So I am looking for the proof of the generalization of the Sawayama-Thebault theorem as follows:
2. Generalization of the Sawayama-Thebault theorem
Problem 2:( See Figure 2). Let A B C A B C be a triangle the red circle through B B, C C, the blue circle tangent to A B A B, A C A C and the red circle. Let P P be arbitrary point in the plane, through P P construct two tangent lines P D P D, P E P E to the blue circle. Let (O 1)(O 1) is the yellow circle tanegnt to the right tangent line P D P D, tangent to the red circle and the sideline B C B C; (O 2)(O 2) is the yellow circle tanegnt to the left tangent line P F P F, tangent to the red circle and tangent to the sideline B C B C. Then three points O 1 O 2 O 1 O 2 through a fixed point when P P be moved on a given line.
3. Construct the two yellow circles?
Problem 3: (The generalization of the Sawayama Lemma)The lines G F G F, H K H K through the incenter of A B C A B C (Figure 2)
4. Construct the point I′I′
See also:
Does this geometry theorem have a name?
mg.metric-geometry
euclidean-geometry
plane-geometry
triangles
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edited Dec 19, 2018 at 10:43
Đào Thanh OaiĐào Thanh Oai
asked Jul 5, 2018 at 8:02
Đào Thanh OaiĐào Thanh Oai
2,174 12 12 silver badges 40 40 bronze badges
5
1 What is the definition of the point I' ?godelian –godelian 2018-12-17 15:29:13 +00:00 Commented Dec 17, 2018 at 15:29
@godelian , Thank You very much I updated Đào Thanh Oai –Đào Thanh Oai 2018-12-17 15:32:44 +00:00 Commented Dec 17, 2018 at 15:32
1 Your green circle looks blue Fedor Petrov –Fedor Petrov 2018-12-17 15:33:28 +00:00 Commented Dec 17, 2018 at 15:33
Thank You dear Professor @FedorPetrov Đào Thanh Oai –Đào Thanh Oai 2018-12-17 15:35:03 +00:00 Commented Dec 17, 2018 at 15:35
That's nice generalization for Sawayama Thebault's Theorem.Tran Quang Hung –Tran Quang Hung 2020-07-15 07:07:51 +00:00 Commented Jul 15, 2020 at 7:07
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7226 | https://www.tiger-algebra.com/drill/a~3-2a~2-3a/ | Copyright Ⓒ 2013-2025
tiger-algebra.com
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Solution - Simplification or other simple results
Other Ways to Solve
Step by Step Solution
Step 1 :
Equation at the end of step 1 :
Step 2 :
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
a3 - 2a2 - 3a = a • (a2 - 2a - 3)
Trying to factor by splitting the middle term
3.2 Factoring a2 - 2a - 3
The first term is, a2 its coefficient is 1 .
The middle term is, -2a its coefficient is -2 .
The last term, "the constant", is -3
Step-1 : Multiply the coefficient of the first term by the constant 1 • -3 = -3
Step-2 : Find two factors of -3 whose sum equals the coefficient of the middle term, which is -2 .
| | | | | | | |
--- --- ---
| | -3 | + | 1 | = | -2 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 1
a2 - 3a + 1a - 3
Step-4 : Add up the first 2 terms, pulling out like factors :
a • (a-3)
Add up the last 2 terms, pulling out common factors :
1 • (a-3)
Step-5 : Add up the four terms of step 4 :
(a+1) • (a-3)
Which is the desired factorization
Final result :
How did we do?
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Copyright Ⓒ 2013-2025
tiger-algebra.com |
7227 | https://www.youtube.com/watch?v=0exBzsexUoI | 1.4.1 Propositional Operators: Video
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Transcript:
we're going to talk about propositions and logical operations in this little clip and let's begin then with the discussion of propositions so to a mathematician and in particular in this class we're going to use the word proposition to refer to something that is either true or false an example would be there are five regular solids this happens to be true and sometimes we even some terms we even prove it it implies for example that if you wanted to place let's say a hundred fixed position satellites around the earth in a uniform way or you can't do it because there isn't any a hundred vertex regular solid the biggest one is 20 vertices okay well if I change it six guess what the assertion there are six regular solids is false so that's a simple-minded example of a well-defined to well-defined propositions one of which is true one of which is false propositions don't have to be true always what are some non examples well wake up is not a proposition because it's an imperative it's not true or false and where am I is a question it's not true or false and it's 3:00 p.m. is not a proposition because it's true or false at any given moment but whether or not it's true or false depends on the time and that's a complication we don't want to get into the idea is a proposition is some fixed assertion that say they're true forever or not true forever now one of the reasons why mathematicians bring up this abstraction of propositions and the operations on them that we're about to say is that ordinary language tends to be ambiguous and that of course will cause problems in mathematical reasoning just as it would in a program the most and one of the most ambiguous of the phrases in English that connects propositions is or so let's look at this example Greeks carry swords or javelins and if I was transcribing this into precise math notation I can say gee for Greeks implies s for swords or J for javelins so this is an assertion if you're Greek then you carry a sword or a Java Java line Greeks implies sword or Jeff learn really I should say Greek soldiers but let that be implicit okay that's how we're gonna translate this sentence into just using these operators to paraphrase what's going on the problem is what is or mean and it turns out that for javelins and swords it's true even if a Greek carries both a sword in the javelin this is an inclusive or a Greek soldier would carry both a sword and a javelin because in fact the Jeff wants a good long distance weapon and a sword is good for defending yourself close in and you certainly want to have both especially after you've thrown your javelin and you don't have anything left but the sword now this some many standard notations for these logical connectives that that build up larger propositions out of component propositions so one of the things is this V symbol or disjunction symbol is used by logicians often instead of or and this arrow means implies or sometimes a double bar arrow also means employees but we're not going to get into that I'm not gonna ask you to memorize these symbols we'll just stick to the words which don't take up that much more room let's look at another example Greeks carry bronze or copper swords syntactically this has the same structure as the previous phrase but we're going to translate it differently and the reason is that we mean here that a Greek soldier is not going to carry both a bronze in a copper sword why is that well bronze swords are just way better than copper swords they'll slice right through copper they're much harder and it's not worth the weight to carry this inferior copper sword when you have a much better one so this time we mean that Greeks carry exactly one of a bronze or a copper so you'd carry a copper so if you didn't have a access to a bronze one and so now we translate that into Greek implies B or bra B for bronze or C for copper but this time we use the X or X or means that one of them is true exactly but not both and not neither again there's this plus sign notation Forex or because as we'll see it acts a little bit like adding numbers bond to where one plus one is zero so let's be more precise about the two definitions of or an XOR and how they work and the the assertion is that if I think of P and Q as placeholders for propositions that are either true or false then the composite proposition P or Q is true if and only if P is true or Q is true or both are true and I could express Express this assertion English this is if and only if by giving you a so-called truth table where in these two columns are in these all these rows I've enumerated all the possible pairs of values of P and Q so P and Q might both be true that one P might be true and Q false P false Q true and both of them are false and for each of those possible combinations of the truth values of P and Q I tell you the truth value of P or Q and the notable thing is this last row where I'm telling you that the only way that P or Q is false is if both P and Q are false okay for X or as we said the P or X or Q is true if and only if exactly one of P and Q is true so if I was expressing that as a truth table we see that where this TT is false and where this FF is false because it's not the case in either of those two rows that exactly one is true but the middle rows exactly one is true and so px or Q is true so this truth table is just a precise way of defining how XOR acts on truth values there's another connective and which works even more straightforwardly the baloff value of P and Q is true if and only if both P and Q are true and there's is truth table again the salient Row is that it's true only if and only if both P and Q are true another crucial logical operation is the negation operation or not a sum not P just flips the truth value of P if P is true and true then not P is false if not P is truth and P is false and there's it's very trivial truth table chubby because there's only two values to be concerned about when P is true not P is false when P is false not P is true now one of the places that this notion of combining basic propositions into using logical operations to build up more complicated composite proposition is in programming here's a typical kind of phrase that comes from Java Java uses this double vertical bar to mean or inclusive by the way and double ampersand to mean and so in Java this is a piece of legitimate Java code that's doing a test to evaluate this expression if X is greater than 0 or X is less than or equal to 0 and Y is greater than 100 then if that test comes out to be true then you do a bunch of code that follows the test upto some of the limiter that tells you where to stop and if it's false you just skip all that stuff and go on so we use sort of boolean expressions or logical expressions like this in a very standard way in most programming languages the other place where these operations comes up in is in digital logic and the digital circuit designers have their own notation which I'll just mention but we're not going to again impose on you but you should be aware of what a notation that u is that we'll use as well because it's so economical is that not X can be abbreviated by writing a bar over the X more generally not a complicated expression can be abbreviated by writing a bar over the complicated expression I just saved some space and so we'll use it but not the following in digital logic on the idea is that you're talking about circuits where the only distinction of the signal that's coming in electrically is whether it's a high voltage or low voltage with high voltages denoted by 1 and low voltage denoted by 0 and the way the digital logic behaves is the one corresponds to true and the 0 corresponds to false then the end operation is simply multiplication because 1 or 0 times 1 or 0 is 1 when both of number one for ordinary multiplication which is exactly the way and behaves when one means true and zero means false unfortunately the digital designers use plus when they mean or they do not mean that one plus one is two they mean that one plus one is one and just the thing to watch out for part that's part of the reason we're not going to use this notation let's just stick to our ordinary word notation that we give on the right |
7228 | https://www.scribd.com/document/335428294/6690-Apollonius-Circle | 6690 Apollonius Circle | PDF | Circle | Perpendicular
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6690 Apollonius Circle
The document discusses the Circle of Apollonius, which is defined as the locus of points whose distances from two fixed points (A and B) are in a constant ratio k. It then proves that: 1) F…
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The Circle of Apollonius By BJ Kim
Who is Apollonius? (History of mathematician)
Apollonius of Perga (about 262 B.C- about 190 B.C.) was a Greek mathematician known as 'The Great Geometer'. His works had a very great influence on the development of mathematics and his famous book Conics introduced the terms parabola, ellipse and hyperbola.
What is the Circle of Apollonius?
The locus of a point P whose distance from a fixed point A is a multiple of its distance from another fixed point B. If the multiple is equal to 1, then the locus is a line-- the perpendicular bisector of the segment AB. If the multiple is not equal to 1, then it is a circle. The locus is called the Circle of Apollonius.
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In the diagram, AP:BP = m:n. When the point P moves keeping this ratio the locus of P is a circle. We call it the circle of Apollonius. This circle connects interior and exterior division points of A and B. The locus of a variable point whose distances from two fixed points are at a constant ratio k, is a circle for k
"
1 and the perpendicular bisector of the two points for k = 1. The family of such loci for all real values of e forms a coaxial family of circles with the two fixed points as limit circles.
Proof
Let L be the line through the two fixed points A, B. Let P be a variable point such that PA/PB = k is a constant. WLOG [without loss of generality] assume k > 1. Let P1 and P2 be two points on L, P1 between A and B, and P2 outside, such that PiA/PiB = k for i = 1, 2. Then since PA/PB = P1A/P1B = area (
!
PAP1) / area (
!
PAP2), P1 is equidistant from PA and PB, so that PP1 bisects <APB internally. Similarly, P2 is equidistant from lines PA and PB, hence PP2 bisects <APB externally.
!
PP1
"
PP2. The internal and external bisectors of at an angle of a triangle are perpendicular. The sum of the internal angle and the external angle is 180 degrees. The angle between the internal and external bisectors is the sum of one-half of each. Therefore the angle measures 90 degrees.
!
P lies on the circle Ze on P1P2 as diameter. Let M be the midpoint of AB.
!
MP1/MB = (
) (AP1-BP1) / (
) (AP1 + BP1) = (k-1)/ (k+1) = (
) (AP2-BP2) / (
) (AP2 + BP2) = MB/MP2.
!
MB2 = MP1
$
MP2.
!
The tangent from M to Ze has length MB, or AB/2, which is independent of k. So for each real value of k, the different circles Ze all have L as diameter
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and have tangents from M having AB/2 as length, thus forming a radical family of circles with the perpendicular bisector L' of AB as radical axis. The points A and B corresponds to the cases when k = 0 and
"
and are null circles. For k = 1, the locus is the perpendicular bisector L' of AB. The proof of this result is based on the following theorems: (i) The angle bisectors of a triangle (ii) The proportion resulting from a line drawn parallel to a side of a triangle. In fact, the proof connects these theorems in a nice way, and provides are refreshing idea about circles. The students can use Circle of Apollonius to tackle interesting, sometimes challenging, problems. Let A and B be two fixed points, and let P be a point where PA: PB = k, a positive constant not equal to one. Next, bisect the angle APB internally and externally by PI and PE respectively, where I and E are on AB. Since PI bisects <APB and PE bisects its supplementary angle BPC, then by the bisector angle theorem, I and E divide AB internally and externally in the ratio k. Hence I and E are fixed points. Since m<IPE = 90°, the locus of P is a circle with IE being a diameter,
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Conclusion: Apollonius's Alternative Definition of a circle
A circle is usually defined as the set of points P at a given distance r (the circle's radius) from a given point (the circle's cen ter). However, there are oth er, equivalent definitions o f a circle. Apollonius discovered that a circle could also be defined as the set of points P that have a given ratio of distances k = d1/d2 to two given points (labeled A and B in below Figure). These two points are sometimes called the foci.
Apollonian Circles Theorem
1
The locus is a circle, unless of course r = 1, in which case it's the perpendicular bisector of AB. The proof explo its the properties o f angle bisectors: internal and extern al. Construct poin ts C and D on the line AB such that AC/BC = AD/BD= r. For r 1, C and D always exist. Note that both C and D lie on the sought locus. The circle at hand has CD as a diameter. Indeed, C serves as the feet of the internal bisector of triangle APB at apex P, PD serves as the external bisector. Therefore, PC PD. Chords PC and PD are perpendicular an d therefore def ine a 90
inscribed angle. The angle subtends a 180
arc, which means that CD is a diameter of the circle. For any P on the circle, the internal and external bisectors of angle APB pass through (the fixed points) C and D. Since, for the same A and B, each of the Apollonian circles corresponds to a different r, no two Apollonian circles intersect. For 0 < r < 1, the circles are closer to A and surround it. For smaller values of r, they are increasingly close to A. For r > 1, the circles surround B and, as r grows become
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7229 | https://en.wiktionary.org/wiki/pertinent | pertinent - Wiktionary, the free dictionary
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Beginning
1 EnglishToggle English subsection
1.1 Etymology
1.2 Pronunciation
1.3 Noun
1.4 Adjective
1.4.1 Related terms
1.4.2 Translations
1.5 Further reading
2 CatalanToggle Catalan subsection
2.1 Etymology
2.2 Pronunciation
2.3 Adjective
2.3.1 Derived terms
2.3.2 Related terms
2.4 Further reading
3 FrenchToggle French subsection
3.1 Etymology
3.2 Pronunciation
3.3 Adjective
3.3.1 Derived terms
3.3.2 Related terms
3.4 Further reading
4 LatinToggle Latin subsection
4.1 Verb
5 RomanianToggle Romanian subsection
5.1 Etymology
5.2 Adjective
5.2.1 Declension
pertinent
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From Wiktionary, the free dictionary
English
[edit]
Etymology
[edit]
Inherited from Middle Englishpertinent, borrowed from Old Frenchpertinent and Latinpertinēns, pertinentem.
Pronunciation
[edit]
(UK)IPA(key): /ˈpətɪnənt/
(US)IPA(key): /ˈpɚtɪnənt/
Audio (US):Duration: 2 seconds.0:02(file)
Noun
[edit]
pertinent (pluralpertinents)
(law) A right that attaches to land, in Scotslaw.
Adjective
[edit]
pertinent (comparativemore pertinent, superlativemost pertinent)
Important with regard to (a subject or matter); pertaining; relevant. synonym, antonym, coordinate term▲quotations▼Synonyms:seeThesaurus:pertinentAntonym:nonpertinentCoordinate term:semipertinent
1992, Rudolf M[athias] Schuster, The Hepaticae and Anthocerotae of North America: East of the Hundredth Meridian, volume V, Chicago, Ill.: Field Museum of Natural History, →ISBN, page viii:Fourthly, I have made an effort to call the attention of the reader to the pertinent literature.
Related terms
[edit]
impertinent
pertain
pertinence
pertinency
unpertinent
Translations
[edit]
show ▼±pertaining
[Select preferred languages] [Clear all]
Bulgarian: уместен, подходящ(umesten, podhodjašt)
Catalan: pertinent
Chinese: Cantonese: 相關嘅/ 相关嘅(soeng 1 gwaan 1 ge 3)
Czech: odpovídající(cs), relevantní(cs), vztahující se(cs), příslušný(cs), patřičný(cs)m, vhodný(cs)m
Dutch: relevant(nl)
Esperanto: laŭtema
Finnish: osuva(fi), relevantti(fi), olennainen(fi)
French: pertinent(fr)
Georgian: შესაბამისი(šesabamisi), რელევანტური(relevanṭuri), შესაფერისი(šesaperisi), მართებული(martebuli), საქმესთან დაკავშირებული(sakmestan daḳavširebuli)
German: maßgeblich(de), wesentlich(de), bedeutend(de), relevant(de), in diesem Zusammenhang Bedeutung habend
Greek: σχετικός(el)(schetikós), συναφής με ένα θέμα(synafís me éna théma)
Hungarian: illő(hu), találó(hu)
Italian: pertinente(it)
Japanese: 当てはまる(ja)(Atehamaru)
Manx: bentynagh
Portuguese: pertinente(pt)m
Romanian: pertinent(ro)
Russian: уме́стный(ru)(uméstnyj), подходя́щий(ru)(podxodjáščij)
Slovak: príslušný, relevantný(sk)
Spanish: atinado(es), pertinente(es)
Swedish: relevant(sv), tillämplig(sv)
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
Further reading
[edit]
“pertinent”, in Webster’s Revised Unabridged Dictionary, Springfield, Mass.: G. & C. Merriam, 1913, →OCLC.
William Dwight Whitney, Benjamin E[li] Smith, editors (1911), “pertinent”, in The Century Dictionary[…], New York, N.Y.: The Century Co., →OCLC.
“pertinent”, in OneLook Dictionary Search.
Catalan
[edit]
Etymology
[edit]
Learned borrowing from Latinpertinēns, pertinentem.
Pronunciation
[edit]
IPA(key): (Central)[pər.tiˈnen]
IPA(key): (Balearic)[pər.tiˈnent]
IPA(key): (Valencia)[peɾ.tiˈnent]
Audio (Barcelona):Duration: 2 seconds.0:02(file)
Adjective
[edit]
pertinentm or f (masculine and feminine pluralpertinents)
relevant, pertinent
Derived terms
[edit]
pertinentment
Related terms
[edit]
pertànyer
pertinença
pertinència
Further reading
[edit]
“pertinent”, in Diccionari de la llengua catalana [Dictionary of the Catalan Language] (in Catalan), second edition, Institute of Catalan Studies [Catalan: Institut d'Estudis Catalans], April 2007
“pertinent”, in Gran Diccionari de la Llengua Catalana, Grup Enciclopèdia Catalana, 2025
“pertinent” in Diccionari normatiu valencià, Acadèmia Valenciana de la Llengua.
“pertinent” in Diccionari català-valencià-balear, Antoni Maria Alcover and Francesc de Borja Moll, 1962.
French
[edit]
Etymology
[edit]
Inherited from Old Frenchpertinent (first attested circa 1300), borrowed from Latinpertinēns, pertinentem.
Pronunciation
[edit]
IPA(key): /pɛʁ.ti.nɑ̃/
Audio (Switzerland (Valais)):Duration: 2 seconds.0:02(file)
Audio (France (Brétigny-sur-Orge)):Duration: 2 seconds.0:02(file)
Audio (France (Vosges)):Duration: 2 seconds.0:02(file)
Adjective
[edit]
pertinent (femininepertinente, masculine pluralpertinents, feminine pluralpertinentes)
pertinent; relevantraisons pertinentes ― relevant reasons
judicious; justified
Derived terms
[edit]
pertinemment
pertinence
Related terms
[edit]
appartenir
impertinent
Further reading
[edit]
“pertinent”, in Trésor de la langue française informatisé[Digitized Treasury of the French Language], 2012.
Latin
[edit]
Verb
[edit]
pertinent
third-personpluralpresentactiveindicative of pertineō
Romanian
[edit]
Etymology
[edit]
Borrowed from Frenchpertinent, ultimately from Latinpertinēns, pertinentem.
Adjective
[edit]
pertinentm or n (feminine singularpertinentă, masculine pluralpertinenți, feminine and neuter pluralpertinente)
relevant
Declension
[edit]
show ▼Declension of pertinent| | singular | | plural |
---
| | masculine | neuter | feminine | masculine | neuter | feminine |
| nominative- accusative | indefinite | pertinent | pertinentă | pertinenți | pertinente |
| definite | pertinentul | pertinenta | pertinenții | pertinentele |
| genitive- dative | indefinite | pertinent | pertinente | pertinenți | pertinente |
| definite | pertinentului | pertinentei | pertinenților | pertinentelor |
Retrieved from "
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English terms derived from the Proto-Indo-European root ten-
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en:Law
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pertinent
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7230 | https://www.quora.com/What-do-people-mean-by-times-less-What-is-5-times-less-than-100 | What do people mean by 'times less'? What is 5 times less than 100? - Quora
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What do people mean by "times less"? What is 5 times less than 100?
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Tom Bell
Works at Self-Employment (2019–present) · Author has 6.2K answers and 3.8M answer views
·2y
I can offer what I would consider the speaker is trying to communicate with “5 times less than 100.”
First off, the speaker is saying “much less than a hundred” or “a fraction of a hundred.”
We can express that mathematically as
(1/5) x 100 = 20.
I have heard people say “times less” in conversation, but not in many years. I suspect “five times less” is a phrase that is not formal or literary, but rather a verbal style used in ordinary or familiar conversation.
Upvote ·
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Related questions
More answers below
Which number is 10% less than 100?
What does it mean to say x is 3 times less than y?
What is the meaning of "times less"? It makes no sense to me, but is increasingly seen in expressions like "A weighs ten times less than B".
What is the meaning of less than 100 words?
How many times is 5 in (1 to 100)?
Charlie Hess
The words you use cause you to win or lose. · Author has 324 answers and 820.3K answer views
·1y
What it means really is they are trying to be cool by imitating someone they heard say it. Like Mark Rober or Neil deGrasse Tyson.
It's wrong for many reasons. Similar to the phrase 100 times smaller or a billion times dimmer these are not real measurements. You can't measure how small something is, only how large it is. There's no dimness scale. There's a brightness scale . You could compare two t
Continue Reading
What it means really is they are trying to be cool by imitating someone they heard say it. Like Mark Rober or Neil deGrasse Tyson.
It's wrong for many reasons. Similar to the phrase 100 times smaller or a billion times dimmer these are not real measurements. You can't measure how small something is, only how large it is. There's no dimness scale. There's a brightness scale . You could compare two things using adjectives but once you start using numbers all is lost.
For example, if you're weighing something and one thing is heavier than the other, then you might say this is three times heavier. But if you said this one is three times lighter that makes no sense. Lightness is not a thing. Mass is how heavy (I...
Upvote ·
Sweetkatie
Dangerous Amateur. You ask but no answer? Check my profile. · Author has 374 answers and 151.6K answer views
·2y
It is possible, in English, to construct a phrase with no obvious meaning. I don’t think you would hear this expression from a mathematician. Your best hope is to ask the person who has used this expression.
One sometimes hears “25% less”. This makes sense. If you start with $40m and then have 25% less you have 75% of $40. That is $30. 25% is another way of saying a quarter, (1/4).
So “25% less that x” means x - 0.25x . If you apply the same reasoning to “5 times less than 100”, you might expect it to mean 100–5100 which is -400. On the other hand, some might argue that 100 is five times more t
Continue Reading
It is possible, in English, to construct a phrase with no obvious meaning. I don’t think you would hear this expression from a mathematician. Your best hope is to ask the person who has used this expression.
One sometimes hears “25% less”. This makes sense. If you start with $40m and then have 25% less you have 75% of $40. That is $30. 25% is another way of saying a quarter, (1/4).
So “25% less that x” means x - 0.25x . If you apply the same reasoning to “5 times less than 100”, you might expect it to mean 100–5100 which is -400. On the other hand, some might argue that 100 is five times more than 20, so that 20 is five times less than 100.
The confusion comes from using English to express a numerical idea. Five times 20 is 100. That is clear. “Five time more than 20” causes confusion because the word “more” is simply out of place. If you increase 20 by five times 20, you would get 20 + 520 = 120.
If you want to express a mathematical idea in plain language without recourse to math notation, then you should use only standard expressions.
Hope this helps. K
Upvote ·
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9 1
Santtu Suihkonen
Software Developer (2004–present) · Author has 504 answers and 158.6K answer views
·2y
he answer can be found near the bottom end of the list of mathematically sound explanations.
It’s not 100 minus (5 times hundred), which is 100 – 500 = -400.
It’s not the negative of five hundred either, -500.
The word “times” is also not a constant, for example 3, so 5 times less than 100 does not become 100-(5 times 3) = 85.
“Times” and “times less” are actually used to mean multiplication and the inverse, respectively.
So, 5 times less than 100 means the multiplicative inverse of 5 multiplied by 100, which is 20.
People use this kind of expression in everyday conversation to be understood. The wo
Continue Reading
he answer can be found near the bottom end of the list of mathematically sound explanations.
It’s not 100 minus (5 times hundred), which is 100 – 500 = -400.
It’s not the negative of five hundred either, -500.
The word “times” is also not a constant, for example 3, so 5 times less than 100 does not become 100-(5 times 3) = 85.
“Times” and “times less” are actually used to mean multiplication and the inverse, respectively.
So, 5 times less than 100 means the multiplicative inverse of 5 multiplied by 100, which is 20.
People use this kind of expression in everyday conversation to be understood. The words “times” and “less” are used together to indicate that a number is decreasing from a starting point.
For example, if someone has a deal for tires for $100 and they show up with $20, the owner might ask, “What do you have there, pal? The deal was for $100! That’s 5 times less!”
If there only was a word for it..
Fifth! 20 is fifth of hundred!
Could that work?
Upvote ·
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Most drivers only think about their insurance rate when it’s time to renew — and even then, many just accept the renewal notice and keep the same policy. But car insurance rates are not static. They can change for a variety of reasons, some of which have nothing to do with you personally.
For example, rates may shift because of changes in state regulations, overall accident trends, or insurers adjusting their risk models. Your own life changes — moving to a new ZIP code, improving your credit score, paying off your car, or maintaining a clean driving record — can all lower what you should be paying. The problem is that many people never check, so they keep paying the same (or more) when they might actually qualify for a better rate.
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What is the meaning of "more than four times"?
What does “I think so” mean?
What is the meaning of “much less”?
What should we say instead of “three times less than” or “three times more than?”
Is 300, three times more than 100, or 2 times more than 100?
Hal Mickelson
Former Corporate Attorney; AB, History, JD, Law · Author has 15K answers and 16.6M answer views
·2y
People may not be clear about what they mean. If someone has been talking about a 10% reduction in 100 units, five times that reduction would be a 50% reduction. “A” says “I can bring down the cost by 10%.” “B” intends to say “I can bring it down by five times that amount,” meaning 50%; instead “B” says, “I can make it five times less,” which would not be quite right.
More than 100% less than 100 i
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People may not be clear about what they mean. If someone has been talking about a 10% reduction in 100 units, five times that reduction would be a 50% reduction. “A” says “I can bring down the cost by 10%.” “B” intends to say “I can bring it down by five times that amount,” meaning 50%; instead “B” says, “I can make it five times less,” which would not be quite right.
More than 100% less than 100 is a negative number. What they may mean by “five times less than 100” could be either “minus 400” or “m...
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Hazel Gold
Knows English · Author has 185 answers and 102.4K answer views
·2y
Newtown's water tank is better than Oldtown's, because Oldtown's water tank holds five times less water.
Newtown's tank holds 100 liters of water. That's five times more than Oldtown's tank.
Five times less: 100 / 5 = 20. Oldtown's tank only holds 20 liters of water.
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Johnny M
Master's Degree from Harvard University (Graduated 2011)
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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.
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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.
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Curt Weinstein
HS in Physics&Mathematics, The Bronx High School of Science (Graduated 1969) · Author has 42K answers and 27.4M answer views
·2y
Well, five times more than 10 is 50. Does 5 times less mean you divide by 5? It’s a phrase with a meaning that is not clear. Most would say that 10 is five times less than 50 (50/5=10).
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Kevin Solari
Author has 7K answers and 1.8M answer views
·2y
It's poor use of English - probably trying to say a fifth of 100 in this example.
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Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success.
The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickly and maintaining operations.
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Chris J Vines
Studied at Chadron State College
·5y
Related
What exactly does "three times less" of something equate to, or what is meant by it?
Using the phrase “x times less” is not a mathematically factual way to state something. Instead, the correct way to state the information would be to say 1/x the amount.
Here is an example.
Incorrect way: Yesterday was a sunny day, I had 300 dandelions in my yard. Today is a rainy day so I have 3 times less dandelions in my yard.
Correct way: Yesterday was a sunny day, I had 300 dandelions in my yard. Today is a rainy day, so I have 1/3 as many dandelions in my yard. (Which is 100 dandelions).
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John Fish
Engineer at Consultants (2005–present)
·4y
Related
What exactly does "three times less" of something equate to, or what is meant by it?
“three times less” is nonsense, if I hear such a term I assume the speaker is an idiot.
To me three times less than 10 would be -20 (that’s 3x10 subtracted from the original 10), the correct term that people are looking for when they say “three times less” is “a third”.
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Skip Shekhani
BS in Industrial Engineering (college major)&Psychology, Iowa State University (Graduated 1985) · Author has 25.4K answers and 7.3M answer views
·2y
That means it is less than the number in the way of multiple of it. So, in your example, it means multiple of 5 (times less). So, you may divide or Multiply by the number depends if you are going “More than” or “Less than” in the works.
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Daniel Lindsäth
Scientifically minded skeptic with more than five years of STEM univ studies · Author has 5.7K answers and 13.2M answer views
·10y
Related
What exactly does "three times less" of something equate to, or what is meant by it?
I have absolutely no idea how you managed to get a minus sign in there. I'd be genuinely interested to hear why, because it might show a difference in how we analyse language and that kind of thing is always fun to learn about.
I would definitely interpret "three times less than X" as "one third of X".
I have three times less than you, so you must consequently have three times more than me:
Me = You / 3
You = Me 3
Some people interpret "three times more than me" as "what I have, and then three times that again," which is "You = Me 4" but that always seemed a biy strange to me since the expres
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I have absolutely no idea how you managed to get a minus sign in there. I'd be genuinely interested to hear why, because it might show a difference in how we analyse language and that kind of thing is always fun to learn about.
I would definitely interpret "three times less than X" as "one third of X".
I have three times less than you, so you must consequently have three times more than me:
Me = You / 3
You = Me 3
Some people interpret "three times more than me" as "what I have, and then three times that again," which is "You = Me 4" but that always seemed a biy strange to me since the expression is so similar to "this is three times that" which obviously doesn't involve the number four.
Look at me rambling...
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Peter Upton
BA in Physics&Mathematics, The Open University · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 14.5K answers and 10.8M answer views
·May 6
Related
Why do some people interpret "10 times less" as subtracting 10 times the amount, and how can we clarify this misunderstanding?
Easy by not using confusing language. ‘Times’ mean multiply. When you were young I hoped you learnt your 5 times table, which is a multiplication table.
So ‘10 times less means’ multiply by 10 but have a smaller answer. How does that work? You are complaining people misunderstand what you mean, this is because you are talking nonsense.
What you really mean is 1/10 of the original amount not ‘10 times less’.
If you don’t like that then you could always use the correct mathematical operator - division.
The correct wording is easy to understand and has the benefit of being correct. Example, “The pric
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Easy by not using confusing language. ‘Times’ mean multiply. When you were young I hoped you learnt your 5 times table, which is a multiplication table.
So ‘10 times less means’ multiply by 10 but have a smaller answer. How does that work? You are complaining people misunderstand what you mean, this is because you are talking nonsense.
What you really mean is 1/10 of the original amount not ‘10 times less’.
If you don’t like that then you could always use the correct mathematical operator - division.
The correct wording is easy to understand and has the benefit of being correct. Example, “The price of computer memory chips is a tenth of what it was in the year 2008.”
Upvote ·
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Which number is 10% less than 100?
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What is the meaning of less than 100 words?
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7232 | https://www.scirp.org/journal/paperinformation?paperid=101351 | Is Titration as Accurate as HPLC for Determination of Vitamin C in Supplements? —Titration versus HPLC for Vitamin C Analysis
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American Journal of Analytical Chemistry>Vol.11 No.7, July 2020
Is Titration as Accurate as HPLC for Determination of Vitamin C in Supplements?
—Titration versus HPLC for Vitamin C Analysis ()
Lucile Tiemi Abe-Matsumoto1, Geni Rodrigues Sampaio2, Deborah Helena Markowicz Bastos2
1Physical, Chemical and Sensory Department, Food Center, Adolfo Lutz Institute, São Paulo, Brazil.
2Nutrition Department, School of Public Health, University of São Paulo, São Paulo, Brazil.
DOI:10.4236/ajac.2020.117021PDFHTMLXML 1,748 Downloads 11,835 ViewsCitations
Abstract
Herein, the iodometric titration and HPLC-RP methods were compared for the determination of vitamin C in vitamin and mineral supplements. The methods were validated in terms of linearity, limits of detection (LOD), limits of quantification (LOQ), precision, and recovery by using vitamin standards and a reference material SRM 3280 (multivitamin/multimineral tablets) obtained from the National Institute of Standards and Technology (NIST). A total of 22 samples of vitamin supplements, randomly acquired in local markets of Sao Paulo (Brazil) were evaluated for content of vitamin C by these two methodologies. The precision expressed as RSD was lower than 5% for both methods. LOD was 3.6 μg/mL for HPLC and 1.0 mg for titration, while LOQ were 12.0 μg/mL and 3.0 mg for HPLC and titration, respectively. Percent recoveries (%) of spiked samples ranged from 98.7 to 100.5 for HPLC and from 98 to 104 for titration. The contents of vitamin C in SRM 3280 (Certified value = 42.2 ± 3.7 mg/g) were 40.2 and 42.1 mg/g when determined by HPLC and titration respectively. Statistically, there was no difference in the analysed vitamin C content for half of the samples, irrespective of the method applied. HPLC was more sensitive, but the titrimetric method was faster and consumed less reagent. Although both methods were accurate in determining the vitamin C content in SRM 3280, the matrix constituents of some vitamin supplements may have interfered with the analysis.
Keywords
Vitamin C, Ascorbic Acid, Dietary Supplements, HPLC, Titration
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Abe-Matsumoto, L. , Sampaio, G. and Bastos, D. (2020) Is Titration as Accurate as HPLC for Determination of Vitamin C in Supplements?
—Titration versus HPLC for Vitamin C Analysis. American Journal of Analytical Chemistry, 11, 269-279. doi: 10.4236/ajac.2020.117021.
1. Introduction
Vitamin C is known as ascorbic acid in its reduced form and as dehydroascorbic acid in its oxidized form. The conversion to the oxidized form is reversible and occurs in the presence of metal ions, heat, light, or mildly alkaline conditions; however, vitamin C activity decreases. Ascorbic acid can donate and receive electrons, and it supports specific enzyme reactions and plays an essential role as an antioxidant in the body. It is an important water-soluble vitamin that is not stored by the body . Some studies report, that vitamin C can protect lipid peroxidation, especially that of HDL cholesterol, inducing a cardioprotective and atherogenesis inhibitory effect .
The vitamin C recommended daily allowance (RDA) for a healthy adult is 45 mg, although some studies suggest, that the intake of 80 to 120 mg vitamin C may reduce the risk of non-infectious chronic diseases. It was reported that smokers need more than 140 mg per day . The initial clinical manifestations of vitamin C deficiency are fatigue, loss of appetite, drowsiness, pallor, lack of energy in the limbs and joints, irritability, slow healing of minor injuries, and minor bleeding. The severe deficiency of ascorbic acid leads to scurvy, a disease characterized by general weakness, anaemia, gingivitis and cutaneous haemorrhages .
The main source of dietary vitamin C is fruits and vegetables; however, due to changes in the dietary patterns of the population, supplements did become important sources of vitamins, and therefore, consumption of vitamin and mineral nutraceutical products is common worldwide . The media widely advertise the benefits of antioxidant vitamins (e.g. vitamins C and E), targeting different socioeconomic levels of the global population . The accuracy of the supplement’s label information with regards to the vitamin content is essential to prevent consumers from suffering adverse health consequences and economic losses.
Several methods have been used for determination of vitamin C in foods, including spectrophotometry, electrophoresis, titration, and high performance liquid chromatography (HPLC) . The oxidation-reduction titration method using indophenol dye as indicator was stablished as the official method for vitamin C determination by the Association of Official Analytical Chemists (AOAC) for many years. In this method, ascorbic acid is oxidized to dehydroascorbic acid and the indophenol dye is reduced to a colorless compound, indicating the end point of the reaction . The iodometric titration for vitamin C determination was the official method for Public Health Laboratories in Brazil. The endpoint of this titration is determined by the first excess of iodine in the solution, that reacts with the starch indicator, forming a complex with an intense dark blue-violet color . This method is single, fast, and reliable, however, just as for the AOAC method, the end point of the titration cannot be easily detected when the food sample has intense color. The alternative to exclude color interference was to transfer the traditional iodometric titration to automatic potentiometric titrator, as described in this work. Thus, the end point of titration is indicated by platinum ring electrode, and no color interference occurs. In the last decades, HPLC and UHPLC methods were developed in substitution to titration method, showing high accuracy, but these equipment cost is high .
New regulations on dietary supplements were published by the Brazilian Health Regulatory Agency (ANVISA) in July 2018, with a 5-year period for industries to comply with the new rules. The new legislation changed, among the other requirements, the labelling rules, and the minimum and maximum vitamin limits allowed in supplements. This means that some products considered to be medicines and that were required to be registered with the Ministry of Health will be exempted from registration as they will be considered as dietary supplements. With these changes, fiscalisation of supplements will be increasingly needed, as more products will be exempt from registration with the Ministry of Health, facilitating their commercialization . To evaluate the content of vitamin C in supplements, a fast, simple, and low-cost analytical method is essential. Therefore, the objective of this work was to compare two methods, automatic potentiometric titration and HPLC, for vitamin C determination in vitamin supplements.
2. Materials and Methods
2.1. Samples
A total of 22 vitamin supplements from different brands, available in tablet form, sugar-coated tablets, hard and soft gelatine capsules, and solutions, were acquired in trade from Sao Paulo, Brazil. Their vitamin C content was evaluated using HPLC and titrimetric methods. Acquisition of supplements was made randomly according to market availability in Sao Paulo, Brazil. In order to obtain a representative sample of marketed products, the following criteria were used to select the samples: different matrices and compositions, national and imported samples, supplements from multinational and from small national companies. Most of the samples (73%) were of national origin (Table 1).
2.2. Standards and Reagents
The L-ascorbic acid standard was purchased from Sigma-Aldrich (St Louis, USA); the standard reference material SRM 3280—Multivitamin/Multielement tablet was obtained from the National Institute of Standards and Technology (NIST); analytical grade reagents potassium iodide (KI) and potassium iodate (KIO 3) were purchased from Synth (Sao Paulo, Brazil); sulfuric acid (H 2 SO 4) and ortho-phosphoric acid (H 3 PO 4) were from Merck (Darmstadt, Germany); sodium phosphate monobasic (NaH 2 PO 4) was from Calbiochem (San Diego, EUA), and HPLC grade methanol from Carlo Erba (Milan, Italy).
A 0.05 M sodium phosphate buffer solution (pH 3.0) was prepared weekly by dissolving 6.9 g of monobasic sodium phosphate in 1 L of Milli-Q water and the pH adjusted to 3.0 ± 0.1 with addition of 85% orthophosphoric acid. The solution
Sample Supplement Form Origin Composition
Vitamin/Mineral
1 Tablet Canada 13/10
2 Tablet Argentina 13/10
3 Tablet USA 13/11
4 Tablet Brazil 13/10
5 Tablet Brazil 10/2
6 Tablet USA 13/10
7 Hard capsule Brazil 11/9
8 Tablet Brazil 1/0
9 Tablet Brazil 3/3
10 Tablet Canada 13/10
11 Tablet France 8/4
12 Sugar coated tablet Brazil 9/3
13 Sugar coated tablet Brazil 9/3
14 Hard capsule Brazil 12/10
15 Hard capsule Brazil 13/7
16 Hard capsule Brazil 3/4
17 Soft Gelatine capsule Brazil 13/10
18 Soft Gelatine capsule Brazil 8/2
19 Soft Gelatine capsule Brazil 13/10
20 Soft Gelatine capsule Brazil 10/2
21 Solution Brazil 4/2
22 Solution Brazil 11/1
Table 1. Sample forms, origin, and quantity of vitamin/mineral contents of supplements evaluated.
was placed in an ultrasonic bath for 10 min and filtered through 0.45 μm membranes prior to use.
2.3. Chromatographic Determination of Vitamin C
The determination of vitamin C was conducted according to Abe-Matsumoto . Standard: The vitamin C standard was dissolved in 0.05 M sodium phosphate buffer and the calibration curve was constructed at five levels in the concentration range 29.8 to 476.9 µg∙mL−1. About 350 mg of the grounded sample was weighed into a 50 mL screw capped polypropylene tube. 35 mL of 0.05 M sodium phosphate buffer (pH 3.0) was added to the tubes, which were vortexed for 20 s, first filtered through a quantitative filter paper and then through a 0.22 µm membranes (Millipore) into amber vials. Chromatographic analyses were performed immediately after the preparation. Chromatographic conditions: Samples were analysed with high performance liquid chromatography using a LC-20AT with an autosampler SIL-20AC, controller CBM-20A, column oven CTO-20A, coupled with diode array detector PDA-20A, supplied by Shimadzu. The separation of compounds was performed using a Shim-pack C18 VP-ODS-2 reversed phase column (150 mm, 4.6 mm, 5 µm particles) from Shimadzu, operating at 35˚C. The mobile phase consisted of methanol (A) and 0.05 M sodium phosphate buffer (B). A gradient mode was applied at a flow rate of 0.8 mL∙min−1, starting at 98% A, rising to 40% A at 9 min, returning to the baseline at 15 min, and maintaining this ratio up to 25 min to reach mobile phase equilibrium. The injection volume was 10 μL and the temperatures in the automatic injector and column oven were maintained at 15˚C and 26˚C, respectively. Vitamin C was detected at 254 nm. Results were expressed in milligrams (mg) of vitamin per serving of supplement. A serving portion refers to the daily intake recommended by the supplement manufacturer.
2.4. Potentiometric Determination of Vitamin C
The potentiometric determination of vitamin C was performed on the same day as the HPLC analysis to avoid potential interferences in the sample comparison due to degradation. The iodometric titration for vitamin C quantification was conducted with an automatic potentiometric titrator Titrando 905 (Metrohm Pensalab, Herisau, Switzerland). The platinum ring electrode was controlled by Tiamo® Software . Between 200 and 600 mg of powdered sample was weighed into a 150 mL glass beaker. For oily matrices, amounts of samples between 150 and 500 mg were weighed in 50-mL polypropylene tubes to which 5 mL of 5% metaphosphoric acid and 5 mL of hexane were added and were vortexed for 1 min. An aliquot of the aqueous phase was pipetted into a 150-mL beaker for titration. For liquid samples, volumes between 2 and 5 mL were pipetted into a 150-mL beaker. A volume of 10 mL of 20% H 2 SO 4 solution, 90 mL of distilled water and 1 mL of 10% KI solution were added to the beaker. After 30 s of homogenization with a magnetic stirrer, the sample was titrated with a 0.002 M KIO 3 solution in an automatic titrator. The results were expressed in mg of vitamin C per serving.
2.5. Validation of the Methods
The validation of the methods was carried out according to the International Conference on Harmonisation (ICH) guide, evaluating the following parameters: linearity, intraday (repeatability) and interday (intermediate precision) precision, accuracy, limits of detection (LOD) and limits of quantification (LOQ). The linearity was verified by the correlation coefficient (R 2) of the calibration curve with samples prepared in triplicate and 5 concentration levels. The intermediate precision was determined as the relative standard deviation (% RSD) with 6 replicates in SRM 3280 reference material on two different days by the same analyst. Vitamin C analyses were performed in triplicate in three concentration with standard-spiked samples, analysed on two different days. The repeatability was evaluated by the % RSD recovery test. Accuracy was assessed with recovery standards spiked in the powder, oily and liquid matrix at three different concentration levels . In addition, the standard reference material SRM 3280 was analysed in 6 replicates, and Z-scores were calculated. For determining the LOD a small concentration of analytes were spiked in blank samples, and the concentration equivalent to a peak area about three times larger than the noise was established as the LOD. By using the formula LOD = LOQ/3.3, the LOQ was established.
2.6. Statistical Analysis
The vitamin C content in vitamin supplements was expressed as mean ± standard deviation. The one-way analysis of variance (ANOVA) and Tukey’s multiple comparison test was used to differentiate between the means of the vitamin C content (p < 0.05) determined with the HPLC and titration method. All statistical analyses were conducted using Microsoft Office Excel (2010) and Action Software . The level of statistical significance was set at 5% for all analyses.
3. Results and Discussion
3.1. Validation
The linearity of the detector response was checked by linear regression. The coefficient of determination (R 2) was higher than 0.99 in the tested range, for both methods (Table 2). Recovery (%) ranged from 98.7% to 100.5 and 98 to 104% for the HPLC and titration method, respectively. These results were considered as satisfactory. The values obtained for the reference material SRM 3280 are also within the expected range, being Z-score values smaller than 2 (Table 2).
Intraday and interday precision, expressed as the relative standard deviation (% RSD), were estimated at three different concentrations for each analyte. Samples were spiked with standard solutions and analysed independently on two different days. The corresponding results are summarized in Table 2 and show the good precision, with RSD values lower than 5% for both methods. The LOD was 3.6 µg∙mL−1 and 1.0 mg, and the LOQ was 12.0 µg∙mL−1 and 3.0 mg for the HPLC and titration methods, respectively.
3.2. Vitamin C Content in the Vitamin Supplements
According to the Brazilian legislation RDC No. 360/2003, the declared label value has a tolerance of 20%, thus the nutrient content determined is not allowed to be less than 80% or more than 120% of the declared value .
HPLC Titration
Range 29.8 - 476.9 µg/mL 3.0 - 40.0 mg
Linearity (R 2)0.99901 0.99999
Accuracy
Recovery (%)98.7 ± 3.4 - 100.5 ± 4.2 98 ± 1 - 104 ±2
SRM (mg/g) (RV)42.2 ± 3.7 42.2 ± 3.7
SRM (mg/g) (AV)40.2 ± 0.3 42.1 ± 0.5
Z-Score 0.4629 0.1353
Precision (%RSD)
Intraday 0.5 - 4.2 0.4 - 3.3
Interday 0.6 - 4.5 0.6 - 3.9
LOD3.6 µg/mL (0.36 mg/g in the sample)1.0 mg
(1.7 mg/g in the sample)
LOQ12.0 µg/mL
(1.2 mg/g in the sample)3.0 mg
(5.0 mg/g in the sample)
Table 2. Validation parameters for HPLC and titration methods.
SRM: Standard reference material; RV: Real value; AV: Analysed value; RSD: Relative standard deviation; LOQ: Limit of quantification; LOD: Limit of detection; Mean ± Standard deviation (n = 6).
Vitamin C levels analysed with the HPLC and titration method in supplement samples are shown in Table 3. The vitamin C value stated on the nutritional information for the supplement label on tablets, capsules and sugar-coated tablets was 45 mg/serving, while the liquid samples presented a value of 30 mg/serving.
For half of the samples there was no statistical difference in the vitamin C content analysed with both methods. For the other half, there were differences of up to 20% observed in the vitamin C content. When comparing the analysed with the declared value on the label, four samples (2, 17, 19, and 20) presented divergent results, considering the 20% tolerance allowed by the legislation (Table 3). Taking legislation into account, samples 13 and 22 showed lower vitamin C concentrations than stated on the label, irrespective of the applied analysis method. On the other hand, samples 2 and 17 presented contents above the declared value only with the titrimetric method, while samples 19 and 20 presented values above the declared value only with the chromatographic method (Figure 1). A higher quantity of micronutrients is permitted by law to ensure the labelling stated by the expiry date. Thus, there would be no major problems in determining vitamin contents above the stated value, but there is a need to improve the extraction of compounds to limit the differences between results.
3.3. Comparison of methods
HPLC requires a vitamin extraction step, while in the titration method the sample is crushed, dissolved and analysed directly without the need for extraction, making the analysis faster. The titration analysis takes about 3 min for a run, while the required HPLC run time is 25 min. Considering that the analysis is always performed in triplicate, HPLC takes at least 90 min per sample, while the titration method needs only 10 minutes to analyse a sample. With regards to the sensitivity of both methods, HPLC has a detection and quantification limit about 4 times lower than that of titration. In multivitamin supplements, vitamin C has the highest concentration compared to other vitamins, and therefore, is not usually analysed at low concentrations, i.e. close to the limit of quantification. Therefore, the LOD and LOQ of the methods are not critical for choosing the best method.
Sample Declared
(mg/per serving)Analysed (mg/per serving)Conformity with legislation
HPLC Titration HPLC Titration
1 45 37.8 ± 0.7 a 45.7 ± 0.4 b c c
2 45 49.2 ± 0.8 a 55.9 ± 0.7 b c above
3 45 35.9 ± 1.7 a 37.5 ± 1.0 a c c
4 45 36.8 ± 0.5 a 39.5 ± 1.7 b c c
5 45 55.4 ± 0.9 a 50.5 ± 0.8 b c c
6 45 51.2 ± 1.6 a 50.6 ± 2.5 a c c
7 45 39.9 ± 0.7 a 45.0 ± 0.7 b c c
8 45 44.2 ± 0.8 a 48.4 ± 0.4 b c c
9 45 39.1 ± 0.0 a 45.7 ± 1.1 a c c
10 45 40.2 ± 1.7 a 49.0 ± 1.6 b c c
11 45 45.7 ± 4.3 a 52.7 ± 0.8 b c c
12 45 39.0 ± 0.3 a 39.5 ± 1.4 a c c
13 45 33.7 ± 0.3 a 32.9 ± 0.5 a lower lower
14 45 36.0 ± 1.8 a 49.8 ± 0.7 a c c
15 45 42.4 ± 1.8 a 47.5 ± 1.2 b c c
16 45 39.5 ± 0.4 a 37.2 ± 3.62 a c c
17 45 45.9 ± 0.95 a 56.9 ± 1.85 b c above
18 45 43.2 ± 1.36 a 43.3 ± 1.17 a c c
19 45 57.0 ± 2.16 a 46.9 ± 0.18 b above c
20 45 55.4 ± 0.91 a 45.0 ± 0.75 b above c
21 30 34.5 ± 0.78 a 31.9 ± 1.16 a c c
22 30 20.5 ± 0.08 a 21.6 ± 0.01 a lower lower
Table 3. Vitamin C content analysed by HPLC and titration.
Different superscript letters in the same line indicate statistical difference (p < 0.05), according to Tukey’s Test. C: in compliance with declared value; lower: vitamin C content lower than declared; above: vitamin C content above that declared.
Figure 1. Percentage of relative variation of the analysed value in relation to the value declared on the label for vitamin C content. Red lines are the 20% tolerance limit permitted by the legislation.
One of the advantages of analysing vitamin C with chromatographic methods is that if the supplement contains other water-soluble vitamins such as B1, B6, niacin and pantothenic acid, they can be analysed simultaneously.
Najwa Fatin and Azrina did analyses on the vitamin C content of citrus fruits and observed significant differences between the HPLC-PDA and indophenol titration method. They indicated that at certain conditions the oxidation-reduction titration may overestimate the vitamin C content of the fruits, because the titration end point was difficult to detect, specifically when intense coloured fruits were used. In addition, reducing substances in the fruit samples can react with the indophenol dye and cause overestimation of the vitamin C content in fruit samples. In this work, the sample colour was not a problem because the end point of the titration was indicated by the platinum ring electrode.
With fruits, there may be differences in the determination of vitamin C, because fruits contain vitamin C in the form of ascorbic acid and dehydroascorbic acid. If the method does not quantify dehydroascorbic acid, the content of vitamin C may be underestimated . Odriozola-Serrano et al. found significant amounts of dehydroascorbic acid in strawberry, tomato, and apple, showing that quantification of this compound is important when fruits and vegetables are analysed. In supplements, the amount of dehydroascorbic acid is insignificant, as there is no mention of this compound in the NIST certified reference material SRM 3280.
4. Conclusion
Both HPLC and titration methods accurately analysed vitamin C in SRM 3280, and no statistically significant difference was observed for the results. However, in the commercial samples, the matrix constituents of some vitamin supplements seem to interfere with the analysis, causing differences in the HPLC and titrimetric results. In samples containing other water-soluble vitamins in addition to vitamin C, such as multivitamin supplements, the HPLC method is recommended, because there is the possibility of simultaneous analyzis of other water-soluble vitamins. In samples containing only vitamin C, potentiometric titration may be more advantageous. While HPLC was more sensitive, the titrimetric method was faster and consumed less reagent. The main advantage of the titration method is its simplicity, requiring only a single equipment and inexpensive chemicals.
Funding
This work was supported by the São Paulo Research Foundation (FAPESP) [grant numbers 2013/23006-4 and 2019/17839-0].
Conflicts of Interest
The authors declare that they have no conflict of interest.
References
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Brasil (2003) Resolução RDC n° 360, de 23 de dezembro de (2003) Aprova o Regulamento Técnico sobre Rotulagem Nutricional de Alimentos Embalados, tornando obrigatória a rotulagem nutricional. Ministério da Saúde, Brasília, DF.
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Bailey, R.L., Gahche, J.J., Miller, P.E., Thomas, P.R. and Dwyer, J.T. (2013) Why US Adults Use Dietary Supplements. JAMA Internal Medicine, 173, 355-361.
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Shaw, G., Lee-Barthel, A., Ross, M.L.R., Wang, B. and Baar, K. (2017) Vitamin C-Enriched Gelatin Supplementation before Intermittent Activity Augments Collagen Synthesis. American Journal of Clinical Nutrition, 105, 136-143.
Spínola, V., Llorent-Martínez, E.J. and Castilho, P.C.J. (2014) Determination of Vitamin C in Foods: Current State of Method Validation. Journal of Chromatography A, 1369, 2-17.
Baghizadeh, A., Karimi-Maleh, H., Khoshnama, Z., Hassankhani, A. and Abbasghorbani, M. (2015) A Voltammetric Sensor for Simultaneous Determination of Vitamin C and Vitamin B6 in Food Samples Using ZrO2 Nanoparticle/Ionic Liquids Carbon Paste Electrode. Food Analytical Methods, 8, 549-557.
Skrovankova, S., Mlcek, J., Sochor, J., Baron, M., Kynicky, J. and Jurikova T. (2015) Determination of Ascorbic Acid by Electrochemical Techniques and Other Methods. International Journal of Electrochemical Sciences, 10, 2421-2431.
AOAC (Association of Official Analytical Chemists) (2012) Official Methods of Analysis. 19th Edition, Association of Official Analytical Chemists, Gaithersburg.
Instituto Adolfo Lutz and Ministério da Saúde (2005) Agência Nacional de Vigil&acirencia Sanitária. IAL. Métodos físico-químicos para análise de alimentos, 4th Edition, 1018 p.
Klimczak, I. and Gliszczyńska-Swiglo, A. (2015) Comparison of UPLC and HPLC Methods for Determination of Vitamin C. Food Chemistry, 175, 100-105.
Brasil (2018) Resolução RDC n° 243, de 26 de julho de 2018. Dispõe sobre os requisitos sanitários dos suplementos alimentares. Ministério da Saúde, Brasília, DF.
Abe-Matsumoto, L.T., Sampaio, G. R. and Bastos, D.H.M. (2018) Do the Labels of Vitamin A, C, and E Supplements Reflect Actual Vitamin Content in Commercial Supplements. Journal of Food Composition and Analysis, 72, 141-149.
International Conference on Harmonization (ICH) (1996) Validation of Analytical Procedures: Methodology, Q2B.
Equipe Estatcamp (2014) Software Action. Estatcamp—Consultoria em estatística e qualidade.
Najwa Fatin, R. and Azrina, A. (2017) Comparison of Vitamin C Content in Citrus Fruits by Titration and High-Performance Liquid Chromatography (HPLC) Methods. International Food Research Journal, 24, 726-733.
Phillips, K.M., Council-Troche, M., McGinty, R., Rasor, A.M. and Tarrago-Trani, M.T. (2017) Stability of Vitamin C in Fruit and Vegetable Homogenates Stored at Different Temperatures. Journal of Food Composition and Analysis, 45, 147-162.
Odriozola-Serrano, I., Hernández-Jover, T. and Martín-Belloso, O. (2007) Comparative Evaluation of UV-HPLC Methods and Reducing Agents to Determine Vitamin C in Fruits. Food Chemistry, 105, 1151-1158.
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Underwriting Manual: Tenancy In Common
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Table of Contents
19.12.1 In General
19.12.2 Title Considerations Relative To A Tenancy In Common
19.12.3 Title Insurance Considerations Pertaining To Tenancies In Common
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19.12.1 In General
19.12.2 Title Considerations Relative To A Tenancy In Common
19.12.3 Title Insurance Considerations Pertaining To Tenancies In Common
V 2
19.12.1
V 2
In General
When two people own the same property at the same time, they own it as joint tenants, tenants by the entirety, tenants in common or as community property. In most jurisdictions, a tenancy in common is a form of joint ownership that comes into existence when real property is transferred, whether by conveyance, inheritance or operation of law, to two or more persons in their own right, unless acquired in partnership, in joint tenancy or by husband and wife as community property or as tenants by the entirety.
The interests of each tenants may accrue under different titles, or accrue under the same title but at different periods, or be conferred by words or limitations. These interests do not need to be equal in quality or duration.
The only necessary unity for the existence of a tenancy in comman is the unity of possession. Tenants in common enjoy no rights of survivorship and their interests as tenants in common pass, at time of death, as assets of their estates. Tenants in common may sell, encumber or devise their interests any way they like. On the transfer of the interest of a tenant in common, the new owner becomes a tenant in common with the remaining owners.
In case of obscurity or vagueness in the wording used for the creation of a joint estate, the normal rule of construction is to presume the creation and existence of a "tenancy in common."
19.12.2
V 2
Title Considerations Relative To A Tenancy In Common
The interest of the co-tenants may be equal or unequal.
If the grant does not specify the extent or interest of each co-tenant, there is a rebuttable presumption that the shares are equal. It is considered good drafting to set forth in every real estate instrument the percentage of ownership pertaining to each co-tenant.
Any tenant in common can sell his interest in the property without the consent of the co-tenants, but no co-tenant can attempt to transfer the entire property without the consent of all the co-tenants.
Each percentage of ownership must be accounted for. The interest of each tenant can be attached by creditors, divested in bankruptcy or conveyed and is subject to encumbrance as a separate and distinct interest.
Where title is being acquired by two or more persons as tenants in common, the deed should clearly define their respective interests. If their respective interests are not set forth in the instrument of acquisition or otherwise disclosed, there is a presumption that their interests are equal; but this presumption is not conclusive and may be overcome by evidence showing that by virtue of unequal contributions to the purchase price or otherwise the co-tenants hold unequal interests.
One co-tenant can never, acting alone in his individual capacity, sell or convey more than his own interest in property held in common with other parties. If attempted, a conveyance of more than that interest will be effective only as to the interest of the grantor.
One co-tenant can never, acting alone in his individual capacity, mortgage more than his own interest in property held in common with other parties. The mortgagee in such a mortgage may foreclose in the usual manner but can foreclose only the undivided interest held as security. Upon completion of foreclosure (unless redeemed), the mortgagee will succeed to the interest of mortgagor and become a tenant in common in his place.
One co-tenant can never, acting alone in his individual capacity, impose or grant an easement on the entire commonly held property. One co-tenant can grant an easement binding his own interest in the common property but such will not create in the grantee rights against the other tenants in common.
Although a foreclosure of most liens affecting the interest of one tenant in common will result only in a foreclosure of that undivided interest certain federal court decisions have held that the United States may enforce a federal tax lien by a sale of the full fee simple title in which the delinquent taxpayer holds an undivided interest. By so doing, the United States may subject all the other owners of undivided interest to the foreclosure by notice and by providing in the judgment that their proportionate part of the net proceeds of the foreclosure sale is to be paid over to them. The government may subject the whole property to sale with proportionate reimbursement to the other co-tenants.
19.12.3
V 2
Title Insurance Considerations Pertaining To Tenancies In Common
If only a percentage interest of a tenancy in common is to be insured in one or more but not all of the tenants:
The policy should specify the source of title substantially as follows:
A and B as to undivided 60% interest as tenants in common, such interest having been derived by deed dated _ , from _ , recorded __ in Book __ , Page __ , in the Recorder's Office for _ County, ____.
Any federal tax lien against another co-tenant must be shown as an exception in Schedule B since a lien against one co-tenant's interest may affect all the property.
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Deriving Chi-Square formula for contingency table 2x2
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Given a 2x2 contingency table, with a, b, c, and d as the observed values and n as the total, I have to show from the fact that
Chi-Square = sum(((O-E)^2)/E)
Where O is the observed value, and E is the expected value of the observed, that:
Chi-Square= (n(ad-bc)^2)/(a+b)(a+c)(b+d)(c+d)
I just can't seem to get there. Any links or tips about how this formula is derived?
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edited Feb 12, 2019 at 4:59
Suraj Kumar
5,664 8 8 gold badges 23 23 silver badges 45 45 bronze badges
asked Feb 26, 2018 at 2:58
Suhar PrasetyoSuhar Prasetyo
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This is something I was looking for a long time. My major was in Biology. It doesn't affect my work, but I am curious about how to get to the final equation. This question was asked before here math.stackexchange.com/questions/1704169/… with more details.Raj006 –Raj006 2019-02-15 16:37:11 +00:00 Commented Feb 15, 2019 at 16:37
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7235 | https://mathstrek.blog/2011/11/06/order-of-an-element-modulo-m-and-applications-part-i/ | Mathematics and Such
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← Homework (29 Oct 2011)
Order of an Element Modulo m and Applications – Part II →
Order of an Element Modulo m and Applications – Part I
Posted on November 6, 2011 by limsup
Background required : modular arithmetic
If you’ve any experience observing powers of numbers, you’d have noticed that the last digit runs in cycles: e.g. if you take the last digits of successive powers of 7, you get 7 → 9 → 3 → 1 → 7 → 9 → 3 → 1 → … . More generally, if you start with a positive integer m and take an a which is coprime to m then it’s not hard to see that the successive powers 1, a, a2, a3, … modulo m must run in a cycle eventually. Why? Because since each of these powers is congruent to some c mod m, 0 ≤ c ≤ m-1, and there are only finitely many possibilities of c, eventually we have ai ≡ aj (mod m) for some 0 < i < j. And since (a, m) = 1, we can perform cancellation by ai to obtain aj-i ≡ 1 (mod m).
E.g. if we pick mod 11, and a = 2, the cycle is given by:
1 → 2 → 4 → 8 → 5 → 10 → 9 → 7 → 3 → 6 → 1
Or if we pick mod 35 and a = 3, we get:
1 → 3 → 9 → 27 → 11 → 33 → 29 → 17 → 16 → 13 → 4 → 12 → 1
Recall that Fermat’s Little Theorem gave us an upper bound for the length of the cycle when p is prime, since ap-1 ≡ 1 (mod p) if a is not divisible by p. For the general case, we have Euler’s Theorem:
Theorem. Let m be a positive integer and (a, m) = 1. Then , where φ(m) = Euler’s totient function = the number of i between 0 and m-1 inclusive such that i and m are coprime.
Proof. Let be the distinct elements in [0, m-1] which are coprime to m. Consider the products . For each i = 1, 2, …, φ(m), we can pick ci between 0 and m-1 such that . Now all the ci‘s are distinct: indeed if ci = cj then and since a is coprime to m, we get , which is a contradiction. Hence, the ci‘s are just a permutation of the bi‘s. If we take the product of them all, we get:
Since each bi is coprime to m, we can apply cancellation to obtain . ♦
Thus, Euler’s theorem gives us an upper bound on the length of a cycle when we take the successive powers of an element modulo m.
Let a be an integer coprime to m – we write ordm(a) for the smallest positive integer d for which . In other words, ordm(a) is the length of the cycle when we consider the successive powers of a modulo m. Hence, the above examples give us ord11(2) = 10, ord35(3) = 12. Note that we always have ordm(1) = 1.
Caution. The notation ordm(a) is sometimes used for another wholly different concept. When p is a prime, ordp(a) is also used to denote the highest r for which pr divides a. Thus, ord5(500) = 3 for example. It’s thus safer to describe in words: let d be the multiplicative order of a modulo m …
The key property we shall make use of is as below.
Let d = ordm(a). Suppose k is some positive integer such that . Then k must be a multiple of d.
From the cycle diagram this should be obvious, but let’s give a quick proof anyway.
Proof. Dividing k by d, we can write k = qd + r, where 0 ≤ r ≤ d-1. Then we get:
and so . Since r < d and we assumed d to be the smallest positive integer satisfying , we must have r = 0, and so k is a multiple of d. ♦
Coming up next: applications in problem solving …
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7236 | https://users.ece.cmu.edu/~xinli/classes/cmu_18660/Lec17.pdf | Slide 1 18-660: Numerical Methods for Engineering Design and Optimization Xin Li Department of ECE Carnegie Mellon University Pittsburgh, PA 15213 Slide 2 Overview Constrained Optimization Linear equality constraint Lagrange multiplier Slide 3 Constrained Nonlinear Optimization Equality constraint can be written as two inequality constraints ( ) ( ) ( ) ≤ ≤ 0 0 S.T.
min 2 1 X g X g X f X ( ) ( ) 0 S.T.
min = X g X f X ( ) ( ) ( ) ≤ − ≤ 0 0 S.T.
min X g X g X f X Slide 4 Linear Equality Constraint Linear equality constraint can be efficiently handled by a number of optimization algorithms We do not write AX = B as two inequality constraints It can be directly solved with high efficiency ( ) B AX X f X = S.T.
min Slide 5 Subspace Reduction Eliminate linear equality constraint X = FZ+D is the (non-unique) solution of under-determined linear equation AX = B For any Z value, AX is equal to B B AX = ( ) P N R Z D FZ X − ∈ ∀ + = P×N P×1 N×(N−P) (N−P)×1 Slide 6 A Simple Example 1 2 1 = + x x + ⋅ − = 1 0 1 1 2 1 z x x 1 2 1 + − = = z x z x B AX = D FZ X + = Slide 7 Subspace Reduction Solve the optimal value Z by unconstrained optimization – minimizing f(FZ+D) Calculate the optimum X = FZ+D ( ) B AX X f X = S.T.
min D FZ X + = ( ) D FZ f Z + min Slide 8 Lagrange Multiplier Equality constraint can also be handled by Lagrange multiplier If X is a local minimum of there exist λ1, λ2, ..., λP, called Lagrange multipliers, such that ( ) ( ) ( ) P i X g X f i X , , 2 , 1 0 S.T.
min = = ( ) ( ) 0 1 = ∇ ⋅ + ∇ ∑ = P i i i X g X f λ Slide 9 A Simple Example ( ) ( ) 0 2 S.T.
min 2 2 2 1 2 1 , 2 1 = − + = + = x x X g x x X f x x x1 x2 Gradient ( ) X f ∇ ( ) X g ∇ ( ) ( ) − − = ∇ = ∇ − − = 2 2 1 1 1 1 X g X f X Slide 10 Lagrange Multiplier Optimality condition for linear constraints ( ) ( ) ( ) P i X g X f i X , , 2 , 1 0 S.T.
min = = ( ) ( ) 0 1 = ∇ ⋅ + ∇ ∑ = P i i i X g X f λ ( ) B AX X f X = S.T.
min ( ) ( ) ( ) 0 : , = − ⋅ = i B X i A X gi ( ) ( ) T i i A X g : , = ∇ 0 = − ⋅ B X A i-th row Slide 11 Lagrange Multiplier Optimality condition for linear constraints ( ) ( ) ( ) P i X g X f i X , , 2 , 1 0 S.T.
min = = ( ) ( ) 0 1 = ∇ ⋅ + ∇ ∑ = P i i i X g X f λ ( ) ( ) T i i A X g : , = ∇ ( ) ( ) V A i A X g T P i T i P i i i = ⋅ = ∇ ⋅ ∑ ∑ = = 1 1 : , λ λ ( ) ( ) ⋅ 2 1 : , 2 : , 1 λ λ T T A A AT V ( ) 0 = + ∇ V A X f T Slide 12 Linear Equality Constrained Quadratic Programming We first consider quadratic cost function Any smooth nonlinear cost function can be locally approximated as a quadratic function (2nd-order Taylor expansion) ( ) B AX X f X = S.T.
min ( ) 0 = + ∇ V A X f T ( ) B AX C X R QX X X f T T X = + + = S.T.
2 1 min ( ) R QX X f + = ∇ 0 = + + V A R QX T Slide 13 Linear Equality Constrained Quadratic Programming Optimality condition for quadratic programming B AX C X R QX X T T X = + + S.T.
2 1 min B AX V A R QX T = = + + 0 − = ⋅ B R V X A A Q T 0 Solve linear equation to determine X (optimal solution) and V (Lagrange multipliers) Slide 14 Linear Equality Constrained Nonlinear Programming Minimize nonlinear function f(X) given linear constraint AX = B B AX C X R QX X T T X = + + S.T.
2 1 min − = ⋅ B R V X A A Q T 0 ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) ( ) [ ] ( ) ( ) ( ) k k k k k k T k k T k AX B AX X A X X A X A X f X X f X X f X X f − = − ⋅ = − ⋅ = ∆ ⋅ + ∆ ⋅ ∇ + ∆ ⋅ ∇ ⋅ ∆ ⋅ ≈ + + + 1 1 2 1 2 1 ( ) B AX X f X = S.T.
min Slide 15 Linear Equality Constrained Nonlinear Programming ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) k k T k k T k AX B X A X f X X f X X f X X f − = ∆ ⋅ + ∆ ⋅ ∇ + ∆ ⋅ ∇ ⋅ ∆ ⋅ ≈ + 2 1 2 1 B AX C X R QX X T T X = + + S.T.
2 1 min − = ⋅ B R V X A A Q T 0 ( ) [ ] ( ) [ ] ( ) − ∇ − = ∆ ⋅ ∇ k k T k AX B X f V X A A X f 0 2 ( ) ( ) X X X k k ∆ + = +1 Solve linear equation to determine ∆X Update X(k+1) Slide 16 Linear Equality Constrained Nonlinear Programming If X(k) is a feasible solution We can start from an initial solution X(0) that is feasible Even if X(0) is not feasible, X(1) is feasible after one iteration ( ) B AX X f X = S.T.
min ( ) [ ] ( ) [ ] ( ) − ∇ − = ∆ ⋅ ∇ k k T k AX B X f V X A A X f 0 2 ( ) 0 = − k AX B ( ) [ ] ( ) [ ] ∇ − = ∆ ⋅ ∇ 0 0 2 k T k X f V X A A X f ( ) B AX k = Slide 17 A Simple Example 1 S.T.
min 2 1 4 2 4 1 , 2 1 = + + x x x x x x ( ) [ ] 1 1 1 4 2 4 1 = = + = B A x x X f ( ) B AX X f X = S.T.
min ( ) ( ) = ∇ = ∇ 3 2 3 1 2 2 2 1 2 4 4 12 0 0 12 x x X f x x X f ( ) = 0 1 0 X (Feasible solution) Slide 18 A Simple Example 1 S.T.
min 2 1 4 2 4 1 , 2 1 = + + x x x x x x [ ] 1 1 = A ( ) ( ) = ∇ = ∇ 3 2 3 1 2 2 2 1 2 4 4 12 0 0 12 x x X f x x X f ( ) = 0 1 0 X ( ) [ ] ( ) [ ] ∇ − = ∆ ⋅ ∇ 0 0 2 k T k X f V X A A X f ( ) [ ] ( ) [ ] = ∇ = ∇ 0 4 0 0 0 12 0 0 2 X f X f − = ∆ ⋅ 0 0 4 0 1 1 1 0 0 1 0 12 V X ( ) = − = ∆ 33 .
0 67 .
0 33 .
0 33 .
0 1 X X Slide 19 A Simple Example 1 S.T.
min 2 1 4 2 4 1 , 2 1 = + + x x x x x x [ ] 1 1 = A ( ) ( ) = ∇ = ∇ 3 2 3 1 2 2 2 1 2 4 4 12 0 0 12 x x X f x x X f ( ) = 33 .
0 67 .
0 1 X ( ) [ ] ( ) [ ] ∇ − = ∆ ⋅ ∇ 0 0 2 k T k X f V X A A X f ( ) [ ] ( ) [ ] = ∇ = ∇ 14 .
0 20 .
1 31 .
1 0 0 39 .
5 1 1 2 X f X f − − = ∆ ⋅ 0 14 .
0 20 .
1 0 1 1 1 31 .
1 0 1 0 39 .
5 V X ( ) = − = ∆ 49 .
0 51 .
0 16 .
0 16 .
0 2 X X Slide 20 Linear Equality Constrained Nonlinear Programming Linear equality constraints can be efficiently handled by subspace reduction or Lagrange multiplier Nonlinear equality constraints and inequality constraints must be handled by a different algorithm Interior point method (also referred to as barrier method) More details in future lectures Slide 21 Summary Constrained optimization Linear equality constraint Lagrange multiplier |
7237 | https://www.m-hikari.com/imf-2010/17-20-2010/krzywkowskiIMF17-20-2010.pdf | International Mathematical Forum, 5, 2010, no. 18, 869 - 874 New Proofs of Some Fibonacci Identities Marcin Krzywkowski Faculty of Applied Physics and Mathematics Gdańsk University of Technology Narutowicza 11/12, 80952 Gdansk, Poland fevernova@wp.pl Abstract Lucas proved in 1876 several identities for Fibonacci numbers. We give elementary and short proofs of them.
Mathematics Subject Classification: 11B39 Keywords: Fibonacci number By Fibonacci sequence we mean the sequence {Fn}∞ n=1 such that F1 = 1, F2 = 1, and Fn = Fn−2 + Fn−1, for n ≥3. The elements of this sequence are called Fibonacci numbers. Lucas proved in 1876 that for every positive integer n we have F2n+1 = F 2 n + F 2 n+1, F2n = F 2 n+1 −F 2 n−1, Pn i=1 Fi = Fn+2 −1, Pn i=1 F2i−1 = F2n, Pn i=1 F2i = F2n+1 −1, see , pages 69, 71, and 79. We give combinatorial proofs of these identities which are elementary and short.
Let us consider dominoes of dimensions 2 × 1 and an area of dimensions 2 × n, where n is a positive integer. Squares of our area are signed as follows: upper from left to right by integers from 1 to n, and lower from left to right by symbols from 1′ to n′. By the i-th column we mean the pair of squares i and i′. By the position of a domino we mean the set of squares on which this domino lies. The covering of the area is the set of positions of dominoes which cover this area. Two coverings are distinguish if and only if proper sets of positions are different. Let the sequence {an}∞ n=1 be such that an is the number of distinguish coverings of the area of dimensions 2 × n. For example, a1 = 1, a2 = 2, and a3 = 3, see Figure 1. We also define a0 = 1.
Figure 1 870 M. Krzywkowski In the following lemma we give a recursive formula for an.
Lemma 1 For every positive integer n ≥3 we have an = an−2 + an−1.
Proof. If we want to cover the area of dimensions 2 × n, then the domino on the square 1 lies horizontally or vertically, see Figure 2.
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Figure 2 If it lies horizontally, then we have to cover the rest area of dimensions 2× (n −2). There are an−2 possibilities of doing that. If it lies vertically, then we have to cover the rest area of dimensions 2 × (n −1).
There are an−1 possibilities of doing that. Adding these numbers, we get an = an−2 + an−1.
Now we prove a relation between elements of sequences {an}∞ n=1 and {Fn}∞ n=1.
Lemma 2 If n is a positive integer, then an = Fn+1.
Proof. We have a1 = 1 = F2, a2 = 2 = F3, and the same recurrent formula effects for both sequences {an}∞ n=1 and {Fn}∞ n=1, so an = Fn+1.
Now let us prove the following formula for a Fibonacci number with an odd index.
Theorem 3 (Lucas, 1876) For every positive integer n we have F2n+1 = F 2 n + F 2 n+1.
Proof. By Lemma 2, the identity above is equivalent to a2n = a2 n−1 + a2 n. Let us consider two halfs of the area of dimensions 2×2n (two areas of dimensions 2 × n each). If we cover the area of dimensions 2 × 2n, then its halfs have common dominoes or do not have common dominoes, see Figure 3.
In the first case it remains to cover independently two areas of dimensions 2× (n −1) each, there are a2 n−1 possibilities of doing that. In the second case we have to cover independently two areas of dimensions 2×n each, there are a2 n possibilities of doing that. Adding these numbers, we get a2n = a2 n−1 + a2 n.
New proofs of some Fibonacci identities 871 .....
.....
.....
.....
Figure 3 Now we prove a formula for a Fibonacci number with an even index.
Theorem 4 (Lucas, 1876) If n is a positive integer, then F2n = F 2 n+1 −F 2 n−1.
Proof. By Lemma 2, the identity above is equivalent to a2n−1 = a2 n −a2 n−2.
Since a2 n −a2 n−2 = (an−2 +an−1)2 −a2 n−2 = a2 n−1 +2an−2an−1, it suffices to prove that a2n−1 = a2 n−1 +2an−2an−1. If we cover the area of dimensions 2×(2n−1), then there are the following three possibilities respecting the column n: it is covered by a domino lying vertically, or it is covered (together with column n −1) by pair of dominoes lying horizontally, or it is covered (together with column n + 1) by pair of dominoes lying horizontally, see Figure 4.
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.....
.....
.....
.....
.....
Figure 4 In the first case it remains to cover independently two areas of dimensions 2× (n −1) each, there are a2 n−1 possibilities of doing that. In the second case it remains to cover independently two areas of dimensions 2×(n−2) and 2×n, there are an−2an possibilities of doing that. In the third case, by symmetry to 872 M. Krzywkowski the previous case, there are also an−2an possibilities. Adding these numbers, we get a2n−1 = a2 n−1 + 2an−2an−1.
Now let us observe that Theorem 4 can be also easily proved using Theorem 3. We have F2n = F2n+1−F2n−1 = F 2 n +F 2 n+1−F 2 n−1−F 2 n = F 2 n+1−F 2 n−1. The-orem 3 similarly follows from Theorem 4, as F2n+1 = F2n+2−F2n = F 2 n+2−F 2 n −F 2 n+1 + F 2 n−1 = (Fn + Fn+1)2 −F 2 n −F 2 n+1 + (Fn+1 −Fn)2 = F 2 n + 2FnFn+1 +F 2 n+1 −F 2 n −F 2 n+1 + F 2 n+1 −2FnFn+1 + F 2 n = F 2 n + F 2 n+1.
Now let us prove a formula for n first Fibonacci numbers.
Theorem 5 (Lucas, 1876) For every positive integer n we have n X i=1 Fi = Fn+2 −1.
Proof. By Lemma 2, the identity above is equivalent to Pn−1 i=0 ai = an+1 −1.
Let us consider all possible coverings of the area of dimensions 2 × (n + 1) ex-cluding the covering in which every domino lies vertically. There are an+1 −1 such coverings. Now let us count these coverings in different way. Since we exclude the covering in which every domino lies vertically, every considered covering has at least one pair of dominoes lying horizontally (one above an-other). Let us consider the pair of indices of columns which are covered by first (considering from left side) pair of dominoes lying horizontally. ”The small-est” such possible pair is (1, 2), ”the next” possible is (2, 3), and ”the greatest” possible is (n, n + 1), see Figure 5.
.....
.....
.....
Figure 5 There are ai possibilities of covering the area of dimensions 2 × (n + 1) in such way that the first pair of dominoes lying horizontally covers the columns n −i and n −i + 1. Thus the number of distinguish coverings of the area New proofs of some Fibonacci identities 873 of dimensions 2 × (n + 1) excluding the covering in which every domino lies vertically is equal to Pn−1 i=0 ai.
Now we prove the following formula for n first Fibonacci numbers with even indices.
Theorem 6 (Lucas, 1876) If n is a positive integer, then n X i=1 F2i = F2n+1 −1.
Proof.
The identity above is equivalent to Pn i=1 a2i−1 = a2n −1.
Let us consider all possible coverings of the area of dimensions 2 × 2n excluding the covering in which every domino lies horizontally. There are a2n −1 such cov-erings. Let us count these coverings in different way. Since we exclude the covering in which every domino lies horizontally, every considered covering has a domino lying vertically. Let us consider the column which is covered by the first (considering from the left side) domino lying vertically. The index of this column is odd, because the part of the area on the left side of that col-umn has an even length, as it is covered only by dominoes lying horizontally.
The smallest such possible index is 1, the next possible is 3, and the greatest possible is 2n −1, see Figure 6.
.....
.....
.....
Figure 6 There are a2i−1 possibilities of covering the area of dimensions 2 × 2n in such way that the first (considering from the left side) domino lying vertically covers the column 2n−2i−1. Thus the number of coverings of the area of dimensions 2 × 2n excluding the covering in which every domino lies horizontally is equal to Pn i=1 a2i−1.
Now we prove a formula for n first Fibonacci numbers with odd indices.
874 M. Krzywkowski Theorem 7 (Lucas, 1876) For every positive integer n we have n X i=1 F2i−1 = F2n.
Proof. The identity above is equivalent to Pn−1 i=0 a2i = a2n−1. Let us consider all possible coverings of the area of dimensions 2 × (2n −1). There are a2n−1 such coverings. Let us count these coverings in different way. Every counted covering has a domino lying vertically, because our area has an odd length.
Let us consider the column which is covered by first (considering from the left side) domino lying vertically. The index of this column is odd, because the part of the area on the left side of that column has an even length, as it is covered only by dominoes lying horizontally. The smallest such possible index is 1, the next possible is 3, and the greatest possible is 2n −1, see Figure 7.
.....
.....
.....
Figure 7 There are a2i possibilities of covering the area of dimensions 2 × (2n −1) in such way that the first (considering from the left side) domino lying vertically covers the column 2n −2i −1. Thus there are Pn−1 i=0 a2i distinguish coverings of the area of dimensions 2 × (2n −1).
It is easy to see that Theorem 5 follows from Theorems 6 and 7, as P2n i=1 Fi = Pn i=1 F2i−1 + Pn i=1 F2i = F2n + F2n+1 −1 = F2n+2 −1. Similarly, Theorems 6 and 7 follow from each other.
References T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley-Interscience, Canada, 2001.
Received: November, 2009 |
7238 | https://www.youtube.com/watch?v=NCpY9YnMzbQ | MPMP8: Folded Paper Puzzle
James Explains
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Posted: 11 Jun 2020
In this video we take a look at the folded paper puzzle from Stand-up Maths. We take a look at how the physical properties of the folding can help us determine whether a certain order of the sheets is possible, and go on to work out what the different possible orders are for a sheet of 8 sections.
Original Video:
More Puzzles:
If you have any suggestions on what I should make a video on, let me know in the comments down below!
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Transcript:
Introduction welcome to another episode of James explains where today we are again taking a look at one of Matt Parker's math puzzles in this puzzle we have a sheet of paper folded into eight sections where the aim of the puzzle is to fold it in such a way that the stack of folded sections are in a particular order from top to bottom to see the examples given and for the original puzzle check out the original video on the stand-up maths channel the extended puzzle was to find out how many ways a stack can be folded to arrange the sections into different orders to make Folding a Stack things easier we will number all of our sections from 0 to 7 let's start by looking at what happens when we make a fold as we fold the line between 0 & 1 the section numbered 1 disappears below the 0 if we are to remove the 0 we can see the section number 1 is now face down if we limit section 0 to always being face-up no matter what other folds happen before or after the only possible direction for section 1 to face is down we can continue this pattern and see that with section 1 facing down section 2 must face up and therefore section 3 must face down 2 if we make a fold along the horizontal line you can see that sections 4 & 6 will be face down and sections 5 & 7 will be face up as well as this all these lower sections will also be upside down try folding a piece of paper randomly for yourself and you'll see that no matter what order you make the folds in the individual sections will always face in these directions now that we know this let's take a look at section 5 we can see that it will end up face up and upside down so it should look somewhat like this once all the folds have been made we can also see that it will also have a folded join with section 6 on the left edge of the stack I folded join with section 1 on the bottom edge of the stack and a folded join with section 4 on the right edge of the stack if we Checking the Folds have a look at our whole stack we can check which folds will be on which edges this can be checked by either taking each individual section from the earlier layout rotating them and noting the edges or by folding a stack and just checking where the folds end up if you try this you'll see that on the Left we have the fold between sections one and two as well as the fold between sections five and six on the right edge we have the folds between 0 & 1 2 & 3 4 & 5 & 6 & 7 on the bottom edge we have the folds between 0 & 4 1 & 5 2 & 6 & 3 & 7 so why is knowing where the folds are important to us well let's take a look at a Theoretical Example theoretical example with four pages a B C and D we will say there is a fold between a and B and one between C and D we will also say that both these folds are on the left side of the stack if we arrange the stack so the order is ABCD we can show that there is one fold between a and B and another fold between C and D these folds are one atop another so this should work fine if we rearrange the stack so the order is now a DC B we can see the folds between a and B as well as the fold between C and D these folds are nested but it still should work fine now let's rearrange it again so the stack is ordered a d B C we can show the fold between a and B but when we try and show the fold between D and C we can see that the folds must overlap due to the limitation of paper in three dimensions this is not possible to achieve however this could be something that is fun to make or do in the fourth dimension going back to our original Permutations stack of eight sections we know that the zero section is locked into the first position so what we need to do is check all permutations of the remaining seven positions luckily this is only 5040 different permutations to check it might take a while to do by hand that should be much quicker using your computer but what exactly do we have to check about each permutation let's take a look at the Example example given by Matt Parker taking the section number of each letter in turn we can see the order of the stack we are looking for is 0 4 5 3 7 6 2 1 we also know that there are 2 folds on the left edge 4 on the right edge and 4 on the bottom edge 2 starting with a left edge we can see there's a fold between 1 and 2 and another fold between 5 and 6 as these folds are separated so far we don't have any problems with the order of this stack checking the bottom edge we can see there are folds between 0 & 4 1 & 5 2 & 6 & 3 & 7 while some of the folds are nested inside other folds none of them cross one another so this stack order is still looking good finally with the right edge we have a fold between 0 & 1 2 & 3 4 & 5 & 6 & 7 in this case we have even more nested folds but again none of the folds cross so now that we have checked all the faults we can say that this stack order is definitely possible now for a contrary example let's change around some of the positions with a new order of 0 1 5 6 7 4 2 3 we can demonstrate the default between 0 & 4 & the fold between 2 & 6 must cross each other whilst there are more folds in this permutation that are crossing once we have found one we know that the order is not possible and can discard it so now that we know what we're looking for we can check all 5040 permutations that are possible to see how many of them have no crossing folds whilst this is probably Conclusion way too much work to do by hand writing a simple program was able to find the following results pretty much instantly if we substitute the numbers for letters with zero being a one being B and so on we are left with the following possible starting positions for the folded paper challenge to try to verify that we have found valid orders we can take a look for the different orders presented in Matt Parker's video first we have the sample sheet shown at the start with an order of Adah B CFG then we have the first puzzle of a HGD bc f a and the second puzzle of a HB g da c f so if you've already managed to make the folds for the puzzles given in the original video why not try to see how many more of these you can achieve if you would like to see the original puzzle or many more like it visit think math co uk slash math puzzles and if you have any suggestions on what else I should make a video on or if you found any flaws in the results I have shown make sure you let me know in the comments section down below as always thanks for watching |
7239 | https://www.napf.ru/news/napf_news_market/byudzhet-na-tri-s-polovinoy/ | ===============
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Бюджет на три с половиной
В 2024 году расходы опередили доходы на 3,5 трлн рублей
Дефицит федерального бюджета в 2024 году составил, по предварительной оценке, 1,7% ВВП, сообщил Минфин. Это почти вдвое больше, чем предполагалось изначально, но укладывается в осенние ожидания ведомства. Доходы по сравнению с 2023 годом выросли на 26%, расходы — на 24%. Нефтегазовые поступления поддержали некоторое ослабление рубля и относительно высокая цена нефти, прочие доходы — сборы НДС, которые в первой трети года заметно росли на восстановлении деловой активности.
Фото: Анатолий Жданов, Коммерсантъ
Минфин наконецподвелпубличные итоги исполнения федерального бюджета в 2024 году. При доходах в 36,707 трлн руб. расходы составили 40,192 трлн руб. Дефицит — 3,485 трлн руб., или предварительно 1,7% ВВП. Дефицитным бюджет складывается все три военных года: в 2022-м доходы отстали от расходов на 2,1%, в 2023-м — на 1,9% ВВП. В профиците «госказна» в последний раз закрывалась в 2021 году (см. график).
Как и в предыдущие годы, основные параметры фактического исполнения отличаются от того, что было изначально заложено Минфином в законе о бюджете-2024: доходы оказались на 5%, а расходы — на 10% выше запланированных. Это при том, что жестким бюджет не был: при его верстке уже был заложен заметный рост как поступлений, так и трат — на 20% и 13% соответственно к итогу 2023 года.
Из-за увеличения расходов в конце года дефицит-2024 получился почти вдвое выше начального плана в 0,9% ВВП (в середине года поправками к закону о бюджете он было увеличен до 1,1% ВВП). Итоговые 3,5 трлн руб. дефицита, как и прежде, закрыты из двух главных источников. 1,3 трлн руб. были взяты из Фонда национального благосостояния — его ликвидная часть (юани и золото) в результате «похудела» еще на треть и на 1 января 2025 года составляла 3,8 трлн руб. Остальной дефицит был закрыт внутренними заимствованиями через размещение ОФЗ.
Сравнивая итоги исполнения бюджета-2024 уже не с начальным планом, а с результатами 2023 года, отметим получившийся почти равномерным прирост обеих его частей.
Доходы приросли к уровню предыдущего года на 26% (как нефтегазовые, так и ненефтегазовые), расходы — на 24%.
Подробностей про доходную часть Минфин давно не раскрывает. Впрочем, про два налога, составляющих основу федерального бюджета (70% всех поступлений), известно. НДС, внутреннего и внешнего (то есть с производства и с импорта), собрано в минувшем году на 13,5 трлн руб.
Это на 16% больше, чем было в 2023-м— помогли восстановление деловой активности и, соответственно, рост оборотов бизнеса (прежде всего в первой трети года). НДПИ собрано 12,2 трлн руб. (по факту несколько меньше из-за возвратных выплат нефтяникам). Прирост — 29%.
Отметим, что увеличение нефтегазовых доходов произошло на фоне некоторого снижения в 2024 году российской нефтедобычи и ее экспорта (по данным Международного энергетического агентства) и при несколько «отставшей» от первоначального плана цене Urals. По данным Минэкономики, баррель российской нефти в среднем стоил в 2024 году $67,85 при изначально заложенной в закон о бюджете цене $70,3 (сниженной в середине года поправками до $65). Проседания доходов удалось избежать благодаря некоторому ослаблению рубля. При запланированном при верстке бюджета курсе 90,1 руб./$ по факту за 2024 год, по данным ЦБ, он сложился на среднем уровне 92,44 руб./$.
Структуру фактических расходов Минфин с мая 2022 года не раскрывает, ссылаясь на желание снизить риски введения новых санкций. На что пойдут деньги налогоплательщиков, сообщается только при принятии бюджета. Так, было известно, что военные ассигнования в 2024 году должны были быть увеличены в 1,7 раза по сравнению с 2023 годом, достигнув 25% всех трат (32,5% в 2025-м).
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7240 | https://www.andrews.edu/~rwright/Precalculus-RLW/Text/05-02.html | Published Time: Thu, 07 Nov 2024 17:54:06 GMT
5-02 Fundamental Trigonometric Identities, Part B
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Precalculus by Richard Wright
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Blessed are those who hunger and thirst for righteousness, for they will be filled. Matthew 5:6 NIV
5-02 Fundamental Trigonometric Identities, Part B
Mr. Wright teaches the lesson.
Summary: In this section, you will:
Factor and multiply trigonometric expressions.
Use trigonometric identities with rational expressions.
Use trigonometric substitution.
SDA NAD Content Standards (2018): PC.5.1
Waterslide. (pixabay/Tania Dimas)
The coefficient of friction is a measure of how slippery a surface is and is used to calculate force of friction. The fundamental trigonometric identities can be used to simplify expressions for the coefficient of friction.
More Applications of the Fundamental Trigonometric Identities
Review the fundamental trigonometric identities in lesson 5-01. This lesson will continue using the identities to simplify trigonometric expressions.
Factor Trigonometric Expressions
Factor the expression, then use fundamental trigonometric identities to simplify.
csc 4 x − cot 4 x
Solution
This is a difference of squares. (a 2 − b 2) = (a − b)(a + b)
csc 4 x − cot 4 x
(csc 2 x − cot 2 x)(csc 2 x + cot 2 x)
The Pythagorean identity, 1 + cot 2 x = csc 2 x, lets us substitute for csc 2 x.
((1 + cot 2 x) − cot 2 x)((1 + cot 2 x) + cot 2 x)
1 + 2 cot 2 x
Factor and simplify 1 + cos x − cos 2 x − cos 3 x.
Answer
sin 2 x (1 + cos x)
Multiply Trigonometric Expressions
Multiply and use trigonometric identities to simplify.
(3 sin x − 3)(3 sin x + 3)
Solution
Multiply.
(3 sin x − 3)(3 sin x + 3) = 9 sin 2 x − 9
Factor out the 9.
9(sin 2 x − 1)
A Pythagorean identity relates sin 2 x and 1. Solve sin 2 x + cos 2 x = 1 for sin 2 x to get sin 2 x = 1 − cos 2 x and substitute.
9(1 − cos 2 x − 1)
9(−cos 2 x)
−9 cos 2 x
Simplify (cos x − 1)(cos x + 1).
Answer
−sin 2 x
Add or Subtract Fractions
Add or subtract the expressions and use trigonometric identities to simplify.
1 cos x+1−1 cos x−1 1 cosx+1−1 cosx−1
Solution
Use a common denominator to add the fractions.
1 cos x+1−1 cos x−1 1 cosx+1−1 cosx−1
cos x−1(cos x+1)(cos x−1)−cos x+1(cos x+1)(cos x−1)cosx−1(cosx+1)(cosx−1)−cosx+1(cosx+1)(cosx−1)
–2 cos 2 x−1–2 cos 2x−1
Use the Pythagorean identity, sin 2 x + cos 2 x = 1, solved for cos 2 x to substitute. cos 2 x = 1 − sin 2 x
–2 1−sin 2 x−1–2 1−sin 2x−1
–2−sin 2 x–2−sin 2x
2 csc 2 x
Simplify 1 1+sec x−1 1−sec x 1 1+secx−1 1−secx
Answer
−2 csc x cot x
Rewrite Fractions
Rewrite the fraction so it is not in fractional form.
1 csc x−cot x 1 cscx−cotx
Solution
Multiply the numerator and denominator by the conjugate of the denominator. This will make the denominator have csc 2 x and cot 2 x which are related by a Pythagorean identity.
1 csc x−cot x 1 cscx−cotx
csc x+cot x(csc x−cot x)(csc x+cot x)cscx+cotx(cscx−cotx)(cscx+cotx)
csc x+cot x csc 2 x−cot 2 x cscx+cotx csc 2x−cot 2x
Substitute the Pythagorean identity, 1 + cot 2 x = csc 2 x.
csc x+cot x(1+cot 2 x)−cot 2 x cscx+cotx(1+cot 2x)−cot 2x
csc x + cot x
Rewrite the fraction so that it is not in fractional form: sin 2 x 1−cos x sin 2x 1−cosx.
Answer
1 + cos x
Use Trigonometric Substitution
Use trigonometric substitution to write the algebraic expression as a trigonometric function.
16−x 2−−−−−−√16−x 2; x = 4 sin θ
Solution
Substitute the trigonometric expression for x.
16−x 2−−−−−−√=16−(4 sin θ)2−−−−−−−−−−−√16−x 2=16−(4 sinθ)2
16−16 sin 2 θ−−−−−−−−−−−√16−16 sin 2θ
16(1−sin 2 θ)−−−−−−−−−−−√16(1−sin 2θ)
The Pythagorean identity, sin 2 x + cos 2 x = 1, relates sin 2 x and 1. Solve this for sin 2 x and substitute. sin 2 x = 1 − cos 2 x
16(1−(1−cos 2 θ))−−−−−−−−−−−−−−−−√16(1−(1−cos 2θ))
16 cos 2 θ−−−−−−−√16 cos 2θ
4 cos θ
Use trigonometric substitution to write the algebraic expression as a trigonometric function.
4+x 2−−−−−√;x=2 cot α 4+x 2;x=2 cotα
Answer
2 csc α
Lesson Summary
Fundamental Trigonometric Identities
Reciprocal Identities
sin u=1 csc u sinu=1 cscu csc u=1 sin u cscu=1 sinu
cos u=1 sec u cosu=1 secu sec u=1 cos u secu=1 cosu
tan u=1 cot u tanu=1 cotu cot u=1 tan u cotu=1 tanu
Quotient Identities
tan u=sin u cos u tanu=sinu cosu cot u=cos u sin u cotu=cosu sinu
Pythagorean Identities
sin 2 u+cos 2 u=1 sin 2u+cos 2u=1
tan 2 u+1=sec 2 u tan 2u+1=sec 2u
1+cot 2 u=csc 2 u 1+cot 2u=csc 2u
Even/Odd Identities
Even
cos(–u) = cos u sec(–u) = sec u
Odd
sin(–u) = –sin u csc(–u) = –csc u
tan(–u) = –tan u cot(–u) = –cot u
Cofunction Identities
sin(π 2−u)=cos u sin(π 2−u)=cosu cos(π 2−u)=sin u cos(π 2−u)=sinu
tan(π 2−u)=cot u tan(π 2−u)=cotu cot(π 2−u)=tan u cot(π 2−u)=tanu
sec(π 2−u)=csc u sec(π 2−u)=cscu csc(π 2−u)=sec u csc(π 2−u)=secu
Helpful videos about this lesson.
Mr. Wright Teaches the Lesson (
Fundamental Trigonometric Identities (
Practice Exercises
Factor the expression, then use fundamental trigonometric identities to simplify.
sin 2 x csc 2 x − sin 2 x
cot 4 x + 2cot 2 x + 1
tan 4 x − sec 4 x
Multiply and use trigonometric identities to simplify.
(cos x + sin x)2
(2sec x + 2)(2sec x − 2)
(csc x − cot x)(csc x + cot x)
Add or subtract the expressions and use trigonometric identities to simplify.
1 1+sin x+1 1−sin x 1 1+sinx+1 1−sinx
cos x 1−sin x−sin x cos x cosx 1−sinx−sinx cosx
cot x−csc 2 x cot x cotx−csc 2x cotx
Rewrite the fraction so it is not in fractional form.
cos 2 x 1−sin x cos 2x 1−sinx
4 sec x+tan x 4 secx+tanx
Use trigonometric substitution to write the algebraic expression as a trigonometric expression.
4−x 2−−−−−√;x=2 sin θ 4−x 2;x=2 sinθ
x 2−36−−−−−−√;x=6 csc θ x 2−36;x=6 cscθ
Rewrite the expression as a single logarithm.
ln |sin x| + ln |csc x|
Problem Solving
If an object slides down an inclined surface at a constant speed, the force of friction has to equal the component of the object's weight pulling it down the surface. The equation becomes
μW cos θ = W sin θ
Object on inclined surface. (wikimedia/J.Spudeman)
where μ is the coefficient of friction, W is the weight, and θ is the incline of the surface. Solve the equation for μ and use fundamental trigonometric identities to simplify the expression.
Mixed Review
(5-01) Simplify (1 − cos 2 x) csc x.
(5-01) Given sin θ=−5√5 sinθ=−5 5 and tan θ< 0, evaluate cos θ and cot θ.
(4-05) Evaluate tan 7 π 4 tan7 π 4 using reference angles.
(3-05) A substance has a half-life of 5.00 minutes. If the initial amount of the substance was 20.0 grams, how many half-lives will have passed before the substance decays to 13.0 grams?
(2-06) Find all the solutions of 0 = x 3 − 7 x 2 + 17 x − 15.
Answers
cos 2 x
csc 4 x
−2 tan 2 x − 1
2 cos x sin x + 1
4 tan 2 x
1
2 sec 2 x
sec x
−tan x
1 + sin x
4 sec x − 4 tan x
2 cos θ
6 cot θ
0
μ = tan θ
sin x
cos θ=2 5√5 cosθ=2 5 5, cot θ = −2
−1
0.621
3, 2 + i, 2 − i
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7241 | https://arxiv.org/pdf/2504.03448 | Published Time: Mon, 07 Apr 2025 00:44:54 GMT
arXiv:2504.03448v1 [math.CO] 4 Apr 2025
A New Dominating Set Game on Graphs
Sean Fiscus ∗ Glenn Hurlbert † Eric Myzelev ‡ Travis Pence §
Abstract
We introduce a new two-player game on graphs, in which players alternate choosing vertices until the set of chosen vertices forms a dominating set. The last player to choose a vertex is the winner. The game fits into the scheme of several other known games on graphs. We characterize the paths and cycles for which the first player has the winning strategy. We also create tools for combining graphs in various ways (via graph powers, Cartesian products, graph joins, and other methods) for building a variety of graphs whose games are won by the second player, including cubes, multidimensional grids with an odd number of vertices, most multidimensional toroidal grids, various trees such as specialized caterpillars, the Petersen graph, and others. Finally, we extend the game to groups and show that the second player wins the game on abelian groups of even order with canonical generating set, among others.
Keywords: dominating set, games on graphs, graph power, Cartesian product, graph join, Cayley graph
MSC 2020: 91A43, 05C57
1
Department of Mathematics, Duke University, Durham, North Carolina, USA, sean.fiscus@duke.edu
2
Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, Richmond, Virginia, USA,
ghurlbert@vcu.edu
3
Department of Mathematics, University of Pennsylvania, Philadelphia, Pennsylvania, USA, myzelev@sas.upenn.edu
4
Department of Computer Science, University of Wisconsin, Madison, Wisconsin, USA, tnpence@wisc.edu
11 Introduction
In this work we introduce a new combinatorial game, which we call the graph domination (GD ) game . On a given graph, two players take turns choosing vertices until the set of chosen vertices forms a dominating set. The winner is the player who was last to choose a vertex. This game is related to a wide range of other games played on graphs, and generalizes the following natural game on groups. For a group Γ with generating set Σ, players alternate choosing elements of Γ until every unchosen element h equals h′g for some chosen h′ and some g ∈ Σ. In this instance, the game played on a group is equivalent to the game played on the Cayley graph of that group. All graphs G = ( V, E ) are assumed to be finite and connected. We denote the path on n vertices by
Pn and the cycle on n vertices by Cn, and refer to the degree 1 vertices of a path as endpoints, and the degree 2 vertices as interior vertices. The open (closed) neighborhoods of a vertex u and of a set of vertices
U are denoted N (u) (resp. N [u] = N (u) ∪ { u}) and N (U ) = ∪u∈U N (u), respectively. We also define the set notation NX (u) = N (u) ∩ X. For u, v ∈ V (G), the function dist G(u, v ) measures the distance between them; i.e., the number of edges in a shortest uv -path. This allows us to define the closed distance-d neighborhood
of u: N d[u] = {v ∈ V | dist (u, v ) ≤ d}. A set of vertices D ⊆ V is called a dominating set if V = N [D]. More generally, D is a distance-d dominating set if V = N d[D].
1.1 History
There is a large body of work in graph theory that studies, not the graphs themselves, but the activities or games involving the placement of objects on the vertices of a graph and movements of them along edges under certain rules for various purposes. Some versions of these model the spread of information or disease, supply-demand optimization, network search, and other applications, while others are studied purely for game-theoretic research. Autonoma such as Conway’s Game of Life and bootstrap percolation can be thought of as 0-person games, while others such as chip firing , zero forcing , and graph burning can be thought of as 1-person games, or puzzles. The objects of some 2-person games like cops & robbers and graph pebbling [10, 18] are moved around by players, while some like Go and Hex are not. Our game is of this last category. An interesting and extremely popular game is the k-person board game RISK ® . Its initial setup of shares some of the above characteristics, as pieces are placed on vertices of an implicit graph in order to eventually dominate it. Several games like ours have been studied. For example, one can play a game in which two players alternate choosing vertices so that the set of chosen vertices is independent. Phillips and Slater initiated
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2the study of the case in which the last player to chose a vertex wins, while Phillips and Slater and Goddard and Henning have studied the case of one player trying to maximize and the other player trying to minimize the size of the final chosen set. Huggan, Huntemann, and Stevens studied the game in which players alternate choosing vertices of a hypergraph H (in their case a block design) so that the chosen set does not contain any edge of H, with the winner being last player able to choose a vertex. As mentioned, in the game we study here, the players alternate choosing vertices until the chosen vertices form a dominating set. In this context, Breˇ sar, et al. [8, 9] have studied the “maximizer-minimizer” version of the game, and Henning, Klavˇ zar, and Rall did the same for total dominating sets. (Another interesting maximize-minimize variation is studied by Alon, et al. , in which players alternate orienting the edges of a graph until the entire graph is oriented, observing the size of the resulting oriented domination number.) Here we study the winner-loser version, the winner being the player who first creates a dominating set from the chosen vertices.
1.2 Game Theory Basics
We follow the traditional language and notation of [4, 11]. Because a GD game is a finite, perfect information game with no ties, we know that one of the players has a winning strategy. At every stage of the game, the player who is next to play is denoted Next with the other player denoted Prev (thought of as “previous”). At the beginning of the game, we say that Player 1 is next, and write P1 = Next , with Player 2 being P2 = Prev .For clarity, notice that after the first move we have Next = P2 and Prev = P1 ; That is, P1 and P2 are the permanent names of the players, while Next and Prev describe which of them is about to play. Then the set
N is defined to be the set of all positions of the game for which Next has the winning strategy, while P is defined to be the set of all positions of the game for which Prev has the winning strategy. By identifying a game with its initial position, we can say that the game is in N (or P) if its initial position is in N (or P). Every position Q corresponds to a set B of chosen vertices, so we define the size of Q to be |B|. Because
Next must always choose a vertex that has not yet been chosen, we will abuse notation slightly by writing
u ∈ Q to mean that u is a move available for N to play from position Q. The position created from Q by
Next choosing the vertex u ∈ Q is denoted Q(u), and corresponds to the chosen set B ∪ { u}. Thus, P and N
can be calculated recursively by first placing the position corresponding to V (G) in P. Then we repeatedly consider all positions Q of size one less than prior: if Q corresponds to a dominating set then place it in P;otherwise, if Q(u) ∈ P for some u then place it in N; otherwise, place it in P. A common technique in game theory for proving that a position is in P is to display a pairing strategy : a function φ from the possible moves for N to those for P, along with a condition C such that all winning positions satisfy C and, for every 3position Q that satisfies C and every move u from Q, Q(u) does not satisfy condition C and Q(u, φ (u)) does satisfy C. We make use of this idea below. Let GD to refer to domination games on general graphs; for games on paths and cycles, we refer to them by PD and CD , respectively.
1.3 GD and PD Game Notations
To assist with the analysis of game positions and player strategies, it will be useful to keep track of more than just the chosen vertices B; in particular, we will also distinguish vertices that are or are not dominated by B. When the graph G is understood (in our case G = Pn or Cn), we will leave it out of the positional notation. With this finer perspective, a GD position Q on G is a partition {B, S, W } of V (G) (for black ,
shaded , and white ) such that no white vertex is adjacent to a black vertex. The interpretation here is that
B is the set of chosen vertices, S = N (B) − B is the set of non-chosen vertices dominated by B, and B is a dominating set of G if and only if W = ∅. GD positions with W = ∅ are called trivial . Thus all trivial positions are in P. We will use the notations BQ, (resp. SQ, WQ) when necessary to indicate the black (resp. shaded, white) vertices of position Q.Define two GD positions Q1 = {B1, S 1, W 1} and Q2 = {B2, S 2, W 2} on graphs G1 and G2 to be isomor-phic , written Q1 ∼= Q2, if there is a graph isomophism ϕ : G1 − B1 → G2 − B2 such that ϕ(S1) = S2 and
ϕ(W1) = W2. Clearly, if Q1 ∼= Q2 then Q1 ∈ P if and only if Q2 ∈ P.Now define the game D = D(G) with initial position Q0 having W = V (G). At any stage of the game with position Q, if WQ 6 = ∅ then Next chooses some vertex u ∈ SQ ∪ WQ, resulting in the position Q(u)having BQ(u) = BQ ∪ { u}, SQ(u) = ( SQ − { u}) ∪ NW (u), and WQ(u) = WQ − ({u} ∪ N (u)).
1.4 Involutions
For a graph G, an involution of G is an automorphism of G of order two. We say that an involution φ
is d-involutionary if, for all v ∈ V (G), dist (v, φ (v)) ≥ d. Define a graph to be d-involutionary if it has
d-involution. Observe that an involution φ is 1-involutionary if and only if it has no fixed points, and is 3-involutionary if and only if N [v] ∩ N [φ(v)] = ∅ for all v ∈ V (G). The definition can be extended to a position Q by requiring that φ be an automorphism of Q. The importance of a 3-involution is that it gives rise to a pairing strategy, as we show in Lemma 1. Suppose that Γ is a group with elements Λ and generating set Σ (which we consider as closed under inverses). For the purposes of this paper, it is not important what set of relations Γ has; consequently, we will use the notation Γ = (Λ , Σ) to denote any such group. The Cayley graph C(Γ , Σ) is defined to be the 4undirected graph G = ( V, E ) with V = Λ and E = {{ g, h } ∈ (Λ2
) | gh −1 ∈ Σ}. We say that (Γ , Σ) is
involutionary if there is some g ∈ Λ − Σ of order 2, and 3-involutionary if g also cannot be expressed as the product of two generators. Thus, if (Γ , Σ) is 3-involutionary then so is C(Γ , Σ). Additionally, we define the game D(Γ , Σ) to be the game described above, in which players alternate choosing elements of Γ until every unchosen element h equals h′g for some chosen h′ and some g ∈ Σ. The last person able to choose an element is the winner. It is easy to see that D(Γ , Σ) = D(G), where G = C(Γ , Σ).
1.5 Graph constructions
Among our results are statements involving the following graph constructions.
Graph power. Given a graph G = ( V, E ) and positive integer d, the dth power of G is denoted G(d) = ( V, F ), where F = {{ x, y } ∈ (V
2
) | dist G(x, y ) ≤ d}. Observe that N dG[u] = NG(d) [u]. For example, if M is a perfect matching in K2t, then C(k−1) 2k ∼= K2k − M .
Cartesian product. Given graphs G1 = ( V1, E 1) and G2 = ( V2, E 2), the Cartesian product of G1 and G2 is denoted G1 G2 = ( V, E ), where V = V1 ×V2 and E = {{ (u, v 1), (u, v 2)} | { v1, v 2} ∈ G2}∪{{ (u1, v ), (u2, v )} | {u1, u 2} ∈ G1}. For example, the 2-dimensional m × k grid on mk vertices is isomorphic to Pm Pk . Also, the d-dimensional cube equals P d
2
= P2 P2 · · · P2 (d copies of P2).
Join. Given graphs G = ( V, E ) and H = ( W, F ), each with distinguished vertices x ∈ V and y ∈ F , the join of (G, x ) and (H, y ) is denoted ( G, x ) · (H, y ) = ( V ′, E ′), where V ′ = V ∪ W , setting x = y, and E′ = E ∪ F .For example, for any such x and y, the join ( Cn, x ) · (Cm, y ) can be drawn to resemble a figure eight or infinity symbol.
Dangling. Given a graph G = ( V, E ) with a sequence of distinct, distinguished vertices x = ( x1, . . . , x k) and a sequence of graphs H = ( H1, . . . , H k) (each Hi = ( Wi, F i)) with corresponding sequence of distinguished vertices y = ( y1, . . . , y k) (each yi ∈ Wi), a dangling of H on G is denoted ( G, x) · (H, y) = ( V ′, E ′), where
V ′ = V ∪ki=1 Wi, setting each xi = yi, and E′ = E ∪ki=1 Fi. One can think of dangling as a sequence of joins done in parallel. For example, if x is any ordering of V and each Hi ∼= K2 with any choice of yi, then (Cn, x) · (H, y) is known as the k-sunlet graph Sk, shown in Figure 1, below.
Bridging. Given a graph G = ( V, E ) with a distinguished vertex x and a sequence of graphs H =(H1, . . . , H k) (each Hi = ( Wi, F i)) with corresponding sequence of distinguished vertices y = ( y1, . . . , y k)(each yi ∈ Wi), a bridging of (G, x ) and (H, y) is denoted ( G, x ) · (H, y) = ( V ′, E ′), where V ′ = V ∪ki=1 Wi
and E′ = E ∪ki=1 (Fi ∪ {{ x, y i}} . If each ( Hi, y i) = ( H, y ) we denote this operation by ( G, x )–( H, y )k , with 5Figure 1: The 7-sunlet graph S7.the exponent suppressed when k = 1. For example, for degree 2 vertices x and y, the bridging ( P3, x ) · (P3, y )yields the internally 3-regular caterpillar shown in Figure 2.
x y
Figure 2: The internally 3-regular caterpillar ( P3, x ) · (P3, y ).
1.6 Our results
The following lemma is fundamental to carving out many results.
Lemma 1 (Involution Lemma) . If Q is a 3-involutionary position, then Q ∈ P. In particular, if the graph
G is 3-involutionary then D(G) ∈ P.Proof. We show that Q 6 ∈ N by proving that if Q is a 3-involutionary position then, for all u ∈ Q we have (1)
Q(u) is not trivial and (2) Q(u, v ) is 3-involutionary for some v ∈ Q (u). Thus P1 can never end the game. Indeed, let φ be 3-involution of Q, suppose that Next chooses u ∈ Q , and define v = φ(u). Because φ
is an involution, it partitions the vertices into pairs {z, φ (z)} such that the elements of each pair are of the same “color” (white or shaded). Thus, if u ∈ WQ then v ∈ WQ. Because φ is 3-involutionary, v ∈ WQ(u).Also, if u ∈ SQ then v ∈ SQ. So, if Q(u) is trivial, then there is some x ∈ NWQ (u), which implies that
φ(x) ∈ NWQ (v). But because φ is 3-involutionary, φ(x) 6 ∈ N [u], and so f (x) ∈ WQ(u), a contradiction. This proves part (1). For part (2), we observe that φ is a 3-involution of Q(u, v ).
The Involution Lemma yields several immediate results.
Corollary 2. For all d ≥ 2 we have D(P d
2
) ∈ P, and D(P2) ∈ N.
6Proof. Label the two vertices of P2 by 0 and 1; then this labeling naturally extends to the vertices of G = P d
2
by labels in {0, 1}d. Define the antipodal mapping φ on G by φ(x1, . . . , x d) = (1 − x1, . . . , 1 − xd). For
d = 1, the statement is obvious. For d = 2, if P1 plays u then P2 wins by playing φ(u). For d ≥ 3, φ is 3-involutionary, so D(G) ∈ P by Lemma 1.
Corollary 3. For every 3-involutionary group Γ with generating set Σ, we have D(Γ , Σ) ∈ P.Proof. Let (Γ , Σ) be a 3-involutionary group, and g be an order 2 element of Λ − Σ that cannot be expressed as the product of two generators. Define G = C(Γ , Σ) = ( V, E ), and let φ : G → G be an automorphism given by φ(v) = gv . Then φ is also an involution on G. To see that φ is 3-involutionary, first notice that φ(u) =
gu / ∈ N [u] since g / ∈ Σ. Then supposing, for contradiction, that there is an element v ∈ N (u) ∩ N [φ(u)], we see that there are generators a, b ∈ Σ such that abu = φ(u) = gu . Canceling u yields ab = g, contradicting the fact that g cannot be written as a product of at most two generators. This shows that ϕ is 3-involutionary.
Example 4. Let n ≥ 6 be even and let k < n/ 4 be a positive integer. Consider the game in which two players alternate choosing numbers from {0, 1, 2, . . . , n − 1} until every unchosen number is at distance at most k from some chosen number, where the distance between two numbers a and b equals |a − b| mod n,and the winner is the last player to move. This game can be modeled by the game on the group Γ = Zn,with generating set Σ = {± 1, ..., ±k}, where Zn denotes the cyclic group of order n. Then n/ 2 /∈ Σ and cannot be expressed as the sum of two elements in the generating set, which implies that the automorphism
φ(x) = x + n/ 2 mod n is 3-involutionary, and so, by Corollary 3, D(Γ , Σ) ∈ P. We can also observe that
C(Γ , Σ) ∼= C(k)
n
, and thus this also shows that C(k)
n
∈ P.
Example 5. Let n1, ..., n d be positive integers, each at least 3, and suppose that n1 is even and greater than or equal to 6. Consider the group Γ = Zn1 × · · · × Znd . Let ej ∈ Γ be the element with 1 in coordinate j and 0 in every other coordinate, and let the canonical generating set of Γ be Σ = {± ej | 1 ≤ j ≤ d}. Then, since
(n1/2, 0, ..., 0) /∈ Σ and cannot be expressed as ej + eℓ for any 1 ≤ j, ℓ ≤ d, by Corollary 3 we conclude that
D(Γ , Σ) ∈ P. Furthermore, we can observe that C(Γ , Σ) ∼= Cn1 · · · Cnd , which shows that Cn1 · · · Cnd ∈ P.
For an abelian group Γ, we say that the representation Γ ∼= Zn1 × · · · × Znd is good if n1 ≥ 6 or if Γ ∼= Zk
2
× Zm
4
for some k and m. Notice that every even-order abelian group has a good representation. Then Example 5 proves the following proposition unless Γ ∼= Zk
2
× Zm
4
, for which case Corollary 2 applies because
C(Zk
2
× Zm
4
, Σ) = P k+2 m
2
, where Σ is the canonical generating set. 7Proposition 6. Let Γ 6 ∼= Z2 be an even-order abelian group with good representation and corresponding canonical generating set Σ. Then D(Γ , Σ) ∈ P.
We will defer the proof of the following Theorem to Section 3.1.
Theorem 7. The game D(G) ∈ P for the following graphs G:1. C(k)
n
, for even n ≥ 6 and 1 ≤ k < n/ 4.2. G0 H, whenever (a) G0 is 3-involutionary and H is any graph, or (b) G0 is 2-involutionary and H
is 1-involutionary. 3. (G0, (x, φ (x)) · (( H, H ), (y, y )) , such that φ is a 3-involution of G0, H = ( W, F ) is any graph, and
y ∈ F .4. (G0, x)·(H, y), |V (G0)| ≥ 2, x = V (G0), and each Hi ∼= K1,m i with odd mi ≥ 1 and yi is a dominating vertex of Hi.5. (G0, x )–(G0 , x )k, such that x is a cut-vertex of G0, some component of G0 − x is a singleton, and k is odd.
Paths and odd cycles are examples of graphs that are not 3-involutionary, so they require additional techniques. We are able to characterize which path and cycle games are won by Player 1, as follows.
Theorem 8. For n ≥ 3, D(Cn) ∈ N if and only if n is odd but not equal to 5.
Theorem 9. For n ≥ 2, D(Pn) ∈ N if and only if n is odd or n ∈ { 2, 6, 8, 10 , 12 }.
In Section 2 we develop the lemmas necessary to prove Theorems 7–9, which we do in Section 3. Section 4 contains theorems about games on specific graphs such as caterpillars, Cartesian products of paths and cycles, and the Petersen graph. We offer several interesting questions, open problems, and a conjecture in Section 5.
2 Key Lemmas
2.1 GD games
For a set of GD positions Qi on corresponding graphs Gi (1 ≤ i ≤ k), define the game sum Q = ⊕ki=1 Qi to have BQ = ∪ki=1 BQi , SQ = ∪ki=1 SQi , WQ = ∪ki=1 WQi on the disjoint graph union + ki=1 Gi so that, for each
i 6 = j, no move in Qi affects Qj . Each Qi is called a component of Q. While game sums are normally fairly 8simple to interpret, the complexity in this case is that players can continue to play on the shaded vertices of a component that has been “won” (is trivial).
Fact 10. For any position Q and any vertex u ∈ Q , we have |SQ(u)| + |WQ(u)| = |SQ| + |WQ| − 1.Proof. This holds because |BQ(u)| = |BQ| + 1.
Lemma 11. For GD games A and B, if A ∼= B then A ⊕ B ∈ P.Proof. Let A and B be GD games, f : A→B be an isomorphism. Define φ : A ⊕ B → A ⊕ B by setting
φ(a) = f (a) for all a ∈ A and φ(b) = f −1(b) for all b ∈ B . Then φ is 3-involution of A ⊕ B and so, by Lemma 1, A ⊕ B ∈ P.
2.2 PD games
To describe PD games, we first make an observation that will yield simpler notation that will facilitate our analysis. The observation is that every interior vertex of a path is a cut vertex, so when Next chooses an interior vertex u, the game on the original path P becomes a sum of games on two paths defined by P − u.Thus we define the position Pik to be the path on k vertices having i shaded endpoints. Then, for example, if
u has distance at least two from each endpoint of Pik then Pik(u) ∼= Pha ⊕ P jb , for some a, b ≥ 2, a + b = k − 1, 1 ≤ { h, j } ≤ 2, and h + j = 2 + i. When u is an endpoint or neighbor of one, it is slightly trickier to write a general formula, but simple to calculate a particular instance; for example, if u is a neighbor of the shaded endpoint of P17 then P17 (u) ∼= P11 ⊕ P 15 . Notice that, with this notation, there is no mention of B since it no longer exists — the chosen vertices are not colored black but instead are removed — so the original definition of S no longer applies. Instead we know now that S is a subset of the endpoints of paths, and endpoints next to a chosen vertex become shaded. We define a PD position Q to be a finite sum of such Pik positions. We say a PD position or game
Q = ⊕mj=1 Pij
kj
is standard if ij = 2 when kj > 1, and ij = 1 otherwise; that is, every endpoint is shaded. Thus, if Q is standard then, for every move u in Q, we have that Q(u) is standard. To simplify the notation of standard positions for most of this paper, we write Pk in place of P2
k
for k > 1 and P1 in place of P11 ,with P0 denoting the empty position (having no vertices). Given standard Q = ⊕mi=1 Qi where Qi ∼= Pni , we define the functions one (Q) = |{ i | ni = 1 }| , four (Q) =
|{ i | ni = 4 }| , and odd (Q) = |{ i | ni is odd }| . We say that Q is even if |WQ| + |SQ| is even. Equivalently,
odd (Q) is even. We say that Q is totally even if both |WQ| and |SQ| are even and four (Q) is even. Note that 9|SQ| is even if and only if one (Q) is even. Thus, an even Q is totally even if and only if one (Q) and four (Q)are both even. Denote by xi the vertex x ∈ Q i. Let φ denote the automorphism of any path that swaps its endpoints. We say that xi is the center of Qi if φ(xi) = xi. Note that if xi is the center of Qi then ni is odd.
Lemma 12. Let Q = ⊕mi=1 Pni be nontrivial such that |SQ| + |WQ| is odd. Then Next has a move u ∈ Q
such that Q(u) is totally even. Proof. By Fact 10, Q(u) is even for any move u, and we therefore recall that Q(u) is totally even if and only if one (Q(u)) and four (Q(u)) are both even. We consider the following four cases:
• one (Q) and four (Q) are both odd: Since four (Q) is odd we can write Q ∼= A ⊕ P 4 for some game A. Then Next plays an interior vertex u ∈ P 4, so that one (Q(u)) = one (Q) + 1 and
four (Q(u)) = four (Q) − 1.
• one (Q) is even and four (Q) is odd: Since four (Q) is odd we can again write Q ∼= A⊕P 4. Then
Next plays an endpoint u ∈ P 4, so that one (Q(u)) = one (Q) and four (Q(u)) = four (Q) − 1.
• one (Q) is odd and four (Q) is even: Since one (Q) is odd we can again write Q ∼= A ⊕ P 1.Then Next plays the unique vertex u ∈ P 1, so that one (Q(u)) = one (Q)−1 and four (Q(u)) =
four (Q).
• one (Q) and four (Q) are both even: Since one (Q) is even, so is |SQ|, which makes |WQ| odd. Thus there is some i for which ni ≥ 3 is odd, and so we write Q ∼= A ⊕ P ni . Then Next
plays the center u of Pni . By symmetry, one (Q(u)) and four (Q(u)) are both even. Therefore, in all cases there is a move u such that Q(u) is totally even.
Fact 13. Let Q = ⊕mi=1 Pni be nontrivial and totally even. Then for all vertices u ∈ Q , the position Q(u) is nontrivial. Proof. The only even positions that can be won in one move are of the form ( ⊕mi=1 P1) ⊕ P 4, where m is even. Such positions are not totally even.
Theorem 14.
Version A: Let Q = ⊕mi=1 Pni be nontrivial. Then Q ∈ P if and only if Q is even and |WQ| ≥ 4, or Q is isomorphic to Q ∼= ( ⊕mi=1 P1) ⊕ P 3 ⊕ A , where m is odd and A is either P4 or P1 ⊕ P 3.
10 Version B: Let Q = ⊕mi=1 Pni be nontrivial. Then Q ∈ P if and only if Q is even and |WQ| ≥ 4, or Q is isomorphic to one of the following:
• (⊕mi=1 P1) ⊕ P 3 ⊕ P 3, where m is even
• (⊕mi=1 P1) ⊕ P 4 ⊕ P 3, where m is odd
We recall that Q = ⊕mi=1 Pni is a finite position and therefore must lead to a trivial position in finitely many steps. Thus, Q is in exactly one of P and N.
Proof of Theorem 14. Say that Q has condition C if Q is totally even or Q is even and |WQ| ≥ 4. We argue that if Q has condition C then Q(u) is nontrivial for all u ∈ Q , and then use Lemma 12 to conclude that
Q(u, v ) has condition C. Thus P1 can never win Q; i.e. Q ∈ P. Indeed, Fact 13 takes care of the totally even case, and |WQ| ≥ 4 implies that |WQ(u)| ≥ 1 for all u ∈ Q . This shows that if Q is even and |WQ| ≥ 4, then Q ∈ P.Next, we gh at the two cases listed in the statement of the theorem. First, suppose that Q ∼= ( ⊕mi=1 P1) ⊕P3 ⊕ P 3, where m is even. Then we can write it as Q ∼= A ⊕ A where A ∼=
(
⊕m/ 2
i=1
P1
)
⊕ P 3. Then Q ∈ P by Lemma 11. Second, suppose that Q ∼= ( ⊕mi=1 P1) ⊕ P 4 ⊕ P 3, where m is odd. Without loss of generality, we can assume that m = 1. If P1 chooses u ∈ P 3 or v ∈ WP4 , then P2 will choose v or u, respectively, to win. If P1 chooses u ∈ SP1 or v ∈ SP4 , then P2 will choose v or u, respectively, to yield Q(u, v ) ∼= P3 ⊕ P 3, which
P2 wins by Lemma 11. Hence Q ∈ P.Conversely, by contrapositive, we first suppose that Q is odd. By Lemma 12, P1 has a move u such that
Q(u) is totally even and thus in P. Therefore, Q ∈ N.The only remaining case is Q is even, |WQ| < 4, and Q is not isomorphic to one of the games listed in the statement of the theorem. One can see that Q ∼= ( ⊕mi=1 P1) ⊕ P k, where k ∈ { 3, 4, 5} and m ≡ k (mod 2). In each of the cases, there are at most three white vertices and they are all consecutive. Thus, P1 can win in one move.
Remark 15. In the notation of Theorem 14, suppose that |WQ| ≥ 10 . Then, after any move u of P1 , any response v by P2 keeps the game even by Fact 10. Also, |WQ(u,v )| ≥ 4 and so Q(u, v ) ∈ P by Theorem 14. Therefore, while there remain at least 10 white vertices, P2 ’s response to P1 can be arbitrary.
Corollary 16. The position Pk ∈ P if and only if k is even and k 6 = 4 .Proof. This follows as a special case of Theorem 14.
11 3 Proofs
3.1 Proof of Theorem 7
For the first 3 cases, we show that the graph under consideration has a 3-involution. Then the results follow from Lemma 1. 1. This was proved in Example 4. ⋄
For case (a), let φ′ : G0 → G0 be a 3-involution, and define the involution φ on G0 H by setting
φ(g, h ) = ( φ′(g), h ). For case (b), let φ′ : G0 → G0 be a 2-involution and φ′′ : H → H be a 1-involution, and define the involution φ on G0 H by setting φ(g, h ) = ( φ′(g), φ ′′ (h)). In both cases, φ
is 3-involutionary. ⋄
Let G0 = ( V, E ) and H = ( W, F ), and set G = ( G0, (x, φ (x)) · (( H, H ), (y, y )). We will say that
W = {w1, ..., w n}, and define Wi = {wi
1
, ..., w in} for i = 1 , 2. Then we can write V (G) = V ∪ W1 ∪ W2.Let φ : G0 → G0 be a 3-involution. Then define φ : G → G by setting φ(u) = φ(u) for u ∈ V (G),
φ(w1
j
) = w2
j
for w1
j
∈ W1, and φ(w2
j
) = w1
j
for w2
j
∈ W2. Then φ is a 3-involution. ⋄
We prove this statement for a broader class of graphs and for a broader set of positions on them. Let G
be the set of positions Q on graphs G = (( G0, x) · (H, y)) ∪ G1 ∪ G2, where G1 ∪ G2 is an independent set, with the following properties:
• |V (G0)| ≥ 2;
• x = V (G0);
• no xi ∈ Q B ;
• each Hi ∼= K1,m i , with odd mi ≥ 1;
• each yi is a dominating vertex of Hi;
• G1 is a set of evenly many isolated vertices in QW ; and
• G2 is a set of evenly many isolated vertices in QS .For such Q ∈ G , we prove by induction on |Q| that Q ∈ P. At each stage, when a player plays a vertex
v in some graph G, we remove black vertices so that G(v) = G − v. By doing so, when isolated vertices occur, they get moved in G1 ∪ G2 according to their color. Let Q ∈ G and k = |V (G0)|. We write NHi (yi) = {zi, 1, . . . , z i,m i } and suppose that P1 plays u ∈ Hi.Since k ≥ 2, there is some j 6 = i such that zj, 1 ∈ Q W (u), and so u is not a winning move. We are done 12 if P2 has a winning move, so we assume otherwise and show that P2 has a move v ∈ Q (u) that makes
Q(u, v ) ∈ G , which will complete the induction. For ease of notation, relabel the {Hi} so that P1 plays in Hk.First consider the case that u = xk . If k = 2 then P2 has a winning move v = x1, so we must have
k > 2. Thus P2 plays v = zk,m k , so that Q(u, v ) ∈ G . This is because an even number of vertices, namely {zk, 1, . . . , z k,m k −1}, have been moved into G2.Next consider the case that u = zk,m k . If mk = 1 and k = 2, then P2 has the winning move v = x1,so we must have either mk ≥ 2 or k ≥ 3. If mk ≥ 2, then P2 can play v = zk,m k−1 , while if k ≥ 3then P2 can play v = xk. In either case, we have Q(u, v ) ∈ G . In the former case, this is because no new isolated vertices were created and the oddness of leaves was maintained. In the latter case, this is because an even number of shaded leaves were moved to G2. ⋄
Rewrite the bridging G = ( G0, x )–( G0, x )k as ( G0, x 0)·(G, x), where G = ( G1, . . . , G k), x = ( x1, . . . , x k ), and each Gi ∼= G0. Since k is odd, we may pair Gj with Gj+1 for all even j < k , which allows us to extend the identity map on G0 to the automorphism φ of G composed of the natural isomorphisms between each pair Gj and Gj+1 . Now we define the corresponding mirroring strategy for P2 that plays
φ(u) for each play u by P1 . We claim that this is a winning strategy. Write zi for a singleton in Gi − xi. Suppose that P1 wins and let u be the winning move by P1 . That is, there is a position Q on G such that WQ 6 = ∅ and WQ(u) = ∅. This means that there is some
w ∈ WQ ∩ N [u], which implies that φ(w) ∈ WQ ∩ N [φ(u)]. Because φ(w) 6 ∈ WQ(u), it must be that
φ(w) ∈ N (u), and so u = xi, for some i. But this implies that zi ∈ WQ since zi is a leaf, and so
φ(zi) ∈ WQ. Since φ(zi) 6 ∈ N (xi) , we arrive at the contradiction φ(zi) ∈ WQ(u). Hence P2 wins. ⋄
This completes the proof of Theorem 7.
3.2 Proof of Theorem 8
Recall the statement of Theorem 8: For n ≥ 3, D(Cn) ∈ N if and only if n is odd but not equal to 5. Let
Dn = D(Cn). Because of the symmetry of Cn, for any u we have Dn(u) ∼= Pn−1, and so Dn ∈ N if and only if Pn−1 ∈ P. The result follows from Corollary 16.
13 3.3 Proof of Theorem 9
Recall the statement of Theorem 9: For n ≥ 2, D(Pn) ∈ N if and only if n is odd or n ∈ { 2, 6, 8, 10 , 12 }. Let
Dn = D(Pn) and P1 = Next , and write {0, 1, . . . , n − 1} for the vertex labels of Dn.We first suppose that n is odd. P1 plays the center vertex u = ( n − 1) /2 ∈ D n. Then Dn(u) ∼= A ⊕ B
where A ∼= B, and thus Dn(u) ∈ P by Lemma 11. Thus Dn ∈ N.At this point we recall the more specific notation Pik for any path on k vertices with exactly i shaded endpoints. Thus, Dk can also be written as P0
k
.Next we suppose that n is even and consider the following cases:
n = 2: Here P1 wins with any move so D2 ∈ N.
n = 4: Let P1 play u ∈ D 4. By symmetry, we can assume that u = 0 or u = 1. In both cases, P2
wins by playing vertex 3. Thus D4 ∈ P.
n = 6: In this case P1 plays an endpoint u, so that D6(u) ∼= P15 . From this, it is easy to see that, for any move v by P2 the at most three white vertices of D6(u, v ) are consecutive, and subsequently can be dominated on the next move by P1 . Hence D6 ∈ N (and P15 ∈ P).
n = 8: Now P1 plays a neighbor u of an endpoint. If P2 plays one neighbor v of u then P1 plays the other w, so that D8(u, v, w ) ∼= P15 , which is in P by the argument in the case n = 6 above. So we assume otherwise. If P2 plays v so that the at most 3 white vertices of D8(u, v ) are consecutive, then P1 plays to dominate them on the next move. The only remaining case is that P2 plays the vertex v at distance two from the unshaded endpoint of D8(u). Now
D8(u, v ) ∼= P11 ⊕ P 23 ⊕ P 12 , and P1 plays the unique vertex w in P11 . At this point, note that any move in either component P23 or P12 dominates that component, so that, when P2 plays in one of them, P1 plays in the other to win. Therefore D8 ∈ N.
n ∈ { 10 , 12 }: For these values of n, the case analysis is more extensive, but not insightful. We verified that Dn ∈ N for each such n by computer.
n ≥ 14: The following strategy can be used by P2 to win this game. Irrespective of the first two moves of P1 , the first two moves of P2 should be the endpoints of P0
n
. If P2 cannot choose an endpoint because P1 has already played it, then P2 can play any other vertex. Thus, after each player has made two moves, we arrive at the even position Q = Pa1 ⊕ P a2 ⊕ P a3 ,where a1 + a2 + a3 = n − 4 and each ai ≥ 0. Because each path has at most 2 shaded endpoints, |WQ| ≥ (n − 4) − 6 ≥ 4. Hence Q ∈ P by Theorem 14, and so Dn ∈ P.14 4 Other specific graphs
The path Pn can be described as the internally 2-regular tree on n vertices. A natural extension of this class of trees is the set of internally r-regular trees. (Internally 3-regular trees can be thought of as rooted binary trees with an additional pendant vertex attached to the root.) One instance in this class is the set of internally r-regular caterpillars. A caterpillar is a tree such that every vertex not on a longest path Pk
(called the spine ) is adjacent to some vertex of Pk. Such a caterpillar is called even (odd ) if k is even (odd ). The bridging construction yields the following theorem.
Theorem 17. If G is an even internally r-regular caterpillar with r ≥ 3, then D(G) ∈ P.Proof. Let e = {x1, x 2} be the middle edge of the spine P of G; then G − e is the disjoint union of two isomorphic copies of some tree T1 ∼= T2. Then G = ( T1, x 1)–( T2, x 2) and each Ti − xi has a singleton component, so D(G) ∈ P by Theorem 7.5.
Theorem 18. If G is the k-sunlet graph Sk, with even k ≥ 2, then D(G) ∈ P.Proof. As noted after the dangling definition, Sk = ( Ck, x) · (H, y). Hence D(Sk) ∈ P by Theorem 7.4.
Another interesting class of graphs to consider are the grids Pk Pm. Because of the complex pattern of
D(Pn) winner given in Theorem 9, one might guess that no simple pattern describes the winner of grids in general. However, evidence suggests otherwise.
Theorem 19. If km is odd then D(Pk Pm) ∈ N.Proof. Define s = ( k − 1) /2 and t = ( m − 1) /2, and use the coordinate system {− s, . . . , 0, . . . , s } × {− t, . . . , 0, . . . , t } for the vertices of G = Pk Pm. Define the involution φ on G by φ(a, b ) = ( −a, −b). Then
φ is 3-involutionary on G − (0 , 0). Hence, P1 wins D(G) by playing u = (0 , 0) because G(u) = G − (0 , 0) ∈ P
by Lemma 1.
Notice that the same argument yields that D(Pk1 Pk1 · · · Pkd ) ∈ N whenever k1k2 · · · kd is odd. Similarly, toroidal grids are equally interesting. As pointed out in Example 5, which follows from Theorem 7.2, even higher-dimensional products of cycles are in P, provided that some cycle has length at least 6. We record this below.
Theorem 20. Suppose that n1 ≥ 6 is even, 3 ≤ n2 ≤ · · · ≤ nd are integers, and G = Cn1 Cn2 · · · Cnd .Then D(G) ∈ P.
15 12
34 35 45
15 25 24 14 13 23
Figure 3: The Petersen graph P = K(5 , 2) with its vertex labels shown. For m ≥ 2t + 1, the Kneser graph K(m, t ) has vertex set ([m]
t
), with edges between disjoint pairs of vertices. The Petersen graph equals K(5 , 2) and is of great importance in many areas of graph theory. It is natural to investigate which player wins the games on these graphs.
Theorem 21. For the Petersen graph P , we have D(P ) ∈ P.Proof. We simplify the notation of the vertices of P by writing ij instead of {i, j } (see Figure 3). By symmetry, we may assume that P1 chooses 12. Because P − N is a 6-cycle C, which is 3-involutionary,
P2 will lose by choosing any vertex u of C (P1 responds by playing the vertex of C opposite from u). By symmetry again, we may assume that P2 chooses 35. Similarly, P1 loses by choosing a vertex of C, and so chooses, by symmetry, 34. But now P2 wins by choosing 45.
5 Final Comments
Given the results of Section 4, we offer the following open problems and conjectures.
Problem 22. Find the winning player for D(G) when G is an odd internally regular caterpillar.
The first several cases of the following conjecture are easy to verify by hand.
Conjecture 23. For k, m > 1 with mk even, we have D(Pk Pm) ∈ P.
Problem 24. For integers 3 ≤ n1 ≤ . . . ≤ nd, with every even ni = 4 , determine the winner of D(G) for
G = Cn1 · · · Cnd .
16 A general statement one might hope to prove about Cartesian products involves the case in which the winner of one of the graphs is known.
Question 25. Let G, H be graphs such that D(G) ∈ P. Under what conditions is it true that G H ∈ P?
For example, Theorem 7.23 states that D(G H) ∈ P if G is 3-involutionary, so the question is open for graphs in P that are not 3-involutionary, such as D(P4). If Conjecture 23 is true, though, D(P4 Pm) ∈ P
for any m ≥ 2. A similar question arises when considering graph powers, along the lines of Theorem 7.1.
Question 26. For what conditions on a graph G can we determine the winner of D(G(k))? We listed five graph constructions in Section 1.5 that produce P games by Theorem 7. There are many other graph constructions one might consider.
Question 27. What other graph constructions produce games in P?
Finally, we offer a reachable problem on problem on Kneser graphs.
Problem 28. Find the winning player for D(G) when G is the Kneser graph K(m, 2) and m ≥ 6.
While solving this game on a general graph appears to be out of reach, the game can be played on a number of other classes of graphs that we believe to be natural directions for future study. Some of the techniques described above may shed light on the case of trees. In particular, spiders, other caterpillars, and other internally 3-regular trees seem to be approachable cases to work on in the future; complete bipartite graphs are also a natural choice, as are non-Cartesian products of graphs and higher order Kneser graphs. Finally, there are several related variants of the graph domination game. The first is the Mis` ere version in which the first player to create a dominating set loses rather than wins. Another version that can be played is that players take turns choosing vertices until the complement of the set of chosen vertices is not a dominating set. Additionally, all of these variants can be played with “dominating sets” replaced by “total dominating sets”, which are subsets D ⊆ V of vertices of a graph with the property that N (D) = V .
Acknowledgement
This material is based upon work supported by the U.S. National Science Foundation under Grant No. 2015425. 17 References
AIM Minimum Rank-Special Graphs Work Group . Zero forcing sets and the minimum rank of graphs. Linear Algebra Appl. 428 , 7 (2008), 1628–1648. Alon, N., Balogh, J., Bollob´ as, B., and Szab´ o, T. Game domination number. Discrete Math. 256 , 1-2 (2002), 23–33. Berlekamp, E., and Wolfe, D. Mathematical Go . A K Peters, Ltd., Wellesley, MA, 1994. Chilling gets the last point, With a foreword by James Davies. Berlekamp, E. R., Conway, J. H., and Guy, R. K. Winning ways for your mathematical plays. Vol. 1 , second ed. A K Peters, Ltd., Natick, MA, 2001. Bj¨ orner, A., Lov´ asz, L., and Shor, P. W. Chip-firing games on graphs. European J. Combin. 12 ,4 (1991), 283–291. Bollob´ as, B., and Riordan, O. Percolation . Cambridge University Press, New York, 2006. Bonato, A. A survey of graph burning. Contrib. Discrete Math. 16 , 1 (2021), 185–197. Breˇ sar, B., Henning, M. A., Klavˇ zar, S., and Rall, D. F. Domination games played on graphs .Springer Briefs Math. Springer, Cham, 2021. Breˇ sar, B., Klavˇ zar, S., and Rall, D. F. Domination game and an imagination strategy. SIAM J. Discrete Math. 24 , 3 (2010), 979–991. Chung, F. R. K. Pebbling in hypercubes. SIAM J. Discrete Math. 2 , 4 (1989), 467–472. Fraenkel, A. S. Combinatorial games: selected bibliography with a succinct gourmet introduction. In More games of no chance (Berkeley, CA, 2000) , vol. 42 of Math. Sci. Res. Inst. Publ. Cambridge Univ. Press, Cambridge, 2002, pp. 475–535. Gardner, M. Mathematical games. Scientific American 223 , 4 (1970), 120–123. Goddard, W., and Henning, M. A. The competition-independence game in trees. J. Combin. Math. Combin. Comput. 104 (2018), 161–170. Hayward, R. B. Hex—a playful introduction , vol. 54 of Anneli Lax New Mathematical Library . Math-ematical Association of America, Washington, DC, ©2022. 18 Henning, M. A., Klavˇ zar, S., and Rall, D. F. Total version of the domination game. Graphs Combin. 31 , 5 (2015), 1453–1462. Honary, E. Total Diplomacy: The Art of Winning RISK . BookSurge LLC, Charleston, SC,
©2007. Huggan, M. A., Huntemann, S., and Stevens, B. The combinatorial game nofil played on steiner triple systems. J. Combin. Des. 30 , 1 (2022), 19–47. Hurlbert, G., and Kenter, F. Graph pebbling: A blend of graph theory, number theory, and optimization. Notices Amer. Math. Soc. 68 , 11 (2021), 1900–1913. Nowakowski, R., and Winkler, P. Vertex-to-vertex pursuit in a graph. Discrete Math. 43 , 2-3 (1983), 235–239. Phillips, J. B., and Slater, P. J. An introduction to graph competition independence and enclave-less parameters. Graph Theory Notes N. Y. 41 (2001), 37–41. Phillips, J. B., and Slater, P. J. Graph competition independence and enclaveless parameters. In
Proceedings of the Thirty-third Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2002) (2002), vol. 154, pp. 79–100. 19 |
7242 | https://tasks.illustrativemathematics.org/content-standards/tasks/1471 | Typesetting math: 100%
Symmetries of a quadrilateral I
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Task
Suppose ABCD is a quadrilateral for which there is exactly one rotation, through an angle larger than 0 degrees and less than 360 degrees, which maps it to itself. Further, no reflections map ABCD to itself. What shape
is ABCD?
IM Commentary
This task provides an opportunity to examine the taxonomy of quadrilaterals from the point of view of rigid motions. As is the case with triangles, the different types of quadrilaterals are often characterized by their symmetries: the square is the most symmetric, followed by the rectangle and rhombus, then come the regular trapezoid and parallelogram: other quadrilaterals lack symmetry. Students will need to think about this characterization of quadrilaterals in terms of rotational symmetry in order to identify ABCD
as a parallelogram which is not a rhombus and not a rectangle.
If students are aware that a quadrilateral is a parallelogram when its diagonals bisect each other, then they can use this criterion: once it is shown that the rotation which takes ABCD to itself is a 180 degree rotation, this implies that the diagonals bisect each other as their point of intersection is the center of the rotation.
If appropriate geometry software is available,
this task provides an ideal opportunity for students to engage in MP5, Use Appropriate Tools Strategically, as students can experiment with different quadrilaterals in order to determine which ones have rotational symmetry. Once they have reached a conclusion about what type of quadrilateral ABCD is, they will need to Reason Abstractly and Quantitatively (MP2) in order to show both that ABCD is a parallelogram and that it is neither a rhombus nor a rectangle. In particular, the beginning of the argument will need to identify that the rotation must be a 180 degree rotation and this will require examining the composition of two rotations of the plane.
Solution
Suppose r is the rotation which maps ABCD to itself. Applying r twice will then also carry ABCD to itself. Since there was only one non-identity rotation that mapped ABCD to itself, it must be the case that applying r twice is the identity. This means that r rotates through exactly half of a full circle, or 180 degrees.
Next we examine the effect of r on the sides of the quadrilateral.
A rotation by 180 degrees takes each line ℓ to a line r(ℓ) which is parallel to ℓ.
Suppose O is the center of the rotation. This point could lie on zero, one, or two of the lines containing the sides of ABCD: in all cases, there is at least one pair of these lines, through a vertex of ABCD, neither of which contains O. The 180 degree rotation takes each of these lines to a distinct parallel line. This gives two different pairs of parallel lines containing opposite sides of ABCD (and also shows that O does not lie on any of the four lines containing the sides of ABCD). So ABCD is a parallelogram.
Before proceeding to examine ABCD further, we provide a picture:
A 180 degree rotation interchanges A and C and also interchanges B and D. Moreover, O is the midpoint of both of AC¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯. This particular parallelogram has no other symmetries, save for the identity. Certain special parallelograms do admit other symmetries: for example a rectangle has two lines of reflective symmetry in addition to a 180 degree rotation (plus additional symmetries if the rectangle happens to be a square). Another special class of parallelograms which admit additional symmetries are are the rhombuses: these also have two lines of reflective symmetry (passing through opposite vertices) like a rectangle. So we conclude with a final answer of:
Quadrilateral ABCD is a parallelogram which is neither a rectangle nor a rhombus.
Below are pictures of a rectangle and a rhombus with their lines of symmetry:
Note that a square is both a rectangle and a rhombus and has 4 lines of symmetry as a result. |
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SAT Math Multiple-Choice Question 127: Answer and Explanation
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Question: 127
The scatterplot below shows the pH of seven well water samples in West Texas with respect to the bicarbonate concentration in ppm (parts per million). The line of best fit is also shown.
A new well is discovered in West Texas with a bicarbonate concentration of 225 ppm. According to the line of best fit, which of the following best approximates the pH of the well water?
A. 7.1
B. 7.3
C. 7.4
D. 8.4
Correct Answer:B
Explanation:
B
The question asks for the pH of a well with a bicarbonate concentration of 225 ppm, so look it up. Go along the horizontal axis to 225 ppm, and go up to the line of best fit. Now trace across the grid line to the vertical axis. It hits the vertical axis between 7.2 and 7.4, so the pH is approximately 7.3. The correct answer is (B).
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7244 | https://math.stackexchange.com/questions/1849081/if-you-take-the-reciprocal-in-an-inequality-would-it-change-the-signs | Skip to main content
If you take the reciprocal in an inequality, would it change the >/< signs?
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Example:
−16<1x−14<16
In the example above, if you take the reciprocal of
1x−14=x1−41
would that flip the < to > or not?
In another words, if you take the reciprocal of
−16<1x−14<16
would it be like this:
1−16>x1−41>116
algebra-precalculus
inequality
fractions
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edited Jul 4, 2016 at 20:37
Matthew Conroy
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asked Jul 4, 2016 at 20:24
BlueMagic1923BlueMagic1923
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The reciprocal of 1/x−1/4 is not x−4.
– Will Fisher
Commented
Jul 4, 2016 at 20:27
4
Look at the last line of what's written and notice how this implies −1/16>1/16
– Nathaniel B
Commented
Jul 4, 2016 at 20:28
Almso you can take the reciprocal in an inequality, only under the condition both sides have the same side, and swapping < and >.
– Bernard
Commented
Jul 4, 2016 at 20:30
You should check by yourself before asking. 1<2 vs. 1>12.
– user65203
Commented
Jul 4, 2016 at 23:00
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If a and b have the same nonzero sign, then
a1b
(i.e., taking reciprocals reverses the inequality).
If a and b have opposite (nonzero) signs, then
a<b⟺1a<1b
(i.e., taking reciprocals preserves the inequality).
These follow from the fact that the function f(x)=1/x defined on the nonzero reals is strictly decreasing and sign-preserving on each component (−∞,0) and (0,∞) of its domain.
If either of a or b is zero, then you can't take reciprocals.
Finally, compound inequalities like a<b<c should be separated into "a<b and b<c" and each component considered separately.
I also remark that inverting a sum is not the same as inverting the addends separately.
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edited Dec 9, 2019 at 14:42
answered Jul 4, 2016 at 20:44
MPWMPW
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1/x<=1 is tricky. Answer is x > 1 or x < 0. But what is the general rule?
– Tuntable
Commented
Jun 28, 2021 at 3:17
Are you asking how to solve that inequality?
– MPW
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Jun 28, 2021 at 17:33
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It depends if x and y are the same sign.
Case 1: 0<x<y then 0<x(1/y)<y(1/y) and 0<x/y<1 and 0<x/y(1/x)<1(1/x) so 0<1/y<1/x.
If both positive, flip.
Case 2: x<0<y then x/y<0<1. Then as x<0 we flip when we do x/y∗(1/x)>0>1∗(1/x) so 1/y>0>1/x so 1/x<0<1/y. Don't flip.
Case 3: x<y<0 then x/y>1>0 and 1/y<1/x<0. Flip if they are the same sign.
But FOR THE LOVE OF GOD!!!!!!! the reciprical of 1/x−1/4 is !!!!!!!NOT!!!!!! x/1−4/1!!!!!!!!
It is 11/x−1/4=4x4−x.
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answered Jul 4, 2016 at 20:47
fleabloodfleablood
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If you have
x>y(1)
Then you only have to flip the inequality when you take the reciprocal if x and y are of the same signs. Notice that this is because we can multiply both sides of (1) by (xy)−1 to get
xxy>yxy→1y>1x
Which naturally flips the signs, however, if (xy)−1 is negative (which occurs when x and y are different signs) then we have to flip the inequality sign back because we multiplied by negative number.
Note, however, in your question you misused the reciprocal.
1x−14=4−x4x
Which has reciprocal 4x4−x. Also you can apply this to a "multistaged" inequality by breaking it into a series of inequalities of the form (1).
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edited Jul 4, 2016 at 22:49
answered Jul 4, 2016 at 20:42
Will FisherWill Fisher
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Here are some ideas for how reciprocals work.
A different kind of reciprocal
From a picture point of view, reciprocals have to do with "turning fractions upside down"— like turning ab into ba.
From your point of view, which is reasonable, turning ab+cd into ba+dc makes a certain kind of reciprocal.
But in most algebra classes, people use the term reciprocal to talk about starting with a term like x and replacing it with one-divided-by-x. Because 1ab+cd is not the same as ba+dc, your way of talking about reciprocals and the way people talk about reciprocals in algebra classes are different.
You can take reciprocals using only multiplication and division
In algebra, there are rules for what you can do to an equation or inequality to make sure it stays true. For example, you can add the same number to both sides of the equation, or you can multiply both sides of the equation by the same number (except zero).
You can take the reciprocal of an equation just by performing several multiplication and division operations in a row. Here's why:
If you have an equation written like:
ab=cd,
you can turn it into its reciprocal equation by taking the following steps:
You can multiply both sides by the denominators b and d:
a×b×db=c×b×dd
Some factors cancel, leaving you with:
a×d=c×b.
You can divide both sides by a and c (the numerators from the original equation), giving you:
a×da×c=c×ba×c.
Some factors cancel, leaving you with
dc=ba.
which is the reciprocal of your original equation! Because all we did was perform multiplication and division, we know that whenever the original equation is true a/b=c/d, the reciprocal equation is also true b/a=d/c.
If you had an inequality ab<cd, the steps for taking the reciprocal would be the same — but you'd additionally have to keep track of whether a,b,c,and d were negative so you would know when to flip the < sign.
You can't take the reciprocal of a sum by taking the sum of the reciprocals
In general, you can't start with a sum of fractions like ab+cd=ef+gh, and convert it into ba+dc=fe+hg by following the rules of algebra for manipulating equations.
Of course, you already know that if you have ab+cd=ef+gh, you can safely write:
1ab+cd=1ef+gh
— but in general, 1ab+cd will be a very different number from ba+dc.
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edited Jul 4, 2016 at 21:00
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7245 | https://ocw.mit.edu/courses/res-8-005-vibrations-and-waves-problem-solving-fall-2012/resources/simple-harmonic-motion-and-introduction-to-problem-solving/ | Simple Harmonic Motion and Introduction to Problem Solving | Vibrations and Waves Problem Solving | Physics | MIT OpenCourseWare
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Vibrations and Waves Problem Solving
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Articulating the Two Parts of Physics Problems
Coming Full Circle in Problem Solving
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0:21 PROFESSOR: I'm Wit Busza, professor of physics at MIT.
0:25 I'm joining my colleague, Professor Walter Lewin,
0:29 to help you understand the physics of waves
0:31 and vibrations.
0:33 Now you may well ask, why spend so much effort on waves
0:37 and vibrations.
0:38 And the answer is very simple.
0:41 If you take any system, disturb it from equilibrium,
0:48 from a stable equilibrium, the resultant motion
0:52 is waves and vibrations.
0:54 So it's a very common phenomenon.
0:57 Not only is it that very common, understanding waves
1:01 and vibrations have very important
1:04 practical applications.
1:05 And furthermore, the fact that they exist,
1:09 that this phenomenon exists, has tremendous consequences
1:12 on our world.
1:14 If waves and vibrations were different or didn't exist,
1:18 you wouldn't recognize our universe.
1:23 What is the role I am playing in this course?
1:27 To answer that question, I have to remind you
1:31 what is the scientific method.
1:34 In essence, the scientific method has two components.
1:38 The first, you look around and you
1:43 describe what you see in the one and only language that
1:48 can be used, or we find that can be
1:50 used for the description of nature.
1:52 That this mathematics, in terms of mathematical equations.
1:59 The second aspect is since the universe
2:02 is describable in terms of mathematical equations,
2:05 we can solve those equations.
2:09 And that means predict result of situations,
2:14 of experiments, which we've never seen before.
2:19 Again, this is important for two reasons.
2:23 One, practical-- to be able to predict what will happen.
2:27 But the other far more important is
2:30 that it is the way we have, the objective way we have
2:35 of checking whether our understanding of the universe
2:40 is correct or not.
2:42 If the predictions do not give the right-- do not correspond
2:49 to what one actually sees, you know, your theory,
2:51 your understanding is wrong.
2:55 My role is related to the second part.
2:59 In other words, what I would want to help you learn,
3:05 take a given situation, convert it into mathematics,
3:10 solve it, and predict what will happen.
3:15 We call that problem solving.
3:18 OK.
3:19 Let me immediately start with a concrete example.
3:30 What I have here, describing a situation which
3:36 we would like to understand.
3:39 Imagine you have an ideal spring,
3:42 a spring that obeys Hooke's law.
3:46 As I've shown here is the spring constant k, length,
3:50 natural length l0, and your suspend it from the ceiling.
3:54 Imagine that you take a mass, a small mass,
3:59 m, and you attach it to that spring.
4:03 At some instant of time, and in the proceeds of attaching it,
4:08 you may stretch the string.
4:10 So the spring may at this instant not be stretched.
4:13 But let's assume while you're attaching
4:15 and you've stretched the spring a little bit,
4:17 you're holding it, all right.
4:20 At that instant, it's velocity is 0, stationary.
4:24 You let go.
4:26 The question is, what will happen?
4:28 Can you predict what will be the motion of that particle?
4:33 You know, you've seen this often.
4:34 But a priori, it's not obvious what will happen.
4:37 The spring may pull the mass up.
4:39 The mass may pull the spring more down.
4:41 It may oscillate.
4:43 Everything, until you've understood what's going on, you
4:48 cannot predict the outcome.
4:51 So let's assume that at some instant of time,
4:57 we call that time t, it's as shown on the right.
5:03 In order to be able to describe this,
5:05 I have to tell you where these masses are
5:09 at these various times.
5:11 So I will define a coordinate system.
5:14 This is a one dimensional situation.
5:16 So I only need one coordinate.
5:17 And I'll call it the y.
5:19 My y will be up.
5:21 All right.
5:23 Now I also have to measure things from some location.
5:26 So I need to define what I mean by y equals 0.
5:30 And I will define y equals 0, the position
5:34 where if the mass is at that location, the force of gravity
5:40 pulling it down and the spring force pulling it up
5:44 cancel, so that there is no net force on the particle.
5:49 So y equals 0 is the equilibrium position.
5:53 And then the position of the spring
5:55 when it has no mass attached is the distance y0 from that
6:00 at t equals 0, the position I will say is y initial.
6:04 That's some number.
6:05 So that's a known quantity.
6:07 All right.
6:07 And that any other instant time, I defined it as y equals t.
6:12 That is the physical situation I wish to understand.
6:16 I want to know what happened with that spring.
6:20 So now I will translate that into mathematics.
6:25 I will now try to give you a mathematical description
6:28 of that situation.
6:32 So we know that we are dealing with forces and masses.
6:40 So to describe that, we use Newtonian mechanics.
6:45 So here is now my mathematical description.
6:49 The mass is a point m, of mass m, on which two forces act.
6:56 There is the force fs due to the spring and the force fg
7:02 due to the fact that this mass is in the gravitational field,
7:06 and therefore there is a gravitational force
7:10 on this mass, fg.
7:11 OK.
7:13 We call this a force diagram.
7:15 Or some people call it the free body diagram.
7:20 Now this mass, because of the force acting it,
7:26 its motion will change.
7:28 And it will have an acceleration,
7:31 which I will call a of t.
7:34 And by the way, that of course is the second derivative
7:39 of y with respect to dt.
7:42 It's a vector.
7:42 It's in the y direction.
7:46 And in order so I don't have to write things over
7:49 and too many things in these equations,
7:52 I will define the symbol y with two dots on it
7:56 as the second derivative of y with respect to time.
8:01 y dot is the first derivative-- in other words,
8:04 the velocity, et cetera.
8:07 And by the way, you may notice I'm going very slowly here.
8:11 I'm doing that intentionally.
8:13 I'm going to go here in gory detail every part, you know,
8:19 because often I know that when one goes
8:23 to a lecture, or studies in a book, et cetera,
8:26 you look at some step from one step to another.
8:31 And you can't figure it out.
8:33 The reason for it often is not that you are not smart enough
8:37 to do it, is but the because the teacher
8:41 or whoever wrote the book, et cetera,
8:43 is so familiar with the material he will
8:45 do several steps in his head or her head,
8:49 and you don't know about it.
8:51 For this first example, I will try
8:53 to avoid anything of that kind.
8:55 Later on, I'll go faster.
8:57 And I'll do the same as everybody else.
8:59 But the moment, as I say, I'm going in gory detail.
9:02 OK.
9:03 So this is the diagram, this free body diagram,
9:09 of the situation.
9:11 And I know from Newtonian mechanics
9:15 that if there are forces acting on that mass,
9:20 that mass will have an acceleration, which
9:24 will be equal to the net force acting on it,
9:27 divided by the mass, the inertia, of that system.
9:32 So that is what a will be.
9:35 I further know the force is vector,
9:38 so the net force acting on this mass
9:41 is the sum of those, the vectorial sum,
9:44 of those two forces.
9:45 So f is the sum of the force due to the spring
9:51 and due to gravity.
9:54 Next, we also know something about the spring.
9:57 I told you at the beginning that I'm
9:59 considering an ideal spring.
10:01 So for the purpose of this problem,
10:03 I'm assuming I have this fictitious thing, a spring
10:07 which essentially has no mass, massless,
10:10 which obeys exactly Hooke's law.
10:13 And here I can't help digress and point out to you
10:17 that that's a terrible misnomer.
10:19 There is no Hooke's law of nature.
10:23 It is an empirical relation which tells you the force
10:27 that the spring exerts when you stretch it a certain distance,
10:31 all right.
10:32 But anyway, it's stuck historically.
10:34 It's Hooke's law.
10:35 So Hooke's law, from Hooke's law,
10:38 I know what will be the force, fs, when the situation is as
10:45 shown over there, all right, at time t.
10:49 So at this time, this extension of this spring
10:53 will be, of course, y0 minus yt.
10:57 OK.
10:58 And so I get that the force due to the spring
11:01 will be the spring constant times its extension
11:05 at that instant of time.
11:07 It is a vector.
11:08 And y0 is a bigger number than yt, this is a positive number.
11:18 Therefore, the stretched spring will pull the mass up.
11:22 So this is in the y direction.
11:24 This is plus.
11:26 How about the gravitational force?
11:28 Well, that is, of course, the minus mg,
11:34 the force of the gravitational field on that.
11:37 And it's minus in the y direction
11:40 here, because it's pulling this mass down.
11:45 OK.
11:45 Now what else do we know?
11:49 We know that we could get everything done very carefully.
11:53 We know that we defined y equals 0
11:56 to be the equilibrium position.
11:58 Therefore, when y is 0, we know that the second derivative of y
12:05 is 0.
12:06 It's not accelerating.
12:08 So that's a condition we must not forget.
12:10 Another thing we know that initially, in other words,
12:14 at t equals 0, the position of that mass is y initial.
12:20 Finally, I told you that the velocity of that mass
12:26 was 0 at t equals 0, stationary.
12:30 So this is the beginning of our translating all the information
12:35 we gathered here into mathematics.
12:48 Let me continue now using this information
12:52 and try to reduce it to the minimum set of equations.
12:57 From a equals fm, from this, I get that the acceleration
13:03 is the total force divided by m.
13:06 I can now replace these two forces
13:10 from the information I wrote over there.
13:12 And so that is equal, 1 over-- this is f over m.
13:17 It's 1 of m times the net force, which
13:19 is the force due to the spring minus the force
13:22 due to the gravity, OK.
13:24 So from this, I can now actually write an algebraic equation,
13:33 rather than the vector one.
13:34 Notice - ha ha.
13:37 I have noticed myself even something.
13:39 Here this has to be in the direction of y.
13:45 OK.
13:47 This is a vector equation, but all the parts
13:51 are in the same direction, in the y direction.
13:55 Therefore, I can rewrite this just the equation for the one
14:01 component and not bother to write the y hats throughout.
14:07 So this equation I've rewritten now just removing y hat.
14:12 So this is how the mass will be accelerating.
14:17 Unfortunately, it's a single equation,
14:20 but I have more than one unknown in it.
14:22 Because I don't know y0.
14:24 And I don't know y of t.
14:26 Clearly, I won't be able to solve that equation, all right.
14:29 But at that stage, I go back to the information
14:33 I told you at the beginning.
14:35 We defined y equals 0 to be the place where
14:40 a y double dot, the second derivative, is 0.
14:44 Therefore, I can write that at position.
14:50 When y of t is 0, this is 0.
14:54 So 0 is equal to 1 over m, into ky minus mg.
15:00 And I immediately from this get that ky0 is mg.
15:04 Therefore, I have found what y0 is.
15:09 OK.
15:10 Great.
15:11 So there's only one unknown here.
15:14 So using this information in here,
15:18 I end up immediately with this equation,
15:22 that the second derivative of y with respect
15:24 to time, the acceleration of the mass
15:27 is equal to minus k over m times y of t.
15:34 At this stage, I will really find this quantity, k over m.
15:41 For the time being, you can look at it as just for convenience,
15:46 less to write on the board.
15:48 But later, you see this will help us understand
15:54 how to deal with different situations.
15:56 But for the time being, you can just
15:58 think of this as a convenience, so I
16:01 can write less on the board.
16:03 And I end up finally with one equation.
16:08 The second derivative of y with respect to time
16:11 is equal to minus a constant, that's k over m, right,
16:15 times the value of y times t.
16:18 This is the equation of motion for this mass.
16:22 It tells me in mathematical form how the motion of that mass
16:28 changes with time.
16:31 I can now actually predict what will
16:35 happen in this particular situation.
16:39 Because I know what was the motion of it at time 0.
16:46 I know that at time 0, the position was y initial.
16:52 And the velocity was 0.
16:55 These three lines are completely equivalent from the point
17:03 of view of understanding the motion of the mass
17:06 to our original description.
17:09 This is a physical description of the situation.
17:14 This is a mathematical description
17:17 of the same situation.
17:22 So we've achieved step one.
17:24 We've translated a physical situation
17:27 into a mathematical one.
17:30 Let me now try from this, I should
17:33 be able to predict what this mass will do.
17:42 OK.
17:47 I'm now switching into the world the mathematics.
17:51 As I just am repeating here, I've
17:54 gone away from a physical description
17:57 to a mathematical description.
17:59 This is pure mathematics.
18:02 I have an equation, a mathematical equation,
18:05 for y of t.
18:07 It's a second order differential equation.
18:10 I had the boundary conditions, or initial conditions.
18:15 I can solve that using mathematics.
18:19 OK.
18:20 Let's do that.
18:20 So I'm now doing pure mathematics.
18:22 I don't want to teach you math.
18:25 That's the role of the math department, all right.
18:28 So how do I solve that equation?
18:32 And let me tell you how I solve it.
18:35 I am make use of the so-called uniqueness theorem.
18:39 I know, or the mathematicians have told me,
18:43 that if I find a solution to that equation which satisfies--
18:58 if I find a solution which satisfies that equation,
19:03 and if it has the right number of arbitrary constant,
19:09 then I have found the one and only general equation,
19:14 which is a solution of that.
19:18 Let me be concrete.
19:23 y of t equals to a cosine omega t plus phi, where A and phi are
19:39 arbitrary, are arbitrary.
19:47 They are some arbitrary numbers.
19:49 But a number is there.
19:50 This can be 7 and this can be 21 degrees,
19:53 or whatever, but any number.
19:55 This equation satisfies my differential equation.
20:01 If you don't believe me, try it.
20:02 Differentiate this twice, all right,
20:06 for any value of A and phi and you'll satisfy that equation.
20:11 So this is a solution which satisfy that equation.
20:16 It has the right number of arbitrary constants,
20:19 that two arbitrary constants in here.
20:23 And therefore, this is the only solution
20:25 in the universe of that equation.
20:29 OK.
20:30 Now, so being a physicist, I don't
20:33 care how I got the solution.
20:36 Once I had the solution, if I know
20:39 it's the only one that exists, I'm home.
20:42 Now you can say well, I suppose I didn't guess it.
20:46 Well, there are many ways.
20:47 You go on the web and find it.
20:48 You go in the book and find it.
20:49 You ask your friends what it is, all right.
20:52 That's mathematics.
20:54 And once you've found the solution, we can go on.
20:58 All right, so this is the solution of that equation.
21:02 Next, if that's y, what is y dot?
21:05 What is the rate of change of t?
21:08 That's going to equal minus omega 0 A sine omega 0 t
21:15 plus phi, OK.
21:20 Can I predict what will happen?
21:23 All right.
21:24 I still need, in order to be able to predict what
21:28 will happen, I need to find out what
21:30 are the values of A and phi which
21:36 satisfy the other information right here.
21:40 See, I told you that we reduce that physical situation
21:44 to a differential equation, the equation of motion
21:47 for this mass, including the information about where it was
21:53 at some instant of time, how it was moving, et cetera.
21:57 So I need to make sure that this equation satisfies
22:05 these boundary conditions.
22:08 In other words, it its thees boundary conditions
22:12 which will determine what are the A's and phi
22:17 for the particular problem that I had there, OK.
22:21 And so what I do is-- let's, for example, takes here.
22:26 Because I see 0.
22:28 Y dot t is 0, all right, at t equals 0.
22:36 Well, when t equals 0, this is minus omega 0 A sine phi.
22:45 OK.
22:46 Therefore, I immediately conclude that phi is 0.
22:55 OK.
22:56 Next, I know-- so now that I know that phi is 0,
23:00 I can go back to this equation.
23:03 This is now 0.
23:05 And we know that y at t equals 0 is y initial.
23:12 But that t equals 0 cosine of 0 is 1.
23:17 Therefore, A is y initial.
23:21 And so I get finitely y of t is equal to y initial, all right,
23:32 times cosine omega 0.
23:38 Let me now replace it with the 1-- well,
23:41 let me leave it as omega 0 t pluse 0.
23:45 That, and I can rewrite this, putting all the numbers
23:50 that I have, y initial cosine.
23:58 And I'll now even replace omega 0 by k
24:02 over n, square root of k over n, times t.
24:10 Notice there are no unknown quantities in here.
24:17 This tells me two things.
24:22 At any instant of time, I can calculate
24:28 where this mass will be.
24:32 It's given by this equation.
24:35 Secondly, I can describe the kind of motion it does.
24:39 What is this equation?
24:41 As a function of time, this corresponds
24:44 to an oscillating position y.
24:48 So this mass, when I let go, will oscillate.
24:55 What will be the period?
24:56 How long will it take before it comes back to where it started?
25:01 Well, the period t will be how much time
25:06 do I have to add to this t, so that the angle here
25:12 changes by 2 pi?
25:14 Well, that's obviously 2 pi root of m over k.
25:21 OK.
25:23 So I've achieved what I wanted to do.
25:26 I've taken a physical situation.
25:29 And I have predicted if I let go what will happen.
25:35 This is the motion it will experience.
25:40 This is the period.
25:42 I can predict the time, et cetera.
25:44 At this stage, let's stop for a second
25:48 and consider what we've done.
25:52 Because it's the essence of-- this
25:56 is a good example of the essence of the scientific method.
26:04 We have taken a physical situation.
26:08 We've described it in terms of mathematics.
26:20 Then we made an act of faith that if I
26:26 take the mathematical equations and I solve them,
26:33 that the resultant answer will actually
26:37 correspond to what nature will do.
26:41 If you stop to think about that, it's amazing.
26:45 Nobody understands that fact.
26:48 Why that's true.
26:49 Why it happened.
26:51 In other words, nobody understands
26:53 why nature can be described in terms of mathematics.
26:59 OK.
27:00 But it is that fact which makes the scientific method possible.
27:09 Finally on this note, let me give a quotation from Einstein
27:15 which beautifully summarizes what I've just said.
27:20 And that is the following, "The most incomprehensible thing
27:24 about the universe is that it is comprehensible."
27:29 The fact that we can follow this procedure is amazing.
27:38 OK.
27:39 Let me at this stage go and take another example, all right.
27:54 So let's take another example.
28:05 Consider the following situation.
28:08 I take something like a ruler, a uniform rod.
28:14 And I put a nail through it, some kind of a pivot.
28:19 There is some pivot.
28:21 I pivot the ruler on it.
28:24 And it's hanging like this.
28:27 OK.
28:28 Let's assume the mass is m of the ruler.
28:34 The length is l.
28:41 It's a uniform ruler, a rod of some kind.
28:46 And at t equals 0, I give it an impulse.
28:54 I give it a little impulse, so we are now at t equals 0.
29:05 We give it an impulse.
29:07 At that instant, the ruler is still hanging vertically.
29:15 Let me, just so that when you look on the board,
29:18 you may be confused in which plane I am.
29:21 This is the vertical plane.
29:23 So this is up.
29:29 So I give it an impulse.
29:31 So at that instant of time, this ruler
29:38 will have an angular velocity which I will call theta dot.
29:53 This is at time equals 0.
29:56 And it has some number as a result,
29:58 depends how big an impulse I gave it.
30:01 And so that you remember what I'm talking about,
30:04 I like to give this, instead of using a symbol,
30:07 I'll call this angular velocity at t equals 0.
30:16 So this is some number, so many radians per second.
30:20 That's at t equals 0.
30:22 And I'm now going to follow this method again.
30:25 I want to know what will be the motion of this.
30:29 What's going to happen to this ruler.
30:31 Is it going to start spinning around this, like this forever?
30:36 What will happen?
30:38 So I will try to translate this problem into mathematics.
30:47 Because of the mechanical constraint,
30:50 at some instant of time, the ruler may be doing this.
30:55 Let's call this the time t.
30:57 This is time t.
31:01 And time t is like this.
31:04 And I've got to define some coordinate system.
31:07 So I'll take this angle from the vertical.
31:11 And I call that theta at time t.
31:17 That's why I call this theta dot.
31:19 This is the rate of change of that.
31:22 So at some instance of time, it will be at this position,
31:27 all right.
31:28 At that instant of time, it'll have
31:31 a velocity in this direction.
31:35 And we'll have an acceleration in that direction.
31:37 So for example, the acceleration will
31:40 be theta double dot at time t.
31:49 And just so that at this stage, I will still to remind you
31:54 that's alpha, alpha time t.
32:01 Because often alpha is used as the acceleration.
32:04 So at the moment, I just want you
32:06 so you can easy for you to see what I'm talking about.
32:09 So at some instance of time, that is the physical situation.
32:14 I would like to now convert this into mathematics.
32:21 Follow the same procedure as before.
32:28 I need to write the equation of motion for this.
32:33 And I need to write down the initial conditions.
32:39 So how do I do with that?
32:42 So now I start off by the free body diagram.
32:47 Here is the pivot.
32:50 That's the route.
32:55 This angle is theta t.
33:05 There will be a force acting.
33:08 We're now dealing with rigid body motion.
33:11 So today we did Newtonian mechanics for masses and forces
33:18 through a single point mass and forces.
33:20 Now we are doing a Newtonian dynamics for rigid body motion.
33:26 You know that if a rigid body is in the gravitational field,
33:30 the gravity acts force fg.
33:36 We can analyze it, as if there was a force fg
33:38 g acting through the center of mass here of the body.
33:44 So this length now is l over 2.
33:52 So there will be a force fg acting.
33:55 And as a result, there will be torques about this point.
34:03 Now let me say the following.
34:05 We are dealing here with motion, rotations, in a single plane.
34:13 And so we are dealing about rotations,
34:17 about an axis through this point p.
34:22 We're not dealing with three dimensional rotations,
34:26 but simple situation where all the motion
34:30 is about a single axis, which is perpendicular to this point, p.
34:40 There will be a torque about p because
34:43 of the gravitational force.
34:46 And as a result, there's going to be the acceleration, which
34:50 as we've said over there, is theta double dot of t.
35:00 Now, we know that torques gives rise to angular acceleration.
35:16 Let me define that we will take clockwise motion, clockwise
35:23 motion, clockwise rotations to be positive.
35:42 So any rotation, this angle, for example, I am sorry.
35:48 I meant anti-clockwise.
35:53 Anti-clockwise is positive.
35:55 Look at this.
35:56 If this rotates like that to this angle, this
36:01 I take to be a positive number, it's
36:03 an anti-clockwise rotation.
36:06 Similarly, if this acceleration is a positive number,
36:09 it's accelerating in this direction.
36:13 Since we are dealing with rotations about a single axis,
36:18 we don't have to go to the vector formulation.
36:22 We can consider it just the magnitude.
36:27 And we know that the acceleration
36:36 is equal to the torque divided by the moment of inertia.
36:43 Or you may have seen it the other way.
36:45 Torque Equals I alpha.
36:56 I prefer it this way.
36:58 For me, it's more logical.
36:59 The angular acceleration is a consequence of the torque.
37:03 So I write it like that.
37:05 So this is the dynamic equation, which tells you
37:12 how the motion of this mass changes with time.
37:17 All right.
37:18 So alpha is theta double dot of t.
37:26 OK.
37:28 What is the torque at that instant of time?
37:32 Well, you know general torque is r cross f, all right.
37:42 That's true in three dimension.
37:44 So it will apply here.
37:46 So the torque is going to be this force times this distance.
37:58 OK.
38:00 So it's going to be-- let's write it.
38:01 The force is mg right, times l over 2 sine theta,
38:15 theta of time t.
38:16 OK.
38:18 That's the torque about this axis p on this rod, all right.
38:27 And it's divided by I, where I is the moment of inertia
38:35 of this rod about an axis through p
38:39 perpendicular to the board.
38:41 OK.
38:43 Now we need to calculate the moment,
38:45 in order to continue further, we need to calculate I.
38:49 Since we know this mass of the rod.
38:52 And we know it's a uniform rod.
38:54 And we know it's length l, we can calculate it.
38:59 You know how to do it.
39:00 If you don't, you can look it up in the book on mechanics,
39:04 all right.
39:05 Or just look up the moments of inertia.
39:07 And you will find that the moment of inertia,
39:16 you will find that the moment of inertia I for a rod like that
39:22 is 1/3 the mass times the length squared.
39:29 OK.
39:30 So now I have to continue.
39:32 But I've run out of board space.
39:35 So I'm going to erase the board at the far end.
39:38 And we'll continue from there.
39:42 So I erased the board.
39:45 And then so that you don't have to look backwards and forwards,
39:48 I've started rewriting it and I realized that I actually
39:53 missed the negative sign.
39:55 So I'm going to correct it here.
39:56 So that's why it's completely written out.
39:58 So let me just remind you.
40:01 The situation we have is this rod,
40:05 which at time t, we define this angle
40:07 to be theta t, the rotation of the rod.
40:10 It has an acceleration, theta double dot t.
40:13 And we are considering rotations about an axis perpendicular
40:18 to the board through this point here.
40:20 OK.
40:21 We know, that was the last thing we did,
40:24 that the acceleration is given by the torque divided
40:29 by the moment of inertia.
40:31 All right.
40:33 The torque is mg, l over 2 sine theta, I derived it
40:38 for you before, divided by I.
40:40 But what I neglected to put a negative sign.
40:44 And that you could do in your head, right.
40:46 Consider we've taken all the rotations to be positive
40:53 if they're anti-clockwise.
40:55 So this angle is a positive rotation.
40:58 This would be, this direction would be a positive rotation.
41:02 But the torque if you look at this,
41:05 there is a force acting down on this.
41:08 So about this point, it's trying to rotate
41:11 this in the clockwise direction.
41:14 And so it's minus.
41:15 And I didn't-- it would have naturally come out if I did
41:19 the full vector calculation, the torque is R times F.
41:25 It would have come out, the sign would have come out.
41:28 So that's where this minus sign comes in.
41:30 OK, so this is where we got on the board over there.
41:33 And now let's continue.
41:35 We can replace I from here.
41:38 And we get that theta double dot of t
41:45 is equal to minus, all right, 3 halves, 3 l, 3 halves, not l.
42:00 It's divided by 2.
42:01 l is at the bottom.
42:02 Sorry.
42:04 3 halves g over l times sine theta, theta of t.
42:13 OK.
42:17 I'm sorry.
42:20 Sine theta of t.
42:25 OK, as before, to simplify it, I will write omega, I'll define.
42:33 Let's define omega 0 squared to be equal to 3 halves g over l.
42:47 With this definition, we get that theta double
42:53 dot t is equal to minus omega 0 squared sine theta of t.
43:05 OK.
43:06 So this is our equation of motion for this problem.
43:13 That's the equation of motion.
43:16 And these are the boundary conditions.
43:20 So these three equations are a translation
43:25 of this problem in the language of mathematics.
43:33 If we now want to predict what will happen
43:37 to this rod at some other time, we
43:41 have to solve these equations.
43:43 And admit now, I have a problem.
43:47 If you remember, when we did it for the spring,
43:52 the equation of motion was one where I guessed the answer.
43:58 I don't know what the answer is of this.
44:04 If you go into books, you will find
44:07 that this is not one of the differential equations
44:11 which you can analytically solve.
44:13 It's, in fact, a second order differential equations
44:16 with transcendental functions in it.
44:19 So this is not something we know the answer to.
44:23 So the only thing if I want to now predict what will happen,
44:28 I have to numerically solve this.
44:31 And then I can-- I have enough information.
44:35 I can numerically solve this equation with these boundary
44:39 conditions and predict what will happen.
44:41 That's not very instructive for the purpose of course
44:44 at the moment.
44:46 So let me do something else.
44:48 OK, let me modify the problem.
44:50 Rather than take the problem we took, let me say,
44:55 how about if I took this rod and gave it
44:59 only a very tiny impulse.
45:01 So this angle is small.
45:04 Let me make the angles sufficiently small, such
45:11 that sine theta of t is always approximately
45:18 equal to theta of t.
45:24 Depends how well you want to approximate this.
45:27 But typically, if you use your calculator or computer,
45:31 up to about 10 degrees, that approximation is pretty good.
45:36 So I will now change my problem.
45:40 And I said OK, let's see whether we can predict analytically
45:44 the motion of the rod where I give the impulse, which
45:47 is sufficiently small, that this angle is always small.
45:52 Under those conditions, note that my equation of motion
45:58 becomes theta double dot of t is equal to minus omega squared
46:09 times theta at t.
46:12 Because sine theta t is always approximately equal to theta t
46:17 if I take the angle small enough.
46:26 And eureka, I can solve that one.
46:30 Because that's exactly the same equation we solved before.
46:43 OK, so we get the solution to that equation,
46:46 is theta of t is some constant cosine omega 0 t plus phi.
47:02 As before, A and phi are some arbitrary constants.
47:07 And clearly, if it worked over there for that same equation,
47:11 it works here.
47:12 The only difference here is we have theta of t instead y of t.
47:16 That's just different symbols, but the solution
47:19 is exactly the same.
47:21 So we know that's the solution of this equation.
47:25 We know the boundary conditions.
47:27 Therefore, we can predict what will happen.
47:28 Let's continue and do that.
47:30 So from here, you get theta dot of t is equal minus omega 0
47:38 A sine omega 0 t plus phase.
47:46 OK.
47:48 And we have to put in the boundary conditions.
47:52 OK.
47:54 Now at t equals 0, OK, we get that this is at t equals 0.
48:06 So at t equals 0, we get theta of t 0
48:13 is equal to A cosine phi.
48:22 OK.
48:22 Therefore, phi is pi over 2.
48:30 That's a possible value of phi.
48:35 Now that gives me that if I is pi over 2 here,
48:43 we get to that theta dot of t, which is equal to the angular,
48:55 angular velocity at t equal to 0, all right,
49:04 will be equal to minus omega 0 A sine omega ) t plus pi over 2.
49:26 OK, from which I can get that A is
49:31 angular velocity over t plus 0.
49:33 And so my final solution is that theta of t
49:39 is equal to angular velocity at t equal 0
49:52 divided omega 0 sine, sine omega O t.
50:03 OK.
50:03 And I want to make sure I'm not making a sign mistake again.
50:06 I'm not.
50:07 All right.
50:08 And so, and omega 0 we know, and so the
50:12 in terms of knowing quantities, the answer
50:16 is angular velocity of t equals 0 over.
50:25 And omega 0, we have found defined to be that,
50:30 so the square root of 3 g over 2 l times
50:39 sine square root 3g over 2l times t.
50:50 Now this is theta t.
50:54 So we have completely solved the problem.
50:56 And we have predicted the motion.
50:58 So as before, following this process
51:04 of taking the physical situation,
51:07 describing it in terms of mathematics,
51:10 solving the mathematical equations,
51:12 including all the information we have
51:14 about the problem, the boundary--
51:16 initial conditions or boundary conditions,
51:19 we can predict what will happen to this angle
51:24 as a function of time, and also the kind of motion
51:28 this is an accelerating motion.
51:31 I can also predict, as before, that the period of this
51:35 will be 2 pi, 2 pi square root of 2l over 3g, et cetera.
51:46 OK.
51:49 Now, one of the things you'll notice, that in some ways,
52:01 it seems I'm repeating myself.
52:06 We took completely different situations,
52:09 and yet the result, the equations of motion,
52:13 and the results, have good very similar form.
52:17 Now this is part of the beauty of the scientific method.
52:22 Because it turns out that very many
52:29 different physical situations can
52:32 be described by the same mathematical equations.
52:37 So once you've solved the problem for one
52:43 physical situation, you have automatically have solved it
52:48 for an almost infinite number of other situations which
52:53 are described by this same mathematics.
52:57 Finally, let me do just more as a question of practice,
53:01 one more problem of this kind that apparently
53:06 seems to be completely different.
53:07 I'll take a problem from electricity and magnetism.
53:14 Let me consider the following situation.
53:25 So now we're going to a different problem.
53:29 The physical situation is suppose I have two plates, two
53:34 metal plates, and I connect them with a wire.
53:38 Schematically, it consists of a capacitor C
53:46 connected to an inductor.
53:53 This is a schematic representation
53:56 of two parallel plates connected, short
54:00 circuited by a wire.
54:03 I will assume for simplicity here that these wires have
54:10 no resistance, superconducting, all right.
54:14 Any loop like has an inductant L.
54:21 And the capacity between these is C.
54:24 So this is an L C circuit.
54:27 And I'm going to assume that at time equal 0, so this is now
54:32 time equals 0, I have a charge here, minus Q0 plus Q0
54:41 here, OK.
54:44 And let's assume that at time there's even a current flowing,
54:47 so I is 0 here.
54:54 So this is a system which is disturbed from equilibrium.
54:59 And what will happen is a function of time.
55:05 I will do the same and almost boring you to tears,
55:10 I'm going to, you'll see I'm essentially
55:13 doing the same problem again.
55:15 I will now consider this circuit at some arbitrary time t,
55:23 derive the equation of motion for the charges in the current,
55:29 therefore translate this physical situation,
55:32 or describe this physical situation
55:34 in terms of mathematics, deriving
55:37 mathematical equations, solve them,
55:40 and predict what will happen.
55:42 So I just follow what I just did a second ago.
55:46 So at some instant of time, that same circuit L
55:55 will have some current I of t to the charge
56:00 minus Q of t plus Q of t, all right.
56:06 This at time t.
56:12 So from this, I can derive the equation of motion.
56:22 Let me remind you about Faraday's law.
56:32 You know that if you have current coil in the loop,
56:41 it produces magnetic flux in that loop.
56:48 The changing flux gives rise to an EMF around that loop.
56:54 To be specific, Faraday's law I can write.
56:57 If I take this circuit of the wire, the integral of E
57:06 dot dl around a closed loop, that is equal to minus du phi.
57:16 5.
57:17 Now watch out.
57:18 The Greek alphabet has a limited number of letters
57:21 so you'll find one constant is reusing the letters.
57:27 But at the moment not to confuse you,
57:29 I'm going to put here magnetic flux,
57:32 total magnetic, total magnetic.
57:41 So phi is the total magnetic flux
57:44 linking this circuit, all right, dt.
58:00 So Faraday's law tells us that the integral of ED
58:05 all around this loop will be equal two
58:09 minus the rate of change magnetic flux.
58:12 This is the dynamic equation which
58:17 tells you how this behaves.
58:18 It is the analogous to Newton's law f equals ma
58:24 in the case of our mass, or 2 torque is I
58:30 alpha in the case of rotations, et cetera.
58:33 This is the non dynamic equations.
58:35 So let me calculate this.
58:38 And now I'm going through the-- around this.
58:41 And you find since this wire I'm assuming is superconducting,
58:46 there can be no electric field inside it.
58:49 So the contribution to this line integral
58:52 is 0 when I go through the wire.
58:56 So the only place where this line integral is non-zero
58:59 is between the plates, all right.
59:02 And that is simply the potential difference
59:04 between those, which is q over c.
59:09 OK.
59:10 Q at time t over c, where c is the capacitance is.
59:17 That is the integral of EDL around that loop.
59:22 And that's going to be equal to minus D dI at time t dt.
59:39 All right.
59:39 Because the magnetic flux, this is
59:44 by definition of the inductance or first inductance
59:50 is that the total flux linking the circuit when
59:55 the current flowing in it is I of I, the total flux
59:59 is L times I. Here, I have the rate of change of that flux,
1:00:04 so it's equal to this.
1:00:07 OK.
1:00:08 Now we know by charge conservation,
1:00:13 that the current I of t.
1:00:19 What is the current?
1:00:20 It's the charge is flowing per second
1:00:22 will be equal to the number of charges per second that arrive
1:00:28 at this plate here or the part from there is equal to dQ dt.
1:00:39 OK.
1:00:39 Or in other words, Q dot.
1:00:46 OK, let me continue.
1:00:57 So from these two equations, right, the dIdt therefore,
1:01:04 this is the second thing, so I end up from there
1:01:07 that Q double dot, second derivative of t,
1:01:13 is equal to minus right 1 over LC times Q of t.
1:01:24 Eureka.
1:01:25 We have once again the same equation.
1:01:28 This I can define as before, omega O squared.
1:01:32 If I define that as one over LC, all right,
1:01:37 then what I have here is Q of t double dot is equal to minus
1:01:45 omega ) squared Q of t.
1:01:52 Again, we have come to the same equation.
1:01:55 This is the same equation of motion
1:01:59 as we came in the other two situations.
1:02:02 So the answer will be the same.
1:02:05 The variables will be different here.
1:02:07 It'll be the charge that will be changing with time,
1:02:12 while there in the one case was the angle.
1:02:15 In the other case was the position of the mass.
1:02:21 OK, and the solution to this problem,
1:02:24 I can now write immediately, is Q of t
1:02:29 is A cosine omega 0 t plus fe.
1:02:36 Note that this is the fe is nothing to do with that phi.
1:02:40 OK.
1:02:40 And Q dot of t, which by the way is I of t,
1:02:50 is equal to minus omega o A sine omega o t plus phi.
1:03:02 OK.
1:03:03 Now as before, what actually happens
1:03:09 depends on the initial conditions.
1:03:14 And we look at that picture on the top board.
1:03:19 We know that initially Q is Q 0.
1:03:23 And we know that initially Q dot t is I0.
1:03:28 To save time, I'll just immediately write.
1:03:30 You can do that in your head.
1:03:36 And if you the write that out, you
1:03:41 find that if you've used those two conditions,
1:03:48 you find that time phi this time is equal to minus IO over
1:03:59 Q0 omega 0 and A is equal to Q0 over cosine phi.
1:04:11 I saved time without just solving algebraic equations.
1:04:15 Take these two equations.
1:04:17 Consider t equals 0, the values of those quantities,
1:04:22 and just solve for the two unknowns and you get this.
1:04:25 And so once again, we have predicted what will happen.
1:04:30 And what I would like to just at this stage
1:04:35 emphasize that although we have taken three
1:04:40 different physical situations, in each case,
1:04:45 we took the system, displaced it from equilibrium, let go
1:04:52 and we wanted to see what will happen.
1:04:56 In all the cases, it turned out that
1:04:59 the mathematical description, the mathematical equations
1:05:03 are identical in form.
1:05:05 And so they gave, not surprisingly,
1:05:08 the same kind of motion.
1:05:12 This motion that we see in all those cases,
1:05:15 we called simple harmonic motion.
1:05:21 It has the characteristic that if you displace the system
1:05:24 from equilibrium, it oscillates with harmonic motion,
1:05:29 meaning it oscillates as sine or a cosine
1:05:34 of different phases, et cetera.
1:05:37 If you tell me any one of these systems
1:05:40 where it was at any instant of time,
1:05:43 I can predict it forever in the future.
1:05:48 Now finally, the last few minutes.
1:05:55 Some of you may have noticed that in each of these--
1:06:00 or I told you at the beginning that I
1:06:01 can take a physical situation and describe it
1:06:08 in terms of mathematics, and thereby predict the future.
1:06:13 But in each case, I in some ways almost cheated.
1:06:18 I said let's consider an ideal spring.
1:06:21 We'll assume it has no mass, that it exactly
1:06:25 obeys Hooke's law.
1:06:27 Or when it came to that rod, I assumed
1:06:31 that it's only displaced by small amounts,
1:06:36 so that's sine theta equals-- I can approximate with theta.
1:06:44 In the case of the electrical circuit,
1:06:48 it maybe not so obvious what I assumed,
1:06:51 but I certainly made assumptions about that the wire is
1:06:56 perfectly conducting, et cetera.
1:06:58 And I didn't discuss in detail what
1:07:02 happens in between the plates or the capacitor
1:07:04 where the fields are, et cetera.
1:07:06 One is doing approximations.
1:07:10 In reality, if you look at any physical situation
1:07:14 in the world, it's always incredibly complicated.
1:07:23 It's never that you have an idealized situation like this.
1:07:27 So to what extent does what we have just done
1:07:32 correspond-- is it useful at all.
1:07:35 And the way I'll answer it is by another example.
1:07:40 This is the last thing I'll do on the top of a simple harmonic
1:07:45 motion, last problem.
1:07:47 Suppose I'm looking out of the window.
1:07:51 And I see there is a tree and a branch and a bird lands on it.
1:08:00 Do I understand what will happen?
1:08:04 It's clearly extremely complicated.
1:08:07 The mechanics of the branch is complicated.
1:08:10 There is air friction.
1:08:13 Nothing is simple about it.
1:08:15 And yet you and I can predict what will happen.
1:08:19 You know what will happen.
1:08:21 As the bird lands, it'll start oscillating and finally
1:08:27 come to rest, very much like harmonic motion.
1:08:34 I claim I can use the word, I understand what's going on.
1:08:41 And the reason why I claim that is the following.
1:08:44 That to understand something, all
1:08:49 I would like to understand the general features of what's
1:08:54 going on.
1:08:54 I don't need to know what every atom in the branch
1:08:58 is going on in the process of trying
1:09:02 to understand what the bird is doing.
1:09:04 If I want to understand what atoms are doing,
1:09:06 that's a different story.
1:09:09 And so one of the important abilities we have to develop
1:09:20 is to be able to, when you see some situation,
1:09:26 model it in terms of the most important aspects
1:09:33 of the situation.
1:09:35 And let me be concrete.
1:09:38 In this case, I can say look, I can model this approximately
1:09:45 as the branch I'll treat as a spring,
1:09:49 of some spring constant k.
1:09:51 The bird I'm going to treat as a mass m.
1:09:54 And I'm going to consider this situation
1:09:57 to be modeled by a mass being placed on a spring and let go.
1:10:07 Now is that going to be exactly this?
1:10:09 No.
1:10:11 But from the point of view of understanding
1:10:15 the general features of this, it will
1:10:18 be a reasonable approximation.
1:10:23 Now how can I check, this is the scientific method,
1:10:28 that this is a good approximation is the following.
1:10:33 Make a prediction.
1:10:35 Suppose when I see the bird landing,
1:10:41 it makes five oscillations, five oscillations in ten seconds.
1:10:55 OK.
1:10:57 I can predict approximately once the oscillations have
1:11:03 died out how much the bird has compressed, distorted,
1:11:10 this branch.
1:11:11 In other words, from the moment it landed,
1:11:15 what distance will the branch, its position,
1:11:19 be lowered when it comes to rest.
1:11:26 So let's model it.
1:11:28 I'll model it first of all, as I did here,
1:11:32 as a mass and a spring problem using-- this is now
1:11:38 the same problem we did at the beginning that is still
1:11:40 on the board here.
1:11:42 I can calculate that for this idealized situation,
1:11:46 the period will be equal to 2 pi square root m
1:11:50 over k, for this idealized model.
1:11:54 In reality, there's friction.
1:11:57 So this oscillation will be dumped out.
1:12:02 And you would have learned from Professor Walter Lewin
1:12:06 that if you have a damped oscillator,
1:12:12 the frequency of oscillations does not depend significantly
1:12:20 on how much damping it is, provided it is weak damping.
1:12:24 So I would make the assumption that the period
1:12:30 will be given by that.
1:12:33 I also know that at the time when the motion has been damped
1:12:38 out and the bird has come to rest, at that instant,
1:12:44 there's no net force on the bird.
1:12:46 And so the force of gravity on it
1:12:49 will be equal to the restoring force due to the spring.
1:12:54 So mg is equal to kl.
1:12:57 From this, I get that m over k is l over g.
1:13:03 All right.
1:13:04 But we know what the period is.
1:13:07 We said five oscillations in 10 seconds.
1:13:10 So the period is two seconds.
1:13:14 So two seconds will equal 2 pi divided
1:13:17 by but this, which is l over g, the square root of l over g.
1:13:25 Square this and calculate the one unknown l
1:13:29 and you get one meter.
1:13:33 So my prediction is that this bird
1:13:36 will, after it's settled down, roughly be lower by one meter.
1:13:44 It's certainly not going to be one millimeter.
1:13:46 It's not going to be one centimeter.
1:13:48 It's not going be 10 meters.
1:13:50 And if you go and you measure it,
1:13:53 you find this is approximately correct.
1:13:56 The fact that I can predict it is for me
1:14:01 the same as saying I understand what's going on.
1:14:05 I realize it's not exact.
1:14:08 But with the approximations that I've made,
1:14:13 I get an answer which is consistent with what
1:14:17 is observed.
1:14:19 Today I have tried to tell you what my role in this course
1:14:25 is, what I'm trying to help you learn.
1:14:32 I intentionally went very slowly.
1:14:35 I used the word gory detail.
1:14:39 I tried, in particularly in the first problem,
1:14:42 not to miss any steps.
1:14:46 And what we covered today is the phenomenon
1:14:51 of simple harmonic motion.
1:14:54 It occurs whenever you have any system which
1:15:01 is displaced from equilibrium where the restoring force is
1:15:07 proportional to the displacement.
1:15:11 And it illustrated that you could
1:15:13 have very, very different physical situations which,
1:15:18 when translated into mathematics,
1:15:20 give essentially the same problem.
1:15:24 So it's a beautiful example of the scientific method where
1:15:32 we utilize this same-- well, once we've learned it
1:15:40 for one system, we can apply the results to another system.
1:15:46 So as I said, today I did simple harmonic motion.
1:15:50 Next time, we would be considering problems
1:15:53 to do with simple harmonic motion,
1:15:56 but which includes friction, damping.
1:15:59 We'll then go on to talk about harmonic oscillators which
1:16:04 are driven.
1:16:05 So we have driven harmonic motion.
1:16:07 And gradually in the course, we'll
1:16:09 go to more and more decrease of freedom, waves, et cetera.
Course Info
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Prof. Wit Busza
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Physics
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Fall 2012
Level
Undergraduate
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7246 | https://www.gauthmath.com/solution/1803569111825414/Creative-Section-A-4-Let-s-take-a-collection-of-any-three-high-mountains-of-Nepa | Solved: Creative Section - A 4. Let's take a collection of any three 'high mountains of Nepal' an [Math]
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Creative Section - A 4. Let's take a collection of any three 'high mountains of Nepal' and answer following questions. a) Is it a well-defined collection? Give reason. b) Is it a set? Give reason. c) Let's express it as a well-defined collection.
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Answer
a) The collection is not well-defined due to the vague and subjective nature of the term 'high mountains of Nepal.'
b) The collection is not a set because it lacks clearly defined elements.
c) To express it as a well-defined collection, specific criteria such as naming three mountains above a certain altitude in Nepal can be used.
Explanation
a) The collection of any three 'high mountains of Nepal' is not well-defined because the term 'high mountains of Nepal' is vague and subjective. Without specific criteria for what constitutes a 'high mountain,' the collection lacks clarity and precision.
b) The collection is not a set because a set must have clearly defined elements, and the term 'high mountains of Nepal' does not provide specific mountain names or criteria for inclusion. Sets require distinct and unambiguous members.
c) To express the collection as a well-defined set, specific criteria need to be established, such as naming three mountains above a certain altitude in Nepal, like Mount Everest, Kanchenjunga, and Lhotse. This would provide clarity and precision to the collection, making it a well-defined set.
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Creauve Secuon 4. Let's take a collection of any three 'high mountains of Nepal' and answer th following questions. a Is it a well-defined collection? Give reason. b Is it a set? Give reason. It's c Let's express it as a well-defined collection. 5. Let's take a collection of 'smaller odd numbers less than 10 . Answer the following 10. questions. a Is it a well-defined collection? Give reason. b Is it a set? Give reason. c Let's express it as a well-defined collection and list the member of the set so formed. d What is the cardinal number of this set?
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Creative Section - A 4. Let's take a collection of any three 'high mountains of Nepal' and answer th following questions. a Is it a well-defined collection? Give reason. b Is it a set? Give reason. c Let's express it as a well-defined collection. 5. Let's take a collection of 'smaller odd numbers less than 10' . Answer the followin questions. a Is it a well-defined collection? Give reason. b Is it a set? Give reason. c Let's express it as a well-defined collection and list the member of the set so formed. d What is the cardinal number of this set? 6. The factor-cards given below show the all possible factors of 12 and 18. 1 2 3 4 6 12 1 2 3 6 9 18 a Let's write the set of factors of 12F_12 in listing method. b Let's write the set of factors of 18F_18 in set-builder method. c List the common elements of F_12 and F_18 in a separate set P and write its cardinal number. 7. The multiple-cards given below show the first five multiples of 4 and 7. 4 8 12 16 20 7 14 21 28 35 a Let's write the set of the first five multiples of 4M_4 and 7M_7 in listng method. b Rewrite the sets M_4 and M, ub set builder method. c Let's write a set A taking the common elements of M_4 and M_7. d What type of set is the set A? Give reason.
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1.1 Set - Looking back Classwork - Exercise 1. Let's tick√ the well-defined collections. a A collection of High mountains of Nepal. b A collection of mountains of Nepal with more than 6000 m altitude. c A collection of fruits. d A collection of tasty fruits. 2. Let's write the objects of the following well-defined collections as the members of sets. a A collection of stationery items inside your school bag. _ b A collection of furniture inside your classroom. _ c A collection of clours in our national flag. _ d A collection of clours in a rainbow. _ 3. If W= 0,1,2,3,4,5 and A= 2,4,6 , let's say and write 'true' or 'false'. b a A ∈ W _ 2 ∈ W _ c 6 ∉ W _d 2,4 ∉ W _ 4. Let's say and list the elements of these sets. a letters of the word SUCCESS' = _ b x:x<10,x ∈ odd numbers = _ Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 7 Vedanta Excel in Mathematics - Book 7
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Why Use a Zero-ohm Resistor?
Why Use a Zero-ohm Resistor?
by: Mia Aug 10,2023 24294 Views 0 Comments Posted in PCB Basic Information
ResistorZero-ohm Resistor
What is Zero-ohm Resistor？
A zero-ohm resistor, also known as a jumper resistor, is a specialized resistor with a nominal resistance value of zero ohms. It is commonly used in PCB design and other applications as an idealized component. Despite being labeled as 'zero ohms,' the actual resistance value of a zero-ohm resistor is not exactly zero, but very close to it. There are two common packaging forms for zero-ohm resistors: axial lead zero-ohm resistors and surface-mount zero-ohm resistors.
The Functions of Zero-ohm Resistor
Debugging and Compatibility
When hardware engineers design PCB boards, it's important to consider compatibility issues. For example, a certain pin of a chip might have two functions: driving a buzzer or driving an LED. However, these two functions cannot work simultaneously. In order to enable the selection of which module to drive on the same circuit board, a zero-ohm resistor can be added to the circuits connecting the buzzer and the LED. By soldering the zero-ohm resistor on either the buzzer's path or the LED's path, engineers can choose to drive either the buzzer or the LED.
As shown in the diagram below, signal usage can be selected, and the corresponding component can be soldered, and leave the unused one empty.
During configuration, especially for embedded systems, different boot sources can be chosen through on-board configurations when selecting the boot source signal. At this point, zero-ohm resistors are used.
Used as Jumpers
To reduce wiring complexity, enhance aesthetics, and simplify installation, while avoiding high-frequency interference caused by jump wires.
Bridging circuits during wiring
During the PCB layout and routing phase, there are instances where certain connections cannot be established easily, especially on small boards with numerous connections. In such cases, if a connection requires a lengthy detour to establish, a zero-ohm resistor can be used to bridge the previous traces. This doesn't add extra layers and helps reduce the production cost of the PCB.
Reserved for debugging
When circuit parameters cannot be ascertained during parameter matching and practical testing requires removing a resistor or trying different resistance values to achieve an optimal solution, a zero-ohm resistor is typically employed. The zero-ohm resistor is used as a placeholder until the specific parameter values are determined during actual testing, after which it's replaced with actual components of the precise values.
For instance, when there are two different supply voltage options, 5V and 10V, but uncertainty exists regarding whether the different power supply voltages would affect the entire circuit in an unknown way, a zero-ohm resistor is often used to connect them. This allows selecting the power supply connection during circuit debugging.
Measuring circuit power consumption
When unsure about the power consumption of a chip or an entire circuit, a zero-ohm resistor can be placed in series with the power supply terminal. After the prototype board is completed, the zero-ohm resistor can be removed, and the two solder points can be directly measured using a multimeter. This measurement helps determine the actual working current, aiding in the calculation of chip or overall circuit power consumption. If you want to measure the current consumption of a specific circuit section, you can attach a zero-ohm resistor and a current meter to facilitate current measurement, which is useful for measuring high currents.
As shown in the diagram below, a zero-ohm resistor is placed between the power supply terminal and the circuit being measured.
Acting as Capacitors or Inductors in High-Frequency Signals
Under high-frequency signals, a zero-ohm resistor, when matched with the external circuit characteristics, can function as a small capacitor or inductor. This can effectively address EMC issues, such as between ground connections or between power supply and chip pins.
Serving as Fuses for Overcurrent Protection
Due to the high fusing current of traces on a PCB, it's challenging for them to fuse during short circuits or overcurrent faults. This might lead to more significant accidents. However, the current-carrying capacity of a zero-ohm resistor is relatively weak, and it would melt first during overcurrent conditions. This action can disconnect the circuit, preventing more substantial accidents from occurring.
Single-Point Grounding for Analog and Digital Grounds
During circuit design, particularly in mixed circuits involving digital and analog components, it's often required to have separate grounds with a single-point connection. A zero-ohm resistor can be employed to connect these two grounds instead of directly joining them. This approach divides the ground lines into two networks, which proves beneficial during processes like large-scale copper pouring, preventing the accumulation of charge due to "floating ground" and static electricity.
In similar scenarios, the following methods might also be used to address this issue: 1) using ferrite beads; 2) using capacitors; 3) using inductors; 4) using zero-ohm resistors.
Used in Current Loop for Bypassing
When a ground plane is split, leading to the interruption of the shortest return path for a signal, the signal path has to take a detour, resulting in a significant loop area. This can amplify the influence of electric and magnetic fields, making it susceptible to interference or causing interference. Placing a zero-ohm resistor across the split area provides a shorter return path, reducing interference.
As shown in the diagram below, when two signals are communicating and situated on different ground planes, adding a zero-ohm resistor between these two signals forces the signals to take the shortest path.
Configuration Circuit
This function is similar to jumpers or DIP switches, but it's soldered in place to prevent casual modification by regular users. By installing resistors in different positions, the circuit's functionality or address settings can be altered.
Noise Suppression
Due to the characteristics of a 0-ohm resistor, it can effectively suppress loop currents, thereby attenuating noise. In reality, a 0-ohm resistor doesn't truly have zero impedance; only superconductors can achieve true zero impedance. Therefore, a 0-ohm resistor actually provides attenuation across all frequency bands.
What is the difference between a zero-ohm resistor and a wire?
Certainly, there are differences. As evident from the functions I mentioned earlier, the capabilities of a zero-ohm resistor are quite diverse. Furthermore, in some cases, the roles of a zero-ohm resistor and a wire are equivalent. For instance, during circuit bridging, zero-ohm resistors are used instead of wires because in the manufacturing process, surface mount technology (SMT) is employed, and SMT machines can recognize zero-ohm resistors while identifying wires can be challenging. Hence, for production efficiency and convenience, zero-ohm resistors are often chosen.
Is there a difference in the accuracy of zero-ohm resistors?
Yes, there is a distinction. For instance, when the accuracy of a zero-ohm resistor is specified as ±1% or ±5%, it comes in various package sizes such as 0603, 0805, 1206, 1210, 1812, and more. Different accuracies lead to varying dimensions and forms, as depicted in the diagram below.
How much current can a zero-ohm resistor handle?
In circuit design, zero-ohm resistors are often used, and the selection of the resistor's rated power is based on the circuit current. The actual resistance value of a typical zero-ohm resistor is around 50 milliohms with a deviation of approximately ±5%. Based on the rated power, you can calculate the rated current for the zero-ohm resistor.
For various package sizes, the calculation of the rated current is as follows:
0402 1/16W: 1/16 = I I 0.05, which gives I = 1.118A
0603 1/8W: 1/8 = I I 0.05, which gives I = 1.58A
0805 1/4W: 1/4 = I I 0.05, which gives I = 2.236A
However, the specific current-carrying capacity of a zero-ohm resistor for each package size also depends on the resistor's heat dissipation on the PCB.
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7249 | https://www.youtube.com/watch?v=jwuiVb84Xx4 | Quotient Rule Made Easier (NancyPi)
NancyPi
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Posted: 23 Sep 2018
MIT grad shows an easy way to use the Quotient Rule to differentiate rational functions and a shortcut to remember the formula. The calculus Quotient Rule derivative rule is one of the derivative rules for differentiation. It's used to take the derivative of a rational function. To skip ahead: 1) For an easy way to remember the Quotient Rule formula, skip to time 0:21. 2) For an example of how to use the Quotient Rule to take the derivative of a fraction or quotient of functions (rational function), skip to 1:41. This video is a basic introduction to the Quotient Rule for taking derivatives in calculus. Nancy formerly of MathBFF explains the steps.
For more help with Quotient Rule derivatives and HOW TO TAKE THE DERIVATIVE of a function using the DERIVATIVE RULES (Power Rule, Product Rule, Quotient Rule), jump to:
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The Quotient Rule (calculus) tells you how to find the derivative of rational functions (a fraction, or one function divided by another function). The formal definition (textbook definition) of the Quotient Rule is often unnecessarily complex and intimidating.
There is a memory trick, or mnemonic, for how to remember the Quotient Rule formula. All you need to remember is the song "LO dee-HI minus HI dee-LO, over LO LO," where "dee" means the "derivative of." "HI" means your top function in the numerator, and "LO" means your bottom function in the denominator.
In other words, multiply the bottom function times the derivative of the top function MINUS the top function times the derivative of the bottom function, then DIVIDED by the bottom function times itself. After you differentiate the function with the Quotient Rule, remember to simplify the expression as much as possible using algebra.
This video is a basic intro to the Quotient Rule. For more of my calculus math videos and examples of taking derivatives, differentiation rules like the chain rule, differential calculus, basic calculus, integral calculus, common derivatives, and calculus problems (including Calculus 1, AP Calculus AB, AP Calculus BC, and Calculus 2), as well as precalculus and algebra math help, check out:
233 comments
Transcript:
Hi! I'm Nancy. I'm going to show you a shortcut to remember the quotient rule. So what the quotient rule is is a formula for taking the derivative of one function divided by another function. So let's take a look at the formal definition of the quotient rule. So this mess is the formal definition of the quotient rule. So I don't know about you but this is hard to remember. Let me show you a way that makes it a lot easier. OK. Say you have this function and you need to take the derivative of it. Let's call the top function your "HI" function. And the bottom function the "LO" function. Alright. So here's the shortcut you can remember the quotient rule by just remembering LO dee HI minus HI dee LO over LO LO. "dee" just means 'the derivative of'. So we have "LO" function times derivative of "HI" function minus "HI" function times derivative of "LO" function all over "LO" function times "LO" function. That's bottom times derivative of top. minus top times derivative of bottom all over bottom times bottom. So the shortcut to remember is just LO dee HI - HI dee LO over LO LO. So let's try it. Our function derivative f'(x) will equal "LO" 3x - 1 times "dee Hi" derivative of x squared. 2x. minus "HI" just x squared function. times "dee LO" derivative of 3x - 1. Which is 3. All divided by "LO LO" Just your bottom function times itself. (3x - 1) (3x - 1) Which can also be written as (3x - 1) ^ 2 and that is the answer to your function derivative. using the quotient rule. Of course you would need to simplify this answer. You can simplify the numerator. The top. But this is basically it. So you can either remember this... or you can just remember "LO dee HI minus HI dee LO over LO LO" Sounds kinda dumb when you say it in your head, but it is super helpful. So I hope this trick helps. Look, anything that makes calculus easier is a good thing. So if you found this helpful please click 'Like' on my video or subscribe! |
7250 | https://proofwiki.org/wiki/Product_of_Three_Consecutive_Integers_is_never_Perfect_Power | Product of Three Consecutive Integers is never Perfect Power
From ProofWiki
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Theorem
Let $n \in \Z_{> 1}$ be a (strictly) positive integer.
Then:
: $\paren {n - 1} n \paren {n + 1}$
cannot be expressed in the form $a^k$ for $a, k \in \Z$ where $k \ge 2$.
That is, the product of $3$ consecutive (strictly) positive integers can never be a perfect power.
Proof
Aiming for a contradiction, suppose $\paren {n - 1} n \paren {n + 1} = a^k$ for $a, k \in \Z$ where $k \ge 2$.
We have that:
: $\gcd \set {n − 1, n} = 1 = \gcd \set {n, n + 1}$
Thus $n$ must itself be a perfect power of the form $z^k$ for some $z \in \Z$.
That means $\paren {n - 1} \paren {n + 1} = n^2 - 1$ must also be a perfect power of the same form.
Let:
: $n = r^k$ and $n^2 − 1 = s^k$
for $r, s \in \Z$.
Then:
: $\paren {r^2}^k = 1 + s^k$
But the only consecutive integers that are $k$th powers are (trivially) $0$ and $1$.
Hence by Proof by Contradiction there can be no such $n$.
$\blacksquare$
Sources
1982: Donald J. Newman: A Problem Seminar
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $3$
Retrieved from "
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7251 | https://www.quora.com/If-we-take-a-rectangle-with-dimensions-r-and-h-and-rotate-it-360-then-a-cylinder-will-be-formed-the-area-of-which-can-be-rh-2%CF%80r-2%CF%80r%C2%B2h-I-know-this-is-wrong-but-still-where-I-am-doing-a-mistake | If we take a rectangle with dimensions 'r' and 'h' and rotate it 360° then a cylinder will be formed, the area of which can be (rh) 2πr = 2πr²h I know this is wrong but still where I am doing a mistake? - Quora
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If we take a rectangle with dimensions 'r' and 'h' and rotate it 360° then a cylinder will be formed, the area of which can be (rh) 2πr = 2πr²h I know this is wrong but still where I am doing a mistake?
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Janak Arora
BSc ( Chemical Engg.) from Institute of Technology, BHU, Varanasi (Graduated 1970) · Author has 180 answers and 330.3K answer views
·3y
The area of the curved surface of a cylinder formed by rotating a rectangle of dimensions r and h through 360° will be equal to the area of the rectangle which was rotated, ie r h.
Let us look at it another way. Let r be length of the rectangle and h be its width. Let us rotate the rectangle through 360° along its length r. Let x be the radius of the cylinder formed.
Then, 2 pi x = r (the length of the rectangle)
So, x = r / 2 pi = radius of the cylinder formed.
The height of the cylinder formed will be h.
The curved surface area of a cylinder is given by
2 pi (radius of the base of the cylinde
Continue Reading
The area of the curved surface of a cylinder formed by rotating a rectangle of dimensions r and h through 360° will be equal to the area of the rectangle which was rotated, ie r h.
Let us look at it another way. Let r be length of the rectangle and h be its width. Let us rotate the rectangle through 360° along its length r. Let x be the radius of the cylinder formed.
Then, 2 pi x = r (the length of the rectangle)
So, x = r / 2 pi = radius of the cylinder formed.
The height of the cylinder formed will be h.
The curved surface area of a cylinder is given by
2 pi (radius of the base of the cylinder) height
But, radius x of the base of the cylinder as calculated
above = r / 2 pi
So, the curved surface area of the cylinder =
( 2 pi r/ 2 pi) h = r h
Hope, I have been able to explain.
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More answers below
A cylinder is generated by rotating a rectangle about one of its edges. The perimeter of the rectangle is 30 cm. To obtain a cylinder of maximum volume, what dimensions should the rectangle have?
How can a cylinder change into rectangle?
Why does a rectangle that is rotated on its side become a cylinder?
How can you slice any rectangle twice to form a square with the same area?
How did you compute the dimension of each fundamental rectangle?
Girija Warrier
Author of the book ‘Madeira Math 500’ (2018–present) · Author has 5.9K answers and 13.9M answer views
·3y
GIVEN: a rectangle with the length = ‘r’ unit , breadth or height = ‘h’ unit
let me make it clear that standard formula of the curved surface area of a cylinder is derived by making a cylinder by FOLDING a piece of rectangle, the length of which = 2pi r & breadth = ‘h’ , ‘r’ is not the length of the rectangle but ‘r’ is the radius of the base of the cylinder.. that is why the standard formula , you
Continue Reading
GIVEN: a rectangle with the length = ‘r’ unit , breadth or height = ‘h’ unit
let me make it clear that standard formula of the curved surface area of a cylinder is derived by making a cylinder by FOLDING a piece of rectangle, the length of which = 2pi r & breadth = ‘h’ , ‘r’ is not the length of the rectangle but ‘r’ is the radius of the base of the cylinder.. that is why the standard formula , you get is 2pi r h
There are 2 points here, where you are utterly confused…
(1): In this particular case, you are making the cylinder by rotating the rectangle through 360 degree. (See the fig no 1)
(2): Second way of making the cylinder (the standard way) is just by folding the rectangle length wise or breadth wise , so that 2 edges join. (See the fig no 2)
In the (1)st case , Cylinder formed is larger & its base radius = r
& in the (2) nd case cylider formed is smaller & its base radius < ‘r’ lets assume it as ‘s’ & its circumference = r , i...
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Parimal Kumar Ghosh
Studied West Bengal BOARD of Secondary Education from RANESWAR UNION ADARSHA VIDYALAYA, PCM Biology · Author has 1K answers and 463.8K answer views
·3y
YOU have a rectangle whose 2D are r and h, in either way of treating them the length and breadth.
The AREA of the rectangle is r.h square Units. If you rotate the rectangle 360° a circular flat or plane Surface Area will be formed. There is NO QUESTION OF FORMING A CYLINDRICAL SHAPE OF THREE DIMENSIONAL CUBICLE FIGURE.
The surface area of the circle will be π.r^2. Now, the QUESTION is what the LENGTH OF THE RADIUS OF THIS CIRCLE! YOU need to fix a corner of the so-called rectangle as the CENTRE, and rotate by the DIAGONAL LENGTH OF THE RECTANGLE. To cover maximum Area the Diagonal i.e. the RADIU
Continue Reading
YOU have a rectangle whose 2D are r and h, in either way of treating them the length and breadth.
The AREA of the rectangle is r.h square Units. If you rotate the rectangle 360° a circular flat or plane Surface Area will be formed. There is NO QUESTION OF FORMING A CYLINDRICAL SHAPE OF THREE DIMENSIONAL CUBICLE FIGURE.
The surface area of the circle will be π.r^2. Now, the QUESTION is what the LENGTH OF THE RADIUS OF THIS CIRCLE! YOU need to fix a corner of the so-called rectangle as the CENTRE, and rotate by the DIAGONAL LENGTH OF THE RECTANGLE. To cover maximum Area the Diagonal i.e. the RADIUS can be used which works out to be …
✓(r^2 + h^2) Units.
Now evaluate π.r^2 replacing r by the above Radius. You will get the area of the desired circular figure in square units. Thanks!
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Pranav
Someone who will dig a bit deeper into any topic · Author has 468 answers and 2.8M answer views
·3y
If we take a rectangle with dimensions 'r' and 'h' and rotate it 360° then a cylinder will be formed, the area of which can be (rh) 2πr = 2πr²h I know this is wrong but still where I am doing a mistake?
The mistake you are making is area will have two dimensions - you are taking 3.
You took area as = r(1) x h(2) x 2πr(3) - hence 3 dimensions
You nee only 2 = h x 2πr = 2πrh.
if you want the additional area of the two circles in the bottom, then add 2 x πr² = 2πr²
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More answers below
How can we find the angle of rotation of a rectangle?
What is the starting area of a rectangle?
What two things form a cylinder shape?
How do I rotate a rectangle relative to its parent rectangle?
What is a 3-dimensional rectangle called?
Exam Question Answer
Answered by
Ravi Sharma
· Author has 15.1K answers and 3.8M answer views
·May 12, 2022
YOU HAVE NOT CLEARED WHETHE THE RECTANGLE IS BEING ROTATED AROUND r OR h. IN BOTH CASES ANSWER WILL BE DIFFERENT.
CURVED SURFACE AREA OF CYLINDER= 2πRH
TOTAL SURFACE AREA= 2πR²+2πRH= 2πR(R+H)
ROTATION AROUND r
RADIUS= h
HEIGHT= r
CURVED SURFACE= 2πrh
TOTAL SURFACE= 2πh(r+h)
ROTATION AROUND h
RADIUS= r
HEIGHT= h
CURVED SURFACE= 2πrh
TOTAL SURFACE= 2πr(h+r)
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Srinivasan Narasimhan
Retired from Govt.,service after serving for 34 years in varc · Author has 4K answers and 2.3M answer views
·3y
A rectangle can be formed only with a length and width. If consider the width of rectangle as height, then the length of the rectangle will be the circumference of the cylinder by formation of the rotation of the width side by 360 degrees with a central point. Therefore the area of recangle = L x W. L = 2 Pi. r where Pi=22/7 and ‘r’ is the radius of formed circle by rotation of the width side by 360 degrees abot the central point. and W = H. The surface area of cylinder = 2 x 22/7 x r x H and will be equal to L x W square units.
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Exam Question Answer
Answered by
Gary Russell
· Author has 6K answers and 3.1M answer views
·May 11, 2022
I assume that you want the surface area.
The area of the bottom is (pi)r^2. You have a top and a bottom. So, this gives you 2(pi)r^2.
The area of the side is 2(pi)rh.
Putting this together, you get
Surface Area = 2(pi)r^2 + 2(pi)rh = 2(pi)r(r+h)
Upvote ·
Muthusamy Piramanayagam ( முத்துசாமி பிரமநாயகம்)
M.Sc Physics, Diploma in science teaching. H.O.D in Science · Author has 11K answers and 7.8M answer views
·3y
It is enough to multiply by h . What is the reason to multiply by rh .
More over the axis of rotation must be specified .
curved surface area = circumference of the base ( circle for cylinder) . height of the cylinder.
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99 83
Bhalchandra Khare
Former Self Employed Chemical Engineer (1975–2016) · Author has 4.3K answers and 1.4M answer views
·3y
Your answer shows the volume generated by revolution around vertical edge of rectangle (dimension h).
For area the answer is 2pirh.
Upvote ·
Awnon Bhowmik
Studied at University of Dhaka · Author has 3.7K answers and 11.2M answer views
·7y
Related
A cylinder is generated by rotating a rectangle about one of its edges. The perimeter of the rectangle is 30 cm. To obtain a cylinder of maximum volume, what dimensions should the rectangle have?
A2A
Given, the perimeter…
2 x+2 y=30 x+y=15 y=15−x 2 x+2 y=30 x+y=15 y=15−x
Volume of the cylinder…
V=π r 2 h V(x)=π x 2(15−x)V(x)=π(15 x 2−x 3)V=π r 2 h V(x)=π x 2(15−x)V(x)=π(15 x 2−x 3)
Critical number(s):
V′(x)=0 π(30 x−3 x 2)=0 3 x(10−x)=0 x=0,10 But x>0 So x=10 V′(x)=0 π(30 x−3 x 2)=0 3 x(10−x)=0 x=0,10 But x>0 So x=10
This gives us
y=5 y=5
Evaluating the second derivative at x=10 x=10 will prove that this value actually maximizes the volume.
Continue Reading
A2A
Given, the perimeter…
2 x+2 y=30 x+y=15 y=15−x 2 x+2 y=30 x+y=15 y=15−x
Volume of the cylinder…
V=π r 2 h V(x)=π x 2(15−x)V(x)=π(15 x 2−x 3)V=π r 2 h V(x)=π x 2(15−x)V(x)=π(15 x 2−x 3)
Critical number(s):
V′(x)=0 π(30 x−3 x 2)=0 3 x(10−x)=0 x=0,10 But x>0 So x=10 V′(x)=0 π(30 x−3 x 2)=0 3 x(10−x)=0 x=0,10 But x>0 So x=10
This gives us
y=5 y=5
Evaluating the second derivative at x=10 x=10 will prove that this value actually maximizes the volume.
Upvote ·
99 20
9 2
Dave Benson
trying to make maths easy. · Author has 6.1K answers and 2.1M answer views
·Updated 3y
Related
The sides of the rectangle are 3 and 4. This rectangle rotates around its diagonal. What is the area and volume of the body created by rotation?
These are 2 cones of slant height 3 & 4 so rotate diagonals are 5 (3,4,5 ∆)
Area half rectangle = 34/2 = 6 = area of ∆ = 5r/2 ⟹ r = 12/5 = 2.4 & r is radius
H & h: Cone heights so volume = 𝝿r²/3(H+h) = 5𝝿(2.4)²/3 ≈ 30.16 units³ Answer
Surface area A = 𝝿r{√(H²+r²) +√(h²+r²) & r² = 5.76
H² = 16-5.76 = 10.24 and h² = 9–5.76 = 3.24 plug in r, H² and h²
A = 𝝿(2.4){√(10.24+5.76)+√(3.24+5.76)} = 𝝿(2.4){√(16)+√(9)}
A = 𝝿(2.4){4+3} = 𝝿(2.4){7} ≈ 52.78 units² Answer
Continue Reading
These are 2 cones of slant height 3 & 4 so rotate diagonals are 5 (3,4,5 ∆)
Area half rectangle = 34/2 = 6 = area of ∆ = 5r/2 ⟹ r = 12/5 = 2.4 & r is radius
H & h: Cone heights so volume = 𝝿r²/3(H+h) = 5𝝿(2.4)²/3 ≈ 30.16 units³ Answer
Surface area A = 𝝿r{√(H²+r²) +√(h²+r²) & r² = 5.76
H² = 16-5.76 = 10.24 and h² = 9–5.76 = 3.24 plug in r, H² and h²
A = 𝝿(2.4){√(10.24+5.76)+√(3.24+5.76)} = 𝝿(2.4){√(16)+√(9)}
A = 𝝿(2.4){4+3} = 𝝿(2.4){7} ≈ 52.78 units² Answer
Upvote ·
9 3
Adham Ghoname
High school senior from Future Language School
·7y
Related
A cylinder is generated by rotating a rectangle about one of its edges. The perimeter of the rectangle is 30 cm. To obtain a cylinder of maximum volume, what dimensions should the rectangle have?
Here's the answer using high school calculus.
Continue Reading
Here's the answer using high school calculus.
Upvote ·
9 3
9 4
Gary Ward
MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views
·Updated 3y
Related
The sides of the rectangle are 3 and 4. This rectangle rotates around its diagonal. What is the area and volume of the body created by rotation?
The sides of the rectangle are 3 and 4. This rectangle rotates around its diagonal. What is the area and volume of the body created by rotation?
By this rotation we get a base to base cone and truncated cone reflected over the z = 2.5 plane. We can just find the area and volume of the left half and double it.
Since it is a 3–4–5 triangle the equation of the lateral face of the whole cone is y = (4/3)x. The other is y = -(3/4)x + 15/4
(4/3)x = -(3/4)x + 15/4 times 12 → 16x = -9x + 45 ; x = 45/25 = 9/5 = 1.8
1.8 is the height. (4/3)(9/5) = 36/15 = 2.4 is the radius.
The intersection of the top of the
Continue Reading
The sides of the rectangle are 3 and 4. This rectangle rotates around its diagonal. What is the area and volume of the body created by rotation?
By this rotation we get a base to base cone and truncated cone reflected over the z = 2.5 plane. We can just find the area and volume of the left half and double it.
Since it is a 3–4–5 triangle the equation of the lateral face of the whole cone is y = (4/3)x. The other is y = -(3/4)x + 15/4
(4/3)x = -(3/4)x + 15/4 times 12 → 16x = -9x + 45 ; x = 45/25 = 9/5 = 1.8
1.8 is the height. (4/3)(9/5) = 36/15 = 2.4 is the radius.
The intersection of the top of the truncated cone is at x = 2.5 units where y = (3/4)x = (3/4)(5/2) = 15/8 = 1.875 units for the radius.
Slant height for the whole cone is 3, radius is 2.4
Slant area A = ½ · (2 · π · 2.4) · 3 = 7.2π ≈ 22.619 units² for whole cone.
Slant height for the part cone whole is 4 (since it is a rectangle side) and the radius is 2.4 the inner cone slant height is √(2.5²+1.875²) ≈ 3.125 and the radius is 1.875
Slant area A = ½ · 2 · π (2.4 · 4 - 1.875 · 3.125) ≈ 11.752 units² for the truncated cone.
Total area is 2 times the sum or Atotal = 68.742 units²
V = (1/3) · π · 2.4² · 1.8 ≈ 10.857 units^3 volume for full cone.
V = (1/3) · π (2.4² · 3.2 - 1.875² · 2.5) ≈ 10.098 units^3
Add them together and double it for the complete volume
Vtotal = 2(10.857 + 10.098) ≈ 41.91 units^3
2 π(∫1.8 0(4 3 x)2 d x+∫2.5 1.8(−3 4 x+15 4)2 d x)≈41.91 u n i t s 3 2 π(∫0 1.8(4 3 x)2 d x+∫1.8 2.5(−3 4 x+15 4)2 d x)≈41.91 u n i t s 3
2⋅2 π(5 3∫1.8 0 4 3 x d x+5 4∫2.5 1.8−3 4 x+15 4 d x)=68.742 u n i t s 2 2⋅2 π(5 3∫0 1.8 4 3 x d x+5 4∫1.8 2.5−3 4 x+15 4 d x)=68.742 u n i t s 2
I keep coming back to this one because it is interesting. You may wonder where the constants in front of the integrals come from. When finding the length of a curve you are stuck with
√1+(d y d x)2=√1+(4 3)2=√9 9+16 9=5 3 1+(d y d x)2=1+(4 3)2=9 9+16 9=5 3
With a constant slope, you will end up with a constant in the equation equal to the length of the unit slope over the run.
So the length of the slope is 5 and the run is 3 giving 5/3 for the constant.
Upvote ·
9 9
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7252 | https://www.cuemath.com/numbers/conjugates-and-rationalization/ | Conjugate in Math
The term conjugate means a pair of things joined together. For example, the two smileys: smiley and sad are exactly the same except for one pair of features that are actually opposite of each other. If you look at these smileys, you will notice that they are the same except that they have opposite facial expressions: one has a smile and the other has a frown. Similarly, conjugate in math refers to either the conjugate of a surd or the conjugate of a complex number where there is just a sign change in the number with respect to a few conditions.
Let us see what is conjugate in math in different cases and what is the use of finding it.
| | |
--- |
| 1. | What is Conjugate in Math? |
| 2. | Conjugate of a Surd |
| 3. | Conjugate of a Complex Number |
| 4. | Conjugate and Rational Factor |
| 5. | Conjugates and Rationalization |
| 6. | FAQs on Conjugate in Math |
What is Conjugate in Math?
The conjugate in math is formed by changing the sign between two terms in a binomial with respect to the condition that the sum and product of the binomial and its conjugate are rational. Here, the binomial can be either a surd or a complex number. Observe the following binomials and their conjugates.
| | Binomial | Conjugate | Sum and Product are Rational Numbers |
--- --- |
| Surd | 1 + √3 | 1 - √3 | Sum = (1 + √3) + (1 - √3) = 2 Product = (1 + √3) (1 - √3) = 1 - 3 = -2 |
| Complex Number | 2 + i | 2 - i | Sum = (2 + i) + (2 - i) = 4 Product = (2 + i) (2 - i) = 22 - i2 = 4 - (-1) = 5 |
We study two types of conjugates in math.
Let us study each of these in detail in the upcoming sections.
Conjugate of a Surd
The conjugate of a surd x + y√z is always x - y√z and vice versa. This is because the sum is (x + y√z) + (x - y√z) = 2x, and the product = (x + y√z) (x - y√z) = x2 - (y√z)2 = x2 - y2z (by the formula of a2 - b2) are rational numbers. For example, for the surd 3 + √2, the conjugate surd is 3 - √2, this is because:
Here, both 6 and 7 are rational numbers. Note that to find the conjugate of a surd, just change the sign of the surd (irrational part), but it is not changing the middle sign always. For example, the conjugate of the surd √2 + 3 is -√2 + 3, but NOT √2 - 3. Because, if we assume that √2 - 3 is the conjugate of √2 + 3, then their sum is 2√2, which is NOT a rational number. Here are some more examples of conjugates of surds in math.
| Surd | Conjugate |
--- |
| 2√5 + 3 (write it as 3 + 2√5) | 3 - 2√5 |
| -√7 - 3 (write it as -3 - √7) | -3 + √7 |
| 3 - √2 | 3 + √2 |
The easiest way of writing conjugate surd is to find the conjugate is just to write the given number in the order of rational part first and irrational part next and then change the middle sign.
Conjugate of a Complex Number
The complex conjugate of a complex number z = x + iy is x - iy (and vice versa) and it is represented by (\bar{z}) as their sum (2x) and the product x2 + y2 both are rational numbers. To write the complex conjugate,
Here are some examples of conjugates of complex numbers.
| Complex Number | Conjugate |
--- |
| 2 - i | 2 + i |
| 3i + 5 (write it as 5 + 3i) | 5 - 3i |
| (-1/2) + 5i | (-1/2) - 5i |
Conjugate and Rational Factor
If the product of two surds is a rational number, then each one of them is called the rational factor of the other. For example, the rational factors of 2 + √3 are each of 2 - √3 and -2 + √3. This is because by multiplying 2 + √3 with each of their conjugates result in a rational number as shown below.
Note that -2 + √3 is NOT a conjugate of 2 + √3, its only a rational factor.
Sometimes, rational factor and conjugate of a number are referred to as the same but there is one minor difference between them. The sum of a binomial and its rational factor does NOT need to be a rational number, but the sum of a binomial and its conjugate SHOULD be a rational number. On the other hand, the product of a binomial with each of its rational factor and conjugate should be a rational number.
Conjugates and Rationalization
Rationalization of the denominator is the process of multiplying a fraction (with an irrational or complex denominator) by its rational factor or conjugate to make the denominator a rational number. This is because it is very convenient to have rational numbers in the denominators to compare or make some calculations in math. Here are two examples to understand how to rationalize the denominators by using conjugates.
Rationalization of Surds Example: Rationalize the denominator of 1 / (3 + √2).
Solution: The conjugate of 3 + √2 is 3 - √2. Multiplying and dividing the given fraction by 3 - √2,
1 / (3 + √2) × (3 - √2) / (3 - √2)
= (3 - √2) / (32 - (√2)2)
= (3 - √2) / (9 - 4)
= (3 - √2) / 5
Rationalization of Complex Numbers Example: Rationalize the denominator of 1 / (1 + i).
Solution: The complex conjugate of 1 + i is 1 - i.
1 / (1 + i) × (1 - i) / (1 - i)
= (1 - i) / (12 - i2)
= (1 - i) / (1 + 1) (Because by powers of iota, i2 = -1)
= (1 - i) / 2
Important Notes on Conjugate in Math:
☛ Related Topics:
Examples on Conjugate in Math
Example 1: If x = 2 + √3, then find 1/x.
Solution:
1/x = 1 / (2 + √3). The conjugate of 2 + √3 is 2 - √3. Multiplying and dividing 1/x by the conjugate,
1/x = (2 - √3) / [ (2 + √3) (2 - √3)]
= (2 - √3) / (4 - 3)
= (2 - √3) / 1
= 2 - √3
Answer: 2 - √3.
Example 2: For the value of x given in Example 1, compute the value of x2 + 1/x2.
Solution:
x2 = (2 + √3)2
= 2 + 2(2) √3 + (√3)2 (Using (a+b)^2 Formula)
= 2 + 4√3 + 3
= 5 + 4√3
1/x2 = (1/x)2 = (2 - √3)2
= 2 - 2(2) √3 + (√3)2 (Using (a-b)^2 Formula)
= 2 - 4√3 + 3
= 5 - 4√3
Now, x2 + 1/x2 = (5 + 4√3) + (5 - 4√3) = 10.
Answer:10.
Example 3: Find the values of a and b if (3 + √7) / (3 - √7) = a + b√7.
Solution:
It is given that (3 + √7) / (3 - √7) = a + b√7.
Here the conjugate of the denominator 3 - √7 is 3 + √7. So, we get
= (3 + √7) / (3 - √7) × (3 + √7) / (3 + √7)
= (9 + 6√7 + 7) / (9 - 7)
= (16 + 6√7) / 2
= 8 + 3√7
= a + b√7
Comparing the rational and irrational parts, a = 8 and b = 3.
Answer: a = 8 and b = 3.
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Practice Questions on Conjugate in Math
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FAQs on Conjugate in Math
What are the Math Conjugates?
The math conjugate of a number is a number that when multiplied or added to the given number results in a rational number. For example,
Is Finding Conjugate Means Changing the Middle Sign Always?
No, finding the conjugate does NOT mean changing the middle sign all the time. For example, the conjugate of √5 - 1 is NOT √5 + 1 as their sum is NOT a rational number. After finding the conjugate of a number, check whether both the sum and product of the number and its conjugate are rational or not always.
How to Find Conjugate in Math?
To find the conjugate of a number,
What is an Example of Conjugate?
The sum and product of a number and its conjugate are always rational. For example, the complex conjugate of the complex number 3 - 5i is 3 + 5i as (3 - 5i) + (3 + 5i) = 6 and (3 - 5i) (3 + 5i) = 9 + 25 = 34. Here, both 6 and 34 are rational numbers.
Why it is Called Conjugate in Math?
"Conjugate" means two things that are joined together (or) two things that have common features except for one difference. The conjugate of a number is a mathematical value having a reciprocal relation with the given number.
Does Conjugate Need to Have a Minus Symbol Always?
No, the conjugate doesn't need to have the middle sign to be minus always. For example, the conjugate of 4 - √3 is 4 + √3 and it doesn't have a minus symbol.
Is Rational Factor Same as Conjugate?
Sometimes both are considered as the same but there is one difference. For any number, |
7253 | https://www.scribd.com/document/487032615/Asymptote-Guide | Asymptote Geometry Guide | PDF | Euclidean Geometry | Euclidean Plane Geometry
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HomeRestriction EndonucleasesSalI-HF®
SalI-HF®
SalI-HFhas been reformulated with Recombinant Albumin (rAlbumin) beginning with Lot #10223617.Learn more.
We are excited to announce that all reaction buffers are now BSA-free. NEB began switching our BSA-containing reaction buffers in April 2021 to buffers containing Recombinant Albumin (rAlbumin) for restriction enzymes and some DNA modifying enzymes. Find more details at www.neb.com/BSA-free.
Isoschizomers| Single Letter Code| Pronunciation:
High-Fidelity (HF®) restriction enzymes have the same specificity as native enzymes, but have been engineered for significantly reduced star activity and performance in a single buffer (rCutSmart™ Buffer). All HF-restriction enzymes come with Gel Loading Dye, Purple (6X). Enjoy the enhanced performance and added value of our engineered enzymes at the same price as the native enzyme:
Engineered for improved performance
100% activity in rCutSmart Buffer
Time-Saver™ qualified for digestion in 5-15 minutes
Reduced star activity
Supplied with 1 vial of Gel Loading Dye, Purple (6X)
Restriction Enzyme Cut Site: G/TCGAC
Reagents Supplied Protocols
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Reduce Star Activity with High-Fidelity Restriction Enzymes
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| Catalog # | Concentration | Size | List Price | Your Price | Quantity |
--- --- --- |
| R3138S | 20,000 units/ml | 2,000 units | $72.00 | Sign InSign In | -+ |
| R3138T | 100,000 units/ml | 2,000 units | $72.00 | Sign InSign In | -+ |
| R3138L | 20,000 units/ml | 10,000 units | $296.00 | Sign InSign In | -+ |
| R3138M | 100,000 units/ml | 10,000 units | $296.00 | Sign InSign In | -+ |
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Product InformationProtocols, Manuals & UsageTools & ResourcesFAQs & TroubleshootingCitations & Technical LiteratureQuality, Safety & LegalTOP
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Product Information
High Fidelity (HF) Restriction Enzymes have 100% activity in rCutSmart Buffer; single-buffer simplicity means more straightforward and streamlined sample processing. HF enzymes also exhibit dramatically reduced star activity. HF enzymes are all Time-Saver qualified and can therefore cut substrate DNA in 5-15 with the flexibility to digest overnight without degradation to DNA. Engineered with performance in mind, HF restriction enzymes are fully active under a broader range of conditions, minimizing off-target products, while offering flexibility in experimental design.
NEB extensively performs quality controls on all standard and high-fidelity (HF) restriction enzymes. Examples of nuclease contamination studies for some of our HF restriction enzymes are shown below.
Restriction Enzyme Competitor Study: Nuclease Contamination
EcoRI, NotI, and BamHI from multiple suppliers were tested in reactions containing a fluorescent labeled single stranded, double stranded blunt, 3’overhang or 5’ overhang containing oligonucleotides. The percent degradation is determined by capillary electrophoresis and peak analysis. The resolution is at the single nucleotide level.
Product Source
An E. coli strain that carries the cloned and modified SalI gene from Streptomyces albus G (ATCC 49789) This product is related to the following categories:Restriction Endonucleases S,High-Fidelity (HF®) Restriction Endonucleases,Time-Saver Qualified Restriction EnzymesThis product can be used in the following applications: Restriction Enzyme Digestion
### Reagents Supplied
The following reagents are supplied with this product:
| NEB # | Component Name | Component # | Stored at (°C) | Amount | Concentration |
--- --- --- |
| R3138S-20SalI-HF®R3138SVIAL-20 1 x 0.1 ml 20,000 units/ml rCutSmart™ BufferB6004SVIAL-20 1 x 1.25 ml 10 X Gel Loading Dye, Purple (6X)B7024AVIAL 25 1 x 0.5 ml 6 X |
| R3138T-20SalI-HF®R3138TVIAL-20 1 x 0.02 ml 100,000 units/ml rCutSmart™ BufferB6004SVIAL-20 1 x 1.25 ml 10 X Gel Loading Dye, Purple (6X)B7024AVIAL 25 1 x 0.5 ml 6 X |
| R3138L-20SalI-HF®R3138LVIAL-20 1 x 0.5 ml 20,000 units/ml rCutSmart™ BufferB6004SVIAL-20 1 x 1.25 ml 10 X Gel Loading Dye, Purple (6X)B7024AVIAL 25 1 x 0.5 ml 6 X |
| R3138M-20SalI-HF®R3138MVIAL-20 1 x 0.1 ml 100,000 units/ml rCutSmart™ BufferB6004SVIAL-20 1 x 1.25 ml 10 X Gel Loading Dye, Purple (6X)B7024AVIAL 25 1 x 0.5 ml 6 X |
### Properties & Usage
Unit Definition
One unit is defined as the amount of enzyme required to digest 1 μg of λ DNA (HindIII digest) in 1 hour at 37°C in a total reaction volume of 50 μl.
Reaction Conditions
1X rCutSmart™ Buffer
Incubate at 37°C
1X rCutSmart™ Buffer
50 mM Potassium Acetate
20 mM Tris-acetate
10 mM Magnesium Acetate
100 µg/ml Recombinant Albumin
(pH 7.9 @ 25°C)
Activity in NEBuffers
NEBuffer™ r1.1: 10%
NEBuffer™ r2.1: 100%
NEBuffer™ r3.1: 100%
rCutSmart™ Buffer: 100%
Diluent Compatibility
Diluent A
Storage Buffer
10 mM Tris-HCl
50 mM KCl
1 mM DTT
0.1 mM EDTA
50% Glycerol
300 µg/ml Recombinant Albumin
pH 7.5 @ 25°C
Heat Inactivation
65°C for 20 minutes
Methylation Sensitivity
dam methylation: Not Sensitive
dcm methylation: Not Sensitive
CpG Methylation: Blocked
Isoschizomers
SalI
### Related Products
Companion Products
Nuclease-free Water
Materials Sold Separately
rCutSmart™Buffer
Gel Loading Dye, Purple (6X)
### Product Notes
1. Blocked by CpG methylation.
Protocols, Manuals & Usage
### Protocols
1. Optimizing Restriction Endonuclease Reactions
2. Restriction Digest Protocol
3. Double Digest Protocol with Standard Restriction Enzymes
### Usage Guidelines
Activity at 37°C for Restriction Enzymes with Alternate Incubation Temperatures
Activity of Restriction Enzymes in PCR Buffers
Cleavage Close to the End of DNA Fragments
Digestion of Agarose-Embedded DNA: Info for Specific Enzymes
Double Digests
Effects of CpG Methylation on Restriction Enzyme Cleavage
Heat Inactivation
NEBuffer Activity/Performance Chart with Restriction Enzymes
Optimizing Restriction Endonuclease Reactions
Restriction Endonucleases - Survival in a Reaction
Restriction Enzyme Diluent Buffer Compatibility
Restriction Enzyme Tips
Single Letter Codes
Star Activity
Traditional Cloning Quick Guide
Tools & Resources
### Selection Charts
Alphabetized List of Recognition Sequences
Cleavage of Supercoiled DNA
Compatible Cohesive Ends and Generation of New Restriction Sites
Dam-Dcm and CpG Methylation
Frequencies of Restriction Sites
Isoelectric Points (pI) for Restriction Enzymes
Isoschizomers
NEB Diluent and Buffer Table
Recleavable Filled-in 5' Overhangs
Time-Saver™ Qualified Enzymes
Why Choose Recombinant Enzymes?
### Web Tools
Competitor Cross-Reference Tool
DNA Sequences and Maps Tool
Double Digest Finder
Enzyme Finder
NEBcutter™ v3.0
NEBioCalculator®
REBASE®
FAQs & Troubleshooting
### FAQs
1. Is there any difference in the methylation sensitivity between SalI-HF and SalI?
2. How does the level of star activity of SalI-HF compare to SalI?
3. What is the difference between SalI-HF and SalI?
4. Why does the HF version of the enzyme have a different recommended buffer than the wild type enzyme?
5. When should I choose the HF version of an enzyme?
6. When is star activity a concern?
7. When should I choose the High Fidelity (HF®) version of the enzyme?
8. Can the change in buffer preference of the HF enzyme be advantageous?
9. Will the HF enzyme produce elevated star activity when used in a buffer other than the one recommended?
10. How is the improvement in fidelity of HF restriction endonucleases quantitated?
11. What is the Fidelity Index (FI)?
12. What does HF® refer to following the name of a restriction enzyme?
13. Do I have to set-up digests with Time-Saver™ qualified enzymes for 5-15 minutes? Can I digest longer?
14. Can Gel Loading Dye, Purple 6X (B7024) be stored in cold temperatures?
15. What does it mean to be Time-Saver™ qualified?
16. Can you tell me more about the switch from BSA to Recombinant Albumin (rAlbumin) in NEBuffers?
### Troubleshooting
Restriction Enzyme Troubleshooting Guide
Citations & Technical Literature
### Citations #### Product Citation Tool
Powered by Bioz © 2025See more details on Bioz
Quality, Safety & Legal
### Quality Control Assays Quality Control tests are performed on each new lot of NEB product to meet the specifications designated for it. Specifications and individual lot data from the tests that are performed for this particular product can be found and downloaded on the Product Specification Sheet, Certificate of Analysis, data card or product manual. Further information regarding NEB product quality can be found here.
### Specifications & Change Notifications
SalI-HFhas been reformulated with Recombinant Albumin (rAlbumin) beginning with Lot #10223617. All subsequent (higher number) lots will contain rAlbumin. For additional information, please visit atNEB Restriction Enzyme formulations with Recombinant Albumin (rAlbumin). The Specification sheet is a document that includes the storage temperature, shelf life and the specifications designated for the product. The following file naming structure is used to name these document files: [Product Number][Size][Version]
R3138S_L_v2
R3138T_M_v3
R3138S_L_E_v3
### Certificate of Analysis
The Certificate of Analysis (COA) is a signed document that includes the storage temperature, expiration date and quality controls for an individual lot. The following file naming structure is used to name these document files: [Product Number][Size][Version]_[Lot Number]
R3138S_L_v1_0041301
R3138S_L_v1_0041302
R3138S_L_v1_0051305
R3138S_L_v1_0051306
R3138S_L_v1_0051312
R3138S_L_v1_0061405
R3138S_L_v1_0061406
R3138T_M_v1_0041302
R3138T_M_v1_0041306
R3138T_M_v1_0051312
R3138T_M_v1_0061405
R3138S_L_v1_0061410
R3138T_M_v1_0061410
R3138T_M_v1_0071504
R3138S_L_v1_0071504
R3138T_M_v1_0071509
R3138S_L_v1_0071509
R3138T_M_v1_0071603
R3138T_M_v1_0071604
R3138S_L_v1_0071603
R3138S_L_v1_0081608
R3138T_M_v1_0071610
R3138S_L_v1_0081701
R3138S_L_v1_0101705
R3138S_L_v1_0141709
R3138T_M_v1_0141707
R3138T_M_v1_0141801
R3138S_L_v1_0141802
R3138S_v1_10016069
R3138T_v1_10016072
R3138M_v1_10016070
R3138S_v1_10009919
R3138S_v1_10021231
R3138L_v1_10021229
R3138S_v1_10027079
R3138L_v1_10030758
R3138M_v1_10030760
R3138S_v1_10030762
R3138S_v1_10034666
R3138S_v1_10038177
R3138T_v1_10040926
R3138S_v1_10042668
R3138M_v1_10040928
R3138L_v1_10044874
R3138S_v1_10046692
R3138S_v1_10050302
R3138S_v1_10054594
R3138S_v1_10057009
R3138T_v1_10057316
R3138S_v1_10059860
R3138S_v1_10059863
R3138M_v1_10060455
R3138L_v1_10063476
R3138S_v1_10063191
R3138S_v1_10067334
R3138T_v1_10073650
R3138S_v1_10072224
R3138L_v1_10076275
R3138S_v1_10078082
R3138L_v1_10078777
R3138M_v1_10076529
R3138S_v1_10079962
R3138L_v1_10080431
R3138M_v1_10082613
R3138S_v1_10083526
R3138T_v1_10082614
R3138L_v1_10090252
R3138M_v1_10090254
R3138S_v1_10090255
R3138T_v1_10090256
R3138S_v1_10087054
R3138L_v1_10091956
R3138S_v1_10091882
R3138M_v1_10095195
R3138S_v1_10095194
R3138L_v1_10095198
R3138T_v1_10095196
R3138S_v1_10101358
R3138S_v1_10106993
R3138S_v1_10110974
R3138M_v2_10109499
R3138L_v2_10116248
R3138S_v2_10117682
R3138S_v2_10126550
R3138T_v2_10126510
R3138L_v2_10129289
R3138S_v2_10133248
R3138S_v2_10139670
R3138M_v2_10140975
R3138L_v2_10139673
R3138S_v2_10143732
R3138S_v2_10146921
R3138T_v2_10150025
R3138S_v2_10148748
R3138M_v2_10152978
R3138T_v2_10157692
R3138L_v2_10159235
R3138S_v2_10159236
R3138S_v2_10165638
R3138S_v2_10171946
R3138M_v2_10175026
R3138T_v2_10175028
R3138S_v2_10178134
R3138L_v2_10183140
R3138S_v2_10183139
R3138L_v2_10193701
R3138S_v2_10193702
R3138L_v2_10204953
R3138S_v2_10204957
R3138M_v2_10207698
R3138T_v2_10207697
R3138S_v2_10212397
R3138S_v2_10221482
R3138L_v3_10223624
R3138M_v3_10223622
R3138S_v3_10223618
R3138T_v3_10223621
R3138S_v3_10233788
R3138S_v3_10237442
R3138S_v3_10241473
R3138S_v3_10244399
R3138M_v3_10250725
R3138T_v3_10250723
R3138S_v3_10251170
R3138S_v3_10256974
R3138L_v3_10264787
R3138S_v3_10264804
R3138T_v3_10273754
R3138S_v3_10274787
R3138M_v3_10279753
R3138L_v3_10282149
R3138M_v3_10287274
R3138S_v3_10289544
R3138S_v3_10298070
R3138T_v3_10300918
R3138S_v3_10305327
R3138M_v3_10305338
### Safety Data Sheets
The following is a list of Safety Data Sheet (SDS) that apply to this product to help you use it safely.
SalI-HF®Choose format Download
rCutSmart™ BufferChoose format Download
Gel Loading Dye, Purple (6X)Choose format Download
### Legal and Disclaimers Products and content are covered by one or more patents, trademarks and/or copyrights owned or controlled by New England Biolabs, Inc (NEB). The use of trademark symbols does not necessarily indicate that the name is trademarked in the country where it is being read; it indicates where the content was originally developed. The use of this product may require the buyer to obtain additional third-party intellectual property rights for certain applications. For more information, please email busdev@neb.com.
This product is intended for research purposes only. This product is not intended to be used for therapeutic or diagnostic purposes in humans or animals.
New England Biolabs (NEB) is committed to practicing ethical science – we believe it is our job as researchers to ask the important questions that when answered will help preserve our quality of life and the world that we live in. However, this research should always be done in safe and ethical manner. Learn more.
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7255 | https://wisc.pb.unizin.org/minimisgenchem/chapter/kmt-and-temperature-m5q5/ | Skip to content
24 Gas Behavior, Kinetic Molecular Theory, and Temperature (M5Q5)
Introduction
Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships. This section explores the behavior of ideal gases along with the kinetic molecular theory (KMT) of gases. The kinetic molecular theory will explain these relationships as well as the behavior of ideal gases at the sub-microscopic level. This section includes worked problems, examples, and a glossary.
Learning Objectives for Gas Behavior, Kinetic Molecular Theory, and Temperature
Explain the tenets of Kinetic Molecular Theory and the microscopic relationship between temperature and kinetic energy.
| Kinetic Molecular Theory of Gases | Molecular Velocities and Kinetic Energy | Diffusion of Gases | Kinetic Molecular Theory Practice Problems | Diffusion Practice Problems |
| Key Concepts and Summary | Key Equations | Glossary | End of Section Exercises |
Kinetic Molecular Theory of Gases
Kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous section of this module. This theory is based on the following postulates, which you have seen before in our definition of an ideal gas:
Gases consist of tiny particles that move at different speeds and in every possible direction; they are in constant motion.
The size of gas particles is much smaller than the distances between them. Most of the volume of a gas is empty space.
There is no force of attraction or repulsion between gas particles. This is because the molecules are so far apart from each other.
They undergo billiard-ball-like collisions (elastic collisions) with the walls of the container and with each other.
The various gas laws can be derived from the assumptions of the KMT. We will first look at the individual gas laws conceptually to see how KMT explains them. Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows and in Figure 1:
Figure 1a: Relationship between Pressure and Temperature (Amonton or Gay-Lussac’s Law, P ∝ T; when V, n are constant)
If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure.
Relationship between Volume and Temperature (Charles’s Law, V ∝ T; when P, n are constant)
If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls. One example is a cylinder like that shown in Figure 1 with a piston that moves to increase the volume. These conditions will decrease the frequency of molecule-wall collisions.
Figure 1b: Relationship between Volume and Pressure (Boyle’s Law, P ∝ V; when T, n are constant)If the gas volume is decreased, the distance to the container walls decreases, and the frequency of collisions with the wall increases, thereby increasing the pressure.
Figure 1c: Relation between Volume and Mole Number (Avogadro’s Law, V∝ n; when T, P are constant)
At the atomic scale, each particle gets a certain volume at fixed pressure and temperature. The more molecules, the more volume.
Relation between Pressure and Number of Different Molecules (Dalton’s Law, Ptotal = P1 + P2 + P3 …)
Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.
Molecular Velocities and Kinetic Energy
When we insert a thermometer into a gas sample, it will read a single temperature. In reality, though, this is a measure of the average kinetic energy of the entire sample of gas molecules. Some particles will have a kinetic energy (and therefore speed, which we will discuss later) that is close to this average value. However, most particles will have a kinetic energy significantly greater or significantly less than this average value. The energy of particle motion is called kinetic energy (KE). The value of this energy averaged over all of the gas molecules in the sample is labeled as KE. For any gas, the temperature (T) and amount of gas (n, moles) determines the sample’s kinetic energy: KE
KE = RT
where R = 8.314
Notice that the identity (or molar mass) of the gas is not part of this relationship. Thus, the same amount of any ideal gas at the same temperature will have the same KE. We’ll examine later in this section how the distribution of the particle speeds, however, varies depending on the identity of the gas. At the same temperature, heavier gas will move more slowly (on average) than lighter gases. For simplicity, we’ll start by examining the distribution of particle speeds for one gas at one temperature.
Note that the three dimensional nature of the speed distribution is taken into account in using KE = RT. Particles in three-dimensions have speeds in the x, y, and z directions, thus the 3 × ½.
In a gas sample, individual molecules have widely varying speeds. However, because of the large number of molecules and large number of collisions involved, the molecular speed distribution and average speed are constant at a given temperature for that gas sample. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative number of molecules in a bulk sample of gas that possesses a given speed (Figure 2).
Figure 2. The molecular speed distribution for oxygen gas at 298 K is shown here. Very few molecules move at either very low or very high speeds. The relative number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, vp, is 394 m/s, while the root mean square speed, vrms, is closer to 500 m/s.
Let’s examine the graph in more detail. The speed of an individual gas particle is on the x-axis. Slow particles are on the left; fast particles are on the right. The relative number of particles possessing a given speed is on the y-axis. Any given point on the line is a measure of the quantity of particles possessing that speed. The most probably velocity (vp) is the speed at the peak of the curve; this is the speed of the largest number of particles. The Maxwell-Boltzmann distribution shows that in a collection of gas particles, a wide range of speeds exist. Some particles will have a speed that is close to this average value. However, most particles will have speeds significantly greater and significantly less than this average value. Also, notice that the curve is not symmetrical. There is a longer tail at the higher speeds than at the lower speeds. Particles cannot have speed less than zero, but a small number of particles can have extremely high speeds.
Another characteristic speed of molecules in Figure 2 is the root mean square velocity of a particle, vrms. It is easy to calculate from our relationship between kinetic energy and temperature from above, which we will now derive.
The kinetic energy (KE) of a single particle with mass (m, in kg) and speed (v, in m/s) is given by:
KE = ½mv2
Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m2 s–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square velocity of a particle is defined as the square root of the average of the squares of the velocities and n, the number of particles (in moles):
vrms = =
The average kinetic energy, KE, of 1 mole of particles with molar mass (Μ, in kg/mol) and average speed (, in m/s) is given by:
KE = Μ = RT
Using these equations and solving for gives:
Μ = RT
Taking square root of gives:
= =
where R is the gas constant, T is the kelvin temperature, and M is the mass in kg/mol. When used in this equation, the appropriate form of the gas constant is 8.314 (8.314 ).
Now you can see where “root mean square” comes from: it is the square root of the average (mean) value of the square of the velocity. It is a useful quantity because it is so easy to calculate and because it sits about in the middle of the Maxwell-Boltzmann distribution. You should know the relation between KE and T and the expression for vrms without having to look at your notes.
Demonstration: Speed of Br2(g) in air vs. a vacuum
Set up. In this demonstration, we will observe the speed of Br2(g) diffusing. To begin, Br2(l) is contained in a capsule. In the first part of the experiment, the capsule will be broken open in a tube that contains air, the liquid will evaporate, and the brown Br2(g) will begin to move. In the second part of the experiment, the Br2(l) capsule will be broken in a tube containing no air (i.e., a vacuum).
Prediction. Before watching the demonstration, predict whether the bromine will diffuse faster in the tube that contains air or the tube that contains no air.
Explanation. This demonstration shows that the diffusion of Br2(g) in air is significantly slower than the spreading out of Br2(g) into a vacuum. As shown above, the speed of a gas depends only on the temperature and molecular weight. In this demonstration, the temperature is constant and the molecular weight is constant (since both scenarios use Br2). Furthermore, this cannot be explained simply by the Maxwell-Boltzman distribution of particle speeds, because this distribution is present in both samples. The only difference between the scenarios is the presence of the air molecules in the first part. While the air molecules do not change the speed of the Br2 gas, they do introduce an obstacle. Br2 undergoes many collisions with the surrounding air molecules and therefore it is not able to make as much progress even if the molecules are moving at the same speed as in the second part. Imagine running through a crowded hallway vs. running through a completely empty hallway. Even if you are moving at the same speed in both scenarios, you will have to navigate many more collisions and obstacles in the crowded hallway and it will take you longer to reach the other end.
Example 1
Calculation of vrms
Calculate the root-mean-square velocity for a nitrogen molecule at 30 °C.
Solution
Convert the temperature into Kelvin:
30 °C + 273 = 303 K
Determine the mass of a nitrogen molecule in kilograms:
× = 0.028 kg/mol
Replace the variables and constants in the root-mean-square velocity equation, replacing Joules with the equivalent kg m2s–2:
=
= = 520 m/s
Try converting your answer into miles per hour and see what you get. Molecules move at high speeds!
Check Your Learning
Calculate the root-mean-square velocity for an oxygen molecule at –23 °C.
Answer:
441 m/s
Next, we will consider the distribution of gas speeds in two types of scenarios: either one type of gas at different temperatures or different gases at the same temperature. If the temperature of a gas increases, its KE increases, and more molecules have higher speeds and fewer molecules have lower speeds. Thus, the distribution shifts toward higher speeds overall; that is, to the right. If temperature decreases, KE decreases, and more molecules have lower speeds and fewer molecules have higher speeds. Thus, the distribution shifts toward lower speeds overall; that is, to the left. Notice also that the curve flattens and spreads out as temperature increases. This behavior is illustrated for nitrogen gas in Figure 3.
Figure 3. The molecular speed distribution for nitrogen gas (N2) shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.
At a given temperature, all molecules have the same average kinetic energy. The relation = teaches us that lighter molecules will have higher speeds on average and heavier molecules will have lower speeds on average. In fact, the entire speed distribution shifts to higher or lower speeds just as you expect, as shown in Figure 4. Again, notice also that curve flattens and spreads out as the molar mass of the gas decreases.
Figure 4. Molecular velocity is inversely related to molecular mass. As shown above at 298 K, lighter molecules move faster on average than heavier molecules.
Demonstration: Molecular velocity is inversely related to molecular mass
Set up. In the following demonstration, a long glass tube is set up with bulbs at both ends. The bulb on the left side contains HCl(aq) and the bulb on the right side contains NH4OH(aq), the source of NH3(g). Over time, HCl(g) and NH3(g) evaporate from their respective solutions and when the two gases meet in the tube, they react. Before watching the video, write the balanced chemical reaction that you expect to observe.
Prediction. Based on what you learned above, make a prediction about whether the reaction will take place in the middle of the tube, more to the left, or more to the right.
Explanation. The reaction observed in this demonstration is:
HCl(g) + NH3(g) → NH4Cl(s)
As the reaction occurs, we see the formation of a white solid, the NH4Cl salt. We first see the appearance of the white solid to the left of center. As time progresses, the white starts drifting to the right. To explain this observation, consider the molecular weight of the two reactants. The molecular weight of the HCl is 36.5 g/mol and the molecular weight of NH3 is 17.0 g/mol. The NH3 traveled at a higher velocity down the tube than the HCl. Therefore, at the time that the gases met, the NH3 had traveled further than the HCl and thus began reacting to the left of center. As time continued to progress, the HCl continued to travel down the tube and we see the white product form to the right of the initial meeting spot.
Example 2
What is the kinetic energy of one mole of argon atoms at 25 °C?
Solution
Convert the temperature into Kelvin:
25 °C + 273 = 298 K
Find kinetic energy:
KE = RT
KE = × 8.314 × 1 mol × 298 K
= 3.7 kJ
Remember this number! It will become important when we compare it to the strengths of chemical bonds.
Check Your Learning
What is the kinetic energy of one mole of N2at 25 °C? or of CO2? Or H2O?
Answer:
3.7 kJ
Even though the speed distribution of gases varies with the identity of the gas, the kinetic energy of one mole of any gas at 25 °C is the same!
The gas simulator may be used to examine the effect of temperature on molecular velocities. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures.
Diffusion of Gases
In the previous section, we discussed the motions of molecules in a gas. Here’s a brief review before we consider rates of diffusion in light of the root mean square velocity of gases.
Gaseous molecules travel at tremendous speeds (hundreds of meters per second). In general, we know that when a gas sample is introduced to one part of a closed container, its molecules very quickly disperse throughout the container. This process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 5). In a closed environment, diffusion will ultimately result in equal concentrations of gas molecules throughout, as depicted in Figure 5. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).
Figure 5. (a) Two gases, H2 and O2, are initially separated. (b) When the stopcock is opened, they mix together. The lighter gas, H2, passes through the opening faster than O2, so just after the stopcock is opened, more H2 molecules move to the O2 side than O2 molecules move to the H2 side. (c) After a short time, both the slower-moving O2 molecules and the faster-moving H2 molecules have distributed themselves evenly on both sides of the vessel.
In 1832, Thomas Graham studied the rates of diffusion of different gases and formulated Graham’s law: The rate of diffusion of a gas is inversely proportional to the square root of the mass of its particles.
rate of diffusion ∝
Another way to state Graham’s law is in terms of the root mean square velocity, vrms:
diffusion rate ∝ vrms
From earlier in this section,
vrms =
= = =
This relationship assumes that two gases—A and B—are at the same temperature and pressure. The following examples illustrate how to apply Graham’s law and root mean square velocity to solve interesting problems. For example, if we know the identities of gas A and B, then we can determine the ratio of their diffusion rates. Similarly, if we know the ratio of their diffusion rates and the identity of gas A, then we can calculate the molar mass of gas B to identify an unknown gas.
Example 3
Applying Graham’s law to Rates of Diffusion
Calculate the ratio of the rate of diffusion of hydrogen to the rate of diffusion of oxygen.
Solution
Using molar masses:
= = =
Hydrogen diffuses four times as rapidly as oxygen.
Check Your Learning
At a particular pressure and temperature, nitrogen gas diffuses at the rate of 79 mL/s. Using the same apparatus at the same temperature and pressure, at what rate will sulfur dioxide diffuse (in mL/s)?
Answer:
52 mL/s
Here is one more example showing how to calculate molar mass from diffusion rate data.
Example 4
Determining Molar Mass Using Graham’s Law
An unknown gas diffuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas? What is the identity of the unknown gas, given? Possible choices include NO2, NH3, CH4, and HCl.
Solution
From Graham’s law, we have:
=
Plug in known data:
=
Μunknown = = 16.0 g/mol
The unknown gas is CH4, a gas with this molar mass.
Check Your Learning
Hydrogen gas passes through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Answer:
161 g/mol
Chemistry in Real Life: Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235U, the kind of uranium that is fissile, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF6, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235UF6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238UF6 molecules. The gas that has passed through the barrier is slightly enriched in 235UF6 and the residual gas (238UF6) are slightly depleted. The small difference in molecular weights between 235UF6 and 238UF6 causes only about 0.4% enrichment in one diffuser (Figure 6). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
Figure 6. In a diffuser, gaseous UF6 is pumped through a porous barrier, which partially separates 235UF6 from 238UF6 The UF6 must pass through many large diffuser units to achieve sufficient enrichment in 235U.
The large scale separation of gaseous 235UF6 from 238UF6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required overcoming many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF6.
Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy.
Key Concepts and Summary
The kinetic molecular theory is a simple but very effective model that explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.
Key Equations
KE = RT
where R = 8.314 = 8.314
=
where Μ is molar mass in kg/mol
=
Glossary
: diffusion
: movement of an atom or molecule from a region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)
Graham’s law
: rates of diffusion of gases are inversely proportional to the square roots of their molecular masses
kinetic molecular theory
: theory based on simple principles and assumptions that effectively explains the behavior of ideal gases
rate of diffusion
: amount of gas diffusing through a given area over a given time
root mean square velocity (Vrms)
: measure of average velocity for a group of particles calculated as the square root of the average squared velocity
Chemistry End of Section Exercises
Kinetic Molecular Theory
Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
Can the average speed of molecules in a constant volume of gas double at constant temperature? Explain your answer.
Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:
The pressure of the gas is increased by reducing the volume at constant temperature.
The pressure of the gas is increased by increasing the temperature at constant volume.
The average velocity of the molecules is increased by a factor of 2.
The distribution of molecular velocities in a sample of helium is shown in Figure 4. If the sample is cooled, will the distribution of velocities look more like that of H2 or of H2O? Explain your answer.
What is the ratio of the average kinetic energy of a SO2 molecule to that of an O2 molecule in a mixture of two gases? What is the ratio of the root mean square speeds, vrms, of the two gases?
A 1-L sample of CO initially at STP is heated to 546 °C, and its volume is increased to 2 L.
What effect do the changes have (increase or decrease) on the frequency of collisions of molecules per unit area with the wall?
What is the effect on the average kinetic energy of the molecules?
What is the effect on the root mean square speed of the molecules?
The root mean square speed of H2 molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N2 molecule at 25 °C?
Answer the following questions:
Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?
The average temperature of the gas in a hot air balloon is 1.30 × 102 °F. Calculate its density, assuming the molar mass equals that of dry air.
The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
An average hot air balloon has a diameter of 60 feet and a volume of 1.1 × 105 ft3. How much mass can the balloon lift? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?
Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, , is the same at 0 °C and 100 °C.
Diffusion
A balloon filled with helium gas is found to take 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?
Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.
Calculate the relative rate of diffusion of H2 (molar mass = 2.0 g/mol) compared to that of He (molar mass = 4.0 g/mol) and the relative rate of diffusion of O2 (molar mass = 32 g/mol) compared to that of O3 (molar mass = 48 g/mol).
A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH4Cl forms where gaseous NH3 and gaseous HCl first come into contact. At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both NH3 and HCl, and find out how much faster NH3 diffuses than HCl.) NH3(g) + HCl(g) → NH4Cl(s)
Answers to Chemistry End of Section Exercises
Gases consist of tiny particles that move at different speeds and in every possible direction; they are in constant motion.
Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. However, the average velocity of all the gas molecules stays constant at constant temperature.
(a) The average kinetic energy remains unchanged.
(b) As T increases, the average kinetic energy increases by the same factor.
(c) The average kinetic energy increases by a factor of 4.
H2O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier.
Average kinetic energy ratio: 1:1. vrms ratio: SO2:O2 = 1:√2
(a) The frequency of collisions per unit area of the container wall increases. (b) The average kinetic energy triples. (c) The root mean squared speed (vrms) increases to √3 times its initial value.
515 m/s
(a) equal; (b) less than; (c) 29.48 g mol−1; (d) 1.0966 g L−1; (e) 0.129 g/L; (f) 4.01 × 105 g; net lifting capacity = 384 lb; (g) 39.1 kJ min−1
From Graham’s law, we know that the rate of diffusion of a gas is inversely proportional to the square root of the mass of its particles. It is not temperature dependent.
4.2 hours
1.05
1.4; 1.2
66.0 g/mol
51.7 cm |
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Contents
1 History
2 Properties and uses
3 Square roots of positive integers
3.1 As decimal expansions
3.2 As expansions in other numeral systems
3.3 As periodic continued fractions
4 Computation
5 Square roots of negative and complex numbers
5.1 Principal square root of a complex number
5.2 Algebraic formula
5.3 Notes
6 nth roots and polynomial roots
7 Square roots of matrices and operators
8 In integral domains, including fields
9 In rings in general
10 Geometric construction of the square root
11 See also
12 Notes
13 References
14 External links
Square root
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From Wikipedia, the free encyclopedia
Number whose square is a given number
"Square roots" redirects here. For other uses, see Square Roots (disambiguation).
In mathematics, a square root of a number x is a number y such that ; in other words, a number y whose square (the result of multiplying the number by itself, or ) is x. For example, 4 and −4 are square roots of 16 because .
Every nonnegative real number x has a unique nonnegative square root, called the principal square root or simply the square root (with a definite article, see below), which is denoted by where the symbol "" is called the radical sign or radix. For example, to express the fact that the principal square root of 9 is 3, we write . The term (or number) whose square root is being considered is known as the radicand. The radicand is the number or expression underneath the radical sign, in this case, 9. For non-negative x, the principal square root can also be written in exponent notation, as .
Every positive number x has two square roots: (which is positive) and (which is negative). The two roots can be written more concisely using the ± sign as . Although the principal square root of a positive number is only one of its two square roots, the designation "the square root" is often used to refer to the principal square root.
Square roots of negative numbers can be discussed within the framework of complex numbers. More generally, square roots can be considered in any context in which a notion of the "square" of a mathematical object is defined. These include function spaces and square matrices, among other mathematical structures.
History
The Yale Babylonian Collection clay tablet YBC 7289 was created between 1800 BC and 1600 BC, showing and respectively as 1;24,51,10 and 0;42,25,35 base 60 numbers on a square crossed by two diagonals. (1;24,51,10) base 60 corresponds to 1.41421296, which is correct to 5 decimal places (1.41421356...).
The Rhind Mathematical Papyrus is a copy from 1650 BC of an earlier Berlin Papyrus and other texts – possibly the Kahun Papyrus – that shows how the Egyptians extracted square roots by an inverse proportion method.
In Ancient India, the knowledge of theoretical and applied aspects of square and square root was at least as old as the Sulba Sutras, dated around 800–500 BC (possibly much earlier). A method for finding very good approximations to the square roots of 2 and 3 are given in the Baudhayana Sulba Sutra. Apastamba who was dated around 600 BCE has given a strikingly accurate value for which is correct up to five decimal places as . Aryabhata, in the Aryabhatiya (section 2.4), has given a method for finding the square root of numbers having many digits.
It was known to the ancient Greeks that square roots of positive integers that are not perfect squares are always irrational numbers: numbers not expressible as a ratio of two integers (that is, they cannot be written exactly as , where m and n are integers). This is the theorem Euclid X, 9, almost certainly due to Theaetetus dating back to c. 380 BC. The discovery of irrational numbers, including the particular case of the square root of 2, is widely associated with the Pythagorean school. Although some accounts attribute the discovery to Hippasus, the specific contributor remains uncertain due to the scarcity of primary sources and the secretive nature of the brotherhood. It is exactly the length of the diagonal of a square with side length 1.
In the Chinese mathematical work Writings on Reckoning, written between 202 BC and 186 BC during the early Han dynasty, the square root is approximated by using an "excess and deficiency" method, which says to "...combine the excess and deficiency as the divisor; (taking) the deficiency numerator multiplied by the excess denominator and the excess numerator times the deficiency denominator, combine them as the dividend."
A symbol for square roots, written as an elaborate R, was invented by Regiomontanus (1436–1476). An R was also used for radix to indicate square roots in Gerolamo Cardano's Ars Magna.
According to historian of mathematics D.E. Smith, Aryabhata's method for finding the square root was first introduced in Europe by Cataneo—in 1546.
According to Jeffrey A. Oaks, Arabs used the letter jīm/ĝīm (ج), the first letter of the word "جذر" (variously transliterated as jaḏr, jiḏr, ǧaḏr or ǧiḏr, "root"), placed in its initial form (ﺟ) over a number to indicate its square root. The letter jīm resembles the present square root shape. Its usage goes as far as the end of the twelfth century in the works of the Moroccan mathematician Ibn al-Yasamin.
The symbol "√" for the square root was first used in print in 1525, in Christoph Rudolff's Coss.
Properties and uses
The principal square root function (usually just referred to as the "square root function") is a function that maps the set of nonnegative real numbers onto itself. In geometrical terms, the square root function maps the area of a square to its side length.
The square root of x is rational if and only if x is a rational number that can be represented as a ratio of two perfect squares. (See square root of 2 for proofs that this is an irrational number, and quadratic irrational for a proof for all non-square natural numbers.) The square root function maps rational numbers into algebraic numbers, the latter being a superset of the rational numbers).
For all real numbers x, (see absolute value).
For all nonnegative real numbers x and y, and
The square root function is continuous for all nonnegative x, and differentiable for all positive x. If f denotes the square root function, whose derivative is given by:
The Taylor series of about x = 0 converges for |x| ≤ 1, and is given by
The square root of a nonnegative number is used in the definition of Euclidean norm (and distance), as well as in generalizations such as Hilbert spaces. It defines an important concept of standard deviation used in probability theory and statistics. It has a major use in the formula for solutions of a quadratic equation. Quadratic fields and rings of quadratic integers, which are based on square roots, are important in algebra and have uses in geometry. Square roots frequently appear in mathematical formulas elsewhere, as well as in many physical laws.
Square roots of positive integers
A positive number has two square roots, one positive, and one negative, which are opposite to each other. When talking of the square root of a positive integer, it is usually the positive square root that is meant.
The square roots of an integer are algebraic integers—more specifically quadratic integers.
The square root of a positive integer is the product of the roots of its prime factors, because the square root of a product is the product of the square roots of the factors. Since only roots of those primes having an odd power in the factorization are necessary. More precisely, the square root of a prime factorization is
As decimal expansions
The square roots of the perfect squares (e.g., 0, 1, 4, 9, 16) are integers. In all other cases, the square roots of positive integers are irrational numbers, and hence have non-repeating decimals in their decimal representations. Decimal approximations of the square roots of the first few natural numbers are given in the following table.
| n | truncated to 50 decimal places |
--- |
| 0 | 0 |
| 1 | 1 |
| 2 | 1.41421356237309504880168872420969807856967187537694 |
| 3 | 1.73205080756887729352744634150587236694280525381038 |
| 4 | 2 |
| 5 | 2.23606797749978969640917366873127623544061835961152 |
| 6 | 2.44948974278317809819728407470589139196594748065667 |
| 7 | 2.64575131106459059050161575363926042571025918308245 |
| 8 | 2.82842712474619009760337744841939615713934375075389 |
| 9 | 3 |
| 10 | 3.16227766016837933199889354443271853371955513932521 |
As expansions in other numeral systems
As with before, the square roots of the perfect squares (e.g., 0, 1, 4, 9, 16) are integers. In all other cases, the square roots of positive integers are irrational numbers, and therefore have non-repeating digits in any standard positional notation system.
The square roots of small integers are used in both the SHA-1 and SHA-2 hash function designs to provide nothing up my sleeve numbers.
As periodic continued fractions
A result from the study of irrational numbers as simple continued fractions was obtained by Joseph Louis Lagrange c. 1780. Lagrange found that the representation of the square root of any non-square positive integer as a continued fraction is periodic. That is, a certain pattern of partial denominators repeats indefinitely in the continued fraction. In a sense these square roots are the very simplest irrational numbers, because they can be represented with a simple repeating pattern of integers.
| | |
--- |
| | = [1; 2, 2, ...] |
| | = [1; 1, 2, 1, 2, ...] |
| | = |
| | = [2; 4, 4, ...] |
| | = [2; 2, 4, 2, 4, ...] |
| | = [2; 1, 1, 1, 4, 1, 1, 1, 4, ...] |
| | = [2; 1, 4, 1, 4, ...] |
| | = |
| | = [3; 6, 6, ...] |
| | = [3; 3, 6, 3, 6, ...] |
| | = [3; 2, 6, 2, 6, ...] |
| | = [3; 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, ...] |
| | = [3; 1, 2, 1, 6, 1, 2, 1, 6, ...] |
| | = [3; 1, 6, 1, 6, ...] |
| | = |
| | = [4; 8, 8, ...] |
| | = [4; 4, 8, 4, 8, ...] |
| | = [4; 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, ...] |
| | = [4; 2, 8, 2, 8, ...] |
The square bracket notation used above is a short form for a continued fraction. Written in the more suggestive algebraic form, the simple continued fraction for the square root of 11, [3; 3, 6, 3, 6, ...], looks like this:
where the two-digit pattern {3, 6} repeats over and over again in the partial denominators. Since 11 = 32 + 2, the above is also identical to the following generalized continued fractions:
Computation
Main article: Methods of computing square roots
Square roots of positive numbers are not in general rational numbers, and so cannot be written as a terminating or recurring decimal expression. Therefore in general any attempt to compute a square root expressed in decimal form can only yield an approximation, though a sequence of increasingly accurate approximations can be obtained.
Most pocket calculators have a square root key. Computer spreadsheets and other software are also frequently used to calculate square roots. Pocket calculators typically implement efficient routines, such as the Newton's method (frequently with an initial guess of 1), to compute the square root of a positive real number. When computing square roots with logarithm tables or slide rules, one can exploit the identities where ln and log10 are the natural and base-10 logarithms.
By trial-and-error, one can square an estimate for and raise or lower the estimate until it agrees to sufficient accuracy. For this technique it is prudent to use the identity as it allows one to adjust the estimate x by some amount c and measure the square of the adjustment in terms of the original estimate and its square.
The most common iterative method of square root calculation by hand is known as the "Babylonian method" or "Heron's method" after the first-century Greek philosopher Heron of Alexandria, who first described it. The method uses the same iterative scheme as the Newton–Raphson method yields when applied to the function y = f(x) = x2 − a, using the fact that its slope at any point is dy/dx = f′(x) = 2x, but predates it by many centuries. The algorithm is to repeat a simple calculation that results in a number closer to the actual square root each time it is repeated with its result as the new input. The motivation is that if x is an overestimate to the square root of a nonnegative real number a then a/x will be an underestimate and so the average of these two numbers is a better approximation than either of them. However, the inequality of arithmetic and geometric means shows this average is always an overestimate of the square root (as noted below), and so it can serve as a new overestimate with which to repeat the process, which converges as a consequence of the successive overestimates and underestimates being closer to each other after each iteration. To find x:
Start with an arbitrary positive start value x. The closer to the square root of a, the fewer the iterations that will be needed to achieve the desired precision.
Replace x by the average (x + a/x) / 2 between x and a/x.
Repeat from step 2, using this average as the new value of x.
That is, if an arbitrary guess for is x0, and xn + 1 = (xn + a/xn) / 2, then each xn is an approximation of which is better for large n than for small n. If a is positive, the convergence is quadratic, which means that in approaching the limit, the number of correct digits roughly doubles in each next iteration. If a = 0, the convergence is only linear; however, so in this case no iteration is needed.
Using the identity the computation of the square root of a positive number can be reduced to that of a number in the range [1, 4). This simplifies finding a start value for the iterative method that is close to the square root, for which a polynomial or piecewise-linear approximation can be used.
The time complexity for computing a square root with n digits of precision is equivalent to that of multiplying two n-digit numbers.
Another useful method for calculating the square root is the shifting nth root algorithm, applied for n = 2.
The name of the square root function varies from programming language to programming language, with sqrt (often pronounced "squirt") being common, used in C and derived languages such as C++, JavaScript, PHP, and Python.
Square roots of negative and complex numbers
First leaf of the complex square root
Second leaf of the complex square root
Using the Riemann surface of the square root, it is shown how the two leaves fit together
The square of any positive or negative number is positive, and the square of 0 is 0. Therefore, no negative number can have a real square root. However, it is possible to work with a more inclusive set of numbers, called the complex numbers, that does contain solutions to the square root of a negative number. This is done by introducing a new number, denoted by i (sometimes by j, especially in the context of electricity where i traditionally represents electric current) and called the imaginary unit, which is defined such that i2 = −1. Using this notation, we can think of i as the square root of −1, but we also have (−i)2 = i2 = −1 and so −i is also a square root of −1. By convention, the principal square root of −1 is i, or more generally, if x is any nonnegative number, then the principal square root of −x is
The right side (as well as its negative) is indeed a square root of −x, since
For every non-zero complex number z there exist precisely two numbers w such that w2 = z: the principal square root of z (defined below), and its negative.
Principal square root of a complex number
To find a definition for the square root that allows us to consistently choose a single value, called the principal value, we start by observing that any complex number can be viewed as a point in the plane, expressed using Cartesian coordinates. The same point may be reinterpreted using polar coordinates as the pair where is the distance of the point from the origin, and is the angle that the line from the origin to the point makes with the positive real () axis. In complex analysis, the location of this point is conventionally written If then the principal square root of is defined to be the following: The principal square root function is thus defined using the non-positive real axis as a branch cut. If is a non-negative real number (which happens if and only if ) then the principal square root of is in other words, the principal square root of a non-negative real number is just the usual non-negative square root. It is important that because if, for example, (so ) then the principal square root is but using would instead produce the other square root
The principal square root function is holomorphic everywhere except on the set of non-positive real numbers (on strictly negative reals it is not even continuous). The above Taylor series for remains valid for complex numbers with
The above can also be expressed in terms of trigonometric functions:
Algebraic formula
When the number is expressed using its real and imaginary parts, the following formula can be used for the principal square root:
where sgn(y) = 1 if y ≥ 0 and sgn(y) = −1 otherwise. In particular, the imaginary parts of the original number and the principal value of its square root have the same sign. The real part of the principal value of the square root is always nonnegative.
For example, the principal square roots of ±i are given by:
Notes
In the following, the complex z and w may be expressed as:
where and .
Because of the discontinuous nature of the square root function in the complex plane, the following laws are not true in general.
Counterexample for the principal square root: z = −1 and w = −1 This equality is valid only when
Counterexample for the principal square root: w = 1 and z = −1 This equality is valid only when
Counterexample for the principal square root: z = −1) This equality is valid only when
A similar problem appears with other complex functions with branch cuts, e.g., the complex logarithm and the relations logz + logw = log(zw) or log(z) = log(z) which are not true in general.
Wrongly assuming one of these laws underlies several faulty "proofs", for instance the following one showing that −1 = 1:
The third equality cannot be justified (see invalid proof).: Chapter VI, Section I, Subsection 2 The fallacy that +1 = −1 It can be made to hold by changing the meaning of √ so that this no longer represents the principal square root (see above) but selects a branch for the square root that contains The left-hand side becomes either if the branch includes +i or if the branch includes −i, while the right-hand side becomes where the last equality, is a consequence of the choice of branch in the redefinition of √.
nth roots and polynomial roots
The definition of a square root of as a number such that has been generalized in the following way.
A cube root of is a number such that ; it is denoted
If n is an integer greater than two, a n-th root of is a number such that ; it is denoted
Given any polynomial p, a root of p is a number y such that p(y) = 0. For example, the nth roots of x are the roots of the polynomial (in y)
Abel–Ruffini theorem states that, in general, the roots of a polynomial of degree five or higher cannot be expressed in terms of nth roots.
Square roots of matrices and operators
Main article: Square root of a matrix
See also: Square root of a 2 by 2 matrix
If A is a positive-definite matrix or operator, then there exists precisely one positive definite matrix or operator B with B2 = A; we then define A1/2 = B. In general matrices may have multiple square roots or even an infinitude of them. For example, the 2 × 2 identity matrix has an infinity of square roots, though only one of them is positive definite.
In integral domains, including fields
Each element of an integral domain has no more than 2 square roots. The difference of two squares identity u2 − v2 = (u − v)(u + v) is proved using the commutativity of multiplication. If u and v are square roots of the same element, then u2 − v2 = 0. Because there are no zero divisors this implies u = v or u + v = 0, where the latter means that two roots are additive inverses of each other. In other words if an element a square root u of an element a exists, then the only square roots of a are u and −u. The only square root of 0 in an integral domain is 0 itself.
In a field of characteristic 2, an element either has one square root or does not have any at all, because each element is its own additive inverse, so that −u = u. If the field is finite of characteristic 2 then every element has a unique square root. In a field of any other characteristic, any non-zero element either has two square roots, as explained above, or does not have any.
Given an odd prime number p, let q = pe for some positive integer e. A non-zero element of the field Fq with q elements is a quadratic residue if it has a square root in Fq. Otherwise, it is a quadratic non-residue. There are (q − 1)/2 quadratic residues and (q − 1)/2 quadratic non-residues; zero is not counted in either class. The quadratic residues form a group under multiplication. The properties of quadratic residues are widely used in number theory.
In rings in general
Unlike in an integral domain, a square root in an arbitrary (unital) ring need not be unique up to sign. For example, in the ring of integers modulo 8 (which is commutative, but has zero divisors), the element 1 has four distinct square roots: ±1 and ±3.
Another example is provided by the ring of quaternions which has no zero divisors, but is not commutative. Here, the element −1 has infinitely many square roots, including ±i, ±j, and ±k. In fact, the set of square roots of −1 is exactly
A square root of 0 is either 0 or a zero divisor. Thus in rings where zero divisors do not exist, it is uniquely 0. However, rings with zero divisors may have multiple square roots of 0. For example, in any multiple of n is a square root of 0.
Geometric construction of the square root
The square root of a positive number is usually defined as the side length of a square with the area equal to the given number. But the square shape is not necessary for it: if one of two similar planar Euclidean objects has the area a times greater than another, then the ratio of their linear sizes is .
A square root can be constructed with a compass and straightedge. In his Elements, Euclid (fl. 300 BC) gave the construction of the geometric mean of two quantities in two different places: Proposition II.14 and Proposition VI.13. Since the geometric mean of a and b is , one can construct simply by taking b = 1.
The construction is also given by Descartes in his La Géométrie, see figure 2 on page 2. However, Descartes made no claim to originality and his audience would have been quite familiar with Euclid.
Euclid's second proof in Book VI depends on the theory of similar triangles. Let AHB be a line segment of length a + b with AH = a and HB = b. Construct the circle with AB as diameter and let C be one of the two intersections of the perpendicular chord at H with the circle and denote the length CH as h. Then, using Thales' theorem and, as in the proof of Pythagoras' theorem by similar triangles, triangle AHC is similar to triangle CHB (as indeed both are to triangle ACB, though we don't need that, but it is the essence of the proof of Pythagoras' theorem) so that AH:CH is as HC:HB, i.e. a/h = h/b, from which we conclude by cross-multiplication that h2 = ab, and finally that . When marking the midpoint O of the line segment AB and drawing the radius OC of length (a + b)/2, then clearly OC > CH, i.e. (with equality if and only if a = b), which is the arithmetic–geometric mean inequality for two variables and, as noted above, is the basis of the Ancient Greek understanding of "Heron's method".
Another method of geometric construction uses right triangles and induction: can be constructed, and once has been constructed, the right triangle with legs 1 and has a hypotenuse of . Constructing successive square roots in this manner yields the Spiral of Theodorus depicted above.
See also
Apotome (mathematics)
Cube root
Functional square root
Integer square root
Nested radical
Nth root
Root of unity
Solving quadratic equations with continued fractions
Square-root sum problem
Square-root method – Method of allocating voting weight by populationPages displaying short descriptions of redirect targets
Quantum gate § Square root of NOT gate (√NOT)
Notes
^ Gel'fand, p. 120 Archived 2016-09-02 at the Wayback Machine
^ "Squares and Square Roots". www.mathsisfun.com. Retrieved 2020-08-28.
^ Zill, Dennis G.; Shanahan, Patrick (2008). A First Course in Complex Analysis With Applications (2nd ed.). Jones & Bartlett Learning. p. 78. ISBN 978-0-7637-5772-4. Archived from the original on 2016-09-01. Extract of page 78 Archived 2016-09-01 at the Wayback Machine
^ Weisstein, Eric W. "Square Root". mathworld.wolfram.com. Retrieved 2020-08-28.
^ "Analysis of YBC 7289". ubc.ca. Retrieved 19 January 2015.
^ Anglin, W.S. (1994). Mathematics: A Concise History and Philosophy. New York: Springer-Verlag.
^ Seidenberg, A. (1961). "The ritual origin of geometry". Archive for History of Exact Sciences. 1 (5): 488–527. doi:10.1007/bf00327767. ISSN 0003-9519. S2CID 119992603. Seidenberg (pp. 501-505) proposes: "It is the distinction between use and origin." [By analogy] "KEPLER needed the ellipse to describe the paths of the planets around the sun; he did not, however invent the ellipse, but made use of a curve that had been lying around for nearly 2000 years". In this manner Seidenberg argues: "Although the date of a manuscript or text cannot give us the age of the practices it discloses, nonetheless the evidence is contained in manuscripts." Seidenberg quotes Thibaut from 1875: "Regarding the time in which the Sulvasutras may have been composed, it is impossible to give more accurate information than we are able to give about the date of the Kalpasutras. But whatever the period may have been during which Kalpasutras and Sulvasutras were composed in the form now before us, we must keep in view that they only give a systematically arranged description of sacrificial rites, which had been practiced during long preceding ages." Lastly, Seidenberg summarizes: "In 1899, THIBAUT ventured to assign the fourth or the third centuries B.C. as the latest possible date for the composition of the Sulvasutras (it being understood that this refers to a codification of far older material)."
^ Joseph, ch.8.
^ Dutta, Bibhutibhusan (1931). "On the Origin of the Hindu Terms for "Root"". The American Mathematical Monthly. 38 (7): 371–376. doi:10.2307/2300909. JSTOR 2300909. Retrieved 30 March 2024.
^ Cynthia J. Huffman; Scott V. Thuong (2015). "Ancient Indian Rope Geometry in the Classroom - Approximating the Square Root of 2". www.maa.org. Retrieved 30 March 2024. Increase the measure by its third and this third by its own fourth, less the thirty-fourth part of that fourth. This is the value with a special quantity in excess.
^ J J O'Connor; E F Robertson (November 2020). "Apastamba". www.mathshistory.st-andrews.ac.uk. School of Mathematics and Statistics, University of St Andrews, Scotland. Retrieved 30 March 2024.
^ Heath, Sir Thomas L. (1908). The Thirteen Books of The Elements, Vol. 3. Cambridge University Press. p. 3.
^ Craig Smorynski (2007). History of Mathematics: A Supplement (illustrated, annotated ed.). Springer Science & Business Media. p. 49. ISBN 978-0-387-75480-2. Extract of page 49
^ Brian E. Blank; Steven George Krantz (2006). Calculus: Single Variable, Volume 1 (illustrated ed.). Springer Science & Business Media. p. 71. ISBN 978-1-931914-59-8. Extract of page 71
^ Boyer, Carl B.; Merzbach, Uta C. (2011). A History of Mathematics (3rd ed.). Hoboken, NJ: John Wiley & Sons. pp. 51–53. ISBN 978-0470525487.
^ Stillwell, John (2010). Mathematics and Its History (3rd ed.). New York, NY: Springer. pp. 14–15. ISBN 978-1441960528.
^ Dauben (2007), p. 210.
^ "The Development of Algebra - 2". maths.org. Archived from the original on 24 November 2014. Retrieved 19 January 2015.
^ Oaks, Jeffrey A. (2012). Algebraic Symbolism in Medieval Arabic Algebra (PDF) (Thesis). Philosophica. p. 36. Archived (PDF) from the original on 2016-12-03.
^ Manguel, Alberto (2006). "Done on paper: the dual nature of numbers and the page". The Life of Numbers. Taric, S.A. ISBN 84-86882-14-1.
^ Parkhurst, David F. (2006). Introduction to Applied Mathematics for Environmental Science. Springer. pp. 241. ISBN 9780387342283.
^ Solow, Anita E. (1993). Learning by Discovery: A Lab Manual for Calculus. Cambridge University Press. pp. 48. ISBN 9780883850831.
^ Aitken, Mike; Broadhurst, Bill; Hladky, Stephen (2009). Mathematics for Biological Scientists. Garland Science. p. 41. ISBN 978-1-136-84393-8. Archived from the original on 2017-03-01. Extract of page 41 Archived 2017-03-01 at the Wayback Machine
^ Heath, Sir Thomas L. (1921). A History of Greek Mathematics, Vol. 2. Oxford: Clarendon Press. pp. 323–324.
^ Muller, Jean-Mic (2006). Elementary functions: algorithms and implementation. Springer. pp. 92–93. ISBN 0-8176-4372-9., Chapter 5, p 92 Archived 2016-09-01 at the Wayback Machine
^ "Function sqrt". CPlusPlus.com. The C++ Resources Network. 2016. Archived from the original on November 22, 2012. Retrieved June 24, 2016.
^ Overland, Brian (2013). C++ for the Impatient. Addison-Wesley. p. 338. ISBN 9780133257120. OCLC 850705706. Archived from the original on September 1, 2016. Retrieved June 24, 2016.
^ Abramowitz, Milton; Stegun, Irene A. (1964). Handbook of mathematical functions with formulas, graphs, and mathematical tables. Courier Dover Publications. p. 17. ISBN 0-486-61272-4. Archived from the original on 2016-04-23. {{cite book}}: ISBN / Date incompatibility (help), Section 3.7.27, p. 17 Archived 2009-09-10 at the Wayback Machine
^ Cooke, Roger (2008). Classical algebra: its nature, origins, and uses. John Wiley and Sons. p. 59. ISBN 978-0-470-25952-8. Archived from the original on 2016-04-23.
^ This sign function differs from the usual sign function by its value at 0.
^ Maxwell, E. A. (1959). Fallacies in Mathematics. Cambridge University Press. ISBN 9780511569739. {{cite book}}: ISBN / Date incompatibility (help)
^ Mitchell, Douglas W., "Using Pythagorean triples to generate square roots of I2", Mathematical Gazette 87, November 2003, 499–500.
Dauben, Joseph W. (2007). "Chinese Mathematics I". In Katz, Victor J. (ed.). The Mathematics of Egypt, Mesopotamia, China, India, and Islam. Princeton: Princeton University Press. ISBN 978-0-691-11485-9.
Gel'fand, Izrael M.; Shen, Alexander (1993). Algebra (3rd ed.). Birkhäuser. p. 120. ISBN 0-8176-3677-3.
Joseph, George (2000). The Crest of the Peacock. Princeton: Princeton University Press. ISBN 0-691-00659-8.
Smith, David (1958). History of Mathematics. Vol. 2. New York: Dover Publications. ISBN 978-0-486-20430-7. {{cite book}}: ISBN / Date incompatibility (help)
Selin, Helaine (2008), Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures, Springer, Bibcode:2008ehst.book.....S, ISBN 978-1-4020-4559-2.
External links
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Algorithms, implementations, and more – Paul Hsieh's square roots webpage
How to manually find a square root
AMS Featured Column, Galileo's Arithmetic by Tony Philips – includes a section on how Galileo found square roots
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7257 | https://www.ccjm.org/content/ccjom/75/7_suppl_5/S17.full.pdf | CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 75 • SUPPLEMENT 5 JULY 2008 S17 ROBIN K. DORE, MD Clinical Professor of Medicine, Division of Rheumatology, David Geffen School of Medicine, University of California at Los Angeles, Los Angeles, CA The gout diagnosis ■ABSTRACT Synovial fluid aspiration and analysis is the gold standard for making the diagnosis of gout but is not always per-formed when indicated in clinical practice. In clinical situa-tions when joint aspiration simply cannot be performed, a presumptive (or clinical) diagnosis of gout may be made in consultation with published recommendations and criteria from expert societies. A thorough patient history and phys-ical examination are critical to a presumptive diagnosis of gout, as is serum urate measurement at the time of an acute attack and at follow-up 2 weeks later.
■KEY POINTS If the serum urate level was not elevated when measured during an acute attack of arthritis, it will likely be elevated at 2-week follow-up if the patient does indeed have gout.
Gouty tophi are typically found in the olecranon bursa, whereas rheumatoid nodules are usually located on the extensor surface of the forearm.
Urate crystals of gout are negatively bifringent and fine and needlelike in shape, whereas the crystals of pseudo-gout are weakly positively birefringent and rhomboid.
Gout and septic arthritis can coexist; when the differential diagnosis includes septic arthritis, joint aspiration is required.
Until criteria for the presumptive diagnosis of gout are validated, clinicians should become familiar with the technique of joint aspiration.
T he presence of urate crystals in synovial fluid is the gold standard for diagnosing gout,1 yet clinicians⎯both primary care physicians and rheumatologists⎯may not routinely perform syn-ovial fluid analysis even when evaluating a patient who presents with an acute inflammatory arthritis.2 This paper discusses the various reasons why this is so and reviews several important resulting clinical issues: how a pre-sumptive diagnosis of gout is made, when to measure the serum urate level, and special considerations in the dif-ferential diagnosis. ■SYNOVIAL FLUID ANALYSIS: WHY IS THE GOLD STANDARD NOT MORE ROUTINE?
When synovial fluid containing monosodium urate crys-tals of gout is viewed under a polarizing microscope, bright yellow needlelike negatively birefringent crystals are seen3 (Figure 1A). Since synovial fluid analysis is the definitive method for diagnosing gout, why then is synovial fluid aspiration not performed routinely in clin-ical practice? Occasionally, the aspirated joint does not appear to contain any joint fluid and the clinician may be concerned about the possibility of a “dry tap.” Other possible reasons include lack of experience with synovial fluid aspiration and evaluation, or limited access to the polarizing micro-scopes used to examine synovial fluid. Time is another fac-tor; in a busy primary care practice, where patients are usu-ally seen approximately every 7 to 11 minutes, there may not be time to aspirate a joint. The urgency of fluid exam-ination is another issue, as synovial fluid must be exam-ined immediately, since the crystals can become smaller, less numerous, and less birefringent with time.4 ■THE CLINICAL, OR PRESUMPTIVE, DIAGNOSIS In the appropriate clinical scenario, a presumptive diag-nosis of gout can be made on the basis of typical clinical features and the presence of hyperuricemia.1,2 Expert societies offer guidance, but no validation studies to date Evidence-based recommendations for the diagnosis of gout from the European League Against Rheumatism (EULAR) state that in acute attacks, the rapid develop-ment of severe pain, swelling, and tenderness that peaks within 6 to 12 hours, especially with overlying erythema, Dr. Dore reported that she has received research grant support, consulting/advi-sory fees, and honoraria for speaking/teaching from TAP Pharmaceutical Products.
Dr. Dore received honoraria for participating in the symposium that formed the basis of this supplement and for writing this article.The honoraria were paid by Fallon Medica LLC, a medical communication company, on behalf of TAP Phar-maceutical Products, the underwriter of this supplement. TAP had no input on the content of presentations at the symposium or on this article.
This article is based on Dr.Dore’s lecture on this subject at the symposium that formed the basis of this supplement. Dr. Dore reported that she prepared her lecture, Fallon Medica transcribed her lecture,and she alone developed the transcript into this arti-cle without assistance from undeclared contributors.The article underwent format-ting and nonsubstantive copyediting by Fallon prior to submission to the Journal. on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from is highly suggestive of crystal inflammation although not specific for gout.5 These recommendations further state that for typical presentations of gout (such as recurrent podagra [gouty pain in the great toe] with hyperuricemia), a clinical diagnosis alone is reasonably accurate.5 In 1977, the American College of Rheumatology (ACR) published its preliminary criteria for the diagno-sis of acute gout, as outlined in Table 1.6 It concluded that any of the following is highly suggestive of gout:6 • The presence of urate crystals in joint fluid • A tophus containing urate crystals • Fulfillment of 6 or more of the criteria in Table 1.
No subsequent studies have been published on the validity or usefulness of any of these diagnostic criteria. What must inform the presumptive diagnosis Both the EULAR recommendations and the ACR cri-teria state that although the gold standard for diagnos-ing gout is the presence of urate crystals on synovial fluid analysis, a clinical diagnosis of gout can be made on the basis of certain patient criteria. This clinical, or pre-sumptive, diagnosis of gout should be made based on the following: • A careful patient and family history, including questions regarding comorbid conditions frequently associated with gout (such as hypertriglyceridemia, dia-betes, coronary heart disease, hypertension, and the metabolic syndrome) and whether the patient has had previous similar episodes of acute joint pain and swelling in the absence of trauma • Thorough identification of all current medications, some of which may be associated with hyperuricemia • A thorough physical examination.
■THE PHYSICAL EXAMINATION FOR GOUT Examination of patients with a history suggestive of gout should include not only the joints but also the extensor surface of the forearms and feet. When patients are seen for a visit and gout is suspected, they should be instructed to remove their shoes and socks and roll up their sleeves to allow examination for evidence of tophi, which would suggest a past history of gouty arthritis. The ear, knee, and olecranon bursa are other common sites for tophi,3 so patients should also be asked to roll up their pants and sleeves and remove any head coverings.7 In the late stages of gouty arthritis, multiple joints may be involved, which can cause the condition to be confused with other diagnoses such as psoriatic arthritis or erosive osteoarthritis.7 ■ACUTE PRESENTATIONS OF GOUT The typical gout presentation is remarkable for very intense pain that often occurs at night when the extremities are colder. Precipitation of urate in the dis-tal extremities can occur when the extremities are hori-zontal and tend to become cold.8 Approximately 90% of initial gout attacks are mono-articular, leaving only 10% of cases that are oligoarticular or polyarticular.7 If more than one joint is involved, espe-cially if the patient has a family history suggestive of gout or takes a medication that causes hyperuricemia, gout should be considered in the differential diagnosis even if the patient denies having a prior gout attack. Frequently, patients will call their primary care physi-cian during a gout attack but are not be able to schedule an appointment until after the attack has resolved. When possible, patients should be seen during the attack to con-firm whether the attack is due to gout. A diagnosis of gout should not be made over the phone when a patient describes pain in the great toe, as only 50% of initial gout attacks occur in the great toe7 and it is not known what proportion of acute pain episodes in the great toe are attributable to gout. The most common cause of pain in the great toe is osteoarthritis. THE GOUT DIAGNOSIS S18 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 75 • SUPPLEMENT 5 JULY 2008 FIGURE 1. (A) Monosodium urate crystals of gout appear as fine yellow needlelike crystals that are negatively birefringent under compensated polarized light. (B) In contrast, crystals of calcium pyrophosphate dihydrate (CPPD) crystal deposition disease are rhomboid in shape and weakly positively birefringent under compensated polarized light. Arrows alongside the crystals indicate the direction of the compensator.
Images courtesy of Brian F. Mandell, MD, PhD.
B A on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from Gout can also occur in the ankle or forefoot7 (Figure 2) and may appear to be cellulitis.1 In this instance, a prior history of a gouty attack, a family history of gout, an exposure to cold, binge drinking, or a history of hyperuricemia is suggestive of a gout diagnosis, but not definitively so. ■SPECIAL CONSIDERATIONS FOR THE PRESUMPTIVE DIAGNOSIS OF GOUT How long have acute attacks been occurring?
In a clinical scenario in which synovial fluid aspiration cannot be performed, the appropriateness of a presump-tive diagnosis can be assessed by a discussion with the patient about how long he or she has been experiencing acute attacks of joint pain. If the attacks have occurred for more than 10 years, tophi will likely be present.3 After even longer periods, gout may become poly-articular.7 In postmenopausal women, the distal inter-phalangeal joints may be involved,3 which may lead to a misdiagnosis of osteoarthritis, as these joints are typically affected by osteoarthritis.
Is the patient taking a urate-raising medication?
Certain medications have been associated with hyper-uricemia, including cyclosporine and thiazide diuretics.9 If a patient has been taking one of these medications, gout should be considered in the differential diagnosis if the patient presents with acute joint pain.
It has been argued that a reduction in joint pain and swelling after the use of colchicine confirms a diagnosis of gout. However, other conditions⎯such as tendonitis, calcium pyrophosphate dihydrate (CPPD) crystal depo-sition disease (pseudogout),3 and rheumatoid arthritis (RA)⎯can also improve after treatment with colchicine.1 Be vigilant for fever Another consideration in making a clinical diagnosis of gout is the association with a low-grade fever; these patients may feel as if they have the flu.8 Acute gout may also cause a high fever and an elevated white blood cell (WBC) count;3 in this situation, synovial fluid aspiration must be performed to exclude septic arthritis, either alone or in the presence of gouty arthritis. In situations where septic arthritis is suspected, an emergency visit to a rheumatologist is indicated for synovial fluid aspiration to be performed, as gout and sepsis can coexist.5 In such instances, Gram staining and culture of the synovial fluid should still be performed even if monosodium urate crys-tals are identified.5 ■MEASUREMENT OF SERUM URATE LEVELS Measuring serum urate levels during an acute attack, treating the acute attack with anti-inflammatory med-ications, and reevaluating the patient in the office 2 weeks after the acute attack are all recommended in the management of a patient with gout. If the serum urate level was not elevated during the acute attack, it is likely to be elevated 2 weeks later if the patient has gout.10 Elevated levels of serum urate during the intercritical periods are predictive of future gout attacks.11 Measuring serum urate during the initial attack and then 2 weeks later yields two serum urate levels that can be compared to assist in considering a presumed diagnosis of gout. A study by Rigby and Wood concluded that in patients CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 75 • SUPPLEMENT 5 JULY 2008 S19 DORE FIGURE 2. Swollen, erythematous ankle and first metatarsopha-langeal joints during an acute attack of gout.
© 1972–2004 American College of Rheumatology Clinical Slide Collection.
Used with permission.
TABLE 1 American College of Rheumatology criteria for the classification of acute arthritis of primary gout 1 More than 1 attack of acute arthritis 2 Maximum inflammation develops within 1 day 3 Monoarthritis attack 4 Redness over joints 5 First metatarsophalangeal joint painful or swollen 6 Unilateral first metatarsophalangeal joint attack 7 Unilateral tarsal joint attack 8 Tophus (proven or suspected) 9 Hyperuricemia 10 Asymmetric swelling within a joint on radiography 11 Subcortical cysts without erosions on radiography 12 Monosodium urate monohydrate microcrystals in joint fluid during attack 13 Joint fluid culture negative for organisms during attack Reprinted from Arthritis and Rheumatism (Wallace SL, et al. Preliminary criteria for the classification of the acute arthritis of primary gout. Arthritis Rheum 1977; 20:895–900), copyright 1977, with permission of Wiley-Liss, Inc., a subsidiary of John Wiley & Sons, Inc. on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from with low serum urate levels (< 4 mg/dL) 2 weeks follow-ing an inflammatory arthritis attack, a diagnosis of gout is unlikely.12 ■DIFFERENTIAL DIAGNOSIS OF GOUT Rheumatoid arthritis Patients with RA may present with nodules on their elbows, which can be mistaken for gouty tophi.3 However, the differences between RA and gout are appreciated on careful physical examination. Rheumatoid nodules are firm and nontender on physical exam,13 and usually are present on the extensor surface of the forearm (Figure 3A), whereas gouty tophi are usually located in the ole-cranon bursa (Figure 3B). In later stages of both RA and gout, the presentation can be that of a polyarticular inflammatory symmetric arthritis.14 A misdiagnosis of RA may be made if the serum urate level is normal at initial presentation,3 underscoring the importance of the follow-up visit 2 weeks after the attack. Serum urate levels are likely to be elevated after an attack, suggesting a clinical diagnosis of gout, per the EULAR recommendations,5 if the attack occurred in the great toe. An elevated serum urate level alone is not sufficient to support a presumed diagnosis of gout. CPPD crystal deposition disease (pseudogout) CPPD crystal deposition disease, or pseudogout, must also be included in the differential diagnosis of gout.
This disease usually occurs in joints previously affected by osteoarthritis or joints that have been injured in the past.15 Attacks of CPPD crystal deposition disease com-monly occur in the knee, in the wrist at the base of the thumb, or in the shoulder.15 Radiographic examination may reveal a line of calcification along the cartilage out-lining the joint.15 Like gout, pseudogout attacks can occur spontaneously or after trauma, surgery, or a severe illness such as myocardial infarction or stroke.16 The presentation of pseudogout can be very similar to an acute attack of gout. The difference is seen when evaluating the crystals through a polarizing microscope.
CPPD crystals are weakly positively birefringent (Figure 1B), in contrast to the negatively birefringent crystals seen with gout (Figure 1A).7 If a polarizing microscope is not available, the crystals usually can be distinguished by their differing shapes: urate crystals are fine and needlelike, whereas CPPD crystals are rhomboid (Figure 1). Septic arthritis When the differential diagnosis includes septic arthritis, the joint must be aspirated; a presumed diagnosis cannot be made. Among patients with an acute gouty attack, low-grade fever is reported during the attack in 29% of gout patients and 38% of patients with CPPD crystal deposi-tion disease.14 Temperatures of 101°F or higher are not usually seen in patients with gout or CPPD crystal deposi-tion disease and suggest an infection, although patients with septic arthritis may be afebrile, especially if they are taking immunosuppressive therapy or glucocorticoids, which can inhibit a febrile response. Synovial fluid analy-sis in patients with gout and septic arthritis can reveal WBC counts above 100,000 per mm3, whereas synovial fluid WBC counts above 50,000 per mm3 are more com-mon in infection. As noted earlier, gout and septic arthritis can coexist.
In a patient presenting with a fever and a warm erythe-matous swollen joint, synovial fluid aspiration must be performed and evaluated for the presence of crystals and bacteria. The patient may require treatment for both causes of acute monoarticular arthritis.
In a patient undergoing renal dialysis, where gout or pseudogout can occur and where there is frequent intravascular manipulation, a septic joint can occur simultaneously.3,14 In this situation, not only must joint aspiration be performed, but the synovial fluid also needs to be evaluated for both crystals and bacteria.
Again, the patient may require treatment for both causes of acute monoarticular arthritis.
THE GOUT DIAGNOSIS FIGURE 3. (A) Rheumatoid nodules are firm and usually not movable but rather are attached to the extensor surface of the forearm.
(B) Tophi appear as firm, gritty particles in the olecranon bursa.
© 1972–2004 American College of Rheumatology Clinical Slide Collection. Used with permission.
A B S20 CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 75 • SUPPLEMENT 5 JULY 2008 on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from ■CONCLUSIONS The gold standard for diagnosing gout remains synovial fluid aspiration and analysis. In clinical situations when joint aspiration cannot be performed, the EULAR rec-ommendations5 and the ACR criteria6 provide guidance for making a clinical or presumptive diagnosis of gout. A thorough patient history⎯both personal and family⎯and physical examination are critical in making a presumed diagnosis of gout. If the patient presents dur-ing an acute attack, serum urate measurement may be useful in making a clinical diagnosis if it reveals an ele-vated level. When the patient returns for follow-up 2 weeks later, a second serum urate measurement should be taken to allow comparison of the two levels. If the serum urate level is elevated at the follow-up visit, the EULAR recommendations state that a clinical diagnosis of gout can be made if the patient had an acute attack of arthritis in the great toe.
As noted in the EULAR recommendations, the future research agenda should include validating the clinical manifestations of gout against a diagnosis estab-lished by identification of urate crystals on synovial fluid analysis.5 Until this task can be completed, clinicians should become familiarized with the technique of joint aspiration so that in situations where a clinical or pre-sumptive diagnosis of gout cannot be made⎯including cases where the differential diagnosis includes a septic joint⎯clinicians will be able to perform aspiration with confidence.
■REFERENCES 1. Jelley MJ, Wortmann R. Practical steps in the diagnosis and man-agement of gout. BioDrugs 2000; 14:99–107.
2. Eggebeen AT. Gout: an update. Am Fam Physician 2007; 76:801–808, 811–812.
3. Rott KT, Agudelo CA. Gout. JAMA 2003; 289:2857–2860.
4. Poór G, Mituszova M. History, classification and epidemiology of crystal-related arthropathies. In: Hochberg MC, Silman AJ, Smolen JS, Weinblatt ME, Weisman MH, eds. Rheumatology. 3rd ed, vol 2.
Edinburgh: Mosby; 2003:1893–1901.
5. Zhang W, Doherty M, Pascual E, et al. EULAR evidence based rec-ommendations for gout. Part I: Diagnosis. Report of a task force of the Standing Committee for International Clinical Studies Including Therapeutics (ESCISIT). Ann Rheum Dis 2006; 65:1301–1311.
6. Wallace SL, Robinson H, Masi AT, Decker JL, McCarty DJ, Yü TF. Preliminary criteria for the classification of the acute arthritis of primary gout. Arthritis Rheum 1977; 20:895–900.
7. Harris MD, Siegel LB, Alloway JA. Gout and hyperuricemia. Am Fam Physician 1999; 59:925–934.
8. Saag KG, Mikuls TR. Recent advances in the epidemiology of gout. Curr Rheumatol Rep 2005; 7:235–241. 9. Edwards NL. Gout. B. Clinical and laboratory features. In: Klippel JH, Crofford LJ, Stone JH, Weyand CM, eds. Primer on the Rheumatic Diseases. 12th ed. Atlanta, GA: Arthritis Foundation; 2001:313–319. 10. Urano W, Yamanaka H, Tsutani H, et al. The inflammatory process in the mechanism of decreased serum uric acid concentrations dur-ing acute gouty arthritis. J Rheumatol 2002; 29:1950–1953.
11. Shoji A, Yamanaka H, Kamatani N. A retrospective study of the relationship between serum urate level and recurrent attacks of gouty arthritis: evidence for reduction of recurrent gouty arthritis with anti-hyperuricemic therapy. Arthritis Rheum 2004; 51:321–325.
12. Rigby AS, Wood PH. Serum uric acid levels and gout: what does this herald for the population? Clin Exp Rheumatol 1994; 12:395–400.
13. George DL. Skin and rheumatic disease. In: Hochberg MC, Silman AJ, Smolen JS, Weinblatt ME, Weisman MH, eds. Rheumatology.
3rd ed, vol 1. Edinburgh: Mosby; 2003:279–292.
14. Ho G Jr, DeNuccio M. Gout and pseudogout in hospitalized patients.
Arch Intern Med 1993; 153:2787–2790.
15. Agarwal AK. Gout and pseudogout. Prim Care 1993; 20:839–855.
16. Halverson PB, Ryan LM. Arthritis associated with calcium-con-taining crystals. In: Klippel JH, Crofford LJ, Stone JH, Weyand CM, eds. Primer on the Rheumatic Diseases. 12th ed. Atlanta, GA: Arthritis Foundation; 2001:299–306.
Correspondence: Robin K. Dore, MD, Division of Rheumatology, David Geffen School of Medicine, University of California at Los Angeles, 18102 Irvine Boulevard, #104, Tustin, CA 92780; rkdmail@sbcglobal.net.
CLEVELAND CLINIC JOURNAL OF MEDICINE VOLUME 75 • SUPPLEMENT 5 JULY 2008 S21 DORE on September 28, 2025. For personal use only. All other uses require permission.
www.ccjm.org Downloaded from |
7258 | https://www.tutorela.com/math/increasing-function/examples-exercises | Increasing and Decreasing Function Intervals Practice Problems | Tutorela
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Increasing and Decreasing Function Intervals Practice Problems
Master identifying increasing and decreasing intervals with step-by-step practice problems. Learn to analyze function behavior using graphs and algebraic methods.
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📚Master Increasing and Decreasing Function Intervals
Identify increasing intervals where f(x₂) > f(x₁) when x₂ > x₁
Find decreasing intervals by analyzing quadratic function vertices
Determine strictly increasing vs. non-decreasing function behavior
Solve vertex formula x = -b/(2a) for parabolic functions
Analyze function behavior using coefficient signs (a > 0 or a < 0)
Apply increasing/decreasing concepts to real-world scenarios
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Functions for Seventh Grade
Increasing and Decreasing Intervals (Functions)
Increasing functions
Increasing and Decreasing Function Intervals Practice Problems
Understanding Increasing functions
Complete explanation with examples
Increasing functions
What is an Increasing Function?
An increasing function is a type of relationship where, as you move to the right on the graph (increasing the x x x-value), the y y y-value also gets bigger. It’s like climbing a hill—the higher you go (the more you increase x x x), the more your height (the y y y-value) increases.
We will say that a function is increasing when, as the value of the independent variableX X Xincreases, the value of the functionY Y Yincreases.
How to Spot an Increasing Function:
On a Graph: The line or curve goes upwards as you move from left to right.
In Numbers: For any two xxx-values, if the second number is larger than the first x 2>x 1x_2 > x_1x 2>x 1, then the second y y y-value will also be larger than the first f(x 2)>f(x 1)f(x_2) > f(x_1)f(x 2)>f(x 1).
Real-Life Example:
Think about saving money in a piggy bank. Every day you add more coins, and the total amount of money keeps going up. That’s an increasing function in action—your savings are the y y y-values, and the number of days is the x x x-values.
Fun Fact:
If the line or curve never stops going up, it's called strictly increasing. If it flattens for a bit before going up again, it's just increasing.
let's see an example of strictly increasing linear function:
Detailed explanation
Practice Increasing functions
Test your knowledge with 17 quizzes
Determine which domain corresponds to the function described below:
The function represents the amount of fuel in a car's tank according to the distance traveled by the car.
a
Always increasing
b
Always decreasing
c
Partly increasing and partly decreasing
d
Impossible to know
Incorrect
Correct Answer:
Always decreasing
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Continue
Examples with solutions for Increasing functions
Step-by-step solutions included
Exercise #1
Is the function in the graph decreasing?
Step-by-Step Solution
To solve this problem, we'll follow these steps:
Step 1: Verify the graph's overall path direction
Step 2: Confirm if the y-values are decreasing as we proceed from the left side of the graph to the right side (increasing x-values).
Now, let's work through each step:
Step 1: By examining the graph, the red line starts at a higher point on the y-axis and moves downward to a lower point as it moves horizontally across the x-axis from left to right.
Step 2: Since for every point, the red line descends as it progresses from the leftmost point to the rightmost, this indicates a consistent decrease in the y-values.
Therefore, the solution to the problem is Yes, the function in the graph is decreasing.
Answer:
Yes
Exercise #2
Is the function in the graph decreasing?
Step-by-Step Solution
To analyze whether the function in the graph is decreasing, we must understand how the function's behavior is defined by its graph:
Step 1: Examine the graph. The graph presented is a horizontal line.
Step 2: Recognize the properties of a horizontal line. Horizontally aligned lines correspond to constant functions because the y y y-value remains the same for all x x x-values.
Step 3: Define the criteria for a function to be decreasing. A function decreases when, as x x x increases, the value of f(x) f(x) f(x) decreases.
Step 4: Apply this criterion to the horizontal line. Since the y y y-value is constant and does not decrease as x x x moves rightward, the function is not decreasing.
Therefore, the function represented by the graph is not decreasing.
Answer:
No
Exercise #3
Is the function in the graph below decreasing?
Step-by-Step Solution
To determine if the function is decreasing, we will analyze the graph visually:
The graph shows a line connecting from the bottom-left to the top-right of the graph area, indicating the line has a positive slope. This type of graph indicates the function is increasing, not decreasing.
A decreasing function means its value goes down as x x x increases, which is equivalent to having a negative slope.
Since the graph appears with a positive slope, the function is not decreasing.
Thus, the correct choice to the problem, which asks if the function in the graph is decreasing, is No.
Answer:
No
Exercise #4
Does the function in the graph decrease throughout?
Step-by-Step Solution
To solve this problem, we'll begin by examining the graph of the function provided:
Step 1: Observe the graph from left to right along the x-axis.
Step 2: Look for any intervals where the function value (y-coordinate) does not decrease as the x-value increases.
Step 3: Pay special attention to segments where the graph might look horizontal or rising.
Upon inspecting the graph, we find:
There are sections where the function's y-values appear to remain constant or potentially rise as the x-values increase. Specifically, even if the function decreases in major portions, any interval where it doesn't means the function cannot be classified as decreasing throughout.
Thus, the function does not strictly decrease on the entire interval shown. Therefore, the solution to the problem is No.
Answer:
No
Exercise #5
Is the function shown in the graph below decreasing?
Step-by-Step Solution
The graph presented is a straight line. To determine whether the function is decreasing, we need to examine the slope of this line.
The line has a negative slope, as it moves downward from left to right. A function is considered decreasing when its slope is negative.
In formal terms, for a linear function expressed as y=m x+c y = mx + c y=m x+c, if the slope m m m is negative, the function is decreasing over its entire domain.
From the graph, it's evident that the line has a negative slope, thus indicating that the function is indeed decreasing.
Therefore, the answer to the problem is Yes.
Answer:
Yes
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Frequently Asked Questions
Everything you need to know about Increasing functions
What is the difference between increasing and decreasing functions?
+
An increasing function has y-values that get larger as x-values increase (moving left to right on a graph, the line goes up). A decreasing function has y-values that get smaller as x-values increase (the line goes down from left to right). Mathematically, a function is increasing when f(x₂) > f(x₁) whenever x₂ > x₁.
How do you find increasing intervals of a quadratic function?
+
For quadratic functions, find the vertex using x = -b/(2a). If a > 0 (parabola opens up), the function decreases before the vertex and increases after. If a < 0 (parabola opens down), the function increases before the vertex and decreases after. The vertex marks the transition point between increasing and decreasing intervals.
What does strictly increasing mean in mathematics?
+
A strictly increasing function continuously rises without any flat sections - it never stays constant or decreases. In contrast, a regular increasing function can have flat portions where it remains constant before continuing to increase. Strictly increasing functions have a steeper, more consistent upward trend.
How can you tell if a function is increasing from its graph?
+
Look at the direction of the line or curve as you move from left to right: • If it slopes upward, the function is increasing • If it slopes downward, the function is decreasing • If it's horizontal, the function is constant • Check specific intervals, as functions can increase in some areas and decrease in others
What are some real-world examples of increasing functions?
+
Common examples include: savings account balance over time (with regular deposits), height of a growing plant over months, temperature rising throughout a spring day, or distance traveled at constant speed over time. These all show one variable consistently increasing as another variable increases.
How do you solve y = -(x+3)² type problems for increasing intervals?
+
Step-by-step process: 1. Expand: y = -x² - 6x - 9 2. Identify coefficients: a = -1, b = -6, c = -9 3. Find vertex: x = -(-6)/(2×-1) = -3 4. Since a < 0, parabola opens down (maximum at vertex) 5. Function increases for x < -3 and decreases for x > -3
Why is the vertex important for finding increasing and decreasing intervals?
+
The vertex represents the turning point of a parabola - where the function changes from increasing to decreasing (or vice versa). For upward-opening parabolas (a > 0), the function decreases before the vertex and increases after. For downward-opening parabolas (a < 0), the function increases before the vertex and decreases after.
Can a function be both increasing and decreasing?
+
Yes, many functions have different intervals where they increase and decrease. For example, a parabola y = x² decreases for x < 0 and increases for x > 0. When analyzing functions, we identify specific intervals or domains where the function exhibits increasing or decreasing behavior, rather than describing the entire function with one term.
More Increasing functions Questions
Click on any question to see the complete solution with step-by-step explanations
Increasing and Decreasing Intervals of a Function
Finding Decreasing Intervals: Analyzing Function Behavior from GraphIdentifying Decreasing Domains in a Piecewise Function GraphFinding Decreasing Intervals on a Curved Function Graph: [0,6] AnalysisFinding Function Increase Domain with Vertical Line x=-0.8: Graph AnalysisFinding Decreasing Intervals: Graph Analysis with x=0.65 Reference Line
Continue Your Math Journey
Master Increasing functions first, then advance to these related topics that build on your fraction skills
Suggested Topics to Practice in Advance
Ways to Represent a FunctionRepresenting a Function Verbally and with TablesGraphical Representation of a FunctionAlgebraic Representation of a FunctionNotation of a FunctionRate of Change of a FunctionVariation of a FunctionRate of change represented with steps in the graph of the functionRate of change of a function represented graphicallyConstant Rate of ChangeVariable Rate of ChangeRate of Change of a Function Represented by a Table of Values
Topics Learned in Later Sections
Functions for Seventh GradeIncreasing and Decreasing Intervals (Functions)Decreasing functionConstant FunctionDecreasing Interval of a functionIncreasing Intervals of a functionDomain of a FunctionIndefinite integralInputing Values into a Function
Practice by Question Type
Choose your preferred learning style and practice format for fraction problems
Determine whether or not it is possible to create the functionDetermine whether the graph is increasing or decreasingFinding Increasing or Decreasing DomainsIdentifying increase and decrease from a verbal description of the functionIdentifying which domain is applicable for the described functionMatching the graph to the story
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7259 | https://static.kaplanlearn.com/assets/44/63/Unit9_KeyPointReview.pdf | KEY POINT REVIEW Real estate agency relationships are governed by common law, which is established by tradition and court decisions, and statutory law, which is passed by state legislatures and other governing bodies.
An agent is hired by a principal to act on the principal’s behalf. Agency is the fiduciary relationship in which the agent is held in a position of special trust and confidence by the principal. In a real estate transaction, the principal is the client. A customer is the nonrepresented party for whom some level of service is provided and who is entitled to fairness and honesty.
A subagent is authorized to work with the agent on behalf of the client. A nonagent (also known as a facilitator, intermediary, transaction broker, transaction coordinator, or contract broker) assists one or both parties with the transaction without representing either party’s interests and often is subject to specific statutory responsibilities.
The relationship between principal and agent must be consensual. As express agency is based on an express agreement between agent and principal, while an implied agency will be created when the actions of the parties indicate that they have mutually consented to an agency. The source of the agent’s compensation does not determine the agency, because the agent may be compensated by someone other than the client, or the agency may exist even if no compensation is involved—a gratuitous agency.
An agent has a fiduciary relationship of trust and confidence with the principal. The six common-law fiduciary duties can be remembered as COLD-AC, which is an acronym for the following: Care Obedience Loyalty Disclosure Accounting Confidentiality A universal agent is empowered to do anything the principal could do personally.
A general agent represents the principal in a broad range of matters.
A special agent represents the principal in one specific act or business transaction only, under detailed instruction.
A single agency is one in which an agent represents only one party in a transaction. A real estate broker becomes an agent of the seller by entering into a listing agreement for the seller’s property. A buyer’s broker represents a buyer as an agent to find property that meets the buyer’s specifications, as set out in the buyer representation agreement.
A dual agency is one in which an agent represents two principals in the same transaction. Dual agency, where allowed by state law, requires the informed consent of both principals. An undisclosed dual agency, which may occur inadvertently, can result in rescission of the sales contract, forfeiture of a commission, a lawsuit for damages, and possibly license suspension or revocation.
Unit 9 Unit 9 Real Estate Agency 153 A designated agent (or designated representative) is a sales associate authorized by the broker to represent one party to a transaction, while a different sales associate in the same firm represents the other party to the transaction.
Termination of agency may be accomplished by the completion, performance, or fulfillment of purpose of agency; destruction or condemnation of the property; expiration of the terms of the agency; mutual agreement of all parties to the contract; breach by one of the parties, who may be liable for damages; or operation of law, as in the bankruptcy of the principal.
An agency coupled with an interest cannot be revoked by the principal or terminated upon the principal’s death.
Statements to clients and customers should be clearly identified as opinion or fact. Puffing is legal exaggeration of a property’s benefits, while fraud is the intentional misrepresentation of a material fact to harm or take advantage of another person.
A negligent misrepresentation occurs when a real estate professional should have known that a statement about a material fact was false and the real estate professional’s misrepresentation was due to culpable (careless) negligence rather than simple (accidental) negligence.
The seller of residential property may have the duty to disclose any known latent (i.e., hidden) defects that threaten a building’s structural soundness or an occupant’s personal safety. In some states, an agent has an independent duty to conduct a reasonably competent and diligent inspection of the property and to disclose defects to prospective buyers. Disclosure of environmental hazards may be required.
Stigmatized properties may require an agent to consult an attorney. Megan’s Law requires states to make available to the public information about how they can determine where persons convicted of sexual offenses live in the community.
154 Unit 9 Real Estate Agency |
7260 | https://fiveable.me/key-terms/college-bio/pseudopod | Pseudopod - (General Biology I) - Vocab, Definition, Explanations | Fiveable | Fiveable
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General Biology I
Pseudopod
🔬general biology i review
key term - Pseudopod
Citation:
MLA
Definition
A pseudopod is a temporary, foot-like extension of a cell that is used for movement and feeding. This unique structure allows certain protists, such as amoebas, to navigate through their environment by flowing their cytoplasm into these extensions. Pseudopods also play a crucial role in the process of phagocytosis, enabling these organisms to engulf food particles and other cells.
5 Must Know Facts For Your Next Test
Pseudopods are essential for the locomotion of many protists, allowing them to move toward food sources or escape from predators.
In amoebas, pseudopods can change shape rapidly, making them highly adaptable for navigating various environments.
Pseudopods can also help in the process of cell division in some protists, aiding in the formation of daughter cells.
Some protists use their pseudopods to trap and engulf small organisms or particles, showcasing a form of predatory behavior.
The ability to form pseudopods is an evolutionary advantage for certain protists, allowing them to thrive in diverse habitats.
Review Questions
How do pseudopods contribute to the movement and feeding mechanisms of protists like amoebas?
Pseudopods allow amoebas to move and feed by extending their cytoplasm outward to form temporary projections. This enables them to creep along surfaces or swim through water by alternating between extending and retracting these structures. Additionally, pseudopods aid in feeding by surrounding and engulfing food particles through phagocytosis, providing a means for nutrition intake.
Discuss the relationship between pseudopods and phagocytosis in protists.
Pseudopods play a vital role in phagocytosis by allowing protists to capture and engulf solid particles. When an amoeba detects food, it extends its pseudopods around the particle, ultimately enclosing it within a food vacuole. This process not only helps in obtaining nutrients but also showcases how pseudopods enable protists to interact dynamically with their environment, illustrating an efficient means of nutrient acquisition.
Evaluate the ecological significance of pseudopod formation in protists and its impact on their survival in various environments.
Pseudopod formation is ecologically significant as it enhances the adaptability and survival of protists across diverse environments. By enabling movement toward food sources and facilitating prey capture through phagocytosis, pseudopods allow these organisms to thrive in nutrient-scarce conditions. The ability to maneuver through different substrates and fluids contributes to their role as primary consumers in aquatic ecosystems, influencing nutrient cycling and energy flow within food webs.
Related terms
Amoeba:
A type of single-celled protist characterized by its irregular shape and the presence of pseudopods for movement and feeding.
Phagocytosis:
The process by which a cell engulfs solid particles, using pseudopods to form a food vacuole for digestion.
Cytoplasm:
The gel-like substance within the cell membrane where organelles are suspended and where processes like pseudopod formation occur.
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7261 | https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence?srsltid=AfmBOopDVxDnr1KD_MGjZjQA3WoMC-PEnig_zITyLPPsuksg6pscfImT | Art of Problem Solving
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Arithmetic sequence
In algebra, an arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence.
For example, is an arithmetic sequence with common difference and is an arithmetic sequence with common difference ; however, and are not arithmetic sequences, as the difference between consecutive terms varies.
More formally, the sequence is an arithmetic progression if and only if . A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in arithmetic progression if and only if .
Contents
1 Properties
2 Sum
3 Problems
3.1 Introductory problems
3.2 Intermediate problems
4 See Also
Properties
Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let be the first term, be the th term, and be the common difference of any arithmetic sequence; then, .
A common lemma is that given the th term and th term of an arithmetic sequence, the common difference is equal to .
Proof: Let the sequence have first term and common difference . Then using the above result, as desired.
Another common lemma is that a sequence is in arithmetic progression if and only if is the arithmetic mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of arithmetic sequences.
Sum
An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value.
The first is that if an arithmetic series has first term , last term , and total terms, then its value is equal to .
Proof: Let the series be equal to , and let its common difference be . Then, we can write in two ways: Adding these two equations cancels all terms involving ; and so , as required.
The second is that if an arithmetic series has first term , common difference , and terms, it has value .
Proof: The final term has value . Then by the above formula, the series has value This completes the proof.
Problems
Here are some problems with solutions that utilize arithmetic sequences and series.
Introductory problems
2005 AMC 10A Problem 17
2006 AMC 10A Problem 19
2012 AIME I Problems/Problem 2
2004 AMC 10B Problems/Problem 10
2006 AMC 10A, Problem 9
2006 AMC 12A, Problem 12
Intermediate problems
2003 AIME I, Problem 2
Find the roots of the polynomial , given that the roots form an arithmetic progression.
See Also
Geometric sequence
Harmonic sequence
Sequence
Series
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Sequences and series
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7262 | https://web.williams.edu/Mathematics/sjmiller/public_html/math/papers/MSTD_FiniteGroups96.pdf | , MOST SUBSETS ARE BALANCED IN FINITE GROUPS STEVEN J. MILLER AND KEVIN VISSUET ABSTRACT. The sumset is one of the most basic and central objects in additive number theory. Many of the most important problems (such as Goldbach’s conjecture and Fermat’s Last theorem) can be formulated in terms of the sumset S + S = {x + y : x, y ∈S} of a set of integers S. A finite set of integers A is sum-dominated if |A + A| > |A −A|. Though it was believed that the percentage of subsets of {0, . . . , n} that are sum-dominated tends to zero, in 2006 Martin and O’Bryant proved a very small positive percentage are sum-dominated if the sets are chosen uniformly at random (through work of Zhao we know this percentage is approximately 4.5 · 10−4).
While most sets are difference-dominated in the integer case, this is not the case when we take subsets of many finite groups. We show that if we take subsets of larger and larger finite groups uniformly at random, then not only does the probability of a set being sum-dominated tend to zero but the probability that |A + A| = |A −A| tends to one, and hence a typical set is balanced in this case. The cause of this marked difference in behavior is that subsets of {0, . . ., n} have a fringe, whereas finite groups do not. We end with a detailed analysis of dihedral groups, where the results are in striking contrast to what occurs for subsets of integers. Specifically, even though almost all subsets of dihedral groups are balanced as the size grows, more sets are sum-dominated than difference-dominated.
CONTENTS 1.
Introduction 2 2.
Subsets of Finite Groups 3 3.
Sum Dominated sets in Dihedral Groups 6 3.1.
Cyclic Group Preliminaries 6 3.2.
Dihedral Group Case 8 4.
Conclusion 9 References 10 Date: August 9, 2013.
2010 Mathematics Subject Classification. 11B13, 11P99 (primary), 05B10, 11K99 (secondary).
Key words and phrases. More Sum Than Difference sets, sum-dominated and difference-dominated sets, MSTD, sumsets, finite Abelian groups, dihedral group.
The first named author was partially supported by NSF Grant DMS0970067, and the second named author was partially supported by NSF Grant DMS0850577. We thank the participants of the 2012 SMALL REU program, especially Ginny Hogan and Nicholas Triantafillou, as well as Kevin O’Bryant, for helpful discussions.
1 2 STEVEN J. MILLER AND KEVIN VISSUET 1. INTRODUCTION Given a subset S of a group G, we define its sumset S + S and difference set S −S by S + S = {ai + aj : ai, aj ∈A} S −S = {ai −aj : ai, aj ∈A}, (1.1) and let |X| denote the cardinality of X. Notice that we’re writing the group action as addition, but are not assuming commutativity. If we were to write the action multiplica-tively we would still call these the sumset and the difference set, instead of the product and quotient sets, to match the language from earlier work which studied subsets of the integers.
If |S + S| > |S −S| then S is sum-dominant or an MSTD (more sums than differ-ences) set, while if |S + S| = |S −S| we say S is balanced and if |S + S| < |S −S| then S is difference-dominated. If we let the group G be the integers, then we expect that for a ‘generic’ set S we have |S −S| > |S + S|. This is because addition is com-mutative while subtraction is not, since a typical pair (x, y) contributes one sum and two differences.
Though MSTD sets are rare among all finite subsets of integers, they do exist. Ex-amples of MSTD sets go back to the 1960s. Conway is credited with finding {0, 2, 3, 4, 7, 11, 12, 14}; for other early examples see also Marica [Ma] and Freiman and Pi-garev [FP]. Recently there has been much progress in finding infinite families, either through explicit constructions (see Hegarty [He] and Nathanson [Na1]), and existance arguments via non-constructive methods (see Ruzsa [Ru1, Ru2, Ru3] and Miller-Orosz-Scheinerman [MOS]). The main result in the subject is due to Martin and O’Bryant [MO], who proved a positive percentage of subsets of {0, 1, . . ., N} are sum-dominant, though the percentage is small (work of Zhao [Zh2] suggests it is around 4.5 · 10−4).
Almost all previous research on MSTD sets focused exclusively on subsets of the integers, though recently Zhao [Zh1] extended previous results of Nathanson [Na2], who showed that MSTD sets of integers can be constructed from MSTD sets in fi-nite abelian groups. Zhao provides asymptotics for the number of MSTD sets in finite abelian groups. An immediate corollary of the main theorem in [Zh2] is that if {Gn} is a sequence of finite abelian groups with {Gn} →∞then the percentage of MSTD sets is almost surely 0. In this paper we not only extend this result to difference-dominated sets but to non-abelian finite groups as well.
Theorem 1.1. Let {Gn} be a sequence of finite groups, not necessarily abelian, with |Gn| →∞. Let Sn be a uniformly chosen random subset of a Gn. Then P(Sn + Sn = Sn −Sn = G) →1 as n →∞. In other words, as the size of the finite group grows almost all subsets are balanced (with sumset and difference set the entire group).
While Theorem 1.1 shows that in the limit almost all subsets of finite groups are balanced, it leaves open the relative behavior of sum-dominant and difference-dominant sets. Though the number of such sets are lower order and percentagewise tends to zero, are there more, equal or fewer sum-dominant or difference-dominant sets? For example, Figure 1 shows the result of numerical simulations for 10,000 clock groups Z/nZ for n ∈{10, . . . , 100}.
MOST SUBSETS ARE BALANCED IN FINITE GROUPS 3 20 40 60 80 100 20 40 60 80 100 FIGURE 1.
Numerical simulations on the number of balanced, difference-dominated and sum-dominated subsets of Z/nZ for n ∈ {10, . . . , 100}. For each n we uniformly chose 10,000 random subsets of {1, . . . , n}. Top plot is the percentage of balanced, middle is the per-centage of difference-dominated, and bottom is the percentage of sum-dominated.
In Section 3.2 we explore this question for subsets of dihedral groups, and see very different behavior than in the integers. We conjecture that while almost all subsets of the dihedral group are balanced, there are more MSTD sets than there are difference-dominated sets, in sharp contrast to the prevalence of difference-dominated subsets of the integers.
The paper is organized as follows. We first prove our main result for all finite groups in §2. We then explore the MSTD sets of the dihedral group in §3. We end with some concluding remarks and suggestions for future research.
2. SUBSETS OF FINITE GROUPS Martin and O’Bryant [MO] showed that although MSTD subsets of the integers are rare, they are a positive percentage of subsets. MSTD sets in finite groups are even rarer. We will prove that as the size of a finite group tends to infinity, the probability that a subset chosen uniformly at random is sum-dominant tends to zero. Somewhat surprisingly, this is also true for difference-dominated sets. This is very different than the integer case, where more than 99.99% of all subsets are difference-dominated.
The reason the integers behave differently than finite groups is that a subset of the integers contains fringe elements, which we now define. Let S be a subset of In := {0, 1, . . . , n} chosen uniformly at random. The elements of S near 0 and n are called the fringe elements. Interestingly the notion of nearness is independent of n; the reason is that almost all possible elements of In + In and In −In are realized respectively by S+S and S−S; Martin and O’Bryant [MO] prove that S+S and S−S miss on average 10 and 6 elements, while Lazarev, Miller and O’Bryant [LMO] prove the variance is bounded independent of n. Thus whether or not a set is sum-dominant is essentially controlled by the fringe elements of S, as the ‘middle’ is filled with probability 1 and the presence and absence of fringe elements control the extremes. In a finite group, 4 STEVEN J. MILLER AND KEVIN VISSUET there are no fringe elements since each element can be written as |G| different sums and differences, and thus most elements appear in the sumset or difference set with high probability.
In the proof of Theorem 1.1 we reduce certain probabilities to products of Lucas numbers L(n); these satisfy the recurrence L(n + 2) = L(n + 1) + L(n) with ini-tial conditions L(0) = 2 and L(1) = 1. Note this is the same recurrence relation as the Fibonacci numbers F(n), who differ from the Lucas numbers in that their initial conditions are F(0) = 0 and F(1) = 1.
The following lemma is useful, and is in the spirit of calculations from [LMO]. The interpretation will be that the red vertices correspond to elements chosen to be in an S, and the condition that no neighboring vertices are both colored red will ensure that certain elements are not represented in S + S.
Lemma 2.1. Let Cn = {a1, . . . , an} denote a closed chain of n elements (so a1 is adjacent to a2 and an, and so on). If P(n) is the number of ways to color the vertices of Cn red or blue such that no two neighboring vertices are colored red, then P(n) = L(n).
Proof. We derive a recurrence formula for P(n). We may draw Cn as a regular n-gon with the ai’s as the vertices. Let A(n) denote the number of ways a line with n vertices a1, a2, . . . , an can be colored red or blue so that no two neighboring vertices are colored red. We have P(n) = A(n −1) + A(n −3).
(2.1) To see this, there are two cases. Consider the first vertex, a1. If it is colored blue then we may ‘break’ the chain at a1 and the problem reduces to determining the number of ways to color n −1 vertices on a line red or blue so that no two neighboring ones are both red; by definition this is A(n−1). Alternatively, if a1 is colored red then a2 and an must both be colored blue, and thus we are left with coloring n −3 vertices on a line so that no two consecutive vertices are both red; again, by definition this is just A(n −3).
Thus the lemma is reduced to computing A(n), which satisfies the Fibonacci-Lucas recurrence. To see this, consider n vertices on a line, with A(n) the number of ways to color these red and blue so that no neighbors are both colored red. If the first vertex is colored blue, then by definition there are A(n−1) ways to color the remaining vertices, while if the first vertex is colored red then the second must be colored blue, leaving A(n −2) ways to color the remaining vertices. Thus A(n) = A(n −1) + A(n −2).
(2.2) It is easy to see that A(1) = 2 and A(2) = 3, which implies A(n) = F(n + 2), (2.3) where F(n) is the nth Fibonacci number. As P(n) = A(n −1) + A(n −3), we find P(n) = F(n + 1) + F(n −1).
(2.4) As the nth Lucas number satisfies L(n) = F(n + 1) + F(n −1) (2.5) (this can easily be proved directly, or see for example [BQ]), we find P(n) = L(n) as claimed.
□ MOST SUBSETS ARE BALANCED IN FINITE GROUPS 5 We now prove our main theorem.
Proof of Theorem 1.1. We start by showing that the probability a g ∈G = {g1, g2, . . . , gn} is in S + S approaches 1 exponentially fast. For g ∈G, we have P(g / ∈S + S) = P(x / ∈S ∨y / ∈S ∀x, y ∈G s.t. x + y = g).
(2.6) To determine the probability that S + S is not all of G we will add the probabilities P(g / ∈S + S) for each g. Note these probabilities are not independent, as x ̸∈G affects the probability of several g being in S + S.
We concentrate on a fixed g. If x ∈G then there exist a chain of elements {x1, x2 . . . xn} = X ⊆G such that x + x1 = x2 + x3 = · · · = xn−1 + xn = xn + x = g, (2.7) ; clearly the pairs depend on g. Note that X also depends on the choice of x ∈G. If we denote all distinct chains as X1, . . . , Xn then these sets partition G. If S is a subset of G, for g not to be represented in S +S we need at least one element of each pair in each Xi to fail to be in S. The number of ways this can happen is Q L(|Xi|), where L(n) is the nth Lucas number.
To see this equality we use a method similar to that used by Lazarev, Miller, and O’Bryant in [LMO]. Counting the number of subsets of Xi such that we never take two adjacent elements is equivalent to counting the number of ways the vertices of a regular polygon with |Xi| = n vertices can be colored with two colors (say red and blue) such that no two adjacent vertices are blue. Note that each subset S of vertices with this property is equivalent to a set where g ̸∈S + S, and since the Xi partition G, then by Lemma 2.1 the number of such colorings is Q L(|Xi|). Combining the independence of the Xi with Lemma 2.1, we conclude, P(g / ∈S + S) = Q L(|Xi|) 2|G| .
(2.8) For example, take the element a+ b ∈D6 = ⟨a, b|a+ a+ a, b+ b, a+ b+ a+ b⟩, where D6 is the dihedral group with six elements. Here we have that a + b = (a + b) + (a + a + a) = (a + a + a) + (a + b) (2.9) and a + b = (a + a) + (a + a + b) = (a + a + b) + (a) = (a) + (b) = (b) + (a + a), (2.10) where plus denotes the group operation. The two chains we obtain are X1 = {a+b, a+ a + a} and X2 = {a + a, a + a + b, a, b}. Letting SX1 = S ∩X1 and SX2 = S ∩X2 we have that P(a + b / ∈S + S) = P(a + b / ∈SX1 + SX1)P(a + b / ∈SX2 + SX2) = L(2) 22 L(4) 24 , (2.11) where the latter equality occurs because of Lemma 2.1.
6 STEVEN J. MILLER AND KEVIN VISSUET Note that L(n) = φn + (−φ)−n where φ = 1+ √ 5 2 is the golden ratio. As the Xi’s are disjoint, we obtain for each g ∈G that P(g / ∈S + S) = Q L(|Xi|) 2|G| ≤ Q 1.8|Xi| 2|G| = 1.8 2 |G| .
(2.12) As crude bounds suffice, we use the union bound to bound the contribution from each element in G, and find P(|S+S| < |G|) = P(∪g∈Gg / ∈G) ≤ X g∈S+S P(g / ∈S+S) ≤|G| 1.8 2 |G| . (2.13) As the size of the group approaches infinity, we have that P(|S + S| < |G|) approaches zero. The same argument holds for S −S since there is a one to one bijection between group elements and their inverses. Thus most subsets are balanced.
□ Remark 2.2. The above arguments do not apply to subsets of the integers. The reason is due to the lack of a group structure. In particular, the result from equation (2.7) does not hold and different elements have different number of representations as a sum or a difference. For example, for the integers the number of pairs (x, y) ⊂{0, . . . , n −1}2 such that x + y = k is a triangular function of k, peaking when k = n −1. Thus whether or not small (near 0) or large (near 2n −2) k are in the sumset is controlled by the fringe elements of our set. A similar result holds for differences, and thus if the fringe is carefully chosen then we can force our set to be sum-dominant or difference-dominant. Note such forcing arguments cannot happen with a group structure.
Note that we used 1.8 as a very crude bound. While Q L(|Xi|) is much closer to φn then it is to 1.8n, since φ0 is less than L(0), φn does not provide an inequality for all n.
3. SUM DOMINATED SETS IN DIHEDRAL GROUPS Although sum-dominated sets and and difference-dominated sets are rare in arbitrar-ily large finite groups, we can compare the size of the number of sum-dominated subsets and difference-dominated subsets in any fixed finite group. In this section we first ex-plore the sumset and difference set of cyclic groups. We then apply those results to give intuition on why in any dihedral group, there should be more sum-dominated sets than difference-dominated sets.
3.1. Cyclic Group Preliminaries. Before we look at the dihedral group, we explore two different cases in cyclic groups. In the first case we compute the probability of an element missing in the sumset and difference set. In the second case we compute the probability of missing an element in A + B where A and B are both subsets of Z/nZ.
Lemma 3.1. Let S be a uniformly chosen random subset of Z/nZ. Then P(k / ∈S + S) = O (3/4)n/2 .
(3.1) Proof. Let k ∈Z/nZ. Since addition is commutative, all sets of pairs of elements that sum to k partition the group. Furthermore, the number of pairs of distinct elements in Z/nZ is equal to either n/2, n/2−1 or (n−1)/2. The number of distinct pairs depends MOST SUBSETS ARE BALANCED IN FINITE GROUPS 7 on the parity of n and k. From the independence of the pairs of elements that sum to k, we have P(k / ∈S + S) = Y 0≤i≤⌈(n+1)/2⌉ P(i / ∈S ∨k −i / ∈S).
(3.2) Finally, since counting the number of distinct pairs is straightforward, we conclude P(k / ∈S + S) = (1/2)2(3/4)n/2−1 k even and n even (3/4)n/2 k odd and n even (1/2)(3/4)(n−1)/2 n odd.
(3.3) The factor of 1/2 is due to the number of elements x ∈Z/nZ such that x + x = k.
Again, the number of these elements depends on the parity of n and k.
□ Lemma 3.2. Let S1 and S2 be uniformly chosen random subsets of Z/nZ. Then P(k / ∈S1 + S2) = (3/4)n.
(3.4) Proof. Let k ∈Z/nZ. The claim follows immediately from the fact that P(k / ∈S1 + S2) = Y 0≤i≤n−1 P(i / ∈S1 ∨k −i / ∈S2) (3.5) and the fact that these n products are mutually independent.
□ Lemma 3.3. Let S be a uniformly chosen random subset of Z/nZ. Then P(k / ∈S −S) = L(n/d)d 2n = O ((φ/2)n) , (3.6) where gcd(k, n) = d, L(n) is the nth Lucas number, and φ is the golden ratio.
Proof. Let k ∈Z/nZ. Since the order of k in Z/nZ is equal to n/ gcd(n, k), if we have a set {x1, x2, . . . , xm} such that x1 −x2 = x2 −x3 = · · · = xm −x1 = k then m = n/ gcd(n, k). These sets partition the group and thus, the number of subsets of Z/nZ that satisfy this property is gcd(n, k). Combining the fact that these sets have a pairwise trivial intersection with Lemma 2.1 we have P(k / ∈S −S) = L(n/d)d 2n , (3.7) as desired.
□ Lemma 3.4. Let S1 and S2 be uniformly chosen random subsets of Z/nZ. Then P(k / ∈S1 −S2) = 3 4 n .
(3.8) 8 STEVEN J. MILLER AND KEVIN VISSUET Proof. The proof follows immediately from the following equalities: P(k / ∈S1 −S2) = Y x∈Z/nZ P(x / ∈S1 ∪x −k / ∈S2) = Y x∈Z/nZ (1 −P(x ∈S1 ∩x −k / ∈S2)) = Y x∈Z/nZ (1 −P(x ∈S1)P(x −k / ∈S2)) = 3 4 n .
(3.9) □ Proposition 3.5. Let S be uniformly chosen random subsets of Z/nZ then as n ap-proaches infinity P(|S + S| = |S −S| = n) approaches 1.
Proof. This is immediate from the union bound and Lemmas 3.1, 3.2, 3.3 and 3.4.
□ 3.2. Dihedral Group Case. Let S be a subset of D2n = ⟨a, b|an, b2, abab⟩chosen uniformly at random. We first give a proof for the dihedral group subcase of Theorem 1.1 by using the previous lemmas. Before we do so we need two results. The first looks at the probability of a rotation element (k = ai) not being in the sumset. The second looks at the probability of a reflection element (k = aib) not being in the sumset. We denote the set of all rotation elements by R and the set of all reflection elements by F.
Lemma 3.6. Let S be a uniformly chosen random subset of D2n and let k ∈D2n such that k = ai. Then P(k / ∈S + S) ≤(3/4)n/2(φ/2)n and P(k / ∈S −S) ≤(φ/2)2n.
Proof. An element of the form ai can be written as a product of two rotations, axay where x + y = i, or the product of two reflections, axbayb where x −y = i. Since the set of rotations and the set of reflections can be viewed as cyclic groups the proofs follow immediately from Lemmas 3.1 and 3.3.
□ Lemma 3.7. Let S be a uniformly chosen random subset of D2n and let k ∈D2n such that k = aib. Then P(k / ∈S + S) ≤(3/4)n and P(k / ∈S −S) ≤(3/4)n.
Proof. Since an element of the form aib can be written as a product of a rotation and a reflection the proof follows immediately from Lemma 3.2.
□ Theorem 3.8. Let S be a uniformly random subset of D2n. Then, as n approaches infinity, P(|S + S| = |S −S|) approaches 1.
Proof. The proof follows immediately from applying the union bound to Lemmas 3.6 and 3.7.
□ Note that by Theorem 1.1 we know that the percentage of sum-dominated and difference-dominated sets goes to zero at an exponential rate. However, if we look at any fixed D2n we conjecture that the number of sum-dominated subsets is greater than the number of difference-dominated subsets. For the first few dihedral groups (up to D16) Figure 2 shows an exhaustive comparison of the subsets of D2n. Figure 2 also includes a sample MOST SUBSETS ARE BALANCED IN FINITE GROUPS 9 FIGURE 2.
Relative number of sum-dominated sets (larger values) ver-sus difference-dominated sets (lower values) in dihedral groups.
statistic for larger dihedral groups. Note that it is hard to continue a complete enumer-ation.
As Figure 2 suggests, sum-dominated sets are more likely to appear than difference-dominated sets. Let S = R ∪F where R is the set of rotations in S and F is the set of reflections in S. From Table 1 we note that the difference in what contributes to the sumsets and difference sets is R −R which contributes to the difference set and F −R and R + R which contributes to the sumset. It is due to this that there are more sum-dominated sets than difference-dominated sets.
Set Rotations in the Set Reflections in the Set S R F S+S R + R, F + F R + F, −R + F S-S R −R, F + F R + F TABLE 1.
How elements contribute to the size of S + S versus S −S.
4. CONCLUSION We have shown that finite groups behave differently than the integers in the sense that almost all subsets are balanced. The reason is that finite groups do not have a fringe. As a result, in finite groups almost all sumsets and difference sets are equal to the entire group. The dihedral group case also hints at the importance of the size of the commutator subgroup and the number of order two elements. It is easy to see that the size of the sumset is greater when the commutator subgroup is small while the size of the difference set is lower due to the greater amount of order two elements.
A natural question to ask is what would happen if we no longer weight each subset equally. When each subset is chosen with uniform probability then the probability of the subset being balanced is equal to 1; however, in Z/nZ, if we take subsets of the first half of the group (i.e., ¯ 0, ¯ 1, . . . ,¯ ⌊n 2⌋) then the sumsets and differece sets behave like they would in Z. Thus, the percentage of balanced groups is closer to 0. It would be interesting to explore where the phase transition occurs.
Another question to ask is what happens when we look at non-abelian infinite groups.
One difficulty is how we approach subsets of infinite groups. For example, if we look at (Z/2Z)|N| we have two different ways to limit the size of the subset. One possibility is to 10 STEVEN J. MILLER AND KEVIN VISSUET require S to be a subset of a finite subgroup. This would allow for an easier computation of the limiting behavior, though we would have to determine the probability it lives in each finite subgroup.
REFERENCES [BQ] A. T. Benjamin and J. J. Quinn, Proofs that Really Count: The Art of Combinatorial Proof, the Mathematical Association of America, Washington, 2003.
[FP] G. A. Freiman and V. P. Pigarev, The relation between the invariants R and T, Number theoretic studins in the Markov Spectrum and in the structural theory of set addition (Russian), Kalinin GOs. Univ., Moscow, 1973, 172-174.
[GO] G. Martin and K. O’Bryant, Many sets have more sums than differences, CRM Proceedings & Lecture Notes 43, American Mathematical Society (Providence, RI, 2007), 287–305.
[He] P. V. Hegarty, Some explicit constructions of sets with more sums than differences (2007), Acta Arithmetica 130 (2007), no. 1, 61–77.
[ILMZ] G. Iyer, O. Lazarev, S. J. Miller and L. Zhang, Finding and Counting MSTD sets (2011), to appear in the conference in the conference proceedings of the 2011 Combinatorial and Additive Number Theory Conference.
[LMO] O. Lazarev, S. J. Miller and K. O’Bryant, Distribution of Missing Sums in Sumsets (2013), Experimental Mathematics 22 (2013), no. 2, 132–156.
[Ma] J. Marica, On a conjecture of Conway, Canad. Math. Bull. 12 (1969), 233–234.
[MO] G. Martin and K. O’Bryant, Many sets have more sums than differences, Additive combina-torics, 287–305, CRM Proc. Lecture Notes 43, Amer. Math. Soc., Providence, RI, 2007.
[MOS] S. J. Miller, B. Orosz and D. Scheinerman, Explicit constructions of infinite families of MSTD sets, Journal of Number Theory 130 (2010), 1221–1233.
[Na1] M. B. Nathanson, Sets with more sums than differences, Integers : Electronic Journal of Com-binatorial Number Theory 7 (2007), Paper A5 (24pp).
[Na2] M. B. Nathanson, Problems in additive number theory. I, Additive combinatorics, CRM Proc.
Lecture Notes, vol. 43, Amer. Math. Soc., Providence, RI, 2007, pp. 263–270 [Ru1] I. Z. Ruzsa, On the cardinality of A+ A and A−A, Combinatorics year (Keszthely, 1976), vol.
18, Coll. Math. Soc. J. Bolyai, North-Holland-Bolyai T` arsulat, 1978, 933–938.
[Ru2] I. Z. Ruzsa, Sets of sums and differences, S´ eminaire de Th´ eorie des Nombres de Paris 1982-1983 (Boston), Birkh ¨ user, 1984, 267–273.
[Ru3] I. Z. Ruzsa, On the number of sums and differences, Acta Math. Sci. Hungar. 59 (1992), 439– 447.
[Zh1] Y. Zhao, Counting MSTD Sets in Finite Abelian Groups, J. Number Theory 130 (2010), 2308-2322.
[Zh2] Y. Zhao, Sets Characterized by Missing Sums and Differences, Journal of Number Theory 131 (2011), 2107–2134.
E-mail address: sjm1@williams.edu, Steven.Miller.MC.96@aya.yale.edu DEPARTMENT OF MATHEMATICS AND STATISTICS, WILLIAMS COLLEGE, WILLIAMSTOWN, MA 01267 E-mail address: kvissuet@ucsd.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF CALIFORNIA SAN DIEGO, LA JOLLA, CA 91941 |
7263 | https://www.youtube.com/watch?v=cxJ3fQXmKdM | Scalar Product | Scalar Triple Product Formula and Proof | How to Find Scalar Triple Product
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In this video, we'll be exploring the Scalar Triple Product, vector triple product application, a fundamental concept in vector algebra that plays an important role in physics, engineering, and mathematics. We'll start by defining what the Scalar Triple Product is and why it's useful.
Next, we'll delve into the Scalar Triple Product formula, which expresses the product of three vectors in terms of their scalar components. We'll also discuss the geometric interpretation of the Scalar Triple Product and how it relates to the volume of the parallelepiped formed by the three vectors.
To deepen our understanding, we'll prove the Scalar Triple Product formula using vector algebra and trigonometry. We'll step through the proof, making sure to explain each step clearly and simply.
Finally, we'll show you how to find the Scalar Triple Product using both the formula and the cross product. We'll work through several examples, including finding the volume of a tetrahedron and checking if three vectors are coplanar.
By the end of this video, you'll have a solid understanding of the Scalar Triple Product, its formula, its geometric interpretation, and how to find it in practice. Whether you're a student learning vector algebra or a professional in physics or engineering, this video will provide you with a comprehensive guide to this important concept. So, let's dive in!
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दोस्तों आज का हमारा टेंपर्ड ग्लास है कि इलेक्ट्रॉनिक किसके बराबर है यह देखिए इसकी फिजिकल सिग्निफिकेंस क्या है को सबसे पहले देखने के लिए उस थे किलर प्रोडक्ट का मतलब यह है कि वेक्टर्स के प्रोडक्ट और वेस्ट अतिरिक्त हमेशा पहले पढ़ चुके हैं फैसले के खिलाफ इसमें भी और बीएसई में हम डायरेक्टरी संख्या प्रॉपर जाएंगे इस किले का मतलब यह है कि वेक्टर्स की प्रोडक्ट शीघ्र नियुक्ति फ्रेगनेट आएगा वहीं किया प्रोडक्ट्स से लेकिन इन रिजल्ट आएगा वह एब्स है और यह क्वेश्चन इस तरीके से भी पूछा जा सकता है कि आप कैलकुलेट करें सब्सक्राइब जरूर करें रिजल्ट काया आज के दिन वेक्टर है टि्वटर हिंदी थियेटर और सीरवी का पुष्प करें अब देखिए इनका आप इसका आंसर है वह अपने चैनल है उस छेद जरूरी है कि पहले किसी को सब्सक्राइब नहीं किए और में डॉक्टर नियुक्त तो यह सब्सक्राइब होगी आधे सब्सक्राइब टो कि यह बराबर होता है कि किसके बराबर होता है अगर मैं इसको किसी को पहले लिख लूं सी वेक्टर डॉटर कि टीवी एक्टर ए कैरक्टर कि यह इसके बराबर होगा और यह इसके बराबर भी हो सकता है कि अगर मैं भी को पहले लिखूं twitter.com 23 सेक्टर कैरेट के वेक्टर तो यह तीर वो जो प्रोडक्ट क्वालिटी और हमे एग्जाम में भी पूछा जा सकता है कि आप सब्सक्राइब अब स्कूल का यह पिंपल नियुक्त तो यहां पर देखिए यह तीनों में लिखा हुआ है यह एग सब्सक्राइब होंगे चीफ subscribe and subscribe the Channel को कि उन्हें इस चीज को प्रूव करना है कि राजधानी है आपस में एक ऑप्शन है करेंगे उसके लिए हैं को सबसे पहले ऐड करेंगे कि लड कि यह एक्टर इस इक्वल टू किसी भी व्यक्ति को उसके यह कंपनी में लिखा जा सकता सब्सक्राइब की क्लासेज ए प्लस के चीफ के इस वीडियो के एंड पर हम इसकी स्टिचिंग और सिगरेट पी जाएंगे इसके डिंपल प्रोडक्ट की फिजिकल फिटनेस टिप्स क्या है तो अब हम पहले इसको क्लिक करेंगे फिर सब्सक्राइब बटन और यह सब्सक्राइब टो हूं तो बीच 70 को यहां पर लेट कर लेते हैं जो भी वेक्टर है वह बीच प्लस बी वजह ए प्लस बी सी उम्र के कि यह 300 बच्चों सैम और आई जैसा कि आप जानते हैं कि यह कर सकते हैं एक्सर्साइज को प्रेग्नेंट करते हैं और सीरियल एक्टर थे सी एक्सप्रेस अखबार प्लस सिक्यूरिटी एडम्स पेसिफिक बराबर है कि अब मुझे कैलकुलेट करनी है पिंपल प्रोडक्ट तो पहले मैं लेट कर लेता हूं पहले मैं बीच रोशनी को कैलकुलेट कर सकता हूं कंप्यूटर कर सकता हूं तो कंसीडर करें पहले इस बीच राशि कि Bigg Boss को बालवीर वाला है कौन सी इसे फोल्ड प्रॉब्लम को कैसे सेट करते हैं यह हमने जानना है फर्स्ट क्लास इसके अंदर हम पढ़ते हैं यह हमें पता है कि हम इसका डिटेल में लिख सकते हैं फ्रेंड्स कैसे लिखते हैं को सबसे पहले भिड़ंत कि आई कोट और फिर इसके बाद है क्योंकि हमारे पास डायरेक्टर है है तो गहने भी व्यक्ति के पॉइंट्स लिखेंगे बीएफ तीव्र के लिए बिजली इसी तरह सीन एक्टर्स ने है तो सी सी सी ई कि अगर मैं इसका डिपार्टमेंट ओपन करूं तो इटरनेट कैसे ओपन करते हैं पहले आइए के ट्रांसफार्मर अगर मैं इसको ओपन करूं तो आए यूनिट वैसे-वैसे अब इसके नीचे प्रॉब्लम है और यह जो है यह हम खत्म कर देंगे और उसके बाद यह वाला हमारे पास इंटरनेट बचाएगा टू डू काटकर है यह SMS सिर्फ बटन है उसके पति को पहले हम रिड्यूस करते हैं Tubelight हुए दें और फिर आप उसको दो बेटों को बड़ी आसानी से कैलकुलेट कर लेते हैं तो मैं यहां पर डायरेक्टर लिखूंगा कि दीवारें और चीज यह आपस में मल्टीप्ल होंगे बी सी डी ई एफ - यह चीफ कि रेड्डी व है तो यह आईएस ट्रांसफरड टो हैव एक्सप्लेंड किया और इसके लिए वे लिए भी अब मैं जाकर ट्रांसपोर्ट करूंगा तो फिर - जग थे सेकंड पहला पॉजिटिव फीडबैक डिफिकल्ट एस - संघ ने सकते हैं यह देखिए नीचे जोड़ना है यह जो कलर है यह खत्म हो जाएगा और यह जो है यह भी खत्म हो जाएगी तो क्या लिखा जाएगा 20180 व्हाय ई ए पी एक्स एंड मिसेज़ थे फिफ्थ एंड सी एक्स हां भाई जान सकते हैं देखिए आई विल प्रोवाइड यू जी आई क्वेश्चन को इरेक्शन आएगी और शीघ्र तो subscribe and subscribe ए माइनस बी वर्क कि एंड cox-2 समय कर दिया बिल्कुल उसी कि यह व्यक्ति अब हम इसको कट कर लेते हैं यह व्यक्ति के साथ 9th लुट tv.in इस बीच राष्ट्रपति कि यह एक्टर यह ट्रैक्टर है इसके आई जैसा कि आप जानते हैं कि कैसे ली जाती है डोंट प्रोडक्ट एसिड करते हुए वाला के साथ हो जाता है बाकि सारे पॉइंट्स हैं के साथ और सब्सक्राइब होने की वजह से जाता है तो यहां पर आप जो है के साथ यह सब्सक्राइब सब्सक्राइब ए सी ई ए माइनर ए बीर एंड आईटी वर्क्स 8 - जैन जगह पर किया जाएगा ए ब्वॉय लुक्ड क्वाइट इस कार्यक्रम में डिवाइड है तो वह इसके साथ खड़े हो जाएगा तो यह के पीछे गैस से ज़ है नेक्स्ट 9 न्यूज से और आधे इंच का यह आपस में हो रहे हैं तो आप ब्लुटूथ पीछे गए मेरा डिवाइस - your friends पे तो सही सी एक्स को एक कि अब यहां पर गौर करें तो आई जे और के लिए को रिप्लेस कर दिया है सिर्फ एग्जाम और अगर ऐसा दोबारा यही लिखना चाहूं तो आइए और एक जगह पर जस्ट रिप्लेस हो जाएंगे एग और तो इसका जो विटामिन प्रभाव का यह इजी सब्सक्राइब ए पी व्हाय ए बी सी डी एडिसन वायलेंस एंड सी द रिसीवर लुट और यहां पर है वेक्टर डॉट विक्रम MP3 मैं यहां तक हमें इस बात का पता चल गया है कि अगर हमने न इलेक्ट्रिक प्रोडक्ट कैलकुलेट करनी है तो सिंपली इन्वेस्टर्स और डिपार्टमेंट में अगर हम रखेंगे तो आधे मिनट के बराबर होता है है अब हम यह बात करनी है कि सीधे क्रूज पे भी बराबर होता है तो हम यहां पर डिपार्टमेंट की जो प्रॉपर्टीज है उसको यूज करेंगे विटामिन सी प्रॉपर्टीज होती हैं कि अगर हम किसी भी दूर हो या किसी भी तो कमेंट को वापस मीटर चेंज करते हैं तो एक माइनस साइन जाता है तो हम यहां पर दो रोटी या दो कालम को चेंज करके देखेंगे और हम यह प्लेटफार्म बनाने की कोशिश करेंगे देखें में कि मैं ठीक करता हूं अगर हिंदू लोगों को आपस में चेंज कर दूं इंदौर उसको मैं तुम्हारे - का साइज का यह क्योंकि मैं दो रोज को चेंज कर रहा हूं इंटरनेट यूज कर रहा हूं कि पोजीशन बीच सुपर लिख दें को बेवकूफ बना एक और अश्लील वेबसाइट्स सी अपनी जगह ली है सिर्फ में नियुक्त वजह से अब इसे सब्सक्राइब और सी की पोज़िशन करूं तो यह अ में बीएफ ऊपर वैसे का वैसे है बी वाइड ए एंड बी सी डी है और फिर सी एडिसन अधेड़ व्यक्ति चला गया तो यह त्यौहार है मैं एक और - का साइन आ जाएगा वह इस - से मिलकर पॉजिटिव क्या इंसिडेंट से अगर मैं फेल्ट टिप पर लिखना चाहूं तो कैसे लिखेंगे जैसे यह है तो पहले एग सब्सक्राइब अब स्कूल तो अगर यह सब्सक्राइब तो यह बराबर है कि dont कौन सी कैरेट के अब आधे कोई वाला लिखा है बी सी डी ए बी सी एयरपोर्ट अर्पण यहां पर भी सी इसके साथ तो देखिए नियुक्त कि अगर मैं दोबारा से योनी करना चाहूं तो देख लेते हैं डॉट ड्रॉप्सी कि इस इक्वल टू कि यह तो हम जानते हैं कि एग्जाम पॉइंट से ज़ बॉयज पेट बराबर होता है है अब हम यहां पर पहले हमें कोशिश रहती थी कि एक को हमने मुक्त किया था अब हम सीपियों करते हैं ऊपर की तरफ वापिस की पॉजिशन पर लेकर जाते हैं पोजीशन पर एक बाघ का यह आज के इस अपनी पोजीशन पर यह एग्री विड ऊ इस बीच दोनों चेंज हो जाएगा का या बीच से चला गया और सबस्क्राइब और यहां पर सबस्क्राइब इन थिस ईयर अच्छा ठीक है यह - का टाइम आ गया व्यक्ति को पर लिखा है सीक्वेंस कि अब सीखो फिर एक तरफ एक और इंटरटेन करने तो वह - और यह - पॉजिटिव हो जाएगा तो जो आंसर आएगा वही रहेगा सी और मिडिंग करेंगे जैसे यहां पर मैंने जो इंटरटेन किया था इस पॉइंट पर मैंने इसको चेंज किया था और पहले सब्सक्राइब करें हम करें हम करें सब्सक्राइब कीजिए अब स्कूल में थे तो ऊपर निकल जाएगा सीरियल किसर ब्वॉय सीइंग थे इससे इड है और भी एक्स डिवाइड एंड पीसी है तो यह है कि डॉटर डिक राठौर अगर यहां से यह कैसे लिखेंगे subscribe like फिर एग्जाम में लिख डू कहीं पर भी लगा सकते हैं इससे के अवसर पर भारत नहीं आएगा तो देखें यह सीईए यह वाली पवन की सीधे क्रश बी तो आप इसको इक्वेशन वन कहते हैं है और इसको क्वेश्चन तो कह दें फ्रॉम इक्वेशन कि वह कि एंड इक्वेशन टू क्वेश्चन मार्क क्वेश्चन दूसरा यह देख सकते हैं कि जो ए डाउट कि Bigg Boss स्पीड है ब्लुटूथ यह बराबर है इसके बराबर है मैं तुमसे यह के बराबर है 38 फ्रांस पिछले और क्वेश्चन वंश यह चीज है यह बराबर है कि डोंट ए सी प्राइस अपडेटेड ओं है तो यह हमारा रिक्वायर्ड डेढ प्रूफ हो गया अब बात करते हैं फिजिकल सिगरेट या चाहिए इसकी फिजिकल एविडेंस है और जैसे मैंने पहले बताया कि फिजिकल सकें और सबसे इंपोर्टेंट पॉइंट होता है किसी भी आर्टिकल का फिजिकल सिगरेट कैंसर है लुट लुट लुट हु इज द सिग्निफिकेंस पेट कैसे इसके लिए है कि इसकी मदद से हम वॉल्यूम कैलकुलेट कर सकते हैं टेलर स्विफ्ट का टैलेंट वॉल्यूम कैसे अपडेट कर सकते हैं किसी भी चीज वॉल्यूम किसके बराबर होता है पैनल जो है वह एक ही है लेकिन उसकी साइड है तो किसी चीज को अगर सब्स्क्राइब करते हैं और देखते हैं तो यह घृणित इस तरीके से प्लेट किया जा सकता है तो यहां पर मैंने सेट करना है पर मेरे पास सेव है किए हुए को है यह शर्त है और इसकी अप का अगर मैंने वॉल्यूम कैलकुलेट करना है सर्च करेंगे यह जो साइड है इस दौरे की यह साईट और यह साइड इज्जत इनसाइड यह तस्वीर वह कौन-कौन से सेट करें कि यह घृणित और सीरवी कि अगर में इसके वॉल्यूम कैलकुलेट कर रहा है तो वॉल्यूम इस प्रकार से कंप्लीट किया जा सकता है कि पहले इसकी जो बेस इसका एरिया मालूम कर लिया जाए और फिर इसकी हाइट से प्रॉपर्टी प्लस साइज जो है इसे मल्टीप्ल कर दिया जाए लेकिन जब अ पर्टिकुलर है वह इसी के बराबर होगी वह हम कैलकुलेट कर लेते हैं कि उसकी जो परपेंडिकुलर हाइट है वह किस के बराबर होगी बी मुक्तिपथ सी और बी और सी के चेयरमैन में अगर प्रेस प्रेस करूं तो इससे जो रिजल्ट आता है वह बी व सी दोनों के ऊपर पर्टिकुलर होता है और बीएससी इन दोनों के कुछ स्पेसिफिक लो है वह यह है यह घ्र नेटवर्क आ जाता हूं कि बीडीओ एंड केंद्र में जो एंगल है दूसरी ओर केंद्र में जो एंगल है उसको मैं ठीक उलट कर लेता हूं हैं अगर यहां से मैं नोट करूं कि के अगर मैं बीच क्रॉस करता हूं तो यह नीचे जो है प्रोग्राम है पैनल ग्राम पर जो एरिया होता है वह बीच राशि होता है यह हम ऐप राशि में बैठे हैं कि दो व्यस्क तोड़ कर लिया जाए तो इसकी जो मैग्नीट्यूड होती है वह मैग्निट्यूड लोग राम सब्सक्राइब टो क्या हुआ अब यह जो एरिया है इसको हाइट से अगर हम मल्टीप्लाई कर दें तो वह लिया जाएगा अब उसकी हाइट नहीं किया कि वह आप इसके बराबर होगी हाईट सी वेबसाइट नहीं है अ हाइट क्योंकि वह अ पर्टिकुलर लेंथ होती है वह पर्टिकुलर लेंस है वह यह वाली है कि यह एडवांस कि अभी हाईट किसके बराबर होगी यह वाला जो पॉइंट है यह इसके इसका पॉइंट के बराबर है जो इस तरफ है इसके साथ और यह जो पॉइंट होगा इसका वह component1 चौक स्थित है एंगल के साथ को जो कमांड होता है वह ए कंप्लेंट एक हॉस्पिटल के बराबर होगा तो कि यह वाला वेक्टर वेक्टर है और इसका यह वाला जो पॉइंट होगा वह एक और पिता के बराबर होगा यहां से अगर मैं वॉल्यूम कैलकुलेट करूं इसका वॉल्यूम यह उसकी हाइट है और यह जो अ लुट अ हाइट मल्टीप्लाई वेल्थ एरिया थे जो वाइट है वह बराबर है एक हॉस्पिटल के और जो एरिया है वह बराबर है बेबी जिगरा इससे पिछले महीने यह मैग्नेट उन्हें कि अगर डॉक्टर है तो वे में डाउनलोड ओपन करो वह कार्तिक आता है और अगर मैं इस वास्तविकता को अब अगर मैं इस पावन सीता को बात करूं तो जो वॉल्यूम आएगा यह क्रॉस का योग नॉट अट ऑल मेरे प्रिय वॉल्यूम है पहलुओं है आज से दूसरी ओर फिजिकल चिकेन सकती है कि अगर जो दूसरी सीट फिजिकल सीखने फैंस है वह यह है कि इस थे हेड क्वाटर अश्लील अगर इक्वल टू जीरो आ जाइए नित्य पॉइंट होता है कि उसका चेक पॉइंट होता है कि अगर वेक्टर्स प्लेयर है या नहीं है अगर तो ऐड भी प्रश्न का आंसर जीरो आ जाए तो इसका मतलब यह है कि यह जो तीनों बैठकर चाहिए को लेकर है प्लेन का मतलब स्कूल में वीरवार को तो इसका मतलब यह है कि यह जो तीसरे सब्सक्राइब नहीं है यह सीधे ही Play और सब्सक्राइब वह तो यह ट्रैक |
7264 | https://www.intmath.com/functions-and-graphs/what-is-the-converse-of-pythagoras-theorem.php | What is the Converse of Pythagoras' Theorem?
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Introduction to Geometry
Introduction to Geometry
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Introduction to Geometry
Functions and Graphs
1. Introduction to Functions
2. Functions from Verbal Statements
3. Rectangular Coordinates
4. The Graph of a Function
4a. Domain and Range of a Function
4b. Domain and Range interactive applet
4c. Comparison calculator BMI - BAI
5. Graphing Using a Computer Algebra System
5a. Online graphing calculator (1): Plot your own graph (JSXGraph)
5b. Online graphing calculator (2): Plot your own graph (SVG)
6. Graphs of Functions Defined by Tables of Data
7. Continuous and Discontinuous Functions
8. Split Functions
9. Even and Odd Functions
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What is the Converse of Pythagoras' Theorem?
Pythagoras' Theorem is one of the most important and widely used theorems in geometry. It states that in any right triangle, the square of the hypotenuse (the side opposite to the right angle) is equal to the sum of squares of two other sides. In other words, c2 = a2 + b2. But have you ever heard about its converse? Let's find out what it is.
The Converse of Pythagoras' Theorem
The converse of Pythagoras' theorem states that if a triangle has sides such that the square of one side is equal to the sum of squares of other two sides, then it must be a right triangle. Mathematically, this can be written as a2 + b2 = c2 → it’s a right triangle. This means that if we know all three sides of a triangle and their squares’ sums are equal then it must be a right-angled triangle.
Applying the Converse
The converse to Pythagoras theorem can be used to prove whether a given triangle is right-angled or not, without actually measuring any angles or lengths. All we need are three side lengths – one side length and two adjacent lengths - and compare them with each other using the formula stated above. If they do match, we can conclude that the given triangle is indeed a right-angled one; otherwise not. We can also use this theorem to solve word problems related to triangles in geometry by finding out whether they are possible or impossible scenarios for forming a certain shaped triangle when given its dimensions or angles.
Conclusion
In conclusion, knowing about both Pythagoras' theorem and its converse plays an important role in understanding more complex geometry topics like trigonometric ratios, circles and transformations. Both these statements can easily be used to solve various types of problem based on triangles, which makes them essential for students studying geometry at any level. So make sure you understand these concepts thoroughly before moving forward with your studies!
FAQ
What is the converse of Pythagoras Theorem?
The converse of Pythagoras' Theorem states that if a triangle has sides such that the square of one side is equal to the sum of squares of other two sides, then it must be a right triangle. Mathematically, this can be written as a2 + b2 = c2 → it’s a right triangle
What is a converse theorem in geometry?
A converse theorem in geometry is a statement that follows from the original theorem but with the hypothesis and conclusion switched. For example, the converse of Pythagoras Theorem states that if a triangle has sides such that the square of one side is equal to the sum of squares of other two sides, then it must be a right triangle.
What is the definition of Pythagorean theorem?
Pythagoras' Theorem states that in any right triangle, the square of the hypotenuse (the side opposite to the right angle) is equal to the sum of squares of two other sides. In other words, c2 = a2 + b2. It is named after the Ancient Greek philosopher and mathematician Pythagoras.
What is the Pythagorean Theorem converse in your own words?
The Pythagorean Theorem converse states that if the square of one side of a triangle is equal to the sum of squares of two other sides, then the triangle must be a right triangle. Mathematically, this can be written as a2 + b2 = c2 → it’s a right triangle.
It is the inverse of the Pythagoras Theorem.
What is the formula for Pythagorean theorem?
The formula for Pythagoras' Theorem is c2 = a2 + b2, where c is the length of the hypotenuse (the side opposite to the right angle) and a and b are lengths of two other sides. This theorem can be used to calculate the length of any side in a right triangle when given two other sides. It is named after the Ancient Greek philosopher and mathematician Pythagoras.
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7265 | https://www.atsdr.cdc.gov/hac/hair_analysis/hairanalysis-appC.pdf | APPENDIX C Pre-Meeting Comments Hair Analysis Panel Discussion: Exploring the State of the Science Pre-Meeting Comments June 6, 2001 Notice This booklet includes the panelists’ pre-meeting responses to the charge questions. It should be noted that the pre-meeting comments are preliminary in nature. The purpose of these comments is to stimulate meeting discussions. Some panelists’ technical findings might change based on discussions during the meeting; therefore, pre-meeting comments should not necessarily be considered the panelists' final opinions.
Any mention of trade names or commercial products does not constitute endorsement or recommendation for use.
Table of Contents Panelist Comments LuAnn White, Chair . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-1 Robert Baratz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-5 Thomas Clarkson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-19 Michael Greenberg . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-29 Michael Kosnett . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-39 Dan Paschal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-47 Sharon Seidel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C-57 Note: Comments have been printed as received C-1 LuAnn White, Chair LuAnn White C-3 Overview/Perspective from the Panel Chair The overriding question for the use of hair analysis in environmental public health is the need to find reliable methods for assessing chemical exposure of people living in communities near hazardous waste sites. Hair sampling is tantalizing because it is a biological material that is readily available, noninvasive, and easy to collect. However, much controversy exists regarding the use of hair samples as an indicator for environmental exposure, health status, or disease state. The use and misuse of results from hair sampling has stimulated debate and at times, cast a shadow over the issue.
Complex questions linger regarding three overarching issues: 1) accuracy and reliability because of laboratory methods; 2) toxicokinetics of compounds and the biological variability among individuals; and 3) the relationship of the results to exposure and/or potential disease. Within each of these issues, multiple questions arise that include, but are not limited to, the reliability and reproducibility of the analytical methods; interlaboratory variability; types of compounds suitable for hair analysis; baseline of elements and compounds found in hair—for an individual and/or populations; influence of distribution, metabolism, storage and excretion on incorporation of compounds and elements into hair; and duration and level of exposure. Even if all of the methodological and toxicological questions can be answered, there are still great gaps in our knowledge as to the relationship between the concentration of a compound/element in hair and environmental exposure, and then between exposure and disease or reduced health status. Indeed, a lack of knowledge of these complex interrelationships exists with any biological sample and prevents full answers to many questions.
While there is much we do not know, there is a body of knowledge on hair analysis. The challenge is to define the parameters whereby hair analysis can be a valuable tool to assist in exposure investigations, but to guard against overinterpretations beyond our knowledge and experiences.
Perhaps, identifying the issues will open the door to stimulate research to answer questions and fill our knowledge gaps.
C-5 Robert Baratz Robert S. Baratz C-7 Hair Analysis By Robert S. Baratz, MD, PhD, DDS Introduction When physicians study a disease or process, they look for ways to evaluate that process in the body. Blood and urine are taken for testing because they are easily obtained and can be readily standardized. Normal values for populations can be set easily with such testing. Blood generally represents what is inside the body, and urine represents what is excreted from the body.
Hair testing has very limited usefulness in medical practice, because it does not represent either the tissues inside the body or what is excreted. Hair analysis is only useful for detecting exotic compounds that are not normally found in the body. Thus, for example, a medicine that someone is taking, might be detected in the hair. Poisons, such as arsenic, also show up in the hair.
Elements normally found in the body -- such as copper, chromium, zinc, and even lead, mercury, and uranium-- will show up in the hair, but the levels are quite variable and have little or no practical or clinical significance.
Analysis of hair won’t tell you about the source of an element found in the hair. Most minerals obtained by the body come from food or water. Foods are grown all over the country and thus, their constituents, more likely than not, have come from another region, in some cases, from another country. People are more commonly drinking bottled water and juices which also are coming from other regions. Thus, finding something in the hair or body in no way indicates the source of the material. This is especially problematic when dealing with elements that are somewhat ubiquitous in the environment. The most common source of lead, for example, can be from the solder joints of household plumbing. However, lead could also be introduced through any number of foods, and/or beverages. In some cases, even unglazed pottery used for serving food can be a source of lead contamination. When hair analyses have been done rigorously in quantitative laboratory settings, it has been pointed out that great care must be taken to avoid possible sources of contamination. First, the hair itself must be processed in a uniform fashion to avoid introducing any exogenous material.
Metal cutting instruments in sampling hair should be avoided. The hair sample must be standardized as to region of the scalp, length from the scalp and any washing done of the hair during processing. Even so, because hair grows at different rates in different people, there is still a great deal of uncertainty regarding even hair obtained close to the scalp. Many contend that such hair is of more recent vintage and thus more “representative” of the “body composition”. There are no data, however, that confirm this idea. Hair seems to grow at an average rate of about 1 cm per month. However, a considerable portion of the hair shaft lies within the skin and thus hair that has been sampled that has already grown out represents hair that may be as many as several months old. Robert S. Baratz C-8 Even if hair analysis was a highly reproducible and valid test, done properly, there are virtually no data for correlation of findings with levels of elemental minerals found in other tissues or organs.
Given the element of interest in the Colorado plateau region, it is important to point out that radioactive compounds from tailings are unlikely to go into hair. The agent of most interest is radon which ends up in the body as lead. Because radon exists primarily as a gas, the major organ that is affected is the lung. This reviewer is unaware of any studies that correlate amounts of material found in lungs with amounts found in hair. As hair is handled in the laboratory, a number of possible contaminants can enter the hair from solutions used in processing. Rigorous care must be taken to check each and every reagent used in the laboratory. Even acid reagents can have significant amounts of trace elements within them when parts per billion are at issue. Water used in the laboratory for washing, hand washing or even wiping down counters may contaminate samples. Use of vaporization techniques such as atomic absorption spectroscopy can release agents into the laboratory air which would then end up contaminating other samples or solutions or both. Laboratory dust must be excluded since it too can act as a source of contamination. Even powder used to cover gloves of laboratory workers can result in significant contamination of the laboratory environment. Analyses are often done at the level of parts per billion and it takes very little contaminating material to change findings dramatically. Many laboratories that handle hair fail to take into account that exogenous contaminants such as hair shampoos, swimming pools, shower water and the like can all add exogenous agents to hair.
These include: selenium, bromine, zinc, copper and even arsenic. Some elements are removed by pre-washing before hair analysis. Acetone, a common washing agent, has been shown to remove sodium, bromine, and calcium. The same solution is known to add copper, iron, manganese, zinc, and mercury. Even the pH of washing solutions can effect the amounts of lead, mercury and cadmium found in hair samples.
Topic I – Analytic Methods Types of Analytical Methods The principal means for analysis of hair depends on the object of the analysis(es). Simple small molecules such as trace minerals, can be analyzed either using atomic absorption spectroscopy or mass spectroscopy. Analysis for organic compounds would depend on the specific compound being tested. Some of these methods are exquisitely sensitive and small quantities of contaminants found in laboratory air from vaporization, dust, or coatings on lab-ware and/or contamination of test solutions can significantly affect results. Robert S. Baratz C-9 For trace minerals, results are commonly in the range of parts per billion, or smaller. Thus, exquisite attention to detail and lab cleanliness must be followed. Reliable analytical methods exist for detection of most trace minerals, however, quantification becomes an issue particularly when multiple overlapping peaks with spectroscopy occur. Moreover, sampling errors, and the nature of the starting material, often inhibit precise quantification. More commonly, qualitative results can be observed reliably.
For reasons to be discussed in later sections numeric quantitative evaluation of trace mineral substances is not clinically, forensically, or for matters of industrial hygiene, useful. This is largely due to the fact that normal ranges have not been established, cannot be established, or are irrelevant. A finding of an exotic substance that is never normally present is significant. Similarly, a change in order of magnitude of a trace substance that is normally present, and may, at high doses, be poisonous, often has clinical and other relevancies. Hair analysis has been shown to be quantitatively useful for the detection of arsenic, and methyl mercury. Other validated uses of hair analysis are for the finding the presence of drugs of abuse or the presence of certain pharmacological agents. In this reviewer’s experience, commercial laboratories, as opposed to research laboratories, have been observed to have considerable variation in their performance. This variability is the result of inconsistent specimen preparation, source, and handling, inconsistent use of standards, and lack of multiple runs of the same material. Typically, only single samples are run and thus any variability within the laboratory and/or method are often unknown. Where multiple samples have been observed from the same laboratory on the same material, wide variations have been shown to exist. In the case of commercial laboratories, interpretation of results suggests that results are often misleading, inappropriate, and lack sufficient information to make them useful. Two reports by Barrett ( 1985) and Seidel and colleagues (2001) show that, at least in the case of commercial laboratories, reference ranges, results and interpretations vary considerably from laboratory to laboratory. This is not surprising considering the milieu in which this work is done, and the factors described above.
Topic II- Factors Influencing the Interpretation of Analytical Results Regionally, there can be marked differences in elemental composition of hair even for the same element. For example, in 16 different regions of the scalp, antimony content was shown to vary considerably. Even with a person with a standardized diet and living conditions, the composition of hair at different distances out from the scalp itself can vary. This has been shown, in particular, for copper and zinc. Additional problems in doing hair analysis show that there are difficulties in trying to measure more than two or three elements at the same time. In atomic absorption spectroscopy, one of the Robert S. Baratz C-10 more common methods for hair mineral content analysis, many elements give off multiple peaks which overlap. These absorption peaks obscure each other, negating the ability to do quantitative analyses accurately.
There is often a lack of precision and standardization in the amount of hair taken from any particular subject. Unless a uniform sample was taken from which all analyses were done, the validity of the analysis can be called to question. Racial differences among subjects have also been found. There is considerable variability in calcium, iron, nickel, chromium, manganese, arsenic and lead levels between Caucasian subjects and blacks. Similarly, age is a significant factor in metal composition of hair. Paschal and co-workers (1989) found marked differences in concentrations of 28 different metals in hair samples of 199 children compared with 322 adults. Age-dependent increases in calcium, barium, magnesium, zinc and strontium all occur up to about 12-14 years of age. Aluminum is shown to decrease with age. It has been hypothesized that metal composition of hair is related to skeletal and bony growth. Thus, adults undergoing osteoporosis would have differences in their hair composition related to those who did not have this problem. Similarly, anyone with any kind of bone abnormality would have findings that are non-standard. Most analyses on hair do not correlate positively with concentrations found in organs (Yoshinaga and co-workers 1990). It is intuitively obvious why this is likely so, since tissue concentrations involve both uptake and release which vary over time. Hair is essentially a one-way path out of the body. Likewise, some elements have significant diurnal variation. A good example is chromium (Sheard and co-workers 1980).
Many elements, when analyzed in the presence of other elements, can give false readings. The interaction of chromium with other anions and cations in hair may affect analytical results (Sheard and coworkers 1980). Merely painting the laboratory with particular types of paint, failure to use HEPA filters on the air intake and the presence of dust can easily affect sensitive analytical measurements. A variety of hair treatments have been shown to alter hair trace element concentrations.
(McKenzie, 1978). Other common issues are dyeing and permanent waving, shampooing, hair color, sex, seasonal variations, age, and growth rates. It is generally assumed that hair grows approximately 1 cm per month, however this must be verified in each individual tested. Hair growth is a function of individual factors as well as protein in the diet.
Many commercial laboratories claim to be able to detect and measure more than 20 elements in a single sample of hair, however, this is often accomplished without any specific knowledge of the patient’s medical history. What is more troubling, is that there is no definition of a normal range Robert S. Baratz C-11 (Hambridge 1982, Rivlin 1983; Manson and Zlotkin 1985, Barrett, 1985, Druyan and others 1998 and Seidel and others 2001;). Quality control in the laboratory is essential towards having useful data. Rigorous attention to detail, methodology, and sampling techniques must be followed. Even when known standards were used, because of the sensitivity of instrumentation, data varied commonly by up to 10%. (Nowak and Kozkowski 1998). Variations in Sample Collection and Preparation Methods A number of compounding variables limit interpretation of results from hair analysis. It is well known from the literature that the rate of hair growth varies from person to person, with nutritional and disease states, with the presence of particular drugs, with gender, age, ethnicity/race, with site on the scalp and/or other body parts. While some of the factors may be known in an individual case, others are unknown, or cannot be known. Thus, hair analysis from a particular individual is fraught with a series of uncontrolled variables and unknown data. It should be obvious that these belie making any precise quantitative diagnostic or forensic analysis. This becomes even more of a problem when dealing with trace minerals that are normally found ubiquitously in the environment and characteristically in foods, water, and air. Many trace minerals occur in human hair normally. Thus, finding them there is expected. Making interpretations based upon the quantitative analysis of these is fraught with uncertainty due to the unreliability of the data regarding exposure, timing, hair growth, treatment of the hair, diet, nutrition, and a host of other factors mentioned above. Similarly, considerable variation exists from laboratory to laboratory in terms of sample preparation, whether a sample is washed, how it is digested, how long it is digested, and how it is handled after digestion. Even more problematic is the development of “normal ranges” or “reference standards” (“reference ranges”). In most cases, population standards have not been developed. Thus, each laboratory has developed its own “reference range”. The major problem with this is that the source of the specimens used to create the “reference range” in a particular lab may be biased.
Many commercial laboratories accept samples from a variety of practitioners, patients, and other sources. From reports this reviewer has seen, the precision of medical knowledge and facts regarding the source material is often poorly documented. Careful attention to uniform sample collection techniques is often also a problem.
So-called “normal” reference ranges do not exist for most trace minerals found in hair. The reasons for this are obvious. Considerable variation exists from person to person and the variety of unknown variables enter into the equation. Thus, there are no standardized “reference ranges” for most normal trace minerals. This has to do, in part, with the composition of hair. In essence, hair consists of keratinized or cornified cells packed into tight arrays in the hair shaft. These cells are fundamentally similar to the epidermis however contain proportionally more keratin fibrils and somewhat different materials in the thickened cell membrane that is left when the cells keratinize.
Robert S. Baratz C-12 Hair, nails, horn, and some portions of the filiform papillae of some animal tongues are a so-called “hard” keratin compared with so-called “soft” keratins found in epidermis and oral and other, mucosae. All keratinized cells contain virtually no aqueous phase after keratinization. It is unclear if minerals are removed from these cells when they mature, or merely remain in the intracellular matrix. Along these lines, even if some minerals were found to be left “inside” such cells, each and every individual trace mineral would have to be studied in pre-keratinized cells and keratinized cells to see how it was handled. Moreover, epithelium and its derivatives (hair) is an a vascular tissue with little intercellular space or material. What little extracellular material exists is primarily a lipid that forms a barrier to diffusion. The epithelium is neither a gland nor excretory organ but merely forms a protective layer. Thus, substances that would normally be excreted into various body fluids are normally not present in epidermal epithelium. Regionally, there is variability in the thickness of the epidermal epithelium and indeed there is some variability in the consistency of thickness of hair in different regions of the scalp. Hair in other body locations, axillary, pubic, limb, peri-anal, eyebrows, and eyelashes, all vary considerably in their structure, function, and growth rates.
Since hair is principally protein in nature, there is little need for trace minerals in the hair cells themselves. Trace minerals in the body are usually present as co-factors for enzymes. Keratinized cells are generally non-metabollic. After the filaments of keratine aggregate and are coated by other proteinatious material, the cell contents become essentially inert. Nuclear materials, enzymes, carbohydrates, and even lipids are essentially not present in the internal milieu of keratinized cells. Consequently, there would be no need for a regular array of minerals present from a functional point of view.
Some heavy metals may distribute into hair and become complexed with hair proteins. This would be due largely to interaction with free side chains on amino acids and/or forming crosslinks among protein chains as they may be denatured by heavy metals. Some heavy metals are well known as protein denaturants, e.g. mercuric chloride. They may become trapped in hair cells before they become completely keratinized. Whether or not this happens is largely unknown.
Finding trace minerals in hair is neither surprising nor a consistent finding. Because hair shafts consist of essentially of two portions, intra-epithelial and extra-epithelial, the possible absorption of extraneous material is possible in the extra-epithelial portion. The extra-epithelial portion is essentially free in the environment. Thus it is subject to washing, drying, chemical alteration, cosmetics, environmental pollutants present in the water or air, and a host of other chemical and physical insults. Not only may things be adsorbed and absorbed by the hair, but, substances may also be leached from the hair as well. Prolonged immersion and wetting of hair can cause some swelling of the cells of which hair is composed. This can diminish the barriers to diffusion of things both from the outside in, and from the inside out. Moreover, hair is being constantly exposed to scalp oils, and other glandular products excreted into the hair shaft space by sebaceous Robert S. Baratz C-13 and other glands present in skin. These provide an additional source of extraneous material to be adsorbed or absorbed onto or into the hair.
A common and highly variable factor in hair is hair growth. Hair growth can occur in several different ways. First, is the fact that hair undergoes a cycle in its normal growth. That is, hair is regularly shed from the scalp and other locations, and replaced by “new hairs”. The stages of the growth (catagen, anagen, telogen) each have unknown times associated with them in particular subjects. Further complicating an understanding of growth is the fact that hair in humans is known to grow in a mosaic across the scalp. That is, any particular hair may be in a different state than its neighbors. A long list of drugs, hormones, and other factors can either accelerate or prolong the time a hair stays in a particular part of its growth cycle. Moreover, a number of other factors such as diet, nutrition, age, sex, hair color, and other factors are known to influence growth rates. Growth can occur both longitudinally and in diameter. Hair in general varies from individual to individual in shape and not all individuals have a circular cross section of their hair. In particular, individuals with highly curled or “kinky” hair have hair that is somewhat flattened to a ribbon-like shape. In general, hair growth in length is often described at approximately 1 cm per month. However, there is considerable variation in this from individual to individual and results can vary by a factor of 2 either in increase or decrease in rate of growth.
Topic #3: Toxicologic Considerations As previously mentioned, among the mineral toxic agents studied only arsenic and methyl mercury have been shown to have reliable information on their presence and distribution in hair when viewed in comparison to their distribution in other organs.
To have predictive value, the values obtained from analysis of hair of a particular subject must be capable of yielding data that would be predictive for disease in general. This may prove to be considerably problematic in the case of heavy metals as the agents themselves may affect hair growth directly.
This reviewer is less well versed in arsenic and methyl mercury studies than others on the panel and wishes to defer to their knowledge and experience.
Topic #4: Data Gaps and Research Needs Many of the data gaps in our knowledge of hair physiology and growth have been discussed earlier.
Robert S. Baratz C-14 In one sense each trace mineral must be independently studied with regard to the best source for analytical material. In most cases it is likely that hair will prove to be a problematic source. While hair theoretically gives a longitudinal history of prior events, the speed of that history is largely unknown, and may even change over time. Whether this theory meets practice is also unknown. A better understanding of the physiology of hair growth is obviously an important area of research.
This, of course, begs the question that hair may be useful at all for mineral analyses. This later point is yet unproven. A variety of data would suggest that hair is not useful for mineral analyses for most minerals, and that other body sources would be better—e.g. bone, teeth. Knowledge of the dynamics of incorporation of a variety of environmental toxins, principally organic compounds, into hair would be desirable. Attendant to such a study would be studies of absorption, adsorption and leaching of such compounds. Studies of the nature of differences in incorporation of materials into hair at different ages, by different sexes, different ethnic groups, and different hair colors would also be useful.
Topic #5: Identifying Scenarios for Which Hair Analysis May Be Appropriate Hair analysis appears useful only for population studies where much of the individual variability can be eliminated. If a number of factors were known—duration of exposure, rates of incorporation into hair, effects on growth, amounts of leaching, sources of material that were found in hair, etc.—then useful data on exposure could be extracted. Correlating these with clinical findings is more problematic, since such are best done on the individual level, where hair analyses are likely more useful only for population studies.
Particularly for small molecules such as trace minerals, hair is unlikely to prove a reliable source of material for meaningful study. Organic compounds that can be shown to incorporate into hair may be an area where hair analysis could be appropriate for following exposures to environmental toxins.
Robert S. Baratz C-15 Selected References Anderson RA et al. “Designing a Biological Monitoring Program to Assess Community Exposure to Chromium: Conclusions of an Expert Panel” Journal of Toxicology and Environmental Heath, 40:555-583, 1993.
Barrett S. “Commercial Hair Analysis: Science or Scam?”; JAMA; 254(8): 1041-1045; 1985.
Barrett, S. “Commercial Hair Analysis: A Cardinal Sign of Quackery”, www.quackwatch.com, 2000.
Centers for Disease Control, “Blood and Hair Mercury Levels in Young Children and Women of Childbearing Age—United States, 1999”, MMWR, 50 (08); 140-3; March 2, 2001.
Chittleborough G. “A Chemist’s View of the Analysis of Human Hair for Trace Elements”, The Science of the Total Environment, 14: 53-75, 1980.
Clarkson TW. “The Toxicology of Mercury”, Crit Rev Clin Lab Sci, 34(4):369-403; 1997.
Cornelis R. “Neutron activation analysis of hair failure of a mission” J Radioanal Chem; 15:305-316; 1973.
Deening SB and Wever CW; “Hair Analysis of trace minerals in human subjects as influenced by age, sex and contraceptive drug” Am J Clin Nutr ; 31:1175-1180; 1978.
Druyan ME et al. “Determination of Reference Ranges for Elements in Human Scalp Hair”, Biological Trace Element Research, Vol 62, pp 183-197, 1998.
Fletcher DJ. “Hair Analysis: Proven and Problematic Applications” Postgraduate Medicine, Vol 72, No 5, pp 79-81,84,87-88, 1982.
Gaillard Y and Pepin G. “Testing Hair for Pharmaceuticals”; J Chromatogr B Biomed Sci Appl, 733(1-2):231-46; 1999.
Hambidge KM. “Hair Analyses: worthless for vitamins, limited for minerals”, Am J Clin Nutr; 36(5): 943-9; 1982 Hopps HC. “The Biologic Bases for Using Hair and Nail for Analysis of Trace Elements”, The Science of the Toxic Environment, 7: 71-89; 1977.
Kidwell DA, Lee EH, and DeLauder SF. “Evidence for bias in hair testing and procedures to correct bias”; Forensic Sci Int; 107(1-3):39-61; 2000.
Robert S. Baratz C-16 Klevay LM et al. “Hair Analysis in clinical and experimental medicine” Am J Clin Nutr; 46:233-6; 1987.
Lamand M., Faviert A., and Pineau A. “La détermination des oligoéléments dans les poils et dans les cheveux: intérêt et limites”, Annales de Biologie Clinique, 48, 433-442, 1990.
Manson P and Zlotkin S. “Hair Analysis- a critical review”; Can Med Assoc J; 133:186-188; 1985.
McKenzie JM.. “Alteration of the zinc and copper concentration of hair”; Am J Clin Nutr, 31:470-476; 1978.
Mertz W. “Confirmation: Chromium levels in serum, hair and sweat decline with age”; Nutr Rev; 55(10): 373-5; 1997.
Nowak B and Chmielnicka J. “Relationship of lead and cadmium to Essential Elements in Hair, Teeth, and Nails of Environmentally Exposed People”; Ecotoxicol Environ Saf; 46(3):265-274; 2000.
Nowak B and Kozlowski H. “Heavy Metals in Human Hair and Teeth: The Correlation with Metal Concentration in the Environment”, Biological Trace Element Research, Vol 62, 1998.
Paschal DC et al. “Age Dependence of Metals in Hair in a Selected U.S. Population”, Environmental Research, 48, 17-28, 1989.
Ponce RA et al. “Uncertainty analysis methods for comparing predictive models and biomarkers: A Case study of dietary methyl mercury exposure”; Regul Toxicol Pharmacol; 28(2): 96-105; 1998.
Rivlin RS. “Misuse of Hair Analysis for Nutritional Assessment”; Amer Jour of Med; Vol 75: 489-493; 1983.
Seidel S. personal communication to John L. Watson, March 6, 2001.
Sheard EA, Johnson MK, and Carter RJ. “The Determination of Chromium in Hair and other Biological Materials” Hair, Trace Elements and Human Illness, Praeger, NY, 1980.
Sky-Peck HH. “Distribution of Trace Elements in Human Hair”, Clin Physiol Biochem:8:70-80, 1990.
Robert S. Baratz C-17 Teresa M, Vasconcelos SD, and Tavares HMF. “Trace Element Concentration in Blood and Hair of Young Apprentices of a Technical-professional School”, The Science of the Total Environment, 205, 189-199, 1997.
Vir SC and Love AHG. “Zinc and copper nutriture of women taking oral contraceptive agents” Am J Clin Nutr; 34:1479-1483; 1981 Wennig R. “Potential problems with the interpretation of hair analysis results”; Forensic Sci Int 107(1-3); 5-12; 2000.
Willhelm M and Idel H. “Hair Analysis in Environmental Medicine”, Zeutralblatt for Hygiene and Unweltmedizin; 198:485-501; 1996.
Yoshinaga J et al. “Lack of Significantly Positive Correlations Between Elemental Concentrations in Hair and in Organs”; The Science of the Total Environment, 99: 125-135; 1990.
Zlotkin SH, “Hair Analysis: A Useful Tool or a Waste of Money?” International Journal of Dermatology, Vol 24, 161-164; 1985.
C-19 Thomas Clarkson C-21 Hopps (1977) He provides background physiology and histology of human hair formation and growth General questions He gives no information Topic 1 Analytical methods No information Topic 2 Factors influencing the interpretation of analytical results He notes various pathways of metal into hair: 1) via the follicle into the hair matrix 2) secretion of metals in the sebum on to the hair surface 3) secretion of metal in exocrine sweat on to the surface of the hair 4) secretion of metals in apocrine sweat on to the surface of the hair.
He notes that apocrine sweat may not be important for scalp hair.
He discusses the relative merits of head versus pubic hair and concludes that scalp hair is to be preferred He discusses some reports where lead and arsenic have been measured in scalp hair. The metal level depend on the distance from the scalp/ Lead tends to increase towards the tip of the hair strand.
Arsenic appears to be accumulated in hair and may present a historical record of tissue levels.
However hair can accumulate external arsenic in the form of arsenite. Animal experiments indicate arsenic is excreted in sweat.
Variable data have been obtained with cadmium He gives a table of normal levels of metals in hair.
He notes that attempts to distinguish external versus internal uptake of metals have usually been unsuccessful Topic 4 Toxicological consideration No information C-22 C-23 Topic 5 Data gaps and research needs No information Miekelay et al (1998) Compared two methods of measuring metal in samples of scalp hair taken from 1,091 adults living in Rio de Janeiro. They also sent a test sample to commercial laboratories for comparison.
General questions The article indicates the need to revise reference interval for normal levels of metals in hair Topic 1 Analytical methods The article claims that ICP-AES (inductively coupled plasma atomic emission spectrometry is out of date with poor detection limits but is still used by most commercial laboratories. The article claims that ICP-MS (inductively coupled mass spectrometry) is the method of choice.
Tables are presented comparing reference limits published by five commercial laboratories indicating wide differences between laboratories for certain metals. Tables are also presented indicating wide differences in results for certain metals on two hair samples circulated blind to the same five commercial laboratories. However, results for some metals yielded reasonable agreement. These metals included Na, Ca, Mg, Mn, Cu, Fe, Zn. The following metals gave reasonable agreement if results from one of the laboratories were excluded: Pb, Cd, Ba, Ni, Li, P, B, Cr, Mo.
Topic 2 Factors influencing the interpretation of analytical results No information Topic 3 Toxicological consideration No information Topic 4 Data gaps and research needs The study indicates the need to revise reference limits for some metals.
Sky-Peck (1990) He performed X-ray fluorescence analysis in six carefully aligned samples of hair from 987 employees and their families at a major medical center in Cook County, Illinois. The purpose was to elucidate factors that might affect concentrations of trace elements in human scalp hair C-24 General questions He concludes that hair analysis should only be used as a screening method along with other measures of the nutritional status of the patient.
More data are needed on factors affecting trace elements in hair before hair can be used as a quantitative tool to assess the nutritional status of any trace element.
Topic 1 Analytical methods He used only X-ray fluorescence analysis. He did not describe how the weight of hair was obtained. Usually Compton scattering is used to measure the hair mass. This does not appear to be the method used in this report Topic 2 Factors influencing the interpretation of analytical results The method of washing the hair sample can influence the levels of certain trace elements. The mild washing procedure used in the report did not affect levels of 14 selected trace elements.
Treatment with peroxide produced a statistically significant reduction in S, Ca, Fe, and Zn. The reduction in Ca was almost complete and Zn was reduced substantially. Hg levels were not affected.
Permanent waving produced a statistically significant increase in levels of 6 trace elements. Levels of Ca, Ni and As were more than doubled. Mercury was unaffected.
Brunettes and blondes differed significantly in only three trace elements, F (slightly lower in blondes); Mn (slightly lower in blondes); and Pb (almost double in blondes). Compared to brunettes and redheads differed statistically in 5 trace elements. Iron was almost doubled in red heads. Mercury was slightly reduced.
Blacks differed from Caucasians in 10 trace elements. Ni, As, and Pb in blacks were more than twice as high as in Caucasians. Orientals differed from Caucasians in 9 trace elements. Ca and Pb in Orientals were a factor of 2 below corresponding levels in Caucasians. Mercury was the same.
Note: Elements differed according to age. Ca in the older group was less than 50% of the younger group. Br was five times higher. Hg was unaffected.
The longitudinal profiles differ according to the trace element. The levels of As, Hg, Cu, Fe, Zn, S and Se were steady and unaffected by distance from the root end. On the other hand, the levels of Pb, Ni and Mn rose sharply towards the tip of the hair strands suggestion external contamination.
Ca and Sr showed less pronounced changes.
The results indicate that the levels of certain trace elements are influenced by a number of factors.
It would appear that Pb, Ni and Mn are affected by external contamination.
C-25 On the other hand, levels of Hg appear to be robust and unaffected by all but one of the factors tested in this report. For one factor, natural hair color of redheads versus brunettes, there was a statistically significant difference in mercury levels, but this difference was quantitatively small.
The most stable trace elements were S, Cu, Zn, Se, Cr, and Rb because these were not changed by more than a factor of 2 by any of the factors tested in this study. The most unstable elements were Ca, which was affected by more than a factor of 2 by five of the six factors tested. Pb was affected by four factors, and Ni. Br, and Sr by three factors.
Topic 3 Toxicological consideration No information Topic 4 Data gaps and research needs The paper stresses the need for more data on factors affecting levels of trace elements in hair Seidel et al. (2001) The authors sent a common hair sample to six commercial laboratories for trace element analysis Different levels were obtained However, it is difficult to evaluate the data without knowing the correct level. These levels can be compared to the normal ranges for each laboratory. The authors also checked on the accreditation of the labs and on the dietary advice given on the basis on the findings General questions The authors argue that there are few if any trace elements that have been validated as indicators of dietary sufficiency or of toxicity. Methyl mercury may be the only substance for which toxic dose response relationships have been established.
Topic 1 Analytical methods The labs tests used atomic fluorescence or mass spectrometry detection methods. The authors note that the mass spec. method is much lower detection limits Topic 2 Factors influencing the interpretation of analytical results The labs can be compared in terms of identifying with elements are outside their normal range. All six labs agreed that the following elements were with their normal range: Ba, Be, B, Cd, S and Ti.
All labs agreed that Mn and Mo were outside their normal range. For the following elements all C-26 labs except one agreed on classifying according to their normal range: Al, As, Pb, Mg, Hg, Ni, and Zn.
Thus, for approximately half the elements tested, there was reasonable agreement between the commercial labs.
Topic 3 Toxicological consideration There is lack of toxicological information of the value of hair element concentration as a biomarker for tissue levels, especially levels in the target tissue. This information is available only for methyl mercury.
Topic 4 Data gaps and research needs As mentioned above, the data gaps are in dose response information and in relating hair levels to levels in the target tissue Steindel & Howanitz (2001) The authors provide editorial comment on the paper by Seidel and provide a discussion of proficiency testing in clinical chemistry laboratories.
General questions They point out that the current lack of normal ranges for trace elements in hair make interpretation of results impossible. They comment of the difficulty of making nutritional conclusions from hair data Topic 1 Analytical methods No information Topic 2 Factors influencing the interpretation of analytical results The authors listed many problems in interpretation of hair data including external contamination and the absence of reliable reference standards and uniform methods for processing the hair samples.
Topic 3 Toxicological considerations No information Topic 4 Data gaps and research needs C-27 More data are needed on inter-laboratory comparisons C-28 Wennig (2000) This is a review article on the incorporation of drugs into hair. It presents a useful review of hair physiology and biochemistry. It gives recommendations for collection and storage of hair samples.
It gives no information on trace elements in hair.
Yoshinaga et al (1990) The paper compares the concentration of a number of trace elements in hair with corresponding concentrations in several organs and tissues obtained at autopsy.
Unfortunately, little detail was given on how the hair samples were collected or on the length of the hair samples.
General questions Topic 1 Analytical methods A commonly used analytical method was used (ICP-AES). Quality control tests were made.
Topic 2 Factors influencing the interpretation of analytical results The varying length of the hair samples may have influenced the result and accounted for the poor correlations.
Topic 3 Toxicological consideration The main finding was that levels of Ca, Mg, P, and Zn in hair did not correlate with tissue levels or body burden They were not able to draw any conclusions about Fe, Cu or Se as the appropriate tissues were not available for analysis Topic 4 Data gaps and research needs More information is needed on hair versus levels in autopsy tissues. The hair length should be restricted to a short segment close to the scalp.
C-29 Hopps HC. (1977). The biologic bases for using hair and nail for analyses of trace elements. The Science of the Total Environment, 7:71-89.
Miekeley N, Dias Carneiro MW, Porto da Silveira CL. (1998). How reliable are human hair reference intervals for trace elements? The Science of the Total Environment, 218:9-17.
Sky-Peck HH (1990). Distribution of trace elements in human hair. Clin. Physiol. Biochem. 8:70-80.
Seidel S, Kreutzer R, Smith D, McNeel S, Gilliss D. (2001). Assessment of commercial laboratories performing hair mineral analysis. JAMA 285(1):67-72.
Steindel SJ, Howanitz PJ. (2001). The uncertainty of hair analysis for trace metals. JAMA 285(2):83-85.
Wennig R. (2000). Potential problems with the interpretation of hair analysis results. Forensic Science International 107:5-12.
Yoshinaga J, et al.(1990). The Science of the Total Environment 99:125-135.
C-31 Michael Greenberg Michael I. Greenberg, M.D., MPH C-33 Topic #1: Analytical Methods Comments: The laboratory analytical methods available are capable of defining the qualitative existence of a variety of pharmaceuticals, drugs of abuse, and occupational/environmental toxicants. The operative word here is “qualitative”. Quantitation of specific levels are not, in my opinion and experience, either generally reliably reproducible and/or clinically useful. Specific analyte levels are essentially of little or no value in the determination of so-called cut-off levels (e.g., PELs, TLVs, “safe levels”), “normal levels,” or other designators which rely on reference levels. In addition, the analytical techniques currently in use are capable of providing “segmental analysis” of hair, which in turn can provide a historical picture of various qualitative (not quantitative) exposures over time. In addition, hair analysis may help to derive an essential time frame which may indicate, based on the average rate of hair growth, the time of inception for various exposures.
The amount of hair needed for analysis may be dependent on the specific analyte sought as well as the temporal relationship between exposure and hair harvest.
One of the most important shortcomings for hair analysis, as it currently exists, is the fact that reference ranges may often be unreliable. Laboratories frequently base their reference ranges for specific analytes on limited case reports in the medical literature or exclusively on data derived from animals, which has limited applicability to humans. These facts contribute to substantial limitations with regard to interpretation of results. Variability undoubtedly exists from one laboratory to another. Certainly these facts limit the clinician’s ability to interpret and utilize hair-derived values beyond the potential qualitative information that might come from hair testing of any individual.
Michael I. Greenberg, M.D., MPH C-34 Thus, at this time, it may be prudent to recommend that hair testing for all substances (drugs of abuse, occupational toxicants, environmental toxicants) be limited to qualitative determinations as opposed to quantitative determinations. The goal of quantitation for any laboratory analyte is to derive clinical algorithms that translate into levels that indicate disease, dysfunction, or specific risks for disease or dysfunction. With regard to hair testing in its current state, there is little evidence that there is sufficient reliability to use quantitation for these purposes. Laboratory washing procedures prior to digestion may significantly alter the hair content of various analytes. For example, when hair is tested for THC, if it is washed with methanol, THC concentrations may be reduced by as much as 85% (Forensic Drug Abuse Advisor, 1996) by virtue of this process. It is reasonable to expect that similar degradations in analyte concentration occur when other analytes are involved.
Hair pigmentation is a critical factor in the interpretation of the concentration of certain compounds and their metabolites incorporated into hair. Melanin is responsible for the pigmentation. The color and the melanin content of human hair samples differs over a wide range.
Once deposited into hair, chemicals may remain detectable for a period of months to years.
However, if disposition into hair is influenced by those properties attributed to hair color, then certain persons may test positive more frequently than other persons. Removal of the melanin from hair digests prior to hair analysis may reduce the effect of melanin on the total chemical concentration by excluding the drug bound to the pigment. In one study (Hold KM, et al), the effect of melanin removal by centrifugation of hair digests on cocaine concentrations was investigated. Two sets of hair samples from five cocaine users were analyzed for cocaine and metabolites. A solution consisting of 10 mL of 0.5M Tris buffer (pH 6.4) to which is added 60 mg D,L-dithiothreitol, 200 mg SDS, and 200 U Proteinase K, was used to digest the hair. Two milliliters of this solution was added to 20 mg of hair and incubated at 37 degrees in a shaking water bath (90 oscillations/min) overnight. The samples were removed from the water bath and mixed. One set was centrifuged at 2000 rpm and divided into supernatant and melanin pellet. The Michael I. Greenberg, M.D., MPH C-35 other set was not centrifuged. Internal standards were added to all tubes. The samples were further extracted, derivatized, and analyzed by gas chromatography-mass spectrometry. A mean of 8.8% (standard deviation [SD] 7.0%) of the total cocaine concentration (supernatant and pellet) was left behind in the pellet. The same experiment was repeated—except that the melanin pellet was redigested with 0.1 N HCl. After redigestion of the melanin pellet, the mean cocaine concentration in the pellet was 3.8% +/- 4.0% (mean +/- SD) of the total cocaine concentration in hair. These investigators felt that their data demonstrate that removal of melanin from hair digests by centrifugation does not eliminate hair color bias when interpreting cocaine concentrations. TOPIC #2: FACTORS INFLUENCING THE INTERPRETATION OF ANALYTICAL RESULTS Comments: Exposure of hair sample to the external environment could be an important factor in confounding results on both a quantitative and qualitative basis. By way of example, many over the counter hair coloring preparations contain lead acetate (e.g., Grecian formula).
This may persist for long periods on hair shafts and thus confound hair testing results for lead. It is also unclear if the use of coloring agents containing lead acetate alters or enhances the hairs ability to bind other analytes or potential toxicants. Based on the medical literature that describes the use of hair testing for substances of abuse there are differences in hair uptake of various substances based on ethnicity. For example, negroid hair has been suggested to bind cocaine residues with greater affinity than caucasoid hair.
There are also reports in the literature that the ability to bind various chemicals and drugs may depend on endogenous hair color as well as if hair has undergone bleaching. For example, bleached hair radically lowers the drugs [of abuse] content of hair. This may explain the Michael I. Greenberg, M.D., MPH C-36 observation that many competitors on the professional biking circuit sport bleach blond hair (Kintz). Blond hair has been shown to not bind cocaine or its metabolites as well as pigmented hair (Hubbard). In addition, there was no evidence of a dose-related incorporation of these drugs and metabolites into non-pigmented hair. The concern is that similar circumstances may occur with regard to specific occupational or environmental toxicants and chemicals. Based on a study presented by Reid et al. indicating that gray hair takes up less cocaine than non-gray hair, it is possible that gray hair may also alter the utility of hair analysis in other settings.
The Reid study evaluated cocaine levels in the same individuals by comparing the levels in gray and non gray hairs from the same person. In a similar study (Rothe et al.), hair samples from 15 patients receiving medical treatment with amitriptyline, carbamazepine, chlorprothixene, diclofenac, doxepine, indomethacine, maprotiline, or metoclopramide, or with a chronic heroin and cocaine abuse, were separated into white and pigmented fibers and both fractions were independently investigated by GC-MS. The drugs were found in pigmented fibers as well as in white fibers, but the concentrations in the white fibers were smaller than in the pigmented ones for most of the samples investigated. The concentration ratio of the drugs or their metabolites in both hair fractions (white/pigmented) was found to be between 0.09 and 1.57 (mean 0.70, 30 concentration pairs). There are large differences in this ratio between different subjects with the same drug; whereas for different drugs in the same subject—in many cases—similar ratios were measured. As reasons, a different grade of pigmentation of the hair and the influence of the drug structure are discussed. From these results it follows that the natural hair color is an important parameter in the evaluation of drug concentration in hair. Again, similar effects may be seen when dealing with occupational and environmental toxicants. The rate of hair growth may be an important factor in the ability to identify the presence of various materials based on time of exposure. Sources usually indicate that head hair grows at the rate of 1-2 cm per month. That in itself represents a range encompassing up to a 100% difference Michael I. Greenberg, M.D., MPH C-37 in hair growth rate. Obviously, comparisons of individuals whose hair growth rates differ by a factor of 100% is problematic.
TOPIC #3: TOXICOLOGIC CONSIDERATIONS Relatively little is known about the biological uptake of specific substances with regard to the concentration delivered to and incorporated into hair. There is essentially no data that reliably establishes the relationship between chemical concentrations in the hair and blood or other target organs for most chemcials. More specifically, and more importantly, no dose-response data currently exists with regard to chemical concentrations in the hair and blood or other target organs. In addition, no disease predictive value exists for any quantatative data that has been derived to date with regard to the hair concentration of drugs or chemicals. Rollins et al have suggested that the ionization state of any given chemical is what determines whether or not it will bind with hair melanin. These investigators reported that catiomnic drugs are more likely to bind with melanin when compared with anionic drugs. This study may provide some guidance with regard to the binding ability of other toxicants of concern.
TOPIC # 4: DATA GAPS AND RESEARCH NEEDS Comments: The data gaps that most significantly limit the use of hair testing in public health evaluations are 1) the lack of accurate and reliable reference range data and 2) the lack of specific information about dose response relationships with regard to the relationship between chemical concentrations in the hair and blood or other target organs. In my estimation, these two items constitute the most pressing research needs with regard to hair testing. Michael I. Greenberg, M.D., MPH C-38 Future studies must address these basic data gaps in order to even begin to decide if hair testing has clinical screening or other clinical usefulness.
TOPIC # 5: IDENTIFYING SCENARIOS FOR WHICH HAIR ANALYSIS MAY BE APPROPRIATE Comments: Hair testing for acute exposures is clearly not the best alternative for determination of either dose or exposure with regard to any potential toxicant. If acute exposure is defined as the pre-distribution time frame, then blood or urine testing would be far superior to hair testing in any scenario. However, in the event of a single exposure (as opposed to an ongoing exposure) the use of hair testing after the completion of the pre-distribution phase of kinetics may be helpful in qualitatively identifying the fact that exposure has indeed occurred and/or generally timing that exposure. The use of hair analysis in this setting may have forensic as well as public health value.
In the setting of chronic exposures, hair analysis may have value in identifying and documenting a given exposure. This, again, may have forensic, civil-legal, and risk assessment value for individuals as well as communities and populations. Obviously, the length of any given individual’s hair may limit the use of hair analysis, as well as how frequently the hair is cut. In any scenario, however, the state of the art is such that specific and measurable health effects will generally not be uncovered by hair analysis. In addition, public health and/or individual risk assessment determinations will be limited by whatever conclusions may be drawn by what is essentially a qualitative and not quantitative toxicological evaluation.
Michael I. Greenberg, M.D., MPH C-39 ADDITIONAL CONSIDERATIONS: Comments: One interesting study (Al-Delaimy, et al) used hair analysis to measure the relation between workplace smoking policies and exposures to environmental tobacco smoke (ETS) of workers in bars and restaurants. In this study, 114 workers were questioned about sources of exposure to ETS and smoking habits, and details of the smoke-free policy in their work place were recorded.
A hair sample was collected from each participant and tested for nicotine. Among non-smoking workers, hair nicotine levels varied strongly according to the smoke-free policy at their place of work. Those working in 100% smoke-free restaurants had much lower levels than staff working in bars with no restrictions on smoking, and levels were intermediate for staff working in places with a partial smoking ban. Hair nicotine levels among nonsmokers working in places with no restriction on smoking were similar to hair nicotine levels of active smokers. The findings from this study highlight the substantial levels of exposure of bar and restaurant staff from patrons' smoking. The potential sources for confounding variables in the hair testing arena are truly legion. This fact is demonstrated in one instance by a paper from Japan wherein investigators sought to draw a relationship between head hair mercury and health. However, in the end, these investigators discovered that “some subjects who showed a high total mercury level made habitual use of toilet soap containing much mercury.” Thus, the confounding effect of an unusual source for a heavy metal can interfere with effective hair analysis. Additional References: Kintz P et al. Abstract presented at American Academy of Forensic Sciences meeting, Reno, Nevada, February 2000.
Michael I. Greenberg, M.D., MPH C-40 Hubbard D et al. Society of Forensic Toxicologists meeting, Snowbird, Utah, 2000.
Reid R. et al. Society of Forensic Toxicologists meeting, Snowbird, Utah, 2000.
Rollins D et al. Society of Forensic Toxicologists meeting, Snowbird, Utah, 2000.
Rothe M et al. Effect of pigmentation on the drug deposition in hair of grey-haired subjects.
Forensic Sci Int 1997 Jan 17;84(1-3):53-60.
Harada M et al. Monitoring of mercury pollution in Tanzania: relation between head hair mercury and health. Sci Total Environ 1999 Mar 9;227(2-3):249-56.
Holde et al. Quantitation of cocaine in human hair: the effect of centrifugation of hair digests. J Anal Toxicol 1998 Oct;22(6):414-7.
Al-Delaimy W et al. Nicotine in hair of bar and restaurant workers. N Z Med J 2001 Mar 9;114(1127):80-3.
C-41 Michael Kosnett Michael J. Kosnett, MD, MPH C-43 June 4, 2001 The following are preliminary comments regarding some topics that constitute the charge to the panel. However, I am still in the process of reviewing some relevant studies and therefore may revise or amend this material in a subsequent submission. Topic #1: Analytical methods The key analytical methods currently used by clinical laboratories to measure trace elements in hair appear to be inductively coupled plasma atomic emission spectrometry, and inductively coupled plasma mass spectrometry (Miekeley et al, 1998; Seidel et al, 2001). Graphite furnace atomic absorption spectrometry has been used to measure arsenic in hair, with reported limits of detection of 0.005 to 0.01 ? g/g (Rebel et al, 1998; Hewlett et al, 1995). Total and inorganic mercury in hair has been determined by cold vapor atomic absorption (Boischio and Cernichiari, 1998; NRC, 2000), and the difference between total and inorganic Hg yielded by this method has been used as a surrogate for the methyl mercury hair content. Methyl mercury in hair has also been determined directly by gas chromatography using a tritium foil electron capture detector (Smith et al, 1997). Selenium in hair has been measured fluorometrically after complexation with 2,3-diaminonaphthalene (Yoshinaga et al, 1990). The preceding methods appear to have generally required a hair specimen size on the order of 50 mg or more. Although commercial laboratories commonly measure the submitted hair sample in bulk, the methodology is sufficiently sensitive to allow investigators to yield segmental analysis (³ 1 cm) on bundles of hair for which information on the alignment and distance from the root has been preserved. Segmental analysis may potentially offer information on the temporal pattern of exposure to the element in question that is of value in epidemiological and forensic investigations. Neutron activation analysis (NAA) has been used in forensic investigations and occasionally in epidemiological or clinical studies for the sensitive determination of certain trace elements in minute quantities of hair. For example, neutron activation analysis has been used to measure arsenic in 2 mm segments of an individual hair, each segment weighing approximately 3 ? g (Smith, 1964; Curry and Pounds, 1977). NAA has also been used to measure the hair content of Zn, Au, Cu, Mn, Hg, Sb, and Th (Jervis, 1968; Cornelius, 1973). The distribution of mercury in 2 mm segments along the length of a single strand of hair may be determined by nondestructive x-ray fluorescence (Cox et al, 1989, cited by NRC, 2000). Proton induced x-ray emission has been used to measure the spatial distribution of multiple elements in 10 micron increments across axial cross section of a single shaft of hair (Cookson and Pilling, 1975; Hindmarsh et al, 1999). A multitude of factors influence the quality control of laboratory hair analysis. These include the finite limitations of the assay method (ideal method recovery and precision), and the variability associated with within-run and day to day operation of the assay (actual method recovery and precision). Although not necessarily reflective of a systematic review of the literature, a few references may be cited as offering examples of operational precision in research investigations.
Michael J. Kosnett, MD, MPH C-44 Using NAA to measure 7 elements in a single specimen of hair, the coefficient of variation ranged from 5.92% in the case of Mn (mean concentration 1.65 ppm) to 15.7% in the case of Sb (mean concentration 0.18 ppm) (Cornelius, 1973). Wilhelm et al (1989) reported a day to day coefficient of variation of approximately 6% for atomic absorption measurement of Zn, Pb, Cu and Cd in hair. The issue of inter-laboratory variability of multi-element hair analysis for trace elements provided by commercial laboratories using ICP-AES and ICP-MS has recently been addressed by Miekeley et al (1998) and Seidel et al (2001), both of whom obtained widely discrepant results from split samples sent to 4 to 6 different commercial laboratories. Topic #2: Factors Influencing the Interpretation of Analytical Results One of the most fundamental factors impacting the potential utility of hair analysis as an exposure assessment tool in public health evaluations is the limited capacity of such measurements to distinguish external contamination from internal incorporation. In particular, multiple studies have noted that toxic metals may become incorporated into hair following external contact with metal containing dust, soil, water or hair care products. There is no reliable analytical approach that can distinguish this external contamination from elevations in hair metal content that result from metal ingestion or inhalation (Chittleborough, 1980). Although pre-analysis washing or rinsing methods are often used in an attempt to selectively remove external contamination, there is no standardized approach that has been shown to achieve the desired result. The experience with arsenic, a toxic metalloid that is often encountered through environmental exposures, is a case in point. In vitro studies have demonstrated that hair incorporates appreciable amounts of arsenate and arsenite from aqueous solutions, and that the extent of absorption increases with duration of contact time and moderate decrements in pH (e.g. pH 3 to 5) (Atalla et al, 1965; Bate, 1966; Van den Berg et al, 1967; Fergusson et al, 1983). Adsorption of arsenic to hair may also be substantial following contact with arsenic containing dust (Atalla et al, 1965).
The extent of adsorption may vary significantly along the length of a single hair (Maes and Pate, 1977). Adsorption-desorption experiments demonstrate that externally deposited arsenic cannot be completely removed from hair by a variety of washing and rinsing techniques (Smith, 1964; Atalla et al, 1965; Van den Berg et al, 1968). Moreover, washing may complicate interpretation further by partially removing arsenic present in hair as a result of internal incorporation (Atalla et al, 1965; Van den Berg et al, 1968; Young and Rice, 1944). Studies with other metals have reported similar findings with respect to adsorption onto hair from external contamination, and variable removal of both internal and externally derived traces by washing regimens (Chittleborough, 1980; Fergusson et al, 1983; Wilhelm et al, 1989). The problems posed by this inability to distinguish external adsorption from internal incorporation places substantial constraints on what can be learned from the results of hair analysis for an environmental toxin where the suspected route of human exposure is via contact with contaminated dust, soil, airborne particulate, or tap water. Although these routes of exposure might result in ingestion or inhalation of an environmental toxin and its subsequent appearance in Michael J. Kosnett, MD, MPH C-45 hair through incorporation at the hair follicle, they also create ample opportunity for the agent to become externally adsorbed onto hair via airborne deposition, hand to hair contact, or bathing. In such settings (which are probably characteristic of the majority of sites subject to ATSDR health assessments), the finding of elevated levels of a environmental toxin in the hair of a given subject or a study population is limited at best to establishing the potential for that subject or population to have come into contact with the agent in a manner that may have resulted in ingestion or inhalation. In addition to being a test of low specificity, the information on potential exposure gleaned from an elevated hair level in such settings is likely to be qualitative in nature. That is because with the notable exception of methyl mercury, quantitative information on the relationship between ingestion or inhalation of a environmental toxin and its concentration in hair is limited, and appears to be subject to considerable inter-subject and inter-population variability. Again, an example derived from the measurement of arsenic in hair is instructive. Although several epidemiological studies have noted a correlation between levels of arsenic in hair and arsenic in dust, soil, or water, (e.g. Bencko and Symon, 1977; Hartwell et al, 1983; Valentine et al, 1979), the hair arsenic levels may not correlate with levels of arsenic in urine (Harrington et al, 1978; Hewlett et al, 1995). For example, Harrington et al (1978) studied hair and urine arsenic levels in a community near Fairbanks, Alaska, where the arsenic concentration of water obtained from domestic wells averaged 224 ? g/L (range 1.0 to 2450 ? g/L). A subset of subjects whose wells contained arsenic averaging 345 ? g/L consumed only bottled water. Although they had relatively low arsenic levels in urine (average 43 ? g/L), the arsenic concentration of their hair was high, averaging 5.74 ppm. Subjects consuming water from domestic wells with the lowest levels of arsenic (less than 50 ? g/L in water) had hair arsenic concentrations averaging 0.46 ppm, and urine arsenic levels averaging 38 ? g/L. Thus, the arsenic level in hair varied by 14-fold, despite similar levels of arsenic in urine. The authors noted the likely implication that the elevated hair arsenic levels were probably due to external contamination derived from bathing in, but not drinking, the high arsenic well water. Topic #3 To what extent may hair analysis be used to predict adverse health outcomes? and Topic #5, Under what scenarios may hair analysis be appropriate for evaluating exposures to environmental contaminants?
From a medical standpoint, there appears to be no disease or illness caused by an environmental toxin for which there is a general medical consensus that the results of hair analysis would form the basis for specific medical treatment. In the case of methyl mercury, segmental maternal hair analysis may have diagnostic value as a biomarker of fetal exposure to levels of this neurotoxin that are associated with a postnatal risk of adverse neurobehavioral development (NRC, 2000). Some data suggest that the level of hair methylmercury in children and adults may also be a biomarker of exposure associated with adverse effects on neurological function and other health endpoints (NRC, 2000). Because most contemporary exposure to methylmercury is confined to ingestion via seafood, there is little Michael J. Kosnett, MD, MPH C-46 potential for high hair levels of methylmercury to be a result of external contamination. In most populations whose level of seafood ingestion is of a sufficient magnitude to pose a potential health risk from methylmercury, measurement of total mercury in hair may be an acceptable surrogate for measurement of methylmercury in hair.
In certain settings, segmental hair measurement of arsenic (and potentially other toxins such as thallium) may be of diagnostic and/or forensic value in identifying or confirming a high dose toxic exposure or poisoning that terminated months (but not years) in the past. For example, segmental analysis of a sufficiently long hair might help to confirm a suspicion that an episode or outbreak of severe gastroenteritis followed by peripheral neuropathy that occurred 8 to 10 months in the past was likely to have been the consequence of acute arsenic or thallium poisoning. Months after the exposure ended, levels of arsenic or thallium in the urine may have fallen to normal values, and high peak levels in the hair (or nails) may offer the only remaining confirmatory forensic evidence.
It should be noted that although the hair measurements in such scenarios might conceivably be of value in confirming past poisoning, the epidemiological database on hair analysis is insufficient to use these measurements to predict the risk of latent diseases such as cancer.
Supplemental comments from Michael J. Kosnett, MD, MPH (submitted June 21, 2001) 1.
A key factor to be addressed prior to ATSDR’s use or interpretation of hair testing is the predictive value of a positive or negative test with respect to detecting an exposure and/or internally absorbed dose of a toxic substance of sufficient magnitude to be of pathological or public health significance.
2.
One of the inherent limitations of hair analysis arises from the fact that hair represents a matrix that is in direct contact with the external environment and as such may be subject to greater contamination than other analytes traditionally used in biological monitoring, such as blood, urine, or even expired air.
Supplemental references submitted by Michael J. Kosnett, MD, MPH (June 21, 2001) Atalla L, Silva CM, Lima FW. Activation Analysis of Arsenic in Human Hair—Some Observations on the Problems of External Contamination. Ann. Acad. Bras. Cien., 1965, 37:432-441.
Bate LC. Adsorption and Elution of Trace Elements on Human Hair. Int. J. Appl. Rad. Isot., 1966, 17:417-423.
Bencko V, Symon K. Health Aspects of Burning Coal with a High Arsenic Content. I. Arsenic in Hair, Urine, and Blood in Children Residing in a Polluted Area. Environ. Res., 1977, 13:378-385.
Michael J. Kosnett, MD, MPH C-47 Chittleborough G. A Chemist’s View of the Analysis of Human Hair for Trace Elements. Sci. Tot.
Envir. 1980, 14:-75.
Cornelis R. Neutron Activation Analysis of Hair: Failure of a Mission. J. Radioanal. Chem., 1973, 15:305-316.
Cox C, Clarkson TW, Marsh DO, Amin-Zaki L, Tikriti S, and Myers GG. 1989. Dose-response analysis of infants prenatally exposed to methyl mercury: An application of a single compartment model to single-strand hair analysis. Environ. Res. 49(2):318-322.
Curry AS, Pounds CA. Arsenic in Hair. J. For. Sci. Soc., 1977, 17:37-44.
Ferguson JE, Holzbecher J, Ryan DE. The Sorption of Copper [II], Manganese [II], Zinc [II], and Arsenic [III] into Human Hair and Their Desorption. Sci. Tot. Envir., 1983, 26:121-135.
Gebel TW, Suchenwirth RHR, Bolten C et al. Human biomonitoring of arsenic and antimony in case of an elevated geogenic exposure. Environ Health Persp 1998; 106:33-39 Harrington JM, Middaugh JP, Morse DL et al. A Survey of a Population Exposed to High Concentrations of Arsenic in Well Water in Fairbanks, Alaska. Am. J. Epid., 1978, 108(5):377-385.
Hartwell TD, Handy RW, Harris BS et al. Heavy Metal Exposure in Populations Living Around Zinc and Copper Smelters. Arch. Environ. Health, 1983, 38(5):284-295.
Henke G, Nucci A, Queiroz LS. Detection of Repeated Arsenical Poisoning by Neutron Activation Analysis of Foot Nail Segments. Arch. Toxicol, 1982, 50:125-131.
Hewitt DJ, Millner GC, Nye AC, et al. Investigation of arsenic exposure from soil at a Superfund site. Environ Research 1995; 68:73-81 Houtman, JPW, de Bruin M., de Goeij JIM: Arsenic Levels of Human Hair as an Indicator for Environmental Exposure, in Nuclear Activation Techniques in the Life Sciences, Vienna, IAEA, 1978, pp. 599-614.
Jervis RE, Present Status of Activation Analysis Applications in Criminalistics. Isot. Rad. Tech., 1968, 6(1):57-70.
Leslie ACD, Smith H. Napoleon Bonaparte’s Exposure to Arsenic During 1816. Arch. Toxicol., 1978, 41:163-167.
Michael J. Kosnett, MD, MPH C-48 Maes D, Pate BD.: The Absorption of Arsenic into Single Human Head Hairs. J. For. Sci., 1977.
22:89-99.
National Research Council. Toxicological Effects of Methyl Mercury. National Academy Press: Washington, DC, 2000 Pounds CA, Pearson EF, Turner TD. Arsenic in Fingernails. J. For. Sci. Soc., 1979, 19:165-173.
Smith H. The Interpretation of the Arsenic Content of Human Hair. J. For. Sci. Soc., 1964, 4:192-199.
Valentine JL, King HK, Spivey G. Arsenic Levels in Human Blood, Urine, and Hair in Response to Exposure via Drinking Water. Environ. Res., 1979, 20:24-32.
Van den Berg AJ, de Bruin M, Hortman JPW. Sorption Behavior of Trace Elements in Human Hair, in Nuclear Activation Techniques in the Life Sciences, Vienna, IAEA, 1967, pp. 661-674.
Van den Berg AJ, de Geoij JJM, Houtman JPW et al. Arsenic Content of Human Hair After Washing as Determined by Activation Analysis, in DeVoe JR [ed.] Modern Trends in Activation Analysis, Vol. I. Washington, DC: NBS, 1968, pp. 272-282.
Wilhelm M, Ohnesorge FK, Lombeck I et al. Uptake of aluminum, cadmium, copper, lead, and zinc by human scalp hair and elution of the absorbed dose. J Anal Toxicol 1989; 13:17-21 Young EG, Rice FAH. On the Occurrence of Arsenic in Human Hair and Its Medicological Significance. J. Lab. Clin. Med., 1944, 29:439-446.
C-49 Dan Paschal Michael J. Kosnett, MD, MPH C-50 Daniel C. Paschal–CDC C-52 ATSDR Hair Analysis Workshop June 12–13, 2001 Atlanta, GA Charge Questions for Panelists: Analytical Methods 1) What analytical methods currently exist?
Analytical methods for hair analysis include cold vapor atomic absorption analysis (1); graphite furnace atomic absorption (2); inductively coupled argon plasma optical emission spectrometry (3,4); inductively coupled argon plasma mass spectrometry (5); proton induced X-Ray emission (PIXE) spectrometry (6) ; X-Ray analysis (7); and neutron activation analysis (8).
2) Substances/elements for which reliable analyses exist include: a) mercury- methyl and inorganic (1); b) arsenic (2,8); c) aluminum (3,4); d) gold (3,4); e) boron (3,4); f) barium (3,4); g) beryllium (2,3,4); h) calcium (3,4); i) cadmium (2,3,4); j) cobalt (3,4); k) chromium (2,3,4); l) copper (2,3,4); m) iron (3,4); n) lithium (2,3,4); o) magnesium (2,3,4); Daniel C. Paschal–CDC C-53 p) manganese (2,3,4); q) molybdenum (3,4,5 ); r) sodium (3,4); s) nickel (2,3,4); t) phosphorous (3,4); u) lead (2,3,4,5); v) antimony (3,4); w) selenium (2,3,4,5); x) strontium (3,4); y) titanium (3,4); z) thallium (2,3,4,5); aa) vanadium (2,3,4); bb) zinc (2,3,4); cc) drugs of abuse -cocaine, PCP, opiates (9,10) 3) For what purposes are these methods typically used?
Forensics- As Exposure evaluation- As, Cd, Cr, Hg, Mn, Pb, Se, Al Diet/Nutrition Status- Ca, Mg, Na, Se, Sr, V, Zn, Cu, Co 4) What amount (g) of hair is needed?
0.1-0.5g (4,5)- Amount depends on type (occipital or other) and detection limit (4,5,9,10).
5) Intralaboratory variability (within-lab/run precision and accuracy)- MUST be evaluated with a stable, homogeneous, well-characterized pooled material.
Daniel C. Paschal–CDC C-54 6) Interlaboratory variability-(among laboratories accuracy and precision)- evaluation can be by regulation (CLIA or state/county/city licenses) or voluntary participation in Quality Assurance/Quality Control programs- e.g. Center for Toxicology of Quebec (
Factors Influencing the Interpretation of Analytical Results Variations in sample collection A variety of sample preparations have been suggested to sort exogenous (presumable contamination from exposure to the external environment) and endogenous metals and drugs from collected hair specimens. These vary from no treatment, washing with deionized/distilled/ultrapure water only to washing with ionic or non-ionic detergents, either alone or in concert with organic solvent washes. For details and references, see (2).
Sampling methods CDC has standardized the specimen collection and washing for hair, based on studies conducted internally and reported (4,5) in the literature. We obtain about 0.5 grams of occipital hair, and wash with a non-ionic detergent. Quality control is preformed by analysis of reference materials from NIST (SRM 1643d-Trace Elements in Water; SRM 1641d Mercury in Water), and a digested hair sample characterized by our operational method(s). Normal or “reference” ranges for 28 elements were published (4). “Abnormal” ranges would be those outside (generally higher than) the 95% upper limits for these analytes- toxic levels vary considerable depending on the adverse health outcome for each individual toxicant.
Exposure of hair to external environment includes copper from certain chlorinated swimming pools, lead from lead acetate “Grecian Formula”, selenium from dandruff shampoo (“Selsun”); zinc from “herbal” shampoos (Herbal Daniel C. Paschal–CDC C-55 Essence; Head and Shoulders), lead , cadmium, mercury and arsenic from dust, dirt, smoke, etc (4,5,11).
Exogenous and endogenous hair levels are difficult to distinguish, due to the high porosity of hair, and ineffective and non-standard “washing” procedures. The ideal washing/cleaning procedure would remove ONLY exogenous metals or other analytes- unfortunately, none have been reported (4,5,12,13,14).
Hair color pigmentation (melanin?) (15) and location (4,5,11) have been demonstrated to affect hair concentrations of several analytes.
Gender, ethnicity affect hair metals concentrations due to presence or absence of gender-linked hair treatment activities (e.g. coloring, permanent) and pigmentation (4,5,11).
Rate of Growth of hair has been assumed by many investigators to be relatively “constant” at about 1 cm/month (4,5,11) but is known to vary somewhat with age/gender/season (4,5,11).
Toxicologic Considerations Biological uptake of metals (4,5,11,16,17) and drugs of abuse have been extensively studied and described.
Relationship between hair and other tissue concentration levels, including urine (18) , whole blood (1,19) and serum (20) as well as other tissues (21) has been studied and described to some degree. The most complete and compelling evidence exist for hair mercury/blood mercury Daniel C. Paschal–CDC C-56 (methylmercury) and for arsenic in hair/urine/fingernail/tissue (1,21,23,24). Other metals and drugs of abuse are less well characterized (17).
Dose response relationships have been demonstrated in very few recognized studies—only hair mercury and arsenic have been clearly associated with body burden and health (adverse) effects (25,26). Other evidence, e.g. correlation between the concentration of manganese in hair and behavioral disorder or violence, is less compelling (27). Data Gaps Methodological- Quality control/quality assurance- although some laboratories are licensed for trace metals determinations, there are very few (28) proficiency testing programs or reference materials available (29,30) for evaluation and documentation of precision and accuracy of laboratory analytical systems.
Toxicological- Serious disagreement exists as to “reference” (normal or expected) values for a large number of elements. Drugs of abuse can often be detected at low concentrations; there is some disagreement as to the correlation between results of hair testing for abused drugs and more conventional determinations of drugs in urine, exhaled breath, or other (29).
Research Needs- Simply stated, carefully designed studies of exposure, body burden, and hair concentrations are needed to move beyond “anecdotal” levels of documentation. These studies, will, unfortunately, be limited by available funds and other resources.
Scenarios Where Hair Analysis May Be Appropriate Exposure Pathway Chronology Exposure Measurable Health Duration Effects (Y/N) Individual Ingestion Past Chronic Yes (if high) (MeHg) Daniel C. Paschal–CDC C-57 Individual Ingestion Past Chronic Yes (if high) Inhalation (As 3/5) Individual Ingestion Past Chronic Yes (if elevated) Lead REFERENCES 1) Greenwood MR, Dhahir P, Clarkson TW, Farant JP, Chatrand A, and Khayat AJ. Analyt.
Toxicol. 1, 265 (1977).
2) Tsalev DL. Atomic Absorption Spectrometry in Occupational and Environmental Health Practice. Boca Raton FL: CRC,1995. 3) DiPietro DS et al. Biological Trace Element Research, 22, pp. 83-100 (1989).
4) Paschal DC et al. Environmental Research 48, pp. 17-28 (1989). 5) Miekeley N, Dias Carniero MTW, and Porto de Silviera CL. Science of the Total Environment, 218, pp. 9-17 (1998).
6) Du Y, Mangelson NF, Rees LB, and Matheny RT, "PIXE Elemental Analysis of South American Mummy Hair," Nucl. Instru. Meth. Phys. Res. B, (1996).
7) Stephenson,J. JAMA Vol. 281 No. 17,May 5, 1999.
8) Cornelis R., Speecke A. Neutron activation analysis of human hair collected at regular intervals for 25 years J. Forensic Sci. Soc., 11/1 (1971), 29-46.
9) Tomoaki Sakamoto, Akira Tanaka, and Yuji Nakahara Journal of Analytical Toxicology, 20, pp. 124-130 (1996).
10) Baumgartner WA, Hill VA, Blahd WH. Journal of Forensic Sciences , 34 , pp. 1433-1453 (1989).
11) Esteban E, Rubin C, Jones R, and Noonan G. Archives of Environmental Health, 54, pp. 436-440 (1999).
12) Assarian GS and Oberleas D. Clin. Chem. Vol. 23, pp. 1771-2 (1977).
13) Buckley RA and Dreosoti IE. Am. J. Clin. Nutr. Vol. 50., pp. 840-6 (1984).
14) Wilhelm M., Ohnesorge FK, Lombeck I, and Hafner,D. J. Anal. Toxicol. Vol. 13, pp. 17-21 (1989). Daniel C. Paschal–CDC C-58 15) Sky-Peck HH. Clin. Phys. Biochem., Vol. 8, pp. 70-80 (1990).
16) Koren G. Forensic Sci Int. Vol 70 (1-3) pp 77-82 (1995).
17) Sachs H. Forensic Sci Int. Vol 70 (1-3) pp 53-61 (1995).
18) Taneja SK, Mohajan M, Gupta S, Singh. KP, Biol Trace Elem Res , 62, pp. 255-264 (1998).
19) Pihl RO, Drake H, Vrana F. Department of Psychology, McGill University, Montreal, Quebec, Canada.: Hair Analysis in Learning and Behavior Problems. Hair, Trace Elements, and Human Illness. Brown, A. C.; Crounse, R. G., eds. Praeger Publications, 1980.
20) Yoshinaga J, Imai H, Nakazawa M,. and Suzuki T. Sci Total Environ. Vol. 99, pp. 125-135 (1990).
21) Sherlock JD, Lindsay DG, Hislop JE, Evans WH, and Collier TR. Archives of Environmental Health, 37, pp. 271-8 (1982).
22) Wibowo AE, Herber RM, Das HA, Roeleveld N, and Zielhuis RL. Environmental Research, 40, pp. 346-56 (1986).
23) MMWR, March 02, 2001 / 50(08);140-3.
24) "High hair manganese in children with attention deficit-hyperactivity disorder," EJ Cordova, FM Crinella, and JE Ericson, unpublished study. Address: F. M. Crinella, UC Irvine, Child Development Center, 19722 MacArthur Blvd., Irvine, CA 92612.
25) Kirschmann G & J. Nutrition Almanac, 4th ed. New York: McGraw Hill, 1996.
26) (PT) 27) (CTQ ICP-MS hair specimens) 28) Steindel SJ, Howanitz, PJ. JAMA, 285(1), (2001).
29) Wennig R. Forensic Science International, 107, pp. 5-12 (2000).
30) (IAEA reference materials) Daniel C. Paschal–CDC C-59 C-61 Sharon Seidel Sharon Seidel C-63 ATSDR Hair Analysis Workshop - Charge questions: Topic #1: Analytical Methods.
Atomic absorption spectroscopy (AAS) is commonly used for individual elements, and can now do more than one element at a time. Lead, for example, is commonly measured by graphite furnace AAS. A well-established conventional laboratory with forensic services typically measures individual elements or a small panel of elements in hair for chronic exposure (e.g., first panel -mercury by cold vapor AAS; lead, arsenic, chromium and cadmium by graphite furnace (GF)-AAS; second panel - cadmium, manganese, nickel and thallium, all by GF-AAS). The AAS methods are considered well-established methods. The amount of hair required for either AAS panel (above) is 0.5 gram. Other analytical methods have the potential to measure a number of elements simultaneously, including inductively coupled plasma-atomic emission spectroscopy (ICP-AES) and mass spectroscopy (ICP-MS). Newer ICP-AES instruments can attain a sensitivity equivalent to single element AAS. ICP-MS is a more sensitive method than AES. In a carefully conducted study, a major research laboratory at the Centers for Disease Control and Prevention (CDC) reported the determination of 28 elements in hair from non-occupationally exposed U.S. populations.1 These investigators used ICP-AES for all elements except mercury, which was measured with an LDC mercury monitor. The required minimum hair sample weight was 0.5 gram. Miekeley et al. more recently reported results for ICP-MS analysis of a suite of elements from hair in a Brazilian population, with improved sensitivity compared to ICP-AES.2 The amount of hair required was approximately 0.3 gram.
Of the 9 commercial “nutritional hair analysis” laboratories currently operating in the United States, 3 indicate that they primarily use ICP-MS, 4 primarily use ICP-AES, and 1 reports use of directly coupled plasma (DCP)-AES. DCP-AES is an older technique that is potentially less stable than ICP-AES. On average, these laboratories measure 26 elements per hair sample.
Nutritional hair analysis laboratories require between 0.3 and 1 gram for the AES methods, and Sharon Seidel C-64 0.25-1 gram for ICP-MS. Puchyr et al. also discuss preparation of hair for elemental analysis by ICP-MS from a nutritional hair analysis laboratory perspective. 3 Other investigative techniques for measuring elements in hair are reported in the scientific literature. A general discussion of common methods is provided by Jacobs and by Haraguchi et al.
4,5 Various other methods and example references, e.g: Differential pulse voltametric (DPV); 6 Instrumental Neutron Activation Analysis (INAA); 7,8 Microwave-Induced Plasma Mass Spectroscopy (MIP-MS); 9 Capillary electrophoresis (CE) and High Performance Liquid Chromatography (HPLC); 10 and Particle Induced X-ray Emission (PIXE).11 Laboratory variability has been investigated for the commercial “nutritional hair analysis” laboratories on several occasions.2,12-14 Inter-laboratory variability was high for reference ranges, results, interpretations and health advice. For example, for one hair sample that was split and sent to six of the laboratories, there was a difference of an order of magnitude or more between laboratories in reported results for over 10 elements, including arsenic, lead, and mercury.13 In the same split hair sample, no two laboratories flagged the same element as high, and laboratories had conflicting health interpretations and dietary recommendations based on their analysis of the sample. When intra-laboratory variability was investigated for nutritional hair analysis laboratories, results were similarly discrepant.12 Topic #2: Factors Influencing the Interpretation of Analytical Results.
A.) Sample collection and analysis: Sample collection and preparation methods can have a significant impact on the data collected. Hopps notes that scalp hair has about 90% of follicles in the growth phase at any given time, growing at about 0.45 mm/day.15 Scalp hair grows in a mosaic pattern over the scalp, with similar growth activity in the various regions of the scalp. However, sampling near the face is usually avoided due to increased likelihood of contamination from sebaceous secretions and facial hygiene products/cosmetics. Miekeley et al. note that larger samples of scalp hair (50 g.), cut into Sharon Seidel C-65 <1cm pieces and manually homogenized, showed homogeneity in repeated analyses of aliquots of the samples.2 Commercial nutritional hair analysis laboratories frequently offer the option of collecting samples of axillary or pubic hair. Hair from these regions of the body grows more slowly, with a much greater proportion in the resting phase, and is likely to be subject to external contamination from apocrine gland secretions, in addition to use of personal hygiene products, clothing, etc.
There are no published reference ranges for elements from non-scalp hair. A lack of correlation has been shown between scalp and pubic hair for Ca, Cu, Fe, Mg and Zn.16 Homogenization can be a concern, particularly if long lengths of hair are collected.
Concentrations in hair of a number of environmentally-important elements have been shown to increase from the proximal to distal end of hair, e.g. Pb, Cu, Fe, Mn, and Zn.17,18 Contamination is also a concern if the laboratory uses a grinding tool that introduces contaminants, as occurred in the preparation of one hair reference material, where Al, Fe, Ti, Mn, and Mg contamination were introduced through use of an agate ball grinding mill.19 Sample preparation and washing methods vary greatly and can cause different analytical results. Chittleborough provides a detailed review of these issues.20 Various washing recommendations include: no-wash;20 use of a standardized washing procedure recommended by the International Atomic Energy Agency (IAEA) which uses a nonpolar solvent-acetone and deionized water;21 a mild ionic detergent-sodium lauryl sulfate emulating a detergent shampoo;1 and more extreme methods including a chelating agent, EDTA;22 and others (see review by Chittleborough).20 There is no washing method presently available which is capable of reliably removing external contaminants without also affecting endogenously-deposited elements.20,23-25 While a no-wash approach offers the least disturbance to endogenous elements, the demonstration by scanning electron microscopy of dust, dead skin, etc., adhering to much of the length of unwashed hair samples discourages use of this approach.26 Other aspects of laboratory sample Sharon Seidel C-66 preparation that may be critical include procedures which minimize loss of more volatile elements, such as mercury, during sample dissolution.
A major stumbling block in interpreting metals data for hair is laboratory analytical error.
The World Health Organization recommends the following quality assurance methods for laboratory analyses. 1) Reference samples of the same matrix (hair) with known concentrations of the metal (element) should be used as standards. 2) Reference samples should contain the metal (element) at about the same concentration as the samples. 3) If such reference materials are not available, analysis of quality-control samples at different laboratories by different analytical methods must be used (i.e., split samples). 4) Since results may vary over time and for different metals (elements), results should be presented for the corresponding time periods and elements.27 There are various certified reference materials (CRM), for one (mercury) or multiple elements in hair, which meet certification requirements including certified values with a stated level of confidence in each value.19,28-30 There is no certified hair reference material for all elements currently analyzed by commercial “nutritional hair analysis” laboratories. The Chinese hair CRM, reportedly used by four of these laboratories, certifies 17 elements: Al, As, Ca, Cd, Co, Cr, Cu, Fe, Hg, Mg, Mn, Na, Ni, Pb, Se, Sr, and Zn - about half the elements tested by these laboratories.
A common practice among these and other laboratories is to use aqueous element standards, or other non-hair standards such as bovine liver. The difficulty with this is the possibility of complex matrix interferences in the hair sample that are not accounted for by the calibration standard.
Reference ranges cited by commercial U.S. nutritional hair analysis laboratories show some rather broad inter-laboratory variations, e.g. arsenic (<0.06 vs. <5 ppm), lead (<0.8 vs. 2-20 ppm), and lithium (0.0035-0.025 vs. 1.25-3 ppm).13 Investigations of “nutritional hair analysis” laboratory practices using split samples have shown wide discrepancies.2,12,13 An approved proficiency testing program for hair element analysis is not available under the Clinical Laboratory Improvement Act (CLIA). This type of analysis is Sharon Seidel C-67 classified as a high-complexity test, with method and accuracy verification left up to the individual laboratory.
“Normal” reference ranges are largely undefined, due to the wide variation in elemental hair concentrations in presumed healthy populations. Contributing factors include geography, age, sex, ethnicity, hair type, hair treatments and other exogenous exposures. Examples of U.S. studies follow. DiPietro et al. published reference intervals for 28 elements in a non-occupationally exposed U.S. adult population.1 These investigators used extensive questionnaire data to control for many of these factors. A partial list of geometric means for healthy adults in this study includes: arsenic (0.15 ppm); cadmium (<0.15 ppm); nickel (0.39 ppm); and lead (2.43 ppm). A number of population studies have been conducted for mercury in hair. For methylmercury, the geometric mean hair concentration for U.S. women reporting some seafood consumption was 0.36 ppm, and 0.24 ppm for no seafood consumption.31 Published clinical references for biomonitoring for metals/elements in hair are sparse. These include arsenic (<1 ppm) and thallium (?5-10 ppb)32 and mercury (<1ppm) and nickel (0.01-1.8 ppm).33 These are secondary to the established blood and/or urine reference levels, and the problem of external contamination is noted as a major stumbling block which limits the use of the hair references. Generally speaking the use of the term “normal” is misleading. What is being estimated is a background or baseline level for a population, typically by geographic region, rather than a state of health. Methylmercury data are an exception. Methylmercury exposures commonly occur through consumption of fish and seafood. Clear dose-response relationships have been demonstrated between dietary consumption of mercury-contaminated fish and concentrations in human hair. Methylmercury is the only metal (compound) which has a health benchmark based on hair concentrations. The U.S. EPA has a reference dose (RfD) for methylmercury of 0.0001 mg/kg body wt/day. This is based on a benchmark dose of 11 ppm in maternal hair, equivalent to a maternal blood level of 44 micrograms/L, for developmental neurological abnormalities in infants.34 Several reference range studies for methylmercury are available.31,35,36 Sharon Seidel C-68 B.) Other factors influencing analytical results: Hopps notes the sources of elements in hair as: 1) papilla (contacted by blood and lymph) during hair formation; 2) sebaceous glands, sweat glands, desquamating skin cells (endogenous exposures not necessarily related to blood/organ concentrations); 3) and exogenous materials.2 Salts of sodium, potassium, and calcium predominate in sweat, but minor amounts of other elements are also found, e.g., zinc.15 There is evidence for an extra input route from the root sheaths into the hair shaft, other than longitudinal growth, complicating the picture of a simple blood compartment:hair relationship.11 Finally, the lipids and waxes in sebum and skin may contribute to sealing exogenous contaminants into the hair shaft.
Exogenous contaminants can range from personal care products to elements present in air, water, soil, occupational environments, etc. As mentioned above, there is currently no washing method capable of removing exogenous elemental contaminants while leaving endogenous elements undisturbed. Chemicals such as methylmercury, which are generally from dietary sources, suffer less from this drawback, provided unusual sources of inorganic mercury do not complicate the picture, e.g., mercury vapor in occupational settings. Practically speaking, public health concerns are often related to exposure, and hair can serve as an index of overall exposure, if not of biological uptake.
Examples of external contaminants of hair include both personal care products and environmental sources. Hair is a porous material (witness the rapid uptake of water and increase in weight during washing) and may bind through weak ion-exchange sites (e.g., Na, K, Ca, Mg), and through stronger bonds, particularly with sulfur, (e.g, arsenic). Arsenic binds avidly to hair, due to the sulfur content of keratin. Exogenous arsenic is readily taken up by hair and cannot be differentiated from endogenous arsenic.15 It has been shown that adsorption of other metals such as Al, Cd, Cu, Pb and Zn into scalp hair from aqueous solutions cannot be reversed even by extreme washing methods.25 Hair treatments such as permanents can alter such binding.37 Dandruff shampoos containing selenium can contaminate hair.38 DiPietro et al. noted significant Sharon Seidel C-69 difference between dandruff shampoo vs. regular shampoo for Na, Se, and Ti for men, and between permanents/color and any shampoo for Ba, Ca, Cu, Mg, Na, and Sr for women.1 Hair dyes may contain metals, e.g., lead in “Grecian Formula.”39 Sky-Peck also found that peroxide bleaches and permanents altered S, Ca, Fe, and Ni in hair, peroxide altered Zn, and permanents increased Cu and As concentrations.39 Soil, house dust, and water may contribute contaminants.40 Air serves as a contamination source.41 This is a major concern in occupational settings. Cadmium is an example of a metal where environmental sources contribute to concentrations in hair, e.g., drinking water and dust levels and seasonal influences.42 As noted, gender, ethnicity, diet, age, geographic location, and season are capable of influencing hair reference ranges in populations. Sky-Peck found the following for a healthy midwestern U.S. population: 1) gender – females had higher Ca and Ni and lower Pb, Br and Se compared to males; 2) hair color - blondes had less Fe than brunettes, red-heads had more Fe and Cu; 3) ethnicity/race - Blacks had increased Ca, Fe, Ni, Cr, Mn, As, and Pb, and decreased Hg, compared to Caucasians; Orientals had decreased Ca, Fe, Cu, Mn and Pb; 4) age – a decrease in S, Ca and Sr, and an increase in Pb with age; 5) geography – increased hair strontium in areas with elevated strontium in drinking water, and increased hair lead in industrial/older residential areas.39 Sky-Peck notes that some of the differences in gender and ethnicity may be due to differences in hair treatment and/or environmental exposure. While Sky-Peck found no differences between gray hair and natural hair, other investigators have noted pigmentation effects,43 and it is known that various chemicals, including metals, will bind to melanin.44 Other investigators have studied age-related differences in hair elements. Paschal et al.
observed age-dependent increases in Ca, Ba, Mg and Sr (Group 2A alkali elements) and Zn up to 12-14 yrs in U.S. residents.45 In comparison, an Italian study showed increases in Cu, Zn, Cr and Br, and decreases in Fe, Mn and Sr up to 8 yrs.46 In Japanese children, Zn decreased up to 12-14 Sharon Seidel C-70 yrs, and Cu showed a similar trend.47 The reason for differences between laboratories and/or populations is not presently known.
Baseline reference values for elements in clinical specimens, including hair, have been referenced by international location.48,49 International differences are identified for hair Zn, Cd, Cu, Mn and Pb. Some of this geographical difference may be due to differences in environmental metal concentrations, industrialization, etc. Seasonal differences in hair element concentrations, e.g. cadmium, may be due to time spent outdoors and contact with soil, dust, etc.
Topic #3: Toxicological Considerations.
As discussed above, methylmercury is the only element (compound) for which sufficient data exist to define the relationship between concentrations in blood, concentrations in hair, and effects on the target (the developing fetus). It is also the only element (compound) with a health benchmark, the U.S. EPA reference dose, based on a threshold concentration in human hair. It should be noted that this threshold was identified based on massive poisoning incidents in human populations and not on typical (dietary) exposures.34,50 Forensic medicine has used hair to assess poisoning by other elements, e.g, arsenic and lead. However, these document overwhelming poisoning exposures, rather than a threshold for earliest/most subtle adverse health effects. Nor is there a need in these instances to differentiate between a “normal” background and subtle increases in exposures. Such a distinction is difficult due to the wide variations in background reference ranges. This has caused a number of investigators to conclude that results for an individual are not likely to be meaningful with respect to less drastic environmental/dietary exposures, and that statistical analyses of group data must be employed.13,42,51,52 Finally, if the goal is also to provide an index of body burden, rather than simply document exposure to environmental contaminants, the lack of a washing technique capable of reliably separating exogenous contaminants from biologically-deposited elements is a substantial concern and must be addressed. Sharon Seidel C-71 Of the trace elements that have been tested in hair, only a few have research data relating hair concentrations to blood levels and/or tissue concentrations. Aside from mercury, the focus has largely been on aluminum, arsenic, cadmium, chromium, copper, lead, selenium, and zinc.
Data highlights are summarized below.
Aluminum (Al) – Aluminum is elevated in hair only in extreme exposures (and even then is inconsistent), and is unrelated to serum or bone aluminum.53-56 Aluminum dietary intake is unrelated to aluminum in hair, even with controlled dietary intake.57 Aluminum in hair is not a useful biological indicator of exposure.
Arsenic (As) – Arsenic is well taken up in hair. Animals show a dose-related increase in hair arsenic.42 Forensic hair tests can determine the time-course of chronic arsenic poisoning.58 Increased arsenic in soil (<20 to 370 ppm soil As) show a slight correlation with slightly elevated hair arsenic using group statistics (0.02 ppm to 0.06 ppm hair As).59 Consumption of drinking water with elevated arsenic concentrations showed a correlation with hair arsenic, using group comparisons.60-62 This correlation was not seen in a study where drinking water exposure was only modestly above a legal threshold.63 Group statistics show elevated hair As in patients with Blackfoot disease.64 Cadmium (Cd) – Animal studies show conflicting results with respect to any correlation between cadmium in hair and the target organ, the kidney.42,52 The most significant non-occupational exposure to cadmium occurs through tobacco smoke. Smokers have elevated blood cadmium levels compared to nonsmokers. Studies show conflicting results with respect to hair cadmium concentrations in smokers versus non-smokers.65-67 A nationwide German environmental survey found little correlation with cadmium in hair and active cigarette smoking, although it was the major predictor for blood and urine cadmium concentrations. In contrast, outdoor activities, seasonality, and cadmium in tap water were more important predictors in hair cadmium concentrations, emphasizing the role of exogenous deposition of cadmium into hair.67 Copper (Cu) – Taylor’s review notes that animal studies showed a proportional relationship between copper in hair and liver.52 Yoshinaga et al. found no significant correlation between hair copper and various internal organs, including the liver, in autopsy samples.68 Sharon Seidel C-72 Literature studies of human populations show conflicting results with respect to hair versus serum copper.52,69,70 Serum copper is generally higher in women than in men.33,49 However, hair copper is inconsistent with respect to sex. Contiera and Folin found no effect of sex on hair copper.71 Sky-Peck found a modest correlation (p<0.025) for higher hair copper in women compared to men (24 vs. 20 ppm).39 In human patients with biliary cirrhosis, or Wilson’s disease (systemic copper intoxication), with increased liver copper, hair copper was typically not increased.52 Further studies of Wilson’s disease confirmed these findings, with no increase in hair copper in patients with this disease.72 In copper deficiencies (malnutrition or Menkes syndrome), hair copper was not significantly reduced.52 Chromium (Cr) – Studies of hair chromium are somewhat limited. A large study (40,872 patients) in England found age-related decreases in hair chromium for males and females [0.98 ppm (mean at age 1-4 yrs) to 0.5 ppm (mean at age 70 plus yrs)], slightly lower hair chromium in males ages 25-49 years, and a correlation between hair and serum chromium, all statistically significant.73 In comparison, a U.S. study found no difference in hair chromium by sex or age in 987 individuals.39 Hair chromium has been hypothesized to increase in gestational diabetes (in early pregnancy), compared to non-diabetic pregnant women.74 Hair chromium measurements have been used in monitoring occupational exposures, although blood and urine chromium are the standard biological indices.75 Lead (Pb) – There are a number of studies relating lead exposure to tissue concentrations, including hair. Animal studies show a dose-dependent correlated increase in lead in bone and hair during the exposure period.76 Isotopic tracer studies have shown the deposition of lead into human facial hair, interpreted as the integral of a blood lead pool over approximately 3 months.77 In humans, hair analysis can be used to demonstrate lead poisoning.72 Occupational exposures show a correlation between blood and hair lead.52,78 Lower-level exposures have more variable results,52 but larger studies appear to support a relationship between hair and blood lead.42 Exogenous deposition of lead onto scalp hair may be influential, e.g., season, dust exposure, and hair treatment.42 Centers for Disease Control (CDC) investigators compared hair and blood samples from 189 children to gauge the accuracy of using hair to screen for lead poisoning (mean Sharon Seidel C-73 blood lead 9.8 ug/dl; mean hair lead 7.2 ppm).79 Hair lead as a screening method had a 57% sensitivity and 18% false-negative rate. The investigators concluded that hair lead measurements are NOT an adequate method of screening for childhood lead poisoning. The reliable measure of individual lead exposure is a blood lead test.
Selenium (Se) - Animal studies show that: 1) hair selenium is strongly influenced by the chemical form of selenium and the level in the diet, with a greater increase for L-selenomethionine than sodium selenate 2) sodium selenate increases hair selenium but not muscle selenium (the largest body Se pool); and 3) dietary methionine deficiency increases selenium deposition in hair.80 These observations suggest caution when evaluating environmental selenium exposures.
Population measurements have shown a correlation between low hair selenium and selenium-deficient soils.81 The hair-to-blood selenium ratio is calculated to be ?3 in dietary selenium deficiency, increasing to 10 as toxic levels are approached. Hair selenium will continue to rise far beyond the plasma saturation concentration, indicating contribution from another body pool.82 A hair concentration of >5 ppm Se is reported to be associated with elevated exposure, while a concentration <0.12 ppm Se is reported to be associated with chronic selenium deficiency.83 However, most population studies have preferred blood or urine to indicate selenium exposure.84 Exogenous contamination with selenium-containing dandruff shampoos is a serious confounding factor in developed countries.1 Yoshinaga et al. found no significant correlation between selenium concentrations in hair and in internal organs.68 Zinc (Zn) – Zinc in hair has been reviewed by several authors.52,85-88 These reviewers note that hair is a difficult medium for interpretation of zinc status. The interpretation of zinc concentrations in hair can be obscured by confounders such as sex, body composition, and hair treatment.89 In severe zinc deficiency, hair growth slows, producing normal or even elevated hair zinc concentrations.87 Yoshinaga et al. found no significant correlation between concentrations of zinc in hair and in various internal organs.68 Administration of zinc in the diet did not increase zinc in beard hair.90 Serum zinc is typically decreased in dialysis patients. Hair zinc in these patients is not consistent with serum findings.91 Sharon Seidel C-74 In conclusion, with the exception of methylmercury, there is no good indication that hair analysis offers any improvement over currently available clinical tests to determine individual biological exposure to metals/metalloids of concern.92 Occupational texts note that hair analysis is unproven to detect toxic chemicals in the body to account for symptoms and inappropriate in the diagnosis of “environmental” illness.93 Group statistics on hair data, preferably geometric means, may be useful in population screening for exposure to some of these metals (e.g., arsenic).
Confounding factors, such as hair treatments, must be controlled for in these studies. Analysis of hair minerals to predict nutritional status is a practice not supported by the state of the science.
Topic #4. Data Gaps and Research Needs.
Generally speaking, further information is needed on concentrations of elements in the hair of individuals with known exposures to trace elements, particularly where environmental exposures are of concern. Laboratory studies of elemental concentrations in blood and target tissues compared to hair concentrations are needed. Such data are important if one is to hypothesize that there is a relationship between hair element concentrations and critical/target organ effects. Clinical studies correlating hair concentrations with clinical conditions (deficiencies or elevations) may also be helpful. Further work is needed on sample washing methods.
Standardization on one washing method is important for comparison of studies.
Specific recommendations: ? Do not use hair analysis for individual nutritional assessment. The state of the science does not support this application.
? If hair analysis is undertaken for comparison of groups, choose element(s) for which the literature supports such an approach, e.g., methylmercury, e.g., NOT aluminum.
? When studying control versus exposed groups, chose a group size of sufficient statistical power to determine differences between group means, based on current literature findings.
? Use geometric means in analyzing group data.
Sharon Seidel C-75 ? Collect blood and/or urine samples for comparison with the hair results in the analysis of group data. If this is not feasible for the entire study population, choose a subset of sufficient size to provide statistically meaningful comparison data.
? A questionnaire should be administered to each individual in the study, determining: age, sex, ethnicity, hair wash and hair treatment history including products used on hair, swimming habits, time spent outdoors, occupation, smoking history, etc. (e.g, DiPietro et al., 1989).
Sharon Seidel C-76 Topic #5: Identifying scenarios for which hair analysis may be appropriate.
Exposure Scenario Chemical/ Exposure Pathway Exposure Chronology Exposure Duration Measurable Health Effects (Y/N) Individual – severe poisoning/forensic Group/population: Mercury, Arsenic, Lead Methylmercury-diet (fish, seafood) Arsenic, Cadmium, Lead Past / present Past / present Past / present Acute (1-2 months min.); chronic Acute (1-2 months min.); chronic Acute (1-2 months min.); chronic Possible with very high exposure Unlikely unless very high exposure Unlikely unless very high exposure References 1.
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7266 | https://www.youtube.com/watch?v=NhrPI7utUrQ | Power of a Point Theorem in Geometry
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Posted: 23 May 2021
The power of a point theorem is a useful result in Euclidean geometry about lengths. Without proof, we present the theorem in a couple of ways and mention the various cases that come up in practice. This theorem is very useful on math competitions and olympiads.
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Transcript:
Definition [Music] hi everyone let's talk about the power of a point theorem in geometry first we need to define what the power of a point means so let's say we have a circle we're going to call it gamma capital gamma and let's say we've got a point p somewhere on the plane then what we do is that we take the center of the circle and we draw the line through these two points so we've got the segment po and what we see is that the power of p with respect to gamma is equal to op squared minus the radius squared which then is equal to by difference of squares op minus r times op plus r so notice that op plus r is this segment here and op minus r is equal to this segment here so really what we're doing is that we're taking the first part of the segment and we're multiplying it by the whole second segment from p to the other end so Three Cases that is called the power of a point p but that's not the theorem in itself that's just a definition what the theorem says is that there are three cases let's let's draw three diagrams here because they're going to correspond to the cases the first one we've got p as an exterior point which cuts through the circle in a couple of spots let's call them x and y the second case p is an interior point and there's a line segment that cuts through the circle at two spots again so p and x and y and the final one is where we have p as an exterior point but there's a line segment that's tangent to the circle we're going to call that z so p and z so here we have an exterior point and a secant here we have an interior point and a cord and here we have an exterior point and a tangent and what we have here is that the p x times p y is equal to the power with respect to gamma of p in the second case we have px times p y is equal to the negative of the power with respect to gamma of p and in the final case we have pz squared so it's sort of like a limiting case of the first one pz times pz is equal to power of gamma with respect to p so that's that's what the power of the power of a point theorem says and unlike pretty much all of our other videos i'm not going to prove this i know proof by uh doing case work and that uses similar triangles and a very nice proof that uses complex numbers but both would take up too much time what instead i'm gonna do is show you the various cases that show up because the theorem is a bit abstract and the practical cases are different Practical Cases here's what the practical cases say there are there are three cases so let's draw three circles again what the practical cases say is that if we have two intersecting chords and not necessarily intersecting in the center we're going to call the intersection p then they have a common value by they i mean a certain expression so hold on a second this is a b and this is c d so p a times pb is equal to pc times pd and the reason is that the common value is the power with respect to gamma of p so that comes up often the second case that comes up often is where we have an external point and two secants let's call this p and this point here a and this b and this one c and this one d then in this case again we have p a times p b is equal to p c times p d and once again the common value is i'm sorry back here it's negative negative power of gamma of p that's the common value over here though the common value of is power with respect to gamma of p and the final case that we have is where we have one secant and the other is a tangent so we're going to call this p a b and c and what we have is p a times p b is equal to p c squared and once again the common value is the power with respect to gamma of p so those are the cases that come up in practice of the power of a point theorem thank you for watching and i'll see you next [Music] time |
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Show more… (100)Creating Sparse Matrices in Oct-FilesCreating StringsCreating StructuresCreation and Manipulation of Sparse MatricesCurrent Working DirectoryCursor MotionCustomizing readlineCustomizing the PromptCustomizing Toolkit BehaviorDActual Bugs We Haven’t Fixed YetData ContainersData RetrievalData Sources in Object GroupsData Structure ObjectsData TypesDebug ModeDebuggingDefault ArgumentsDefining FunctionsDefining Indexing And Indexed AssignmentDelaunay TriangulationDerivatives / Integrals / TransformsDescriptive StatisticsDHave You Found a Bug?DHow to Distinguish Between Octave and MatlabDHow To Get Help with OctaveDHow to Report BugsDiagonal and Permutation MatricesDiagonal Matrix FunctionsDiary and Echo CommandsDifferences in Treatment of Zero ElementsDifferential EquationsDifferential-Algebraic EquationsDisplaying ImagesDistributionsDocumentation and Testing of Oct-FilesDReporting BugsDSending Patches for OctaveDWhere to Report BugsEBuild DependenciesEBuild ToolsECompiling Octave with 64-bit IndexingEExternal PackagesEInstallation ProblemsElement-by-element Boolean OperatorsEmpty MatricesEnabling and Disabling WarningsEnd of File and ErrorsEntering Debug ModeEnvironment VariablesEObtaining the Dependencies AutomaticallyError Bar SeriesError MessagesErrors and WarningsERunning Configure and MakeEscape Sequences in String ConstantsEvaluating PolynomialsEvaluationEvaluation in a Different ContextEvaluation NotationExamplesExamples of UsageException and Error Handling in Oct-FilesExecutable Octave ProgramsExplicit and Implicit ConversionsExponents and LogarithmsExpressionsExpressions Involving Diagonal MatricesExpressions Involving Permutation MatricesFamous MatricesFigure PropertiesFile Archiving UtilitiesFile PositioningFilesystem UtilitiesFinding Elements and Checking ConditionsFinding Information about Sparse MatricesFinding RootsFKeywordsFloating-Point ConversionsFontsFormat of DescriptionsFormatted InputFormatted OutputFParserFTP ObjectsFunction ApplicationFunction FilesFunction HandlesFunction Handles, Anonymous Functions, Inline FunctionsFunction LockingFunction OverloadingFunction PrecedenceFunctions and ScriptsFunctions of a MatrixFunctions of Multiple VariablesFunctions of One VariableFunctions That Are Aware of These MatricesGeometryGetting StartedGetting Started with Mex-FilesShow more… (274)
6 Manual: Appendices4 Manual: Indexes1428 Functions
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28.1 Evaluating Polynomials
The value of a polynomial represented by the vector c can be evaluated at the point x very easily, as the following example shows:
N = length (c) - 1;
val = dot (x.^(N:-1:0), c);
While the above example shows how easy it is to compute the value of a polynomial, it isn’t the most stable algorithm. With larger polynomials you should use more elegant algorithms, such as Horner’s Method, which is exactly what the Octave function polyval does.
In the case where x is a square matrix, the polynomial given by c is still well-defined. As when x is a scalar the obvious implementation is easily expressed in Octave, but also in this case more elegant algorithms perform better. The polyvalm function provides such an algorithm.
y = polyval (p, x)y = polyval (p, x, [], mu)[y, dy] = polyval (p, x, s)[y, dy] = polyval (p, x, s, mu)
Evaluate the polynomial p at the specified values of x.
If x is a vector or matrix, the polynomial is evaluated for each of the elements of x.
When mu is present, evaluate the polynomial for (x-mu(1))/mu(2).
In addition to evaluating the polynomial, the second output represents the prediction interval, y +/- dy, which contains at least 50% of the future predictions. To calculate the prediction interval, the structured variable s, originating from polyfit, must be supplied.
See also:polyvalm, polyaffine, polyfit, roots, poly.
polyvalm (c, x)
Evaluate a polynomial in the matrix sense.
polyvalm (c, x) will evaluate the polynomial in the matrix sense, i.e., matrix multiplication is used instead of element by element multiplication as used in polyval.
The argument x must be a square matrix.
See also:polyval, roots, poly.
© 1996–2018 John W. Eaton
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7268 | https://webbook.nist.gov/cgi/cbook.cgi?ID=C64197&Mask=200 | Jump to content
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Acetic acid
Formula: C2H4O2
Molecular weight: 60.0520
IUPAC Standard InChI: InChI=1S/C2H4O2/c1-2(3)4/h1H3,(H,3,4)
IUPAC Standard InChIKey: QTBSBXVTEAMEQO-UHFFFAOYSA-N
CAS Registry Number: 64-19-7
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Acetic acid-d4
Other names: Ethanoic acid; Ethylic acid; Glacial acetic acid; Methanecarboxylic acid; Vinegar acid; CH3COOH; Acetasol; Acide acetique; Acido acetico; Azijnzuur; Essigsaeure; Octowy kwas; Acetic acid, glacial; Kyselina octova; UN 2789; Aci-jel; Shotgun; Ethanoic acid monomer; NSC 132953
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7269 | https://www.coconino.edu/resources/files/pdfs/academics/arts-and-sciences/college-mathematics/3rd/chapter-6-graph-theory.pdf | Chapter 6: Graph Theory __________ __________ Page 221 Chapter 6: Graph Theory Graph theory deals with routing and network problems and if it is possible to find a “best” route, whether that means the least expensive, least amount of time or the least distance. Some examples of routing problems are routes covered by postal workers, UPS drivers, police officers, garbage disposal personnel, water meter readers, census takers, tour buses, etc. Some examples of network problems are telephone networks, railway systems, canals, roads, pipelines, and computer chips. Section 6.1: Graph Theory There are several definitions that are important to understand before delving into Graph Theory. They are: • A graph is a picture of dots called vertices and lines called edges. • An edge that starts and ends at the same vertex is called a loop. • If there are two or more edges directly connecting the same two vertices, then these edges are called multiple edges. • If there is a way to get from one vertex of a graph to all the other vertices of the graph, then the graph is connected. • If there is even one vertex of a graph that cannot be reached from every other vertex, then the graph is disconnected. Example 6.1.1: Graph Example 1 Figure 6.1.1: Graph 1 In the above graph, the vertices are U, V, W, and Z and the edges are UV, VV, VW, UW, WZ1, and WZ2. This is a connected graph. VV is a loop. WZ1, and WZ2 are multiple edges. Chapter 6: Graph Theory __________ __________ Page 222 Example 6.1.2: Graph Example 2 Figure 6.1.2: Graph 2 Figure 6.1.3: Graph 3 The graph in Figure 6.1.2 is connected while the graph in Figure 6.1.3 is disconnected. Graph Concepts and Terminology: Order of a Network: the number of vertices in the entire network or graph Adjacent Vertices: two vertices that are connected by an edge Adjacent Edges: two edges that share a common vertex Degree of a Vertex: the number of edges at that vertex Path: a sequence of vertices with each vertex adjacent to the next one that starts and ends at different vertices and travels over any edge only once Circuit: a path that starts and ends at the same vertex Bridge: an edge such that if it were removed from a connected graph, the graph would become disconnected Example 6.1.3: Graph Terminology Figure 6.1.4: Graph 4 In the above graph the following is true: Chapter 6: Graph Theory __________ __________ Page 223 Vertex A is adjacent to vertex B, vertex C, vertex D, and vertex E. Vertex F is adjacent to vertex C, and vertex D. Edge DF is adjacent to edge BD, edge AD, edge CF, and edge DE. The degrees of the vertices: A 4 B 4 C 4 D 4 E 4 F 2 Here are some paths in the above graph: (there are many more than listed) A,B,D A,B,C,E F,D,E,B,C Here are some circuits in the above graph: (there are many more than listed) B,A,D,B B,C,F,D,B F,C, E, D, F The above graph does not have any bridges. Section 6.2: Networks A network is a connection of vertices through edges. The internet is an example of a network with computers as the vertices and the connections between these computers as edges. Spanning Subgraph: a graph that joins all of the vertices of a more complex graph, but does not create a circuit Chapter 6: Graph Theory __________ __________ Page 224 Example 6.2.1: Spanning Subgraph Figure 6.2.1: Map of Connecting Towns This is a graph showing how six cities are linked by roads. This graph has many spanning subgraphs but two examples are shown below. Figure 6.2.2: Spanning Subgraph 1 This graph spans all of the cities (vertices) of the original graph, but does not contain any circuits. Figure 6.2.3: Spanning Subgraph 2 Chapter 6: Graph Theory __________ __________ Page 225 This graph spans all of the cities (vertices) of the original graph, but does not contain any circuits. Tree: A tree is a graph that is connected and has no circuits. Therefore, a spanning subgraph is a tree and the examples of spanning subgraphs in Example 6.2.1 above are also trees. Properties of Trees: 1. If a graph is a tree, there is one and only one path joining any two vertices. Conversely, if there is one and only one path joining any two vertices of a graph, the graph must be a tree. 2. In a tree, every edge is a bridge. Conversely, if every edge of a connected graph is a bridge, then the graph must be a tree. 3. A tree with N vertices must have N-1 edges. 4. A connected graph with N vertices and N-1 edges must be a tree. Example 6.2.2: Tree Properties Figure 6.2.2: Spanning Subgraph 1 Consider the spanning subgraph highlighted in green shown in Figure 6.2.2. a. Tree Property 1 Look at the vertices Appleville and Heavytown. Since the graph is a tree, there is only one path joining these two cities. Also, since there is only one path between any two cities on the whole graph, then the graph must be a tree. b. Tree Property 2 Since the graph is a tree, notice that every edge of the graph is a bridge, which is an edge such that if it were removed the graph would become disconnected. Chapter 6: Graph Theory __________ __________ Page 226 c. Tree Property 3 Since the graph is a tree and it has six vertices, it must have N – 1 or six – 1 = five edges. d. Tree Property 4 Since the graph is connected and has six vertices and five edges, it must be a tree. Example 6.2.3: More Examples of Trees: All of the graphs shown below are trees and they all satisfy the tree properties. Figure 6.2.4: More Examples of Trees Minimum Spanning Tree: A minimum spanning tree is the tree that spans all of the vertices in a problem with the least cost (or time, or distance). Chapter 6: Graph Theory __________ __________ Page 227 Example 6.2.4: Minimum Spanning Tree Figure 6.2.5: Weighted Graph 1 The above is a weighted graph where the numbers on each edge represent the cost of each edge. We want to find the minimum spanning tree of this graph so that we can find a network that will reach all vertices for the least total cost. Figure 6.2.6: Minimum Spanning Tree for Weighted Graph 1 This is the minimum spanning tree for the graph with a total cost of 51. Chapter 6: Graph Theory __________ __________ Page 228 Kruskal’s Algorithm: Since some graphs are much more complicated than the previous example, we can use Kruskal’s Algorithm to always be able to find the minimum spanning tree for any graph. 1. Find the cheapest link in the graph. If there is more than one, pick one at random. Mark it in red. 2. Find the next cheapest link in the graph. If there is more than one, pick one at random. Mark it in red. 3. Continue doing this as long as the next cheapest link does not create a red circuit. 4. You are done when the red edges span every vertex of the graph without any circuits. The red edges are the MST (minimum spanning tree). Example 6.2.5: Using Kruskal’s Algorithm Figure 6.2.7: Weighted Graph 2 Suppose that it is desired to install a new fiber optic cable network between the six cities (A, B, C, D, E, and F) shown above for the least total cost. Also, suppose that the fiber optic cable can only be installed along the roadways shown above. The weighted graph above shows the cost (in millions of dollars) of installing the fiber optic cable along each roadway. We want to find the minimum spanning tree for this graph using Kruskal’s Algorithm. Chapter 6: Graph Theory __________ __________ Page 229 Step 1: Find the cheapest link of the whole graph and mark it in red. The cheapest link is between B and C with a cost of four million dollars. Figure 6.2.8: Kruskal’s Algorithm Step 1 Step 2: Find the next cheapest link of the whole graph and mark it in red. The next cheapest link is between A and C with a cost of six million dollars. Figure 6.2.9: Kruskal’s Algorithm Step 2 Chapter 6: Graph Theory __________ __________ Page 230 Step 3: Find the next cheapest link of the whole graph and mark it in red as long as it does not create a red circuit. The next cheapest link is between C and E with a cost of seven million dollars. Figure 6.2.10: Kruskal’s Algorithm Step 3 Step 4: Find the next cheapest link of the whole graph and mark it in red as long as it does not create a red circuit. The next cheapest link is between B and D with a cost of eight million dollars. Figure 6.2.11: Kruskal’s Algorithm Step 4 Chapter 6: Graph Theory __________ __________ Page 231 Step 5: Find the next cheapest link of the whole graph and mark it in red as long as it does not create a red circuit. The next cheapest link is between A and B with a cost of nine million dollars, but that would create a red circuit so we cannot use it. Therefore, the next cheapest link after that is between E and F with a cost of 12 million dollars, which we are able to use. We cannot use the link between C and D which also has a cost of 12 million dollars because it would create a red circuit. Figure 6.2.12: Kruskal’s Algorithm Step 5 This was the last step and we now have the minimum spanning tree for the weighted graph with a total cost of $37,000,000. Section 6.3: Euler Circuits Leonhard Euler first discussed and used Euler paths and circuits in 1736. Rather than finding a minimum spanning tree that visits every vertex of a graph, an Euler path or circuit can be used to find a way to visit every edge of a graph once and only once. This would be useful for checking parking meters along the streets of a city, patrolling the streets of a city, or delivering mail. Euler Path: a path that travels through every edge of a connected graph once and only once and starts and ends at different vertices Chapter 6: Graph Theory __________ __________ Page 232 Example 6.3.1: Euler Path Figure 6.3.1: Euler Path Example One Euler path for the above graph is F, A, B, C, F, E, C, D, E as shown below. Figure 6.3.2: Euler Path This Euler path travels every edge once and only once and starts and ends at different vertices. This graph cannot have an Euler circuit since no Euler path can start and end at the same vertex without crossing over at least one edge more than once. Chapter 6: Graph Theory __________ __________ Page 233 Euler Circuit: an Euler path that starts and ends at the same vertex Example 6.3.2: Euler Circuit Figure 6.3.3: Euler Circuit Example One Euler circuit for the above graph is E, A, B, F, E, F, D, C, E as shown below. Figure 6.3.4: Euler Circuit This Euler path travels every edge once and only once and starts and ends at the same vertex. Therefore, it is also an Euler circuit. Chapter 6: Graph Theory __________ __________ Page 234 Euler’s Theorems: Euler’s Theorem 1: If a graph has any vertices of odd degree, then it cannot have an Euler circuit. If a graph is connected and every vertex has an even degree, then it has at least one Euler circuit (usually more). Euler’s Theorem 2: If a graph has more than two vertices of odd degree, then it cannot have an Euler path. If a graph is connected and has exactly two vertices of odd degree, then it has at least one Euler path (usually more). Any such path must start at one of the odd-degree vertices and end at the other one. Euler’s Theorem 3: The sum of the degrees of all the vertices of a graph equals twice the number of edges (and therefore must be an even number). Therefore, the number of vertices of odd degree must be even. Finding Euler Circuits: 1. Be sure that every vertex in the network has even degree. 2. Begin the Euler circuit at any vertex in the network. 3. As you choose edges, never use an edge that is the only connection to a part of the network that you have not already visited. 4. Label the edges in the order that you travel them and continue this until you have travelled along every edge exactly once and you end up at the starting vertex. Example 6.3.3: Finding an Euler Circuit Figure 6.3.5: Graph for Finding an Euler Circuit Chapter 6: Graph Theory __________ __________ Page 235 The graph shown above has an Euler circuit since each vertex in the entire graph is even degree. Thus, start at one even vertex, travel over each vertex once and only once, and end at the starting point. One example of an Euler circuit for this graph is A, E, A, B, C, B, E, C, D, E, F, D, F, A. This is a circuit that travels over every edge once and only once and starts and ends in the same place. There are other Euler circuits for this graph. This is just one example. Figure 6.3.6: Euler Circuit The degree of each vertex is labeled in red. The ordering of the edges of the circuit is labeled in blue and the direction of the circuit is shown with the blue arrows. Section 6.4 Hamiltonian Circuits The Traveling Salesman Problem (TSP) is any problem where you must visit every vertex of a weighted graph once and only once, and then end up back at the starting vertex. Examples of TSP situations are package deliveries, fabricating circuit boards, scheduling jobs on a machine and running errands around town. Chapter 6: Graph Theory __________ __________ Page 236 Hamilton Circuit: a circuit that must pass through each vertex of a graph once and only once Hamilton Path: a path that must pass through each vertex of a graph once and only once Example 6.4.1: Hamilton Path: Figure 6.4.1: Examples of Hamilton Paths a. b. c. Not all graphs have a Hamilton circuit or path. There is no way to tell just by looking at a graph if it has a Hamilton circuit or path like you can with an Euler circuit or path. You must do trial and error to determine this. By the way if a graph has a Hamilton circuit then it has a Hamilton path. Just do not go back to home. Graph a. has a Hamilton circuit (one example is ACDBEA) Graph b. has no Hamilton circuits, though it has a Hamilton path (one example is ABCDEJGIFH) Graph c. has a Hamilton circuit (one example is AGFECDBA) Complete Graph: A complete graph is a graph with N vertices in which every pair of vertices is joined by exactly one edge. The symbol used to denote a complete graph is KN. Chapter 6: Graph Theory __________ __________ Page 237 Example 6.4.2: Complete Graphs Figure 6.4.2: Complete Graphs for N = 2, 3, 4, and 5 a. K2 b. K3 c. K4 d. K5 two vertices and one edge three vertices and three edges four vertices and six edges five vertices and ten edges In each complete graph shown above, there is exactly one edge connecting each pair of vertices. There are no loops or multiple edges in complete graphs. Complete graphs do have Hamilton circuits. Reference Point: the starting point of a Hamilton circuit Example 6.4.3: Reference Point in a Complete Graph Many Hamilton circuits in a complete graph are the same circuit with different starting points. For example, in the graph K3, shown below in Figure 6.4.3, ABCA is the same circuit as BCAB, just with a different starting point (reference point). We will typically assume that the reference point is A. Figure 6.4.3: K3 A B C Chapter 6: Graph Theory __________ __________ Page 238 Number of Hamilton Circuits: A complete graph with N vertices is (N-1)! Hamilton circuits. Since half of the circuits are mirror images of the other half, there are actually only half this many unique circuits. Example 6.4.4: Number of Hamilton Circuits How many Hamilton circuits does a graph with five vertices have? (N – 1)! = (5 – 1)! = 4! = 4321 = 24 Hamilton circuits. How to solve a Traveling Salesman Problem (TSP): A traveling salesman problem is a problem where you imagine that a traveling salesman goes on a business trip. He starts in his home city (A) and then needs to travel to several different cities to sell his wares (the other cities are B, C, D, etc.). To solve a TSP, you need to find the cheapest way for the traveling salesman to start at home, A, travel to the other cities, and then return home to A at the end of the trip. This is simply finding the Hamilton circuit in a complete graph that has the smallest overall weight. There are several different algorithms that can be used to solve this type of problem. A. Brute Force Algorithm 1. List all possible Hamilton circuits of the graph. 2. For each circuit find its total weight. 3. The circuit with the least total weight is the optimal Hamilton circuit. Example 6.4.5: Brute Force Algorithm: Figure 6.4.4: Complete Graph for Brute Force Algorithm Chapter 6: Graph Theory __________ __________ Page 239 Suppose a delivery person needs to deliver packages to three locations and return to the home office A. Using the graph shown above in Figure 6.4.4, find the shortest route if the weights on the graph represent distance in miles. Recall the way to find out how many Hamilton circuits this complete graph has. The complete graph above has four vertices, so the number of Hamilton circuits is: (N – 1)! = (4 – 1)! = 3! = 321 = 6 Hamilton circuits. However, three of those Hamilton circuits are the same circuit going the opposite direction (the mirror image). Table 6.4.5: Hamilton Circuits Hamilton Circuit Mirror Image Total Weight (Miles) ABCDA ADCBA 18 ABDCA ACDBA 20 ACBDA ADBCA 20 The solution is ABCDA (or ADCBA) with total weight of 18 mi. This is the optimal solution. B. Nearest-Neighbor Algorithm: 1. Pick a vertex as the starting point. 2. From the starting point go to the vertex with an edge with the smallest weight. If there is more than one choice, choose at random. 3. Continue building the circuit, one vertex at a time from among the vertices that have not been visited yet. 4. From the last vertex, return to the starting point. Example 6.4.6: Nearest-Neighbor Algorithm A delivery person needs to deliver packages to four locations and return to the home office A as shown in Figure 6.4.5 below. Find the shortest route if the weights represent distances in miles. Starting at A, E is the nearest neighbor since it has the least weight, so go to E. From E, B is the nearest neighbor so go to B. From B, C is the nearest neighbor so go to C. From C, the first nearest neighbor is B, but you just came from there. The next nearest neighbor is A, but you do not want to go there yet because that is the Chapter 6: Graph Theory __________ __________ Page 240 starting point. The next nearest neighbor is E, but you already went there. So go to D. From D, go to A since all other vertices have been visited. Figure 6.4.6: Complete Graph for Nearest-Neighbor Algorithm The solution is AEBCDA with a total weight of 26 miles. This is not the optimal solution, but it is close and it was a very efficient method. C. Repetitive Nearest-Neighbor Algorithm: 1. Let X be any vertex. Apply the Nearest-Neighbor Algorithm using X as the starting vertex and calculate the total cost of the circuit obtained. 2. Repeat the process using each of the other vertices of the graph as the starting vertex. 3. Of the Hamilton circuits obtained, keep the best one. If there is a designated starting vertex, rewrite this circuit with that vertex as the reference point. Example 6.4.7: Repetitive Nearest-Neighbor Algorithm Suppose a delivery person needs to deliver packages to four locations and return to the home office A. Find the shortest route if the weights on the graph represent distances in kilometers. Chapter 6: Graph Theory __________ __________ Page 241 Figure 6.4.7: Complete Graph for Repetitive Nearest-Neighbor Algorithm Starting at A, the solution is AEBCDA with total weight of 26 miles as we found in Example 6.4.6. See this solution below in Figure 6.4.7. Figure 6.4.8: Starting at A Chapter 6: Graph Theory __________ __________ Page 242 Starting at B, the solution is BEDACB with total weight of 20 miles. Figure 6.4.9: Starting at B Starting at C, the solution is CBEDAC with total weight of 20 miles. Figure 6.4.10: Starting at C Chapter 6: Graph Theory __________ __________ Page 243 Starting at D, the solution is DEBCAD with total weight of 20 miles. Figure 6.4.11: Starting at D Starting at E, solution is EBCADE with total weight of 20 miles. Figure 6.4.12: Starting at E Chapter 6: Graph Theory __________ __________ Page 244 Now, you can compare all of the solutions to see which one has the lowest overall weight. The solution is any of the circuits starting at B, C, D, or E since they all have the same weight of 20 miles. Now that you know the best solution using this method, you can rewrite the circuit starting with any vertex. Since the home office in this example is A, let’s rewrite the solutions starting with A. Thus, the solution is ACBEDA or ADEBCA. D. Cheapest-Link Algorithm 1. Pick the link with the smallest weight first (if there is a tie, randomly pick one). Mark the corresponding edge in red. 2. Pick the next cheapest link and mark the corresponding edge in red. 3. Continue picking the cheapest link available. Mark the corresponding edge in red except when a) it closes a circuit or b) it results in three edges coming out of a single vertex. 4. When there are no more vertices to link, close the red circuit. Example 6.4.8: Cheapest-Link Algorithm Figure 6.4.13: Complete Graph for Cheapest-Link Algorithm Suppose a delivery person needs to deliver packages to four locations and return to the home office A. Find the shortest route if the weights represent distances in miles. Chapter 6: Graph Theory __________ __________ Page 245 Step 1: Find the cheapest link of the graph and mark it in blue. The cheapest link is between B and E with a weight of one mile. Figure 6.4.14: Step 1 Step 2: Find the next cheapest link of the graph and mark it in blue. The next cheapest link is between B and C with a weight of two miles. Figure 6.4.15: Step 2 Chapter 6: Graph Theory __________ __________ Page 246 Step 3: Find the next cheapest link of the graph and mark it in blue provided it does not make a circuit or it is not a third edge coming out of a single vertex. The next cheapest link is between D and E with a weight of three miles. Figure 6.4.16: Step 3 Step 4: Find the next cheapest link of the graph and mark it in blue provided it does not make a circuit or it is not a third edge coming out of a single vertex. The next cheapest link is between A and E with a weight of four miles, but it would be a third edge coming out of a single vertex. The next cheapest link is between A and C with a weight of five miles. Mark it in blue. Figure 6.4.17: Step 4 Chapter 6: Graph Theory __________ __________ Page 247 Step 5: Since all vertices have been visited, close the circuit with edge DA to get back to the home office, A. This is the only edge we could close the circuit with because AB creates three edges coming out of vertex B and BD also created three edges coming out of vertex B. Figure 6.4.18: Step 5 The solution is ACBEDA or ADEBCA with total weight of 20 miles. Efficient Algorithm: an algorithm for which the number of steps needed to carry it out grows in proportion to the size of the input to the problem. Approximate Algorithm: any algorithm for which the number of steps needed to carry it out grows in proportion to the size of the input to the problem. There is no known algorithm that is efficient and produces the optimal solution. Chapter 6: Graph Theory __________ __________ Page 248 Chapter 6 Homework 1. Answer the following questions based on the graph below. a. What are the vertices? b. Is this graph connected? c. What is the degree of vertex C? d. Edge FE is adjacent to which edges? e. Does this graph have any bridges? 2. Answer the following questions based on the graph below. a. What are the vertices? b. What is the degree of vertex u? c. What is the degree of vertex s? d. What is one circuit in the graph? Chapter 6: Graph Theory __________ __________ Page 249 3. Draw a spanning subgraph in the graph below. 4. Find the minimum spanning tree in the graph below using Kruskal’s Algorithm. Chapter 6: Graph Theory __________ __________ Page 250 5. Find the minimum spanning tree in the graph below using Kruskal’s Algorithm. 6. Find an Euler Path in the graph below. 7. Find an Euler circuit in the graph below. Chapter 6: Graph Theory __________ __________ Page 251 8. Which graphs below have Euler Paths? Which graphs have Euler circuits? 9. Highlight an Euler circuit in the graph below. 10. For each of the graphs below, write the degree of each vertex next to each vertex. Chapter 6: Graph Theory __________ __________ Page 252 11. Circle whether each of the graphs below has an Euler circuit, an Euler path, or neither. a. Euler circuit b. Euler path c. Neither a. Euler circuit b. Euler path c. Neither a. Euler circuit b. Euler path c. Neither 12. How many Hamilton circuits does a complete graph with 6 vertices have? 13. Suppose you need to start at A, visit all three vertices, and return to the starting point A. Using the Brute Force Algorithm, find the shortest route if the weights represent distances in miles. Chapter 6: Graph Theory __________ __________ Page 253 14. If a graph is connected and ___, the graph will have an Euler circuit. a. the graph has an even number of vertices b. the graph has an even number of edges c. the graph has all vertices of even degree d. the graph has only two odd vertices 15. Starting at vertex A, use the Nearest-Neighbor Algorithm to find the shortest route if the weights represent distances in miles. 16. Find a Hamilton circuit using the Repetitive Nearest-Neighbor Algorithm. Chapter 6: Graph Theory ___________ __________ Page 254 17. Find a Hamilton circuit using the Cheapest-Link Algorithm. 18. Which is a circuit that traverses each edge of the graph exactly once? A. Euler circuit b. Hamilton circuit c. Minimum Spanning Tree 19. Which is a circuit that traverses each vertex of the graph exactly once? A. Euler circuit b. Hamilton circuit c. Minimum Spanning Tree 20. For each situation, would you find an Euler circuit or a Hamilton Circuit? a. The department of Public Works must inspect all streets in the city to remove dangerous debris. b. Relief food supplies must be delivered to eight emergency shelters located at different sites in a large city. c. The Department of Public Works must inspect traffic lights at intersections in the city to determine which are still working. d. An insurance claims adjuster must visit 11 homes in various neighborhoods to write reports. |
7270 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9442073/ | Cochlear implant and large vestibular aqueduct syndrome in children - PMC
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Cochlear implant and large vestibular aqueduct syndrome in children
Trissia Maria Farah Vassoler
Trissia Maria Farah Vassoler
1 MD. 3rd year ent resident physician
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1,, Gilberto da Fontoura Rey Bergonse
Gilberto da Fontoura Rey Bergonse
2 MD. 2nd year ent resident physician
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2, Silvio Meira Junior
Silvio Meira Junior
3 M.D. Full member of the Brazilian College of Radiologists; Radiologist - Imagem Diagnósticos Médicos, Hospital Beneficência Portuguesa de Bauru and Centro de Pesquisas Audiológicas do Hospital de Reabilitação de Anomalias Craniofaciais/USP - Campus Bauru
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3, Maria Cecília Bevilacqua
Maria Cecília Bevilacqua
4 USP Full Professor - Campus Bauru, Coordinator - interdisciplinary team for the Cochlear Implant Program - Hospital de Reabilitação de Anomalias Craniofaciais da USP - Bauru
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4, Orozimbo Alves Costa Filho
Orozimbo Alves Costa Filho
5 USP Full Professor - Campus Bauru, Coordinator - Audiologic Research Center - CPA - Hospital de Reabilitação de Anomalias Craniofaciais da USP - Bauru. Hospital de Reabilitação de Anomalias Craniofaciais (HRAC) - Centro de Pesquisas Audiológicas (CPA) - Universidade de São Paulo (USP) - Bauru/SP
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5
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1 MD. 3rd year ent resident physician
2 MD. 2nd year ent resident physician
3 M.D. Full member of the Brazilian College of Radiologists; Radiologist - Imagem Diagnósticos Médicos, Hospital Beneficência Portuguesa de Bauru and Centro de Pesquisas Audiológicas do Hospital de Reabilitação de Anomalias Craniofaciais/USP - Campus Bauru
4 USP Full Professor - Campus Bauru, Coordinator - interdisciplinary team for the Cochlear Implant Program - Hospital de Reabilitação de Anomalias Craniofaciais da USP - Bauru
5 USP Full Professor - Campus Bauru, Coordinator - Audiologic Research Center - CPA - Hospital de Reabilitação de Anomalias Craniofaciais da USP - Bauru. Hospital de Reabilitação de Anomalias Craniofaciais (HRAC) - Centro de Pesquisas Audiológicas (CPA) - Universidade de São Paulo (USP) - Bauru/SP
Send correspondence to: Trissia Maria Farah Vassoler - Avenida Iguaçú 811 apto. 403B Rebouças Curitiba PR 80230-020. Tel. (0xx14) 8125-0250 trissiavassoler@hotmail.com
Received 2007 Jan 29; Accepted 2007 Sep 29; Collection date 2008 Mar-Apr.
.
This is an open access article under the CC BY license (
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PMCID: PMC9442073 PMID: 18568206
Summary
Children with LVAS can develop a severe sensorineural hearing loss early in childhood, but they can be rehabilitated with hearing aids to continue their regular studies and to have a normal life. The problem is that they can deteriorate their hearing capacity, and at this point a cochlear implant can be used to preserve their hearing skills and vocalization.
Aim
to evaluate the hearing skills of 3 children with LVAS referred to cochlear implants.
Material
retrospective study based on medical charts' review.
Results
Speech recognition in open field: patient 1, 80%; patient 2, 87.5%; patient 3, 4 %.
Conclusion
Children with LVAS are considered good candidates for Cochlear implant surgery by the most important centers of the world because most of them can develop good speech recognition, providing them a good social life.
Keywords: vestibular aqueduct, cochlear implant, hearing loss
INTRODUCTION
The vestibular aqueduct is a bony canal within the temporal bone, which goes from the medial wall of the inner ear vestibule all the way to the posterior surface of the petrous bone pyramid. The endolymphatic duct crosses the vestibular aqueduct and ends in the endolymphatic sac. During embryogenesis, the vestibular aqueduct starts as a long and narrow vestibular diverticulum. A development defect before the diverticulum starts to narrow, on the fifth week of gestation, results in an enlarged vestibular aqueduct.1
In 1978, Valvassori and Clemis identified 50 cases of enlarged vestibular aqueduct in a retrospective study of 3,700 patients who were submitted to temporal bone CT scan in order to study inner ear structures. A vestibular aqueduct is considered enlarged when its antero-posterior diameter is equal to 1.5mm. These authors were the first to use the definition of Enlarged Vestibular Aqueduct Syndrome (EVAS). Since this time, many studies were carried out in order to better characterize the syndrome and found that 59% to 94% of the cases were bilateral; 60% to 66% of the patients are female, and sensorineural hearing loss is progressive in 46% to 65% of the patients.1, 2, 3, 4, 5, 6,13,15
The enlarged vestibular aqueduct may happen as an isolated anomaly or in association with other inner ear malformations. The most common is the enlargement of the horizontal semi-circular canal, 60% to 66%, and cochlear hypoplasia, 28%7.
The pathophysiology of this sensorineural hearing loss caused by an enlargement of the vestibular aqueduct is still unknown. Two things may happen, the first suggests a rupture of the labyrinthine membrane or perilymphatic fistula, resulting in a direct transmission of the cerebro-spinal fluid pressure to the middle cochlear turn through the endolymphatic duct and enlarged vestibular aqueduct, and the second suggests a hyperosmolar fluid reflux to the cochlea coming from the endolymphatic sac. More recent studies have mapped the hearing loss associated with the enlarged vestibular aqueduct in the region of chromosome 7q31, however, its clinical meaning still requires more investigation.1,12,15
Children with EVAS may have moderate to severe hearing deficiencies during their early stages of childhood, however their residual hearing allows them to develop oral language with conventional hearing aids and may be completely integrated to regular school conditions. Nonetheless, these children have a worsening in their hearing skills with time and cochlear implants are being offered as an option to keep their hearing and oral communication skills in proper levels4.
This study reports the experience of treating 3 patients with enlarged aqueduct syndrome with cochlear implants.
MATERIALS AND METHODS
This study was analyzed and approved by the Ethics in Research with Human Beings Committee of the HRAC/USP under protocol # 158/2007-SVAPEPE-CEP.
We carried out a retrospective study to identify all the patients using cochlear implants in our population and were pre-operatively diagnosed with enlarged vestibular aqueducts. So far, 504 cochlear implant surgeries have been performed and only 3 (0.6%) of these patients were diagnosed with enlarged vestibular aqueduct through CT scan and MRI, a much lower number when compared to what has been found in other large centers.
Radiology exams were carried out under sedation with 20% chloral hydrate. CT scans were carried out in a spiral Elscint Twin device, with 0.5mm axial sections, with later axial plane reconstruction (Figure 1).
Figure 1.
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Normal bilateral vestibular aqueduct - Temporal bone CT scan - axial view.
The vestibular aqueduct was considered enlarged when its antero-posterior diameter was equal to or greater than 1.5mm (Figure 2).
Figure 2.
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Bilaterally enlarged vestibular aqueduct - temporal bone CT scan - axial view.
MRI exams were carried out in Phillips devices with 1.0 Tesla magnetic fields, following the protocol for patients eligible to receive the cochlear implant:
FLAIR sequence for the encephalon.
Turbo SpinEcho (TSE) sequence, axial plane, T1 weighed images of the posterior fossa (Figure 3)
Figure 3.
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Bilaterally enlarged vestibular aqueduct - temporal bone MRI - axial view in T1 weighed slices.
TSE sequence, axial plane, T2 weighed images (Figure 4, Figure 5).
Figure 4.
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Bilaterally enlarged vestibular aqueduct - temporal bone MRI - axial view - T2 weighed submillimeter slice.
Figure 5.
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Bilaterally enlarged vestibular aqueduct - temporal bone MRI - axial view - T2 weighed submillimeter slice.
MIP sequence, with submillimeter cross-sections (0.6mm) in the axial and sagittal planes, T2 weighed images, to evaluate the membranous labyrinth and the VII and VIII cranial nerves.
MIP 3D reconstruction (Figure 6).
Figure 6.
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Left side enlarged vestibular aqueduct - Temporal bone MRI, 3D reconstruction.
RESULTS
The cochlear implant was done through a transmastoid approach, with posterior tympanotomy and a 2.0mm cochleostomy made in a right angle with the stapes tendon. We had no gusher during cochleostomy, however there was some pulsating oozing of perilymphatic/cerebro-spinal fluid. In all cases the active electrodes were inserted without difficulties, we achieved complete insertion, and the cochlea was sealed with fragments of temporal muscle fascia.
Patient number 1 was male, pre-lingual, and was implanted at 6 years and 5 months of age on the left side.
He received a Nucleus cochlear implant, model CI24RST (Cochlear Corp. Englewood, CO). He had an enlarged vestibular aqueduct and hyperplasic cochlea in the image exam.
Patient 02 was female, post-lingual, and was implanted at the age of 5 years and 9 months, on the right side. She received a Nucleus, CI24RST (Cochlear Corp. Englewood, CO) implant, and did not have other malformations seen in her image exams.
Patient 03 was also a female, pre-lingual, and was implanted at 6 years and 1 month of age, and she was reimplanted three years later, because of a failure in the internal device. She was implanted and reimplanted on the right side and received a MedEl, model C40+ short (MedEl, Innsbruck, Austria) cochlear implant both times, and she did not have other malformations visible in her image exams.
All patients used conventional hearing aids for at least 6 months after the implant and had bilateral enlarged vestibular aqueduct.
Audiometric evaluation for each patient included pure tone recordings without the hearing aid, 6 months of hearing aid use and 1 year after the implant (Table 1).
Table 1.
Evaluation of pure tone averages of 500, 1000 and 2000 Hz without hearing aid, with hearing aid and with CI.
| Patient | 1 | 2 | 3 |
:---: :---: |
| Pure tone averages without hearing aid | 93.3 | 106.6 | 103.3 |
| Pure tone averages with hearing aid | 56.6 | 88.3 | 83.3 |
| Pure tone averages with CI | 26.6 | 26.6 | 40.8 |
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Tests from Delgado (8), GASP (9) and Ling's sounds (10) were used to assess speech perception.
Patient # 1 had 80% word recognition, in open settings, patient # 2 had 87.5% in open settings and patient # 3 had word recognition of 4% in open settings, Ling's sound detection was equal to 100%, with phoneme /a/ discrimination equal to 50%, phoneme /i/ equal to 0% and phoneme /u/ equal to 0% in a closed setting.
DISCUSSION
The Enlarged Vestibular Aqueduct Syndrome (EVAS) represented a treatment challenge for physicians who treated these patients, because of the lack of a protocol able to efficiently prevent the progression of hearing loss in them. Conservative measures such as education to avoid head injuries and barotrauma or pressure fluctuations and treatment with steroids in cases of sudden hearing loss have yielded some success. Endolymphatic sac surgeries were carried out in patients with progressive congenital impairment, however were not considered efficient. Endolymphatic sac obliteration was also described by Wilson et al. as a means to control the progressive hearing loss; however, its results could not be repeated in other studies1,15.
Some few studies have considered the use of a cochlear implant as an option for the patient with EVAS, if bearers of severe-profound or profound sensorineural hearing loss. Harker et al. reported 5 pediatric patients with EVAS, who were implanted and presented excellent results in speech detection. He did not find gusher in any of the surgical procedures. Bent et al. reported a case of 10 patients with EVAS, who received cochlear implant. They noticed a small perilymph pulsating oozing caused by cochleostomy in 5 patients, which were easily controlled with the use of temporal muscle fascia. Seven of the eight children who had already been implanted some time before improved in word recognition in an open setting. Myamoto et al. implanted 23 patients with EVAS, 9 children and 14 adults. He reported gusher in 5 of these patients, however there were no difficulties in inserting electrodes in these patients and the cochleostomy was sealed with a fragment of temporal muscle fascia. This study also found benefits for patients with EVAS1.
These results, as well as our experience, consider the cochlear implant surgery to be safe in patients with enlarged vestibular aqueduct and a complete insertion of electrodes is doable. Despite the oozing of perilymph during cochleostomy being somewhat common, it is easily controlled by sealing it with a fragment of temporal muscle fascia. Searching for the association of other malformations with the enlarged vestibular aqueduct and the diameter of the vestibular aqueduct were not relevant in predicting gusher during surgery.6
In regards of developing speech skills, there is a marked difference between the many types of inner ear malformations. Unanimously, all cochlear implant centers that have performed surgery in children with inner ear malformations report that those with cochlear dysplasia, enlarged vestibular aqueduct and dilated vestibule, in other words, Mondini's malformation and, especially, those with EVAS are the ones that have the greatest gains in their development with cochlear implants in terms of word recognition and speech6,14. We must also stress that the results are even better in those patients whose hearing loss was installed in the post-lingual phase. As it can be seen with our patients, the post-lingual children had a faster speech perception development when compared to those that had an earlier installed hearing loss. Currently, patients 1 and 2 are evolving very well, and in speech recognition tests present good results in the open setting and most of them can use the telephone in their daily lives. Patient 03 did not do well and other clinical syndromes are being investigated.
Based on the audiometric evaluation of these patients and their speech perception, we can see that these are children with great potential to develop speech perception in an open setting, phonemes and phrases, and this allows them to go to regular schools and have a normal social development. When we notice a low development, we must bear in mind that the enlarged vestibular aqueduct is present in other syndromes such as the Pendred and the Otobrachiorenal syndromes13, which impair their neurological development and prevent the children to develop their skills to their maximum.
The major doubt still lies in defining the ideal time for surgery, because hearing loss can behave in different ways such as progressive, sudden or fluctuating, the assessment of these patients require a thorough follow up11. The Children's Hospital Implant Center in Sydney, Australia4, considers a patient with enlarged vestibular aqueduct eligible for the cochlear implant when the hearing impairment is evident, even using the best hearing aids available, if the periods of inadequate hearing start to impair the child's school performance or if the patient has more than 3 episodes of hearing deterioration in 1 year4. We believe the parameters necessary to assess whether or not a patient is eligible for receiving a cochlear implant in cases of enlarged vestibular aqueduct depend on a careful evaluation from a multidisciplinary team, thus, we do not establish specific criteria for this statement.
CONCLUSION
Patients with enlarged vestibular aqueduct are considered eligible for receiving cochlear implants by the major cochlear implant centers in the world, because most of them develop good audiometric and speech recognition performances, and this provides them better social participation.
Nonetheless, we stress that it is important, besides image studies, to investigate the presence of other syndromes that may follow that of the enlarged vestibular aqueduct.
REFERENCES
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Uncited Reference
8.Delgado EMC, Bevilacqua MC. Lista de palavras como procedimento de avaliação da percepção dos sons da fala para crianças deficientes auditivas. Pró-Fono R Atual Cient. 1999;11(1):59–64. [Google Scholar]
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7271 | https://www.wikihow.com/Find-the-Perimeter-of-a-Triangle | Log in
How to Find the Perimeter of a Triangle
Last Updated: April 15, 2025
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Finding the perimeter of a triangle means finding the distance around the triangle.
X
Research source The simplest way to find the perimeter of a triangle is to add up the length of all of its sides, but if you don't know all of the side lengths you will need to calculate them first. This article will first teach you to find the perimeter of a triangle when you do know all three side lengths; this is the easiest and most common way. It will then teach you to find the perimeter of a right triangle when only two of the side lengths are known. Finally, it will teach you to find the perimeter of any triangle for which you know two side lengths and the angle measure between them (an "SAS Triangle"), using the Law of Cosines.
Steps
Finding the Perimeter When Three Side Lengths are Known
Finding the Perimeter of a Right Triangle
Joseph Meyer
Use this visual trick to understand the Pythagorean Theorem. Imagine a right triangle with squares constructed on each leg and the hypotenuse. by rearranging the smaller squares within the larger square, the areas of the smaller squares (a² and b²) will add up visually to the area of the larger square (c²).
Do you have the perimeter and are missing one side? Then you should subtract the sum of the two sides from the perimeter. This number equals the length of the missing side.
Finding the Perimeter Using the Law of Cosines
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To find the perimeter of a triangle, use the formula perimeter = a + b + c, where a, b, and c are the lengths of the sides of the triangle. For example, if the length of each side of the triangle is 5, you would just add 5 + 5 + 5 and get 15. Therefore, the perimeter of the triangle is 15. If you only know the length of 2 of the triangle’s sides, you can still find the perimeter if it’s a right triangle, which means the triangle has one 90-degree angle. Just use the Pythagorean theorem, which is a^2+ b^2 = c^2, where a and b are the lengths of the known sides and c is the length of the unknown hypotenuse. For example, if the length of the known sides are 3 and 4, you would just add 3^2+ 4^2, or 9 + 16, and get 25. Then, you would take the square root of 25 to find c, which is 5. Therefore, the length of the unknown side is 5. Finally, add all of the side lengths together to find the perimeter. In this case you would add 3 + 4 + 5 and get 12. Therefore, the perimeter of the triangle is 12. If you want to learn how to solve the perimeter of your triangle if you only know 2 sides and an angle, keep reading the article!
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7272 | https://www.uni-regensburg.de/assets/wirtschaftswissenschaften/bwl-roeder/Principles_of_Corporate_Finance/Chapter_06.pdf | Corporate Finance Fifth Edition, Global Edition Chapter 6 Valuing Bonds Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 2 Chapter Outline 6.1 Bond Cash Flows, Prices, and Yields 6.2 Dynamic Behavior of Bond Prices 6.3 The Yield Curve and Bond Arbitrage 6.4 Corporate Bonds 6.5 Sovereign Bonds Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 3 Learning Objectives (1 of 4) • Identify the cash flows for both coupon bonds and zero-coupon bonds, and calculate the value for each type of bond.
• Calculate the yield to maturity for both coupon and zero-coupon bonds, and interpret its meaning for each.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 4 Learning Objectives (2 of 4) • Given coupon rate and yield to maturity, determine whether a coupon bond will sell at a premium or a discount; describe the time path the bond’s price will follow as it approaches maturity, assuming prevailing interest rates remain the same over the life of the bond.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 5 Learning Objectives (3 of 4) • Illustrate the change in bond price that will occur as a result of changes in interest rates; differentiate between the effect of such a change on long-term versus short-term bonds.
• Discuss the effect of coupon rate to the sensitivity of a bond price to changes in interest rates.
• Define duration, and discuss its use by finance practitioners.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 6 Learning Objectives (4 of 4) • Calculate the price of a coupon bond using the Law of One Price and a series of zero-coupon bonds.
• Discuss the relation between a corporate bond’s expected return and the yield to maturity; define default risk and explain how these rates incorporate default risk.
• Assess the creditworthiness of a corporate bond using its bond rating; define default risk.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 7 6.1 Bond Cash Flows, Prices, and Yields (1 of 2) • Bond Terminology – Bond Certificate States the terms of the bond – Maturity Date Final repayment date – Term The time remaining until the repayment date – Coupon Promised interest payments 8 Mantel Bogen Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Coverded Bond - Pfandbrief König Friedrich II.
König von Preußen Friedrich der Große Der alte Fritz 24. Januar 1712 - 17. August 1786 Langfristige Schuldverschreibung, die der Finanzierung von Baukrediten dient.
Pfandbriefe werden von Hypothekenbanken, Schiffspfandbriefbanken und öffentlich-rechtlichen Kreditinstituten ausgegeben. Sie sind ähnlich wie Anleihen ausgestattet. Durch die Beleihung von Grundvermögen sind sie jedoch besonders gut besichert. Ein Treuhänder kontrolliert, dass die emittierten Pfandbriefe zu jeder Zeit in gleicher Höhe durch Hypotheken mit mindestens gleichem Zinsertrag gedeckt sind.
Im Sommer 1769 erließ Friedrich II eine Kabinettsorder: Ein Pfandbriefinstitut, damals Landschaft genannt, sollte den schlesischen Rittergütern, die unter den Folgen des Siebenjährigen Krieges litten, günstige langfristige Darlehen verschaffen. 9 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 10 6.1 Bond Cash Flows, Prices, and Yields (2 of 2) • Bond Terminology – Face Value Notional amount used to compute the interest payments – Coupon Rate Determines the amount of each coupon payment, expressed as an A P R – Coupon Payment Coupon Rate × Face Value = Number of Coupon Payments per Year CPN Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 11 Zero-Coupon Bonds (1 of 7) • Zero-Coupon Bond – Does not make coupon payments – Always sells at a discount (a price lower than face value), so they are also called pure discount bonds – Treasury Bills are U.S. government zero-coupon bonds with a maturity of up to one year.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 12 Zero-Coupon Bonds (2 of 7) • Suppose that a one-year, risk-free, zero-coupon bond with a $100,000 face value has an initial price of $96,618.36. The cash flows would be – Although the bond pays no “interest,” your compensation is the difference between the initial price and the face value.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 13 Zero-Coupon Bonds (3 of 7) • Yield to Maturity – The discount rate that sets the present value of the promised bond payments equal to the current market price of the bond Price of a Zero-Coupon bond = (1 + )n n FV P YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 14 Zero-Coupon Bonds (4 of 7) • Yield to Maturity – For the one-year zero coupon bond: 1 100,000 96,618.36 = (1 + ) YTM 1 100,000 1 + = = 1.035 96,618.36 YTM Thus, the Y T M is 3.5% Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 15 Zero-Coupon Bonds (5 of 7) • Yield to Maturity – Yield to Maturity of an n-Year Zero-Coupon Bond 1 = 1 n n FV YTM P Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 16 Textbook Example 6.1 (1 of 2) Yields for Different Maturities Problem Suppose the following zero-coupon bonds are trading at the prices shown below per $100 face value. Determine the corresponding spot interest rates that determine the zero coupon yield curve.
Maturity 1 Year 2 Years 3 Years 4 Years Price $96.62 $92.45 $87.63 $83.06 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 17 Textbook Example 6.1 (2 of 2) Solution Using Equation. 6.3, we have 1 1 1 2 2 2 1 3 3 3 1 4 4 4 100 = = 1 = 3.50% 96.62 100 = = 1 = 4.00% 92.45 100 = = 1 = 4.50% 87.63 100 = = 1 = 4.75% 83.06 r YTM r YTM r YTM r YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 18 Zero-Coupon Bonds (6 of 7) • Risk-Free Interest Rates – A default-free zero-coupon bond that matures on date n provides a risk-free return over the same period.
– Thus, the Law of One Price guarantees that the risk-free interest rate equals the yield to maturity on such a bond.
– Risk-Free Interest Rate with Maturity n. = n n r YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 19 Zero-Coupon Bonds (7 of 7) • Risk-Free Interest Rates – Spot Interest Rate Another term for a default-free, zero-coupon yield – Zero-Coupon Yield Curve A plot of the yield of risk-free zero-coupon bonds as a function of the bond’s maturity date Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 20 Coupon Bonds (1 of 2) • Coupon Bonds – Pay face value at maturity – Pay regular coupon interest payments • Treasury Notes – U.S. Treasury coupon security with original maturities of 1–10 years • Treasury Bonds – U.S. Treasury coupon security with original maturities over 10 years Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 21 Textbook Example 6.2 (1 of 2) The Cash Flows of a Coupon Bond Problem The U.S. Treasury has just issued a five-year, $1000 bond with a 5% coupon rate and semiannual coupons. What cash flows will you receive if you hold this bond until maturity?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 22 Textbook Example 6.2 (2 of 2) Solution The face value of this bond is $1000. Because this bond pays coupons semiannually, from Eq. 6.1, you will receive a coupon payment every six months of Here is the timeline, based on a six-month period: 5% CPN = $1000× = $25.
2 Note that the last payment occurs five years (10 six-month periods) from now and is composed of both a coupon payment of $25 and the face value payment of $1000.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 23 Coupon Bonds (2 of 2) • Yield to Maturity – The Y T M is the single discount rate that equates the present value of the bond’s remaining cash flows to its current price – Yield to Maturity of a Coupon Bond 1 1 = × 1 + (1 + ) (1 + ) N N FV P CPN y y y Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 24 Textbook Example 6.3 (1 of 3) Computing the Yield to Maturity of a Coupon Bond Problem Consider the five-year, $1000 bond with a 5% coupon rate and semiannual coupons described in Example 6.2. If this bond is currently trading for a price of $957.35, what is the bond’s yield to maturity?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 25 Textbook Example 6.3 (2 of 3) Solution Because the bond has 10 remaining coupon payments, we compute its yield y by solving: 10 10 1 1 1000 957.35 = 25 × 1 + (1+ ) (1+ ) y y y We can solve it by trial-and-error or by using the annuity spreadsheet: Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 26 Textbook Example 6.3 (3 of 3) Blank N P E R R A T E P V P M T F V Excel Formula Given 10 blank −957.35 25 1,000 blank Solve for Rate blank 3.00% blank blank blank = RATE (10, 25, −957.35, 1000) Therefore, y = 3%. Because the bond pays coupons semiannually, this yield is for a six-month period. We convert it to an A P R by multiplying by the number of coupon payments per year. Thus the bond has a yield to maturity equal to a 6% A P R with semiannual compounding.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 27 Textbook Example 6.4 (1 of 2) Computing a Bond Price from Its Yield to Maturity Problem Consider again the five-year, $1000 bond with a 5% coupon rate and semiannual coupons presented in Example 6.3. Suppose you are told that its yield to maturity has increased to 6.30% (expressed as an A P R with semiannual compounding). What price is the bond trading for now?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Textbook Example 6.4 (2 of 2) Solution Given the yield, we can compute the price using Eq.65. First, note that a 6.30% A P R is equivalent to a semiannual rate of 3.15%. Therefore, the bond price is 10 10 1 1 1000 = 25× 1 + = $944.98 0.0315 1.0315 1.0315 P We can also use the annuity spreadsheet: Blank N P E R R A T E P V P M T F V Excel Formula Given 10 3.15% blank 25 1,000 blank Solve for P V blank blank −944.98 blank blank = PV (0.0315, 10, 25, 1000) Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 29 6.2 Dynamic Behavior of Bond Prices • Discount – A bond is selling at a discount if the price is less than the face value • Par – A bond is selling at par if the price is equal to the face value • Premium – A bond is selling at a premium if the price is greater than the face value Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 30 Discounts and Premiums (1 of 3) • If a coupon bond trades at a discount, an investor will earn a return both from receiving the coupons and from receiving a face value that exceeds the price paid for the bond.
– If a bond trades at a discount, its yield to maturity will exceed its coupon rate.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 31 Discounts and Premiums (2 of 3) • If a coupon bond trades at a premium, it will earn a return from receiving the coupons, but this return will be diminished by receiving a face value less than the price paid for the bond.
• Most coupon bonds have a coupon rate so that the bonds will initially trade at, or very close to, par.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 32 Discounts and Premiums (3 of 3) Table 6.1 Bond Prices Immediately After a Coupon Payment When the bond price is We say the bond trades This occurs when greater than the face value “above par” or “at a premium” Coupon Rate > Yield to Maturity equal to the face value “at par” Coupon Rate = Yield to Maturity less than the face value “below par” or “at a discount” Coupon Rate < Yield to Maturity Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 33 Textbook Example 6.5 (1 of 2) Determining the Discount or Premium of a Coupon Bond Problem Consider three 30-year bonds with annual coupon payments. One bond has a 10% coupon rate, one has a 5% coupon rate, and one has a 3% coupon rate. If the yield to maturity of each bond is 5%, what is the price of each bond per $100 face value? Which bond trades at a premium, which trades at a discount, and which trades at par?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 34 Textbook Example 6.5 (2 of 2) Solution We can compute the price of each bond using Eq.6.5. Therefore, the bond prices are 30 30 1 1 100 (10% coupon) =10 × 1 + = $176.86 0.05 1.05 1.05 P (trades at a premium) 30 30 1 1 100 (5% coupon) = 5× 1 + = $100.00 0.05 1.05 1.05 P (trades at par) 30 30 1 1 100 (3% coupon) = 3× 1 + = $69.26 0.05 1.05 1.05 P (trades at a discount) Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 35 Time and Bond Prices • Holding all other things constant, a bond’s yield to maturity will not change over time.
• Holding all other things constant, the price of discount or premium bond will move toward par value over time.
• If a bond’s yield to maturity has not changed, then the I R R of an investment in the bond equals its yield to maturity even if you sell the bond early.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 36 Textbook Example 6.6 (1 of 4) The Effect of Time on the Price of a Coupon Bond Problem Consider a 30-year bond with a 10% coupon rate (annual payments) and a $100 face value. What is the initial price of this bond if it has a 5% yield to maturity? If the yield to maturity is unchanged, what will the price be immediately before and after the first coupon is paid?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 37 Textbook Example 6.6 (2 of 4) Solution We computed the price of this bond with 30 years to maturity in Example 6.5: 30 30 1 1 100 = 10× 1 + = $176.86 0.05 1.05 1.05 P Now consider the cash flows of this bond in one year, immediately before the first coupon is paid. The bond now has 29 years until it matures, and the timeline is as follows: Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 38 Textbook Example 6.6 (3 of 4) Again, we compute the price by discounting the cash flows by the yield to maturity. Note that there is a cash flow of $10 at date zero, the coupon that is about to be paid. In this case, we can treat the first coupon separately and value the remaining cash flows as in Eq. 6.5: 29 29 1 1 100 (just before first coupon) = 10 +10× 1 + = $185.71 0.05 1.05 1.05 P Note that the bond price is higher than it was initially. It will make the same total number of coupon payments, but an investor does not need to wait as long to receive the first one. We could also compute the price by noting that because the yield to maturity remains at 5% for the bond, investors in the bond should earn a return of 5% over the year: $176.86 × 1.05 = $185.71.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 39 Textbook Example 6.6 (4 of 4) What happens to the price of the bond just after the first coupon is paid? The timeline is the same as that given earlier, except the new owner of the bond will not receive the coupon at date zero. Thus, just after the coupon is paid, the price of the bond (given the same yield to maturity) will be 29 29 1 1 100 (just after first coupon) = 10× 1 + = $175.71 0.05 1.05 1.05 P The price of the bond will drop by the amount of the coupon ($10) immediately after the coupon is paid, reflecting the fact that the owner will no longer receive the coupon. In this case, the price is lower than the initial price of the bond. Because there are fewer coupon payments remaining, the premium investors will pay for the bond declines. Still, an investor who buys the bond initially, receives the first coupon, and then sells it earns a 5% return if the bond’s yield does not change: (10 +175.71) = 1.05.
176.86 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 40 Figure 6.1 The Effect of Time on Bond Prices Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 41 Accrued Interest (Stückzinsen) Interest Accrued - Price Dirty Price Clean de Kuponperio aktueller in Tage ng Kuponzahlu letzer seit Tage g Kuponbetra n Stückzinse period coupon current in days payment coupon last since days Amount Coupon Int. Accrued Accrued Interest Time (Coupon Periods) Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 42 Interest Rate Changes and Bond Prices (1 of 2) • There is an inverse relationship between interest rates and bond prices.
– As interest rates and bond yields rise, bond prices fall.
– As interest rates and bond yields fall, bond prices rise.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 43 Interest Rate Changes and Bond Prices (2 of 2) • The sensitivity of a bond’s price to changes in interest rates is measured by the bond’s duration.
– Bonds with high durations are highly sensitive to interest rate changes.
– Bonds with low durations are less sensitive to interest rate changes.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 44 Textbook Example 6.7 (1 of 3) The Interest Rate Sensitivity of Bonds Problem Consider a 15-year zero-coupon bond and a 30-year coupon bond with 10% annual coupons. By what percentage will the price of each bond change if its yield to maturity increases from 5% to 6%?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 45 Textbook Example 6.7 (2 of 3) Solution First, we compute the price of each bond for each yield to maturity: Yield to Maturity 15-Year, Zero-Coupon Bond 30-Year, 10% Annual Coupon Bond 5% 6% 15 100 = $48.10 1.05 30 30 1 1 100 10× 1 + = $176.86 0.05 1.05 1.05 15 100 = $41.73 1.06 30 30 1 1 100 10× 1 + = $155.06 0.06 1.06 1.06 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 46 Textbook Example 6.7 (3 of 3) The price of the 15-year zero-coupon bond changes by if its yield to maturity increases from 5% to 6%. For the 30-year bond with 10% annual coupons, the price change is Even though the 30-year bond has a longer maturity, because of its high coupon rate, its sensitivity to a change in yield is actually less than that of the 15-year zero coupon bond.
(41.73 48.10) = 13.2% 48.10 (155.06 176.86) = 12.3%.
176.86 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 47 Figure 6.2 Yield to Maturity and Bond Price Fluctuations over Time Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 48 6.3 The Yield Curve and Bond Arbitrage • Using the Law of One Price and the yields of default-free zero-coupon bonds, one can determine the price and yield of any other default-free bond.
• The yield curve provides sufficient information to evaluate all such bonds.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 49 Replicating a Coupon Bond (1 of 3) • Replicating a three-year $1000 bond that pays 10% annual coupon using three zero-coupon bonds: Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 50 Replicating a Coupon Bond (2 of 3) Table 6.2 Yields and Prices (per $100 Face Value) for Zero-Coupon Bonds Maturity 1 year 2years 3 years 4 years Y T M 3.50% 4.00% 4.50% 4.75% Price $96.62 $92.45 $87.63 $83.06% Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 51 Replicating a Coupon Bond (3 of 3) • By the Law of One Price, the three-year coupon bond must trade for a price of $1153.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 52 Valuing a Coupon Bond Using Zero-Coupon Yields • The price of a coupon bond must equal the present value of its coupon payments and face value.
– Price of a Coupon Bond = (Bond Cash Flows) V PV 2 1 2 + = + + + 1 + (1 + ) (1 + )n n CPN CPN CPN FV YTM YTM YTM 2 3 100 100 100 + 1000 = + + = $1153 1.035 1.04 1.045 P Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 53 Coupon Bond Yields • Given the yields for zero-coupon bonds, we can price a coupon bond 2 3 100 100 100 + 1000 = 1153 = + + (1 + ) (1 + ) (1 + ) P y y y 2 3 100 100 100 + 1000 = + + = $1153 1.0444 1.0444 1.0444 P Blank N P E R R A T E P V P M T F V Excel Formula Given 3 blank −1,153 100 1,000 blank Solve for Rate blank 4.44% blank blank blank = RATE(3, 100, −1153, 1000) Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 54 Textbook Example 6.8 (1 of 3) Yields on Bonds with the Same Maturity Problem Given the following zero-coupon yields, compare the yield to maturity for a three-year, zero-coupon bond; a three-year coupon bond with 4% annual coupons; and a three-year coupon bond with 10% annual coupons. All of these bonds are default free.
Maturity 1 year 2 years 3 years 4 years Zero- coupon Y T M 3.50% 4.00% 4.50% 4.75% Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 55 Textbook Example 6.8 (2 of 3) Solution From the information provided, the yield to maturity of the three-year, zero-coupon bond is 4.50%. Also, because the yields match those in Table 6.2, we already calculated the yield to maturity for the 10% coupon bond as 4.44%. To compute the yield for the 4% coupon bond, we first need to calculate its price. Using Eq. 6.6, we have 2 3 40 40 40 + 1000 = + + = $986.98 1.035 1.04 1.045 P The price of the bond with a 4% coupon is $986.98. From Eq. 6.5, its yield to maturity solves the following equation: 2 3 40 40 40 + 1000 $986.98 = + + (1 + ) (1 + ) (1 + ) y y y Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 56 Textbook Example 6.8 (3 of 3) We can calculate the yield to maturity using the annuity spreadsheet: Blank N P E R R A T E P V P M T F V Excel Formula Given 3 blank −986.98 100 1,000 blank Solve for Rate blank 4.47% blank blank Blank = RATE(3, 40, −986.98, 1000) To summarize, for the three-year bonds considered Coupon rate 0% 4% 10% Y T M 4.50% 4.47% 4.44% Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 57 Treasury Yield Curves • Treasury Coupon-Paying Yield Curve – Often referred to as “the yield curve” • On-the-Run Bonds – Most recently issued bonds – The yield curve is often a plot of the yields on these bonds.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 58 Yield Curve Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 59 Historic Yield Curves Germany Von Henning. H., Thomas Steiner, and Vlado Plaga - CC BY-SA 3.0, Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 60 6.4 Corporate Bonds • Corporate Bonds – Issued by corporations • Credit Risk – Risk of default Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 61 Corporate Bond Yields (1 of 9) • Investors pay less for bonds with credit risk than they would for an otherwise identical default-free bond.
• The yield of bonds with credit risk will be higher than that of otherwise identical default-free bonds.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 62 Corporate Bond Yields (2 of 9) • No Default – Consider a one-year, zero-coupon Treasury Bill with a Y T M of 4%.
What is the price?
1 1000 1000 = = = $961.54 1 + 1.04 P YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 63 Corporate Bond Yields (3 of 9) • Certain Default – Suppose now bond issuer will pay 90% of the obligation.
What is the price?
1 900 900 = = = $865.38 1 + 1.04 P YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 64 Corporate Bond Yields (4 of 9) • Certain Default – When computing the yield to maturity for a bond with certain default, the promised rather than the actual cash flows are used.
1000 = 1 = 1 = 15.56% 865.38 FV YTM P 900 = 1.04 865.38 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 65 Corporate Bond Yields (5 of 9) • Certain Default – The yield to maturity of a certain default bond is not equal to the expected return of investing in the bond.
– The yield to maturity will always be higher than the expected return of investing in the bond.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 66 Corporate Bond Yields (6 of 9) • Risk of Default – Consider a one-year, $1000, zero-coupon bond issued.
– Assume that the bond payoffs are uncertain.
There is a 50% chance that the bond will repay its face value in full and a 50% chance that the bond will default and you will receive $900.
– Thus, you would expect to receive $950.
Because of the uncertainty, the discount rate is 5.1%.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 67 Corporate Bond Yields (7 of 9) • Risk of Default – The price of the bond will be 950 = = $903.90 1.051 P – The yield to maturity will be 1000 = 1 = 1 = 10.63% 903.90 FV YTM P Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 68 Corporate Bond Yields (8 of 9) • Risk of Default – A bond’s expected return will be less than the yield to maturity if there is a risk of default.
– A higher yield to maturity does not necessarily imply that a bond’s expected return is higher.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 69 Corporate Bond Yields (9 of 9) Table 6.3 Price, Expected Return, and Yield to Maturity of a One-Year, Zero-Coupon Avant Bond with Different Likelihoods of Default Avant Bond (1-year, zero-coupon) Bond Price Yield to Maturity Expected Return Default Free $961.54 4.00% 4% 50% Chance of Default $903.90 10.63% 5.1% Certain Default $865.38 15.56% 4% Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 70 Bond Ratings • Investment Grade Bonds • Speculative Bonds – Also known as Junk Bonds or High-Yield Bonds Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 71 Table 6.4 Bond Ratings (1 of 2) Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 72 Table 6.4 Bond Ratings (2 of 2) [Table 6.4 continued] Ratings: Moody’s/Standard & Poor’s Source: www.moodys.com Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 73 Corporate Yield Curves • Default Spread – Also known as Credit Spread – The difference between the yield on corporate bonds and Treasury yields Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 74 Figure 6.3 Corporate Yield Curves for Various Ratings, February 2018 Source: Bloomberg Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 75 Figure 6.4 Yield Spreads and the Financial Crisis Source: Bloomberg.com Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 76 6.5 Sovereign Bonds • Bonds issued by national governments – U.S. Treasury securities are generally considered to be default free – All sovereign bonds are not default-free, e.g., Greece defaulted on its outstanding debt in 2012 – Importance of inflation expectations Potential to “inflate away” the debt – European sovereign debt, the E M U, and the E C B Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 77 Figure 6.5 Percent of Debtor Countries in Default or Restructuring Debt, 1800–2006 Source: Data from This Time Is Different, Carmen Reinhart and Kenneth Rogoff, Princeton University Press, 2009.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 78 Figure 6.6 European Government Bond Yields, 1976–2018 Source: Federal Reserve Economic Data, research.stlouisfed.org/fred2.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 79 Chapter Quiz 1. What is the relationship between a bond’s price and its yield to maturity?
2. If a bond’s yield to maturity does not change, how does its cash price change between coupon payments?
3. How does a bond’s coupon rate affect its duration—the bond price’s sensitivity to interest rate changes?
4. Explain why two coupon bonds with the same maturity may each have a different yield to maturity.
5. There are two reasons the yield of a defaultable bond exceeds the yield of an otherwise identical default-free bond. What are they?
Corporate Finance (2 of 2) Fifth Edition, Global Edition Chapter 6 Appendix Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 81 Forward Interest Rates • 6A.1 Computing Forward Rates – A forward interest rate (or forward rate) is an interest rate that we can guarantee today for a loan or investment that will occur in the future.
– In this chapter, we consider interest rate forward contracts for one-year investments, so the forward rate for year 5 means the rate available today on a one-year investment that begins four years from today.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 82 Computing Forward Rates (1 of 5) • By the Law of One price, the forward rate for year 1 is equivalent to an investment in a one-year, zero-coupon bond.
1 1 = f YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 83 Computing Forward Rates (2 of 5) • Consider a two-year forward rate.
• Suppose the one-year, zero-coupon yield is 5.5% and the two-year, zero-coupon yield is 7.0%.
• We can invest in the two-year, zero-coupon bond at 7.0% and earn after two years.
2 $(1.07) • Or, we can invest in the one-year bond and earn $1.055 at the end of the year.
– We can simultaneously enter into a one-year interest rate forward contract for year 2 at a rate of f2.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 84 Computing Forward Rates (3 of 5) • At then end of two years, we will have $(1.055)(1 + f2).
• Since both strategies are risk free, by the Law of One Price, they should have the same return: 2 2 (1.07) = (1.055)(1+ ) f Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 85 Computing Forward Rates (4 of 5) • Rearranging, we have 2 2 1.07 (1+ ) = = 1.0852 1.055 f • The forward rate for year 2 is f2 = 8.52%.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 86 Computing Forward Rates (5 of 5) • In general, 1 1 (1+ ) = (1+ ) (1+ ) n n n n n YTM YTM f • Rearranging, we get the general formula for the forward interest rate: 1 1 (1+ ) = 1 (1+ ) n n n n n YTM f YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 87 Textbook Example 6A.1 (1 of 2) Computing Forward Rates Problem Calculate the forward rates for years 1 through 5 from the following zero-coupon yields: Maturity 1 2 3 4 Y T M 5.00% 6.00% 6.00% 5.75% Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 88 Textbook Example 6A.1 (2 of 2) Solution Using Eqs. 6A.1 and 6A.2: 1 1 2 2 2 2 1 3 3 3 3 2 2 2 4 4 4 4 3 3 3 = = 5.00% (1+ ) 1.06 = 1 = 1 = 7.01% (1+ ) 1.05 (1+ ) 1.06 = 1 = 1 = 6.00% (1+ ) 106 (1+ ) 1.0575 = 1 = 1 = 5.00% (1+ ) 1.06 f YTM YTM f YTM YTM f YTM YTM f YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 89 Alternative Example 6.A1 (1 of 2) • Problem – At the end of 2015, yields on Canadian government bonds were Maturity 1 Year 2 Years 3 Years 4 Years 5 Years 6 Years 7 Years 8 Years 9 Years 10 Years Yield 0.51% 0.47% 0.54% 0.67% 0.82% 0.97% 1.13% 1.28% 1.42% 1.55% – Based on this yield curve, what is f2? f10?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 90 Alternative Example 6.A1 (2 of 2) • Solution 2 2 1 10 10 9 (1.0047) = 1 = 0.0043 = 0.43% (1.0051) (1.0155) = 1 = 0.02728 = 2.27% (1.0142) f f Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 91 6A.2 Computing Bond Yields from Forward Rates • It is also possible to compute the zero-coupon yields from the forward interest rates: 1 2 n (1+ )×(1+ )×...×(1+ ) = (1+ ) n n f f f YTM • For example, using the forward rates from Example 8A.1, the four-year zero-coupon, yield is 14 4 1 2 3 4 14 1+ = (1+ )(1+ )(1+ )(1+ ) = (1.05)(1.0701)(1.06)(1.05) = 1.0575 YTM f f f f Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 92 6A.3 Forward Rates and Future Interest Rates (1 of 2) • How does the forward rate compare to the interest rate that will actually prevail in the future?
• It is a good predictor only when investors do not care about risk.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 93 Textbook Example 6A.2 (1 of 2) Forward Rates and Future Spot Rates Problem JoAnne Wilford is corporate treasurer for Wafer Thin Semiconductor. She must invest some of the cash on hand for two years in risk-free bonds. The current one-year, zero-coupon yield is 5%. The one-year forward rate is 6%. She is trying to decide between three possible strategies: (1) buy a two-year bond, (2) buy a one-year bond and enter into an interest rate forward contract to guarantee the rate in the second year, or (3) buy a one-year bond and forgo the forward contract, reinvesting at whatever rate prevails next year. Under what scenarios would she be better off following the risky strategy?
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 94 Textbook Example 6A.2 (2 of 2) Solution From Eq. 6A.3, both strategies (1) and (2) lead to the same risk-free return of 2 2 1 2 (1+ ) = (1+ )(1+ ) = (1.05)(1.06).
YTM YTM f The third strategy returns (1.05)(1 + r), where r is the one-year interest rate next year. If the future interest rate turns out to be 6%, then the two strategies will offer the same return. Otherwise Wafer Thin Semiconductor is better off with strategy (3) if the interest rate next year is greater than the forward rate—6%—and worse off if the interest rate is lower than 6%.
Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 95 Forward Rates and Future Interest Rates (2 of 2) • We can think of the forward rate as a break-even rate.
• Since investors do care about risk, Expected Future Spot Interest Rate = Forward Interest Rate + Risk Premium Finanzmanagement Kapitel 6: Die Bewertung von Anleihen 96 96 Finanzmanagement Kapitel 6: Die Bewertung von Anleihen 97 97 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 1981 Einführung Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 17.04.2015 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. 02.04.2019 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. -0,2% 2015 -0,4% 2016 -0,5% 2019 Satz der Einlagefazilität 26.7.2012: „Die EZB wird alles Notwendige tun, um den Euro zu erhalten.
Und glauben Sie mir, es wird ausreichen.“ Finanzmanagement Kapitel 6: Die Bewertung von Anleihen 102 102 Credit-Spread im Zeitablauf -1 1 3 5 7 9 11 12.01.1982 21.06.1983 27.11.1984 06.05.1986 13.10.1987 21.03.1989 28.08.1990 04.02.1992 13.07.1993 20.12.1994 28.05.1996 04.11.1997 13.04.1999 19.09.2000 26.02.2002 05.08.2003 11.01.2005 20.06.2006 27.11.2007 05.05.2009 12.10.2010 20.03.2012 27.08.2013 03.02.2015 12.07.2016 19.12.2017 Rendite in Prozent/Spread in Prozentpunkt Corporate Bonds Gov. Bonds 3-5 Jahre Credit-Spread Finanzmanagement Kapitel 6: Die Bewertung von Anleihen 103 103 Credit Default Swap Ein Credit Default Swap (CDS) ist ein Kreditderivat zum Handeln von Ausfallrisiken von Krediten, Anleihen oder Schuldnernamen.
Der Sicherungsnehmer bezahlt normalerweise eine regelmäßige (häufig vierteljährliche oder halbjährliche) Gebühr und erhält bei Eintritt des bei Vertragsabschluss definierten Kreditereignisses, also beispielsweise dem Ausfall der Rückzahlung aufgrund Insolvenz des Schuldners, eine Ausgleichszahlung.
Der CDS ähnelt einer Kreditversicherung. Dadurch erhalten Banken und Investoren ein flexibles Instrument, um Kreditrisiken zu handeln.
Finanzmanagement Kapitel 6: Die Bewertung von Anleihen 104 104 Ausgewählte CDS Senior Debt 5 Jahre 0 100 200 300 400 500 600 29.09.06 29.09.07 29.09.08 29.09.09 29.09.10 29.09.11 29.09.12 29.09.13 29.09.14 29.09.15 29.09.16 29.09.17 29.09.18 Daimler BMW VW Finanzmanagement Kapitel 6: Die Bewertung von Anleihen 105 105 Spanien, BRD; Italien Griechenland 5 Jahre CDS in EUR 0 50 100 150 200 250 300 350 400 450 500 14.12.07 14.04.08 14.08.08 14.12.08 14.04.09 14.08.09 14.12.09 14.04.10 14.08.10 14.12.10 14.04.11 14.08.11 14.12.11 14.04.12 14.08.12 14.12.12 14.04.13 14.08.13 14.12.13 14.04.14 14.08.14 14.12.14 14.04.15 14.08.15 14.12.15 14.04.16 14.08.16 14.12.16 14.04.17 14.08.17 14.12.17 14.04.18 14.08.18 14.12.18 Spanien BRD Italien Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Deutsche Bank ITL Zero Bond 15.10.2021 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. formel 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑌𝑖𝑒𝑙𝑑ൌ ଵ ଵଵ,ସଽ మఱ െ1 ൌ8,95% Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Legal Facts Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Price History 0 20 40 60 80 100 120 10.10.2006 06.07.2009 01.04.2012 27.12.2014 22.09.2017 18.06.2020 Price per face value = 100 Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Yield to Maturity History 0 2 4 6 8 10 12 14 16 10.10.2006 06.07.2009 01.04.2012 27.12.2014 22.09.2017 18.06.2020 YTM Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Stock Exchanges Copyright © 2020 Pearson Education Ltd. All Rights Reserved. comdirect Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Realtime Quotes Berlin Exchange Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Trades Berlin Exchange Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Realtime Quotes Stuttgart Exchange Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Quotes Stuttgart Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Realtime Quotes Frankfurt Exchange Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Trade 16.10.2019 Hinweis: Maklercourtage 0,075% vom Nennwert.
10.000.000 ITL / 1.936,26 ITL/EUR = 5.164,57 EUR 5164,57 EUR 0,075/100 = 3,87 EUR Copyright © 2020 Pearson Education Ltd. All Rights Reserved. Trade 16.10.2019 |
7273 | https://math.stackexchange.com/questions/56847/angle-between-lines-joining-tetrahedron-center-to-vertices | geometry - Angle between lines joining tetrahedron center to vertices - Mathematics Stack Exchange
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Angle between lines joining tetrahedron center to vertices
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Asked 14 years, 1 month ago
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What are the angles formed at the center of a tetrahedron if you draw lines to the vertices?
I'm trying to make these:
I need to know what angles to bend the metal.
geometry
polyhedra
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edited Dec 13, 2011 at 16:38
t.b.
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asked Aug 11, 2011 at 4:48
AndrewFerraraAndrewFerrara
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I've taken the liberty of adding more information to the title, so that the question is clear at first glance.user856 –user856 2011-08-24 04:44:27 +00:00 Commented Aug 24, 2011 at 4:44
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The required tetrahedral angle is arccos(−1 3)≈109.5∘arccos(−1 3)≈109.5∘. You can use the law of cosines to show this... or more transparently, you can exploit the fact that a tetrahedron is easily embedded inside a cube:
I suppose now's as good a time as any to post the synthetic proof.
One can use the Pythagorean theorem to show that a square with unit edge length has a diagonal of length √2 2–√. The Pythagorean theorem can be used again to show that a right triangle with leg lengths 1 1 and √2 2–√ will have a hypotenuse of length √3 3–√ (corresponding to the triangle formed by an edge, a face diagonal, and a cube diagonal). We know that the diagonals of a rectangle bisect each other; this can be used to show that the diagonals of a cube bisect each other. From this, we find that the side lengths of the (isosceles!) triangle formed by two half-diagonals of the cube (corresponding to two of the arms of your caltrops) and a face diagonal are √3 2 3√2, √3 2 3√2, and √2 2–√. From the law of cosines, we have
2=3 4+3 4−2 3 4 cos θ
2=3 4+3 4−2 3 4 cos θ
where θ θ is the obtuse angle whose measure we are seeking. Algebraic manipulation yields cos θ=−1 3 cos θ=−1 3.
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edited Aug 24, 2011 at 3:19
answered Aug 11, 2011 at 5:01
J. M. ain't a mathematicianJ. M. ain't a mathematician
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5 Anecdotal addendum: that 109.5° bit is stuck in my head, as that fact seems to be drummed into every chemistry student (tetrahedra do pop up a lot in chemistry).J. M. ain't a mathematician –J. M. ain't a mathematician 2011-08-11 15:09:48 +00:00 Commented Aug 11, 2011 at 15:09
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One way is to write the vertices as vectors a,b,c,d a,b,c,d with norm ‖⋅‖=1∥⋅∥=1. Then a+b+c+d=0 a+b+c+d=0. But
0=‖a+b+c+d‖2=4+2(4 2)cos θ,
0=∥a+b+c+d∥2=4+2(4 2)cos θ,
so θ=arccos(−1/3)θ=arccos(−1/3).
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edited Aug 23, 2011 at 8:09
answered Aug 11, 2011 at 5:18
anonanon
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+1 A neat argument! Even though it relies on the existence of the obvious symmetries.Jyrki Lahtonen –Jyrki Lahtonen 2011-08-23 08:15:44 +00:00 Commented Aug 23, 2011 at 8:15
10 @Jyrki: Even though? Because! :-)joriki –joriki 2011-08-24 07:18:51 +00:00 Commented Aug 24, 2011 at 7:18
1 @joriki: :-) :-)Jyrki Lahtonen –Jyrki Lahtonen 2011-08-24 07:37:13 +00:00 Commented Aug 24, 2011 at 7:37
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(Assuming the tetrahedron is supposed to be regular) Take the tetrahedron with vertices (1,1,1),(1,−1,−1),(−1,1,−1),(−1,−1,1)(1,1,1),(1,−1,−1),(−1,1,−1),(−1,−1,1), which has centre at the origin, and use the dot product formula:
a⋅b=|a||b|cos θ a⋅b=|a||b|cos θ
which gives cos θ=−1 3 cos θ=−1 3
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edited Aug 11, 2011 at 19:11
J. M. ain't a mathematician
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answered Aug 11, 2011 at 5:12
Mark BennetMark Bennet
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This is essentially J.M.'s second observation, realised with a convenient cube.Mark Bennet –Mark Bennet 2011-08-11 05:14:25 +00:00 Commented Aug 11, 2011 at 5:14
Using coordinates indeed makes for a shorter derivation. :) The "synthetic" route would rely on repeated use of the law of cosines (which is of course equivalent to your use of the dot product) and the Pythagorean theorem...J. M. ain't a mathematician –J. M. ain't a mathematician 2011-08-11 05:19:08 +00:00 Commented Aug 11, 2011 at 5:19
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The angle is θ=arccos(−1 3)≈109.47.
θ=arccos(−1 3)≈109.47.
You can find this angle by writing the volume of a regular tetrahedron in two different ways.
(this is an adaptation of a blog post I wrote for a website that has since been deleted)
For setup, here is a picture of the regular tetrahedron in question:
If we knew the vertical distance, h h, from the base of the tetraheron to the center, then we could find the angle to be θ=arccos(−h).
θ=arccos(−h).
This follows from basic 2D trigonmetry, as shown in the following diagram which views the tetrahedron from the side:
We can find h h in a roundabout way by, taking advantage of the symmetry of the tetrahedron. Break the big tetrahedron into 4 smaller tetrahedra:
Now we may use the formula volume=1 3×base×height volume=1 3×base×height on the smaller tetrahedra, and the larger tetrahedra, to come up with two different expressions for the volume. Then, we can set these two expressions equal, solve for h h, and plug the result into the previous formula to get the desired angle:
One may trivially extend this argument to regular simplices of arbitrary dimension. There one would break an n n-simplex into n+1 n+1 equal pieces, therefore yielding the angle θ=arccos(−1 n).
θ=arccos(−1 n).
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edited Apr 23, 2019 at 5:36
answered Apr 23, 2019 at 5:29
Nick AlgerNick Alger
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[+1] Have you already seen this proof elsewhere ?Jean Marie –Jean Marie 2019-09-08 03:58:51 +00:00 Commented Sep 8, 2019 at 3:58
1 @JeanMarie I thought of this proof on my own when I was in highschool, around year 2000. But unsurprisingly it appears that other people have come up with it independently. After I made a blog post about this in the early 2010's, the author of the following article emailed me saying he had published a similar proof here: "C. Giomini, G. Marrosu, M.E. Cardinali. The exploded tetrahedron."Nick Alger –Nick Alger 2019-09-08 04:58:40 +00:00 Commented Sep 8, 2019 at 4:58
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If the tetrahedron is regular, we can use statical equilibrium of four equal isotropic forces with angle θ θ between any two of them. Referencing with reference to any one force,
F cos θ+..+..+F=0
F cos θ+..+..+F=0
cos θ=−1 3
cos θ=−1 3
To generalize for all i i direction forces on a particular Z direction the vector dot product sum can be used:
Σ F i.Z=0
Σ F i.Z=0
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answered May 26, 2016 at 6:33
NarasimhamNarasimham
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Using the illustration of the tetrahedron embedded into the cube, you can lay out some dimensions. Assign a length of 2 to the sides of the cube, and then according to an old guy name Pythagoras, the hypotenuse of each side will be d 2√2. Since the center of the tetrahedron is also the center of the cube, then you can draw an isosceles triangle using the hypotenuse and the center that will be 2√2 at the base and 1 unit high. Split that in half to make two right triangles, each with a base of √2 and a height of 1. Now comes the real trig. Tan = Opposite / Adjacent, or √2/1, or just √2. Grab your calculator and you'll find that the angle you're looking for a the top of the right triangle is 54.735°. Put those two triangles back together, and you get an angle of 109.47°. For fabrication, I think you'd be fine with an angle of 110°.
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answered Mar 24, 2017 at 2:52
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Consider a sphere passing through the four vertices of the regular tetrapod with it's centre at the centre of the tetrapod. Each set of three vertices form four congruent equilateral spherical triangles on the surface of the sphere. For a spherical triangle ABC (unlike plane triangles),
The sides a,b,c are also measured as angles subtended at the centre of the sphere.
π<A+B+C<3 π π<A+B+C<3 π Here A=B=C=120°A=B=C=120° and a=b=c a=b=c The required angle between any two arms is thus the measure of any side of any of the triangles. Cosine rule for spherical triangles cos A=−cos B.cos C+sin B.sin C.cos a cos A=−cos B.cos C+sin B.sin C.cos a gives cos a=−1/3 cos a=−1/3 a=109.47°a=109.47°
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edited Sep 5, 2020 at 14:44
user577215664
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answered Sep 5, 2020 at 14:23
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Another possible solution:
Consider the equilateral triangle, edge length x, formed by any 3 tips of a regular tetrapod (a face on it's corresponding tetrahedron).
The 3 medians of this equilateral triangle section it into 6 congruent right triangles with hypotenuse length y, and angles of 60° between lines around the centroid.
sin60° = (1/2x) / y
therefore
y = x / 2sin60°
Now consider a line running orthogonal to the previously identified triangle, from its centroid to the tip of the opposing tetrapod leg. This line is coincident with that opposing tetrapod leg and forms a right triangle with the previously identified y, having a hypotenuse equal in length to x.
Back to our tetrapod, any two legs form an isosceles triangle with base length x. Therefore, the angle between any two legs, A, can be expressed:
A = 180° - 2𝜃
where
𝜃 = arcsin (y / x)
𝜃 = arcsin ([x / 2sin60°] / x)
𝜃 = arcsin (1 / 2sin60°)
therefore
A = 180° - 2[arcsin (1 / 2sin60°)]
A ≈ 109.47°
A more spatial approach, and certainly not the most direct, but it got me out of a pinch during a grade 12 chemistry test years ago. We were supposed to have the angles of molecular structures memorised and of course I blanked mid-test so I found myself deriving something to this effect in the margins.
This was on mobile so please forgive the formatting and lack of diagrams which surely would have made this explanation easier to follow.
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answered Mar 1, 2021 at 16:57
RoyRoy
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Watch this Khan Academy video of Tetrahedral bond angle proof. In the video, Mathematical proof of the bond angles in methane (a tetrahedral molecule);
We have the tetrahedron structure, for example methane, which is a three-dimensional structure shown above. (It's very difficult to see tetrahedral geometry on a two-dimensional, But it's much easier to see it over here on the right with the three- dimensional representation of methane.)
The bond angle of sp3 hybridized carbon in methane is 109.5 degrees, so you could say that this angle is the same all the way around. orienting the tetrahedron in this way allows us to find the bond angle that we are going for.
And we don't know what bond angle yet, but we can figure out this angle right here. So I'm going to call this theta for this triangle that's formed. And I know that this distance down here is positive/negative square root of 2, and then we go up 1 on the y-axis and then 0 on the z-axis. And if I want to find my bond angle in here, those angles have to add up to equal 180 degrees.
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answered Oct 30, 2021 at 17:26
user845875 user845875
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Extreme Value Theorem proof help
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Modified 7 years, 9 months ago
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10
$\begingroup$
Extreme Value Theorem: If $f$ is a continuous function on an interval [a,b], then $f$ attains its maximum and minimum values on [a,b].
Proof from my book: Since $f$ is continuous, then $f$ has the least upper bound, call it $M$. Assume there is no value $c \in [a,b]$ for which $f(c)=M$. Therefore, $f(x)
$g(x)=\frac{1}{M-f(x)}$
Observe $g(x)>0$ for every $x\in[a,b]$ and that $g$ is continuous and bounded on [a,b]. Therefore there exists $K>0$ such that $g(x)\le K$ for every $x\in [a,b]$. Since for each $x \in [a,b]$,
$g(x)= \frac{1}{M-f(x)} \le K$ is equivalent to $f(x)\le M-\frac{1}{K}$,
we have contradicted the fact that $M$ was assumed to be the least upper bound of $f$ on [a,b].Hence, there must be a balue $c\in[a,b]$ such that $f(c)=M$.
Q: Where does the function $g$ come from? Is there a popular alternative proof?
analysis
alternative-proof
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edited Aug 11, 2013 at 10:24
user53259
asked Apr 29, 2013 at 20:13
user65384user65384
61433 gold badges1111 silver badges2222 bronze badges
$\endgroup$
4
1
$\begingroup$ using Bolzano-weierstrass ,there is an alternate nicer proof.see :en.wikipedia.org/wiki/Extreme_value_theorem $\endgroup$
Halil Duru
– Halil Duru
2013-04-29 20:46:37 +00:00
Commented Apr 29, 2013 at 20:46
$\begingroup$ @StefanSmith, $f$ is bounded by the boundedness theorem $\endgroup$
Elimination
– Elimination
2014-07-17 15:52:33 +00:00
Commented Jul 17, 2014 at 15:52
3
$\begingroup$ @Elimination : Thanks, you're right. I had to Google "the boundedness theorem". The proof should mention why $f$ must be bounded, unless it is clear from context, from something immediately before it in the OP's book. I still prefer the proof that Halil Duru cites. $\endgroup$
Stefan Smith
– Stefan Smith
2014-07-19 00:24:01 +00:00
Commented Jul 19, 2014 at 0:24
$\begingroup$ How come we assumed that $g(x) \le K$, when what we need to prove in the first place is that a continuous function like this will be less that equal to its least upper bound, and not just strictly less than that. That's where the proof began right? With assuming $f(x) \lt M$. $\endgroup$
Aniruddh
– Aniruddh
2022-06-26 08:14:44 +00:00
Commented Jun 26, 2022 at 8:14
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4 Answers 4
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$\begingroup$
The "simplest" proof I know goes something like this : If $M$ is the supremum of $f$, then there is a sequence $(x_n)$ such that $f(x_n) \to M$. Now, $(x_n)$ itself may not be convergent, but since $[a,b]$ is compact, it will have a convergent subsequence $(x_{n_k})$. Suppose $x_{n_k} \to c \in [a,b]$, then $f(x_{n_k}) \to f(c)$. But $f(x_{n_k})$ is a subsequence of $f(x_n)$, and hence must converge to $M$. Hence, $f(c) = M$.
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answered Aug 11, 2013 at 10:33
Prahlad VaidyanathanPrahlad Vaidyanathan
32.4k11 gold badge4040 silver badges8585 bronze badges
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10
$\begingroup$
This is quite a simple proof, isn't it? Why do you want a 'popular alternative proof'?
The proof can't be too simple, because the result is not true if $f$ is defined over $\mathbb Q$ instead of $\mathbb R$. For instance, define $f:\mathbb Q \to \mathbb Q$ by $f(x) = x^3 - x$. Then $f$ doesn't attain its maximum in $[-1,0]$, because $-\sqrt\frac{1}{3} \notin \mathbb Q$. Hence any proof of your theorem must use the properties of the real numbers in an essential way.
As an illuminating exercise, try to see where the proof breaks down if $f$ is only defined over the rational numbers.
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answered Apr 29, 2013 at 20:35
TonyKTonyK
68.3k55 gold badges9595 silver badges189189 bronze badges
$\endgroup$
2
$\begingroup$ May you guide where the proof breaks Sir ? @TonyK $\endgroup$
Orion_Pax
– Orion_Pax
2022-05-06 23:55:31 +00:00
Commented May 6, 2022 at 23:55
$\begingroup$ @Orion_Pax The least upper bound property fails to hold, so M isn't actually in the domain of f $\endgroup$
Anshuman Agrawal
– Anshuman Agrawal
2023-05-03 06:53:53 +00:00
Commented May 3, 2023 at 6:53
Add a comment |
5
$\begingroup$
Where does the function $g$ come from?
We need to show that $f(x)=M$ for some $x$. A natural move is to consider the difference between $f$ and $M$. Let $d(x)=M-f(x)$. $f(x)=M \leftrightarrow d(x)=0$. The reason that the definition $g(x)=(d(x))^{-1}$ uses the inverse of the difference is that if $g$ is bounded from above by $K>0$, than $d$ is bounded from below by $K^{-1}>0$. $g$ is bounded by the boundedness theorem, thus we know a positive lower bound of $d$. Applying the boundedness theorem directly to $d$ is useless because the lower bound of $d$ can be $0$. This is the intuition behind $g$.
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answered May 18, 2016 at 15:03
beroalberoal
2,5732020 silver badges2525 bronze badges
$\endgroup$
5
$\begingroup$ Sir the proof shows the maximum bound which is being achived that is f(x) = M for some c . But what about the minimum value which the function is taking how to show it takes the minimum value too ? ( I am referring to a situation where maxium of function is at y= 5 but minimum is at y= -3 ) that is not symmetric about y axis $\endgroup$
Orion_Pax
– Orion_Pax
2022-04-20 04:08:32 +00:00
Commented Apr 20, 2022 at 4:08
$\begingroup$ @Orion_Pax If I understood you correctly, you want the proof for the minimum value. It's similar to the proof for the maximum value, but is mirrored with respect to the Y axis. You start with proving that $f$ has the greatest lower bound. $\endgroup$
beroal
– beroal
2022-05-06 20:47:18 +00:00
Commented May 6, 2022 at 20:47
$\begingroup$ Actually i already told whats wrong with that mirror part , may you once check this : @beroal $\endgroup$
Orion_Pax
– Orion_Pax
2022-05-06 23:49:39 +00:00
Commented May 6, 2022 at 23:49
$\begingroup$ math.stackexchange.com/q/4431692/922054 this @beroal $\endgroup$
Orion_Pax
– Orion_Pax
2022-05-06 23:50:58 +00:00
Commented May 6, 2022 at 23:50
$\begingroup$ @Orion_Pax Okay, I wrote the mirror proof in full math.stackexchange.com/a/4445009/7011 . Comment there if your think there is something wrong with it. $\endgroup$
beroal
– beroal
2022-05-07 08:04:38 +00:00
Commented May 7, 2022 at 8:04
Add a comment |
1
$\begingroup$
You asked for a "popular alternative proof". This is an alternative proof. I don't know how popular it is, but I like it. It uses the Bolzano-Weierstrass theorem (convergent subsequences) but hardly anything else, no least upper bounds; it skips the step of proving boundedness, going straight for the maximum. It could be shortened by using the fact that the set of rational numbers is countable, but that seems unnecessarily sophisticated.
Given a continuous real-valued function $f$ on $[a,b]$, we will show that the set $Y = f([a,b])$ has a greatest element.
For each positive integer $n$, define a finite set $Q_n = {\frac{p}{q}: p,q \text{ integers, } 0 < q \le n, |p| \le n}$.
Choose $y_n\in Y$ so as to maximize the number of elements in the set ${r\in Q_n: y_n > r}$, and choose $x_n\in[a,b]$ with $f(x_n) = y_n$.
The sequence ${x_n}$ has a subsequence converging to a point $c\in[a,b]$. Since $f$ is continuous, the corresponding subsequence of ${y_n}$ converges to $f(c)$. We will show that $f(c)$ is the greatest element of $Y$.
Assume for a contradiction that $f(c) r$ whenever $r\in Q_n$. Since $r\in Q_n$ for all sufficiently large $n$, we have $y_n > r > f(c)$ for all sufficiently large $n$. But this is absurd, since ${y_n}$ has a subsequence converging to $f(c)$.
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answered May 14, 2013 at 1:01
user75900user75900
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Grade 5: Computation and Estimation
5.CE.1 ------ The student will estimate, represent, solve, and justify solutions to single-step and multistep contextual problems using addition, subtraction, multiplication, and division with whole numbers. -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
5.CE.1.a
Estimate the sum, difference, product, and quotient of whole numbers in contextual problems.
Fractions as division word problems
### 5.CE.1.b ### Represent, solve, and justify solutions to single-step and multistep contextual problems by applying strategies (e.g., estimation, properties of addition and multiplication) and algorithms, including the standard algorithm, involving addition, subtraction, multiplication, and division of whole numbers, with and without remainders, in which:
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sums, differences, and products do not exceed five digits;
Multiply and divide whole numbers by 10, 100, and 1000
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factors do not exceed two digits by three digits;
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divisors do not exceed two digits; or
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5.CE.1.b.iv
dividends do not exceed four digits.
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5.CE.2 ------ The student will estimate, represent, solve, and justify solutions to single-step and multistep problems, including those in context, using addition and subtraction of fractions with like and unlike denominators (with and without models), and solve single-step contextual problems involving multiplication of a whole number and a proper fraction, with models. -----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
5.CE.2.a
Determine the least common multiple of two numbers to find the least common denominator for two fractions.
(Content unavailable)
5.CE.2.b
Estimate and determine the sum or difference of two fractions (proper or improper) and/or mixed numbers, having like and unlike denominators limited to 2, 3, 4, 5, 6, 8, 10, and 12 (e.g., 5/8 + 1/4, 4/5 – 2/3, 3 3/4 + 2 5/12), and simplify the resulting fraction.
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Estimate to add and subtract fractions with different denominators
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Estimate and solve single-step and multistep contextual problems involving addition and subtraction with fractions (proper or improper) and/or mixed numbers having like and unlike denominators, with and without models. Denominators should be limited to 2, 3, 4, 5, 6, 8, 10, and 12. Answers should be expressed in simplest form.
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Fraction multiplication as scaling examples
5.CE.3 ------ The student will estimate, represent, solve, and justify solutions to single-step and multistep problems, including those in context, using addition, subtraction, multiplication, and division with decimal numbers. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
5.CE.3.a
Apply estimation strategies (e.g., rounding to the nearest whole number, tenth or hundredth; compatible numbers, place value) to determine a reasonable solution for single-step and multistep contextual problems involving addition, subtraction, and multiplication of decimals, and single-step contextual problems involving division of decimals.
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### 5.CE.3.b ### Estimate and determine the product of two numbers using strategies and algorithms, including the standard algorithm, when given:
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a two-digit factor and a one-digit factor (e.g., 2.3 × 4; 0.08 × 0.9; .16 × 5);
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a three-digit factor and a one-digit factor (e.g., 0.156 × 4, 3.28 × 7, 8.09 × 0.2); and
Multiply and divide whole numbers by 10, 100, and 1000
Multiplying and dividing decimals by 10
Multiplying and dividing decimals by 10, 100, 1000
5.CE.3.b.iii
a two-digit factor and a two-digit factor (e.g., 0.85 × 3.7, 14 × 1.6, 9.2 × 3.5).
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### 5.CE.3.c ### Estimate and determine the quotient of two numbers using strategies and algorithms, including the standard algorithm, in which:
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quotients do not exceed four digits with or without a decimal point;
Basic multi-digit division
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Dividing by 2-digits: 9815÷65
Dividing by a 2-digits: 4781÷32
Division by 2-digits
Estimating multi-digit division
Introduction to dividing by 2-digits
Long division with remainders: 2292÷4
Long division with remainders: 3771÷8
Multiply and divide decimals by 10, 100, and 1000
Multiplying and dividing decimals by 10
Multiplying and dividing decimals by 10, 100, 1000
Strategies for dividing multiples of 10, 100, and 1000
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quotients may include whole numbers, tenths, hundredths, or thousandths;
Divide whole numbers by 0.1 or 0.01
Divide whole numbers to get a decimal (1-digit divisors)
Division strategies for decimal quotients
Multiply and divide decimals by 10, 100, and 1000
Multiplying and dividing decimals by 10
Multiplying and dividing decimals by 10, 100, 1000
5.CE.3.c.iii
divisors are limited to a single digit whole number or a decimal expressed as tenths; and
Divide whole numbers by 0.1 or 0.01
5.CE.3.c.iv
no more than one additional zero will need to be annexed.
(Content unavailable)
5.CE.3.d
Solve single-step and multistep contextual problems involving addition, subtraction, and multiplication of decimals by applying strategies (e.g., estimation, modeling) and algorithms, including the standard algorithm.
Volume word problems
5.CE.3.e
Solve single-step contextual problems involving division with decimals by applying strategies (e.g., estimation, modeling) and algorithms, including the standard algorithm.
(Content unavailable)
5.CE.4 ------ The student will simplify numerical expressions with whole numbers using the order of operations. -------------------------------------------------------------------------------------------------
### 5.CE.4.a ### Use order of operations to simplify numerical expressions with whole numbers, limited to addition, subtraction, multiplication, and division in which:
5.CE.4.a.i
expressions may contain no more than one set of parentheses;
Create expressions with parentheses
Evaluate expressions with parentheses
Evaluating expressions with & without parentheses
Translate expressions with parentheses
Translating expressions with parentheses
5.CE.4.a.ii
simplification will be limited to five whole numbers and four operations in any combination of addition, subtraction, multiplication, or division;
Evaluate expressions with parentheses
Evaluating expressions with & without parentheses
5.CE.4.a.iii
whole numbers will be limited to two digits or less; and
Evaluate expressions with parentheses
Evaluating expressions with & without parentheses
5.CE.4.a.iv
expressions should not include braces, brackets, or fraction bars.
Evaluate expressions with parentheses
Evaluating expressions with & without parentheses
Translating expressions with parentheses
5.CE.4.b
Given a whole number numerical expression involving more than one operation, describe which operation is completed first, which is second, and which is third.
Evaluate expressions with parentheses
Evaluating expressions with & without parentheses
Translating expressions with parentheses
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Complex Borel Measure and Bounded Variation Functions
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3.29 Theorem in Folland states that: If μ μ is complex Borel measure on R R and F(x)=μ(−∞,x])F(x)=μ(−∞,x]), then F∈N B V F∈N B V. Conversely if F∈N B V F∈N B V, there is a unique complex Borel measure μ F μ F such that F(x)=μ(−∞,x])F(x)=μ(−∞,x]). NBV={F is of Bounded Variation: F is right-continuous and F(−∞)=0 F(−∞)=0}.
Now all, of sudden I am confused/completely lost. My professor writes that:
If F(x)=χ[a,b)F(x)=χ[a,b), then it is in BV and right continuous so it is in NBV? and u F=δ a−δ b u F=δ a−δ b. I have no idea how u F=δ a−δ b u F=δ a−δ b. How was this formulated/come up with? I have no idea how u F u F was determined Also I'm not exactly sure what is meant by δ a δ a and -δ b δ b
Does Dirac measure in this case mean, δ a δ a mean
{1 0 x=a x≠a.{1 x=a 0 x≠a.
δ b δ b mean
{1 0 x=b x≠b.{1 x=b 0 x≠b.
? Or is this wrong.
Why then is μ F=δ a−δ b μ F=δ a−δ b? I'm very anxious why I don't understand how this was derived/ why it is true.
Like wise, my professor writes that if F(x)=arctan(x)F(x)=arctan(x), for x>0 x>0, 0 0 otherwise then F∈F∈ NBV. and d μ F=1 1+x 2 χ x>0 d m d μ F=1 1+x 2 χ x>0 d m. (dm meaning respect to Lebesgue measure). Then how do I get μ F μ F? I'm honestly not sure why the expression for d μ F d μ F is true. Now I'm very anxious why I'm not understanding these formulations. Any help would be much appreciated.
real-analysis
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edited Dec 6, 2018 at 11:08
Davide Giraudo
183k 71 71 gold badges 279 279 silver badges 433 433 bronze badges
asked Dec 5, 2018 at 23:14
user431865 user431865
3
As far as guessing what μ μ is it would be very helpful to know some basic probability theory. If you are not familiar with probability theory it is best to just verify that the statement made by the professor are right. BTW, δ a δ a denotes the measure defined by δ a(E)=1 δ a(E)=1 if a∈E a∈E and 0 0 otherwise.Kavi Rama Murthy –Kavi Rama Murthy 2018-12-05 23:22:01 +00:00 Commented Dec 5, 2018 at 23:22
So the measure is 1 at the point a?user431865 –user431865 2018-12-05 23:23:30 +00:00 Commented Dec 5, 2018 at 23:23
I don't understand why μ F=δ a−δ b μ F=δ a−δ b?user431865 –user431865 2018-12-05 23:24:09 +00:00 Commented Dec 5, 2018 at 23:24
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Yes, this F F is NBV. The δ δ's are indeed Dirac measures, however, as such, they take measurable sets as values, rather than function values.
δ a(A):={1,0,A∋a otherwise.(A∈B R)δ a(A):={1,A∋a 0,otherwise.(A∈B R)
μ F μ F is now the Lebesgue-Stieltjes measure based on NBV function F F. In order to show this, we need to show that
μ F((c,d])=δ a((c,d])−δ b((c,d])=F(d)−F(c)(c,d∈R,c<d)μ F((c,d])=δ a((c,d])−δ b((c,d])=F(d)−F(c)(c,d∈R,c<d)
It is probably best to consider different cases:
(i) c<d<a<b⇒μ F((c,d])=0−0=F(d)−F(c)c<d<a<b⇒μ F((c,d])=0−0=F(d)−F(c)
(ii) c<a≤d<b⇒μ F((c,d])=1−0=F(d)−F(c)c<a≤d<b⇒μ F((c,d])=1−0=F(d)−F(c)
(iii) a≤c<d<b⇒μ F((c,d])=1−1=F(d)−F(c)a≤c<d<b⇒μ F((c,d])=1−1=F(d)−F(c)
(iv) a≤c<b≤d⇒μ F((c,d])=0−1=F(d)−F(c)a≤c<b≤d⇒μ F((c,d])=0−1=F(d)−F(c)
(v) a<b≤c<d⇒μ F((c,d])=0−0=F(d)−F(c)a<b≤c<d⇒μ F((c,d])=0−0=F(d)−F(c)
As the set of intervals {(c,d]:c,d∈R}{(c,d]:c,d∈R} is a semiring, we can now apply Carathéodory's extension theorem and know that the
μ F(A)=δ a(A)−δ b(A)(A∈B R).μ F(A)=δ a(A)−δ b(A)(A∈B R).
Maybe one hint to give you an intuition, also towards the second question: If you have an NBV function, you can compute a something like a generalised "derivative". "Derivative" refers to the measure-theoretic fundamental theorem of calculus. Like in the case above, this "derivative" may be represented by a measure rather than a function - especially in cases like in 1., where F F is clearly not differentiable. It actually is a function, if F F is absolutely continuous (with respect to the Lebesgue measure).
The latter is the case here. In particular, we have μ F(A)=∫A F′(x)d x(x∈B R)μ F(A)=∫A F′(x)d x(x∈B R) Also, arctan′(x)=1/(1+x 2)arctan′(x)=1/(1+x 2), (x > 0)
F(x)F′(x):={arctan(x),0,x>0 otherwise.(x∈R)={arctan′(x),0,x>0 otherwise.={1/(x 2+1),0,x>0 otherwise.(x∈R)(x∈R)F(x):={arctan(x),x>0 0,otherwise.(x∈R)F′(x)={arctan′(x),x>0 0,otherwise.={1/(x 2+1),x>0 0,otherwise.(x∈R)(x∈R)
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edited Dec 6, 2018 at 9:12
answered Dec 5, 2018 at 23:54
JonasJonas
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thanks for the very detailed answer. My question is, I sort of get the proof but I'm not too confident in the intuition behind why μ F=δ A=δ B μ F=δ A=δ B. Like if I saw a Question determine: μ F μ F for the characteristic function given in question 1, what is the quick intuition that tells me that it is δ a−δ b δ a−δ b? Is there a quick way to see/know that the answer is δ a−δ b δ a−δ b? My teacher seemed to think it was obvious, does it have to do with support of the measure. Just would like to know the quick intuition, thanks.user431865 –user431865 2018-12-06 00:47:36 +00:00 Commented Dec 6, 2018 at 0:47
Or further I'm not sure why the derivative is represented as the difference of two measures in this case where F is not differentiable? Why is this so?, thanks user431865 –user431865 2018-12-06 00:49:22 +00:00 Commented Dec 6, 2018 at 0:49
The measure is concentrated, where the variation of F F lies. F F is constant aside from the points a,b a,b. Hence, there is no variation whatsoever aside from {a,b}{a,b}. In a a there is positive variation, in b b negative variation. For such piecewise constant functions, this may be a good intuition.Jonas –Jonas 2018-12-06 09:15:38 +00:00 Commented Dec 6, 2018 at 9:15
Thanks @Jonas, one last question , but why is there positive variation in a and negative variation in b. I don't understand what positive variation and negative variation mean in this circumstance, or what variation means in this case user431865 –user431865 2018-12-06 09:18:39 +00:00 Commented Dec 6, 2018 at 9:18
Intuitively: In a a the function increases (F(a−)=0↑1=F(a)F(a−)=0↑1=F(a)), in b b it decreases (F(b−)=1↓0=F(b)F(b−)=1↓0=F(b)).Jonas –Jonas 2018-12-06 11:03:45 +00:00 Commented Dec 6, 2018 at 11:03
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7277 | https://en.wikipedia.org/wiki/Exothermic_reaction | Jump to content
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1 Examples
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Exothermic reaction
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From Wikipedia, the free encyclopedia
Chemical reaction that releases energy as light or heat
In thermochemistry, an exothermic reaction is a "reaction for which the overall standard enthalpy change ΔH⚬ is negative." Exothermic reactions usually release heat. The term is often confused with exergonic reaction, which IUPAC defines as "... a reaction for which the overall standard Gibbs energy change ΔG⚬ is negative." A strongly exothermic reaction will usually also be exergonic because ΔH⚬ makes a major contribution to ΔG⚬. Most of the spectacular chemical reactions that are demonstrated in classrooms are exothermic and exergonic. The opposite is an endothermic reaction, which usually takes up heat and is driven by an entropy increase in the system.
Examples
Examples are numerous: combustion, the thermite reaction, combining strong acids and bases, polymerizations. As an example in everyday life, hand warmers make use of the oxidation of iron to achieve an exothermic reaction:
: 4Fe + 3O2 → 2Fe2O3 ΔH⚬ = - 1648 kJ/mol
A particularly important class of exothermic reactions is combustion of a hydrocarbon fuel, e.g. the burning of natural gas:
: CH4 + 2O2 → CO2 + 2H2O ΔH⚬ = - 890 kJ/mol
These sample reactions are strongly exothermic.
Uncontrolled exothermic reactions, those leading to fires and explosions, are wasteful because it is difficult to capture the released energy. Nature effects combustion reactions under highly controlled conditions, avoiding fires and explosions, in aerobic respiration so as to capture the released energy, e.g. for the formation of ATP.
Measurement
The enthalpy of a chemical system is essentially its energy. The enthalpy change ΔH for a reaction is equal to the heat q transferred out of (or into) a closed system at constant pressure without in- or output of electrical energy. Heat production or absorption in a chemical reaction is measured using calorimetry, e.g. with a bomb calorimeter. One common laboratory instrument is the reaction calorimeter, where the heat flow from or into the reaction vessel is monitored. The heat release and corresponding energy change, ΔH, of a combustion reaction can be measured particularly accurately.
The measured heat energy released in an exothermic reaction is converted to ΔH⚬ in Joule per mole (formerly cal/mol). The standard enthalpy change ΔH⚬ is essentially the enthalpy change when the stoichiometric coefficients in the reaction are considered as the amounts of reactants and products (in mole); usually, the initial and final temperature is assumed to be 25 °C. For gas-phase reactions, ΔH⚬ values are related to bond energies to a good approximation by:
: ΔH⚬ = total bond energy of reactants − total bond energy of products
In an exothermic reaction, by definition, the enthalpy change has a negative value:
: ΔH = Hproducts - Hreactants < 0
where a larger value (the higher energy of the reactants) is subtracted from a smaller value (the lower energy of the products). For example, when hydrogen burns:
: 2H2 (g) + O2 (g) → 2H2O (g)
: ΔH⚬ = −483.6 kJ/mol
See also
Chemical thermodynamics
Differential scanning calorimetry
Endergonic
Exergonic
Endergonic reaction
Exergonic reaction
Exothermic process
Endothermic reaction
Endotherm
References
^ "Exothermic reaction". The IUPAC Compendium of Chemical Terminology. IUPAC. 2014. doi:10.1351/goldbook.E02269.
^ a b Laidler, K. J. (1996). "A glossary of terms used in chemical kinetics, including reaction dynamics (IUPAC Recommendations 1996)". Pure and Applied Chemistry. 68: 149–192. doi:10.1351/pac199668010149. S2CID 98267946.
^ "Enthalpy (Chapter 5)". Archived from the original on 2013-07-08. Retrieved 2013-07-20.
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7278 | https://math.libretexts.org/Courses/Cosumnes_River_College/Math_401%3A_Calculus_II_-_Integral_Calculus/01%3A_Applications_of_Integration/1.03%3A_Volumes_of_Revolution_-_The_Disk_and_Washer_Methods | [1,4]
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1.3: Volumes of Revolution - The Disk and Washer Methods
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1.2.2: Homework
1.3.1: Resources and Key Concepts
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Roy Simpson
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( \newcommand{\kernel}{\mathrm{null}\,})
| | |
--- |
| Note to the Instructor (click to expand) This sections deals solely with the Disk and Washer methods for finding the volume of rotation. --- The following is a list of learning objectives for this section. Learning Objectives (click to expand) Find the volume of a solid of revolution using the disk method. Find the volume of a solid of revolution with a cavity using the washer method. | |
Now that we have the basic idea of how to compute the volume of a solid using slices, we venture into rotational solids.
Solids of Revolution
If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in the following figure.
Figure 1.3.1a1.3.1a: This is the region that is revolved around the xx-axis.
Figure 1.3.1b1.3.1b: As the region begins to revolve around the axis, it sweeps out a solid of revolution.
Figure 1.3.1c1.3.1c: This is the solid that results when the revolution is complete.
Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest of this section looking at solids of this type. The following example uses the slicing method to calculate the volume of a solid of revolution.
Example 1.3.11.3.1
Use the slicing method to find the volume of the solid of revolution bounded by the graphs of f(x)=x2−4x+5f(x)=x2−4x+5, x=1x=1, and x=4,x=4, and rotated about the xx-axis.
Solution
: Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval [1,4] as shown in the following figure.
Figure 1.3.2: A region used to produce a solid of revolution.
Next, revolve the region around the x-axis, as shown in the following figure.
Figure 1.3.3: Two views, (a) and (b), of the solid of revolution produced by revolving the region in Figure 1.3.2 about the x-axis.
The cross-sections are circles since the solid was formed by revolving the region around the x-axis. The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by f(x). Use the formula for the area of the circle:A(x)=πr2=π[f(x)]2=π(x2−4x+5)2.The volume, then, isV=∫baA(x)dx=∫41π(x2−4x+5)2dx=π∫41(x4−8x3+26x2−40x+25)dx=π(x55−2x4+26x33−20x2+25x)|41=785πThe volume is 78π5 units3.
Checkpoint 1.3.11.3.1
Use the method of slicing to find the volume of the solid of revolution formed by revolving the region between the graph of the function f(x)=1xf(x)=1x and the xx-axisover the interval [1,2][1,2] around the xx-axis. See the following figure.
Answer
: π2 units3
The Disk Method
When we use the slicing method with solids of revolution, it is often called the Disk Method because, for solids of revolution, the slices used to approximate the volume of the solid are disks. To see this, consider the solid of revolution generated by revolving the region between the graph of the function f(x)=(x−1)2+1f(x)=(x−1)2+1 and thexx-axis over the interval [−1,3][−1,3] around the xx-axis (called the axis of rotation). The graph of the function and a representative disk are shown in Figure 1.3.111.3.11 (a) and (b). The region of revolution and the resulting solid are shown in Figure 1.3.41.3.4 (c) and (d).
Figure 1.3.4a1.3.4a: A thin rectangle for approximating the area under a curve. The height of this rectangle is the radius of rotation and it is denoted riri.
Figure 1.3.4b1.3.4b: A representative disk formed by revolving the rectangle about the xx-axis.
Figure 1.3.4c1.3.4c: The region under the curve is revolved around the xx-axis, resulting in Figure 1.3.4d1.3.4d the solid of revolution.
Interactive Element: A Solid of Revolution
Figure 1.3.4e1.3.4e: A dynamic version of this solid of revolution generated using CalcPlot3D.
We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We know that V=limn→∞n∑i=1A(x∗i)Δx=∫baA(x)dx.
V=limn→∞∑i=1nA(x∗i)Δx=∫baA(x)dx.
The only difference with the Disk Method is that we know the formula for the cross-sectional area ahead of time; it is the area of a circle. That is,V=limn→∞n∑i=1π[r(x∗i)]2Δx=∫baπ[r(x)]2dx,
V=limn→∞∑i=1nπ[r(x∗i)]2Δx=∫baπ[r(x)]2dx,
where r(x)r(x) is the radius of rotation (and r(x∗i)r(x∗i) is the radius of rotation for the ithith slice).1 For the Disk and Washer (see below) Methods, the radius of rotation is the distance between the axis of rotation and the function. If the axis of rotation is the xx-axis, then our discussion of measuring distances in Section 1.1 reveals thatr(x)={f(x)−0=f(x)iff(x)≥00−f(x)=−f(x)iff(x)<0
r(x)={f(x)−0=f(x)0−f(x)=−f(x)ififf(x)≥0f(x)<0
In either case, and only if the axis of rotation is the xx-axis, we get[r(x)]2=[f(x)]2
[r(x)]2=[f(x)]2
This leads to the following theorem.
Theorem: The Volume of a Rotational Solid About the xx-axis Using the Disk Method
Let f(x)f(x) be continuous. Define R as the region bounded by the graph of f(x), the x-axis, on the left by the line x=a, and on the right by the line x=b. Then, the volume of the solid of revolution formed by revolving R around the x-axisis given byV=∫baπ[f(x)]2dx.
While Equation 1.3.1 is simple to memorize, doing so will result in a massive lack of understanding of how rotational volumes are computed and inevitably will lead to mistakes. It is best to truly understand the underpinnings of the development of Equation 1.3.1 so that you can compute the volume of a rotational solid in more general situations. For example, Equation 1.3.1 does not work if you are rotating a region about the y-axis, the line y=−2, or the region bounded between two curves about the line x=7.
In general, when needing to compute the volume of a rotational solid, you should fall back to the development of the corresponding Riemann sumV=limn→∞n∑i=1K(x∗i)Δx,
where K(x∗i)Δx is the volume of the ith slice found using logic from our previous Geometry course.2 To compute this volume, we will always start by focusing on the volume of the ith slice,Vi=K(x∗i)Δx.
Example 1.3.2
Compute the volume of the rotational solid from Figure 1.3.4d.
Solution
: Just as we said we should, we start by considering the volume of the ith slice,Vi=K(x∗i)Δx.Since a representative slice of this solid of revolution is a disk (a circle with a thickness), we know the volume of the ith slice isVi=π[r(x∗i)]2Δx.Moreover, since r(x∗i) is the radius of revolution, and the axis of revolution is the x-axis, we know r(x∗i)=f(x∗i) (remember, f(x)≥0 on the given interval). Therefore,Vi=π[f(x∗i)]2Δx.Hence,V=∫baπ[f(x)]2dx=∫3−1π[(x−1)2+1]2dx=π∫3−1[(x−1)4+2(x−1)2+1]2dx=π[15(x−1)5+23(x−1)3+x]|3−1=π[(325+163+3)−(−325−163−1)]=412π15 units3.Again, it cannot be overstated that starting our computation of the volume for a solid of revolution by considering the volume of the ith slice is incredibly helpful.3
Let’s look at some other examples.
Example 1.3.3
Use the Disk Method to find the volume of the solid of revolution generated by rotating the region bounded by the graph of f(x)=√x and the x-axisover the interval [1,4] around the x-axis.
Solution
: The graphs of the function and the solid of revolution are shown in the following figure.
Figure 1.3.5a: The function f(x)=√x over the interval [1,4].
Figure 1.3.5b: The solid of revolution obtained by revolving the region under the graph of f(x) about the x-axis.
We have the volume of the ith slice asVi=K(x∗i)Δx;however, we can see that a drawing of this representative slice would be a disk. Using the fact that the axis of rotation is the x-axis and f(x)≥0, we getVi=π[r(x∗i)]2Δx=π[f(x∗i)]2Δx=π[√x∗i]2Δx=πx∗iΔx.Hence,V=∫baπ[f(x)]2dx=∫41πxdx=π∫41xdx=π2x2|41=15π2The volume is 15π2 units3.
Checkpoint 1.3.3
Use the Disk Method to find the volume of the solid of revolution generated by rotating the region between the graph of f(x)=√4−x and the x-axis over the interval [0,4] around the x-axis.
Answer
: 8π units3
So far, our examples have all concerned regions revolved around the x-axis, but we can generate a solid of revolution by revolving a plane region around any horizontal or vertical line. In the following example, we look at a solid of revolution that has been generated by rotating a region around the y-axis. The mechanics of the Disk Method are nearly the same as when the x-axis is the axis of revolution, but we express the function in terms of y and we integrate with respect to y as well. This is summarized in the following theorem.
Theorem: The Volume of a Rotational Solid about the y-axis Using the Disk Method
Let g(y) be continuous. Define Q as the region bounded by the graph of g(y), the y-axis, below by the line y=c, and above by the line y=d. Then, the volume of the solid of revolution formed by revolving Q around the y-axisis given byV=∫dcπ[g(y)]2dy.
Again, memorizing Equation 1.3.3 is a terrible idea. An example showcases that this formula is a result of our theoretical derivation of the volume of the ith slice.
Example 1.3.4
Let R be the region bounded by the graph of g(y)=√4−y and the y-axis over the y-axis interval [0,4]. Use the Disk Method to find the volume of the solid of revolution generated by rotating R about the y-axis.
Solution
: Figure 1.3.13 shows the function and a representative disk that can be used to estimate the volume. Notice that since we are revolving the function around the y-axis,the disks are horizontal, rather than vertical.
Figure 1.3.6a: Shown is a thin rectangle between the curve of the function g(y)=√4−y and the y-axis.
Figure 1.3.6b: The rectangle forms a representative disk after revolving around the y-axis.
The region to be revolved and the full solid of revolution are depicted in the following figure.
Figure 1.3.7a: The region to the left of the function g(y)=√4−y over the y-axis interval [0,4].
Figure 1.3.7b: The solid of revolution formed by revolving the region about the y-axis.
We can see from Figure 1.3.7b that the ith slice is a disk. Hence, its volume isVi=π[r(y∗i)]2Δy,where the thickness, Δy, is a small "wiggle" in y. The axis of rotation is the y-axis and the radius of rotation is r(y∗i)=xR−xL=g(y∗i)−0=g(y∗i).4 Hence,Vi=π[g(y∗i)]2Δy=π[√4−y∗i]2Δy=π(4−y∗i)Δy.Therefore, we obtainV=∫dcπ[g(y)]2dy=π∫40(4−y)dy=π[4y−y22]|40=8π.The volume is 8π units3.
Checkpoint 1.3.4
Use the Disk Method to find the volume of the solid of revolution generated by rotating the region between the graph of g(y)=y and the y-axis over the interval [1,4] around the y-axis.
Answer
: 21π units3
We now showcase how to compute the volume of a solid of revolution if the axis of rotation is off an axis.5
Example 1.3.5
Find the volume of the solid of revolution generated by rotating the region bounded by f(x)=sin(x4), y=−1, x=0, and x=2π about the line y=−1.
Solution
: As always, we begin by sketching the curve, the axis of rotation, and a representative slice.
Figure 1.3.8: A sketch of f(x)=sin(x4), the axis of rotation (y=−1), and a representative slice.
The volume of the ith slice isVi=K(x∗i)Δx,and we can see that this slice is a disk. Hence,Vi=π[r(x∗i)]2Δx.The difference between this problem and the previous problems is that the radius of rotation involves a little more thought. The distance from the function to the axis of rotation is yT−yB=sin(x∗i4)−(−1)=sin(x∗i4)+1. Hence, the radius of rotation is r(x∗i)=sin(x∗i4)+1. Thus, the volume of the ith slice isVi=π[sin(x∗i4)+1]2Δx=π[sin2(x∗i4)+2sin(x∗i4)+1]Δx.We are now ready to find this volume.V=∫baπ[r(x)]2dx=∫2π0π[sin2(x4)+2sin(x4)+1]dx=π∫2π01−cos(x2)2+2sin(x4)+1dx(Power reduction formula)=π2∫2π01−cos(x2)dx+2π∫2π0sin(x4)dx+π∫2π0dx=π2∫2π0dx−π2∫2π0cos(x2)dx+2π∫2π0sin(x4)dx+π∫2π0dx=π2−π2∫2π0cos(x2)dx+2π∫2π0sin(x4)dx+2π2=3π2−π2∫2π0cos(x2)dx+2π∫2π0sin(x4)dx=3π2−π∫u=πu=0cos(u)du+2π∫2π0sin(x4)dx(Let u=x2⟹du=12dx)=3π2−π(sin(u))u=πu=0+2π∫2π0sin(x4)dx=3π2−π(sin(π)−sin(0))+2π∫2π0sin(x4)dx=3π2+2π∫2π0sin(x4)dx=3π2+8π∫u=π/2u=0sin(u)du(Let u=x4⟹du=14dx)=3π2−8π(cos(u))u=π/2u=0=3π2−8π(cos(π2)−cos(0))=3π2−8π(0−1)=3π2+8πThus, the volume is 3π2+8π units3.6
The Washer Method
Some solids of revolution have cavities in the middle; they are not solid near to the axis of revolution. Sometimes, this is just a result of how the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen is when an axis of revolution other than thex-axis or y-axis is selected.
When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers (disks with holes in the center). For example, consider the region bounded above by the graph of the function f(x)=√x and below by the graph of the function g(x)=1 over the interval [1,4]. When this region is revolved around the x-axis, the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representative washer are shown in Figure 1.3.9 (a) and (b). The region of revolution and the resulting solid are shown in Figure 1.3.9 (c) and (d).
Figure 1.3.9a: A thin rectangle in the region between two curves.
Figure 1.3.9b: A representative disk formed by revolving the rectangle about the x-axis.
Figure 1.3.9c: The region between the curves over the given interval.
Figure 1.3.9d: The resulting solid of revolution.
Interactive Element: The Washer Method
Figure 1.3.9e: A dynamic version of this solid of revolution generated using CalcPlot3D.
The volume of the ith slice is still represented byVi=K(x∗i)Δx;
however, we can see that this slice is a disk with a central disk removed. That is,Vi=π[rO(x∗i)]2Δx−π[rI(x∗i)]2Δx,where rO is the radius of rotation to the outer function and rI is the radius of rotation to the inner function. In both cases, the axis of rotation is the x-axis, so its easily found that rO(x∗i)=f(x∗i)=√x∗i and rI(x∗i)=g(x∗i)=1. Hence,Vi=π[√x∗i]2Δx−π2Δx=πx∗iΔx−πΔx.From this formula, we have the option to use two integrals or one to find the volume of the solid of revolution. That is, we could useV=∫41πxdx−∫41πdxorV=∫41π(x−1)dx.Both are acceptable, but while the second form looks simpler, it can lead to mistakes (see the Caution below). Computing the volume, we getV=π∫41xdx−π∫41dx=π(x22)41−π(4−1)=92π units3.Generalizing this process gives the Washer Method. Again, you should "strive to derive" this formula rather than memorize it (this formula can cause issues if you just try to memorize it).
Theorem: The Washer Method About the x-axis
Suppose f(x) and g(x) are continuous, nonnegative functions such that f(x)≥g(x) over [a,b]. Let R denote the region bounded above by the graph of f(x), below by the graph of g(x), on the left by the line x=a, and on the right by the line x=b. Then, the volume of the solid of revolution formed by revolving R around the x-axisis given byV=∫baπ[(f(x))2−(g(x))2]dx.
Caution
A prevalent mistake is to think that the volume in the previous theorem is∫baπ[f(x)−g(x)]2dx.
This is why I strongly encourage my students to not memorize that formula, but instead to derive it until it makes sense. I will demonstrate this process with each remaining problem.
Example 1.3.6
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of f(x)=x and below by the graph of g(x)=1/x over the interval [1,4] around the x-axis.
Solution
: We start by graphing the functions, the axis of rotation, the solid, and a representative slice (Figure 1.3.10c).
Figure 1.3.10a: The region between the graphs of the functions f(x)=x and g(x)=1/x over the interval [1,4].
Figure 1.3.10b: Revolving the region about the x-axis generates a solid of revolution with a cavity in the middle.
The slice is a disk with an inner disk removed. Therefore, the volume of this slice isVi=π[rO(x∗i)]2Δx−π[rI(x∗i)]2Δx,where the outer radius is the distance between the axis of rotation (the x-axis) and f(x)=x, and the inner radius is the distance between the axis of rotation and g(x)=1x. Thus,Vi=π[f(x∗i)]2Δx−π[g(x∗i)]2Δx=π[x∗i]2Δx−π[1x∗i]2Δx=π(x∗i)2Δx−π(x∗i)2Δx=π[(x∗i)2−1(x∗i)2]Δx.Finally, we getV=π∫41[x2−(1x)2]dx=π[x33+1x]|41=81π4 units3.
##### Interactive Element: Solid of Revolution
Figure 1.3.10c: A dynamic version of this solid of revolution generated using CalcPlot3D.
Checkpoint 1.3.6
Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of f(x)=√x and g(x)=1/x over the interval [1,3] around the x-axis.
Answer
: 10π3units3
As with the Disk Method, we can also apply the Washer Method to solids of revolution that result from revolving a region around the y-axis. In this case, the following rule applies (again, you need to understand how this is derived).
Theorem: The Washer Method About the y-axis
Suppose u(y) and v(y) are continuous, nonnegative functions such that v(y)≤u(y) for y∈[c,d]. Let Q denote the region bounded on the right by the graph of u(y), on the left by the graph of v(y), below by the line y=c, and above by the line y=d. Then, the volume of the solid of revolution formed by revolving Q around the y-axis is given byV=∫dcπ[(u(y))2−(v(y))2]dy.
Rather than looking at an example of the Washer Method with the y-axisas the axis of revolution, we now consider an example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method applies, but you may have to visualize just how to describe the cross-sectional area of the volume.
Example 1.3.7
Find the volume of a solid of revolution formed by revolving the region bounded above by f(x)=4−x and below by the x-axis over the interval [0,4] around the line y=−2.
Solution
: We graph the functions, the axis of rotation, the solid of revolution, and a representative slice (not shown).
Figure 1.3.11a: The region between the graph of the function f(x)=4−x and thex-axis over the interval [0,4].
Figure 1.3.11b: Revolving the region about the line y=−2 generates a solid of revolution with a cylindrical hole through its middle.
A representative slice is a disk with an inner disk removed. Therefore, the volume of this slice isVi=π[rO(x∗i)]2Δx−π[rI(x∗i)]2Δx,where the outer radius is the distance between the axis of rotation, y=−2, and f(x)=4−x, and the inner radius is the distance between the axis of rotation and the x-axis (y=0). Thus,Vi=π[f(x∗i)−(−2)]2Δx−π[0−(−2)]2Δx=π[4−x∗i+2]2Δx−π(2)2Δx=π(6−x∗i)2Δx−4πΔx=π(32−12x∗i+(x∗i)2)Δx.Therefore, we haveV=∫40π(x2−12x+32)dx=π[x33−6x2+32x]|40=160π3units3.
##### Interactive Element: Solution of Revolution
Figure 1.3.11c: A dynamic version of this solid of revolution generated using CalcPlot3D.
Checkpoint 1.3.7
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of f(x)=x+2 and below by the x-axis over the interval [0,3] around the line y=−1.
Answer
: 60π units3
Footnotes
1 For the remainder of this course, I will be using ri to denote the radius of rotation for the ith slice of a rotational solid, and r(x) for the "radius function." The distinction between the two should be evident within the context of the discussion.
2 The use of K(x∗i) instead of π[f(x∗i)]2 here is purposeful. As you will see in Section 1.4, we have choices as to the shapes of our slices. The formula for K(x∗i) will change accordingly.
3 If you haven't noticed, you will perform a lot of fractional arithmetic in Integral Calculus. This is unavoidable. When doing homework, try to use technology sparingly (also, trust that your professor will give you something manageable on an exam).
4 Since we are computing a horizontal distance, we use our discussion of measuring distances in Section 1.1 to compute the horizontal distance xR−xL.
5 The axes of rotation of rotation for this course will be restricted to horizontal or vertical lines.
6 This example showcases your need to be very familiar with Trigonometry in Integral Calculus. If you struggle with Trigonometry, please take the time to review the main topics by referencing Section 1.6 of Differential Calculus.
Key Concepts
For solids of revolution, the volume slices are often disks and the cross-sections are circles. The method of disks involves applying the method of slicing in the particular case in which the cross-sections are circles, and using the formula for the area of a circle.
If a solid of revolution has a cavity in the center, the volume slices are washers. With the method of washers, the area of the inner circle is subtracted from the area of the outer circle before integrating.
Key Equations
Disk Method along the x-axis
V=∫baπ[f(x)]2dx
Disk Method along the y-axis
V=∫dcπ[g(y)]2dy
Washer Method
V=∫baπ[(f(x))2−(g(x))2]dx
Glossary
disk method
: a special case of the slicing method used with solids of revolution when the slices are disks
solid of revolution
: a solid generated by revolving a region in a plane around a line in that plane
washer method
: a special case of the slicing method used with solids of revolution when the slices are washers
1.2.2: Homework
1.3.1: Resources and Key Concepts |
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铺砖之斐波那契数列
最新推荐文章于 2025-08-12 11:08:04 发布
freedom098于 2018-01-10 18:48:49 发布
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我们可以用21的小矩形横着或者竖着去覆盖更大的矩形。请问用n个21的小矩形无重叠地覆盖一个2n的大矩形，总共有多少种方法？
这个题的分析详见 剑指offer，这里记录一点实现细节。
测试数据里有0这个数，代表一块地砖也没有，方法数自然为0
事实上，真正的斐波那契序列是从1开始的，有1块地砖，只有一种铺法，2块地砖就有2种了，后面以此类推。
因此这就需要我们代码单独处理0，然后按照常规写斐波那契序列，考虑到效率使用迭代法，而不是递归法。
```
class Solution {
public:
int rectCover(int number) {
if (number== 0)
return 0;
if (number == 1)
return 1;
if (number == 2)
return 2;
int n0 = 1;
int n1 = 2;
int res = n0;
for (int i = 3; i <= number; i++)
{
res = n0 + n1;
n0 = n1;
n1 = res;
}
return res;
}
};
```
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15.算法习题之 斐波那契 数列
sam475259540的博客
04-23 484
如果某个递归，除了初始项之外，具有如下的形式F(N) = C1 F(N) + C2 F(N-1) + … + Ck F(N-k) ( C1…Ck 和k都是常数)并且这个递归的表达式是严格的、不随条件转移的那么都存在类似 斐波那契 数列 的优化，时间复杂度都能优化成O(logN)
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2018.7 问题_javaunresolved compilation problem: the local vari-CS...
9-13
} /Exception in thread "main" java.lang.Error: Unresolved compilation problems: The local variable n1 may not have been initialized/ AI写代码java 运行 2.有两种 砖,分别是11的 砖 和12的 砖,用这两种 砖 铺 1N的地面,问共有多少种 铺 法。输入为N,请输出相应的 铺 法数 为什么是 斐波那契 数列? 【解答...
3×n 砖块 覆盖问题
9-4
及两个直角边都为2的直角三角形 砖。现在问有多少种不同的方案,能够使用这 三种转 铺 满整个路面? 【输入格式】 一行一个正整数?。 【输出格式】 一行一个整数代表答案对109 + 7取模之后的结果。 【样例输入】 1 【样例输出】 1 【数据规模与约定】 对于100%的数据,令 斐波那契 数列 数列 第一项为1第二项为2第?
斐波那契 螺旋线系列一：斐波那契 数列 介绍
低速无人驾驶自动技术
11-20 2535
质数与合数：斐波那契 数列 的质数元素也是该 数列 的质数项，唯一的例外是第4项元素3。人们早就发现，在自然界的晶体中，原子以重复的样式排列，不同的化合物也许会出现不同的排列方式，但都是简单的平移重复而已。但无论如何，五重对称（图d）却不可能得到，因为其中原子间的距离长短不一，这个样式无法实现旋转对称，由此很容易就充分证明了在晶体中找不到五重对称，依此，七重对称或者更高重的对称都是找不到的…答案是，每月兔子的总数可以用以下 数列 表示：1，1，2，3，5，8，13，21，34，55，89，144，233。
EOJ 3297. 铺 瓷 砖【斐波拉契 数列 的变式+记忆化搜索/DP】⭐⭐⭐⭐⭐
yc_cy1999的博客
05-20 460
文章目录题目描述知识点实现码前思考代码实现码后反思 题目描述 铺 瓷 砖 单点时限: 2.0 sec 内存限制: 256 MB 章鱼王的行宫要 铺 瓷 砖 了。 行宫中有一长度为 的地板，给定三种不同瓷 砖：一种长度为1，一种长度为2，另一种长度为3，数目不限。 要将这个长度为 的地板 铺 满，并且要求任意两个相邻的瓷 砖 长度均不等，一共有多少种不同的 铺 法？在所有的 铺 设方法中，一共用了长度为1的瓷 砖 多少块？ 例如，长度为4的地面一共有如下3种 铺 法，并且，一共用了长度为1的瓷 砖 4块： 4=1+2+1 .
「GXOI / GZOI 2019」逼死强迫症_小 a 要 铺 一条 的路,为此他购买了 块...
9-17
ITX351 由此产生了一个邪恶的想法:他想要在这条路上故意把两块1 × 1 1 \times 11×1的 砖块 分开 铺,不让两块 砖 有相邻的边,其他 砖块 可以随意 铺,直到整条路 铺 满。这样一定可以逼死自身强迫症 sea5! 也许下面的剧情你已经猜到了——他为此兴奋不已,以至于无法敲键盘。于是,他请你帮忙计算一下,有多少种方案可...
寒假集训总结1-递推、递归_学习递推策略心得
9-17
1.斐波那契 数列(Fibonacci) 这是一种很简单的递推模型。 这种模型一般都是,此时第x项数据与前面的数据有直接的数值关联(一般来说是很明显的倍数关系)。【能理解那个意思就行】 例题-铺 砖 1 原题链接(BZOJ) 题目描述 有2 ∗ n 2 n2∗n的一个长方形方格道路,只有一种1$ $2的 砖 去 铺,总共有多少种 铺...
斐波那契 数列 相关的问题
weixin_44001860的博客
08-18 250
1、基本的 斐波那契 数列 又称黄金分割 数列，指的是这样一个 数列：0、1、1、2、3、5、8、13、21、34、…… 递推的方法定义：F(1)=1，F(2)=1, F(n)=F(n - 1)+F(n - 2)（n ≥ 3，n ∈ N） #include using namespace std; int fibo(int n){ int reault = {0,1}; if(n < 2) return result[n]; int a = 1; int b =
C语言（03）——斐波那契 数列 的理解和运用（超详细版） 最新发布
ankleless的博客
08-12 1649
本文探讨了 斐波那契 数列 与杨辉三角在数学和编程中的联系。首先介绍了 斐波那契 数列 的背景故事及数学定义，然后分别展示了一维和二维 斐波那契 数列 的迭代和递归实现方法，并比较了两种方式的优劣。特别指出当递归次数过多时可能导致的栈溢出问题，建议使用动态内存分配优化。文章还揭示了 斐波那契 数列 与杨辉三角的数学关系，以及如何通过矩阵快速幂算法高效计算超大n值时的 斐波那契 数。最后强调数学规律与编程实现相结合的重要性，建议根据具体场景选择合适算法，并注意内存管理。
Fibonacci 数列 整除性质的组合证明_组合证明的艺术
9-27
可分隔(breakable): 一个n nn-平 铺 在k kk单位块可分隔, 即该平 铺 在第k kk个单位块处为方 砖块(长度为1), 若该处是长度为2的 砖块, 则不可分隔. 定理 对m ⩾ 1 , n ⩾ 0 m\geqslant1,n\geqslant0m⩾1,n⩾0, 如果m ∣ n m\,|\,nm∣n, 那么f m − 1 ∣ ...
这可能是最通俗最详细的生成函数讲解了吧!!!_物品的生成函数是什么-CSD...
9-13
那么 斐波那契 的生成函数记作G GG G = 1 + x + 2 x 2 + 3 x 3 + 5 x 4 + 8 x 5 . . . . G=1+x+2x^2+3x^3+5x^4+8x^5...G=1+x+2x2+3x3+5x4+8x5... x G = x + x 2 + 2 x 2 + 3 x 4 + 5 x 5 . . . . xG=\ \ \ \ \ x+x^2+2x^2+3x^4+5x^5....
斐波那契 数列 相关简化4
个人专长：数据结构和算法、大数据、数据库、主流框架
05-23 516
斐波那契 数列 相关简化终结篇，题目都比较简单：用12的瓷 砖，把N2的区域填满 返回 铺 瓷 砖 的方法数
编程之美：第四章 数字之趣 4.2瓷 砖 覆盖地板
qingyuanluofeng的专栏
08-04 1668
/ 瓷 砖 覆盖地板: 原来的地板 铺 有NM块正方形瓷 砖，商店只提供长方形瓷 砖，现在一块长方形瓷 砖 相当于于原来的两块正方形瓷 砖，能否用12的瓷 砖 去覆盖NM的地板呢 本质: 斐波那契 递推 数列 公式 f(i) = f(i-1) + f(i-2) 分析: NM的地板有以下几种可能: 1如果N=1，M为偶数的话，显然12的瓷 砖 可以覆盖1M的地板，需要M/2块瓷 砖 2如果NM为奇数，也就是N和M
UVa900 - Brick Wall Patterns_uva 900
8-10
本文介绍了如何利用Java语言通过数组和循环结构来高效计算 斐波那契 数列 的第n项,展示了算法的实现过程和优化策略。 摘要生成于C知道,由 DeepSeek-R1 满血版支持,前往体验 > importjava.util.Scanner; classMain { publicstaticvoidmain(String[] args){
铺 砖 C++题解
xxy20110830的博客
01-26 611
内存限制： 128 MiB 时间限制： 1000 ms 标准输入输出 题目类型： 传统 评测方式： 文本比较。
求 斐波那契 数列 的第N个数的值
行兮夷犹的专栏
06-23 7198
难易程度 简单题目描述：Find the Nth number in Fibonacci sequence.(译：如题)A Fibonacci sequence is defined as follow: 斐波那契 数列 的定义如下：The first two numbers are 0 and 1. 数列 的前两个数从0，1开始 The i th number is the sum of i-1 th
斐波那契 数列 与相关数学概念探索
斐波那契 数列 是数学中最迷人的整数序列之一，其起始为 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, … 。从第 3 项开始，每一项都是前两项之和。对于 n≥0 n≥0，我们用 F n F n ...
【算法&数据结构体系篇class26】：斐波那契 数列 矩阵快速幂技巧 时间复杂度O(logN)
studyday1的博客
04-13 761
|fn,fn-1,fn-2..fn-i+1| (共i项) = |fi,fi-1,fi-2...1| (共i项) 矩阵ii 的 n-i次方。4）斐波那契 数列 1 1 2 3 5 8 13.....其为fn = fn-1 + fn-2 最底层n-2 是二阶矩阵 n-2次方。4.生牛问题 1,2,3,4,6,9,13,19...2.迈台阶问题 1,2,3,5,8,13,21....1.斐波那契 数列 1,1,2,3,5,8,13....求 斐波那契 数列 矩阵乘法的方法。
一道面试过程中的编程题：两行五列的地板，一行两列的 砖，求 砖 铺 满有几种方法，砖 可以横，可以竖
Cgxwl的博客
06-26 250
铺 地 砖 题，比较笨的办法，这个地板是2n，n大于等于1，n为1时，只有一种方法。n为2时是两种方法，3时三种方法，四时五种方法。得出 数列 1，2，3，5，8，…是 斐波那契 数列。解题思路可以使用for循环n为5时代入 public static void main(String[] args) { System.out.println(fun(5)); } publ...
怎么得到 斐波那契 数列？（mn大小的矩形，从左上角走到左下角有多少种走法？包括一些引申题以及总结）
qomoman的专栏
08-02 2924
这是一道微软的面试题。看到这道timu
状态压缩与动态规划(DP)---编程之美---瓷 砖 覆盖地板---POJ2411
cjc211322的专栏
04-01 5897
一、状态压缩 从状态压缩的特点来看，这个算法适用的题目符合以下的条件： 1.解法需要保存一定的状态数据（表示一种状态的一个数据值） ，每个状态数据通常情况下是可以通过 2 进制来表示的。这就要求状态数据的每个单元只有两种状态，比如说棋盘上的格子，放棋子或者不放，或者是硬币的正反两面。这样用 0 或者 1 来表示状态数据的每个单元，而整个状态数据就是一个一串 0 和 1 组成的二进制数。
动态规划：铺 砖 问题
帅比王的博客
07-28 5202
问题描述：给定nm的格子，每个格子被染成了黑色或者白色。现在要用12的 砖块 覆盖这些格子，要求块与块之间互相不重叠，且覆盖了所有白色的格子，但不覆盖任意一个黑色格子。求一共有多少种覆盖方法，输出方案数对M取余后的结果。 限制条件： 1 <= n <= 15 1 <= m <= 15 2 <= M <= 10^9 样例输入： n = 3 m = 3 ...
课堂在线Java程序设计 铺 砖 问题
薄生的博客
06-22 2141
有两种 砖，分别是11的 砖 和12的 砖，用这两种 砖 铺 1N的地面，问共有多少种 铺 法。输入为N，请输出相应的 铺 法数 输入： 3 输出： 3 这道题我的理解可能不正确 -_-! package javatest012; import java.util.Scanner; / 有两种 砖，分别是11的 砖 和12的 砖，用这两种 砖 铺 1N的地面，问共有多少种 铺 法。输入为N，请输出相应
从 铺 砖 问题到排列组合算法的实现
黑羊的创意工坊
08-30 2982
最近在某公众号上看到了一道算法题，在思考它的过程中发现了一些很有意思的新问题，解决后决定写下这篇博客记录一下。 首先来看看问题：现在有两种 砖，分别是是11,的 砖 和12的 砖，用着两种 砖 铺 1n的地面。问共有多少种 铺 法？（输入为n，请输出相应的 铺 法数） 铺 砖 问题 Ok，那么让我们来先来分析一下这个问题吧。 一、用数学思维分析问题 根据题目我们可以知道 砖 一共两种，分别是11和12两种...
ACM实习---------关于 铺 地板问题
attack_5的博客
06-05 1346
1.8【问题描述】有一长度为N(1<=Ｎ<=10)的地板，给定两种不同瓷 砖：一种长度为1，另一种长度为2，数目不限。要将这个长度为N的地板 铺 满，一共有多少种不同的 铺 法？例如，长度为4的地面一共有如下5种 铺 法：4=1+1+1+14=2+1+14=1+2+14=1+1+24=2+2编程用递归的方法求解上述问题。【输入】 只有一个数N，代表地板的长度【输出】 输出一个数，代表所有不同的瓷...
Leetcode 1240：铺 瓷 砖（超详细的解法！！！）
coordinate的博客
10-28 7162
你是一位施工队的工长，根据设计师的要求准备为一套设计风格独特的房子进行室内装修。 房子的客厅大小为 n x m，为保持极简的风格，需要使用尽可能少的 正方形 瓷 砖 来 铺 盖地面。 假设正方形瓷 砖 的规格不限，边长都是整数。 请你帮设计师计算一下，最少需要用到多少块方形瓷 砖？ 示例 1： 输入：n = 2, m = 3 输出：3 解释：3 块地 砖 就可以 铺 满卧室。 2 块 1x1 地...
解决一下 铺 地板问题
03-09
通常，这类问题属于组合数学或动态规划的范畴，尤其是经典的 铺 砖 问题，比如用2x1或1x1的 砖块 铺 满2xN或3xN的区域，计算有多少种 铺 法。 首先，我需要确认用户的问题是否指的是这类经典问题。比如，常见的例子是用2x1...
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7280 | https://www.scribd.com/document/563457162/Carl-P-Simon-Lawrence-E-Blume-Mathematics-for-Economists-W-W-Norton-Company-1994-1 | Carl P. Simon, Lawrence E. Blume - Mathematics For Economists-W. W. Norton & Company (1994) | PDF
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Carl P. Simon, Lawrence E. Blume - Mathematics For Economists-W. W. Norton & Company (1994)
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Thermal Diffusivity
In subject area:Chemistry
Thermal diffusivity is defined as a measure of how quickly heat spreads through a material, which can be determined experimentally using methods such as the laser flash technique or the 3ω method to analyze temperature changes in response to heat application.
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Thermal diffusivity measurement for SiC/C compositionally graded graphite materials
1997, Functionally Graded Materials 1996J. Nakano, ... R. Yamada
4CONCLUSIONS
Thermal diffusivity for the SiC/C materials with different thicknesses of the graded layer and SiC content has been measured by the laser flash method to study the influence of the graded layer on diffusivity. In addition, the specific surface areas for the SiC/C materials have been measured. The results obtained in this study are summarized as follows.
(1)
: The effect of the SiC/C graded layer on thermal diffusivity was slight within the range of SiC contents(0.27-8.52 mass%) used.
(2)
: The temperature dependence of thermal diffusivity and their values themselves for the SiC/C graded materials were similar to those for graphite.
(3)
: There was no effect of sample thickness on the thermal diffusivity when the thickness ranged from 1.5 to 3.0 mm.
(4)
: The calculated thermal diffusivity based on the two-phase model agreed with the measured one.
(5)
: The influence of the porosity in surface region on the thermal diffusivity was small when there was no change of bulk density.
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Book1997, Functionally Graded Materials 1996J. Nakano, ... R. Yamada
Chapter
Metal Matrix Composite Products by Vibration Casting Method
2016, Reference Module in Materials Science and Materials EngineeringM. Sayuti, ... M.K.A. Arifin
7.3.7Thermal properties
Thermal diffusivity of composite materials is measured using photo flash method. The photoflash detection system consists of a light source, sample holder, thermocouple, low noise pre amplifier, oscilloscope, photodiode and a personal computer. The temperature rise at the back surface of the sample is detected by the thermocouple. The detected signal is amplified by a low-noise preamplifier and the processed by a digital oscilloscope (Carter and Norton, 2007; Yu et al., 2002).
Photoflash detection system is not an expensive method and the standard thermal diffusivity value for aluminum is equal to 0.83 cm2 s-1 for thickness greater than 0.366 cm (Muta et al., 2003). In the photo flash system, the excitation source consists of a high intensity camera flash. This method is well suitable for aluminum, aluminum alloys and aluminum–silicon particulate metal matrix composites (Collieu and Powney, 1973). The thermal diffusivity values can be obtained for different thicknesses of the test samples. The thermal diffusivity, a determines the speed of propagation of heat waves by conduction during changes of temperature with time. It can be related to a, the thermal conductivity through the following equation (Michot et al., 2008; Taylor, 1980).
where density ? and specific heat Cp.
The photo flash technique is originally described by Parker and it is one of the most common ways to measure the thermal diffusivity of the solid samples. The computer is programmed to calculate the thermal diffusivity, a, using the equation:
where L is the thickness in mm and t0.5 is the half rise time in seconds.
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Reference work2016, Reference Module in Materials Science and Materials EngineeringM. Sayuti, ... M.K.A. Arifin
Chapter
Metals
2016, Handbook of Friction Materials and their ApplicationsRoberto C. Dante
9.4Thermal diffusivity and other thermal properties
Thermal diffusivity is a parameter to be taken into account; however, a high value does not necessarily mean that heat is better dissipated, since it is the ratio between the thermal conductivity and the volumetric heat capacity at constant pressure. It provides information about the competition between heat conduction and storing, and both can be useful to keep temperature as low as possible:
(9.1)
where in Equation (9.1), a is the thermal diffusivity, λ is the thermal conductivity, ρ is density, and Cp is the heat capacity. Temperature governs the speed of chemical reactions and flash temperature contributes to the rate of tribochemical reactions too. Thermal diffusivity of a material is related to the speed to reach thermal equilibrium under variable thermal conditions. For this reason, low values are better, to reduce the possibility that a high temperature can be reached in depth and damage the binder deeply. The thermal diffusivity says nothing about the energy flows. On the other hand, the thermal effusivity characterizes the ability to exchange thermal energy with its surroundings. Thermal effusivity is given by the following equation:
(9.2)
In summary, thermal effusivity and diffusivity are representative of two phenomena in competition: the former is related to the ability of the material to absorb heat, while the latter to the speed to reach thermal equilibrium, i.e., to adapt itself to the surroundings. Thermal effusivity explains, for example, why at touch metals seem colder than other bodies at the same temperature, since the heat transfer from the skin to the metal is very fast. In other words, diffusivity is related to heat penetration and effusivity to surface heat exchange. During friction a lot of heat is created at the interface, and this heat must be dissipated; thus the most effective friction material is that which can absorb (dissipate) more heat without raising temperature. Therefore, thermal effusivity (a surface property) seems to be more important than thermal diffusivity (a bulk property). In Table 9.1 a list of thermal diffusivity and thermal effusivity values of some materials is reported.
Table 9.1. Thermal diffusivity (a) and thermal effusivity (e) of some engineering materials at room temperature
| Material | a (mm2 s− 1) | e (W cm− 2 K− 1 s− 0.5) |
---
| Fe | 20.4 | 1.59 |
| Cu | 115 | 3.69 |
| Al | 91 | 2.37 |
| Al2O3 | 10.5 | 0.56 |
| SiO2 | 0.85 | 0.15 |
| Polycrystalline graphite | 100–60 | 1–0.7 |
Table 9.1 shows that the handicap of graphite is plane orientation, which reduces the effect of the high in-plane conductivity, leading to a very low thermal effusivity and a medium thermal conductivity. To a certain extent Fe can have a greater effect. However, this effect could be considerably reduced with graphite with low Lc approaching the graphene layer and extended plane development. This strategy may offer an alternative to the massive use of metals. The reduced Lc is really always achieved in the friction interface, since graphite layers slip and are separated. Therefore, we can say that La (i.e., the in-plane extension of crystallites) or, even better, the ratio La/Lc is really the fundamental parameter controlling high thermal effusivity of graphite.
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Book2016, Handbook of Friction Materials and their ApplicationsRoberto C. Dante
Chapter
Mechanical and Thermal Properties of PSZ/Ni-Base Superalloy Composite
1997, Functionally Graded Materials 1996Shinya Akama
3.2Thermal Conductivity Coefficient
The thermal diffusivities of each composite were measured at 673 K, 1073 K, 1273 K, 1373 K and 1473 K by laser-flash method. The thermal conductivity coefficient, the product of thermal diffusivity, specific heat and density, are plotted as a function of the concentration of PSZ and test temperature in Fig. 2 and Fig.3. In Fig. 2, the thermal conductivity coefficient at room temperature is plotted simultaneously 2). From Fig. 2 and 3, it was recognized that the thermal conductivity coefficient became stronger temperature dependence with decrease the concentration of PSZ.
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Book1997, Functionally Graded Materials 1996Shinya Akama
Chapter
Transport phenomena in terms of mass structure
2020, Transport and Surface PhenomenaKamil Wichterle, Marek Večeř
2.6.3Thermal diffusivity and the Prandtl number in liquids and gases
The thermal diffusivity of gases was dealt with earlier (Section 2.2). The range of thermal diffusivities of nonmetallic liquids and solids is relatively narrow, namely between 0.15 × 10− 6 and 2 × 10− 6 m2/s, while the thermal diffusivity of gases is about 20 × 10− 6 m2/s due to their low density, the value for metals is 10 × 10− 6–100 × 10− 6 m2/s. Examples are shown in Table 2.3.
It is interesting to note that the thermal diffusivity of composite heat insulators containing gases are close to the conductivities of the continuous phase because the presence of gas reduces not only the thermal conductivity but also the density while the heat capacity related to the unit of mass will vary only insignificantly.
In some engineering tasks dealing with liquids, the ratio of kinematic viscosity to thermal diffusivity, the so-called Prandtl number (Eq. 2.18) will play an important role.
While in gases the Prandtl number is somewhat smaller than 1, it is around 5 or more in common nonmetallic materials. In highly viscous materials, the Prandtl number approaches infinity while it is meaningless in solids. Substantially, smaller than 1 is the Prandtl number in molten metals because of their low viscosity and high thermal conductivity. For example, in lead and mercury, it is about 0.02 and in molten steel, it is 0.06.
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Book2020, Transport and Surface PhenomenaKamil Wichterle, Marek Večeř
Chapter
Pulsed laser surface treatment of multilayer gold–nickel–copper (Au/Ni/Cu) coatings to improve the corrosion resistance of components in electronics
2012, Laser Surface Modification of Alloys for Corrosion and Erosion ResistanceN. Semmar, C. Boulmer-Leborgne
4.4.2Effect of the multilayer thermal properties
The thermal diffusivity of gold at the bulk state is close to 1.0 cm2/s. As shown in Fig. 4.11, no melting occurs for this bulk value. Curves C1, C2 and C3 correspond to 0.4, 0.2 and 0.1 cm2/s. As shown previously, experiments with this laser fluence (750 mJ/cm2) are enough to initiate the surface melting. As expected, the strong change of the gold microstructure induces a change in the diffusivity value. As an example, to melt 0.25 μm of the layer (KrF laser beam) the surface must absorb 750 mJ/cm2 when thermal diffusivity is close to 0.1 cm2/s. The lifetime of this melting pool is then roughly 20 ns. The diffusivity value is lower; more important are the melting depth and the lifetime. In contrast, the curves C1, C2 and C3 show that the lower the thermal diffusivity, the shorter is the time to start the melting.
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Book2012, Laser Surface Modification of Alloys for Corrosion and Erosion ResistanceN. Semmar, C. Boulmer-Leborgne
Chapter
Plasma-sprayed thermal barrier coatings with segmentation cracks
2011, Thermal Barrier CoatingsH. Guo, ... L. Zhou
8.4Thermophysical and mechanical properties of segmented TBCs
The thermal diffusivities of the YSZ coatings were measured by laser flash apparatus. The thermal diffusivity values for the non-segmented coating range from 0.002–0.006 cm2 s–1 between room temperature and 1200 °C, while the segmented coating exhibits a perceptible increase in thermal dif- fusivity compared to the non-segmented coating. The difference indicates that the values of the thermal diffusivity largely depend on the microstructure of the coatings.
Thermal conductivity is calculated using the equation: K(T) = α(T) · CP · ρ where K(T) is the thermal conductivity, α(T) is the thermal diffusivity, CP is the specific heat and p is the bulk density of the coating. The thermal conductivity of the segmented coating, with an increase of nearly 20% compared to that of the non-segmented coating,14,23 is approximately 1.75 W/ mK.
Plasma-sprayed TBCs reveal a two-dimensional microcrack network: one is defined as a horizontal microcrack, resulting from the flawed contacts between the inter-splats of the coatings; the other is oriented vertically, primarily resulting from the microcracking of the splats during cooling. The former is particularly effective in reducing the thermal conductivity of the sprayed coatings, because the interfaces formed by such cracks are perpendicular to the main heat flux. Among the coatings mentioned above, the segmented coating has a lower porosity than the non-segmented coating, and thus has a better connection between the splats, which contributes to reducing the amount of horizontal microcracks. This explains why the segmented coating tends to have a slightly higher thermal conductivity compared to the non-segmented coating. Moreover, dispersed spheroid microporosity is also helpful in lowering the thermal conductivity of the YSZ coatings. It is worth noting that branching cracks are also introduced when the spraying is performed with thick lamellae, accompanying the generation of segmentation cracks. The branching cracks play a role in lowering the thermal conductivity of the segmented TBC.20
Both the non-segmented and segmented coatings reveal very similar thermal expansion when heating to 1200 °C from room temperature, despite the different processing parameters and microstructure between the coatings. The average values of the thermal expansion coefficient (TEC) are in the range 10~11.2 × 10–6 K–1.
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Book2011, Thermal Barrier CoatingsH. Guo, ... L. Zhou
Chapter
Fluidized Bed Coating and Granulation
1999, Fluidization, Solids Handling, and ProcessingRichard Turton, ... Bryan J. Ennis
Greek Letters
| | | |
---
| α | Thermal diffusivity | m2/s |
| β | Constant in Eq. (34)
| β() | Coalescence kernel, Eq. (40)
| Γ | Shear rate | s−1 |
| γ | Surface tension or interfacial energy | N/m |
| δ | Dimensionless bubble spacing
| δc | Effective increase in crack length | m |
| ɛ | Dimensionless separation of two particles = 2h/a or powder void fraction, Eq. (26) |
| θ | Contact angle between liquid and powder compact |
| λ | Latent heat of vaporization | J/kg |
| μ | Viscosity | kg/ms |
| ρ | Density | kg/m3 |
| σ y | True yield stress of material | Pa |
| σ f | Applied fracture stress | Pa |
| τ | Coating run time | s |
| τ0 | Yield strength of wet granule | N/m2 |
| ϕ | Filling angle defined in Fig. 22 |
| φ | Probability distribution function |
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Book1999, Fluidization, Solids Handling, and ProcessingRichard Turton, ... Bryan J. Ennis
Chapter
Elastomer–carbon nanotube composites
2011, Polymer–Carbon Nanotube CompositesJ. Fritzsche, ... G. Heinrich
7.3.4Thermal transport processes
The thermal diffusivity of the CNT composites was evaluated by measuring the temperature development in SBR/BR samples (S-SBR 2525–0/CB24 70:30) filled with 3 phr CNT (Nanocyl 7000). Therefore two plates of one sample were placed in a heating press with a temperature of Tu = 100 °C. A temperature sensor was placed in between the two sample plates and the temperature increase T(t) is measured in dependence on time t compared to the start temperature T0. To calculate the thermal diffusivity a, the resulting values can be described by Equation 7.5 with δ = 10 cm the thickness of the sample plates:
[7.5]
To evaluate, the measured values were plotted as ln{(T(t)-Tu)/(T0-Tu)} against time t. The slope in that plot is then directly connected to the thermal diffusivity a by using only that part of the temperature curve showing a linear dependence.
The regression statistical error of the calculated values was about 1%. For the unfilled rubber, one obtains a thermal diffusivity of a = 0.83 × 10–7 m2/s, which increased to 1 × 10–7 m2/s with 3 phr CNT mixed 1:10 with ethanol. A similar value of a = 1.01 × 10–7 m2/s is obtained with 3 phr CNT mixed 1:10 with 2-propanol. This results in a thermal conductivity of λ = 0.19 W(m K)–1 for the unfilled polymer and λ = 0.21 W(m K)–1 for the composite with 3 phr CNT, respectively. This means that λ increases by only 20%, while the electrical conductivity increased by about 10 orders of magnitude.
Since the charge transport through the tube network is controlled by hopping or tunneling over gaps, it is clear that the DC conductivity of the composite would be dominated by the gap, rather than the much higher conductivity of the CNT. This also affects the thermal conductivity induced by electron transport. In addition, the transport of heat in isolated CNTs is dominated by phonons showing a characteristic quadratic dependence on temperature in the range between 50 and 300 K. The thermal conductivity exhibits a maximum at about 320 K of more than 3000 W(m K)–1, and for higher temperatures the thermal conductivity decreases due to phonon back-scattering effects. For CNT dispersed in a polymer matrix, further scattering effects, e.g. interfacial boundary and defect scattering, will appear, leading to a drastic reduction of thermal transport properties (Gojny et al., 2006). In addition, the thermal transport through the CNT network by phonons will be strongly hindered by the gaps between adjacent tubes. From these arguments we conclude that the thermal conductivity of the CNT/rubber composites should lie several orders of magnitude below that of the isolated CNT (up to 3000 W(m K)–1 (Berber et al., 2000; Kim et al., 2001). Accordingly, the thermal conductivity of the CNT network in rubber composites should be not much larger or even smaller than that of the pure polymer (about 0.1 W(m K)–1).
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Book2011, Polymer–Carbon Nanotube CompositesJ. Fritzsche, ... G. Heinrich
Chapter
Unsteady Convective Heat Transfer in Channels
1994, Advances in Heat TransferE.K. Kalinin, G.A. Dreitser
Nomenclature
a
: thermal diffusivity
bq
: constant
c, cp
: heat capacity
C = [(cpρF)b/(cρF)w]
: flow: wall heat capacity ratio
d
: diameter
D
: oval diameter of a helical tube
deq
: equivalent channel diameter
F
: cross-sectional area
G
: mass flowrate
g
: gravity acceleration
i
: enthalpy
L, l
: length, height of a gas cushion of a tank
li
: length of a pipeline section
p
: pressure
q
: heat flux density
qv
: volume density of heat release
r
: radius
r0
: tube radius
R = r/r0
T
: temperature
Tb
: mean-mass flow temperature; gas temperature in a gas cushion of a tank
Tb0
: channel inlet liquid temperature
: mean-mass temperature in a cushion
T∞
: ambient temperature
Tf
: fluid surface temperature
Tl
: liquid temperature
Tw
: temperature of an inner tube surface or tank surface
ΔTw = Tw−Tw0
: heat exchange temperature variation
ΔTb = Tb−Tb0
: flow temperature variation
ΔTb1 = Tb(0)−Tb1
: flow temperature variation at pipeline inlet
U
: channel perimeter
w
: velocity
: pulsations of the axial and radial velocity components
wmax
: axial velocity
x, y, z
: coordinates
X = x/d
α
: heat transfer coefficient
αout
: heat transfer coefficient for an outer tube surface
β
: volume expansion coefficient
δ
: thickness
η
: dimensionless distance from wall
: thermal activity coefficient
εq
: turbulent thermal diffusivity
εσ
: kinematic turbulent viscosity coefficient
: relative superheating of a jet to be blown
λ
: thermal conductivity
λT
: turbulent thermal conductivity
μ
: dynamic viscosity coefficient
v
: kinematic viscosity coefficient
ξ
: hydraulic resistance coefficient
ρ
: density
τ
: time
τ0
: initial time
ϕ
: dissipative function
Bi
: Biot number
Fo
: Fourier number
: integral dimensionless time parameter
K
: ratio of the unsteady heat transfer coefficient to its quasisteady value
Kξ
: ratio of the unsteady hydraulic resistance coefficient to its quasisteady value
KTKq, , , KT1
: thermal unsteadiness parameters determined by Eqs. (45), (75), (84), (65) and (202), respectively
KG, KGT
: hydrodynamic unsteadiness
KGg
: parameters determined by Eqs. (46), (43), and (147a), respectively
ΔK1
: K variation due to unsteady heat conduction
ΔK2
: K variation due to a difference of turbulent flow structure under unsteady conditions from that under steady conditions
ΔK3
: K variation due to a variable flowrate
Nu
: Nusselt number
Nu0
: quasisteady value of the Nusselt number
Nux
: vessel wall Nusselt number determined in terms of a size x
Nuf
: fluid Nusselt number determined in terms of a vessel diameter D
Pe
: Peclet number
Pr
: Prandtl number
Ra
: Rayleigh number
RaD
: Rayleigh number determined in terms of a vessel diameter D and a temperature drop (Tb − Tw)
Raf
: fluid surface Rayleigh number determined in terms of a vessel diameter D and a temperature drop ()
Re
: Reynolds number
Red
: throttle outlet Reynolds number determined in terms of its diameter d
St
: Stanton number
Tw/Tb
: temperature factor
Subscripts
eq
: equivalent
b
: quantity to be determined in terms of a flow temperature
m
: quantity to be determined in terms of a mean boundary-layer temperature equal to half the sum of Tw and Tb
w
: quantity to be determined in terms of a heat exchange surface temperature
1,2
: initial and final steady state
0
: quasisteady value
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Book series1994, Advances in Heat TransferE.K. Kalinin, G.A. Dreitser
Related terms:
Hydrogen
Laser Pulse
Nanoparticle
Heat Capacity
Porosity
Thermal Conductivity
Phase Change Material
Kinematic Viscosity
Graphene
Diffusivity
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$\begingroup$
I am studying congruence classes and having a hard time. I have two questions about the example below.
1) In the first step why do they replace $(2 3 4)$ with $2$ ?
2) In each step, how and why are they reducing for example $60$ $mod$ $11$ to $5$ $mod$ $11$ and $35$ $mod$ $11$ to $2$ $mod$ $11$?
elementary-number-theory
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asked Feb 27, 2017 at 19:29
AllanAllan
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1
$\begingroup$ Note that $2\times 3\times 4 =24 = 2\cdot 11 + 2$. Similarly $60=5\cdot 11 + 5$ and $35=3\cdot 11 + 2$. Modular arithmetic, we say that two numbers $x$ and $y$ are equivalent modulo $n$, written as $x\equiv y\pmod{n}$ if and only if $x-y=nk$ for some integer $k$. $\endgroup$
JMoravitz
– JMoravitz
2017-02-27 19:41:52 +00:00
Commented Feb 27, 2017 at 19:41
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2 Answers 2
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That is because congruences are compatible with addition and mulriplication ,exactly like equality:
If $a\equiv b $ and $a'\equiv b'\mod m$, then
$a+a'\equiv b+b'\mod m$,
$aa'\equiv bb'\mod m$,
$ka\equiv kb\mod m$,
$a^r\equiv b^r\mod m$.
Thus for instance, since $4!=24\equiv 2\mod 11$, we have $$6!=4!\cdot5\cdot6\equiv(2\cdot 5)\cdot6=10\cdot 6=60\equiv 5\mod11.$$
The last step is because a number is equivalent, by definition, to its remainder ($5$) upon Euclidean division by $11$.
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answered Feb 27, 2017 at 19:50
BernardBernard
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Maybe you didn't learn the definitions yet ?
$a\equiv b\pmod{n}$ means that $a-b$ is a multiple of $n$ (in other words : there exists an integer $k$ such that $b=ka$).
Now $2\times3\times4=24$ and $24-2=22=2\times11$, hence $2\times3\times4\equiv 2\pmod{11}$
Similarly, we see that $60-5=5\times11$ and $35-2=3\times11$, hence $60\equiv5\pmod{11}$ and $60\equiv5\pmod{11}$
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answered Feb 27, 2017 at 19:42
AdrenAdren
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7283 | https://www.assumption-chgo.org/lastsupper.asp | | | |
--- |
| The Last Supper | [BACK] |
This image is of a mosaic at Assumption Church. It is a representation of the famous painting by Leonardo Da Vinci depicting Jesus and his apostles at the last supper. The names listed above the image identify each of these people.
Hover your mouse over a name to learn more about the person.
| | | | | | | |
--- --- ---
| Bartholomew | Andrew | Peter | JESUS | Thomas | Philip | Jude |
| | | | | | | | |
--- --- --- --- |
| | James (Leser) | Judas | John | | James (greater) | Matthew | Simon |
|
| |
Close
Bartholomew was the son of Talmai. He lived in Cana of Galilee. He was known as a missionary in Armenia. Scripture shares that Jesus called him a man with no deceit. Bartholomew was introduced to Jesus Christ through Saint Philip and is also known as "Nathaniel of Cana in Galilee," notably in John's Gospel. Saint Bartholomew is credited with many miracles related to the weight of objects.
Close
Andrew was the first apostle Jesus called and the first apostle to claim Jesus was the Messiah.
Despite his seemingly important role as an early follower of Christ, Andrew is only mentioned 12 times in the entire New Testament—and four of those times are simply lists of the 12 apostles. He comes onto the scene early in the gospels, but only plays a minor role. However, his prominence in the lists of disciples and the few glimpses we get of him seem to suggest he was one of the main apostles—even if he wasn’t one of the “pillars of the church”
Close
James the Lesser, also known as James the Just, was called "the Lesser" to distinguish him from James, son of Zebedee. James the Lesser was one of the pillars of the early Church and was present for significant events such as the feeding of the five thousand. He was also called the brother of Jesus.
Close
Judas Iscariot is notorious for betraying Jesus. Judas’s surname is more probably a corruption of the Latin sicarius (“murderer” or “assassin”) than an indication of family origin, suggesting that he would have belonged to the Sicarii, the most radical Jewish group, some of whom were terrorists. Other than his apostleship, his betrayal, and his death, little else is revealed about Judas in the Gospels. Always the last on the list of the Apostles, he was their treasurer. John 12:6 introduces Judas’s thievery by saying, “as he had the money box he used to take what was put into it.”
Close
Peter, also known as Peter the Apostle, Simon Peter, Simon, or Cephas, was one of the first leaders of the early Christian Church. He appears repeatedly and prominently in all four New Testament gospels as well as the Acts of the Apostles. Catholic tradition accredits Peter as the first bishop of Rome—or pope—and also as the first bishop of Antioch.
Close
John was the brother of James and the son of Zebedee, and came from a family of fisherman. He was one of Jesus' closest confidants, along with Peter and James, and was present at many important events in Jesus' ministry. He is traditionally believed to be the author of several New Testament works, including the Gospel of John, the three Letters of John, and the Revelation to John. He is also known as John the Divine and St. John
Close
Thomas, also known as Didymus, meaning "twin" in Greek and Hebrew. He was a missionary from Galilee, Roman Empire (modern-day Israel). Thomas is mentioned in the listings of the apostles in the Synoptic Gospels. In the Gospel of John, Thomas plays a particularly distinctive part, and is often condemned for his lack of belief, but was equally courageous and relentlessly sought the Truth.
Close
Philip, the Apostle, is one of four people named Philip in the Bible, and he’s often confused with Philip the Evangelist, who plays a minor role in Acts. (The other two Philips are both sons of King Herod the Great.)
Church tradition identifies Philip as the missionary to Greece, Syria, and Phrygia, but even early on, Philip the Apostle was frequently confused with Philip the Evangelist (also known as Philip the Deacon), which makes it hard to distinguish which details describe which Philip. Add to that the legendary accounts from the Acts of Philip and Letter from Peter to Philip, and it becomes difficult to separate fact from fiction, too.
Close
James the Greater was the brother of St. John the Evangelist. James and John, along with the Apostle Peter, were part of Jesus’ “inner circle,” witnessing key events that the other apostles did not witness.
After James was martyred, legend places his remains in Spain, where they are supposedly enshrined in a cathedral known as the Cathedral of Santiago de Compostela. Since the Middle Ages, people have walked from certain starting points across Europe to his tomb along pilgrimage routes that have become known as the Way of St. James.
Close
Matthew, also known Levi, is traditionally regarded as the author of the Gospel of Matthew. When Jesus called Matthew to follow him, he was a tax collector (or “publican”)—one of the most reviled professions in ancient Judaism.
Little is known about this apostle. Aside from a handful of mentions in the gospels, he’s a surprisingly obscure New Testament figure. And despite the fact that the church has long considered him the author of the Gospel of Matthew, little else is known about him.
While Matthew is honored as a martyr, no one knows for sure where or how he died. Various accounts say he was beheaded, stoned, burned, or stabbed—one even suggests he died of natural causes like John.
Close
Jude is the reputed author of the canonical Letter of Jude that warns against the licentious and blasphemous heretics. The devotion to him as patron saint of desperate causes began in France and Germany in the late 18th century.
St. Jude is distinguished in John 14:22 as Judas but “not Iscariot” to avoid identification with the betrayer of Jesus. Indeed, the tradition of calling him “Jude” rather than the Scriptural “Judas” likely started to avoid such confusion. He is listed in Luke 6:16 and Acts 1:13 as “Judas of James,” and, depending on the Bible consulted, he is probably the son (Revised Standard and New English) or brother (Authorized and Douay) of St. James the Less, son of Alphaeus. Jude is more probably identified with Thaddaeus (Lebbaeus) in Mark 3:18 and Matthew 10:3 and less probably with Jesus’ “brother” Judas (Mark 6:3, Matthew 13:55).
Close
Simon, also called the Zealot, is mentioned in the lists of apostles provided in the Gospels of Matthew, Mark, and Luke, as well as in the Book of Acts. His epithet, "the Zealot," distinguishes him from Simon Peter, another of the twelve apostles.
The term "Zealot" refers to a political movement among Judeans that existed during the 1st century, which advocated for the violent overthrow of Roman rule in Judea. However, it is not definitively established whether Simon was actually a member of this movement. The label "Zealot" could indicate his zealous nature or fervent devotion to the Jewish faith and cause rather than membership in a specific political group. Simon the Zealot is often considered to have preached the Gospel in different regions, including Egypt and Persia, where he is sometimes said to have been martyred. |
7284 | https://link.springer.com/article/10.1007/BF00255603 | Advertisement
Fluid requirements of patients with burns and inhalation injuries in an intensive care unit
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Abstract
We have studied 9 patients with burns (20%–75%) who had inhalation injuries and compared their actual fluid requirements with their requirements calculated from the Muir and Barclay formula. All patients were resuscitated with plasma protein fraction at a rate sufficient to keep their physiological variables within the following range: heart rate <120/min, central venous pressure 8–12 cm H2O, urine output >30–50 ml/h, systolic blood pressure >90 mm Hg and diastolic blood pressure >60 mm Hg. The amount of plasma protein fraction needed was 4.38±1.26 ml/kg/% burn in the first 24 h and 2.15±0.97 ml/kg/% burn in the second 24 h. This is an increase of 75% and 110% respectively above values predicted from the formula. We suggest that the observed difference is due to a combination of the presence of an inhalation injury which increases fluid requirements by approximately 30% in the first 24 h and the use of plasma protein fraction rather than the dried plasma used in the original Muir and Barclay formula.
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Intensive Care Unit, University College Hospital, London, UK
K. R. Hughes, R. F. Armstrong, M. D. Brough & N. Parkhouse
c/o Department of Medicine, Middlesex Hospital, Mortimer Street, W1N 8AA, London, UK
K. R. Hughes
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Hughes, K.R., Armstrong, R.F., Brough, M.D. et al. Fluid requirements of patients with burns and inhalation injuries in an intensive care unit.
Intensive Care Med 15, 464–466 (1989).
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Received: 15 November 1988
Accepted: 09 May 1989
Issue Date: November 1989
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7285 | https://mathoverflow.net/questions/105157/triangular-grid-with-4-edges-per-vertex | Stack Exchange Network
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I am trying to create a triangular grid/mesh for a rectangular domain in $\mathbb{R}^2$ with the property that each vertex is shared by (at most) four edges. Is this possible to accomplish?
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asked Aug 21, 2012 at 13:18
halvorhalvor
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$\begingroup$ Finite or infinite? $\endgroup$
Brendan McKay
– Brendan McKay
2012-08-21 14:04:21 +00:00
Commented Aug 21, 2012 at 14:04
$\begingroup$ As long as you do not need more than 8 regions. Gerhard "Otherwise You Need Different Geometry" Paseman, 2012.08.21 $\endgroup$
Gerhard Paseman
– Gerhard Paseman
2012-08-21 14:59:42 +00:00
Commented Aug 21, 2012 at 14:59
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$\begingroup$ I can only see how to do $6$ regions. An octahedron is a closed manifold. $\endgroup$
Will Sawin
– Will Sawin
2012-08-21 17:46:55 +00:00
Commented Aug 21, 2012 at 17:46
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Since there is no requirement that the outer region must be a triangle, the question is not quite as trivial as indicated in earlier comments. The rectangle $[0,n] \times [0,1]$ can be triangulated by dividing it into unit squares and then inserting the SW-NE diagonal in each square. Still, this might not be the kind of grid/mesh one wants. To see the problem, it might be easier to think in terms of angles than to use Euler's polyhedron formula: If there are interior points in the triangulation, then the angles at those points have to be at least $90^\circ$ on average, while the average angle in a triangle is only $60^\circ$. It follows that most vertices of the triangulation have to be on the boundary of the region.
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edited Aug 21, 2012 at 19:16
answered Aug 21, 2012 at 18:45
Johan WästlundJohan Wästlund
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$\begingroup$ How is your suggestion a triangulation? Is it for only some pairs (a,b)? When I try it I get quadrilaterals. Gerhard "Maybe This Is Different Geometry" Paseman, 2012.08.21 $\endgroup$
Gerhard Paseman
– Gerhard Paseman
2012-08-21 19:04:00 +00:00
Commented Aug 21, 2012 at 19:04
$\begingroup$ Actually I try it and I see more triangles, but also degree 6 vertices. I must be doing something wrong. Gerhard "A Picture Would Really Help" Paseman, 2012.08.21 $\endgroup$
Gerhard Paseman
– Gerhard Paseman
2012-08-21 19:05:53 +00:00
Commented Aug 21, 2012 at 19:05
$\begingroup$ Gerhard, maybe I was a bit sloppy, answer edited. $\endgroup$
Johan Wästlund
– Johan Wästlund
2012-08-21 19:20:44 +00:00
Commented Aug 21, 2012 at 19:20
$\begingroup$ OK,now I understand your edited version. Have you any triangulations with more than 3 vertices in the interior? My guess is that even one vertex in the interior seriously limits the possibilities. Gerhard "Ask Me About System Design" Paseman, 2012.08.21 $\endgroup$
Gerhard Paseman
– Gerhard Paseman
2012-08-21 19:24:26 +00:00
Commented Aug 21, 2012 at 19:24
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This question is too elementary for MO, but here's a hint. (See the FAQ for alternative sites to ask your question.)
Familiarize yourself with the notion of Euler characteristic
Convince yourself that the Euler characteristic of your triangulation of a rectangle is 1.
Deduce some inequalities on the numbers of vertices, edges and faces, assuming all faces are triangles and all vertex valences are less than or equal to 4.
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answered Aug 21, 2012 at 16:23
Kevin WalkerKevin Walker
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7286 | https://focusonmath.wordpress.com/2013/03/05/strike-it-out-a-primary-game-from-nrich/ | Strike it Out: a Primary Game from NRICH | Focus on Math
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Strike it Out: a Primary Game from NRICH March 5, 2013
Filed under: Basic Facts,General Math,Parents,Primary Math Ideas & Problems — Focus on Math @ 12:06 pm
Tags: addition, primary math
One thing that I came across recently on the NRICH math website is an quick primary students’ game for practicing addition and subtraction with numbers to 20. On the particular webpage for “Strike it Out”, they offer a poster (a picture of which is posted here), a short video clip of the game, and a power point file all which give the instructions for the game.
I had each of my grade two math classes playing this game recently as a warm-up activity, and they loved it! The games go quickly for the most part – of course, some of the pairs of students were slower at the game, but those students were still engaged and trying their best.
The rules, simply, are these:
• Using a number line marked 0-20, one student begins by creating and recording an addition or subtraction equation, e.g., 4 + 10 = 14. On the number line he crosses out the 4 and 10 and circles the 14. His turn is over.
• The partner must now create a new addition or subtraction equation, but it must use the number 14 as one of the first two numbers, e.g., 14 – 6 = 8. She would crosses out the circled 14, crosses out the 6, and circles the 8. Her turn is over.
• The 8 must be used now by the first partner in his new equation, with the recording and crossing out and circling continuing.
• Play continues until one of the partners cannot make a correct number sentence, and the player who made the last correct equation wins.
Although there are many possibilities for equations near the beginning of the game, there are fewer possibilities as the game progresses. I watched students doing a lot of mental math trying to come up with appropriate equations. The students who needed support had a set of ten frames on the table to use to help the visualize and calculate.
I am including for download the game board page I made for students to use. (Cut the page in half to use.) Two students play on a single game board at a time.
There are lots of other great ideas on the NRICH math site for many different levels. I hope you will take some time and explore what is there!
Mathematically yours,
Carollee
Link to game on the NRICH math site here
Download the game board here.
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7287 | https://medium.com/@miyoko_shimura/monte-carlo-simulation-using-python-random-sampling-by-calculating-%CF%80-be3cdd326361 | Monte Carlo Simulation Using Python Random Sampling by Calculating π | by Miyoko Shimura | Medium
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Monte Carlo Simulation Using Python Random Sampling by Calculating π
Converting Random Points into Mathematical Solutions.
Miyoko Shimura
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Dec 8, 2024
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The Monte Carlo method, at its core, is a computational technique using repeated random sampling to obtain numerical results. Simply put, it “uses random guesses to find an answer.”
The method is named after the famous casino in Monaco, as it relies on the element of chance and randomness, much like the games of luck found in a casino. Monte Carlo simulations use randomness and probability to tackle problems, applying a probabilistic approach similar to calculating the odds of events in games of chance.
In casino games like roulette, the outcome can be determined by random events (like the spin of a wheel or the roll of a ball), and the probability of each outcome can be calculated based on the setup of the game. For example, in roulette, you can calculate the likelihood of the ball landing on a particular color or number by understanding the number of pockets on the wheel.
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Photo by Free Walking Tour Salzburg on Unsplash
Modern computers are becoming incredibly fast. The development of hardware and cloud platforms has democratized Monte Carlo methods. Problems that once required supercomputers can now be solved on a standard laptop.
Simulations that once took weeks to run can now be completed in just a few minutes. This has made Monte Carlo simulations a powerful tool for a wide range of applications, from scientific research to finance, allowing for faster and more efficient problem-solving.
Monte Carlo simulation is a powerful statistical method that uses random sampling to solve problems that might be difficult to approach analytically. The essence of Monte Carlo method lies in two fundamental principles.
First, Monte Carlo method reframes problems as probabilistic processes. Rather than trying to calculate exact outcomes, it samples many possible scenarios randomly. This approach made previously intractable problems solvable.
Second, Monte Carlo method uses the law of large numbers. With sufficient trials, results converge to true values. This is similar to how rolling a die many times will show each number appearing close to one-sixth of the time.
Monte Carlo Simulation Through π
I’ll introduce an intuitive way to understand Monte Carlo technique through a very simple geometric experiment to estimate the value of π.
The math behind this is elegant. Our square has an area of 1 (=1 × 1), and our quarter circle has an area of π/4.
Press enter or click to view image in full size
If you randomly throw darts at a 1×1 blue square containing a quarter red circle, approximately π/4 of the points will land inside the quarter circle. Mathematically, this is expressed as
Thus, we can estimate π using this formula.
Now, let’s see how we can implement this elegant mathematical idea in Python code.
Step 1: Generating Random Points
import numpy as np
import matplotlib.pyplot as plt
First, we create 10,000 random points
n_points = 10000
x = np.random.uniform(0, 1, n_points)
y = np.random.uniform(0, 1, n_points)
This code generates 10,000 random points where each point has an x and y coordinate between 0 and 1. Imagine it like throwing darts randomly at a 1×1 square 10,000 times.
Step 2: Checking Points Inside the Circle
inside = x2 + y2 <= 1
pi_estimate = 4 np.sum(inside) / n_points
This part checks which points fall inside our quarter circle. Think of it as determining whether each dart lands within the circular part of our target.
This check process is based on the Pythagorean theorem,which states that the square of the hypotenuse (in this case, the distance from the origin) is equal to the sum of the squares of the other two sides.
For each point (x,y), if its distance from the origin (0,0) is less than or equal to 1, the point lies within the quarter circle. Mathematically, this is represented as
If x² + y² ≦ 1, the point is classified as being inside the quarter circle.
If x² + y² ＞ 1, the point is considered to be outside the circle.
Step 3: Create the Scatter Plot
plt.figure(figsize=(10, 5))
plt.subplot(121)
Plot points inside and outside the circle
plt.scatter(x[inside], y[inside], c='blue', s=1, alpha=0.6, label='Inside')
plt.scatter(x[~inside], y[~inside], c='red', s=1, alpha=0.6, label='Outside')
Draw quarter circle boundary
theta = np.linspace(0, np.pi/2, 100)
plt.plot(np.cos(theta), np.sin(theta), 'k-')
Formatting
plt.axis('equal')
plt.xlim(-0.1, 1.1)
plt.ylim(-0.1, 1.1)
plt.title(f'Monte Carlo Simulation\n{n_points:,} points')
plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left')
Step 4: Visualize Results
plt.figure(figsize=(10, 5))
plt.subplot(121)
Plot points inside and outside the circle
plt.scatter(x[inside], y[inside], c='blue', s=1, alpha=0.6, label='Inside')
plt.scatter(x[~inside], y[~inside], c='red', s=1, alpha=0.6, label='Outside')
Draw quarter circle boundary
theta = np.linspace(0, np.pi/2, 100)
plt.plot(np.cos(theta), np.sin(theta), 'k-')
Formatting
plt.axis('equal')
plt.xlim(-0.1, 1.1)
plt.ylim(-0.1, 1.1)
plt.title(f'Monte Carlo Simulation\n{n_points:,} points')
plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left')
plt.subplot(122)
plt.axis('off')
Display text with results
info_text = (
f'Total points: {n_points:,}\n'
f'Points inside circle: {np.sum(inside):,}\n'
f'Estimated π: {pi_estimate:.6 f}\n'
f'Actual π: {np.pi:.6 f}\n'
f'Error: {abs(pi_estimate - np.pi):.6 f}'
)
plt.text(0.1, 0.5, info_text, fontsize=12)
Adjust layout and show plot
plt.tight_layout()
plt.show()
Blue dots represent points inside the quarter-circle.
Red dots represent points outside the circle.
A black curve outlines the perfect quarter circle for reference.
Visualization brings clarity. We also present numerical results, including the total number of points, the count of points inside the circle, our estimated value of π, and the error relative to the true value.
Press enter or click to view image in full size
On the left, we see our random points plotted in a square. Blue dots represent points that fell inside the quarter circle, while red dots show points that landed outside. A black curve shows the perfect quarter circle we’re trying to approximate.
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On the right side, we display our numerical results, including the total number of points used, how many fell inside the quarter circle, our estimate of π, and how close we got to the actual value.
Step 5: Interpreting Results
In our experiment, we got 7,831 hits inside the quarter circle. Using these results, we can calculate our estimate of π: Multiplying 4 times our ratio of hits (7,831/10,000 = 0.7831), we get an estimate of π.
Press enter or click to view image in full size
Thus, our estimate for π is approximately
When we compare this to the actual value of π (3.14159…), we see our estimate has an error of about 0.009. This illustrates how Monte Carlo method provides approximate but remarkably accurate results with just 10,000 random points in this solution.
(Full Python Code)
import numpy as np
import matplotlib.pyplot as plt
Set number of points
n_points = 10000
Generate random points
x = np.random.uniform(0, 1, n_points)
y = np.random.uniform(0, 1, n_points)
Check which points are inside the quarter circle
inside = x2 + y2 <= 1
Calculate pi estimate
pi_estimate = 4 np.sum(inside) / n_points
Create plot
plt.figure(figsize=(10, 5))
Left subplot (point distribution)
plt.subplot(121)
plt.scatter(x[inside], y[inside], c='blue', s=1, alpha=0.6, label='Inside')
plt.scatter(x[~inside], y[~inside], c='red', s=1, alpha=0.6, label='Outside')
Draw quarter circle
theta = np.linspace(0, np.pi/2, 100)
plt.plot(np.cos(theta), np.sin(theta), 'k-')
plt.grid(True, linestyle='--', alpha=0.3)
plt.axis('equal')
plt.xlim(-0.1, 1.1)
plt.ylim(-0.1, 1.1)
plt.title(f'Monte Carlo Simulation\n{n_points:,} points')
Move legend outside the plot
plt.legend(bbox_to_anchor=(1.05, 1), loc='upper left')
Right subplot (text display)
plt.subplot(122)
plt.axis('off')
info_text = (
f'Total points: {n_points:,}\n'
f'Points inside circle: {np.sum(inside):,}\n'
f'Estimated π: {pi_estimate:.6 f}\n'
f'Actual π: {np.pi:.6 f}\n'
f'Error: {abs(pi_estimate - np.pi):.6 f}'
)
plt.text(0.1, 0.5, info_text, fontsize=12)
Adjust layout to prevent legend overlap
plt.tight_layout()
plt.show()
The Power of More Points
With 100 points, the error might be pretty big
With 100,000 points, we get much closer to the real value of π
Each time you run the program, we get slightly different results because we use random numbers
The accuracy of our estimation improves as we increase the number of points. With 10,000 points, we typically get quite close to π, usually within a few decimal places. We could get even closer by using more points, though this would require more computation time.
Results obtained through the Monte Carlo method always contain some degree of error, especially when the number of generated random numbers (or trials) is small. However, with the rapid advancements in computing power, we can now achieve highly accurate results with significantly smaller errors.
What Monte Carlo Simulations Can and Cannot Tell Us
Monte Carlo simulations are like running lots of “what-if” scenarios to predict possible outcomes. For example, if you’re planning a long-term investment strategy, you can simulate how your portfolio could grow or shrink under different market conditions, like good years, bad years, and everything in between. But remember, it’s not perfect.
While this might seem comprehensive, there’s a crucial caveat. The simulation can only reflect the patterns and relationships we program into it. Monte Carlo simulations rely on the assumptions and data we provide.
What Monte Carlo Captures
Monte Carlo simulations excel at capturing the quantifiable aspects of market behavior. When simulating stock market returns, they effectively model how prices fluctuate within historical ranges and how different assets correlate with each other. They’re particularly good at representing volatility patterns and the statistical distributions of returns we’ve observed in the past.
What It Misses
Yet these simulations struggle with events and changes that break from historical patterns. The 2008 financial crisis or the 2020 pandemic market crash are perfect examples, events that defied historical patterns and challenged traditional risk models. Monte Carlo methods cannot anticipate unprecedented market events, shifting market structures, or changes in human behavior and psychology. They also struggle to account for technological disruptions and regulatory changes that fundamentally alter how markets function (just like other machine learning predictions).
Consider using Monte Carlo simulation to plan for retirement. The simulation might show you have a 95% chance of maintaining your desired lifestyle with your current savings rate. While this sounds quite reassuring, it’s just based on historical market behavior and does not guarantee future success.
Try running the code yourself and experiment with different numbers of points. This hands-on experience will give you an intuitive feel for how Monte Carlo methods work, and why they’re so useful in so many fields.
The power of Monte Carlo methods lies in its simplicity and scalability. The same principles we’ve used to estimate π can be applied to much more complex problems.
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Further Learning
What Is Monte Carlo Simulation? | IBM ------------------------------------- ### Monte Carlo Simulation is a type of computational algorithm that uses repeated random sampling to obtain the likelihood… www.ibm.com
Bits to Qubits. Quantum Computing Rewrites Global Finance --------------------------------------------------------- ### How quantum computers will finally change economy for everyone medium.com
Python Stock Analysis with 20 & 50-Day Moving Averages ------------------------------------------------------ ### Implementing moving averages with Python medium.com
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7288 | https://tutorial.math.lamar.edu/Solutions/CalcI/AbsExtrema/Prob3.aspx | Calculus I - Finding Absolute Extrema
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Section 4.4 : Finding Absolute Extrema
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Determine the absolute extrema of R(t)=1+80 t 3+5 t 4−2 t 5 R(t)=1+80 t 3+5 t 4−2 t 5 on [−4.5,4][−4.5,4].
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Hint : Just recall the process for finding absolute extrema outlined in the notes for this section and you’ll be able to do this problem!
Start Solution
First, notice that we are working with a polynomial and this is continuous everywhere and so will be continuous on the given interval. Recall that this is important because we now know that absolute extrema will in fact exist by the Extreme Value Theorem!
Now that we know that absolute extrema will in fact exist on the given interval we’ll need to find the critical points of the function.
Given that the purpose of this section is to find absolute extrema we’ll not be putting much work/explanation into the critical point steps. If you need practice finding critical points please go back and work some problems from that section.
Here are the critical points for this function.
R′(t)=240 t 2+20 t 3−10 t 4=−10 t 2(t−6)(t+4)=0⇒t=−4,t=0,t=6 R′(t)=240 t 2+20 t 3−10 t 4=−10 t 2(t−6)(t+4)=0⇒t=−4,t=0,t=6 Show Step 2
Now, recall that we actually are only interested in the critical points that are in the given interval and so, in this case, the critical points that we need are,
t=−4,t=0 t=−4,t=0 Show Step 3
The next step is to evaluate the function at the critical points from the second step and at the end points of the given interval. Here are those function evaluations.
R(−4.5)=−1548.13 R(−4)=−1791 R(0)=1 R(4)=4353 R(−4.5)=−1548.13 R(−4)=−1791 R(0)=1 R(4)=4353
Do not forget to evaluate the function at the end points! This is one of the biggest mistakes that people tend to make with this type of problem.
Show Step 4
The final step is to identify the absolute extrema. So, the answers for this problem are then,
Absolute Maximum :4353 at t=4 Absolute Minimum :−1791 at t=−4 Absolute Maximum :4353 at t=4 Absolute Minimum :−1791 at t=−4
Note the importance of paying attention to the interval with this problem. Had we neglected to exclude t=6 t=6 we would have gotten the wrong answer for the absolute maximum. Also note that if we’d neglected to check the endpoints at all we also would have gotten the wrong absolute maximum.
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7289 | https://chemistry.stackexchange.com/questions/172809/effect-of-inert-gas-on-the-rate-of-reaction | equilibrium - Effect of inert gas on the rate of reaction - Chemistry Stack Exchange
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Effect of inert gas on the rate of reaction
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Is there a general effect on the rate of reaction of a dynamic equilibrium when an inert gas is introduced at a constant volume? I know that the position of equilibrium won't change, but much like a catalyst which increases the rate of both forward and reverse reactions, I'm wondering if the introduction of an inert gas will have a similar effect.
My preliminary thoughts were that the inert gas would 'block' successful collisions by the reactants and products, thereby decreasing both forward and reverse rates. However, these 'blocked' collisions could simply be the result of more collisions taking place since there are more molecules in the same volume so the rate doesn't actually change.
My other thought was that certain inert gases might increase the rate of reaction by acting as a heterogenous catalyst –a surface for the reaction to take place.
Clearly, my thoughts haven't led to a definite answer so I was wondering if anyone could clear up exactly what happens to the rate of reaction when an inert gas is introduced. Thanks!
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asked Apr 12, 2023 at 6:53
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At otherwise the same partial pressure of the reacting gas, an inert gas slows down heterogenous reactions of a gas with condensed phases.
It is caused by local additional decreasing of partial pressure of the reacting gas at the boundary layer, where it gets partly depleted and is not being resupplied fast enough from the bulk volume.
The effect is the most obvious if products stay in the condensed phase. Like
O X 2(g)+A(s)O X 2 0.21 a t m v s a i r→B(s)
O X 2(g)+A(s)−→−−−−−−−−−O X 2 0.21 a t m v s a i r B(s)
Pure oxygen would have boundary pressure equal to bulk pressure. But with nitrogen and fast enough reaction, oxygen boundary partial pressure may drop significantly -- having near pure nitrogen there -- and so would the reaction rate.
This inert gas effect is in breathing or fuel burning context empirically confirmed by better tolerance of lower oxygen partial pressure due altitude, compared to lowered oxygen percentage at normal pressure.
The typical threshold range for burning organic/fossil fuel is 16−18%16−18% of oxygen in air, with the partial pressure 0.16−0.18 a t m 0.16−0.18 a t m . But in mountains, locals burn wood at much lower oxygen pressure, due lower air pressure.
For homogenous reactions, the effect is more subtle, but an extra inert gas generally slows down reactions, by the following effects:
Its molecules take way part of energy released by the reactions, in macro scale via heat capacity and on molecular level. It decreases the probability of reactant collisions leading to the reaction.
It affects diffusion of limiting reactant, especially if the inert molecules are slow or reactants are not in the stoichiometric ratio.
If some reactions steps involves some intermittent reactants in excited state, it may cancel their excitation before they can react.
There is possibly more effects.
"Unimolecular" reactions are already well described by Porphyrin.
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edited Apr 13, 2023 at 6:54
answered Apr 12, 2023 at 7:39
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Do you know any information about what happens for non-hetereogenous reactions? In other words, just gases reacting like in the Haber process.Pen and Paper –Pen and Paper 2023-04-12 07:56:30 +00:00 Commented Apr 12, 2023 at 7:56
@PenandPaper But that IS a heterogenous reaction, gases reacting on surface of solid catalyst. If we added to H2 and N2 e.g. Ar, there wouldb be locally decreased partial pressures of N2 and H2 on catalyst surface. I mean even more than just due formed NH3.Poutnik –Poutnik 2023-04-12 08:02:07 +00:00 Commented Apr 12, 2023 at 8:02
Could you also just explain why the inert gas at the boundary layer stops the reacting gases from being replenished. Wouldn't this happen without the inert gas since the reagents are reacting anyway?Pen and Paper –Pen and Paper 2023-04-13 00:52:19 +00:00 Commented Apr 13, 2023 at 0:52
@PenandPaper see the question update.Poutnik –Poutnik 2023-04-13 05:35:12 +00:00 Commented Apr 13, 2023 at 5:35
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In the gas phase in so called 'Unimolecular Reactions' an inert gas does play a vital part by collisionally exciting a potentially reactive molecule. The scheme is
A+M k 2⇌k−2 A∗+M A X∗k 1→p r o d u c t
where M is the inert gas and A∗ is vibrationally excited or 'energised' A. The rate law depends on [M] and [A] and can be found by applying steady state conditions to [A∗]. Methyl isocyanide C H X 3 N C to acetonitrile C H X 3 C N isomerisation is an example of such a reaction. The simple scheme (above) is due to Lindemann, a more advanced one is called RRKM theory.
Additional comment.
The rate of product formation with steady state on A∗is
d P d t=k 1 k 2[A][M]k 1+k−2[M]
so that the rate is small and bi-molecular at low pressures of [M] i.e. (k 2[A][M]) but becomes larger and unimolecular at high pressure when k−2[M]≫k 1 and is k 1 k 2[A]/k−2. Often a plot of 1[A]d P d t vs [A] is made better to illustrate these effects.
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edited Apr 13, 2023 at 8:08
answered Apr 12, 2023 at 10:14
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So the inert has in this case increases the rate?Pen and Paper –Pen and Paper 2023-04-13 00:50:50 +00:00 Commented Apr 13, 2023 at 0:50
its not quite as simple as that,I've added a comment in my answer due to lack of space here.porphyrin –porphyrin 2023-04-13 07:44:26 +00:00 Commented Apr 13, 2023 at 7:44
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Table of Contents
Table of Contents
Declining Balance Method
How to Calculate
What Does It Tell You?
Declining vs. Double-Declining
FAQs
The Bottom Line
Understanding the Declining Balance Method: Formula and Benefits
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What Is the Declining Balance Method?
The declining balance method is an accelerated way to record larger depreciation in an asset's early years. The system records smaller depreciation expenses during the asset's later years.
Key Takeaways
The declining balance method accelerates depreciation by recording larger expenses during an asset's early years.
This method is ideal for technology products that rapidly depreciate due to obsolescence.
Unlike the straight-line method, declining balance leads to smaller depreciation expenses in later years.
The double-declining method depreciates assets twice as quickly by doubling the rate of the declining balance method.
Accelerated depreciation techniques can reduce taxable income early in an asset's life.
Calculating Declining Balance Depreciation: A Step-by-Step Guide
Depreciation under the declining balance method is calculated with this formula:
Declining Balance Depreciation=CBV×DRwhere:CBV=current book valueDR=depreciation rate (%)
Current book value is the asset's net value at the start of an accounting period. It's calculated by deducting the accumulated depreciation from the cost of the fixed asset.
Residual value is the estimated salvage value at the end of the useful life of the asset.
The rate of depreciation is defined according to the estimated pattern of an asset's use over its useful life. The expense would be $270 in the first year, $189 in the second year, and $132 in the third year if an asset costing $1,000 with a salvage value of $100 and a 10-year life depreciates at 30% each year.
Understanding the Insights of the Declining Balance Method
The declining balance method is also known as the reducing balance method. It's ideal for assets that quickly lose their value or inevitably become obsolete. This is classically true with computer equipment, cell phones, and other high-tech items that are generally useful earlier on but become less so as new models are brought to market. An accelerated method of depreciation ultimately factors in the phase-out of these assets.
Important
Employing the accelerated depreciation technique means there will be lesser taxable income in the earlier years of an asset's life.
The declining balance method contrasts with straight-line depreciation, which suits assets that lose value steadily.
The straight-line depreciation method simply subtracts the salvage value from the cost of the asset and this is then divided by the useful life of the asset. The annual straight-line depreciation expense would be $2,000 ($15,000 minus $5,000 divided by five) if a company shells out $15,000 for a truck with a $5,000 salvage value and a useful life of five years.
Comparing Declining Balance and Double-Declining Methods
It may signal that a company is using accelerated depreciation methods such as the double-declining balance depreciation method if it often recognizes large gains on sales of its assets, Net income will be lower for many years but this ultimately leads to a bigger gain when the asset is sold because book value ends up being lower than market value. Its sale could portray a misleading picture of the company's underlying health if the asset is still valuable.
What Is Accumulated Depreciation?
Accumulated depreciation is total depreciation over an asset's life beginning with the time when it's put into use. Depreciation is typically allocated annually in percentages.
How Does Depreciation Affect Taxes?
Depreciation lets a company deduct an asset's value decline, lowering taxable income. The company must own the asset and use it to generate income. Its anticipated service life must be for more than one year and it must have a determinable useful life expectancy.
How Does the Double-Declining Balance Depreciation Method Work?
The double-declining method involves depreciating an asset more heavily in the early years of its useful life. A business might write off $3,000 of an asset valued at $5,000 in the first year rather than $1,000 a year for five years as with straight-line depreciation. The double-declining method depreciates assets twice as quickly as the declining balance method as the name suggests.
The Bottom Line
The declining balance method is an accelerated depreciation technique that allows businesses to report larger depreciation expenses in the early years of an asset's life, which reduces taxable income during those years. This method is particularly effective for assets like technology products that quickly become obsolete.
While this approach results in smaller depreciation amounts in later years, it is advantageous for managing tax liabilities in the short term. Understanding the right depreciation method can significantly impact a business's financial statements and tax obligations. Businesses must consider the nature of their assets and financial strategy when selecting a depreciation method.
Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
AccountingTools. "Declining Balance Method Definition."
Internal Revenue Service. "Publication 946 (2023), How to Depreciate Property."
CFI Education. "Accumulated Depreciation."
Internal Revenue Service. "What Small Business Owners Should Know About the Depreciation of Property Deduction."
Bench. "Double Declining Balance: A Simple Depreciation Guide."
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verb as in make tired
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annoy
bore
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burden
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pain
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strain
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Weak matches
cause ennui
fall off
grow tired
have had enough
leave one cold
lose interest
make discontented
take it out of
tire out
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People who were fully committed to rebuilding in the immediate aftermath of destruction are now rethinking it, having grown weary of the slog.
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It was an observation some within Labour are weary of hearing about again, recalling the fuss a decade ago when Labour had mugs printed with the promise of "controls on immigration".
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He immediately came under fire within Israel, with opposition leader Yair Lapid writing: "Today saw a weary and whining Israeli prime minister, in a speech overloaded with overused gimmicks"
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The cutesy performance shtick around a shoestring variety show — imagine Pee-wee’s Playhouse without the bewitching oddity — grows wearying.
FromLos Angeles Times
“No more migra,” I said, loud enough to draw a weary smile from my hosts.
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all in
adjective as in exhausted
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fatigued
spent
tired
tired out
tuckered
wearied
weariful
weary
beat
adjective as in very tired
dog-tired
exhausted
fatigued
kaput
wearied
weary
worn-out
blase
adjective as in nonchalant
apathetic
been around twice
bored
cloyed
cool
disenchanted
disentranced
done it all
fed up
glutted
indifferent
jaded
knowing
laid-back
lukewarm
mellow
mundane
offhand
satiated
sick of
sophisticate
sophisticated
surfeited
unconcerned
unexcited
uninterested
unmoved
weary
world-weary
worldly
blasé
adjective as in nonchalant
apathetic
been around twice
bored
cloyed
cool
disenchanted
disentranced
done it all
fed up
glutted
indifferent
jaded
knowing
laid-back
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mellow
mundane
offhand
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sick of
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sophisticated
surfeited
unconcerned
unexcited
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dead tired
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From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group.
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7293 | https://www.youtube.com/watch?v=FnQNbFl72LY | Finding Modular Inverses
Cathy Frey
1470 subscribers
1267 likes
Description
146816 views
Posted: 9 Sep 2017
Step by step instructions to find modular inverses.
65 comments
Transcript:
Intro modular inverses please pause the video and read the definition and then we'll talk about it so we're finding multiplicative inverses and that's two numbers that multiply together to get you one mod M we were really finding multiplicative inverses in the last video when we were solving many or congruences we just didn't give them that name and this theorem is important a has an inverse mod M if and only if the greatest common divisor of a and M is equal to one if the greatest common divisor between a and M is greater than one then there is no inverse mod M so let's take a look at some examples Example suppose we want to find the inverse of 5 mod 7 that's going asking us to find what we can multiply 5 by to get 1 mod 7 so we do as we've done in the past mod 7 1 is congruent to 8 that's not a multiple of 5 so we add 7 again and we get 15 15 is 5 times 3 mod 7 so 3 is the inverse of 5 mod 7 a notation we'll write 5 inverse is congruent to 3 mod 7 now pause the video and you try this one find the inverse of 5 mod 11 so hopefully you said one was congruent to 12 model evan with 1 and add the mod on congruent to 23 adding 1134 and eventually we get to 45 and that's 5 times 9 mod 11 so the inverse of 5 is congratulate my 11 again if you feel comfortable pause the video and try this one so you've got one congruent to 18 because we're adding 17 this time and that's 6 times 3 mod 17 so 6 inverse is congruent to 3 mod 17 now we want to Shortcut find the inverse of 18 mod 19 now we want to find the inverse of 18 mod 19 so we could do 1 is congruent to 20 and that's not a multiple of 18 is congruent to adding 19 we get 39 and then we get 58 and then we get 77 still no multiples of 18 eventually if you keep adding on you're going to get up to 324 which is 18 times 18 so 18 inverse is 18 okay so that will always work but sometimes you end up spending a lot of time trying to find the inverse and there's a little bit of a shortcut you can do for this problem if you realize that 18 is the same thing as negative 1 mod 19 we're used to adding adding 19 but we can subtract 19 as well what's the inverse of negative 1 well we multiply negative 1 to get positive by positive 1 that's negative 1 right 1 is the same thing as negative 1 times negative 1 and since 18 is negative 1 18 is its own inverse okay it would work the same way let's suppose I asked you to find the inverse of 1790 17 is the same thing as negative 2 mod 19 all right if I had 19 2 negative 2 I get 17 looking back up here 20 is negative 2 times negative 10 so negative 2 the inverse of negative 2 is negative 10 mod 19 negative 2 adding 19 gives us 17 so the inverse of 17 is congruent to I'm going to add 19 to that negative 10 because I want positive numbers is equal to 9 mod 19 okay if you did this a long way eventually you would get up to in your list of congruences up here 17 times 9 which would be 153 so you get there it just takes a little bit longer ok so now what if I was asked to find Problem the inverse of 5 mod 10 you could start chugging along and add 10 to 1 and you get 11 and 21 and 31 and you could spend the rest of your life adding ten to one or multiples of one and you would never get a multiple of five so what's the problem here well the problem is that the greatest common divisor of five and ten is five which is not one so the answer is that there is no inverse so if there's a greatest common divisor greater than one between those two values then there is no inverse so I hope that helps and I think you're ready to practice so have a great day [Music] |
7294 | https://www.sciencedirect.com/topics/neuroscience/upper-motor-neuron-lesion | Skip to main content
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Outline
1. Introduction
2. Anatomy and Physiology of Upper Motor Neurons
3. Pathophysiology of Upper Motor Neuron Lesions
4. Clinical Manifestations and Diagnostic Evaluation
5. Neuroplasticity, Recovery, and Therapeutic Interventions
6. Conclusion
Topic summaryAI
1. Introduction
An upper motor neuron (UMN) lesion refers to damage affecting the neurons that originate in the cerebral cortex and descend through the subcortical white matter, internal capsule, cerebral peduncle, brainstem, and spinal cord, ultimately synapsing with the anterior horn cells of the lower motor neuron (LMN) throughout the spinal cord. The UMN system comprises the corticospinal and indirect pathways, and pathology generally involves both, resulting in a mixture of corticospinal deficits along with antigravity weakness, spasticity, and hyperreflexia attributed to indirect pathway damage. The UMN consists of the corticospinal, corticobulbar, and pyramidal systems, and together with the LMN, comprises the classic two-neuron system responsible for voluntary activation of somatic musculature. Lesions of the UMN are common features of neurological disorders affecting the central nervous system. Typical motor dysfunctions resulting from an UMN lesion include muscle weakness, spasticity, clonus, hyperreflexia, and pathological reflexes such as the Babinski sign. Damage to the UMN leads to disinhibition of the stretch reflex arc, resulting in hyperexcitable reflexes and the clinical state known as spasticity.
2. Anatomy and Physiology of Upper Motor Neurons
Most axons of the corticospinal tract originate from neurons in the precentral gyrus (Brodmann area 4), with significant contributions from other areas of the cortex, especially the premotor area 6 and supplementary motor area. Layer 5 in the primary motor cortex contains the distinctive giant pyramidal neurons known as Betz cells, whose axons become part of the corticospinal tract. These neurons send their axons through the subcortical white matter, entering the posterior limb of the internal capsule, and then descending into the cerebral peduncle of the midbrain, the basis pontis of the pons, and the pyramid of the medulla oblongata. At the level of the medullary-spinal cord junction, most corticospinal tract fibers decussate to the contralateral side, forming the lateral corticospinal tract, while a smaller portion continues ipsilaterally as the anterior corticospinal tract. The axons of UMNs synapse either directly or indirectly (via interneurons) with LMNs in the anterior horn of the spinal cord.
The corticospinal tract is responsible for voluntary motor control, particularly of distal limb muscles. The UMN system also integrates with extrapyramidal pathways, which include the rubrospinal, vestibulospinal, reticulospinal, and tectospinal tracts. These extrapyramidal tracts originate from various nuclei in the brain and descend outside the pyramids of the medulla, modulating involuntary movements, muscle tone, and reflexes. The rubrospinal tract sends information to flexor and extensor muscles, vestibulospinal tracts regulate posture and balance via input from the inner ear, reticulospinal tracts influence eye movements and respiratory muscles, and tectospinal tracts coordinate head, neck, and upper limb responses to sensory stimuli.
Histaminergic projections from the tuberomamillary nucleus of the posterior hypothalamus reach the primary motor cortex, which receives abundant histaminergic input. Histamine directly excites spinal motor neurons through H1 and H2 receptors, increasing their excitability and influencing repetitive firing behavior. This histaminergic modulation affects final motor outputs and regulates ongoing motor execution and spinal motor reflexes.
3. Pathophysiology of Upper Motor Neuron Lesions
Upper motor neuron lesions result in damage to descending motor pathways within the central nervous system. Following UMN damage, there is a withdrawal of inhibitory descending control, particularly from the corticoreticulospinal tract, which leads to disinhibition and hyperexcitability of the stretch reflex arc. This disinhibition manifests clinically as spasticity, defined as a velocity-dependent increase in tonic stretch reflexes and exaggerated tendon jerks, resulting from hyperexcitability of the stretch reflex and forming one component of the upper motor neuron syndrome. Spasticity should be distinguished from rigidity and dystonia, as the term is not applicable to these states even if they coexist with features of an UMN lesion.
Long-term cellular and molecular changes after an acute UMN lesion may include plastic changes in neural structure, such as sprouting of terminals and formation of new synaptic connections, which further contribute to the progressive nature of spasticity. The complexity of spasticity involves both the loss of inhibitory influences and the presence of facilitatory descending inputs, amplifying the reactivity of motor segments to sensory input. Associated phenomena include altered muscle tone (hypertonia), commonly observed in the upper motor neuron syndrome.
4. Clinical Manifestations and Diagnostic Evaluation
UMN lesion is characterized by initial muscle weakness and flaccidity, which subsequently progresses to spastic paralysis, increased muscle tone (hypertonia), and exaggerated muscle stretch reflexes (hyperreflexia). Sustained rhythmic muscle contractions, known as clonus, may be observed when a tendon is held in extension, and this is considered a pathologic indicator of hyperreflexia in the context of UMN lesion. Pathological reflexes such as the Babinski sign, which involves dorsiflexion of the great toe and fanning of the other toes upon firm stroking of the sole, are highly reliable markers of UMN lesion and are associated with lesions of the cortex or corticospinal tract.
In contrast, LMN lesions present with flaccid paralysis, muscle atrophy, fasciculations, hypotonia, and hyporeflexia. The distribution of weakness in UMN lesion typically affects groups of muscles, often following a pyramidal pattern, with sparing of certain muscles due to bilateral innervation, such as the forehead muscles in facial paralysis.
Neurological examination for UMN lesion includes assessment of deep tendon reflexes, which are hyperactive, and testing for pathologic reflexes such as the Babinski sign. Muscle tone is evaluated for spasticity and resistance to passive movement, with hypertonia being a hallmark of UMN lesion. Electrophysiological studies, including electromyography (EMG), nerve conduction studies, and transcranial magnetic stimulation (TMS), are utilized to confirm UMN involvement and differentiate from LMN lesions. Neuroimaging techniques such as magnetic resonance imaging (MRI) are important for lesion localization and differential diagnosis in UMN lesion.
5. Neuroplasticity, Recovery, and Therapeutic Interventions
Recovery following UMN lesion is largely attributed to plastic changes in the sensorimotor cortex, including unmasking of latent horizontal connections, activation of silent synapses, modulation of activity-dependent synaptic plasticity, and generalized changes in the excitability of postsynaptic neurons. Functional electrical stimulation (FES) has been shown to facilitate sensory motor integration and perfusion of the ipsilesional sensorimotor cortex, resulting in functional improvement of hemiparetic upper extremity in chronic stroke patients. FES induces rapid plastic change in the sensorimotor regions of the cortex and may facilitate relearning and contribute to restoration of voluntary control following stroke or brain injury. FES may be applied successfully for different forms of paralytic conditions caused by UMN lesions, such as stroke, traumatic brain injury, and cerebral palsy. Clinical practice guidelines recommend FES for patients with lower limb impairments due to UMN lesions if the individual has enough passive movement of their ankle to allow for walking but are unable to wear a splint and/or if the individual has trouble controlling lower limb movement or keeping balance when walking, and for acute or subacute cervical spinal cord injury to improve hand and upper extremity function.
Transcranial magnetic stimulation (TMS) can be used to exogenously engage or acquire long-term potentiation and long-term depression-like mechanisms. Injury-based animal studies have demonstrated that TMS-induced neuronal activity results in a decrease of pro-apoptotic factors and upregulation of brain-derived neurotrophic factors, which regulate plasticity in an activity-dependent manner.
Spasticity commonly occurs in the presence of UMN lesions or disconnection of the synaptic connection between the brain and spinal motor neurons, and is a disabling symptom of spinal cord injury, multiple sclerosis, and cerebral palsy. Spasms and spasticity are thought to result from dysfunction of inhibitory signals within the spinal cord that depend on descending motor pathways. Treatments for spasticity include anti-spasticity drugs, botulinum toxin, electrical stimulation, and surgery. In cases where patients do not respond to pharmacological and surgical interventions, gene therapy might be warranted, using a vector capable of neuronal tropism and targeted neuromodulation to achieve focused synaptic inhibition.
6. Conclusion
Upper motor neurons are involved in motor control, volitional activity, muscle tone, and reflexes through their influence on descending motor pathways within the central nervous system. Lesions affecting UMNs result in a distinct clinical syndrome characterized by muscle weakness, spasticity, clonus, hyperreflexia, and pathological reflexes such as the Babinski sign. Spasticity is defined as a velocity-dependent increase in tonic stretch reflexes and exaggerated tendon jerks, reflecting hyperexcitability of the stretch reflex and serving as a key diagnostic indicator of UMN lesion. Recovery of motor function following UMN lesions is attributed to plastic changes in the sensorimotor cortex, including unmasking of latent horizontal connections, activation of silent synapses, and modulation of activity-dependent synaptic plasticity. Functional electrical stimulation has been shown to facilitate cortical perfusion and contribute to functional improvement in hemiparetic limbs, supporting the role of neuroplasticity in rehabilitation.
Reference 1
Reference works Chapter
Upper Motor Neuron Lesions
Devereaux M., Katirji B., Daroff R.
Encyclopedia of the Neurological Sciences , 2014 pp 584-586
View PDFView chapter
Reference 2
Review article
Intraoperative motor evoked potential monitoring - A position statement by the American Society of Neurophysiological Monitoring
MacDonald D.B., Skinner S., Shils J., Yingling C.
Clinical Neurophysiology , 2013 pp 2291-2316
View PDFView article
Related quote(s)1 / 1
"... Upper motor neuron system The corticospinal and indirect pathways together comprise the upper motor neuron (UMN) system. Pathology generally involves both, so that UMN lesions typically cause a mixture of corticospinal deficits along with antigravity weakness, spasticity and hyperreflexia attributed to indirect pathway damage. ..."
Reference 3
Book Chapter
Disordered muscle tone
JAMES W. LANCE, JAMES G. McLEOD
A Physiological Approach to Clinical Neurology , 1981 pp 128-153
View PDFView chapter
Related quote(s)1 / 2
"... If the function of the latter is withdrawn by damage to the ‘upper motor neurone’, the stretch reflex arc is disinhibited and therefore hyperexcitable, giving rise to the state known clinically as spasticity. The progressive changes observed with time after an acute lesion may well depend on plastic changes in neural structure such as the sprouting of terminals and formation of new synaptic connections. At this stage, it would be advisable to define ‘spasticity’ or at least to consider the various components of the syndrome called spasticity. Landau 31 points out that the term spasticity may be applied to increase in activity of stretch reflexes, release of flexor reflexes and defect in motor performance, all of which may be found in the ‘Upper Motor Neurone Syndrome’. The term should not be applied to states of dystonia and rigidity even if, as in most cases of cerebral palsy, they are associated with some features of an upper motor neurone lesion. Spasticity may be defined as disinhibition of the stretch reflex, resulting in velocity-dependent increase in tonic stretch reflexes and increased tendon jerks. Other features of the upper motor neurone syndrome are commonly associated with spasticity. If the upper motor neurone remains untarnished but disease prevents the normal inhibitory inflow from the basal ganglia to cortex on the one hand and to the reticular formation on the other, an unrestrained outflow down pyramidal, parapyramidal and extrapyramidal fibres bombards the segmental mechanisms of the spinal cord indiscriminately. This results in activity persisting beyond that necessary for voluntary movement, giving rise to movement disorders such as athetosis, dystonia and Parkinsonian rigidity. A patient with Parkinson's disease is quite unable to modulate the later components (M 2 and M 3 ) of the stretch reflex, which are increased to as much as five times those of normal subjects 33 . A Parkinsonian patient is in a state of continuous alpha-gamma coactivation which fluctuates in level with mental and physical exertion, resembling a tense person who is unable to relax completely of his own volition. ..."
Related quote(s)2 / 2
"... Alpha-gamma excitability is influenced from within the spinal cord by internuncial cells and propriospinal fibres which have facilitatory or inhibitory effects (designated + and – in Figure 6.3 ). Spinal mechanisms in turn are under brainstem control from the vestibulospinal tract ( Figures 5.1 , 5.2 , 6.3 ) and facilitatory reticulospinal pathways ( Figure 5.2 ), which augment the excitability of motor neurones, particularly those controlling extensor muscles in the lower limbs of man. These brainstem tracts are balanced by the inhibitory effect of the corticoreticulospinal tract ( Figure 6.3 ). If the function of the latter is withdrawn by damage to the ‘upper motor neurone’, the stretch reflex arc is disinhibited and therefore hyperexcitable, giving rise to the state known clinically as spasticity. The progressive changes observed with time after an acute lesion may well depend on plastic changes in neural structure such as the sprouting of terminals and formation of new synaptic connections. At this stage, it would be advisable to define ‘spasticity’ or at least to consider the various components of the syndrome called spasticity. Landau 31 points out that the term spasticity may be applied to increase in activity of stretch reflexes, release of flexor reflexes and defect in motor performance, all of which may be found in the ‘Upper Motor Neurone Syndrome’. The term should not be applied to states of dystonia and rigidity even if, as in most cases of cerebral palsy, they are associated with some features of an upper motor neurone lesion. Spasticity may be defined as disinhibition of the stretch reflex, resulting in velocity-dependent increase in tonic stretch reflexes and increased tendon jerks. Other features of the upper motor neurone syndrome are commonly associated with spasticity. 1 ..."
Reference 4
Book Chapter
Strength and Reflexes
Hammerstad J.P.
Textbook of Clinical Neurology , 2007 pp 243-287
View PDFView chapter
Related quote(s)1 / 1
"... Most of the axons of the corticospinal tract come from neurons in the precentral gyrus (Brodmann area 4), 20 although significant contributions come from many other areas of the cortex, especially premotor area 6 and supplementary motor area. This is the only motor tract that passes directly from the cerebral cortex to the spinal cord without making any intermediate synaptic connections. The cytoarchitecture of the primary motor area, supplementary motor area, and premotor cortex is distinct from that of the adjacent sensory and prefrontal regions. Layer 4, the major afferent layer for the sensory cortex, is absent in the motor areas, and hence these motor regions are often termed agranular. Layer 5 in the primary motor cortex contains the distinctive giant pyramidal neurons known as Betz cells . The axons of these cells become the corticospinal or pyramidal tract and represent one of several descending influences on the motor neurons of the brain stem and spinal cord. The remaining cells mostly come from the supplementary motor and premotor cortex. Most of these axons terminate on interneurons in the internuncial pool of the ventral gray of the spinal cord. Axons that connect monosynaptically with the lower motor neurons are involved in control of the distal muscles of the extremities, especially the intrinsic muscles of the hand that control finger movements. 3 Upper motor neurons in the cerebral cortex also send axons to the motor nuclei of the cranial nerves and form the corticobulbar tract. The motor neurons in the primary motor area have a somatotopic organization. The homunculus was mapped by Penfield and colleagues 21 by electrically stimulating the surface of the cortex during epilepsy surgery in humans. Stimulation of most of the lateral surface of the primary motor area produced movement in the mouth, tongue, face, and hand muscles, which is consistent with the importance of fine voluntary control of these structures ( Fig. 15‐17 ). ..."
Reference 5
Book Chapter
Motor Systems
David L. Felten, M. Kerry O'Banion, Mary Summo Maida
Netter's Atlas of Neuroscience , 2016 pp 391-420
View PDFView chapter
Related quote(s)1 / 1
"... Corticospinal Tract The motor portion of the corticospinal tract (CST) originates from neurons of many sizes, mainly from the primary motor cortex (area 4) and the supplemental and premotor cortices (area 6). The primary sensory cortex (areas 3, 1, 2) contributes axons into the CST, but these axons terminate mainly in secondary sensory nuclei to regulate the processing of incoming lemniscal sensory information. The CST travels through the posterior limb of the internal capsule, the middle region of the cerebral peduncle, numerous fascicles of axons in the basis pontis, and the medullary pyramid on the ipsilateral side. Most of the CST axons (approximately 80%, but variable from individual to individual) cross the midline in the decussation of the pyramids at the medullary-spinal cord junction. These crossed fibers descend in the lateral CST in the lateral funiculus of the spinal cord and synapse on alpha and gamma LMNs, both directly and indirectly through interneurons. CST axons that do not decussate continue as the anterior CST in the anterior funiculus of the spinal cord and then decussate at the appropriate level through the anterior white commissure to terminate directly and indirectly on alpha and gamma LMNs contralateral to the cortical cells of origin. Only a very small portion of the motor connections of the corticospinal tract terminate on LMNs on the ipsilateral side of the spinal cord. Clinical Point The motor portion of the CST arises mainly from neurons in the primary motor cortex (area 4) and the supplemental and premotor cortices (area 6). The primary sensory cortex and superior parietal lobule contribute corticospinal axons (corticonuclear fibers) to secondary sensory nuclei in the lower brain stem and spinal cord. Approximately 80% of the CST axons cross in the decussation of the pyramids and terminate directly and indirectly with alpha and gamma LMNs that control movements of the distal extremities, especially the hands and fingers. ..."
Reference 6
Book Chapter
Introduction to human neuromusculoskeletal systems
Jorge Garza-Ulloa
Applied Biomechatronics using Mathematical Models , 2018 pp 53-118
View PDFView chapter
Related quote(s)1 / 1
"... It is indicated a pyramidal using lateral corticospinal tracts; the pathway starting at the cerebral cortex showing the location of the upper motor neural (LMN) cell body, pass through the midbrain, pons, in the medulla oblongata there is the decussation (it is a phenomenon in which the nerve fibers found in one lateral portion of the organ cross another) of UMN axon in the pyramid, pass through a lateral cervical spinal cord, the synapse of UMN and LMN occurs in anterior horn of the lumbar spinal cord, where the LMN exits via anterior root to the skeletal muscle in the limb. Subconscious tracts or extrapyramidal system is part of the motor system causing involuntary movements. The extrapyramidal system represents part of the motor pathway system that has some synapses with the brainstem, which is in contrast to those of the pyramidal system. The extrapyramidal tract is distinguishable from the pyramidal system as tracts do not run within the pyramids of the medulla oblongata and instead run outside. This extrapyramidal pathway contains various multisynaptic pathways that rely within several nuclei in the brain, and these nuclei are dispersed from the telencephalon to the medulla oblongata. This system includes parts of the following tracts: rubrospinal, vestibulospinal , reticulospinal , and tectospinal , where: Rubrospinal tracts send information to the flexor and extensor muscles. Vestibulospinal tracts send information from the inner ear to monitor position of head. The vestibular nuclei respond by altering muscle tone, neck muscle contraction, and limbs for posture and balance. Reticulospinal tract s send information to cause eye movements and activate respiratory muscles. Tectospinal tracts send information to the head, neck, and upper limbs in response to bright and sudden movements and loud noises. The tectum area consists of superior colliculi (receives visual information) and inferior colliculi (receives auditory information). 1 ..."
Reference 7
Review article
Histamine beyond its effects on allergy: Potential therapeutic benefits for the treatment of Amyotrophic Lateral Sclerosis (ALS)
Volonte C., Apolloni S., Sabatelli M.
Pharmacology & Therapeutics , 2019 pp 120-131
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"... Activation of H1R causes concentration-dependent release of [Ca2+] from internal stores and concentration-dependent increase in glutamate release. H2R activation increases cyclic adenosine monophosphate levels and phosphorylation of transcription factor cAMP response-element binding protein, thus emphasizing a role for HA in neuron-glia communication . HA neurons are found merely in the tuberomamillary nucleus of the posterior hypothalamus, from where they project their axons all over the CNS and spinal cord in rats, mice, and humans . Two main bundles of axons from the tuberomamillary nucleus send a large arborization, for instance, to the cortex, and most cortical regions receive a moderately dense and sparse histaminergic input. Among these regions, the primary motor cortex (containing giant pyramidal Betz cells or upper motor neurons) is a key structure that receives abundant histaminergic projections. Upper motor neurons then send their axons down to the spinal cord through the corticospinal tract, where in humans, they synapse directly with anterior horn cells (the lower motor neurons), which, in turn, synapse directly with their target muscles. Importantly, HA directly excites rat spinal motor neurons through H1R and H2R by increasing their excitability, affecting membrane input resistance, and potentiating their repetitive firing behavior. In this way, the hypothalamus-spinal cord histaminergic fibers directly modulate final motor outputs and actively regulate ongoing motor execution and spinal motor reflexes . The elaboration of motor responses into different behavioral states, motor plans, voluntary movements, and motor skills actually relies on the ability of selecting appropriate motor outcomes according to specific environmental inputs, and HA with its receptors actively participates to this modulation of motor control. It is now important to recall and emphasize that both upper and lower motor neurons are the exclusive targets of neurodegeneration and motor impairment during ALS. In addition to motor circuits, histaminergic neurons provide a variety of different signaling mechanisms to the brain. 1 ..."
Reference 8
Handbook Chapter
Neurological Rehabilitation
Sudha Balakrishnan, Anthony B. Ward
Handbook of Clinical Neurology , 2013 pp 145-160
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"... Definition of spasticity Over time there have been many different attempts to define spasticity. The difficulty in defining spasticity reflects the complex features of the syndrome . Lance's definition of 1980 is still relevant and is widely accepted, but does not necessarily apply to the features of the UMN syndrome that require treatment . Lance states “Spasticity is a motor disorder characterized by a velocity-dependent increase in tonic stretch reflexes (muscle tone) with exaggerated tendon jerks, resulting from hyperexcitability of the stretch reflex, as one component of the upper motor neuron syndrome.” More recently, this definition was broadened to include other signs of UMN syndrome and described spasticity as “a motor disorder characterized by a velocity-dependent increase in tonic stretch reflexes that results from abnormal intra-spinal processing of primary afferent input” . Applying Lance's definition to patients in clinical settings has been difficult because UMN lesions produce an array of responses. The pattern depends on the age and onset of the lesion and its location and size. Patients with diffuse lesions produce, for instance, different characteristics to those with localized pathology and the speed of onset changes this again . The SPASM Consortium (A European Thematic Network to Develop Standardized Measures of Spasticity) has attempted to adapt the accepted definition to a more practical base and make it more relevant to clinical practice and to clinical research . 1 ..."
Reference 9
Book Chapter
The symptoms and signs of multiple sclerosis
McDonald I., Compston A.
McAlpine's Multiple Sclerosis , 2006 pp 287-346
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"... Spasticity Spasticity, defined as an inappropriate increase in velocity-dependent tonic stretch reflexes, is associated with sprouting of descending motor pathways to form new synaptic connections with spinal neurons, and with denervation hypersensitivity ( Figure 6.5 ). Together, these physiological and structural changes amplify the reactivity of motor segments to sensory input ( R.R. Young and Delwade 1981a ; 1981b ). Spasticity forms only one component of the upper motor neuron syndrome, since damage to descending motor pathways also results in slowness of movement and weakness, but it often dominates the clinical picture. In practice, the concept of spasticity has extended far beyond the resistance to passive movement by which it is usually estimated and includes a complex disorder of voluntary movement and physical changes in muscle and tendons. This, rather than weakness, no doubt accounts for much disability resulting from the upper motor neuron lesion in multiple sclerosis, particularly that affecting the lower limbs. Nevertheless, increased tone is an important factor, sometimes resulting in extreme resistance to passive flexion of the knees. Tone is often increased in the upper limb but the flexed spastic arm posture of hemiplegia, typical of stroke, is very uncommon in multiple sclerosis. In the progressive stage of the disease, increased extensor tone may be exaggerated, manifesting as extensor spasms. These are particularly liable to occur in bed at night or on waking in the morning. They may also be so severe as to eject the patient from a wheelchair. The legs are held rigidly extended, usually for several minutes. These spasms are inconvenient but rarely disabling and seldom painful. The mechanisms of the changes in muscle tone in multiple sclerosis have been reviewed by P. Brown (1994) . He points out that normal tone arises from a balance between inhibitory effects on local stretch reflexes (mediated by the dorsal reticulo-spinal tract) and facilitatory effects on extensor tone (mediated by the medial reticulo-spinal and to a lesser extent the vestibulo-spinal tracts). 1 ..."
Reference 10
Book Chapter
Motor System II: Corticofugal Systems and the Control of Movement
Mihailoff G.A., Haines D.E.
Fundamental Neuroscience for Basic and Clinical Applications , 2018 pp 360-376.e1
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"... General Features of Motor Deficits Lower Motor Neuron Signs Lower motor neurons are those cells whose axons synapse directly on skeletal muscle. When these neurons or their axons are damaged, the innervated muscles will show some combination of the following signs: (1) flaccid paralysis followed eventually by atrophy, (2) fibrillations or fasciculations (involuntary contractions of one motor unit or a group of motor units), (3) hypotonia (decreased muscle tone), and (4) weakening or absence of muscle stretch reflexes ( hyporeflexia, areflexia ). Upper Motor Neuron Signs The term upper motor neuron is commonly used in reference to corticospinal or corticonuclear cell bodies and their axons. Other neurons, such as rubrospinal or reticulospinal neurons, can also be included under the strict definition of this term. Corticospinal neurons are also called pyramidal neurons because their axons pass through the medullary pyramid. Therefore the terms upper motor neuron signs and pyramidal tract signs are often used synonymously. However, as described later in this chapter, these characteristic signs of “pyramidal tract damage” are in fact the result of injury to other descending motor systems in combination with damage to fibers of the pyramidal tract. For example, ischemic lesions of the internal capsule can potentially involve corticostriatal, corticothalamic, and corticoreticular fibers in addition to corticospinal axons because of the close proximity of these tracts to one another within the internal capsule. Damage to upper motor neurons results in muscles that (1) are initially weak and flaccid but (2) eventually become spastic, (3) exhibit increased muscle tone ( hypertonia ), seen as an increase in resistance to passive movement of an extremity, and (4) show an increase in muscle stretch reflexes ( hyperreflexia ), as may be seen in clonus. Upper motor neuron lesions usually affect groups of muscles, and certain pathologic reflexes and signs often appear. One of the most common is the inverted plantar reflex, also known as the Babinski sign. 1 ..."
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"... Upper Motor Neuron Signs The term upper motor neuron is commonly used in reference to corticospinal or corticonuclear cell bodies and their axons. Other neurons, such as rubrospinal or reticulospinal neurons, can also be included under the strict definition of this term. Corticospinal neurons are also called pyramidal neurons because their axons pass through the medullary pyramid. Therefore the terms upper motor neuron signs and pyramidal tract signs are often used synonymously. However, as described later in this chapter, these characteristic signs of “pyramidal tract damage” are in fact the result of injury to other descending motor systems in combination with damage to fibers of the pyramidal tract. For example, ischemic lesions of the internal capsule can potentially involve corticostriatal, corticothalamic, and corticoreticular fibers in addition to corticospinal axons because of the close proximity of these tracts to one another within the internal capsule. Damage to upper motor neurons results in muscles that (1) are initially weak and flaccid but (2) eventually become spastic, (3) exhibit increased muscle tone ( hypertonia ), seen as an increase in resistance to passive movement of an extremity, and (4) show an increase in muscle stretch reflexes ( hyperreflexia ), as may be seen in clonus. Upper motor neuron lesions usually affect groups of muscles, and certain pathologic reflexes and signs often appear. One of the most common is the inverted plantar reflex, also known as the Babinski sign. This involves dorsiflexion of the great toe in response to firm stroking of the lateral aspect of the sole of the foot with a blunt instrument. The response in the normal adult is plantar flexion of the great toe. In contrast, stroking the sole of the foot in a normal neonate may also result in a Babinski sign. This is related to the incomplete myelination of the corticospinal tract and will disappear as this tract matures. 1 ..."
Reference 11
Book Chapter
Neuromotor Control of Speech
Wanda G. Webb
Neurology for the Speech-Language Pathologist , 2017 pp 110-139
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"... This reaction, which identifies spastic hypertonicity, is analogous to the resistance felt when a knife blade is first opened, followed by the reduction of resistance when the blade is straightened. This reaction occurs more typically in extension rather than in flexion of the elbow. A short span of no tone is usually present, then a rapid buildup of tone and a sudden release as the joint is moved, just as with opening a clasp knife. Spasticity is also associated with exaggerated muscle stretch reflexes, resulting in hyperreflexia . Reflex action is tested at joints by putting stretch on tendons. This elicits the exaggerated muscle stretch reflex. Spastic paralysis, hypertonia, and hyperreflexia have most often been associated with pyramidal tract damage, particularly lesions of the corticospinal tract. However, the corticonuclear tracts are often also involved when a lesion interrupts the corticospinal tract, and signs of spasticity may be found in the midline speech muscles as well as in the distal limb muscles. Therefore the clinical signs of spasticity , or an upper motor neuron lesion, are of equal interest to the speech-language pathologist and the neurologist. Spastic speech muscles may be weak, slow, and limited in range or movement. Hypertonia may decrease muscle flexibility of the articulators and limit the ability to achieve a full range of motion of the speech muscles. Other Confirmatory Signs Several signs, in addition to a clinical demonstration of clasp knife spasticity, hypertonia, and hyperreflexia, are used by the neurologist to help verify the diagnosis of spasticity and localize the lesion to the pyramidal tract. The Babinski sign , or extensor plantar sign, in particular has been identified as an abnormal reflex sign that develops with corticospinal damage. The lesion causes the release of cortical inhibition. 1 ..."
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"... Another confirmatory sign of spasticity is clonus . Hyperactive muscle stretch reflexes associated with spasticity may show a sustained series of rhythmic beats or jerks when a neurologic examiner maintains one tendon of a muscle in extension. To test for clonus, the Achilles tendon in the ankle is often put under extension. If an upper motor neuron lesion is present, the ankle and the calf show sustained jerks. A few clonic jerks, called abortive clonus, are not clinically significant, but if the clonus is sustained over time, it is considered pathologic and an indicator of hyperreflexia. This sign is part of the clinical syndrome resulting from an upper motor neuron lesion. With bilateral upper motor neuron lesions, a characteristic dysarthria may be present. This motor speech disorder, called spastic dysarthria, is discussed in Chapter 8 . Lower Motor Neuron Signs and Symptoms If a lesion directly damages a cranial or peripheral nerve, or is in the cell bodies of the anterior horn cell in the spinal cord or the nuclei of cranial nerves in the brainstem, neural impulses will not be transmitted to the muscles. This condition is called denervation . The result is that the muscles innervated by the cranial or spinal nerve become soft and flabby from a loss of muscle tone. This is a lower motor neuron paralysis. The loss of muscle tone is called hypotonia . Hypotonia results in flaccid muscles, thus the term flaccid paralysis . Hypotonia may be an acquired condition, with disease or injury affecting the peripheral nervous system pathway, but it may also be a condition found at birth or developing shortly after birth. Infantile hypotonia may result from conditions such as chromosome disorders (e.g., Prader-Willi syndrome), genetic defects, spinal cord disorders, spinal muscular atrophy, muscular dystrophy, metabolic myopathies, or other problems that affect the peripheral pathways. 1 ..."
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"... Damage to the nervous system often releases early reflex behavior that has become inhibited by development of higher centers, so signs of damage at that point in time are signs of immaturity at an earlier time. Clinical neurologists believe that the extensor plantar sign usually reaches stability by age 2 years. Other signs, such as a persisting asymmetric tonic neck reflex and the Moro reflex, can be tested to suggest an upper motor neuron lesion in young children (see Chapter 11 ). Another confirmatory sign of spasticity is clonus . Hyperactive muscle stretch reflexes associated with spasticity may show a sustained series of rhythmic beats or jerks when a neurologic examiner maintains one tendon of a muscle in extension. To test for clonus, the Achilles tendon in the ankle is often put under extension. If an upper motor neuron lesion is present, the ankle and the calf show sustained jerks. A few clonic jerks, called abortive clonus, are not clinically significant, but if the clonus is sustained over time, it is considered pathologic and an indicator of hyperreflexia. This sign is part of the clinical syndrome resulting from an upper motor neuron lesion. With bilateral upper motor neuron lesions, a characteristic dysarthria may be present. This motor speech disorder, called spastic dysarthria, is discussed in Chapter 8 . 1 ..."
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"... With a corticospinal lesion, the great toe extends upward and the other toes fan as the foot withdraws slightly. Physicians test this response several times to convince themselves that the upturning great toe sign can be repeatedly and automatically elicited. Automatic repetition of a given response such as this defines it as a reflex. The presence of a repeatable abnormal reflex sharply increases the probability of predicting with accuracy the possible site or sites of a neurologic lesion, though it may not be present if the patient with a vascular lesion is tested early after onset of the insult. The Babinski sign is more reliable in adults than in infants and children. Normal infants are highly variable in display of the sign. The explanation usually given for this variability is that the immature nervous system and the damaged nervous system often show similar symptoms and signs. Damage to the nervous system often releases early reflex behavior that has become inhibited by development of higher centers, so signs of damage at that point in time are signs of immaturity at an earlier time. Clinical neurologists believe that the extensor plantar sign usually reaches stability by age 2 years. Other signs, such as a persisting asymmetric tonic neck reflex and the Moro reflex, can be tested to suggest an upper motor neuron lesion in young children (see Chapter 11 ). Another confirmatory sign of spasticity is clonus . Hyperactive muscle stretch reflexes associated with spasticity may show a sustained series of rhythmic beats or jerks when a neurologic examiner maintains one tendon of a muscle in extension. To test for clonus, the Achilles tendon in the ankle is often put under extension. If an upper motor neuron lesion is present, the ankle and the calf show sustained jerks. A few clonic jerks, called abortive clonus, are not clinically significant, but if the clonus is sustained over time, it is considered pathologic and an indicator of hyperreflexia. This sign is part of the clinical syndrome resulting from an upper motor neuron lesion. With bilateral upper motor neuron lesions, a characteristic dysarthria may be present. 1 ..."
Related quote(s)5 / 6
"... Spastic paralysis, hypertonia, and hyperreflexia have most often been associated with pyramidal tract damage, particularly lesions of the corticospinal tract. However, the corticonuclear tracts are often also involved when a lesion interrupts the corticospinal tract, and signs of spasticity may be found in the midline speech muscles as well as in the distal limb muscles. Therefore the clinical signs of spasticity , or an upper motor neuron lesion, are of equal interest to the speech-language pathologist and the neurologist. Spastic speech muscles may be weak, slow, and limited in range or movement. Hypertonia may decrease muscle flexibility of the articulators and limit the ability to achieve a full range of motion of the speech muscles. Other Confirmatory Signs Several signs, in addition to a clinical demonstration of clasp knife spasticity, hypertonia, and hyperreflexia, are used by the neurologist to help verify the diagnosis of spasticity and localize the lesion to the pyramidal tract. The Babinski sign , or extensor plantar sign, in particular has been identified as an abnormal reflex sign that develops with corticospinal damage. The lesion causes the release of cortical inhibition. The sign has achieved considerable status in the diagnosis of upper motor neuron lesions because it is a highly reliable abnormal reflex, is new behavior released by the presence of a lesion, and is clearly associated with a relatively specific lesion site—the cortex or the corticospinal tract. The speech-language pathologist is not directly interested in it because it does not involve the midline speech muscles, but its presence as a confirmation of an upper motor neuron lesion of the spastic type is important to all who manage neurologic patients. The Babinski sign is observed as a reflex toe sign. It is elicited by stimulating the sole of the foot with a strong scratching maneuver. The normal response to stimulation of the sole, or plantar portion of the foot, is a slight withdrawal of the foot and downward turning or curling under of the toes. 1 ..."
Related quote(s)6 / 6
"... The sign has achieved considerable status in the diagnosis of upper motor neuron lesions because it is a highly reliable abnormal reflex, is new behavior released by the presence of a lesion, and is clearly associated with a relatively specific lesion site—the cortex or the corticospinal tract. The speech-language pathologist is not directly interested in it because it does not involve the midline speech muscles, but its presence as a confirmation of an upper motor neuron lesion of the spastic type is important to all who manage neurologic patients. The Babinski sign is observed as a reflex toe sign. It is elicited by stimulating the sole of the foot with a strong scratching maneuver. The normal response to stimulation of the sole, or plantar portion of the foot, is a slight withdrawal of the foot and downward turning or curling under of the toes. With a corticospinal lesion, the great toe extends upward and the other toes fan as the foot withdraws slightly. Physicians test this response several times to convince themselves that the upturning great toe sign can be repeatedly and automatically elicited. Automatic repetition of a given response such as this defines it as a reflex. The presence of a repeatable abnormal reflex sharply increases the probability of predicting with accuracy the possible site or sites of a neurologic lesion, though it may not be present if the patient with a vascular lesion is tested early after onset of the insult. The Babinski sign is more reliable in adults than in infants and children. Normal infants are highly variable in display of the sign. The explanation usually given for this variability is that the immature nervous system and the damaged nervous system often show similar symptoms and signs. Damage to the nervous system often releases early reflex behavior that has become inhibited by development of higher centers, so signs of damage at that point in time are signs of immaturity at an earlier time. Clinical neurologists believe that the extensor plantar sign usually reaches stability by age 2 years. Other signs, such as a persisting asymmetric tonic neck reflex and the Moro reflex, can be tested to suggest an upper motor neuron lesion in young children (see Chapter 11 ). 1 ..."
Reference 12
Book Chapter
Diseases of Motor Neurons and Neuromuscular Junctions
Harald Sontheimer
Diseases of the Nervous System , 2015 pp 165-197
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"... As the neuron that innervates a motor unit becomes diseased, its axon goes through a period of unpredictability wherein aberrant action potentials cause twitching or fasciculation of a motor unit. Such twitches are commonly observed in the early stages of ALS, and their electrical activity can be measured through an EMG ( Figure 8 ) that reports voltage changes via fine needles inserted into the muscle. EMG serves as a useful tool to diagnose diseases affecting muscles, motor neurons, and the NMJ. As the axon dies back, the muscle becomes denervated and undergoes atrophy. Lower motor neurons also receive modulatory input from the basal ganglia and a number of cortical pathways that travel through the brain stem, including the corticorubrospinal, corticovestibulospinal, and corticoreticulospinal pathways. The modulatory neurons that give rise to these tracts also are designated as upper motor neurons. In the end lower motor neurons in the spinal cord receive both excitatory input from the upper motor neurons and a mixture of modulatory, excitatory, and inhibitory inputs from sensory muscle spindles and modulatory descending pathways. The modulation of the lower motor neurons explains why upper and lower motor neuron lesions present with opposite paralysis. When lower motor neurons or their axons cease to function, muscle tone is reduced and the muscle shows a complete absence of contraction, even when a reflex pathway is activated (areflexia). This is often called flaccid paralysis. By contrast, when only the upper motor neurons are lesioned, reflex pathways remain intact, and, indeed, show more brisk reflexes since the descending inhibitor modulatory pathways that normally dampen their activity are lost. Muscles have increased tone but are unable to receive excitatory commands from the upper motor neurons, causing spastic paralysis and hyperreflexia. Neurologists often take advantage of this distinction and use a common reflex, the plantar reflex, to determine whether a patient has upper or lower motor neuron signs. 1 ..."
Reference 13
Book Chapter
The Cranial Nerves
Russell J. Love, Wanda G. Webb
Neurology for the Speech-Language Pathologist , 1992 pp 112-136
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"... The patient with an upper motor neuron lesion will show complete involvement of the lips and neck muscles, some degree of involvement of the area around the eyes, and little difficulty with the forehead or frontalis muscle. It should be noted that the paralysis is on voluntary movement. The patient may be seen to have almost normal movement for emotionally initiated movements such as a true smile, but will be unable to lateralize the lips when asked to do so voluntarily. Since the facial nerve innervates the stapedius muscle, it may be paralyzed by a lesion. If this occurs, the patient may report that ordinary sounds seem uncomfortably loud. 1 ..."
Reference 14
Book Chapter
Cranial nerves
J.L. Wilkinson
Neuroanatomy for Medical Students , 1992 pp 110-149
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"... APPLIED ANATOMY Upper and lower motor neuron lesions differ markedly. Upper motor neuron, or supranuclear lesions are usually part of a hemiplegia due to a cerebrovascular accident; only the muscles of the lower half of the face on the opposite side are paralysed; the patient can still close the eyelids and wrinkle the forehead. In cortical or capsular lesions, voluntary and emotional facial activity may be dissociated: the patient may smile if amused but cannot ‘show his teeth’ on demand. By contrast, loss of emotional expression is very characteristic of Parkinson's disease. In lower motor neuron lesions (damage to the nucleus or nerve), the upper and lower facial muscles on the same side as the lesion are paralysed. Inability to close the eyelids may lead to corneal ulceration, especially if reflex lacrimation is also lost. The commonest infranuclear lesion is Bell's palsy , thought to be of viral origin, in which oedema compresses the nerve within its canal. The nerve may be affected by inflammation of the middle ear and mastoid air cells, or compressed in the cerebellopontine angle by an acoustic neuroma. A fracture of petrous temporal bone may involve facial nerve, vestibular apparatus and cochlea. In the newborn the nerve is relatively superficial and can be damaged by obstetric forceps. The extent of loss in a lower motor neuron lesion depends on the level of injury. If this is proximal to the geniculate ganglion, taste is lost in the anterior two-thirds of the tongue and secretion from the submandibular, sublingual and lacrimal glands is impaired. Hyperacusis (increased sound perception) is due to paralysis of stapedius. The prognosis is usually good in more distal lesions; in lesions proximal to geniculate ganglion, sensory regeneration is poor and salivatory parasympathetic fibres may be misdirected into the lacrimal gland, producing ‘crocodile tears’ in response to the smell or taste of food. 1 ..."
Reference 15
Book Chapter
Physical and Neurologic Examination
Blake Staub, Meng Huang, Paul J. Holman, Edward C. Benzel
Benzel's Spine Surgery, 2-Volume Set , 2017 pp 124-131.e1
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"... Reflex Examination The neurologic examination also includes an evaluation of the deep tendon (stretch) and superficial reflexes. Deep Tendon Reflexes The deep tendon reflexes are used to assess the integrity of a monosynaptic reflex arc at various levels of the cord. Table 12-4 depicts the common system for grading deep tendon reflexes. Hyperactive reflexes generally indicate an upper motor neuron lesion, and diminished or absent reflexes can be seen in lower motor neuron lesions. Metabolic abnormalities, such as hypothyroidism or hyperthyroidism, should always be excluded as an etiology for abnormal reflexes. Neuromuscular disorders and neuropathies may also present with abnormal reflexes. Reflexes are compared from one side to another, and reinforcement maneuvers that require isometric contraction of other muscle groups can be used to eliminate cortical modulation of the reflex arc. The Jendrassik maneuver can accentuate lower extremity reflexes and requires the patient to pull interlocked fingers apart while the reflex is tested. Asking the patient to clench the teeth or push down on the examination table with the thighs can accentuate upper extremity reflexes. One uncommonly practiced reflex is the finger jerk or finger-thumb reflex. Mediated by mainly the C8 nerve root, it is elicited with patient's palm upturned and the fingers half-flexed. The surgeon then holds the tops of the fingers with his or her own half-flexed fingers, which are then tapped. The patient's fingers will be felt to flex and, most strikingly, the free thumb will be seen to flex. 10 Although the commonly tested reflexes are truly mediated by multiple nerve roots, Tables 12-2 and 12-3 outline the dominant nerve roots involved. Superficial Reflexes The superficial reflexes are mediated by the cerebral cortex with the afferent limb being supplied by cutaneous stimulation. The absence of a normal cutaneous reflex may signal an underlying upper motor neuron lesion. 1 ..."
Reference 16
Reference works Chapter
Lower Motor Neuron Lesions
Ramahi A.A., Katirji B., Devereaux M.
Encyclopedia of the Neurological Sciences , 2014 pp 918-922
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"... Tone Muscle tone is assessed by examining its response to passive stretch. Lower motor neuron lesions reduce muscle tone, whereas upper motor neuron lesions increase muscle tone resulting in spasticity as seen in pyramidal lesions, or rigidity as observed in extrapyramidal lesions. Regulation of muscle tone is normally mediated by the reticulospinal fibers accompanying the pyramidal tract that exert inhibitory effects on the stretch reflex. Flaccidity, or hypotonia, is a typical feature of lower motor neuron damage, and in the extreme form of total flaccid paralysis, all resistance to passive muscle stretch is lost and the limbs become flail-like. Hypotonia can also be seen in cerebral or spinal shock resulting from acute extensive brain, spinal cord, or cerebellar lesions. It is also seen in peripheral nerve or root lesions affecting sensory afferent pathways. In clinical practice, observation of flaccidity is of limited diagnostic usefulness compared to more important physical signs such as wasting or tendon reflex loss. 1 ..."
Reference 17
Book Chapter
Cranial Nerves XI (Spinal Accessory) and XII (Hypoglossal)
Barboi A.C.
Textbook of Clinical Neurology , 2007 pp 231-241
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"... MOTOR/REFLEXES/CEREBELLAR/GAIT When examining a patient with a suspected lesion of CN XI or XII, it is important to perform a thorough motor examination. The examiner should note any primitive reflexes (e.g., suck, snout, and glabellar) or paratonic rigidity (gegenhalten). These signs, along with pseudobulbar palsy, may be indicative of bilateral frontal lobe lesions. In addition, weakness in other ipsilateral muscles in the face, arm, or leg, accompanied by hypertonia, hyper‐reflexia, or extensor plantar responses, is suggestive of an upper motor neuron lesion, 31 whereas weakness with atrophy, loss of reflexes, or fasciculations are signs of either a lower motor neuron lesion or neuronopathy. The supranuclear fibers for the sternal head of the SCM muscle probably decussate twice—once in the midbrain or pons and a second time in the medulla or cervical spinal cord. 23 Hence, a lesion in the right side of the pons may result in (1) left‐sided weakness (involvement of corticospinal tract fibers before their decussation in the medulla) and (2) weakness of the left SCM muscle and deviation of the head to the left (i.e., the same side as the hemiparesis) because of involvement of the supranuclear input to the left SCM muscle after the first but before the second decussation. In contrast, a lesion in the right cerebral hemisphere or right internal capsule or high in the right midbrain, before the initial decussation of the supranuclear fibers to the SCM, may result in (1) left‐sided weakness due to involvement of the corticospinal tract fibers before their decussation and (2) weakness of the right SCM muscle and deviation of the head to the right (i.e., the side opposite the hemiparesis). 13 When dysfunction of CN XI or XII is present, it is also important to perform a thorough examination of coordination and gait. 1 ..."
Reference 18
Book Chapter
Electrodiagnostic Findings in Neuromuscular Disorders
Bashar Katirji
Electromyography in Clinical Practice , 2007 pp 49-58
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"... UPPER MOTOR NEURON LESIONS Patients with upper motor neuron lesions have normal NCSs and needle EMG including normal insertional activity, no spontaneous activity at rest, and normal MUAP morphology. The only abnormality is a reduced interference pattern with poor activation of MUAPs (slow rate of motor unit discharge). Recruitment, measured by either recruitment frequency or ratio, is normal. Hysterical weakness or poor effort produces a similar pattern, except that motor unit firing may be irregular. 1 ..."
Reference 19
Review article
Electrodiagnostic criteria for diagnosis of ALS
de Carvalho M., Dengler R., Eisen A., England J.D., Kaji R., Kimura J., Mills K., Mitsumoto H., Nodera H., Shefner J., Swash M.
Clinical Neurophysiology , 2008 pp 497-503
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"... Electrophysiological evaluation of upper motor neuron (UMN) dysfunction The diagnosis of ALS requires demonstration of a combination of upper and lower motor neuron abnormalities whether by clinical or electrophysiological criteria, as set out in the Airlie House consensus recommendations ( Table 1 ). Although it has been hoped that TMS might prove a sensitive test for upper motor neuron involvement in ALS this has not been confirmed in the published studies, at least not for patients with clinically missing or equivocal upper motor neuron signs, and expert opinions are equivocal. Central motor conduction time (CMCT), however, is frequently prolonged to muscles of at least one extremity . More complex stimulation paradigms have not been helpful in diagnosis of ALS except for short-interval, paired-pulse stimulation, which revealed reduced intracortical inhibition . However, this technique is not practical and therefore not widely used, and experience in early ALS, the crucial time for diagnosis, is limited. A relatively new TMS technique, the triple stimulation method, has shown a high sensitivity to detect upper motor neuron dysfunction in ALS and yielded abnormal results also in early stages of the disease in two laboratories . Confirmation of these results from wider application in other laboratories is required. Developments in magnetic resonance imaging of brain tracts may yet prove helpful, although this has not so far been demonstrated, and the methodology is complex and expensive. 9.1 Transcranial magnetic stimulation (TMS) TMS requires that the target muscle have a detectable CMAP. The following features suggest abnormality of upper motor neuron function: 1. Increased central motor conduction time (CMCT). 2. Increased absolute latency to a tested muscle, provided that distal motor conduction slowing can be excluded. 3. In patients with bulbar onset disease, an absent response to TMS in a limb is supportive of upper motor neuron lesion. 4. The triple stimulation technique has proven sensitive in detecting impairment of upper motor neuron function in the early stages of the disease as well as later in the disease course . However, experience with this promising test is so far relatively limited. 1 ..."
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"... Transcranial magnetic stimulation (TMS) TMS requires that the target muscle have a detectable CMAP. The following features suggest abnormality of upper motor neuron function: 1. Increased central motor conduction time (CMCT). 2. Increased absolute latency to a tested muscle, provided that distal motor conduction slowing can be excluded. 3. In patients with bulbar onset disease, an absent response to TMS in a limb is supportive of upper motor neuron lesion. 4. The triple stimulation technique has proven sensitive in detecting impairment of upper motor neuron function in the early stages of the disease as well as later in the disease course . However, experience with this promising test is so far relatively limited. 1 ..."
Reference 20
Review article
Progressive Muscular Atrophy
Liewluck T., Saperstein D.S.
Neurologic Clinics , 2015 pp 761-773
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"... Imaging Biomarkers of Upper Motor Neuron Involvement The data on radiographic biomarkers of UMN lesions are inconclusive. Two different techniques have been applied to identify UMN involvement in patients with MND: diffusion tensor MRI and magnetic resonance spectroscopy (MRS). Diffusion tensor MRI studies of the structural integrity of neuronal fibers showed modest reduction of fractional anisotropy, suggesting axonal degeneration and myelin breakdown in the corticospinal tract in all 12 patients with PMA, 7 and may predate the clinical evidence of UMN lesions in some patients. 3 However, another study showed no significant abnormal fractional anisotropy in 8 patients with PMA. 45 White matter involvement in patients with PMA may be involved beyond the corticospinal tract. Voxel-based analysis identified fractional anisotropy reduction in nonmotor white matter including the prefrontal area. 7 MRS showed reduced N -acetyl aspartate (NAA) concentration and ratio of NAA to creatine, markers of neuronal health, in the primary motor cortex, suggesting UMN involvement, in about 60% of patients with PMA. 4,15 The corticospinal tract degeneration was pathologically confirmed when autopsy was available. 4 However, normal results on MRS does not exclude UMN pathology. 4 Another MRS study showed only modest, nonsignificant changes in patients with PMA, which could be due to the slightly different imaging technique. 5 1 ..."
Reference 21
Book Chapter
The differential diagnosis of multiple sclerosis
Miller D., Compston A.
McAlpine's Multiple Sclerosis , 2006 pp 389-437
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"... A few cases that evolve to typical motor neuron disease nevertheless have an initial phase in which the manifestations exclusively affect upper motor neurons. In about 70% of patients, MRI reveals symmetrical high signal confined to the corticospinal tracts in the centrum semiovale, internal capsule, cerebral peduncles, pons and upper cervical cord ( Thorpe et al 1996c) . Such findings are unlike those seen in multiple sclerosis and give a clue to the correct diagnosis. 1 ..."
Reference 22
Book Chapter
Motor Stimulation
Alaa Abd-Elsayed, Andrea M. Trescot, Brent Earls, Eellan Sivanesan
Peripheral Nerve Stimulation , 2023 pp 266-273
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"... 32 However, it is now largely accepted that the recovery is due to plastic changes in the sensorimotor cortex due to unmasking of latent horizontal connections, activation of silent synapses, modulation of activity-dependent synaptic plasticity, and generalized changes in the excitability of postsynaptic neurons. 33 Effects on cortical perfusion were recently studied in combination with FES in chronic stroke patients. This study found that the sensory motor integration due to FES may facilitate perfusion of the ipsilesional sensory motor cortex and result in functional improvement of hemiparetic upper extremity. After FES treatment, only patients who showed improved perfusion to the ipsilesional sensory motor cortex perfusion showed functional improvement of upper extremity. 34 This further supports the idea that recovery of paretic limbs due to upper motor neuron lesions is largely the result of reorganization of the sensorimotor cortex, but that FES can help to facilitate this process. This neuroplasticity may also drive the alleviation of pain symptoms experienced by patients with motor stimulation. 35,36 1 ..."
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"... Peripheral and central mechanisms of benefit The rationale for use of motor stimulation is based on the physiological changes at muscle and spinal levels that occur over time. For example, repeated muscle contraction from peripheral nerve stimulation (PNS) increases the oxidative capacity of muscle, increases the number of microcapillaries, and leads to transformation of muscle fiber types. 30 A small but expanding body of literature demonstrates that FES induces rapid plastic change in the sensorimotor regions of the cortex. 31 This suggests that repetitive movements that are goal oriented and functionally relevant can facilitate relearning following stroke or brain injury. 4 Motor relearning is defined as “the recovery of previously learned motor skills that have been lost following localized damage to the central nervous system.” 31 It was hypothesized that the use of FES, which generates both an orthodromic and an antidromic impulse, may modulate activity in anterior horn cells via antidromic impulses. These impulses were thought to strengthen Hebbian-type synapses between pyramidal tract cells and the anterior horn cells, contributing to restoration of voluntary control. 32 However, it is now largely accepted that the recovery is due to plastic changes in the sensorimotor cortex due to unmasking of latent horizontal connections, activation of silent synapses, modulation of activity-dependent synaptic plasticity, and generalized changes in the excitability of postsynaptic neurons. 33 Effects on cortical perfusion were recently studied in combination with FES in chronic stroke patients. This study found that the sensory motor integration due to FES may facilitate perfusion of the ipsilesional sensory motor cortex and result in functional improvement of hemiparetic upper extremity. After FES treatment, only patients who showed improved perfusion to the ipsilesional sensory motor cortex perfusion showed functional improvement of upper extremity. 34 This further supports the idea that recovery of paretic limbs due to upper motor neuron lesions is largely the result of reorganization of the sensorimotor cortex, but that FES can help to facilitate this process. This neuroplasticity may also drive the alleviation of pain symptoms experienced by patients with motor stimulation. 35,36 1 ..."
Reference 23
Book Chapter
The Physiologic Aspects and Clinical Application of Functional Electrical Stimulation in Rehabilitation
KRISTJAN T. RAGNARSSON
The Physiological Basis of Rehabilitation Medicine , 1994 pp 573-597
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"... Publisher Summary This chapter discusses the physiologic aspects and clinical application of functional electrical stimulation (FES) in rehabilitation. FES may be defined as the application of electrical currents to neural tissue for restoring a degree of control over abnormal or absent body functions. Electricity has been applied either experimentally or clinically for many purposes: (1) to improve hearing and sight, (2) to prevent bladder and bowel incontinence, (3) to control evacuation, (4) to regulate heart rhythm, (5) to reduce spasticity, (6) to allow ventilator-free breathing, (7) to correct scoliosis, and (8) to usefully move paralyzed limbs. The chapter discusses one specific form of FES that is usually referred to by scientists as functional neuromuscular stimulation (FNS). FNS restricts the electrical stimulation to the neuromuscular system for the purpose of controlling skeletal muscle contractions. It may be applied successfully for different forms of paralytic conditions caused by upper motor neuron lesions, for example, stroke, traumatic brain injury, and cerebral palsy. The chapter discusses its application after spinal cord injury. 2 ..."
Reference 24
Reference works Chapter
Volume 2
Theresa L. Bender Pape, Nikki M. Barrington, Elise K. Webber, Grace E. Stutzmann
Encyclopedia of the Human Brain , 2025 pp 730-757
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"... FES CPGs are, however, also in place for patients with lower limb impairments due to upper motor neuron lesions (e.g., Polio, Guillain-Barre, etc.). Specifically, these guidelines recommend using FES if the individual has enough passive movement of their ankle to allow for walking but are unable to wear a splint and/or if the individual has trouble controlling lower limb movement or keeping balance when walking . CPGs recommending FES as a treatment option are also available for participants with acute or subacute cervical spinal cord injury to improve hand and upper extremity function . Despite some evidence suggesting therapeutic benefits, there are currently no CPGs for use of FES for MS symptoms or for the treatment of gait abnormalities and spasticity associated with CP. However, a clinical decision-making tool was recently developed to help clinicians use FES for motor rehabilitation in conditions with no available CPGs . The collective evidence presented here indicates that incorporating FES in motor rehabilitation is clinically beneficial, but that further research is needed to delineate optimal treatment protocols (e.g., FES alone or combined with physical retraining exercises). Furthermore, additional research is needed to elucidate optimal therapeutic windows post-injury as well as the optimal number of FES treatment sessions, aiding in the creation of important FES CPGs . Transcranial magnetic stimulation The recovery of motor skills, despite the presence of various neurological conditions, is possible with transcranial magnetic stimulation (TMS), as these devices can be used to exogenously engage or acquire long-term potentiation (LTP) and long-term depression (LTD) like mechanisms. Specifically, injury-based animal studies have demonstrated that TMS-induced neuronal activity results in a decrease of pro-apoptotic factors and in an upregulation of brain derived neurotrophic factors (BDNF) . These results hold therapeutic promise as neurotrophic signaling regulates plasticity in an activity-dependent manner . 2 ..."
Reference 25
Book Chapter
Viral Vector Axonal Uptake and Retrograde Transport: Mechanisms and Applications
Teng Q., Federici T., Boulis N.M.
Gene Therapy of the Central Nervous System , 2006 pp 253-271
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"... E. Spasticity Spasticity is a motor dysfunction characterized by increased muscular resistance to movement with increased velocity of movement. Severe spasticity is often accompanied by spasms, which are involuntary muscle contractions that are often severely painful. Spasticity commonly occurs in the presence of upper motor neuron lesions or disconnection of the synaptic connection between the brain and spinal motor neurons. Spasticity is a disabling symptom of spinal cord injury, multiple sclerosis, and cerebral palsy. Spasms and spasticity are thought to result from a dysfunction of inhibitory signals within the spinal cord that depend on descending motor pathways. Treatments including anti-spasticity drugs, botulinum toxin, electrical stimulation, and surgery are being currently used to reduce spasticity . In cases where patients do not respond to pharmacological and surgical interventions, gene therapy might be warranted, using a vector capable of neuronal tropism and targeted neuromodulation, in order to achieve focused synaptic inhibition. Combining the retrograde transport property of the viral vectors mentioned above with proteins that modulate neurotransmitter release , injection of these viral vectors into the spastic muscles would specifically disrupt acetycholine release at the neuromuscular junction, relieving the symptoms. 2 ..."
Reference 26
Book Chapter
Neurocontrol of Chronic Upper Motor Neuron Syndromes
Milan R. Dimitrijevic
Electromyography in CNS Disorders , 1984 pp 111-128
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"... Publisher Summary There are numerous structures within the cortex, subcortex, brain stem, and spinal cord, and there are complex interconnections among them via corticosubcortical, corticobulbar, subcorticobulbar, corticospinal, and bulbospinal descending pathways. All these morphologic and functional complexities of the central nervous system are involved in motor control and are referred to as upper motor neuron. The upper motor neuron syndrome is also a complex neurologic entity. This chapter focuses on the upper motor neuron syndromes owing to chronic effects of spinal cord injury. It presents the three functions of the upper motor neuron: (1) the degree of preservation or deterioration of volitional activity, (2) the effects of remote muscle contraction on paretic or paralyzed muscles, and (3) characteristic features of stretch and withdrawal reflexes. Upper motor neuron dysfunctions in patients with chronic neurologic disorders usually are due to lesions of pyramidal and extrapyramidal tract fibers. Such an upper motor neuron lesion can result from the destruction of descending motor pathways in the cerebral cortex, the internal capsule, the cerebral peduncles, the brain stem, or the spinal cord. Clinical features of this upper motor neuron syndrome represent different degrees of impairment of neurocontrol of volitional activity, muscle tone, and brain influence on segmental reflexes. 2 ..."
Reference 27
Book Chapter
Clinical experience and recent advances in the management of gait disorders with botulinum neurotoxin
Esquenazi A., Mayer N.H.
Botulinum Toxin , 2009 pp 192-203
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"... Lance published this frequently cited definition for the phenomenon: “a motor disorder characterized by a velocity dependent increase in tonic stretch reflexes (muscle tone) with exaggerated tendon jerks, resulting from hyperexcitability of the stretch reflex, as one component of the upper motoneuron syndrome.” 5 Signs of spasticity are useful as a diagnostic indicator pointing to the presence of an upper motor neuron lesion, but in general, rehabilitation clinicians are more preoccupied with addressing the functional problems of their patients linked to the consequences of an upper motor neuron lesion. Common terminology, which reflects their clinical practice emphasis, includes spastic gait or spastic equinovarus. For these clinicians, evaluation and treatment of spasticity takes on broader dimensions, reflecting their interest in clinical patterns of motor dysfunction that produce functional impairment and can result in disability. The clinician must be able to see spastic phenomena within the larger context of impaired motor control in order to identify appropriate treatment methods for a patient's functional problems resulting from a central nervous system injury. The National Center for Medical Rehabilitation Research (NCMRR) has suggested that new methods of treatment should be evaluated for effects on multiple dimensions of the disabling process, including not only the dimension of impairment (as exemplified by Ashworth scale measurements), but also other dimensions such as functional limitations and disabilities (i.e., difficulty fulfilling role functions). 6 Lower extremity posturing that affects gait and transfer dysfunctions represents a significant spectrum of functional limitations that are attributable to the positive signs of the UMNS. In a broader sense and from our perspective, evaluation of spasticity focuses on three issues: (1) identifying the clinical pattern of motor dysfunction and its source; (2) identifying the patient's ability to control muscles involved in the clinical pattern; and (3) the differential role of muscle stiffness and contracture as it relates to a functional problem. 2 ..."
References (27)
Reference 1
Reference works Chapter
Upper Motor Neuron Lesions
Devereaux M., Katirji B., Daroff R.
Encyclopedia of the Neurological Sciences , 2014 pp 584-586
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"... Introduction An upper motor neuron (UMN) lesion is the result of numerous disorders that can impact the central nervous system. UMNs make up the corticobulbar/corticospinal/pyramidal system. Although the anatomy of the system is described in greater detail in other entries in this Encyclopedia, briefly stated, the origin of the UMN is the Betz cell in the prerolandic cortical motor strip. The UMN descends through the subcortical white matter into the posterior limb of the internal capsule. From there it extends into the cerebral peduncle of the midbrain, basis pontis of the pons, and pyramid of the medulla oblongata, decussating at the level of the inferior medulla to the contralateral corticospinal tract in the dorsolateral cervical spinal cord. Within the brainstem, corticobulbar fibers leave the tract to innervate cranial nerve motor nuclei. Ultimately, the UMN synapses with the anterior horn cells of the lower motor neuron (LMN) throughout the spinal cord. Therefore, the UMN coupled with LMN comprises the classic two-neuron system responsible for voluntary activation of somatic musculature. Coordinated movement is much more complicated than this, with the extrapyramidal, vestibular, and sensory systems interacting with these neurons, but this two-neuron system is the basic substrate for voluntary movement. Assessment of the integrity of the UMN is a fundamental component of the neurological history and examination. The longest UMN in the corticospinal tract descends from the motor cortex to the anterior horn cell in the conus medullaris of the spinal cord and passes with, and by, a number of other systems and structures that control many nervous system functions. Clinical ‘navigation’ through this system identifying neurological abnormalities in the corticospinal/corticobulbar system and, at the same time, nearby structures can often permit accurate localization of the level of the UMN lesion. ..."
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"... Abstract Lesions of the upper motor neuron (UMN) are common features of neurological disorders affecting the central nervous system (CNS). Primary features of a UMN lesion include muscle weakness, spasticity, clonus, hyperreflexia, and pathological reflexes, such as the Babinski sign. Neurological dysfunction accompanying symptoms and signs of a UMN lesion can often lead to a relatively precise localization of the lesion within the CNS. This, coupled with the neurological history, can also often lead to the diagnosis of the neuropathology of the lesion in question. ..."
Reference 2
Review article
Intraoperative motor evoked potential monitoring - A position statement by the American Society of Neurophysiological Monitoring
MacDonald D.B., Skinner S., Shils J., Yingling C.
Clinical Neurophysiology , 2013 pp 2291-2316
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"... Upper motor neuron system The corticospinal and indirect pathways together comprise the upper motor neuron (UMN) system. Pathology generally involves both, so that UMN lesions typically cause a mixture of corticospinal deficits along with antigravity weakness, spasticity and hyperreflexia attributed to indirect pathway damage. ..."
Reference 3
Book Chapter
Disordered muscle tone
JAMES W. LANCE, JAMES G. McLEOD
A Physiological Approach to Clinical Neurology , 1981 pp 128-153
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"... If the function of the latter is withdrawn by damage to the ‘upper motor neurone’, the stretch reflex arc is disinhibited and therefore hyperexcitable, giving rise to the state known clinically as spasticity. The progressive changes observed with time after an acute lesion may well depend on plastic changes in neural structure such as the sprouting of terminals and formation of new synaptic connections. At this stage, it would be advisable to define ‘spasticity’ or at least to consider the various components of the syndrome called spasticity. Landau 31 points out that the term spasticity may be applied to increase in activity of stretch reflexes, release of flexor reflexes and defect in motor performance, all of which may be found in the ‘Upper Motor Neurone Syndrome’. The term should not be applied to states of dystonia and rigidity even if, as in most cases of cerebral palsy, they are associated with some features of an upper motor neurone lesion. Spasticity may be defined as disinhibition of the stretch reflex, resulting in velocity-dependent increase in tonic stretch reflexes and increased tendon jerks. Other features of the upper motor neurone syndrome are commonly associated with spasticity. If the upper motor neurone remains untarnished but disease prevents the normal inhibitory inflow from the basal ganglia to cortex on the one hand and to the reticular formation on the other, an unrestrained outflow down pyramidal, parapyramidal and extrapyramidal fibres bombards the segmental mechanisms of the spinal cord indiscriminately. This results in activity persisting beyond that necessary for voluntary movement, giving rise to movement disorders such as athetosis, dystonia and Parkinsonian rigidity. A patient with Parkinson's disease is quite unable to modulate the later components (M 2 and M 3 ) of the stretch reflex, which are increased to as much as five times those of normal subjects 33 . A Parkinsonian patient is in a state of continuous alpha-gamma coactivation which fluctuates in level with mental and physical exertion, resembling a tense person who is unable to relax completely of his own volition. ..."
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"... Alpha-gamma excitability is influenced from within the spinal cord by internuncial cells and propriospinal fibres which have facilitatory or inhibitory effects (designated + and – in Figure 6.3 ). Spinal mechanisms in turn are under brainstem control from the vestibulospinal tract ( Figures 5.1 , 5.2 , 6.3 ) and facilitatory reticulospinal pathways ( Figure 5.2 ), which augment the excitability of motor neurones, particularly those controlling extensor muscles in the lower limbs of man. These brainstem tracts are balanced by the inhibitory effect of the corticoreticulospinal tract ( Figure 6.3 ). If the function of the latter is withdrawn by damage to the ‘upper motor neurone’, the stretch reflex arc is disinhibited and therefore hyperexcitable, giving rise to the state known clinically as spasticity. The progressive changes observed with time after an acute lesion may well depend on plastic changes in neural structure such as the sprouting of terminals and formation of new synaptic connections. At this stage, it would be advisable to define ‘spasticity’ or at least to consider the various components of the syndrome called spasticity. Landau 31 points out that the term spasticity may be applied to increase in activity of stretch reflexes, release of flexor reflexes and defect in motor performance, all of which may be found in the ‘Upper Motor Neurone Syndrome’. The term should not be applied to states of dystonia and rigidity even if, as in most cases of cerebral palsy, they are associated with some features of an upper motor neurone lesion. Spasticity may be defined as disinhibition of the stretch reflex, resulting in velocity-dependent increase in tonic stretch reflexes and increased tendon jerks. Other features of the upper motor neurone syndrome are commonly associated with spasticity. 1 ..."
Reference 4
Book Chapter
Strength and Reflexes
Hammerstad J.P.
Textbook of Clinical Neurology , 2007 pp 243-287
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"... Most of the axons of the corticospinal tract come from neurons in the precentral gyrus (Brodmann area 4), 20 although significant contributions come from many other areas of the cortex, especially premotor area 6 and supplementary motor area. This is the only motor tract that passes directly from the cerebral cortex to the spinal cord without making any intermediate synaptic connections. The cytoarchitecture of the primary motor area, supplementary motor area, and premotor cortex is distinct from that of the adjacent sensory and prefrontal regions. Layer 4, the major afferent layer for the sensory cortex, is absent in the motor areas, and hence these motor regions are often termed agranular. Layer 5 in the primary motor cortex contains the distinctive giant pyramidal neurons known as Betz cells . The axons of these cells become the corticospinal or pyramidal tract and represent one of several descending influences on the motor neurons of the brain stem and spinal cord. The remaining cells mostly come from the supplementary motor and premotor cortex. Most of these axons terminate on interneurons in the internuncial pool of the ventral gray of the spinal cord. Axons that connect monosynaptically with the lower motor neurons are involved in control of the distal muscles of the extremities, especially the intrinsic muscles of the hand that control finger movements. 3 Upper motor neurons in the cerebral cortex also send axons to the motor nuclei of the cranial nerves and form the corticobulbar tract. The motor neurons in the primary motor area have a somatotopic organization. The homunculus was mapped by Penfield and colleagues 21 by electrically stimulating the surface of the cortex during epilepsy surgery in humans. Stimulation of most of the lateral surface of the primary motor area produced movement in the mouth, tongue, face, and hand muscles, which is consistent with the importance of fine voluntary control of these structures ( Fig. 15‐17 ). ..."
Reference 5
Book Chapter
Motor Systems
David L. Felten, M. Kerry O'Banion, Mary Summo Maida
Netter's Atlas of Neuroscience , 2016 pp 391-420
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"... Corticospinal Tract The motor portion of the corticospinal tract (CST) originates from neurons of many sizes, mainly from the primary motor cortex (area 4) and the supplemental and premotor cortices (area 6). The primary sensory cortex (areas 3, 1, 2) contributes axons into the CST, but these axons terminate mainly in secondary sensory nuclei to regulate the processing of incoming lemniscal sensory information. The CST travels through the posterior limb of the internal capsule, the middle region of the cerebral peduncle, numerous fascicles of axons in the basis pontis, and the medullary pyramid on the ipsilateral side. Most of the CST axons (approximately 80%, but variable from individual to individual) cross the midline in the decussation of the pyramids at the medullary-spinal cord junction. These crossed fibers descend in the lateral CST in the lateral funiculus of the spinal cord and synapse on alpha and gamma LMNs, both directly and indirectly through interneurons. CST axons that do not decussate continue as the anterior CST in the anterior funiculus of the spinal cord and then decussate at the appropriate level through the anterior white commissure to terminate directly and indirectly on alpha and gamma LMNs contralateral to the cortical cells of origin. Only a very small portion of the motor connections of the corticospinal tract terminate on LMNs on the ipsilateral side of the spinal cord. Clinical Point The motor portion of the CST arises mainly from neurons in the primary motor cortex (area 4) and the supplemental and premotor cortices (area 6). The primary sensory cortex and superior parietal lobule contribute corticospinal axons (corticonuclear fibers) to secondary sensory nuclei in the lower brain stem and spinal cord. Approximately 80% of the CST axons cross in the decussation of the pyramids and terminate directly and indirectly with alpha and gamma LMNs that control movements of the distal extremities, especially the hands and fingers. ..."
Reference 6
Book Chapter
Introduction to human neuromusculoskeletal systems
Jorge Garza-Ulloa
Applied Biomechatronics using Mathematical Models , 2018 pp 53-118
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"... It is indicated a pyramidal using lateral corticospinal tracts; the pathway starting at the cerebral cortex showing the location of the upper motor neural (LMN) cell body, pass through the midbrain, pons, in the medulla oblongata there is the decussation (it is a phenomenon in which the nerve fibers found in one lateral portion of the organ cross another) of UMN axon in the pyramid, pass through a lateral cervical spinal cord, the synapse of UMN and LMN occurs in anterior horn of the lumbar spinal cord, where the LMN exits via anterior root to the skeletal muscle in the limb. Subconscious tracts or extrapyramidal system is part of the motor system causing involuntary movements. The extrapyramidal system represents part of the motor pathway system that has some synapses with the brainstem, which is in contrast to those of the pyramidal system. The extrapyramidal tract is distinguishable from the pyramidal system as tracts do not run within the pyramids of the medulla oblongata and instead run outside. This extrapyramidal pathway contains various multisynaptic pathways that rely within several nuclei in the brain, and these nuclei are dispersed from the telencephalon to the medulla oblongata. This system includes parts of the following tracts: rubrospinal, vestibulospinal , reticulospinal , and tectospinal , where: Rubrospinal tracts send information to the flexor and extensor muscles. Vestibulospinal tracts send information from the inner ear to monitor position of head. The vestibular nuclei respond by altering muscle tone, neck muscle contraction, and limbs for posture and balance. Reticulospinal tract s send information to cause eye movements and activate respiratory muscles. Tectospinal tracts send information to the head, neck, and upper limbs in response to bright and sudden movements and loud noises. The tectum area consists of superior colliculi (receives visual information) and inferior colliculi (receives auditory information). 1 ..."
Reference 7
Review article
Histamine beyond its effects on allergy: Potential therapeutic benefits for the treatment of Amyotrophic Lateral Sclerosis (ALS)
Volonte C., Apolloni S., Sabatelli M.
Pharmacology & Therapeutics , 2019 pp 120-131
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"... Activation of H1R causes concentration-dependent release of [Ca2+] from internal stores and concentration-dependent increase in glutamate release. H2R activation increases cyclic adenosine monophosphate levels and phosphorylation of transcription factor cAMP response-element binding protein, thus emphasizing a role for HA in neuron-glia communication . HA neurons are found merely in the tuberomamillary nucleus of the posterior hypothalamus, from where they project their axons all over the CNS and spinal cord in rats, mice, and humans . Two main bundles of axons from the tuberomamillary nucleus send a large arborization, for instance, to the cortex, and most cortical regions receive a moderately dense and sparse histaminergic input. Among these regions, the primary motor cortex (containing giant pyramidal Betz cells or upper motor neurons) is a key structure that receives abundant histaminergic projections. Upper motor neurons then send their axons down to the spinal cord through the corticospinal tract, where in humans, they synapse directly with anterior horn cells (the lower motor neurons), which, in turn, synapse directly with their target muscles. Importantly, HA directly excites rat spinal motor neurons through H1R and H2R by increasing their excitability, affecting membrane input resistance, and potentiating their repetitive firing behavior. In this way, the hypothalamus-spinal cord histaminergic fibers directly modulate final motor outputs and actively regulate ongoing motor execution and spinal motor reflexes . The elaboration of motor responses into different behavioral states, motor plans, voluntary movements, and motor skills actually relies on the ability of selecting appropriate motor outcomes according to specific environmental inputs, and HA with its receptors actively participates to this modulation of motor control. It is now important to recall and emphasize that both upper and lower motor neurons are the exclusive targets of neurodegeneration and motor impairment during ALS. In addition to motor circuits, histaminergic neurons provide a variety of different signaling mechanisms to the brain. 1 ..."
Reference 8
Handbook Chapter
Neurological Rehabilitation
Sudha Balakrishnan, Anthony B. Ward
Handbook of Clinical Neurology , 2013 pp 145-160
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"... Definition of spasticity Over time there have been many different attempts to define spasticity. The difficulty in defining spasticity reflects the complex features of the syndrome . Lance's definition of 1980 is still relevant and is widely accepted, but does not necessarily apply to the features of the UMN syndrome that require treatment . Lance states “Spasticity is a motor disorder characterized by a velocity-dependent increase in tonic stretch reflexes (muscle tone) with exaggerated tendon jerks, resulting from hyperexcitability of the stretch reflex, as one component of the upper motor neuron syndrome.” More recently, this definition was broadened to include other signs of UMN syndrome and described spasticity as “a motor disorder characterized by a velocity-dependent increase in tonic stretch reflexes that results from abnormal intra-spinal processing of primary afferent input” . Applying Lance's definition to patients in clinical settings has been difficult because UMN lesions produce an array of responses. The pattern depends on the age and onset of the lesion and its location and size. Patients with diffuse lesions produce, for instance, different characteristics to those with localized pathology and the speed of onset changes this again . The SPASM Consortium (A European Thematic Network to Develop Standardized Measures of Spasticity) has attempted to adapt the accepted definition to a more practical base and make it more relevant to clinical practice and to clinical research . 1 ..."
Reference 9
Book Chapter
The symptoms and signs of multiple sclerosis
McDonald I., Compston A.
McAlpine's Multiple Sclerosis , 2006 pp 287-346
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"... Spasticity Spasticity, defined as an inappropriate increase in velocity-dependent tonic stretch reflexes, is associated with sprouting of descending motor pathways to form new synaptic connections with spinal neurons, and with denervation hypersensitivity ( Figure 6.5 ). Together, these physiological and structural changes amplify the reactivity of motor segments to sensory input ( R.R. Young and Delwade 1981a ; 1981b ). Spasticity forms only one component of the upper motor neuron syndrome, since damage to descending motor pathways also results in slowness of movement and weakness, but it often dominates the clinical picture. In practice, the concept of spasticity has extended far beyond the resistance to passive movement by which it is usually estimated and includes a complex disorder of voluntary movement and physical changes in muscle and tendons. This, rather than weakness, no doubt accounts for much disability resulting from the upper motor neuron lesion in multiple sclerosis, particularly that affecting the lower limbs. Nevertheless, increased tone is an important factor, sometimes resulting in extreme resistance to passive flexion of the knees. Tone is often increased in the upper limb but the flexed spastic arm posture of hemiplegia, typical of stroke, is very uncommon in multiple sclerosis. In the progressive stage of the disease, increased extensor tone may be exaggerated, manifesting as extensor spasms. These are particularly liable to occur in bed at night or on waking in the morning. They may also be so severe as to eject the patient from a wheelchair. The legs are held rigidly extended, usually for several minutes. These spasms are inconvenient but rarely disabling and seldom painful. The mechanisms of the changes in muscle tone in multiple sclerosis have been reviewed by P. Brown (1994) . He points out that normal tone arises from a balance between inhibitory effects on local stretch reflexes (mediated by the dorsal reticulo-spinal tract) and facilitatory effects on extensor tone (mediated by the medial reticulo-spinal and to a lesser extent the vestibulo-spinal tracts). 1 ..."
Reference 10
Book Chapter
Motor System II: Corticofugal Systems and the Control of Movement
Mihailoff G.A., Haines D.E.
Fundamental Neuroscience for Basic and Clinical Applications , 2018 pp 360-376.e1
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"... General Features of Motor Deficits Lower Motor Neuron Signs Lower motor neurons are those cells whose axons synapse directly on skeletal muscle. When these neurons or their axons are damaged, the innervated muscles will show some combination of the following signs: (1) flaccid paralysis followed eventually by atrophy, (2) fibrillations or fasciculations (involuntary contractions of one motor unit or a group of motor units), (3) hypotonia (decreased muscle tone), and (4) weakening or absence of muscle stretch reflexes ( hyporeflexia, areflexia ). Upper Motor Neuron Signs The term upper motor neuron is commonly used in reference to corticospinal or corticonuclear cell bodies and their axons. Other neurons, such as rubrospinal or reticulospinal neurons, can also be included under the strict definition of this term. Corticospinal neurons are also called pyramidal neurons because their axons pass through the medullary pyramid. Therefore the terms upper motor neuron signs and pyramidal tract signs are often used synonymously. However, as described later in this chapter, these characteristic signs of “pyramidal tract damage” are in fact the result of injury to other descending motor systems in combination with damage to fibers of the pyramidal tract. For example, ischemic lesions of the internal capsule can potentially involve corticostriatal, corticothalamic, and corticoreticular fibers in addition to corticospinal axons because of the close proximity of these tracts to one another within the internal capsule. Damage to upper motor neurons results in muscles that (1) are initially weak and flaccid but (2) eventually become spastic, (3) exhibit increased muscle tone ( hypertonia ), seen as an increase in resistance to passive movement of an extremity, and (4) show an increase in muscle stretch reflexes ( hyperreflexia ), as may be seen in clonus. Upper motor neuron lesions usually affect groups of muscles, and certain pathologic reflexes and signs often appear. One of the most common is the inverted plantar reflex, also known as the Babinski sign. 1 ..."
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"... Upper Motor Neuron Signs The term upper motor neuron is commonly used in reference to corticospinal or corticonuclear cell bodies and their axons. Other neurons, such as rubrospinal or reticulospinal neurons, can also be included under the strict definition of this term. Corticospinal neurons are also called pyramidal neurons because their axons pass through the medullary pyramid. Therefore the terms upper motor neuron signs and pyramidal tract signs are often used synonymously. However, as described later in this chapter, these characteristic signs of “pyramidal tract damage” are in fact the result of injury to other descending motor systems in combination with damage to fibers of the pyramidal tract. For example, ischemic lesions of the internal capsule can potentially involve corticostriatal, corticothalamic, and corticoreticular fibers in addition to corticospinal axons because of the close proximity of these tracts to one another within the internal capsule. Damage to upper motor neurons results in muscles that (1) are initially weak and flaccid but (2) eventually become spastic, (3) exhibit increased muscle tone ( hypertonia ), seen as an increase in resistance to passive movement of an extremity, and (4) show an increase in muscle stretch reflexes ( hyperreflexia ), as may be seen in clonus. Upper motor neuron lesions usually affect groups of muscles, and certain pathologic reflexes and signs often appear. One of the most common is the inverted plantar reflex, also known as the Babinski sign. This involves dorsiflexion of the great toe in response to firm stroking of the lateral aspect of the sole of the foot with a blunt instrument. The response in the normal adult is plantar flexion of the great toe. In contrast, stroking the sole of the foot in a normal neonate may also result in a Babinski sign. This is related to the incomplete myelination of the corticospinal tract and will disappear as this tract matures. 1 ..."
Reference 11
Book Chapter
Neuromotor Control of Speech
Wanda G. Webb
Neurology for the Speech-Language Pathologist , 2017 pp 110-139
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"... This reaction, which identifies spastic hypertonicity, is analogous to the resistance felt when a knife blade is first opened, followed by the reduction of resistance when the blade is straightened. This reaction occurs more typically in extension rather than in flexion of the elbow. A short span of no tone is usually present, then a rapid buildup of tone and a sudden release as the joint is moved, just as with opening a clasp knife. Spasticity is also associated with exaggerated muscle stretch reflexes, resulting in hyperreflexia . Reflex action is tested at joints by putting stretch on tendons. This elicits the exaggerated muscle stretch reflex. Spastic paralysis, hypertonia, and hyperreflexia have most often been associated with pyramidal tract damage, particularly lesions of the corticospinal tract. However, the corticonuclear tracts are often also involved when a lesion interrupts the corticospinal tract, and signs of spasticity may be found in the midline speech muscles as well as in the distal limb muscles. Therefore the clinical signs of spasticity , or an upper motor neuron lesion, are of equal interest to the speech-language pathologist and the neurologist. Spastic speech muscles may be weak, slow, and limited in range or movement. Hypertonia may decrease muscle flexibility of the articulators and limit the ability to achieve a full range of motion of the speech muscles. Other Confirmatory Signs Several signs, in addition to a clinical demonstration of clasp knife spasticity, hypertonia, and hyperreflexia, are used by the neurologist to help verify the diagnosis of spasticity and localize the lesion to the pyramidal tract. The Babinski sign , or extensor plantar sign, in particular has been identified as an abnormal reflex sign that develops with corticospinal damage. The lesion causes the release of cortical inhibition. 1 ..."
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"... Another confirmatory sign of spasticity is clonus . Hyperactive muscle stretch reflexes associated with spasticity may show a sustained series of rhythmic beats or jerks when a neurologic examiner maintains one tendon of a muscle in extension. To test for clonus, the Achilles tendon in the ankle is often put under extension. If an upper motor neuron lesion is present, the ankle and the calf show sustained jerks. A few clonic jerks, called abortive clonus, are not clinically significant, but if the clonus is sustained over time, it is considered pathologic and an indicator of hyperreflexia. This sign is part of the clinical syndrome resulting from an upper motor neuron lesion. With bilateral upper motor neuron lesions, a characteristic dysarthria may be present. This motor speech disorder, called spastic dysarthria, is discussed in Chapter 8 . Lower Motor Neuron Signs and Symptoms If a lesion directly damages a cranial or peripheral nerve, or is in the cell bodies of the anterior horn cell in the spinal cord or the nuclei of cranial nerves in the brainstem, neural impulses will not be transmitted to the muscles. This condition is called denervation . The result is that the muscles innervated by the cranial or spinal nerve become soft and flabby from a loss of muscle tone. This is a lower motor neuron paralysis. The loss of muscle tone is called hypotonia . Hypotonia results in flaccid muscles, thus the term flaccid paralysis . Hypotonia may be an acquired condition, with disease or injury affecting the peripheral nervous system pathway, but it may also be a condition found at birth or developing shortly after birth. Infantile hypotonia may result from conditions such as chromosome disorders (e.g., Prader-Willi syndrome), genetic defects, spinal cord disorders, spinal muscular atrophy, muscular dystrophy, metabolic myopathies, or other problems that affect the peripheral pathways. 1 ..."
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"... Damage to the nervous system often releases early reflex behavior that has become inhibited by development of higher centers, so signs of damage at that point in time are signs of immaturity at an earlier time. Clinical neurologists believe that the extensor plantar sign usually reaches stability by age 2 years. Other signs, such as a persisting asymmetric tonic neck reflex and the Moro reflex, can be tested to suggest an upper motor neuron lesion in young children (see Chapter 11 ). Another confirmatory sign of spasticity is clonus . Hyperactive muscle stretch reflexes associated with spasticity may show a sustained series of rhythmic beats or jerks when a neurologic examiner maintains one tendon of a muscle in extension. To test for clonus, the Achilles tendon in the ankle is often put under extension. If an upper motor neuron lesion is present, the ankle and the calf show sustained jerks. A few clonic jerks, called abortive clonus, are not clinically significant, but if the clonus is sustained over time, it is considered pathologic and an indicator of hyperreflexia. This sign is part of the clinical syndrome resulting from an upper motor neuron lesion. With bilateral upper motor neuron lesions, a characteristic dysarthria may be present. This motor speech disorder, called spastic dysarthria, is discussed in Chapter 8 . 1 ..."
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"... With a corticospinal lesion, the great toe extends upward and the other toes fan as the foot withdraws slightly. Physicians test this response several times to convince themselves that the upturning great toe sign can be repeatedly and automatically elicited. Automatic repetition of a given response such as this defines it as a reflex. The presence of a repeatable abnormal reflex sharply increases the probability of predicting with accuracy the possible site or sites of a neurologic lesion, though it may not be present if the patient with a vascular lesion is tested early after onset of the insult. The Babinski sign is more reliable in adults than in infants and children. Normal infants are highly variable in display of the sign. The explanation usually given for this variability is that the immature nervous system and the damaged nervous system often show similar symptoms and signs. Damage to the nervous system often releases early reflex behavior that has become inhibited by development of higher centers, so signs of damage at that point in time are signs of immaturity at an earlier time. Clinical neurologists believe that the extensor plantar sign usually reaches stability by age 2 years. Other signs, such as a persisting asymmetric tonic neck reflex and the Moro reflex, can be tested to suggest an upper motor neuron lesion in young children (see Chapter 11 ). Another confirmatory sign of spasticity is clonus . Hyperactive muscle stretch reflexes associated with spasticity may show a sustained series of rhythmic beats or jerks when a neurologic examiner maintains one tendon of a muscle in extension. To test for clonus, the Achilles tendon in the ankle is often put under extension. If an upper motor neuron lesion is present, the ankle and the calf show sustained jerks. A few clonic jerks, called abortive clonus, are not clinically significant, but if the clonus is sustained over time, it is considered pathologic and an indicator of hyperreflexia. This sign is part of the clinical syndrome resulting from an upper motor neuron lesion. With bilateral upper motor neuron lesions, a characteristic dysarthria may be present. 1 ..."
Related quote(s)5 / 6
"... Spastic paralysis, hypertonia, and hyperreflexia have most often been associated with pyramidal tract damage, particularly lesions of the corticospinal tract. However, the corticonuclear tracts are often also involved when a lesion interrupts the corticospinal tract, and signs of spasticity may be found in the midline speech muscles as well as in the distal limb muscles. Therefore the clinical signs of spasticity , or an upper motor neuron lesion, are of equal interest to the speech-language pathologist and the neurologist. Spastic speech muscles may be weak, slow, and limited in range or movement. Hypertonia may decrease muscle flexibility of the articulators and limit the ability to achieve a full range of motion of the speech muscles. Other Confirmatory Signs Several signs, in addition to a clinical demonstration of clasp knife spasticity, hypertonia, and hyperreflexia, are used by the neurologist to help verify the diagnosis of spasticity and localize the lesion to the pyramidal tract. The Babinski sign , or extensor plantar sign, in particular has been identified as an abnormal reflex sign that develops with corticospinal damage. The lesion causes the release of cortical inhibition. The sign has achieved considerable status in the diagnosis of upper motor neuron lesions because it is a highly reliable abnormal reflex, is new behavior released by the presence of a lesion, and is clearly associated with a relatively specific lesion site—the cortex or the corticospinal tract. The speech-language pathologist is not directly interested in it because it does not involve the midline speech muscles, but its presence as a confirmation of an upper motor neuron lesion of the spastic type is important to all who manage neurologic patients. The Babinski sign is observed as a reflex toe sign. It is elicited by stimulating the sole of the foot with a strong scratching maneuver. The normal response to stimulation of the sole, or plantar portion of the foot, is a slight withdrawal of the foot and downward turning or curling under of the toes. 1 ..."
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"... The sign has achieved considerable status in the diagnosis of upper motor neuron lesions because it is a highly reliable abnormal reflex, is new behavior released by the presence of a lesion, and is clearly associated with a relatively specific lesion site—the cortex or the corticospinal tract. The speech-language pathologist is not directly interested in it because it does not involve the midline speech muscles, but its presence as a confirmation of an upper motor neuron lesion of the spastic type is important to all who manage neurologic patients. The Babinski sign is observed as a reflex toe sign. It is elicited by stimulating the sole of the foot with a strong scratching maneuver. The normal response to stimulation of the sole, or plantar portion of the foot, is a slight withdrawal of the foot and downward turning or curling under of the toes. With a corticospinal lesion, the great toe extends upward and the other toes fan as the foot withdraws slightly. Physicians test this response several times to convince themselves that the upturning great toe sign can be repeatedly and automatically elicited. Automatic repetition of a given response such as this defines it as a reflex. The presence of a repeatable abnormal reflex sharply increases the probability of predicting with accuracy the possible site or sites of a neurologic lesion, though it may not be present if the patient with a vascular lesion is tested early after onset of the insult. The Babinski sign is more reliable in adults than in infants and children. Normal infants are highly variable in display of the sign. The explanation usually given for this variability is that the immature nervous system and the damaged nervous system often show similar symptoms and signs. Damage to the nervous system often releases early reflex behavior that has become inhibited by development of higher centers, so signs of damage at that point in time are signs of immaturity at an earlier time. Clinical neurologists believe that the extensor plantar sign usually reaches stability by age 2 years. Other signs, such as a persisting asymmetric tonic neck reflex and the Moro reflex, can be tested to suggest an upper motor neuron lesion in young children (see Chapter 11 ). 1 ..."
Reference 12
Book Chapter
Diseases of Motor Neurons and Neuromuscular Junctions
Harald Sontheimer
Diseases of the Nervous System , 2015 pp 165-197
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"... As the neuron that innervates a motor unit becomes diseased, its axon goes through a period of unpredictability wherein aberrant action potentials cause twitching or fasciculation of a motor unit. Such twitches are commonly observed in the early stages of ALS, and their electrical activity can be measured through an EMG ( Figure 8 ) that reports voltage changes via fine needles inserted into the muscle. EMG serves as a useful tool to diagnose diseases affecting muscles, motor neurons, and the NMJ. As the axon dies back, the muscle becomes denervated and undergoes atrophy. Lower motor neurons also receive modulatory input from the basal ganglia and a number of cortical pathways that travel through the brain stem, including the corticorubrospinal, corticovestibulospinal, and corticoreticulospinal pathways. The modulatory neurons that give rise to these tracts also are designated as upper motor neurons. In the end lower motor neurons in the spinal cord receive both excitatory input from the upper motor neurons and a mixture of modulatory, excitatory, and inhibitory inputs from sensory muscle spindles and modulatory descending pathways. The modulation of the lower motor neurons explains why upper and lower motor neuron lesions present with opposite paralysis. When lower motor neurons or their axons cease to function, muscle tone is reduced and the muscle shows a complete absence of contraction, even when a reflex pathway is activated (areflexia). This is often called flaccid paralysis. By contrast, when only the upper motor neurons are lesioned, reflex pathways remain intact, and, indeed, show more brisk reflexes since the descending inhibitor modulatory pathways that normally dampen their activity are lost. Muscles have increased tone but are unable to receive excitatory commands from the upper motor neurons, causing spastic paralysis and hyperreflexia. Neurologists often take advantage of this distinction and use a common reflex, the plantar reflex, to determine whether a patient has upper or lower motor neuron signs. 1 ..."
Reference 13
Book Chapter
The Cranial Nerves
Russell J. Love, Wanda G. Webb
Neurology for the Speech-Language Pathologist , 1992 pp 112-136
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"... The patient with an upper motor neuron lesion will show complete involvement of the lips and neck muscles, some degree of involvement of the area around the eyes, and little difficulty with the forehead or frontalis muscle. It should be noted that the paralysis is on voluntary movement. The patient may be seen to have almost normal movement for emotionally initiated movements such as a true smile, but will be unable to lateralize the lips when asked to do so voluntarily. Since the facial nerve innervates the stapedius muscle, it may be paralyzed by a lesion. If this occurs, the patient may report that ordinary sounds seem uncomfortably loud. 1 ..."
Reference 14
Book Chapter
Cranial nerves
J.L. Wilkinson
Neuroanatomy for Medical Students , 1992 pp 110-149
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"... APPLIED ANATOMY Upper and lower motor neuron lesions differ markedly. Upper motor neuron, or supranuclear lesions are usually part of a hemiplegia due to a cerebrovascular accident; only the muscles of the lower half of the face on the opposite side are paralysed; the patient can still close the eyelids and wrinkle the forehead. In cortical or capsular lesions, voluntary and emotional facial activity may be dissociated: the patient may smile if amused but cannot ‘show his teeth’ on demand. By contrast, loss of emotional expression is very characteristic of Parkinson's disease. In lower motor neuron lesions (damage to the nucleus or nerve), the upper and lower facial muscles on the same side as the lesion are paralysed. Inability to close the eyelids may lead to corneal ulceration, especially if reflex lacrimation is also lost. The commonest infranuclear lesion is Bell's palsy , thought to be of viral origin, in which oedema compresses the nerve within its canal. The nerve may be affected by inflammation of the middle ear and mastoid air cells, or compressed in the cerebellopontine angle by an acoustic neuroma. A fracture of petrous temporal bone may involve facial nerve, vestibular apparatus and cochlea. In the newborn the nerve is relatively superficial and can be damaged by obstetric forceps. The extent of loss in a lower motor neuron lesion depends on the level of injury. If this is proximal to the geniculate ganglion, taste is lost in the anterior two-thirds of the tongue and secretion from the submandibular, sublingual and lacrimal glands is impaired. Hyperacusis (increased sound perception) is due to paralysis of stapedius. The prognosis is usually good in more distal lesions; in lesions proximal to geniculate ganglion, sensory regeneration is poor and salivatory parasympathetic fibres may be misdirected into the lacrimal gland, producing ‘crocodile tears’ in response to the smell or taste of food. 1 ..."
Reference 15
Book Chapter
Physical and Neurologic Examination
Blake Staub, Meng Huang, Paul J. Holman, Edward C. Benzel
Benzel's Spine Surgery, 2-Volume Set , 2017 pp 124-131.e1
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"... Reflex Examination The neurologic examination also includes an evaluation of the deep tendon (stretch) and superficial reflexes. Deep Tendon Reflexes The deep tendon reflexes are used to assess the integrity of a monosynaptic reflex arc at various levels of the cord. Table 12-4 depicts the common system for grading deep tendon reflexes. Hyperactive reflexes generally indicate an upper motor neuron lesion, and diminished or absent reflexes can be seen in lower motor neuron lesions. Metabolic abnormalities, such as hypothyroidism or hyperthyroidism, should always be excluded as an etiology for abnormal reflexes. Neuromuscular disorders and neuropathies may also present with abnormal reflexes. Reflexes are compared from one side to another, and reinforcement maneuvers that require isometric contraction of other muscle groups can be used to eliminate cortical modulation of the reflex arc. The Jendrassik maneuver can accentuate lower extremity reflexes and requires the patient to pull interlocked fingers apart while the reflex is tested. Asking the patient to clench the teeth or push down on the examination table with the thighs can accentuate upper extremity reflexes. One uncommonly practiced reflex is the finger jerk or finger-thumb reflex. Mediated by mainly the C8 nerve root, it is elicited with patient's palm upturned and the fingers half-flexed. The surgeon then holds the tops of the fingers with his or her own half-flexed fingers, which are then tapped. The patient's fingers will be felt to flex and, most strikingly, the free thumb will be seen to flex. 10 Although the commonly tested reflexes are truly mediated by multiple nerve roots, Tables 12-2 and 12-3 outline the dominant nerve roots involved. Superficial Reflexes The superficial reflexes are mediated by the cerebral cortex with the afferent limb being supplied by cutaneous stimulation. The absence of a normal cutaneous reflex may signal an underlying upper motor neuron lesion. 1 ..."
Reference 16
Reference works Chapter
Lower Motor Neuron Lesions
Ramahi A.A., Katirji B., Devereaux M.
Encyclopedia of the Neurological Sciences , 2014 pp 918-922
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"... Tone Muscle tone is assessed by examining its response to passive stretch. Lower motor neuron lesions reduce muscle tone, whereas upper motor neuron lesions increase muscle tone resulting in spasticity as seen in pyramidal lesions, or rigidity as observed in extrapyramidal lesions. Regulation of muscle tone is normally mediated by the reticulospinal fibers accompanying the pyramidal tract that exert inhibitory effects on the stretch reflex. Flaccidity, or hypotonia, is a typical feature of lower motor neuron damage, and in the extreme form of total flaccid paralysis, all resistance to passive muscle stretch is lost and the limbs become flail-like. Hypotonia can also be seen in cerebral or spinal shock resulting from acute extensive brain, spinal cord, or cerebellar lesions. It is also seen in peripheral nerve or root lesions affecting sensory afferent pathways. In clinical practice, observation of flaccidity is of limited diagnostic usefulness compared to more important physical signs such as wasting or tendon reflex loss. 1 ..."
Reference 17
Book Chapter
Cranial Nerves XI (Spinal Accessory) and XII (Hypoglossal)
Barboi A.C.
Textbook of Clinical Neurology , 2007 pp 231-241
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"... MOTOR/REFLEXES/CEREBELLAR/GAIT When examining a patient with a suspected lesion of CN XI or XII, it is important to perform a thorough motor examination. The examiner should note any primitive reflexes (e.g., suck, snout, and glabellar) or paratonic rigidity (gegenhalten). These signs, along with pseudobulbar palsy, may be indicative of bilateral frontal lobe lesions. In addition, weakness in other ipsilateral muscles in the face, arm, or leg, accompanied by hypertonia, hyper‐reflexia, or extensor plantar responses, is suggestive of an upper motor neuron lesion, 31 whereas weakness with atrophy, loss of reflexes, or fasciculations are signs of either a lower motor neuron lesion or neuronopathy. The supranuclear fibers for the sternal head of the SCM muscle probably decussate twice—once in the midbrain or pons and a second time in the medulla or cervical spinal cord. 23 Hence, a lesion in the right side of the pons may result in (1) left‐sided weakness (involvement of corticospinal tract fibers before their decussation in the medulla) and (2) weakness of the left SCM muscle and deviation of the head to the left (i.e., the same side as the hemiparesis) because of involvement of the supranuclear input to the left SCM muscle after the first but before the second decussation. In contrast, a lesion in the right cerebral hemisphere or right internal capsule or high in the right midbrain, before the initial decussation of the supranuclear fibers to the SCM, may result in (1) left‐sided weakness due to involvement of the corticospinal tract fibers before their decussation and (2) weakness of the right SCM muscle and deviation of the head to the right (i.e., the side opposite the hemiparesis). 13 When dysfunction of CN XI or XII is present, it is also important to perform a thorough examination of coordination and gait. 1 ..."
Reference 18
Book Chapter
Electrodiagnostic Findings in Neuromuscular Disorders
Bashar Katirji
Electromyography in Clinical Practice , 2007 pp 49-58
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"... UPPER MOTOR NEURON LESIONS Patients with upper motor neuron lesions have normal NCSs and needle EMG including normal insertional activity, no spontaneous activity at rest, and normal MUAP morphology. The only abnormality is a reduced interference pattern with poor activation of MUAPs (slow rate of motor unit discharge). Recruitment, measured by either recruitment frequency or ratio, is normal. Hysterical weakness or poor effort produces a similar pattern, except that motor unit firing may be irregular. 1 ..."
Reference 19
Review article
Electrodiagnostic criteria for diagnosis of ALS
de Carvalho M., Dengler R., Eisen A., England J.D., Kaji R., Kimura J., Mills K., Mitsumoto H., Nodera H., Shefner J., Swash M.
Clinical Neurophysiology , 2008 pp 497-503
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"... Electrophysiological evaluation of upper motor neuron (UMN) dysfunction The diagnosis of ALS requires demonstration of a combination of upper and lower motor neuron abnormalities whether by clinical or electrophysiological criteria, as set out in the Airlie House consensus recommendations ( Table 1 ). Although it has been hoped that TMS might prove a sensitive test for upper motor neuron involvement in ALS this has not been confirmed in the published studies, at least not for patients with clinically missing or equivocal upper motor neuron signs, and expert opinions are equivocal. Central motor conduction time (CMCT), however, is frequently prolonged to muscles of at least one extremity . More complex stimulation paradigms have not been helpful in diagnosis of ALS except for short-interval, paired-pulse stimulation, which revealed reduced intracortical inhibition . However, this technique is not practical and therefore not widely used, and experience in early ALS, the crucial time for diagnosis, is limited. A relatively new TMS technique, the triple stimulation method, has shown a high sensitivity to detect upper motor neuron dysfunction in ALS and yielded abnormal results also in early stages of the disease in two laboratories . Confirmation of these results from wider application in other laboratories is required. Developments in magnetic resonance imaging of brain tracts may yet prove helpful, although this has not so far been demonstrated, and the methodology is complex and expensive. 9.1 Transcranial magnetic stimulation (TMS) TMS requires that the target muscle have a detectable CMAP. The following features suggest abnormality of upper motor neuron function: 1. Increased central motor conduction time (CMCT). 2. Increased absolute latency to a tested muscle, provided that distal motor conduction slowing can be excluded. 3. In patients with bulbar onset disease, an absent response to TMS in a limb is supportive of upper motor neuron lesion. 4. The triple stimulation technique has proven sensitive in detecting impairment of upper motor neuron function in the early stages of the disease as well as later in the disease course . However, experience with this promising test is so far relatively limited. 1 ..."
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"... Transcranial magnetic stimulation (TMS) TMS requires that the target muscle have a detectable CMAP. The following features suggest abnormality of upper motor neuron function: 1. Increased central motor conduction time (CMCT). 2. Increased absolute latency to a tested muscle, provided that distal motor conduction slowing can be excluded. 3. In patients with bulbar onset disease, an absent response to TMS in a limb is supportive of upper motor neuron lesion. 4. The triple stimulation technique has proven sensitive in detecting impairment of upper motor neuron function in the early stages of the disease as well as later in the disease course . However, experience with this promising test is so far relatively limited. 1 ..."
Reference 20
Review article
Progressive Muscular Atrophy
Liewluck T., Saperstein D.S.
Neurologic Clinics , 2015 pp 761-773
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"... Imaging Biomarkers of Upper Motor Neuron Involvement The data on radiographic biomarkers of UMN lesions are inconclusive. Two different techniques have been applied to identify UMN involvement in patients with MND: diffusion tensor MRI and magnetic resonance spectroscopy (MRS). Diffusion tensor MRI studies of the structural integrity of neuronal fibers showed modest reduction of fractional anisotropy, suggesting axonal degeneration and myelin breakdown in the corticospinal tract in all 12 patients with PMA, 7 and may predate the clinical evidence of UMN lesions in some patients. 3 However, another study showed no significant abnormal fractional anisotropy in 8 patients with PMA. 45 White matter involvement in patients with PMA may be involved beyond the corticospinal tract. Voxel-based analysis identified fractional anisotropy reduction in nonmotor white matter including the prefrontal area. 7 MRS showed reduced N -acetyl aspartate (NAA) concentration and ratio of NAA to creatine, markers of neuronal health, in the primary motor cortex, suggesting UMN involvement, in about 60% of patients with PMA. 4,15 The corticospinal tract degeneration was pathologically confirmed when autopsy was available. 4 However, normal results on MRS does not exclude UMN pathology. 4 Another MRS study showed only modest, nonsignificant changes in patients with PMA, which could be due to the slightly different imaging technique. 5 1 ..."
Reference 21
Book Chapter
The differential diagnosis of multiple sclerosis
Miller D., Compston A.
McAlpine's Multiple Sclerosis , 2006 pp 389-437
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"... A few cases that evolve to typical motor neuron disease nevertheless have an initial phase in which the manifestations exclusively affect upper motor neurons. In about 70% of patients, MRI reveals symmetrical high signal confined to the corticospinal tracts in the centrum semiovale, internal capsule, cerebral peduncles, pons and upper cervical cord ( Thorpe et al 1996c) . Such findings are unlike those seen in multiple sclerosis and give a clue to the correct diagnosis. 1 ..."
Reference 22
Book Chapter
Motor Stimulation
Alaa Abd-Elsayed, Andrea M. Trescot, Brent Earls, Eellan Sivanesan
Peripheral Nerve Stimulation , 2023 pp 266-273
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"... 32 However, it is now largely accepted that the recovery is due to plastic changes in the sensorimotor cortex due to unmasking of latent horizontal connections, activation of silent synapses, modulation of activity-dependent synaptic plasticity, and generalized changes in the excitability of postsynaptic neurons. 33 Effects on cortical perfusion were recently studied in combination with FES in chronic stroke patients. This study found that the sensory motor integration due to FES may facilitate perfusion of the ipsilesional sensory motor cortex and result in functional improvement of hemiparetic upper extremity. After FES treatment, only patients who showed improved perfusion to the ipsilesional sensory motor cortex perfusion showed functional improvement of upper extremity. 34 This further supports the idea that recovery of paretic limbs due to upper motor neuron lesions is largely the result of reorganization of the sensorimotor cortex, but that FES can help to facilitate this process. This neuroplasticity may also drive the alleviation of pain symptoms experienced by patients with motor stimulation. 35,36 1 ..."
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"... Peripheral and central mechanisms of benefit The rationale for use of motor stimulation is based on the physiological changes at muscle and spinal levels that occur over time. For example, repeated muscle contraction from peripheral nerve stimulation (PNS) increases the oxidative capacity of muscle, increases the number of microcapillaries, and leads to transformation of muscle fiber types. 30 A small but expanding body of literature demonstrates that FES induces rapid plastic change in the sensorimotor regions of the cortex. 31 This suggests that repetitive movements that are goal oriented and functionally relevant can facilitate relearning following stroke or brain injury. 4 Motor relearning is defined as “the recovery of previously learned motor skills that have been lost following localized damage to the central nervous system.” 31 It was hypothesized that the use of FES, which generates both an orthodromic and an antidromic impulse, may modulate activity in anterior horn cells via antidromic impulses. These impulses were thought to strengthen Hebbian-type synapses between pyramidal tract cells and the anterior horn cells, contributing to restoration of voluntary control. 32 However, it is now largely accepted that the recovery is due to plastic changes in the sensorimotor cortex due to unmasking of latent horizontal connections, activation of silent synapses, modulation of activity-dependent synaptic plasticity, and generalized changes in the excitability of postsynaptic neurons. 33 Effects on cortical perfusion were recently studied in combination with FES in chronic stroke patients. This study found that the sensory motor integration due to FES may facilitate perfusion of the ipsilesional sensory motor cortex and result in functional improvement of hemiparetic upper extremity. After FES treatment, only patients who showed improved perfusion to the ipsilesional sensory motor cortex perfusion showed functional improvement of upper extremity. 34 This further supports the idea that recovery of paretic limbs due to upper motor neuron lesions is largely the result of reorganization of the sensorimotor cortex, but that FES can help to facilitate this process. This neuroplasticity may also drive the alleviation of pain symptoms experienced by patients with motor stimulation. 35,36 1 ..."
Reference 23
Book Chapter
The Physiologic Aspects and Clinical Application of Functional Electrical Stimulation in Rehabilitation
KRISTJAN T. RAGNARSSON
The Physiological Basis of Rehabilitation Medicine , 1994 pp 573-597
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"... Publisher Summary This chapter discusses the physiologic aspects and clinical application of functional electrical stimulation (FES) in rehabilitation. FES may be defined as the application of electrical currents to neural tissue for restoring a degree of control over abnormal or absent body functions. Electricity has been applied either experimentally or clinically for many purposes: (1) to improve hearing and sight, (2) to prevent bladder and bowel incontinence, (3) to control evacuation, (4) to regulate heart rhythm, (5) to reduce spasticity, (6) to allow ventilator-free breathing, (7) to correct scoliosis, and (8) to usefully move paralyzed limbs. The chapter discusses one specific form of FES that is usually referred to by scientists as functional neuromuscular stimulation (FNS). FNS restricts the electrical stimulation to the neuromuscular system for the purpose of controlling skeletal muscle contractions. It may be applied successfully for different forms of paralytic conditions caused by upper motor neuron lesions, for example, stroke, traumatic brain injury, and cerebral palsy. The chapter discusses its application after spinal cord injury. 2 ..."
Reference 24
Reference works Chapter
Volume 2
Theresa L. Bender Pape, Nikki M. Barrington, Elise K. Webber, Grace E. Stutzmann
Encyclopedia of the Human Brain , 2025 pp 730-757
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"... FES CPGs are, however, also in place for patients with lower limb impairments due to upper motor neuron lesions (e.g., Polio, Guillain-Barre, etc.). Specifically, these guidelines recommend using FES if the individual has enough passive movement of their ankle to allow for walking but are unable to wear a splint and/or if the individual has trouble controlling lower limb movement or keeping balance when walking . CPGs recommending FES as a treatment option are also available for participants with acute or subacute cervical spinal cord injury to improve hand and upper extremity function . Despite some evidence suggesting therapeutic benefits, there are currently no CPGs for use of FES for MS symptoms or for the treatment of gait abnormalities and spasticity associated with CP. However, a clinical decision-making tool was recently developed to help clinicians use FES for motor rehabilitation in conditions with no available CPGs . The collective evidence presented here indicates that incorporating FES in motor rehabilitation is clinically beneficial, but that further research is needed to delineate optimal treatment protocols (e.g., FES alone or combined with physical retraining exercises). Furthermore, additional research is needed to elucidate optimal therapeutic windows post-injury as well as the optimal number of FES treatment sessions, aiding in the creation of important FES CPGs . Transcranial magnetic stimulation The recovery of motor skills, despite the presence of various neurological conditions, is possible with transcranial magnetic stimulation (TMS), as these devices can be used to exogenously engage or acquire long-term potentiation (LTP) and long-term depression (LTD) like mechanisms. Specifically, injury-based animal studies have demonstrated that TMS-induced neuronal activity results in a decrease of pro-apoptotic factors and in an upregulation of brain derived neurotrophic factors (BDNF) . These results hold therapeutic promise as neurotrophic signaling regulates plasticity in an activity-dependent manner . 2 ..."
Reference 25
Book Chapter
Viral Vector Axonal Uptake and Retrograde Transport: Mechanisms and Applications
Teng Q., Federici T., Boulis N.M.
Gene Therapy of the Central Nervous System , 2006 pp 253-271
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"... E. Spasticity Spasticity is a motor dysfunction characterized by increased muscular resistance to movement with increased velocity of movement. Severe spasticity is often accompanied by spasms, which are involuntary muscle contractions that are often severely painful. Spasticity commonly occurs in the presence of upper motor neuron lesions or disconnection of the synaptic connection between the brain and spinal motor neurons. Spasticity is a disabling symptom of spinal cord injury, multiple sclerosis, and cerebral palsy. Spasms and spasticity are thought to result from a dysfunction of inhibitory signals within the spinal cord that depend on descending motor pathways. Treatments including anti-spasticity drugs, botulinum toxin, electrical stimulation, and surgery are being currently used to reduce spasticity . In cases where patients do not respond to pharmacological and surgical interventions, gene therapy might be warranted, using a vector capable of neuronal tropism and targeted neuromodulation, in order to achieve focused synaptic inhibition. Combining the retrograde transport property of the viral vectors mentioned above with proteins that modulate neurotransmitter release , injection of these viral vectors into the spastic muscles would specifically disrupt acetycholine release at the neuromuscular junction, relieving the symptoms. 2 ..."
Reference 26
Book Chapter
Neurocontrol of Chronic Upper Motor Neuron Syndromes
Milan R. Dimitrijevic
Electromyography in CNS Disorders , 1984 pp 111-128
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"... Publisher Summary There are numerous structures within the cortex, subcortex, brain stem, and spinal cord, and there are complex interconnections among them via corticosubcortical, corticobulbar, subcorticobulbar, corticospinal, and bulbospinal descending pathways. All these morphologic and functional complexities of the central nervous system are involved in motor control and are referred to as upper motor neuron. The upper motor neuron syndrome is also a complex neurologic entity. This chapter focuses on the upper motor neuron syndromes owing to chronic effects of spinal cord injury. It presents the three functions of the upper motor neuron: (1) the degree of preservation or deterioration of volitional activity, (2) the effects of remote muscle contraction on paretic or paralyzed muscles, and (3) characteristic features of stretch and withdrawal reflexes. Upper motor neuron dysfunctions in patients with chronic neurologic disorders usually are due to lesions of pyramidal and extrapyramidal tract fibers. Such an upper motor neuron lesion can result from the destruction of descending motor pathways in the cerebral cortex, the internal capsule, the cerebral peduncles, the brain stem, or the spinal cord. Clinical features of this upper motor neuron syndrome represent different degrees of impairment of neurocontrol of volitional activity, muscle tone, and brain influence on segmental reflexes. 2 ..."
Reference 27
Book Chapter
Clinical experience and recent advances in the management of gait disorders with botulinum neurotoxin
Esquenazi A., Mayer N.H.
Botulinum Toxin , 2009 pp 192-203
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"... Lance published this frequently cited definition for the phenomenon: “a motor disorder characterized by a velocity dependent increase in tonic stretch reflexes (muscle tone) with exaggerated tendon jerks, resulting from hyperexcitability of the stretch reflex, as one component of the upper motoneuron syndrome.” 5 Signs of spasticity are useful as a diagnostic indicator pointing to the presence of an upper motor neuron lesion, but in general, rehabilitation clinicians are more preoccupied with addressing the functional problems of their patients linked to the consequences of an upper motor neuron lesion. Common terminology, which reflects their clinical practice emphasis, includes spastic gait or spastic equinovarus. For these clinicians, evaluation and treatment of spasticity takes on broader dimensions, reflecting their interest in clinical patterns of motor dysfunction that produce functional impairment and can result in disability. The clinician must be able to see spastic phenomena within the larger context of impaired motor control in order to identify appropriate treatment methods for a patient's functional problems resulting from a central nervous system injury. The National Center for Medical Rehabilitation Research (NCMRR) has suggested that new methods of treatment should be evaluated for effects on multiple dimensions of the disabling process, including not only the dimension of impairment (as exemplified by Ashworth scale measurements), but also other dimensions such as functional limitations and disabilities (i.e., difficulty fulfilling role functions). 6 Lower extremity posturing that affects gait and transfer dysfunctions represents a significant spectrum of functional limitations that are attributable to the positive signs of the UMNS. In a broader sense and from our perspective, evaluation of spasticity focuses on three issues: (1) identifying the clinical pattern of motor dysfunction and its source; (2) identifying the patient's ability to control muscles involved in the clinical pattern; and (3) the differential role of muscle stiffness and contracture as it relates to a functional problem. 2 ..."
Related topics (10)
Babinski Reflex
Brainstem
Hyperreflexia
Lower Motor Neuron
Lower Motor Neuron Lesion
Motor Unit
Nerve Root
Spinal Cord Injury
Tendon Reflex
Upper Motor Neuron
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Medicine and Dentistry
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CCSS5.NBT.A.2
Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10.
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7296 | https://www.teachoo.com/2769/634/Example-5---y2--8x-Find-focus--axis--directrix/category/Examples/ | Example 5 - Find focus, axis, directrix, latus rectum of y2 = 8x
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Chapter 10 Class 11 Conic Sections
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Example 5 Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y2 = 8x. Given equation is y2 = 8x. Since the above equation is involves y2 Its axis is x-axis Also coefficient of x is positive (+8) Hence we use equation y2 = 4ax Comparing y2 = 8x & y2 = 4ax From (1) & (2) 8x = 4ax 8𝑥4𝑥 = a 2 = a a = 2 Coordinates of focus is = (a, 0) = (2, 0) Equation of Directix is x = −a x = −2 Latus Rectum is 4a = 4 × 2 = 8
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7297 | https://community.spiceworks.com/t/file-server-folder-structure-and-security/22968 | Skip to main content
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Nov 2012
StrainTheBrainheath0024Guajillo
May 2009
I have been tasked with redesigning the folder and security structure for our File server.
We are in an AD environment and have multiple offices with multiple departments in each office. Approximately 250 users total in 5 offices.
From your experience, what is “Best Practice” when designing this structure?
Thank you in advance for your feedback!
3
2
Kris (Spiceworks)kris
May 2009
This is a really general pass, but here are some things I do:
-use groups/nested groups - don’t assign rights to individual users unless the resource specifically calls out for it (user folder). This allows a much easier management of rights
-set the share at a reasonable level of the structure, meaning, don’t share a folder that is contained within a folder that is already shared and has permissions applied
-use NTFS security permissions instead of share level permissions
KenEismanAurora
May 2009
It all depends on what you want to do… what you need to share… what you need to keep private.
Generally, I start with a file structure that begins with the most general and then subfolders underneath generall getting more specific.
IOW you could have a company-wide folder that contains a shared folder available to everyone in the company and a folder for each of the offices that are only available to the respective offices. Within each office folder you could have a shared folder available to everyone in each office and then a folder for each department that is available only to the respective department. Within each department folder you could have a folder that is shared among everyone in the department and then a folder for each person that is only available to them.
That’s just a basic structure that can, and should, be modified to fit individual requirements within the organization.
Kris has good suggestions for the security structure. Always assign security rights to groups and then put users in those groups. It makes admin much simpler.
HTH
Ken
StrainTheBrainheath0024Guajillo
May 2009
Thank you for taking the time to reply to me Kris. Great security advice…I was thinking that using Groups was defintely a more effecient way of handling permissions.
Do you have any advice as to the folder structure?
Thanks!
Jeff7086jeff70860430
May 2009
I generally like to create my file structure avoiding spaces and special characters, and use fairly descriptive names.
I usually create a top folder that’s the level I share out of, then create each department a folder under that, then under those if needed create a seperate folder with permissions just for managers.
I give advice on how the department should set up their folder, and give them naming guidelines, but leave the rest up to them.
I also might create a misc. folder for sharing between departments (but try to avoid this as that just tends to collect junk) or other folders for special inter-departmental projects.
I like to keep the structure as flat as possible.
User shares go in a different structure and are mapped directly to a drive letter (H: for home or U: user, for example)
For example:
Company (Share level, mapped as drive letter)
-Accounting
–Accounting Managers
-Business
–Business Managers
-Development
–Development Managers
-Etc…
LimeymartinpeverleyReaper
May 2009
We are just about to tackle this at both locations. We are going to try and make it work aloong the lines of our ISO 9001 documentation, but I’ve never read that so I’ll kep you posted.
Where we are now is shares for functional areas, e.g. Engineering, Design, QA, Production, HR, Purchasing. Security groups for controlling the access, I set this at share and NTFS, I have no shares that allow the default “Everyone” group, it’s Domain Users or more specific.
The people using the folder get to decide the underlying structure but there is usually no change in permissions below the top level. KISS, if you forget you have different permissions set three levels down and then check the ‘Apply to subfolders’ then you’ll have to recreate those permissions. BTDT.
I don’t like ‘user’ folders in a functional area, so QA\John’s stuff and QA\Jane’s folder are going to go away. If you are out and someone else needs that file you were working on they need to find it quickly.Put it where it belongs when it is done. That’s not 100% because we do have some files that procedurally need to be kept separate until they are approved but those are closely controlled.
Scott Howardscotthoward2423Hinkelhatz
May 2009
Take the “org chart” that shows who reports to whom and then turn it sideways and thats a good start for building your folder structure. In general, most people only need to see inside their own department sub folder.
LimeymartinpeverleyReaper
May 2009
Scott Howard wrote:
Take the “org chart”
Take the what?? If only we had one…
StrainTheBrainheath0024Guajillo
May 2009
Thank you for all your feedback…it has been very useful.
I dig this Spiceworks community!!!
4 years later
TeBeartimothy
Nov 2012
LOL @ Limey and agree … I am tasked with this project as well.
chenghuach-mHatch
Nov 2012
Apologies for 'breaking in". Talking about security, how does one PREVENT user from mass copying the whole folder if they are given permission even READ only? Is there a better way that we can do to avoid this from happening? I am really desperate to find a solution to this issue. Anyone, please advice.
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7298 | https://www.basic-mathematics.com/conjugate-of-a-complex-number.html | Conjugate of a complex number
The conjugate of a complex number a + bi is the complex number a - bi or a + -bi. The real part of the complex number a + bi is a and the imaginary part is b.
a + bi and a - bi are called complex conjugates.
Notice that to find the complex conjugate, all you need to do is to take the opposite of the imaginary part of the complex number.
Examples showing how to find the complex conjugate of a complex number
Example #1
Find the complex conjugate of 5 + 9i
9 is the imaginary part. The opposite of 9 is -9.
The complex conjugate of 5 + 9i is 5 + -9i or 5 - 9i
Example #2
Find the complex conjugate of -10 - 3i
-10 - 3i = -10 + -3i.
-3 is the imaginary part. The opposite of -3 is 3.
The complex conjugate of -10 - 3i is -10 + 3i
Example #3
Find the complex conjugate of i
i = 1i.
1 is the imaginary part. The opposite of 1 is -1.
The complex conjugate of i is -i
The complex conjugate of a real number is the real number since there is no imaginary part. For example, the complex conjugate of 2 is 2.
Why do we need the conjugate of a complex number?
Let us see what will happen when we multiply a complex number by its complex conjugate.
(a + bi)(a - bi) = a2 - abi + abi - b2i2
(a + bi)(a - bi) = a2- b2i2
(a + bi)(a - bi) = a2 - b2(-1)
(a + bi)(a - bi) = a2 + b2
Since i is not here anymore, a2 + b2 is just a real number
Example
(4 + 3i)(4 - 3i) = 42 + 32
(4 + 3i)(4 - 3i) = 16 + 9
(4 + 3i)(4 - 3i) = 25
This property of the complex conjugate can be very useful when you are trying to simplify rational expressions or when you are trying to get rid of the "i" in the denominator of a rational expression.
For example, simplify 5i / (2 - i).
Multiply the numerator and the denominator of 5i / (2 - i) by the conjugate of 2 - i
The conjugate of 2 - i is 2 + i.
[5i(2 + i)] / (2 - i)(2 + i) = (10i + 5i2) / (22 + 12)
[5i(2 + i)] / (2 - i)(2 + i) = [10i + 5(-1)] / ( 4 + 1)
[5i(2 + i)] / (2 - i)(2 + i) = (10i - 5) / 5
[5i(2 + i)] / (2 - i)(2 + i) = 2i - 1
Therefore, 5i / (2 - i) = 2i - 1
Absolute value of a complex number
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7299 | https://publications.aap.org/pediatrics/article/22/2/259/29340/ECZEMA-VACCINATUM | ECZEMA VACCINATUM | Pediatrics | American Academy of Pediatrics
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Volume 22, Issue 2
August 1958
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Articles|August 01 1958
ECZEMA VACCINATUM Available to Purchase
Audrey H. Reynolds;
Audrey H. Reynolds
Department of Pediatrics, University of Southern California and the Children's Hospital of Los Angeles
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Howard A. Joos
Howard A. Joos
Department of Pediatrics, University of Southern California and the Children's Hospital of Los Angeles
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Pediatrics (1958) 22 (2): 259–267.
Article history
Received:
December 18 1957
Accepted:
February 18 1958
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Audrey H. Reynolds, Howard A. Joos; ECZEMA VACCINATUM. _Pediatrics_ August 1958; 22 (2): 259–267. 10.1542/peds.22.2.259
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Nine cases of eczema vaccinatum are presented, including two fatalities. Seven were caused by contact of a child with eczema with a recently vaccinated sibling.
Suddenly appearing umbilicated vesicles superimposed upon atopic eczema are almost diagnostic of eczema vaccinatum or eczema herpeticum. These do not occur with mere secondary bacterial infection.
Hyperimmune vaccinal gamma-globulin is now available for specific therapy.
Eczema vaccinatum is frequently iatrogenic and uniformly preventable.
The following steps are recommended for prophylaxis: 1) No child with atopic eczema or other skin disorder should be vaccinated. 2) No child should be vaccinated if any member of his family has eczema or other skin disorder. 3) Parents of children with eczema should be notified at the onset of the disease of the danger from vaccination contact. 4) If a sibling of a child with atopic eczema is vaccinated, he must be completely separated from that child for at least 21 days. 5) Forms used by state and local health departments for parents' consent to vaccination should include an appropriate warning of the contraindications. 6) Eczema vaccinatum should be a reportable disease. 7) Patients recently vaccinated must be excluded from pediatric wards containing patients with atopic eczema, other diseases of the skin, burns or healing surgical incisions. 8) Vaccination may be recommended at 2 months of age, especially for babies from strongly allergic families.
Topics:
eczema vaccinatum
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Copyright © 1958 by the American Academy of Pediatrics
1958
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